Student Study Guide and Solutions Manual for Atkins and Jones's
CHEMICAL PRINCIPLES The Quest for Insight
FOURTH
EDITION
JOHN KRENOS / JOSEPH POTENZA LAURENCE LAVELLE / YINFA MA / CARL HOEGER
'- ------
Student Study Guide and Solutions Manual for Atkins and Jones's
CHEMICAL PRINCIPLES The Quest
for
Insight
Rutgers University / JOSEPH POTENZA, Rutgers University LAURENCE LAVELLE, University of California, Los Angeles / YINFA MA, University of Missouri, Rolla CARL HOEGER, University of California, San Diego
JOHN KRENOS,
The goal of Chemical Priniciples is to help you make the journey from student to scientist. To do that, the authors focus on helping you develop "chemical insight" -the ability to see matter through the eyes of a chemist, and to make connections between chemical principles and their applications in the laboratory and the world around us. The Study Guide portion of the book will give you a grasp of the fundamentals while helping you avoid common mistakes. For each chapter of the text, the Study Guide covers: •
•
Section-by-section reinforcement of major points Supplementary tables and graphs which amplify, clarify, or summarize a body of material
•
Notes to point out potential pitfalls and key ideas
•
Highlighted key equations for easy review
I. W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England
The Solutions Manual portion of the book will sharpen your problem-solving skiUs by helping you learn how to strategize your way through problems. It contains step-by-step solutions -to all the odd numbered problems in the text. This combined Student Study Guide and SQlutio s Manual will hone your problem-solving skllls and chemical insight, making you a better student and chemist and helping you earn a better grade in· the course. For additional study resources, please visit our Web site at: www.whfreeman.com!chemicalprinciples4e
,
ISBN-13: 978-1-4292-0099-8
11 1111 11111 1111 11 1 1 1 1 :II ISBN-10: 1-4292-0099-5
90000
"781429 200998
r.
Student Study Guide and Solutions Manual for Atkins and Jones's
CHEMICAL PRINCIPLES The Quest for Insight FOURTH EDITION
John Krenos Rutgers, The State University of New Jersey
Joseph Potenza Rutgers, The StateUniversity of New Jersey
Carl Hoeger University of California, San Diego
Laurence Lavelle University of California, Los Angeles
Yinfa Ma University of Missouri, Rolla
I.
W. H. Freeman and Company New York
ISBN- 13: 978- 1- 4292-0099-8 ISBN-10: 1- 4292- 0099- 5 © 2007 by W. H. Freeman and Company All rights reserved. Printed in the United States of America First Printing W. H. Freeman and Company
41 Madison Avenue
New York, NY 10010
Houndsmills, Basingstoke RG21 6X SEngland www. whfreeman. com
CONTENTS
Preface .................................................. ........................ ...................................... Acknowledgments............ ............ ....................................................................
Vll
SG-
STUDY GUIDE 1
v
Atoms: The Quantum World..............................................................................
1
2 Chemical Bonds .................................................................................................
15
3
Molecular Shape and Structure........ .................................................................
25
4
The Properties of Gases............................................ .......... ..............................
37
5
Liquids and Solids ..............................................................................................
47
6
Thermodynamics: The First Law.......................................................................
59
7
Thermodynamics: The Second and Third Laws................................................
73
.
.
.
8 PhysicalEquilibria ....... ............................................... ...................................... .
9 ChemicalEquilibria ............................................................... ........................... ..
87 105
10
Acids and Bases... ........... ........... ................... ............................. ...... ...... ............
117
11
AqueousEquilibria.............................................................................................
131
12 Electrochemistry.................................................................................................
141
13 Chemical Kinetics..............................................................................................
15 5
14
TheElements: The First Four Main Groups......................................................
171
TheElements: The d Block................................................................................
189
15 16 17
TheElements: The Last Four Main Groups........................... ............................
203
Nuclear Chemistry.................................... .........................................................
219
18
Organic Chemistry I: The Hydrocarbons...........................................................
231
19
Organic Chemistry II: Polymers and Biological Compounds............................
241
.
SOLUTIONS MANUAL
SM-
Fundamentals..................................................................................................... 1
3
Atoms: The Quantum World..............................................................................
50
2 Chemical Bonds.................................................................................................
73
3
Molecular Shape and Structure..........................................................................
94
4
The Properties of Gases......................................................................................
113
5
Liquids and Solids..............................................................................................
139
Thermodynamics: The First Law...... .................. ................................. .............
163
8 PhysicalEquilibria................................................................................... ..........
217
6 7
.
Thermodynamics: The Second and Third Laws................................... ............ .
9 ChemicalEquilibria ............................................................................................
191 248
10
Acids and Bases.................................................................................................
284
11
AqueousEquilibria.............................................................................................
326
12 Electrochemistry .................................................................................................
366
13 Chemical Kinetics..............................................................................................
405
14
430
15 16 17 18 19
TheElements: The First Four Main Groups .............. .......... ..............................
TheElements: The d Block............................................................ ...................
448
Organic Chemistry I: The Hydrocarbons...........................................................
5 08
Organic Chemistry II: Polymers and Biological Compounds.......... ......... .......
5 26
TheElements: The Last Four Main Groups ...................... ......... ..... ....... ............ .
Nuclear Chemistry............................................ .................................................. .
.
466 486
/
PREFACE This Study Guide and Solutions Manual accompanies the textbook
Quest for Insight,
Chemical Principles: The
F ourth Edition, by Peter Atkins and Loretta Jones - an authoritative and
thorough introduction to chemistry for students anticipating careers in science or engineering disciplines. We have followed the order of topics in the textbook chapter for chapter. The parallel between the symbols, concepts, and style of this supplement and the textbook enables the reader to easily move back and forth between the two. Much of the Study Guide is presented in outline style with material highlighted by
bullets, arrows, and tables. In general, bullets offset major items of importance for each
section and arrows provide explanatory material, descriptive material, or some examples to reinforce a concept. This telegraphic style should help the student obtain a broad perspective
of large blocks of material in a relatively short period of time, and may prove particularly useful in preparing for examinations. We believe that the Study Guide will be most useful following a careful reading of the text. Chapter Sections follow those in the text and contain descriptive material (bullets and arrows) as well as numerous examples.
Many examples are
worked out in detail. While we have covered most of the material in the text, we have not attempted to be encyclopedic. To help obtain a broad overview of the material, important equations are highlighted in boxes. Students may find this aspect of the guide particularly useful before exams. Where appropriate, we have introduced supplementary tables either to amplify material in the text, to clarify it further, or to summarize a body of material. Sprinkled throughout the guide are Notes; in the main, these are designed to point out common pitfalls. The Solutions Manual section contains solutions and answers to the odd-numbered exercises in the textbook, including the Fundamentals sections. To display intermediate results we have disregarded rules concerning significant figures, but have adhered to them in reporting the final answers. In exercises with multiple parts, the properly rounded values are used for subsequent calculations. Student answers may differ slightly if only the final values are rounded, a method recommended by some instructors and the text, but not readily implemented in a printed Manual.
v
'-
ACKNOWLEDGMENTS The Study Guide authors, John Krenos and Joseph Potenza, are indebted to many individuals who have made significant contributions to it. First and foremost, we wish to thank Beth Van Assen for a thorough and incisive reading of the drafts. Her critical suggestions led to improved readability and scientific accuracy of the Study Guide. Many examples and important sections of descriptive text were clarified and expanded with her help. Her encouragement, patience, and persistence were essential to the completion of this project. We also thank two of our colleagues at Rutgers for critically reading three chapters. Harvey Schugar carefully reviewed Chapterl6. Our knowledge and appreciation of inorganic chemistry were broadened greatly by his comments and analysis. Spencer Knapp critiqued
Chapters 18 and 19 and gave us a quick tutorial on modern organic chemistry. Many of the chemical structures were prepared with his help.
The copy editors, Jodi Simpson and Alice Allen, did an excellent job in making the material clear, complete, and in harmony with the textbook. In paI1icular, we thank Jodi Simpson for help in developing a workable, consistent format for the telegraphic style. We also wish to acknowledge the help and encouragement of the staff at W. H. Freeman and Company. We especially thank Jessica Fiorillo (chemistry editor) for asking us to author this guide, Jodi Isman (project editor) for guiding us through the laborious process of developing a manuscript from scratch and also for helping with the editing, and Amy Thorne (supplements editor) for encouraging us to keep to a tight schedule. Finally, we extol the efforts of the authors of the textbook, Peter Atkins and Loretta Jones, in creating a new approach to teaching general chemistry. Beginning an introductory chemistry textbook with quantum theory is a logical, but challenging approach. The authors succeed by treating the fundamentals of chemistry (an "as needed" review) in a separate section and by minimizing the coverage of a great deal of material (mostly historical) presented in other books. In our opinion, this approach leads to the development of all the major topics in chemistry enlightened and enlivened in the first instance by the molecular viewpoint.
Vll
Chapter
ATOMS: THE QUANTUM WORLD
1
INVESTIGATING ATOMS (Sections 1 . 1 - 1 .3) 1.1 The N uclear Atom •
Subatomic particles
� �
� •
mass (m = 9. 1 094 x 1 0-3 1 kg) mass (m = 1 .6726 x 1 0-27 kg) mass (m = 1 .6749 x 1 0-27 kg)
charge (-e = -1.602 1 77 x 1 0- 1 9 C) charge ( e = 1 .602 1 77 x 1 0- 19 C) charge = 0
N ucleus
� Nucleons �
•
Electron : Proton: Neutron:
(protons and neutrons) occupy a small volume at the center of the atom. The binding energy of the nucleus is attributed to a strong force (nuclear) acting over a very short distance. The radius of the nucleus (assumed to be spherical) is given roughly by rnue ro A lI3 , where ro:::; 1 .3 x 1 0- 15 m = 1 .3 fm. =
Atom �
�
� �
Atomic number: Z = Np = number of protons in the nucleus Atomic mass number: A = Np + Nn = number of protons and neutrons in the nucleus Uncharged atom : Np = Ne (number of protons equals the number of electrons) Electrons occupy a much larger volume and define the "size" of the atom itself. The binding energy of the electrons is attributed to a weak force (coulomb) acting over a much longer distance.
1.2 The Characteristics of Electromagnetic Radiation •
Oscillating amplitude of electric and magnetic field Distance behavior (fixed time t)
Oscillating Wave
�
Wave characterized by wavelength and frequency
Time behav ior (fixed position x)
- x direction
Oscillating Wave
wavelength A
- I time
period
+-----'-r----f-------'l-..
x
frequency v = (1/period)
Speed of light (distance/time) wavelength / period = wavelength x frequency
Ic
=
AV
SI units :
Note:
I
=
speed of light (m·s- I )
=
wavelength x frequency (m) (S-I )
[ 1 Hz (hertz)
=
1 S-I]
The speed of light c (co in vacuum:::; 3.00 x 1 0 8 m·s- I ) depends on the medium it travels in. Medium effects on wavelength in the visible region are small (beyond three significant fi gures).
SG-l
1.3 Ato m ic Spectra •
• •
Spectral l i nes �
Discharge lamp of hydrogen H 2 + electrical energy - H + H * [ * == asterisk denotes excited atom] ( ( ) [ * ) == denotes a less excited atom] H * - H * + hv
Lines form a d iscrete pattern �
Discrete energy levels
Hydrogen atom s pectral lines
�
�
Johann Rydberg ' s general equation n2 = nupper and n) = n)ower 91 = 3.29
X
1 0 )5 Hz
=
Rydberg constant
Rydberg expression reproduces pattern of lines in H atom emission spectrum. The value of 91 is obtained empirically. Note:
Lines with a common n) can be grouped into a series and some have special names:
n) = I
(Lyman), 2 (Balmer), 3 (Paschen), 4 (Brackett), 5 (Pfund).
QUANTUM THEORY (Sections 1.4-1.7) 1.4 Radiation, Quanta, and Photons •
Black body
� �
Perfect absorber and emitter of radiation Intensity of radiation for a series of temperatures
�
Stefan-Boltzmann law:
�
Wavelength corresponding to maximum intensity = "max
�
Wien's law:
I T"-rnax
Power emitted (watts) Surface area (meter2 )
=
constant
I
=
constant x T4
where constant = 2.88 K-mm
At higher temperature, maximum intensity of radiation shifts to lower wavelength. •
Energy of a quantum (packet) of light (generally called a photon)
�
�
�
Postulated by Max Planck to explain black body radiation Resolved the "ultraviolet catastrophe" of classical physics, which predicted intense ultraviolet radiation for all heated objects ( T > 0) Quantization of electromagnetic radiation photon energy SI units:
(J)
Planck constant x photon frequency (h 6.626 1 x 1 0-:14 J.s) =
=
SG-2
•
Photoelectric effect
�
�
Ejection of electrons from a metal su rface exposed to photons of su fficient energy I ndicates that light behaves as a particle EK = kinetic energy of the ejected electron,
� hv � requ ired •
=
1.6
�
Diffraction and interference effects of su peri mposed waves
(constructive and destructive)
h
� mv
de Brogl; e wavelength fo' a part; cle w; th h ne", momentu m p
�
mv
The Uncerta i n ty Principle
Complementarity of location (x) and momentum (P) �
�
�
1.7
•
Relates photon energy to energy difference between two energy levels in an atom
Su ch matter behaves as a wave with a characteristic wavelength
I I
�
•
I
Matter has wave properties � Proposed by Lou is de Broglie � Consider matter with mass m and velocity v
A
•
Eupper - E lower
The Wave-Particle Duality o f Matter
�
•
for electron ejection
Wave behavior of light
1.5 •
=
Bohr freq uency condition
I hv
•
threshold energy (work fu nction) requ ired for electron ejection from the metal su rface, and hv = photon energy
U ncertainty in x is fu; u ncertainty in p is t:..p Limitation of knowledge
I
I
i3.ptu
17
� h/2
I
Heisenberg u ncertainty principle, where
is called " h bar"
I
Ii
= 1.0546
x
10-34
Ii
=
hl2n
J· s
Refu tes classical physics on the atomic scale
Wavefu nctio n s a n d Energy Levels
Classical trajecto ries � Wavefunction \jJ
Born interpretation
�
�
Precisely defined paths Gives probable position of particle with mass m
Probability of fi nding particle in a region is proportional to \jJ 2
SG-3
•
SchrOd i nger equation �
Allows calcu lation of \1' by solving a differential equ ation
� H\If = E\If; H is
•
called the H amiltonian
Particle in a box
�
Mass m confi ned b etween two rigid walls a distance L apart outside the b ox and at the walls (b oundary condition)
� \1' = 0
\/f17(x) =
() 2
1/2
L
.
Sin
( x) nn
\If n(x) = wavefunction t hat satisfies the Schro dinger equation b etween the b ox l imits. n is a quantum number.
n = l, 2, ...
L
A node is a poi nt in the b ox where \If = 0 and \jJ changes sign.
Note:
I \lf2 � 0, alway sI
En =
allowed energy val ues of a particle in a b ox Note: n = I gives the zero-point energy E • I EI *' 0 implies residual motion. t::"E
=
EIHI
-
En
=
(2n+l)h 2 8117L2
E
Particle in a box
11( 't' n2
E nergy difference b etween two neighb oring levels •
Probability as a fu nction of position in the box
� ->
Plot is shown for the first three levels \lf2 as a function of the dimensionless variab le xlL
//��'"
.
///"'-�"''''''-.
o ________________________�_______________________ ..-..-
"�·:,·:�:__________________________________�_:·:-;cco,�-
__
0.0
- -
..-.-_. _......._---.-
"'''''
0.2
0.4
x/L 06
THE HYDROGEN ATOM (Sections 1 .8-1. 1 1) 1.8 The Principal Quantum N u m ber • •
•
Charge on a n electron � q = - e = -1.602 177 33 Vac u u m permittivity �
Eo
=
8.854 187 817
x
x
10-19 C
2 10-12 C2. N-I . m-
Coulomb potential energy
V
( r)
product of charges on pm1icles
411: x permittivity of free space x distance between charges
=--------�-- --------�--�---------------
SG-4
(SI system)
0.
n
=
2
n
=
1
1.0
]=
Unit s of V(r): •
(C
2
2 c I
2
·N - · m- ) m
=N · m= (k g · m · s-
2)
? -2 m = k g · m-· s
+ + H atom and one-electron ion energy levels (He , Lr, Be3 , etc.)
�
Solut ions t o the Schro dinger equat ion 2=1 & n=I,2,3, ... 9"l =
�
2 EO
=3.289842 x 1015 Hz
Units:
All quant um number s ar e dimensionless. Wit h 91 in unit s of fr equency, the H atom ener gy-level equat ion has t he same for m as the Planck equat ion, E = hv.
� En =
�
4
ny 8h
(n is dimensionless)
91 =
� Z=
ener gy levels (st at es) oft he H at om. N ote t he negative sign. Rydber g constant, calculated exact ly using Bohr theor y or t he Schro di nger equat ion
at omic number, equal t o I for hydr ogen
�
n = pr incipal quantum number
�
I nteger n var ies fr om 1 (gr ound stat e) t o higher int eger s (excit ed st at es) t o 00 (ionizat ion).
�
�
As n incr eases, ener gy incr eases, t he at om becomes less stable, and ener gy stat es become mor e closely spaced (mor e dense) . Ener gies of H at om stat es var y fr om -h91 (n = I) t o 0 (n = 00) . Stat es wit h E > 0 ar e possible and corr espond t o an ionized atom in which the ener gy > 0 equals t he kinetic ener gy oft he electr on.
1 .9 Atomic Orbitals (AOs) •
Definit ion of AO
�
�
�
� •
Wavefunct ion (\jJ, psi) descr ibes an electr on in an at om. Or bit al (\jJ 2) holds 0, 1, or 2 electr ons. Or bit al can be viewed as a cloud wit hin which t he point densit y r epr esent s t he probability of finding the electr on at that point . Or bital is specified by three quant um number s (n, f!, n1(; see below).
Wavefunction � �
�
Fills all space Depends on t he three spher ical coor dinates: r, 8, � Wr it ten as a pr oduct of a r adial [R(r)] and an angular [Y(8,�)] wavefunct ion; mat hemat ical ly, \jJ(r, 8, �) = R(r)Y(8, �)
SG-S
•
Wavefu nction for the H atom 2s orbital � R(r) =
n = 2, I! = 0, and me = °
(2_*}-rI2ao (2ao )312
Units: R(r) =m-
312
ao= 4TImeEeo2fi2 { a=
Y(8,$) = none
0-
11 5.29177X10- m (Bohr radius)
}
2 2 2 1 (C ·W . m- ) (J·S) =m 2 kg. C
Three Quantum Numbers [11, e, mel Specify an Atomic Orbital Name
Sy mbol
•
Constraints
Al lowed Values
Positive integer
1 , 2, 3, . . .
n
Principal quantum number
=
I!
Orbital angular momentum quantum number
= 0 , 1,
me
Magnetic quantum number
2, ... , n-J
= I!, I!-J,1!-2, . . . ,-I! = 0, ±
1 , ± 2, . . .
, ±I!
Each value of n corresponds to n allowed values of I!.
Each value of I! corresponds to (21!+ 1) allowed values of me.
Terminology (nomenclatu re) shell: subshell:
AOs with the same n value AOs with the same n and e values; I! = 0,
1 , 2, 3 equivalent to S-, p-, d-, .f subshells, respectively, or s-orbital => I! = ° me= ° p-orbital => e= 1 me=-i,O,or+i d-orbital => e = 2 me= -2, - I , 0, +1 , or +2 f => e= 3 me = -3, -2, - I , 0, + 1 , +2, or +3 . orbital
•
Physical sign ificance of the wavefu nction \l/(r,e,� ) F or all atoms and molecules, \II 2(r, e , � ) is proportional
to the probability of finding the electron at a point r, e ,� . We can also regard \11 2 (r, e ,� ) as the electron density at point r, e ,� .
•
Plot of electron dens ity for the 2s-orbital of hyd rogen 600
Computer-generated electron density dot diagram for the hydrogen atom 2s-orbital. The nucleus is at the center of the square and the density of dots is proportional to the probability of finding the electron. Notice the location of the spherical node.
400
E .s
_
.� !.,
200 0
�
200
·
400
·
600
14730 Points1
·
.BOO '---�����_��_�---.J ·BOO ·600 ·400 ·200 0 200 400 600 BOO
x-axis (pm)
SG-6
•
Concept of AOs �
Two interpretations are useful: Visualize AO as a cloud ofpoints, with the density of points in a given volume proportional to probabil ity of finding the electron in that volume.
2
•
Visualize AO as a sUiface (boundary surface) within which there is a given probability of finding the electron. For example, the 95% boundary surface, within which the probability of finding an electron is 95%.
Boundary s urfaces � Shapes
s- orbital p-orbital d-orbital
•
of atomic orbitals
spherical dumbbell or peanut four-leafclover or dumbbell with equatorial torus (doughnut)
N u mber and type of orbitals �
Fol low from all owed values of quantum numbers s-orbitals : s means that ! = 0; if ! = 0, then m, = 0. So, for each value of n, there is one s- orbital. f
f
one for each n : I s, 2s , 3 s, 4 s, . . .
p means that f! = I ; if f! I , then mr - I , 0, + I , and n > I . So, for each value of n > I , there are three p- orbitals, px, p )' p o .
p-orbitals:
=
=
three for each n > I : 2px, 2py, 2p : ; 3px, 3p y , 3p : ; . . .
The np- orbitals are referred to col lectively as the 2p-orbitals, 3p-orbitals, and so on.
d means that ! = 2; if f! = 2, then mr = -2, - I , 0, + I , +2, and n > 2. So, for each value of n > 2, there are jive d-orbitals, dxy' dxz ' dyz' dxLy' and dz ' .
£I-orbitals:
f
jive for each n > 2 :
3dxy , 3dxz , 3dYZ' 3dx'_y" and 3dz' ;
The nd-orbitals are referred to collectively as the 3d- orbitals, 4d-orbitals, and so on. 1 . 10 Electron Spin •
Spin states
�
�
�
An electron has two spin states, represented as t and .J, or a and p . Are described by the spin magnetic quantum number, m, For an electron, only two values of m.1 are allowed: ++, -+.
1 . 1 1 The Electronic Structu re of Hyd rogen •
Degeneracy of orb itals In the H atom, orbitals in a given shell are degenerate (have the same energy). For many-electron atoms, this is not true and the energy of a given orbital depends
•
on n and f!.
G round and excited states of H
In the ground state, n = I , and the electron is in the I s-orbital. The first excited state corresponds to n = 2, and the electron occupies one of thefour possible orbitals with n = 2 (2s, 2p-" 2py , or 2p: ). Si milar considerations hold for higher excited states. •
Ionization of H
Absorption of a photon with energy :2: h9� ionizes the atom, creating a free electron and a free proton. Energy in excess of h� is uti lized as kinetic energy of the system.
SG-7
•
Summary
The state of an electron in a H atom is defined by four quantum numbers {n, I!, In" In, } . As the value of n increases, the size of the H atom increases.
MANY�LECTRON ATOMS (Sections 1.12-1.14) 1 . 1 2 Orbital Energies •
M any-electron atoms � Atoms with more than one �
electron Coulomb potential energy equals the sum of nucleus-electron attractions and electron-electron
�
Schro dinger equation cannot be solved exactly. Accurate wavefunctions obtained numerically by using computers
�
•
• •
•
repu lsions.
Variation of energy of orbitals � Shielding and penetration � Shielding
�
Qualitative understanding of orbital energies in atoms
Each electron in an atom is attracted by the nucleus and repelled by all the other electrons. I n effect, each electron feels a reduced nuclear charge (Zelle = effective nuclear charge). The electron (orbital) is shielded to some extent from the nuclear charge and its energy is raised accordingly.
Penetration �
Note:
For orbitals in the same shell but in different subshells, a combination of nucleus-electron attraction and electron-electron repulsion infl uences the orbital energies.
The s-, p-, d-, ... orbitals have different shapes and different electron density distributions. F or a given shell (same value of the principal quantum number n), s-electrons tend to be closer to the nucleus than p-electrons, which are closer than d-electrons. We say that s-electrons are more penetrating than p-electrons, and p-electrons are more penetrating than d-electrons.
Shielding and penetration can be understood qualitatively on the basis of the nucleus-electron potential energy term: -(Ze)e/4m::or, where Ze is the nuclear charge and r the nucleus-electron distance. Shielded electrons have the equivalent of a reduced Z value (Zelj< Z ) and therefore higher energy; penetrating electrons have the equivalent of a reduced value of r and therefore lower energy. �
Z has three equivalent meanings:
•
Review
•
Conseq uen ces of shielding and penetration in many-electron atoms �
Nuclear charge (actually, Ze) Atomic number Number of protons in the nucleus
Orbitals with the same n and different I val ues have different energies.
SG-8
� � �
For a given shell (n), subshell energies increase in the order: ns < np < nd < nf. Example: A 3s-electron is lower in energy than a 3p- electron, which is lower than a 3d-electron. Orbitals within a given subshell have the same energy. Example: The five 3d-orbitals are degenerate for a given atom. Penetrating orbitals of higher shells may be lower in energy than less penetrating orbitals of lower shells. Example: A penetrating 4s-electron may be lower in energy than a less penetrating 3d- electron.
1 . 13 The Bu ilding-Up P rinciple •
Pa uli exclusion principle
�
�
�
•
No more than two electrons per orbital Two electrons occupying a single orbital must have paired spins: I1±l Two electrons in an atom may not have the same four quantum numbers.
Terminology closed shell: valence electrons: electron co nfigu ration:
•
�
�
Order in which electrons are added to subshells and orbitals to yield the electron configuration of atoms Order of fil ling subshells in neutral atoms is determined by filling those with the lowest values of (n + J! ) first. Subshells in a group with the same value of (n + J! ) are filled in the order of increasing n (topic not covered in the text).
Build ing-Up (Aujbau) Pri nciple The (n + f) rule
•
Shell with maximum number of electrons allowed by the exclusion principle Electrons in the outermost occupied shell of an atom; they occupy the shell with the largest value of n and are used to form chemical bonds. List of all occupied subshells or orbitals, with the number of electrons in each indicated as a numerical superscript. Example: Li: I s 22s I.
Usual filling order of su bshells ( n + Note:
J! rule) : I s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < ...
Ionization or subtle differences in shielding and penetration can change the order of the energy of subshells. Thus, energy ordering ofsubshells is not fixed absolutely, but may vary
from atom to atom or from an atom to its ion. •
Orbitals within a su bshell
•
Applicabil ity
�
�
H und' s rule: Electrons add to different orbitals of a subs hell with spins parallel until the subshell is half full.
The buildi ng-up principle in combination with the Pauli exclusion principle and Hund's rule accounts for the ground-state electron configurations of atoms. The principle is generally valid, but there a re exceptions.
1 . 1 4 Electronic Struct u re and the Periodic Table •
Order of filling subshells
�
�
•
Understanding the organization of the periodic table Straightforward determination of (most) electron configurations
Terminology
�
G iven in the following two tables
SG-9
Columns in the periodic table, labeled 1 - 1 8 horizontal ly G roups 1 , 2; s-subshell fills G roups 1 3- 1 8; p-subshell fi l ls G roups 3-1 2; d-subshell fills G roups 3- 1 1 ; d-subshell fil ls G roups 1 , 2 and 1 3- 1 8 4f subshell fills 5f subshell fills
G roups: s-block elements: p-block elements: d-block elements: transition elements: Main-group elements: Lanthan ides: Acti nides: Note:
A transition element has a partially fil led d- subshell either as the element or in any commonly occurring oxidation state. Thus, Zn, Cd, and Hg are not transition elements. The (n - I ) d electrons of the d-block elements are considered to be valence electrons. G roups 1 , 2, and 1 3- 1 8 are alternatively labeled with Roman numerals 1-V 1 1 1 , which correspond to the number of valence electrons in the element. Outermost occupied shell (highest n) Row of the periodic table Principal quantum number of valence shel l H, He; I s-subshell fills Li through Ne; 2s-, 2p-subshells fil l Na through Ar; 3s-, 3p-subshells fi l l K through Kr; 4s-, 3d-, 4p-subshells fi l l
Valence s hell: Period: Period number: Period 1 : Period 2: Period 3:
Period 4: (first long period) •
Periodic table
�
Two forms displayed: One shows the elements, the other the final subshell filled. Look at the tables to understand the terminology. Periodicity of Elements in the Periodic Table
Period
G roup ( 1 - 1 8)
.!-
18
1
�
2
Li
Be
3
Na K Rb Cs Fr
Mg Ca Sr Ba
4 5 '6 7
�
2
Ra
3
4
5
Sc y
Ti Zr Hf Rf
V
Lu Lr
6 C,.
Nb
Mo
Db
Sg
Ta
W
7
8
9
10
Mn Tc Re Bh
Fe
Co
Ni
Os Hs
Ir Mt
Ru
Rh
Pd PI
Ds
Lanthanides Actinides
SG - I O
11
Cu
Ag Au
Uuu
12
Zn Cd Hg
Uub
13
14
15
B AI Ga In TI
C
N
Si Ge Sn Pb
P As Sb Bi
Uuq
,--
16
17
F
He Ne
S Se Te Po
CI Br I At
Ar Kr Xe Rn
0
Uuh
Electron configu rations of the elements �
•
•
See the Appendix section on Ground-State Electron Configurations in the text.
1 7 ital icized elements
�
�
Exceptions to the (n + J!) rule Differ by the placement of one electron 2 2 2 6 2 6 Example: Cr: I s 2s 2p 3s 3p 4s ' 3d5 = [Ar] 4s' 3d5 (not . . .4s 3 d 4 as might be expected)
2 bold, ital icized elements �
•
Differ by the placement of two electrons (Pd, Th) We expect [Kr] 4d 8 si, but actually find [Kr] 4d l O• No Ss-subshell electrons We expect [Rn] s/7i, but actually find [Rn] 6d 27i. No Sj- subshell electrons
Pd Th
Order of Filling Subshells in the Periodic Table G roup ( 1-1 8)
Period
J, 1 2
� 18
� 2s 3s 4s 5s 6s 7s
3 4 5 6 7
13 3
4
5
6
7
3d 4d 5d 6d
8
9
10
11
12
t
14
16
15
17
2p 3p 4p 5p 6p 7p
�
4f 5f
THE PERIODICITY OF ATOMIC PROPERTIES 1.15 Atomic Radius (r) •
Definition of atomic rad ius
� � �
•
(Sections 1.15-1.20)
Half the distance between the centers of neighboring atoms (nuclei) For metallic elements, r is determined for the solid. Example: For solid Zn, 274 pm between nuclei, so r (atomic radius) = 1 37 pm For nometal lic elements, r is determined for diatomic molecules (covalent bond). Example: For 1 2 molecules, 266 pm between nuclei, so r (covalent radius) = 1 33 pm
General tren ds in rad ius with atomic n umber
�
r decreases from left to right across a period (effective nuclear ch arge increases) � r increases from top to bottom down a group (change in valence electron principal quantum
number and size of valence shel l)
SG-II
1 . 1 6 Ionic Radius •
An ion's s hare of the d istance between neigh bors in an ionic solid (cation to an ion)
---'t
---'t
---'t ---'t •
Distance between the nuclei of a neighboring cation and anion is the sum of two ionic radi i Radi us of the oxide anion (02-) i s 1 40 pm. E xample: Distance between Zn and 0 nuclei in zinc oxide is 223 pm; therefore, the ionic radius of Zn2+ is 83 pm [r (Zn2+) = 223 pm - r (0 2-)]. Cations are s maller than parent atoms; for exampl e, Zn ( 1 33 pm) and Zn2+ (83 pm). Anions are larger than parent atoms; for example, 0 (66 pm) and 02-( 1 40 pm). ---'t Atoms and ions with the same number of electrons Ar, and K+ , and Ca2+
Isoelectro nic atoms and ions Example: cr,
1 . 1 7 Ionization Energy (1) ---'t
Energy needed to remove an electron from a gas-phase atom in its lowest energy state
•
I
•
Symbol
•
Number subscripts (e.g. , II, lz) denote removal of successive electrons.
General period ic table trends in II fo r the main-group elements
---'t
---'t •
---'t
I ncreases from left to right across a period Decreases from top to bottom down a group
Exceptions
---'t
(ZefT increases)
(change in principal quantum number n of valence electron)
For Groups 2 and 1 5 in Periods 2-4, I) is larger than the neighboring main group element in Groups 1 3 and 1 6, respectively. Repulsions between electrons in the same orbital and/or extra stability of completed and half completed subshells are responsible for these exceptions to the general trends.
•
Metals toward the lower left of the periodic table
---'t ---'t
•
Have low ionization energies Readily lose electrons to form cations
Non metals toward the upper right of the periodic table
---'t
---'t
Have high ionization energies Do not readily lose electrons
1 .1 8 Electron Affinity (Eea) • •
Energy released w hen an electron is added to a gas-phase atom Periodic table trends in Ecn for the ma in-group elements
---'t
---'t ---'t
•
(ZefT increases) (change in principal quantum number n of valence electron) Generally, the same as for I) with the maj or exceptions displayed in Fig. 1 .49
I ncreases from left to right across a period Decreases from top to bottom down a group
Summary of trends in r, I., and Eo,.
---'t
---'t
All depend on ZefT and n of outer subshell electrons. Recall that there are many exceptions to the general trends, as noted earlier.
SG-12
1 . 1 9 The Inert-Pair Effect •
Tendency to form ions two un its lower i n charge than expected from the gro u p n u m ber relation
�
Due in part to the different energies of the valence p- and s-electrons
�
Valence s-electrons are cal led a "lazy pair."
�
I mportant for the lower two members of Groups 1 3, 1 4, and 1 5; for example, Pb(IV) & Pb (II) or Sb(V) & Sb( I I I)
1 .20 Diagonal Relationships •
Diagonally related pa irs of elements often show similar chem ical properties.
�
�
�
Diagonal band of metal loids dividing metals from nonmetals Similarity of Li and Mg (react directly with N2 to form nitri des) Similarity of Be and AI (both react with acids and bases)
THE IMPACT ON MATERIALS (Sections 1.21-1.22) 1 .2 1 The Main-Group Elements (Groups 1-2, 13- 1 8) •
•
s-block elements (G rou ps 1 and 2) �
All are reactive metals (except H); they form basic oxides (02-), peroxides (0/-), or superoxides (02-).
Compounds of s-block elements are ion ic, except for beryllium. Notes:
Hydrogen (a nonmetal) is placed by itself, or more usually i n Group I , because its electronic configuration (lsi) is similar to those of the alk ali metals: [noble gas] nsl.
Ii, is placed in Group 1 8 because its properties are similar to those of neon, argon, krypton, and xenon.
Helium, with electronic configuration •
p-block elements (G roups 1 3-1 8) � Metals: Note:
Group 1 3 (8); Group 14 (Si, Ge); Group 1 5 (As, Sb); Group 16 (Te, Po)
Metalloids have the physical appe arance and properties of metals but behave chemically as nonmetals.
Non metals: Notes:
Group 1 3 (AI, Ga, In, TI); Group 1 4 (Sn, Pb); Group 1 5 (Bi)
These metals have relatively low 1), but larger than those of the s block and d block.
Metalloids: Note:
Members are metals, metalloids, and nonmetals.
Group 1 4 (C); Group 1 5 (N, P); Group 1 6 (0, S, Se); Groups 1 7 & 1 8 (All)
Hig h Eca in Groups 13-17; these atoms tend to gain electrons to complete their subshells. Group 1 8 noble-gas atoms are generally nonreactive, except for Kr and Xe, which form a few compounds.
SG- 1 3
1 .22 The Transition Metals •
d-block elements (G roups 3- 1 2) � Note:
•
•
Al l are metals (most are transition metals), with properties intermediate to those of s-bl ock and p-bl ock metals.
All G roup 1 2 cations retain the fil led d-subshell. For this reason, these elements are not cl assified as transition metal s. Recall that in this text d-orbital electrons are considered to be val ence electrons for all the d-bl ock elements.
Characteristically, many transition metals form compou nds with a va riety of oxid ation states (oxidation n u m bers). I-block elements (lanthan ides and actin ides) �
�
�
Rare on Earth Lanthanides are incorporated in superconducting material s. Actinides are all radioactive el ements.
SG-14
Chapter 2 CHEMICAL BONDS IONIC BONDS (Sections 2. 1-2.4) 2.1 The Ions That Elements Form �
Remove outermost electrons in the order np, ns, (n- I )d
•
Cations
•
Metallic s-block elements and metallic p-block elements in periods 2 and 3
•
•
•
•
->
Form cations by losing electrons down to the noble-gas core 2 Examples: Mg, [Ne] 3/ � magnesium(l l ), Mg +, [Ne]; 2 AI, [Ne] 3s 3/ � aluminum(III), AI 3+, [Ne]
Metallic p-block elements in periods 4 and higher
�
Form cations with complete, typically unreactive d-subshells Example: Ga, [Ar] 4/ 3d1 o 4p l � gallium( I I I), Ga3+, [Ar] 3d1 o
Metallic t/-block elements
�
Lose s-electrons and often a variable number of d-electrons 2 3 Examples: iron(l l), Fe +, [Ar)3d 6 and iron(l I I ), Fe +, [Ar)3d5
Many metallic p-block elements �
May lose either their p-electrons or all their s - and p-electrons Example: Sn, [Kr] 5/ 4d1 o 5/ � tin( I I), Sn 2+, [Kr] 5/ 4io or tin(IY), Sn4+, [Kr] 4d1o
Anions �
Add electrons until the next noble-gas configuration is reached. Exa mples: Carbide (methanide), C 4- , [He] 2/ 2/ or [Ne] {octet} Hydride, H-, 1 / or [He] {duplet}
2.2 Lewis Symbols (Atoms and Ions) •
Valence electrons � Depicted as dots; a pair of dots represents two paired electrons � Lewis symbols for neutral atoms in the first two periods of the Periodic Table: H·
He:
Lj·
Be: ·B· ·C· :N· . :0·
.
.
.
:F·
:Ne:
Note: Ground-state structures for B and C are : B · and :C· . •
Variable valence
�
�
Abi lity of an element to form two or more ions with different oxidation numbers Displayed by many d-block and p-block elements Examples: Lead in lead(l l ) oxide, PbO, and in lead(IY) oxide, Pb02 Iron in iron( l l ) oxide, FeO, and in iron( l l l ) oxide, Fe2 0 3
2.3 The Energetics of Ionic Bond Formation • •
�
Electrostatic attraction (coulombic) of oppositely charged ions Ionic model � Energy for the formation of ionic bonds is suppl ied mainly by coulombic attraction of oppositely charged ions. This model gives a good description of bonding between the ions of metals (particularly s-block) and those of nonmetals. Ionic bond
SG-J5
•
Ionic sol ids /ionic crystals
�
�
�
�
•
Are assemblies of cations and anions arranged in a regular array The ionic bond is nondirectional; each ion is bound to all its neighbors. Typically have high melting and boiling points and are brittle, NaCI for example Form electrolyte solutions if they dissolve in water
Form u las of compo unds composed of monatomic ions
�
�
�
Formulas are predicted by assuming that atoms forming cations lose all valence electrons and those forming anions gain electrons in the valence subshell(s) until each ion has an octet of electrons or a duplet in the case of H, He, and Be. For cations with variable valence, the oxidation number is used. Relative numbers of cations and anions are chosen to achieve electrical neutrality, using the smal lest possible integers as subscripts.
2.4 Interactions Between Ions •
I n an ionic solid
�
�
�
•
All the cations repel each other, and all the anions repel each other. Each cation is attracted to all the anions to a greater or lesser extent. In this view, an ionic bond is a "global" characteristic of the entire crystal.
Coulomb potential energy between two ions, Ep
e is the elementary charge; ZI and Z2 are the charges on the ions; rl2 is the distance in nanometers (nm) between the ions; Eo is the vacuum permittivity.
•
•
Lattice energy �
Potential energy difference: ions in solid minus ions infinitely far apart A is a positive number; d is the distance between the centers of nearest neighbors Madel u ng constant (A ) in the crystal. �
�
�
A depends on the arrangement of ions in the solid. In the expression above, Ep is always negative and A is positive.
So, ionic crystals with small ions (short interionic distances), large values of A, and h ig h ly tend to have large lattice energies.
charged ions �
1 .747 56 (NaCI structure) [d = 0.2798 nm 1 .762 67 (CsCI structure) [d = 0.35 1 nm
•
Values of A
•
Potential energy of an ionic solid
•
•
�
for NaCI (actual value near 0 K)] for CsCl (sum of ionic radi i)]
Takes attractive and repulsive interionic interactions into account � Is given by the Born-Meyer equation below The constant d* is commonly taken to be 34.5 pm. It derives from the Born-Meyer eq u ation repulsive effects of overlapping electron charge clouds. Refractory material � Substance that can withstand high temperatures; MgO, for example
SG-16
COVALENT BONDS (Sections 2.5-2.8) 2.5 Lewis Structu res •
Covalent bond � Pair of electrons shared between two atoms � Located between two neighboring atoms and binds Examples:
•
•
them together Nonmetallic elements such as H 2 , N 2 , O2 , F 2 , Ch, Br2 , 1 2 , P 4 , and Sg
Rules
�
Atoms attempt to complete duplets or octets by sharing pairs of valence electrons. � Valence of an atom is the number of bonds it can form. � A l ine (-) represents a shared pair of electrons. � A lone pair of non bonding electrons is represented by two dots (:). Example: H-H, (single bond) duplet on each atom (valence of hydrogen = I ) and :N=N:, (triple bond and lone pair) octet on each atom (valence of nitrogen = 3) Lone pairs of electrons � Electron pairs not involved in bonding Example: The electrons ind icated as dots in the Lewis structure of N 2 above
2.6 Lewis Structu res for Polyatom ic Species •
Lewis structures
�
� •
• • • • •
Show which atoms are bonded (atom connectivity) and which contain lone pairs of electrons Do not portray the shape of a molecule or ion
Rules
�
Count total number of valence electrons in the species � Arrange atoms next to bonded neighbors � Use minimum number of electrons to make all single bonds � Count the number of nonbonding electrons required to satisfY octets � Compare to the actual number of electrons left � If lacking a sufficient number of electrons to satisfy octets, make one extra bond for each deficit pair of electrons. � Sharing pairs of electrons with a neighbor completes the octet or duplet. � Each shared pair of electrons counts as one covalent bond (line). One shared pair single bond (-) Bond order = I Two shared pairs double bond (= ) Bond order = 2 (multipl e bond) Three shared pairs triple bond (= ) Bond order = 3 (multiple bond) Bond order � Number of bonds that link a specific pair of atoms Term inal atom � Bonded to only one other atom Central atom � Bonded to at least two other atoms Molecular ions � Contain covalently bonded atoms : NH/, Hg}+, S O/R ules of thumb
�
Usual ly, the element with the lowest II is a central atom, but electronegativity is a better indicator (see Section 2. 12). For example, in HCN, carbon has the lowest II and is the central atom. It is also less electronegative than nitrogen.
SG- 1 7
.� �
Usually, there is a symmetrical arrangement about the central atom. For example, in S02 , OSO is symmetrical, with S as the central atom and the two 0 atoms terminal. Oxoacids have H atoms bonded to 0 atoms; H 2 S04 is actually (HO) 2 S02' Examples: Ethyne (acetylene), C2 H 2 ( 1 0 valence electrons): H-C=C-H Hydrogen cyanide, HCN ( 1 0 valence electrons): H-C=N: Ammonium ion, NIl/ (8 valence electrons):
[H-�-H1
+
2.7 Resonance •
Mu ltiple Lewis structu res �
� •
Some molecules can be represented by different Lewis structures (contributing structures) in which the locations of the electrons, but not the nuclei, vary. M ultiple Lewis structures used to represent a given species are called resonance structures. �
Electron delocal ization
Species requiring mUltiple Lewis structures often involve electron delocalization with some electron pairs distributed over more than
two atoms. •
Blending of structu res
�
Double-headed arrows ( .. .. ) are used to relate contributing Lewis structures, indicating that a blend of the contributing structures is a better representation of the bonding than any one.
•
Reson ance hybrid
�
Blended structure of the contributing Lewis structures
Example:
N 20, nitrous oxide, has 2(5) + 6 = 1 6 valence electrons or eight pairs.
Blending of two structures, both of which obey the octet rule. Note: The central N atom is the least electronegative atom (Section 2. 1 2). 2 Note: A triple bond to an 0 atom is found in species such as BO-, CO, NO+, and 02 +. C6H 6, benzene, has 6(4) + 6( I) = 30 valence electrons, or I S pairs.
Example:
o
..
Kekule structures
Note:
o
or Resonance hybrid
Carbon atoms lie at the vertices of the hexagon (the six C-H bonds radiating from each corner are not shown in the stick structures). The last structure depicts six valence electrons delocalized around the ring.
2.8 Formal Charge •
Formal charge
�
An atom's number of valence electrons ( V) minus the number of electrons assigned to it in a Lewis structure
SG- J 8
•
•
Electron assignment in a Lewis structu re
-+
Atom possesses all of its lone pair electrons (L) and half of its bonding electrons (B) [B means shared bonding electrons]
-+
Formal charge
1
V - (L + "2 B)
=
-+
Contribution of individ u a l Lewis structu res to a resonance hybrid Example: N20 :
Structures with individual formal charges closest to zero usually have the lowest energy and are the major contributors.
:N=N= O : -
0
+1
I
'"
:N: = N- O :
•
o
+1
..
•
-I
:N- N = O: 2
-
+1
+1
(high formal charges) •
Plausibility of isomers Exa mples: N2 0,
-+
The isomer with lowest formal charges is usually preferred.
:N=N = O : -1
+1
0
and :N = O =N :
(low formal charges)
+2
-I
HCN, H-C=N : and H -N=C : 0
0
0
-1
(high formal charges) o
+1
-1
(first structure with lower formal charges preferred)
Summa ry •
Formal charge
-+ -+
-+ -+
•
I ndicates the extent to which atoms have gained or lost electrons in a Lewis structure (covalent bonding) Exaggerates the covalent character of bonds by assuming that electrons are shared equally Structures with the lowest formal charges usually have the lowest energy (major contributors to the resonance hybrid). I somers with lower formal charges are usually favored.
Oxidation n u m ber E xample:
-+
Exaggerates the ionic character of bonds
Carbon disul fide, : S = C = S : Here, C has an oxidation number of +4 (all the electrons in the double bonds are assigned to the more electronegative S atoms) and a formal charge of zero (the electrons in the double bonds are shared equally between the C and S atoms).
EXCEPTIONS TO THE OCTET RULE (Sections 2.9-2. 1 1) 2.9 •
Radicals and Biradicals
Rad icals
-+
-+
-+
Species with an unpaired electron All species with an odd number of electrons are radicals. Are highly reactive, cause rancidity in foods, degradation of plastics in sunlight, and perhaps contribute to human aging
SG- 1 9
Examples:
•
C H3 (methyl radical), OH (hydroxyl), OOH (hydrogenperoxyl), NO (nitric oxide), N02 (nitrogen dioxide), 03- (ozonide ion)
[ :O-O-O: ] -7
.
. .
. .
�
[ :0-0-0: ]� [ :0-0-0: ].. . .
. .
. .
.
.
Species containing two unpaired electrons 0 (oxygen atom) and CH2 (methylene) [unpaired electrons on a single atom] O2 (oxygen molecule) and larger organic molecules [on different atoms] A ntioxidant -7 Species that reacts rapidly with radicals before they have a chance to do damage Examples: V itamins A, C and E; coenzyme Q; substances in coffee, orange j uice, chocolate Biradicals
Examples :
•
2 . 10 Expanded Valence Shells •
Expanded valence s hells -7
-7
-7 •
More than eight electrons associated with an atom in a Lewis structure (expanded octet) A hypervalent compound contains an atom with more atoms attached to it than is permitted by the octet rule. Empty d-orbitals are utilized to permit octet expansion. Electrons may be present as bonding pairs or lone pairs. Are characteristic of nonmetal atoms in Period 3 or higher
First- and second-period elements do not utilize expanded valence shells.
Variable covalence -7
-7
Ability to form different numbers of covalent bonds E lements showing variable covalence include: Valence Shell Occupa ncy Elements
P S Example:
1 0 electrons
8 electrons
PCI3 / PCI/ SF 2 IF
1 2 electrons
PCI6SF6 IFs
PCls SF4 IF3
Major Lewis structures for sulfuric acid, H 2 S04 (32 valence electrons) :0: . . .. I H - O - S - .O. - H .. I :0:
-
:0 II . . • • H-O-S-O-H I :0:
-
:0 II • • H - O. - S - O - H • • • II •
.
:R
I n the structure on the right, all atoms have zero formal charge and the S atom is surrounded by 1 2 electrons (expanded octet). Because it has the lowest
formal charges, this structure is expected to be the most favored one energetically and the one to make the greatest contribution to the resonance hybrid. An additional Lewis structure with one double is bond not shown.
Similar examples include SO/-, S02 , and S 30 (S is the central atom).
2 . 1 1 The Unusual Structu res of Some Group 13/111 Compounds •
Incomplete octet
-7
Fewer than eight valence electrons on an atom in a Lewis structure
SG-20
Example:
� � •
B F 3,
: F -B - F : I : F:
The single-bonded structure with an incomplete octet makes the major contribution. Other compounds with incomplete octets are BCI 3 and A ICI, vapor at high temperature At room temperature, aluminum chloride exists as the dimer, AhCI 6 , which fol lows the octet rule (See structure 33 in the text). �
One in which both electrons come from one atom Example: Donation of a lone pair by one atom to create a bond that completes the octet of another atom with an incomplete octet. In the Lewis structure for the reaction of BF3 with N H3, both electrons in the B-N bond are derived from the N lone pair.
Coord inate cov alent bond
--
IONIC VERSUS COVALENT BONDS (Sections 2.12-2. 1 3) 2. 1 2 Correcting the Covalent Model: E lectronegativity •
Bonds
�
All molecules may be viewed as resonance hybrids of pure covalent and ionic structures.
Example:
•
H 2 The structures are H - H ---- H+ [Hr ---- [: Hr H+ Here, the two ionic structures make equal contributions to the resonance hybrid, but the single covalent structure is of major importance.
Partial charges
� �
When the bonded atoms are different, the ionic structures are not energet ically equivalent. Unequal sharing of electrons results in a polar covalent bond. Example:
HF The structures are H- F
-
Hl r: f
-
[: H n f: r
The electron affinity of F is greater than that of H, resulting in a small negative charge on F and a corresponding smal l positive charge on H: EeaCF) > EeaCH) => O+ H-Fo - (partial charges, 1 0 + 1 = 1 0 - I ) E (H+ F-) « E (H- F +) => H - F+ is a very minor contributor Electric dipole � A partial positive charge separated from an equal but negative partial charge
Note:
• •
Electric dipole moment (I-L)
�
�
�
�
Size of an electric dipole: partial charge ti mes distance between charges Units: debye CD) 4.80 D == an electron (-) separated by 1 00 pm from a proton (+) 1-L = (4.80 D) x o x (distance in pm l 1 00 pm)
SG-2 1
•
E lectronegativity (x)
�
�
�
� �
� •
Electron-attracting power of an atom when it is part of a bond Mulliken scale: X +(1, + Ee.) Follows same periodic table trends as I, and Ee. =
Increasesfrom left to right andfrom bottom to top
Pauling numerical scale based on bond energies [X(F) 4.0] is used in text (qualitatively similar to Mulliken ' s scale). See Section 2. 1 2 in the text for a tabulation of values for various elements. =
Rough ru les of t h u m b
0.5
�
(X A - XB) � 2 (XA - XB) � 1 .5 (XA - XB) � 0.5
bond is essentially ionic bond is polar covalent bond is essentially covalent
2 . 1 3 Correcting the Ionic Model: Pola rizabi lity �
All have some covalent character. A cation's positive charge attracts the electrons of an anion or atom in the direction of the cation (distortion of spherical electron cloud).
•
Ionic bonds
•
H ighly polarizable atoms and ions
•
Readily undergo a large distortion of their electron cloud Examples of polarizable species: large anions and atoms such as 1-, Br- , c r , t, Br, and CI
Polarizing power
�
�
•
�
Property of ions (and atoms) that cause large distortions of electron clouds Increases with decreasing size and increasing charge of a cation Examples of species significant polarizing power: the small and/or highly charged cations Li+, Be2+, Mg2+, and A1 3+
Significant covalent bonding character �
� �
Bonds between highly polarizing cations and highly polarizable anions have significant covalent character. The Be2+ cation is highly polarizing and the Be-CI bond has significant covalent character (even though there is an electronegativity difference of l .6). In the series AgCI to AgI, the bonds become more covalent as the polarizabil ity (and size) of the anion increases (Cr < Br- < 1 -).
THE STRENGTHS AND LENGTHS OF COVALENT BONDS (Sections 2.14-2.16) 2 . 1 4 Bond Strengths •
Dissociation energy (D) �
�
�
Energy required to separate bonded atoms in neutral molecules Bond breaking is homolytic, which means that each atom retains half of the bonding electrons. Determines the strength of a chemical bond The greater the dissociation energy, the stronger the bond.
SG-22
� D
is defined exactly for diatomic (two-atom) molecules. For polyatomic (greater than two-atom) molecules, D also depends on the other bonds in the molecule. However, for many molecules, this dependence is slight.
•
•
Values of 0 (Table 2.3) � Vary from about 1 39
D
kj mor l (I-I single bond) to 1 062 kj mor l ( C=O triple bond)
for several d iatomic molecu les
H(g) + H(g) H-H(g) H(g) + F(g) H-F(g) F-F(g) F(g) + F(g) I-I (g) I (g) + I(g) O=O(g) - O(g) + O(g) N=N(g) - N(g) + N(g)
D = 424 kJ . mor l D 543 kJ·mor i D = 1 46 klmor i D = 1 39 klmor i D = 484 kJ·mor l D = 932 kJ . mor i =
(strong) (strong) (weak) (weak) (strong) (very strong)
2 . 1 5 Variation in Bond Strength •
Bond strength �
Note:
•
For polyatomic molecules, bond strength is defined as the average dissociation energy for one type of bond found in different molecules. For example, the tabulated C-H single bond value is the average strength of such bonds in a selection of organic molecules, such as methane (CH4), ethane (C 2 H 6), and ethene (C2 H4).
Values of average dissociation energies in text Table 2.4 are actually those for the propel1y of bond enthalpy (see Chapter 6), measured at 298. 1 5 K. The bond dissociation energies of diatomic molecules given in Table 2.3 apply at 0 K (absolute zero).
Factors influencing bond strength
�
� �
�
Bond multiplicity (C=C > C=C > C-C) Resonance (C=C > C.:...:....:C (benzene) > C-C) Lone pairs on neighboring atoms (F-F < H-H) (HF > HCI > H Br > H I ) Atomic radii (The smaller the radius, the stronger the bond.)
2 . 1 6 Bond Lengths •
Bond length
�
�
�
� •
I nternuclear distance, at the potential energy minimum, of two atoms linked by a covalent bond Helps determ ine the overall size and shape of a molecule Evaluated by usi ng spectroscopic or x-ray diffraction (for sol ids) methods For bonds between the same elements, length is inversely proportional to strength.
Factors influencing bond length
�
�
� �
(C=C < C=C < C-C) Bond multipl icity Resonance (C=C < C.:...:....:C (benzene) < C-C) Lone pairs on neighboring atoms (F-F > H-H) Atomic radii (HF < HCI < H Br < H I ) (The smaller the radius, the sh0l1er the bond.) Note: These trends are the opposite of the ones for bond strength.
SG-23
•
Covalent rad ius �
�
�
�
�
�
�
Contribution an atom makes to the length of a covalent bond Half the distance between the centers (nuclei) of neighboring atoms joined by a covalent bond (jar like atoms) Covalent radii may be added together to estimate bond lengths in molecules. Tabulated values are averages of radii in polyatomic molecules. Decreases from left to right in the periodic table Increases in going down a group in the periodic table Decreases for a given atom with increasing multiple bond character Example:
Use the covalent radi i given in text Fig. 2.2 1 to estimate the several bond lengths in the acetic acid molecule, CH 3COOH. H :0 I II Lewis Structure: H- C - C - O-H .. I H • •
Bond type
C-H C-C C-O C=O O-H
Bond length estimate
77 pm + 37 pm = 77 pm + 77 pm = 77 pm + 74 pm = 67 pm + 60 pm = 74 pm + 37 pm =
SG-24
1 1 4 pm 1 54 pm 1 5 1 pm 1 27 pm 1 1 1 pm
Actual
� 1 09 pm � 1 50 pm 1 34 pm l 20 pm 97 pm
Chapter 3 MOLECULAR SHAPE AND STRUCTURE THE VSEPR MODEL ( Sections 3.1-3.3) 3.1 The Basic VSEPR (Valence-Shell Electron-Pair Repulsion) Model •
Valence electrons about central atom(s) -;
•
Lewis structure
•
Bonds, lone pairs
•
Bond angle(s)
•
Multiple bonds
•
Electron arrangement
-;
Controls shape of a molecule
Shows distribution of valence electrons in bonding pairs (bonds) and as lone pairs
-; Regions of high electron density that repel each other (Coulomb's law) by rotating about a central atom, thereby maximizing their separation
-; Angle(s) between bonds joining atom centers -; Treated as a single region of high electron concentration in VSEPR -;
Ideal locations of bond pairs and lone pairs about a central atom (angles between electron pairs define the geometry)
Typical Electron Arrangements in Molecules with One Central Atom
L i n e a r ( 1 80°)
Trigo n a l p l a n a r ( 1 20°)
Trigonal b i pyramidal
( 1 200 equ ato/'ial, 900 a x ia l-e q u a torial)
•
Tetra hedral ( 109.5°)
Octa hedra l (90° and 180°)
VSEPR formula, AXIlEI1I (See Section 3.2 in the text) -; A = central atom -; Xn = n atoms (same or different) bonded to central atom -; Em = m lone pairs on central atom -; Useful for generalizing types of structures Note:
Molecular shape is defined by the location of the atoms alone. The bond angles are L. X-A-X.
SG-2S
Typical Shapes of Molecules with One Central Atom*
180 0 '
' •....
� "" . ' .... . "',
.
.
..• ' .
Linear
Seesaw
Angular
Square planar
Trigonal Square pyramidal bipyramidal
T-shaped
Tetrahedral
Octahedral
Pentagonal bipyramidal
* These figures depict the locations of the atoms only, not lone pairs. •
(See Toolbox 3 . 1 in the text) Write the Lewis structure(s). If there are resonance structures, pick any one. Count the number of electron pairs (bonding and nonbonding) around the central atom(s). Treat a multiple bond as a single unit of high electron density. IdentifY the electron arrangement. Place electron pairs as far apart as possible. Locate the atoms and classifY the shape of the molecule. Optimize bond angles for molecules with lone pairs on the central atom(s) with the concept in mind that repulsions are in the order
VSEPR (sometimes pronounced "vesper") method �
�
�
�
�
lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. Example:
Use the VSEPR model to predict the shape of tetrafluoroethene, C2 F 4 • Lewis structure:
:'F: :'F : I
I
:r,- C = C-f.: ..
"
There are three concentrations of electron density around each C atom, so the shape is trigonal planar about each C atom. C2 F 4 is an AX3AX3 species. 3,2 M o l ec u les with L o n e Pairs on the Central Atom •
VSEPR method for molecules with lone pairs (m *- 0)
� �
Lone pair electrons have preferred positions in certain electron arrangements. In the trigonal bipyramidal arrangement, lone pairs prefer the equatorial positions, in which electron repulsions are minimized.
SG-26
�
I n the octahedral arrangement, all positions are identical. The preferred electron arrangement for two lone pairs is the occupation of opposite corners of the octahedron. Example: Use the VSEPR model to predict the shape of trichloroamine, NCI 3 . :CI. - N - C I : , ,
Lewis structure:
.
I
: CI :
The four electron pairs around N yield a tetrahedral
electron arrangement, while the shape is trigonal pyramidal (l ike a badminton birdie). Because lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, the bond angles [L' CI-N-CI ] will be less than 1 09,5°. 3.3 Polar Molecules �
•
Polar molecule
•
Polar bond
•
Molecular dipole moment
•
�
Bond with a nonzero dipole moment �
Representation of dipole � Example:
•
Molecule with a nonzero dipole moment
Lewis structure:
Vector sum of bond dipole moments
Single arrow with head pointed toward the positive end of the dipole H - C l : and polar (dipole) representation: H -- CI
Predicting molecular pola rity
�
�
Determine molecular shape using VSEPR theory. Estimate electric dipole (bond) moments (text Section 2. 1 2) or use text Fig. 3.7 to decide whether the molecular symmeliy leads to a cancellation of bond moments (nonpolar molecule, no dipole moment) or not (polar molecule, nonzero dipole moment). Example:
I s XeF 2 a polar or nonpolar molecule?
Lewis structure: :'F-'X�'- F': The electron arrangement is trigonal bipyramidal; the shape is linear (Io'n'e pa1rs prefer the equatorial positions). The bond angle is 1 80°, the bond dipoles cancel, and the molecule is nonpolar: F --- Xe -- F .
VALENCE-BOND (VB) THEORY (Sections 3.4-3.7) 3.4 Sigma (0) and Pi (n) Bonds • •
Two major types of bonding orbitals �
Ci
and n
Ci-orbital
�
�
�
Has no nodal sUiface containing the interatomic (bond) axis Is cylindrical or "sausage" shaped Formed from overlap of two s-orbitals, an s-orbital and a p-orbital end-to-end, two p-orbitals end to-end, a certain hybrid orbital and an s-orbital, a p-orbital end-to-end with a certain hybrid orbital, or two certain hybrid orbitals end-to-end
SG-27
•
Example of a-orbital formation �
Two I s-orbitals of H atoms combine to form a a-orbital of H 2
atomic orbitals
•
n-orbital
� Nodal plane �
� •
containing the interatomic (bond) axis Two cylindrical shapes (lobes), one above and the other below the nodal plane Formed from side-by-side overlap of two p-orbitals
Example of n-orbital formation
�
Two 2px-orbitals overlap side-by-side
---
•
Electron occupancy �
� •
a- and n-orbitals can hold 0, I, or 2 electrons, corresponding to no bond, a half strength bond, or a ful l covalent bond, respectively. I n any orbital, spins of two electrons must be paired (opposite direction of arrows).
Types of bonds according to valence-bond theory
� Single bond � �
[H2 ' for example] [02 , for example] [N2 ' for example]
is a a-bond. is a a-bond plus one n-bond. Triple bond is a a-bond plus two n-bonds. Double bond
3.5 Electron Promotion and the Hybridization of Orbitals and 3.6 Other Common Types of Hybridization •
Hybrid orbitals
�
�
Produced by mixing (hybridizing) orbitals of a central atom A construct consistent with observed shapes and bonding in molecules
Hybridization Schemes E lectron arrangement a round the central atom
Linear Trigonal planar Tetrahedral Trigonal bipyramidal Octahedral
H ybrid orbitals (number)
Angle(s) between a-bonds
sp (two) Sp 2 (three)
1 80°
BeCb, �;;:O2
1 20°
.�.F3 , �H 3+
Sp 3 (four)
1 09.5°
sp 3d (five) 2 sp 3d (six)
�CI4' NH4+
1 20°, 90°, and 1 80°
.eCl s
90° and 1 80°
.s,F 6
SG-28
Example(s) [ underlined atom)
•
Determination of hybridization schemes
---+ ---+
---+
---+ ---+
•
Write the Lewis structure(s). If there are resonance structures, pick any one. Determine the number of lone pairs and a-bonds on the central atom. Determine the number of orbitals required for hybridization on the central atom to accommodate the lone pairs and a-bonds. Determine the hybridization scheme from the preceding table. Use VSEPR rules to detennine the molecular shape and bond anglers) . (re-bonds are formed from the overlap of atomic orbitals [unhybridized] in a side-by-side arrangement.) Example: Describe the bonding in BF 3 using VB theory and hybridization. ·· : F: .· / The shape is trigonal planar (VSEPR) and the hybridization at B is sl .· F""" 8 (see table above) and the three a-bonds are formed from overlap of , : F: B(sl) orbitals with F(2p) orbitals (B(s p 2)-F(2p) bonds).
Central atoms
Example:
---+
Each central atom can be represented using a hybridization scheme, as in the example below.
Use VB theory to describe the bonding and in and shape of acetaldehyde, a molecule with two central atoms. H I
H-C-C-H I I-I
Lewis structure: Bonds :
Angles:
H
..
:0 II
3-D representation :
:0 '
C- C H� 'H H \
II
C-C a-bonds from overlap of methyl C Sp3 with aldehyde C Sp2 hybrid orbital [Csi CSp2 ] ; c=o a-bond from overlap of aldehyde sl hybrid orbital with methyl C Sp3 orbital; C=O re-bond from side-by-side overlap of C 2p and 0 2p atomic orbitals; C-H single bonds from overlap of H I s AOs and Csi hybrid orbitals (methyl C atom) and H i s AO and Cs/ hybrid orbital
Three L H-C-H 1 09.5°, three L H-C-C 1 09.5°, one L C-C=O L O=C-H 1 20°, and one L C-C-H 1 20° �
�
�
Molecular shape:
�
1 20°, one
�
Tetrahedral at methyl carbon atom, trigonal planar at aldehyde C atom
3.7 Characteristics of M ultiple Bonds
Characteristics: tetrahedral geometry and sp 3 hybridization at C atoms; all C-C and C-H single (a) bonds; rotation allowed about C-C single bonds Examples: Methane, CH4 ; and propane, CH 3 CH 2 CH 3
•
Alkanes (Cn Hzn+z , n
•
Alkenes (CHzn , n
=
1 , 2, 3, ...)
---+
---+
Characteristics: one C=C double bond (a plus re); other C-C bonds and all C-H bonds single (a); trigonal-planar geometry and Sp 2 hybridization at double-bonded C atoms; tetrahedral geometry at other C (Sp3) atoms; rotation not allowed about double C=C bond; rotation al lowed about C-C single bonds Examples: Ethene (ethylene), H 2 C=CH 2 ; and propylene, CH 3CH=CH 2 =
2, 3, 4, ... )
SG-29
•
•
•
Alkynes (CnHZn-Z ' n= 2, 3, 4, ... ) �
Characteristics: one C=C triple bond (a plus two n); other C-C bonds and all C-H bonds (a); linear geometry and sp hybridization at triple-bonded C atoms; tetrahedral geometry at other C (Sp 3) atoms Examples: Ethyne (acetylene), HC=CH ; 2-butyne (dimethylacetylene), H 3 CC=CCH 3
Benzene (C6H6)
�
�
C framework; trigonal-planar geometry and Sp 2 hybridization at C atoms; a framework has C-C and C-H single bonds; 2p-orbitals (one from each C atom) overlap side-by-side form three n-bonds (six n-electrons) delocalized over the entire six C C haracteristics: Planar molecule with hexagonal
atom ring
See Lewis resonance structures in Section 2.7.
Double bonds
�
�
�
�
Consist of one a- and one n-bond; are always stronger than a single a-bond: C=C is weaker than two single C-C a-bonds N=N is stronger than two single N-N a-bonds 0=0 is stronger than two single 0-0 a-bonds Formed readily by Period 2 elements (C, N, 0) Rarely found in Period 3 and higher period elements Impart rigidity to molecules and influence molecular shape Example:
Describe the structure of the carbonate anion in terms of hybrid orbitals, bond angles, and a- and n-bonds. Lewis resonance structures:
[ :0 :0: :0: ]2I II I C --- C --- C . / , . • "y , . • / �
:R· .�:r
:9
·�r
:R·
.
Q
.
Trigonal-planar geometry with s p 2 hybridization at C, 1 200 L O-C-O angles, and Cs/-02p single bonds. Remaining C 2p-orbital can overlap with 0 2p-orbitals to form a n-bond in each of three ways corresponding to the resonance structures. Therefore, each carbon-oxygen bond can be viewed as one a-bond and one-third of a n-bond.
MOLECULAR ORBITAL (MO) THEORY (Sections 3.8-3.12) 3.8 The Limitations of Lewis's Theory •
Valence Bond (VB) deficiencies �
�
�
Cannot explain paramagnetism of O2 (See text Box 3.2 for explanation ofparamagnetism and
diamagnetism)
Difficulty treating electron-deficient compounds sllch as diborane, B2 H6 No simple explanation for spectroscopic properties of compounds such as color
SG-30
•
Molecular Orbital (MO) advantages
�
�
�
�
Addresses all of the above shortcomings of VB theory Provides a deeper understanding of electron-pair bonds Accounts for the structure and properties of metals and semiconductors More facile for computer calculations than VB theory
3.9 Molecular Orbitals (MOs) •
MO theory �
� •
In MO theory, electrons occupy MOs that are delocalized over the entire molecule. In VB theory, bonding electrons are localized between the two atoms.
Molecular Orbitals (MO)
�
Formed by superposition (linea r combination) of atomic orbitals (LCAO-MO) (constructive interference): Increased amplitude or electron density between atoms Antibond ing orbital (destructive interference): Decreased amplitude or electron density between atoms
� Bonding orbital � •
Types of Molecular Orbitals
�
energy lower than that of the constituent AOs energy equal to that of the constituent AOs Antibonding MO: energy greater than that of the constituent AOs Bonding MO:
� Nonbonding MO : �
•
Rules for combining AOs to obtain MOs �
�
•
N atomic orbitals (AOs) yield N molecular orbitals (MOs)
Little overlap of inner shell AOs, and these MOs are usually nonbonding.
MO energy-level d iagrams � �
Relative energies of original AOs and resulting MOs are shown schematically in energy-level diagrams. Arrows used to show electron spin and to indicate location of the electrons in the separated atoms and the molecule Example:
Formation of MOs by LCAO method and MO energy-level diagram for the fonnation of Li 2 from Li atoms . Formation of bonding and antibonding orbitals: \112s
\112s
2s
2s
\112s + \112s
\V2s
-\V2s
\112s - \112s
0 + O -- O D
0 + O -- C) 2scr
2s
Energy-level diagram
2scr
SG-3 l
2s
2scr*
3 . 1 0 The Electron Configura tions of Diatomic Molecules •
Proced u re for determining the electronic configuration of d iatomic molecules
� �
� •
Construct all possible MOs from valence-shell AOs. Place valence electrons in the lowest energy, unoccupied MOs. Follow the Pauli exclusion principle and H u nd's rule as for AOs: Electrons have spins paired in a fully occupied molecular orbital and enter unoccupied degenerate orbitals with parallel spins.
Valence-shell MOs for Period 1 and Period 2 homonuclear d iatomic molecules
Correlation Diagrams Energy 1
Is
----<
,
,
,
"
I S0*
2p
\
\
,
\ S0
\
,
Bond order (b)
�
I
\
Is
2s "
---<
±
2prr*
\ ' ,
}-
Schematic order of MO energy levels for H 2 , He 2 •
,/ '
'---
---::---{
\
\
\
I
" 2p0* '-
1
\
�
2p0 2prr
'---
----.J
2S0* 2S0
\
\
\
' ,\
" "
,
"
"
2p
---::--{:
2s
---<
>---=--
�
Schematic order of MO energy levels for Li 2, Be 2 , B 2 , C 2 , N 2
b = (N - N * )
1 I,
2p
\' \
"
2s "
2p0* '-
� ' 2prr ----;;-"
,
2prr
'--
\
\
\ ,\
2p0
2S0*
2S0
I
,
"
2p
>---=--
______ ',
2s
>---
Schematic order of MO energy levels for O 2 , F 2 , Ne 2
I
N = total number of electrons in bonding MOs N * = total number of electrons in antibonding MOs Examples:
b in Lh = I (two valence electrons in bonding 2scr MO). b in H 2- = .!..2 (two electrons in I scr bonding MO, one electron in I scr * -antibonding MO). 0/ has 1 1 valence electrons, which are placed in the 2scr, 2scr*, 2pcr, two degenerate 2prr., and one of the 2 2 2 degenerate 2prr.* MOs. The valence electron configuration is (2scr) (2scr*) (2pcr) 4 (2prr.) (2prr.*) I. There is one unpaired electron. N = 8 and N * = 3. Bond order = .!..2 ( 8 - 3) = 5/2.
3.1 1 Bonding in Heteron uclear Diatomic Molecules •
Bond properties
�
Polar in nature
�
Atomic orbital from the more electronegative atom is lower in energy.
�
Bonding orbital has a greater contribution from the more electronegative atom.
�
Antibonding orbital has a greater contribution from the less electronegative atom.
SG-32
•
Result of bond properties is modified, unsymmetrical correlation d iagrams. E xample:
HCI molecule
H atom
t
�
Is
cr'
,
CI atom
,
,
cr
3s
3s
Bonding in HCI using MO theory. The correlation diagram shows occupied valence orbitals for H and CI atoms, resulting MOs from overlap of I s H and 3pz Cl orbitals, lone pair nonbonding orbitals 3px and 3Py of Cl, and the nonbonding Cl 3s orbital. The electronic configuration is (CI 3s/ a2 (Cl 3px/ (Cl 3pyf 3 . 1 2 Orbitals in Polyatom ic Molecules •
•
•
•
Bond properties
�
All atoms in the molecule are encompassed by the MOs. � Electrons in MOs bond all the atoms in a molecule. � Electrons are delocalized over all the atoms. � Energies of the MOs are obtained by spectroscopy. While MO theory requires MOs to be delocalized over all atoms, Effectively localized electrons � electrons sometimes are effectively localized on an atom or group of atoms. Example: See the HCI example above with nonbonding Cl 3p, and Cl 3py orbitals effectively localized on Cl. Combining VB and MO theories �
VB and MO theory are sometimes jointly invoked to describe the bonding in complex molecules. Example: In benzene, localized s/ hybridization (hybrids of C 2s-, C 2px-, and C 2Py orbitals on each C) is used to describe the a-bonding framework of the ring, and delocalized MOs (LCAOs of six C 2p:-orbitals) to describe the rc-bonding.
Triumphs of MO theory
� �
Accounts for paramagnetism of O2 Accounts for bonding in electron-deficient molecules Example:
In diborane (B2 H6. with 1 2 valence electrons), six electron pairs bond eight atoms using six MOs. VB theory requires eight electron pairs and fails to describe the bonding.
SG-33
D i borane
�
Provides a framework for interpreting visible and ultraviolet (UV) spectroscopy Example:
Energy
----n*
H
--
-----n*
LUMO
HOMO
7(
H
7(*
__
H
--
hv (light)
n
1t*
t •
7(*
�
----;t*
H
n
--
n
H n
n
Ground state
Excited state
The UV spectrum of benzene is interpreted as the excitation of an electron from the
highest occupied molecular orbital (HOMO, bonding n M O) to the lowest unoccupied molecular orbital (LUMO, antibonding n* M O ). The bond order ofthe
n-system changes from 3 to 2.
IMPACT ON MATERIALS: ELECTRONIC CONDUCTION IN SOLIDS (Sections 3.13-3.14) 3 . 1 3 Bonding in the Solid State and 3 . 1 4 Semiconductors • •
MO theory �
Bands
�
�
�
•
Groups of MOs having closely spaced energy levels (see text Fig. 3 .43) Formed by the spatial overlap of a very large number of AOs Separate into two groups (one mostly bonding, the other mostly antibonding)
Conduction band � Example:
• • •
Accounts for electrical properties of metals and semiconductors
Formation of a conduction band in Na by the overlap of 3s-orbitals
Valence band � Band gap �
Empty or partially filled band
Completely filled band of MOs (insulators such as molecular solids)
An energy range between bands in which there are no orbitals
Electronic cond uctors � I nsulator: Metallic conductor:
Metals, semiconductors, and superconductors Does not conduct electricity: substance with a full valence band far in energy from an empty conduction band (large band gap) Current carried by delocalized electrons in bands Conductivity decreases with increasing temperature Substance with a partially filled conduction band (gap not relevant)
SG -3 4
Semicond uctor:
Current carried by delocalized electrons in bands Conductivity increases with increasing temperature Substance with a full valence band close in energy to an empty conduction band (smal l band gap) : Excitation of some electrons from the valence to the conduction band yields conductivity.
Su percond uctor:
Zero resistance (infinite conductivity) to an electric
Example:
current below a definite transition temperature
The ionic solid K3 (C60) below 1 9.3 K. Ions in the solid are K+ and (C60)'-. The molecule C6Q is buckminsterfullerene (buckyball).
•
Enhanced semiconductors
•
n-type semiconductor
•
p-type semiconductor
•
-;
Prepared by doping, that is, adding a small amount of impurity
-; Adding some valence electrons to the conduction band (n electrons added) Example: A small amount of As (Group 15) added to Si (Group 1 4): Extra electrons are transferred from As into the previously empty conduction band of Si. =
-;
Removing some valence electrons from the valence band (p = positive "holes" in the valence band) Example: A smal l amount of In (Group 1 3) added to Si (Group 1 4): Valence electron deficit is created in the previously full valence band of Si, thereby becoming a conduction band.
p-n junctions
-; p-type semiconductor in contact with an n-type semiconductor -; Solid-state electronic devices Example:
One type of transistor is a p-n sandwich with two separate wires
(source and emitter) connected to the n-side. It acts as a switch in the
following way: When a positive charge is applied to a polysilicon contact (base wire) positioned between the source and emitter wires, a current flows from the source to the emitter. Other examples include diodes and integrated circuits.
SG -3 5
Chapter 4 THE PROPERTIES OF GASES THE NATURE OF GASES (Sections 4.1-4.3) 4.1 Observing Gases �
Examples of bulk matter, forms of matter consisting of large numbers of molecules
•
Gases
•
Atmosp here
•
Physical p roperties
•
Not uniform in composition, temperature (T), or density (d) �
Most gases behave similarly at low pressure and high temperature.
Compressibility
� �
•
�
Ease with which a gas undergoes a volume ( V ) decrease Gas molecules are widely spaced and gases are highly compressible. Gas molecules are in ceaseless chaotic motion.
Expansivity
�
�
The abil ity of a gas to fi l l the space avai lable to it rapidly Atomic and molecular gases move rapidly and respond quickly to changes in the available volume.
4.2 Pressu re
Pressure •
�
�
or
P=£ A
Opposing force from a confined gas
A glass tube, sealed at one end, filled with liquid mercury, and inverted into a beaker also containing l iquid mercury (Torricell i) The height (h) of the liquid col umn is a measure of the atmospheric pressure (P) expressed as the length of the column for a particular l iquid. Common pressure units: mmHg, cmHg, cmH 20
P = F = mg = (d V )g = (dhA)g = dhg A A A A where m = mass of liquid, g = acceleration of gravity (9.806 65 m·s-2 at the Earth's surface), and d = density of l iquid The atmospheric pressure P when the height h of a column of mercury in a barometer is exactly
0.76 m at O°C is 1 0 1 325 kg·m- J ·s-2 , a result which is obtained by using the value of the density of mercury = 1 3 595. 1 kg· m-3 and the gravitational constant at the surface of the earth. 2 2 (SI unit of pressure) Note: 1 Pa (pascal) = 1 kg·m- J ·s- = I N·m-
Manometer (see Fig. 4.5 in the text)
� �
�
•
force area
Barometer
�
•
==
U-shaped tube fil led with liquid connected to an experimental system "open-tube" pressure (system) = atmospheric pressure when levels are equal "closed-tube" pressure (system) = difference in level heights
Ga uge pressure
�
Difference between the pressure inside a vessel containing a compressed gas (e.g., a tire) and the atmospheric pressure
SG-37
4.3 Alternative Units of Pressu re •
Atmosphere -+
I I 1 I
atm = 760 Torr (bold type exact) atm = 1 0 1 325 Pa = 1 0 1 .325 kPa bar = 1 05 Pa = 1 00 kPa Torr = I mmHg (Not exact, but correct to better than 2 x 1 0- 7 Torr) =
THE GAS LAWS (Sections 4.4-4.8) 4.4 The experimental Observations •
Boyle's Law
•
Gas behavior
(fixed amount of gas n at constant temperature T) -+
Boyle's law describes an isothermal system with a fixed amount of gas: 6T = 0,
6n = ° (constant T, n ).
Pressure oc
or
----
volume
I
or
P oc V
PV
=
constant
The functional form is a hyperbola (plot of P versus V), but a plot of P versus V- I gives a straight line. •
Cha nges between two states ( 1 and 2) at consta nt temperature
� Vj
=
constant = P2 V2
or
P2 �
=
IIV2 I/Vj
=
2:L V2
At low temperature and/or high pressure
•
Deviations -+
•
Charles's Law (fixed amount of gas n at constant pressure P)
•
Gas behavior
-+
Charles's law describes an isobaric system with a fixed amount of gas: M = 0, 6n = 0 (constant P, n). Volumeoc temperature
or
V oc T
or
V IT
=
constant
The functional form is linear. •
Kelvin temperatu re scale -+
T = 273. 1 6 K T= 0 K •
Twofixed points define the absolute temperature scale: The triple point of water, where ice, liquid, and vapor are in equil ibrium (see Chapter 8) . Defined from the limiting behavior of Charles ' s law. V approaches 0 at fixed low pressure (extrapolation).
Cels ius temperature scale -+ t (0C) Notes:
=
T(K)
-
273. 1 5
(exactly)
The freezing point of water at I atm external pressure is O°C or 273 . 1 5 K. The symbol t is also used for time.
SG-3 8
•
Changes between two states ( l and 2)
VI /� = constant = �/T2 ---t
or
At high pressure and/or low temperature
•
Deviations
•
Another aspect of gas be havior (fixed amount of gas n at constant vol u me V ) ---t ---t
Pressure oc temperature
or
oc
This relation describes an isochoric system : t-. V = 0 (constant V). The functional form is linear.
•
Cha nges between two cond itions ( 1 and 2)
•
Deviations
•
=
P /T constant
or
P T
---t
For small volume
(gas at constant pressure P and temperature T) A given number of gas molecules at a particular P and T occupies the same volume regardless of chemical identity.
Avogadro ' s princi ple ---t
I Volume
amount
V oc n
or
or
V/n = constant
I
•
Gas behavior
•
Molar volu me
•
Cha nges between two cond itions ( 1 and 2)
•
Deviations
•
A nother aspect of gas behavior (gas at constant vol ume V a nd temperatu re T)
---t
oc
for M = 0 and t-.T = 0
At high pressure and/or low temperature
Pressure oc amount
or
•
Cha nges between two conditions ( 1 and 2)
•
Deviations
•
The ideal gas law, P V = nRT ( R
•
---t
---t
---t
P n oc
or
=
Pin constant
For small volumes =
constant)
Incorporates Boyle's Law, Charles's law, and Avogadro's principle R is the universal gas constant described below.
Gas behavior
---t
As P approaches 0, all gases behave ideally (limiting law).
SG-39
•
Derivation Boyle
v=
P V = constant 1
•
Charles
Avogad ro
constant 2 x T
P = constant 3 x n
P V = (constant 3 x n)(constant 2 x T ) = nRT
equation ofstate
2 R = 8.205 74 x 1 0- L·atm-K- 1 ·mor 1
-7
Gas constant
R = 8.3 1 4 47 L· kPa·K- 1 ·mor l = 8.3 1 4 47 lK- 1 ·mor l
-7
4.5 Applications of the Ideal Gas Law •
P2 V2 n2 T2
lJ Vi nl TJ
_ - - --
Cha nges between two conditions (1 and 2)
Combined gas law Note:
The more general form of the combined gas law includes the case where the amount of gas may change.
Note:
When using this relationship, remember that the actual units used in manipulating ratios of quantities are not important. Consistency in units is essential. A bsolute temperature, however, must be used in all the equations.
I
Vm =
RT P
I
•
Mola r volu me
•
Standard ambient temperature and press ure (SATP)
•
Standard temperature and press u re (STP)
•
Molar volu me at ST P -7
Vm
=
RT
-7
-7
ODC and I atm [or 273. 1 50 K and 1 .0 1 3 25 bar]
2 ( 8.205 8 x I 0- L · atm - K -1 ·mol -1 )(273 . 1 50 K)
=
( I atm)
P
4.6 Gas Density
Molar concen tration
•
Molar concentration at STP
n
--
-
P RT
n = - = -
•
-7
2 9 8. 1 5 K and I bar
V
�n
= --,-
22.4 1 4 1 L ·mol - I
2 4.46 1 48 x 1 0- mol · L- 1
SG-40
= 22.4 1 4 1 L . mol - I
I mn d=
M MP =V=
•
Density of a gas
•
Density of a gas at ST P �
4.7
V
liT
d=
m V
=
I
nM V
= (4.46 l 48x l O-2 mol . L-' ) x M
The Stoichiometry of Reacting Gases
•
Accounting for reaction volu mes �
•
Balanced chemical eq u ation � Example:
4.8
Use mole-to-mole calculations as described in Sections L and M, and convert to gas volume by using the ideal gas law.
See Example 4.7 and the fol lowing self-test exercises in the text.
NH 4N03 (s) - N 2 0 (g) + 2 H 2 0 (g) A total of three moles of gas products is produced from one mole of solid.
M ixtu res of Gases
Mixtures of nonreacting gases behave as if only a single component were present.
•
Behavior �
•
Partial p ress u re �
•
Law of partia l p ress ures �
•
Components at the same T and V �
Pressure a gas would exert if it occupied the container alone
P = nRT V
Total pressure of a gas mixture is the sum of the partial pressures of its components [John Dalton] P = PA + Ps + . . . for mixture containing A, B, . . .
n = nA + ns + . . = total amount of mixture
and
.
_
PA -
nART V
nA PA XA = = = mole fraction of A in the mixture P n Thus, PA = xAP •
H u mid gas �
For air, P = Pdry air + Pwatervapor , and Pwater vapor = 47 Torr at body temperature (37°C). See Section 8.3
Note:
M O LECULAR M OTION (Section 4.9-4.11) 4.9
Diffu sion and Effusion
•
Diffusion �
•
Effusion � �
Gradual dispersal of one substance through another substance
Escape of a gas through a smal l hole (or assembly of microscopic holes) into a vacuum Depends on molar mass and temperature of gas
SG-41
•
G raham's law ---1-
Rate of effusion of a gas at constant temperature is inversely proportional to the square root of its molar mass M. Rate of effusion of A Rate of effusion of B
For two gases A and B ---1-
=
�
MB MA
Rate of effusion is proportional to the average speed of the molecules in a gas at constant
temperature.
A verage speed of A molecules A verage speed of B molecules ---1-
Effusion at different temperatu res
---1-
Rate of effusion at T2 Rate of effusion at 11 •
�
MB MA
Rate of effusion is inversely proportional to the time taken for given volumes or amounts of gas to effuse at constant temperature. Rate of effusion of A Rate of effusion of B
•
=
Combined relationship
---1-
=
=
Time for B to effuse Time for A to effuse
=
�
MB MA
Rate of effusion and average speed increase as the square root of the temperature. Average speed of molecules at T2 Average speed of molecules at 11
=
(fi �r:
Average speed of molecules is directly proportional to the square root of the temperature and inversely proportional to the square root of the molar mass.
A verage speed of molecules in a gas
ex
&
4.1 0 The Kinetic Model of Gases •
Assumptions
I . Gas molecules are in continuous random motion. 2. Gas molecules are infinitesimally small particles. 3. Particles move in straight li nes until they coll ide. 4. Molecules do not influence one another except during collisions. •
Coll is ions with W a l ls
---1-
Consider molecules traveling only in one dimension x with an average of square velocity of ( v}) .
SG -4 2
•
Pressure on wall
�
P
=
Nm (Vx ) 2
, where N is the number of molecules, m is the mass of each
V
molecule, and V is the volume of the container •
Mean square speed �
=
•
Pressure on wall �
Solve for
Note:
•
=
(V}
P
=
Nm vr m s 2
p
=
nNA m vr m s
PV
•
( )
2
+ vy + v} vrm / = v 2 rms root mean square
2
-
nMvr m s 2
3V
nMvr m s 2 " .J
Vnns
Vrms
=
(v}) (V;) (
2 + vz
+
) = 3 ( v})
and N = nNA, where NA is the Avogadro constant
3V
=
)
3V =
nRT
where M = mNA is the molar mass
The term nRT is from the ideal gas law.
is the root mea n sq uare speed of the particle.
Beware afunits. Use R in units of J . K- I · mor l and convert M to kg· mor l . The resulting
units of speed will then be m·s- I .
The equation above demonstrates that the kinetic of a gas is proportional to the temperatu re.
4. 1 1 The Maxwell Distribution of Speeds •
•
Symbols �
Let v represent a particle ' s speed, N the total number of particles,f(v) the Maxwell distribution of speeds, /',. ajinite change, and d an injinitesimal change (not density) in the fol lowing relationships.
For ajinite range of speeds t,.N
•
•
=
Nf(v)/',.v, where f(v) =4 n:
(�) 2 n:RT
3/2
v
2 e-Mv 2 / 2RT
Temperature dependence ofj(v) �
As T increases, the fraction of molecules with speeds greater than a specific speed increases. See Figure 4.27 (hot molecules move faster than cool ones on average).
Dependence ofj(v) on molar mass M
�
As M decreases, the fraction of molecules with speeds greater than a specific speed increases. See Figure 4.26 (l ight molecules move faster than heavy ones on average).
SG-43
•
For a n infinitesimal range of speeds
I �� Note:
f(v)dv
I
Calculus can be used with the Maxwell distribution of speeds to obtain the foll owing properties that are not mentioned in the text.
A verage speed
( v) =
Most probable speed
vm p -
_
rvf(v) dv = )8RT ill
)2RT M
, where
df(v) 0 dv
(maximum in distribution)
THE IMPACT ON MATERIALS: REAL GASES (Sections 4.12-4.14) 4 . 1 2 Deviations from Ideality •
Defi nition �
�
•
•
Evidence
�
Attractions and repulsions between atoms and molecules cause deviations from ideality in gases. Attractions have a longer range than repulsions. Gases condense to liquids when cooled or compressed (attraction). Liquids are difficult to compress (repulsion).
Compression factor (Z) � �
A measure of the effect of intermolecular forces Is the ratio the observed molar volume of a gas to that calculated for an ideal gas For an ideal gas, Z = I For H 2 , Z > 1 at all P (repulsions dominate) For NH3 , Z < I at low P (attractions dominate)
•
For many gases, methane for example (Figure 4.28), attractions dominate at low press u res (Z < 1 ), while rep u ls ive interactions dominate at h igh press u res (Z > 1 ).
4 . 1 3 The Liq uefaction of Gases •
Jou le-T homson effect �
�
When attractive forces dominate, a real gas cools as it expands. (Compression and expansion through a small hole lowers temperature further.) Exceptions are He and H 2 , for which repulsion dominates. The effect is used in some refrigerators and to effect the condensation of gases such as oxygen, nitrogen, and argon.
SG-44
4.1 4 Equations of State of Real Gases
•
Virial equation �
P V = nRT
B
C=
Note:
•
=
[ -B -e 1
+
Vm
+
2
V;n
+
second vi rial coefficient third virial coefficient, etc.
Virial coefficients depend on temperature and are found by fitti ng experimental data to the virial equation. This equation is more general than the van der Waals equation but is more difficult to use to make predictions.
van der Waals equation �
[ ;: ) P+a
(V - nb) = nRT
where a and b are the van der Waals para meters (determined experimentally)
•
Rea rranged form
�
•
van der Waals Z
�
•
where a accounts for the attractive effects and b accounts for the repulsive effects
nRT -a� V - nb V2 V V - nb
an RTV
1 - (nb I V)
an RTV
Repulsion
�
�
� � �
� •
P=
Repulsive forces imply that molecules cannot overlap. Other molecules are excluded from the volume they occupy. The effective volume is then V - nb, not V. Thus, b is a constant corresponding to the volume excluded per mole of molecules. The potential energy describing this situation is that of an impenetrable hard sphere. The units of b are L·mor l .
Attraction
�
�
�
�
�
Attractive forces lead to clustering of molecules in the gas phase. Clustering reduces the total number of gas phase species. The rate of collisions with the wall (pressure) is thereby reduced. Because this effect arises from attractions between pairs of molecules, it should be proportional to the square of the number of molecules per unit volume (N / V ) 2 or equivalently, to the square of the molar concentration (n I V ) 2 . The pressure is predicted to be reduced by an amount a(n I V ) 2 where a is a positive constant that depends on the strength of the attractive forces.
,
SG -4S
� �
The potential energy describing this situation is that of long-range attraction with the form - ( 1 / r 6), where r is the distance between a pair of gas phase molecules. This type of attraction primari ly arises from dispersion or London forces (see Chapter 5). The units of a are atm·L2 ·mor2 .
SG-46
Chapter 5 LIQUIDS AND SOLIDS INTERMOLECULAR FORCES (Sections 5.1-5.5) 5.1 Formation of Condensed Phases -7
Solid, liquid, and gas Form of matter uniform in both chemical composition and physical state
•
Physical states
•
Phase
•
Condensed phase -7
-7 •
-7
Solid or liquid phase Examples: Ag(s); Sn/Pb(s) alloy; H 2 0(s); H 20(l); 1 % NaCl(s) in H 2 0(I) Molecules (or atoms or ions) are close to each other all the time and intermolecular forces are of major importance.
Cou lomb potential energy Ep -7
The interaction between two charges, q l and q2 , separated by distance r is (see Chapter 1 )
-7
::::;>
� q2
Ep ex:: � r
Almost all intermolecular interactions can b e traced back to this fundamental expression. Note: The term intermolecular is used in a general way to include atoms and ions.
5.2 Ion-Dipole Forces •
Hyd ration
-7 -7
-7 •
I on-d ipole i nteraction
-7
-7 -7
-7
•
-7
l on-dipol e interactions are much weaker than ion-ion interactions, but are relatively strong for
small, highly charged cations.
Accounts for the formation of salt hydrates such as CuS04 ·5 H 20 and CrC1 3 ·6 H 20
Size effects -7
•
The potential energy (interaction) between an ion with I z I II charge I z I and a polar molecule with dipole moment 11 at a ::::;> Ep ex:: - 2rdistance r is For proper alignment of the ion and dipole, the interaction is attractive: Cations attract the partial negative charges and anions attract the partial positive charges on the polar molecule. Shorter range (r - 2) interaction than the Coulomb potential (r - I ) Polar molecule needs to be almost in contact with ion for substantial ion-dipole interaction.
Hyd rated compo u nds
-7 •
Attachment of water molecules to ions (cations and anions) Water molecules are polar and have an electric dipole moment 11. A small positive charge on each H atom attracts anions and a small negative charge on the 0 atom attracts cations.
Lt
and Na+ (small) tend to form hydrated compounds. -7 K+ , Rb+ , and Cs + (larger) tend not to form hydrates. -7 NH/ ( 1 43 pm) is simi lar in size to Rb + ( 1 49 pm) and forms anhydrous compounds. Charge effects -7 Ba2+ and K+ are similar in size, yet Ba2+ (larger charge) forms hydrates.
SG-47
5.3 Dipole-Dipole Forces �
I n sol ids, molecules with dipole moments tend to align with partial positive charge on one molecule near the partial negative charge on another.
•
Dipole align ment
•
Dipole-dipole interaction in sol ids �
�
�
� •
Attractive interaction in sol ids for head-to-tail alignment of dipoles 3 Shorter range (r - ) interaction than ion-dipole (r -2) or Coulomb potential (r - I ) For significant interaction, polar molecules need to be almost in contact with each other.
Dipole-d ipole interactions in gas-phase molecules
�
� •
The interaction between two polar molecules with dipole moments II I and 112 aligned and separated by a distance r, is
Dipole-dipole interactions are much weaker in gases than in solids. The potential energy of interaction is Because gas molecules are in motion (rotating as wel l), they experience only a weak net attraction because of occasional alignment.
Dipole-d ipole interactions in liq u id-phase molecules
�
�
�
Same potential energy relationship as in the gas phase, but the interaction is slightly stronger because the molecules are closer. The liquid boiling point is a measure of the strength of the intermolecular forces in a liquid. The boiling point of isomers (see glossary for definition) is often related to the strength of their dipole-dipole interactions. Typical ly, the larger the dipole moment, the higher the boiling point. Example: Cis-dibromoethene (dipole moment � 2.4 D, bp 1 1 2.S°C) vs. trans-dibromoethene (zero dipole moment, bp 1 08°C)
5.4 London Forces •
•
Nonpolar molecu les � London force
�
�
�
�
•
Condensation of nonpolar molecules to form liquids implies the existence of a type of intermolecular interaction other than those described above .
Accounts for the attraction between any pair of ground-state molecules (polar or nonpolar) Arises from instantaneous partial charges (instantaneous dipole moment) in one molecule inducing partial charges (dipole moment) in a neighboring one Exists between atoms and rotating molecules as well Strength depends on polarizability and shape of molecule. Example: Other things equal, rod-shaped molecules tend to have stronger London forces than spherical molecules because they can approach each other more closely.
Polarizability,
�
�
� �
a
Of a molecule is related to the ease of deformation of its electron cloud Is proportional to the total number of electrons in the molecule Because the number of electrons correlates with molar mass (generally increases), polarizability does as well. The potential energy between molecules (polar or nonpolar) with polarizabi lity 0.. 1 and 0..2 , separated by a distance r is
SG -48
Note: •
Same (r -6) interaction as dipole-dipole (r -6) with rotating molecules, but the London interaction is usually of greater strength at normal temperatures.
Liquid boiling poi nt
-j-
-j-
A measure of the strength of intermolecular forces in a liquid In comparing the interactions discussed in Sections 5.3 and 5 .4, the major influence on boi ling point in both polar and nonpolar molecules is the London force. Example:
•
We can predict the relative boiling points of the nonpolar molecules: F 2 , C1 2 , Br2 , and 1 2 • Boiling point correlates with polarizabi lity, which depends on the number of electrons in the molecule. The boil ing points are expected to i ncrease in the order of F 2 ( 1 8 electrons), CI 2 (34) , Br 2 (70) , and 1 2 ( 1 06). The experimental boi ling temperatures are: F 2 (- 1 88°C), Ch (-34°C), Br2 (59°C), and h ( 1 84°C).
Dipole-i nduced d ipole i nteraction -j-
-j-
-j-
The potential energy between a polar molecule with dipole 2 !l l (X 2 moment !ll and a nonpolar molecule with polarizability (X2 => EP oc - -- r6 at a distance r is Also applies to molecules that are both polar, each one can induce a dipole in the other Weaker interaction than dipole-dipole Example: Carbon dioxide (!l = 0) dissolved in water (!l > 0)
5.5 Hydrogen Bon d ing •
Hydrogen bonds
-j-j-j-
-j-j-
-j-
•
I nteraction specific to certain types of molecules (strong attractive forces) Account for unusually high boiling points in ammonia (NH 3 , -33°C), water (H2 0, 1 00°C), and hydrogen fluoride (HF, 20°C) Arise from a H atom, covalently bonded to either an N, 0, or F atom in one molecule, strongly attracted to a lone pair of electrons on an N, 0, or F atom in another molecule Are strongest when the three atoms are in a straight line and the distance between terminal atoms is within a particular range Common symbol for the H bond is three dots : . . . Strongest intermolecular interaction between neutral molecules Examples: N-H · · ·:N hydrogen bonds between N H3 molecules in pure N H 3 O-H · · · : O hydrogen bonds H 2 0 molecules in pure H 20
Hyd rogen bonding in gas-phase molecu les
-j-
-j-
Aggregation of some molecules persists in the vapor phase. In HF, fragments of zigzag chains and (HF) 6 rings ' O : - " H ---: O ' I; \ are formed. In CH 3 COOH (acetic acid), dimers are => H3 C - C C - CH 3 formed. The abbreviated (no C-H bonds shown) \ . /; Lewis structure for the dimer (CH 3 COOH)2 is .O =-- H . . - :O· .
.
•
I mporta nce of hyd rogen bon d i ng
-j-
-j-
-j-
.
.
Accounts for the open structure of sol id water Maintains the shape of biological molecules Binds the two strands of DNA together
SG-49
.
LIQUID STRUCTURE (Sections 5.6-5.7) 5.6 Order in Liquids •
•
•
Liquid phase
� Mobile molecul es with restricted motion � Between the extremes of gas and solid phases Long-range order � Characteristic of a crystalline solid � Atoms or molecules are arranged in an orderly
pattern that is repeated over long distances.
Short-range order
� �
�
Characteristic of the liquid phase Atoms or molecules are positioned in an orderly pattern at nearest-neighbor distances only. Local order is maintained by a continual process of forming and breaking nearestneighbor interactions .
5.7 Viscosity and Surface Tension •
Viscosity �
�
•
•
•
Resistance of a substance to flow: The greater the viscosity, the slower the flow. Viscous liquids include those with hydrogen bonding between molecules [ H3P04 (phosphoric acid) and C3Hg03 (glycerol): many H bonds], liquid phases of metals, and long chain molecules that can be entangled [hydrocarbon oil and greases ].
V iscosity and temperature � Usually viscosity decreases
T
with increasing Exception: unusual behavior of sulfur (see text)
Su rface tension (several definitions) � Tendency of surface molecules in a
liquid to be pul1ed into the interior of the l iquid by an imbalance in the intermolecular forces � Inward pull that determines the resistance of a liquid to an increase in surface area � A measure of the force that must be applied to surface molecules so that they experience the same force as molecules in the interior of the liquid. � A measure of the tightness of the surface layer [Symbol: y (gamma) Units: N·m- 1 or l m- 2 ] Cap i l lary action �
Rise of l iquids up narrow tubes when the adhesion forces are greater than cohesion forces Forces that bind a substance to a surface cohesion: Forces that bind molecules of a substance together to form a bulk material
� adhesion : � •
Men iscus (cu rved surface that a liquid fo rms in a tu be) � Adhesive forces greater than cohesive forces (forms a v shape) � Cohesive forces greater than adhesive forces (forms a " shape) � Glass surfaces have exposed 0 atoms and O-H groups to which �
hydrogen-bonded liquids like H 2 0 can bind. I n this case, the adhesive forces are greater than the cohesive ones. Water wets glass, forms a \..J shape at the surface, and undergoes a capillary rise in a glass tube. Mercury liquid does not bind to glass surfaces. The cohesive forces are greater than the adhesive ones. Mercury does not wet glass, forms a n shape at the surface, and undergoes a capil1ary lowering in a glass tube.
SG -5 0
SOLID STRUCTURES (Sections 5.8-5.13) 5.8 Classification of Solids • •
• •
Amorphous solid �
Atoms or molecules (neutral or charged) lie in random positions.
Crystalline solid
�
Atoms, ions, or molecules are associated with points in a lattice, which is an orderly array of equivalent points in three dimensions. � Structure has long-range order. C rystal faces � Flat, well-defined planar surfaces with definite interplanar angles Classification of crystal l ine solids metallic sol ids: ionic solids: mo lecu lar solids: network solids:
Cations in a sea of electrons Examples: Fe(s), Li(s) Mutual attractions of cations and anions Examples: NaCI(s), Ca(S04Ms) Discrete molecules held together by the intermolecular forces discussed earlier Examples: sucrose, C 1 2 HnO ) ) (s); ice, H 2 0(s) Atoms bonded covalently to their neighbors throughout the entire sol i d Example: diamond, C(d); graphite planes, C(gr)
5.9 Molecular Solids �
�
Solid structures that reflect the nonspherical nature of their molecules and the relatively weak intermolecular forces that hold them together Characterized by low melting temperatures and less hardness than ionic sol ids
5.1 0 Network Solids •
•
•
C rystals
�
�
Atoms joined to neighbors by strong covalent bonds that fonn a network extending throughout the solid Characterized by high melting and boiling temperatures, and hard and rigid structures
Elemental network solids
�
�
Network solids formed from one element only, such as the allotropes graphite and diamond in which the carbon atoms are connected differently Allotropes are forms of an element with different solid structures.
Ceram ics � �
Usually oxides with a network structure having great strength and stability because covalent bonds must break to deform the crystal Tend to shatter rather than bend under stress Examples: Quartz, silicates, and high-temperature superconductors
5. 1 1 Metall ic S o l i d s •
Close-packed structu res
�
�
�
•
Atoms occupy smallest total volume with the least amount of empty space. Metal atoms are treated as spheres with radii r. Type of metal determines whether a close-packed structure is assumed.
Close-packed layer �
Atoms (spheres) in a planar arrangement with the least amount of empty space - contour of the layer has dips or depressions.
SG-SI
-7
-7
Each atom in a layer has six nearest neighbors (hexagonal pattern). Layers stack such that bottoms of spheres in one layer fit in dips of the layer below. -7
I n a close-packed structure, each atom has three nearest neighbors in the layer above, six in its original layer, and three in the layer below for a total of 1 2 nearest neighbors.
•
Stacked layers
•
Coord ination n u m ber
•
Two a rrangements of stac king close-packed layers Pattern :
-7
Number of nearest neighbors of each atom in the solid
ABABABAB . . . A BCABCABC . . .
hcp == hexagonal close-packing ccp == cubic close-packing
•
Occupied space in hcp/ccp structure
•
Tetrahed ral hole -7
-7 •
-7
74% of space is occupied by the spheres and 26% is empty.
Formed when a dip between three atoms in one layer is covered by another atom in an adjacent layer Two tetrahedral holes per atom in a close-packed structure (hcp or ccp)
Octa hedral hole
-7
-7
Space between six atoms, four of which are at the corners of a square plane. The square plane is oriented 45° with respect to two c lose-packed l ayers. Two atoms in the square plane are in layer A and the other two in layer B. There is one additional atom in layer A that is above the square plane and another in layer B that is below the square plane. The result is an octahedral arrangement of 6 atoms with a hole in the center. One octahedral hole per atom in a close-packed structure (hcp or ccp) Figure ofan octahedron and an octahedron rotated to show the relation to two close-packed layers
•
-7
-7
-7
•
Octahedron
Body-centered c u bic (bcc) structure
Coordination number of 8 (not close-packed) Spheres touch along the body diagonal of a cube. Can be converted into close-packed structures under high pressure
Prim itive cu bic (pc) structure
-7
-7
Coordination number of 6 (not close-packed) Spheres touch along the edge of the cube. Only one known example: Po (polonium)
A
Close-packed layer A
B
Close-packed layer B
!�/-� ·\/7· W� �v�
C rystal structure
Coord ination n u m ber
Occupied space
ccp (fcc)
12
74%
Ca, Sr, Ni, Pd, Pt, Cu, Ag, Au, AI, Pb Group 1 8: noble gases at low T, except He
hcp
12
74%
Be, Mg, Ti, Co, Zn, Cd, TI
bcc
8
68%
Group 1 : alkali metals, Ba, Cr, Mo, W, Fe
pc
6
52%
Po (covalent character)
SG-S2
Examples
5. 1 2 Unit Cells •
Lattice
•
U nit cell
•
�
�
A regular array of equivalent points in three dimensions Smallest repeating unit that generates the full array of poi nts
Crystal � Constructed
by associating atoms, ions, or molecules with each lattice point � Meta llic crystals: One metal atom per lattice point [ccp (fcc), bcc, pc] Two metal atoms per lattice point [hcp] •
•
U nit cells for metals � Cubic system: pc, �
bcc, and fcc (ccp) Hexagonal system: primitive unit cell with two atoms per lattice point (hcp)
U nit cel ls in general
Edge lengths: a, b, c Angles between two edges: CJ. (between edges b and c); P (between a and c); y (between a and b) Other crystal systems � Tetragonal, orthorhombic, rhombohedral, monoclinic, and tricIinic
� � • •
Bravais lattices �
1 4 basic patterns of arranging points in three dimensions
�
Each pattern has a different unit cell (see figure below right)
' .:. / @ � @�� . "', . . . t-:Of., . . TETRAGONAL
CUBIC
The 1 4 Bravais Lattices
=>
a
..
a
a
Primitive
•
Vol ume of several unit cells
Cubic: V = a x a x a = a3 [CJ. = P = y = 90 ° ]
Tetragonal : V = a x a x c = a2c [ CJ. = P = y = 90 ° ] Orthorhombic: V = a x b x c = abc [ CJ. = P = y = 9 0 °]
Properties of unit cells
Body-centered
..
.
c
�;; f<'
a
Face-centered
. \ ... . ../ . .
./
a
Primitive
. . �.. \,
"
.....
Body-ce ntered
ORTHORHOMBIC
[]JL!J DD � a
Primitive
Bod y-centered
End face-centered
Face-centered
. I · · ·� · $. dJlJfP · :· · !1ll
HEXAGONAL
!
•
·
. ... / :.... .
::' :
. . . .. ... .. ... . .. . " a 12CY' a
RHOMBOHEDRAL
MONOCLINIC
c
c
a
a
a
a
b
SG-5 3
·
. . .�.. . ... .. . ......
Primitive End face-centered
Each corner point shared among 8 cel ls [ 1 /8 per cell] 8 corner points x 1 /8 per cel l = I corner point per cell [all cells] � Each face-centered point shared between 2 cells [ 1 /2 per cell] 6 face points x 1 /2 per cell = 3 face points per cell flace-centered cells] 2 face points x 1 /2 per cell = I face point per cel l [endface-centered cells] � Each body-centered point unshared [ I per cel l] I body point x 1 per cell = I body point per cel l [body-centered cells] �
TRICLINIC
a
a
e
f
b
•
One lat ice point One [one touch (spr)here)edge: a cornerpoia)]nt (M a r) 8r3 MINA d = --'-( --,-( ------'-)-'-( -- ----:--) --'-) 8r3 One ((sspphere) One here) 8 (1/8)centcorner er point [two r) a)] 4 touch body diagonal: .J3a = 4r a=-r .J3 43 =a3 = ,,3/2 r3 " 68r3
Properties of cubic system unit cells
�
metal atom occupies each
Prim itive
metal atom at each atom (radius per cell (edge length Spheres along an = 2r Volume of the unit cel l : V= 3 = (2 3 = Density of a pc metal = molar mass):
in a unit cell
I atom per cell mass of one atom volume of one unit cell
Body-centered
metal atom at each in a unit cell metal atom at the of the unit cell Total atoms in unit cell = + I (I) = 2 atoms (radius per unit cell (edge length
Spheres
along the
Volume of the unit cell:
or
V
Density of a bcc metal:
d One One [four
_
-
J
� 1 2.J I
(2-"-:) ( MIN,...:-A,-,-) ( 2 atoms per cell ) ( mass of one atom ) (2) ( MINA ) + ( volume of one unit cel l ) ( ..,..-)--, _
-
a3 - 43/33/2 r3 _
->-
here) centcorner pointface ((sspphere) e r r) 8(1/8) 1/ ) 4 a)] = J8r touch face diagonal: J2a = 4r a = �r J2 = a3 = 83/2 r3 A ) = (4)( MINA ) d = (4 ( )( ) ) = (4)( MIN a3 (8312 ) r3 cubiunic t cell atomic radimeasured us density atomic radius densicubityc, atomic radius 175 1 1.3
Face-centered
metal atom at each in a unit cell metal atom at the of each in the unit cell Total atoms in unit cell = + 6( 2 = atoms (radius per unit cell (edge length Spheres
along
or
Vol ume of the unit cel l : V Density of a fcc (ccp) metal:
atoms per cell mass of one atom volume of one unit cel l
•
�
Determining the type from the and of metals in the system; the of a metal atom is determined from the crystal type and the edge length of a unit cel l, which are both obtained by x-ray diffraction (see Major Technique 3 in text). Example: The of lead is 1 1 .34 g·cm-3 and the is pm. Assuming the unit cell is we can determine the type of Bravais lattice. Using the equations given above, calculate the density of Pb assuming pc, bcc, or fcc Bravais lattices. Results are: pc, 8.02 g·cm-3 ; bcc, 1 0.4 g·cm- 3 ; fcc, g. cm-3 . The Bravais lattice is most likely fcc.
Calcu lations
SG-S4
5. 1 3 Ionic Structu res •
Model
�
� •
Rad i us ratio
�
�
<
<
diffe(usual rent ly opposi t e (usually custoholmariesly smalsameler 1 [ Zcatiol1 Zall1iaolrger [zinc-blendeholstersucture larger [rock-salt stholructes ure larger [cesiuholm-chles oride structluarerger opposite
[r (tetrahedral hole)
Spheres of Larger spheres Smaller spheres
r (octa hed ral hole)
r (cu bic hole»)
radii and charges representing cations and anions anions) occupy the unit cell lattice points. cations) fill in the unit cell.
[ Use ru le w ith cau tion, there are many exceptions.)
Ratio of radius of the Defined for ions of the
sphere to the radius of the = I charge number
I]
sphere, symbol
p
Radius Ratio-Crystal Structure Correlations
radius ratio � 0.4 1 4 :
tetrahedral
fil l (fcc for spheres) (one form of ZnS) ]
0.4 1 4 < radi us ratio < 0.732:
octahedral
radius ratio � 0.732:
fil l (fcc for (NaCI) ]
cubic
fill (pc for
spheres)
spheres) (CsCI) ]
�
•
Number of nearest-neighbor ions of charge Coordination of ionic solid � Represented as (cation coordination number, anion coordination number)
•
Properties of the ionic structure of unit cells
•
Coord ination n u m ber
Zinc blende
�
�
� �
Radius ratio (ZnS) = r (Zn2+)/r (S 2-) = (74 pm)/( I S4 pm) = 0.40 a nd/ of a fcc unit cell S 2- ions at the anions (atomic radius per unit cel l (edge length ] Zn2+ ions in/our of the tetrahedral holes in the unit cell cations (atomic radius per unit cell (edge length ) ] Cation and anion spheres in the tetrahedral locations. Each cation has /our nearest-neighbor anions, and each anion has/our nearest-neighbor cations: (4,4)-coordination.
corners aces [four ral1iol1 ) ei g ht [four touch rcatiol1 )
a) a
r [four corners facesral1iol1 ) [four rcruion ) t o uch six touch edge: a rcariol1 ranion
Rock salt
�
Radius ratio (NaCI) = (Na+)/r (Cr) = ( 1 02 pm)/( l S I pm) = 0.564 � cr ions at the and of a fcc unit cell anions (atomic radius per cell (edge length a ) ] � Na+ ions in all/our octahedral holes in the unit cell cations (atomic radius per cell (edge length a ) ] � Cation and anion spheres along the cell edges. Each cation has nearest-neighbor anions, and each anion has nearest-neighbor cations: (6,6)-coordination. = 566 pm +2 =2 along the � Spheres 3 8 3 � Volume of the unit cel l : V = a = l .S I x 1 0 pm = l .S I x 1 O-22 cm3 � Density ofNaCI: M(5S.44 g . mor l) and d(experimental) = 2. 1 7 g . cm-3
SG-5 5
six
d = (4 NaCl per cell)(MINA)
a
3
_
4 ( 58.44INA)
= 2. 1 4 g.cm - ( 1 .8 1 x l O-22 )
-3
Cesiu m ch lo ride �
r (primitive cubic) a) [ one centcorners r a)] nuch) body diagonal [one er rCat1oiocube eight touch body diagonal: .J3a
Radius ratio (CsCI) = (Cs+)/r (Cr) = ( 1 70 pm)/( 1 8 1 pm) = 0.939 � c r ions at the of a pc unit cell anion ( radius anion ) per cell (edge length ] � Cs+ ion at the of the in the unit cell per cel l (edge length cation (radius � Cation and anion spheres along the of the cell. Each cation has nearest-neighbor anions and each anion has cations : (8,8)-coordination. �
� �
Calculations
�
nearest-neighbor
Spheres along the = 2 rcation + 2 rani on = 702 pm and 3 Volume of the unit cel l : V = a = 6.64 x 1 07 pm3 = 6.64 x 1 0-23 cm3 Density of CsCI: M( 1 68.36 g . mor l ) and d(experimental) = 3.99 g·cm-3
d •
eight
=
(I CsCI per cel\) ( M/NJ
I ( 1 68.36/NA )
a
= 405 pm
4.2 1 g . cm-3
a3 idensi onic tystructureionic solcubiid c densityionic radiunii t cell radius-ratio rule (6.64 x 1 0-23 )
=
Determining the in the system from � Determining the of the from the of the Note: Usually within 1 0% of the experimental value
using the
THE IMPACT ON MATERIALS (Sections 5.14-5.17) 5. 1 4 The Properties of Solids •
Mobility of electrons
�
�
�
•
Responsible for characteristic luster and light reflectivity properties of a metal Accounts for and of metals Produces a relatively strong metallic bond that leads to high melting points for most metals
malleability, ductility, electrical conductivity
Slip planes � Plane of atoms that, � � � � �
under stress, may slip or slide along an adjacent plane structures structures (ccp) have eight sets of slip planes in different directions. structures (hcp) have only one set of slip planes. Large number of slip planes accounts for of ccp metallic crystals (may be bent, flattened, or pounded into different shapes). Examples: Coinage metals Cu, Ag, and Au Metallic crystals with the hcp structure tend to be brittle (only one slip plane). Exam ples: Zn, Cd
cl o sep acked Cubic close-clpacked Hexagonal ose-packed Characteristic of
malleability
SG-56
5. 1 5 Alloys •
•
•
Heterogeneous � Homogeneous �
Atoms of different elements are distributed uniformly; for example, brass, bronze, and coinage metals.
Substitutional � Atoms nearly the same size « � �
�
•
Mixture of crystalline phases; various samples have different compositions; for example, solders and mercury amalgams
1 5% difference) can substitute for each other more or less freely ( mainly d-block elements ) Lattice is distorted and electron flow is hindered. Alloy has lower thermal and electrical conductivity than pure element, but is harder and stronger. Homogeneous examples include the Cu-Zn alloy, brass (up to 40% Zn in Cu), and bronze (metal other than Zn or Ni in copper: casting bronze is 1 0% Sn and 5% Pb).
interstices
I n terstitial
�
� �
�
Small atoms (> 60% smal ler than host atoms) can occupy holes or in a lattice. Interstitial atoms interfere with electrical conductivity and the movement of atoms. Restricted motion makes the alloy harder and stronger than the host metal. Examples include low-carbon steel (C in Fe, which is soft enough to be stamped), high-carbon steel (hard and brittle unless subject to heat treatment), and stainless steel (a mixture of Fe with metals such as Cr and Ni, which aid in its resistance to corrosion).
5.1 6 Liq u i d Crystals �
•
•
•
•
Substances that flow like viscous liquids, but molecules form a moderately ordered array simil ar to that in a crystal. � Examples of a an intermediate state of matter with the fluid properties of a liquid and some molecular ordering similar to a crystal Isotropic material � Properties independent of the direction of measurement. Ordinary liquids are isotropic with viscosity values equal in every direction. Anisotropic material � Properties depend on the direction of measurement Certain rod-shaped molecules form liquid crystals, in which molecules are free to slide past one another along their axes but resist motion perpendicular to that direction. Classes of liq u id crystals � Differ in the arrangement of the molecules See text Figs. 5.49 and 5.50.
mesophase,
staggered. layers.
nematic phase:
Molecules lie together in the same direction but are
smectic phase:
Molecules lie together in the same direction in
cholesteric phase:
rotated
nematic/ike layers,
Thermotropic liq uid crystals � �
helical
Molecules form but the molecules of neighboring layers are with respect to each other. The resulting liquid crystal has a arrangement of molecules.
Made by melti ng solid phase Exist over small temperature range between solid and liquid Example: p-azoxyanisole
SG-57
•
Lyotropic l iq u id crystals
�
�
Layered structures produced by the action of a solvent on a solid or a l iquid. Examples include cell membranes and aqueous solutions of detergents and lipids (fats).
5. 1 7 Ionic Liquids �
•
�
Molecular substances tend to have low melting points. I onic, network, and metallic substances tend to have high melting points.
I onic liqu ids �
�
New class of solvents developed to have low vapor pressures but dissolve organic compounds Compounds in which one of the ions (usually the cation) is a large, organic ion that prevents the liquid from crystall izing at ordinary temperatures Example:
�
�
Cation: I -Butyl-3-methylimidazolium (Margin Figure 1 6 in the text) Anion: Tetrafluoroborate: BF4-
One formulation is used to dissolve rubber in old tires for recycl ing. Another is used to extract radioactive waste from groundwater.
SG-58
Chapter 6 THERMODYNAMICS : THE FIRST LAW SYSTEMS, STATES, AND ENERGY (Sections 6. 1-6.7) Overview •
•
Thermodynamics
�
�
Criterion for spontaneity (see Chapter 7) Explains why some chemical reactions occur spontaneously and others do not
Statistical thermodynamics
�
� •
Consequence of the law of conservation of energy Quantitative description of energy changes in physical (e.g. , phase changes) and chemical (e.g. , reactions) processes Energy changes in relation to heat and work: A paddlewheel stirrer immersed in a liquid produces heat from mechanical work (temperature rise in the liquid). A hot gas expanding in an insulated cylinder coupled to a flywheel produces mechanical work from heat (temperature drop in the gas).
Second law of thermodynamics �
•
Generalizations based on experience with bulk matter Not derivable, but understandable without recourse to knowledge of the behavior of atoms and molecules
First law of thermodynamics
�
�
•
Branch of science concerned with the relationship between heat and other forms of energy
Laws of thermodynamics
�
� •
�
Laws of thermodynamics reflected in the behavior of large numbers of atoms or molecules in a sample Link between the atomic level and the behavior of bulk matter
Chapter goals
�
Gain insight into heat and work and their relationship to energy changes in physical and chemical processes
6.1 Systems • •
•
The "universe" or "world" � System �
Composed of a system and its surroundings
Portion of the universe of interest Examples include a beaker of ethanol, a frozen pond, 20 g of CaCI 2 (s), and the earth itsel f.
Surroundings
�
�
Remainder of the universe (everything that is not in the system) Where observations and measurements of the system are made For example, the observation of heat transferred to or from a system is made in the surroundings.
SG-59
• •
Boundary �
Dividing surface between the system and the surroundings
Types of systems Open system
�
�
matter energy
Both and can be exchanged between the system and surroundings. An example is a half-liter of water in an open beaker. Matter can cross the boundary (water evaporates). Energy can cross the boundary (heat from the surroundings may enter the system).
Energy
matter
Closed system �
�
can be exchanged between the system and surroundings, but cannot. An example is the refrigerant in an air-conditioning system. As the refrigerant expands, energy (heat) is withdrawn from the space to be cooled. As the refrigerant is compressed, energy (heat) is supplied to another space that is warmed. The mass of the refrigerant is unchanged.
matter approxi energymately
I solated system
�
�
Neither nor can be exchanged between the system and surroundings. An example of an isolated system is a substance in a well-insulated container, such as ice in a Styrofoam ice-chest or coffee in a covered Styrofoam cup.
6.2 Work and Energy •
Work,
�
�
�
�
•
w
Fundamental thermodynamic property (see Table 6. 1 in the text Movement against an opposing force Definition: work = force x distance N; I N = I kg·m·s-2 Units: (force) (work) J ; I J = 1 N'm = I kg·m2 ·s-2
Internal energy, U
�
�
�
�
Varieties of Work)
joulnewteo, n, total
Internal energy is the capacity of a system to do work. a change in U (f'.. U = Unnal - Uinitial) is measurable. An extensive property (see Section A in the text) For energy transferred to a system by doing work on the system, f'.. U =
Only
6.3 Expansion Work
�
Does not involve a change in volume Examples include electrical current and change of position.
•
Nonexpansion work
•
Expansion work
•
Calculating expansion work with a constant external pressure, Pex �
�
�
�
w.
Change in volume of the system against an external pressure An example includes inflating a tire or balloon.
V,
Consider a gas confined in a volume, by a piston with surface area, A. Work = force x distance F x d = (Fex x A ) x d = Fex (see Section 4.2), where d is the distance the piston is displaced the density). =
I
w
=
- Pex f'..
. Units:
VI
(not
on
f'.. V
By convention, work done the system is positive (+) in sign. I Pa'm-., = I kg·m- I ·s-2 x I m3 = I kg·m2 ·s-2 = I J 3 I L-atm = 1 0-3 m x 1 0 1 325 Pa = 1 0 1 .325 Pa·m3 = 1 0 1 .325 J (exactly) SG-60
• •
Free Expansion
Expansion into a vacuum (Pex = 0 and w 0) =
infinitesimal
Reversible p rocess
�
�
�
�
•
�
A process that can be reversed by an change in a variable For a change in pressure, the external pressure must always be infinitesimally different from the pressure of the system. A net change is effected by a series of infinitesimal changes in the external pressure followed by infinitesimal adjustments of the system. For a change in temperature, the temperature of the system must change by a series of infinitesimal amounts of either heat flow or of work done. Processes in which change occurs by amounts are in nature.
finite
irreversible
Reversible, isothermal expansion or compression of an ideal gas W
�
=
-
nRT
V,f' I !!L = ln --..!
Vinitial
ln P.
nRT � Pfinal
I
The change i n pressure is carried out in infinitesimal steps (text derivation in Section 6.3).
6.4 Heat •
Heat, q
�
�
�
�
•
joule, heat calorie,
T T2 TJ > T2,
�
�
�
nutritional calorie, hermic c qq >q ((eexotndot(adihermi abatic
I f heat flows from the surroundings into the system, 0 process). I f heat flows from the system into the surroundings, < 0 process). I f no heat flows between the system and surroundings, = 0 process). For example, a thermally insulating (adiabatic) wall (Styrofoam or vacuum flask) prevents the passage of energy as heat. (nonadiabatic) walls permit the transfer of energy as heat.
Diathermic
I nternal energy, U
�
�
/').T T2 /'). U q
For energy transferred to a system by the flow of heat, = (w = 0). If the system is initially at temperature TI and - TI > 0, then q > 0 and =
6.5 The Meas u rement of Heat •
Heat capacity
�
� •
TJ •
Sign of q defined in terms of the system �
•
Fundamental thermodynamic property Energy transferred as a result of a temperature difference, /'). = 0 Energy flows as from a region of high temperature, to one of low temperature, S I unit: 1 J = 1 N - m = I kg. m2 ·s-2 Common unit: The 1 cal == 4. 1 84 J, is widely used in biochemistry, organic chemistry, and related fields. The Cal, is I kcal.
C
�
�
o.
of a pure su bstance
Heat capacity of a pure substance is an Amount of heat absorbed
Specific heat capacity,
�
ext e nsi v e by a sample ofthe substance i n t e nsi v e by one gram
/'). U >
Cs ,
property. per degree Celsius rise in temperature
of a pure substance
Specific heat capacity of a pure substance is an property. Amount of heat absorbed of a substance per degree Celsius rise in temperature (values for common materials are given in Table 6.2 in the text) Cs = Clm Units: J . (0C)- J . g- J or J· K-' ·g- ' (commonly,
SG-6 1
theformer)
� •
Molar heat capacity, em , of a pure substance
�
�
�
�
�
•
I q=Cf'I,T =mCsf'l,T
e i n t e nsi v by one mole
property. Molar heat capacity of a pure substance is an Amount of heat absorbed of a substance per degree Celsius rise in temperature (values for some substances are given in Table 6.2 in the text) c'n Units: J . (°q- J ·mor J or J . K- J ·mor J (commonly, the
Cln I q=Cf'I,T =nCm f'l,T I
Values of molar heat capacities increase with increasing molecular complexity. They depend on the temperature and the state of the substance. C.n Oiquid) .n (solid)
>C
Calorimeter � �
�
�
� �
�
lat er)
=
Device used to monitor or measure heat transfer in a system by the temperature changes that take place in the surroundings = In calorimetry, For an exothermic process, 0; the temperature of the calorimeter increases, For an endothermic process, < 0; the temperature of the calorimeter decreases, The heat capacity of a calorimeter, is the amount of heat absorbed degree Celsius rise in temperature.
I qcal
qcal qsurr qcal > qcal CcaI ,
=
f'l,f'l,TT> by the calorimeter
Cca1f'l,T I
o.
< O.
per
A calorimeter has a heat capacity that is determined experimentally by the addition of a known amount of heat supplied electrically and the measurement of the resulting temperature rise in the calorimeter (see Example 6.4 in the text). Units of heat capacity: l(oq- J or J · K- J
6.6 The First Law •
Combining heat and work
�
�
•
•
I f'l,U = q+w I
Change in internal energy of a system is the < 0) the system the work done
(q from
plus isolated
First law of thermodynamics
�
onsum
byadded (w (q > to removed
of the heat (w > 0) or
0)
or
< 0) the system.
The internal energy of an system is constant. � It is an extension of the law of conservation of energy. � It is a generalization based on experience and cannot be proven. If the first law were false, one could construct a "perpetual motion machine" by starting with an isolated system, removing it from isolation, and letting it do work on the surroundings. I t could then be placed in isolation again, and its internal energy allowed to return to the initial value. This feat has never been accomplished despite many attempts. Constant vol ume process, f'l, V = 0 � No expansion work is done � = (no work of any kind is done) � Heat absorbed or released by a system equals the change in internal energy for a constant volume process if no other forms of work are done:
f'l, U q
f'l, U = q.
SG-62
• •
State of a system � State function
�
�
•
Defined when all of its properties are fixed
Property that depends only on the current state of the system I ndependent of the manner in which the state was prepared Examples of state functions include temperature, T; pressure, P; volume, V; and internal energy, (Heat, q, and work, w, are not state functions.)
Path
�
�
Sequence of intermediate steps linking an initial and a final state Consider the transition from state I to state 2. The changes in the state functions are independent of path. The quantities I'1U, M, I'1 V, and I'1T are uniquely determined. The amount of heat transferred or the work done depends on the sequence of steps fol lowed.
6.7 A Molecular Interlude: The Origin of lnternal Energy • • •
I nternal energy, U � Potential energy �
Energy stored as potential and kinetic energy of molecules
Energy an object has by virtue of its position
Kinetic energy
� �
�
Energy an object has by virtue of its motion For atoms, kinetic energy is translational in nature. For molecules, kinetic energy has translational, rotational, and vibrational components. Translation is the motion of the molecule as a whole. (See Section 1 .7 in the text for the particle-in-a-box translational energy levels.) Rotation and vibration are motions within the molecule and do not change the center of gravity (mass) of the molecule. (See Box 2.2 in the text for the rigid-rotor rotational energy levels.) (See Major Technique I in the text for a description of vibrational motion.) Translation al, Rotation al, and V i brational Motion ofF2
• ,
F
F
1F
I-
F
Translation •
,
•
Rotation
V ibration
Degrees of freedom
�
� •
U.
Modes of motion (translational, rotational, and vibrational) Mainly translational and rotational degrees of freedom store internal energy at room temperature.
Equ ipartition theorem
�
Average
�
A molecule moving in three dimensions has three translational degrees of freedom.
kinetic energy of each degree of freedom of a molecule in a sample at temperature T is equal to 2. kT. The Boltzmann constant k = 1 .380 658 x 1 0-23 J . K- J • 2
SG-63
U(translation) = 3 �
x
2
2
� kT = 2 kT 2
2
For 1 mol of molecules, Um(translation) = NA(2 kT) = 2 RT (R = NA k ). A l inear molecule has two rotational degrees of freedom. U (rotation, l inear) = 2
x
2
� kT = kT
For I mol of molecules, Um(rotation, linear) = NA(kT) = RT �
� �
A nonlinear molecule has three rotational degrees of freedom. U(rotation, nonlinear) = 3 x � kT = 2 kT 2
2
2
2
For 1 mol of molecules, Um(rotation, nonlinear) = NA(2 kT) = 2 RT A molecule of an ideal gas does not interact with its neighbors and the potential energy is O. The internal energy is independent of volume and depends only on temperature, Um (T). A molecule of a liquid, solid, or real gas does interact with its neighbors. The potential energy is an important component of the internal energy, which has a significant dependence on volume. The internal energy depends on both temperature and volume, �n(T, V).
ENTHALPY (Sections 6.8-6.12) Overview •
Heat transfer at constant volume, qv (notation not used in the text)
�
�
•
I t,.U
qv
I
for a system in which only expansion work is possible Combustion reactions are studied in a bomb calorimeter of constant volume. =
H eat transfer at constant pressure, qP (notation not used in the text)
� �
I
tJ{
=
qp
I
a new state function (enthalpy) for a system at constant pressure
Most chemical reactions occur at constant pressure largely because reaction vessels are usual ly open to the atmosphere.
6.8 Heat Transfers at Constant Pressu re •
Definition of enthalpy, H
�
--*
•
IH
=
U + PV
I
a state function because U, P, and V are state functions
Change in enthalpy of a system, 6H = 6U + 6(PV) At constant pressure, tJ{ = t,.U + Pt,. V = ql' + W + Pt,.V = ql' - Pext,. V + Pt,. V = ql' for a system open to the atmosphere and for which only expansion work may occur.
H eat change for a constant pressure process
�
�
An exothermic process is one in which heat is released by the system into the surroundings, t,.H = ql' < O. An endothermic process is one in which heat is absorbed by the system from the surroundings, t,.H = ql' > O .
SG-64
6.9 Heat Capacities at Constant Vol u m e and Constant Pressure •
Gene .. ' definition of heat capadty
At constant volume •
-?
I C = 6.qT I
At constant pressure
-?
CVCV,,mm =:::: Cp,Cv/nm liquids andCP,msol= iCp/n ds, Cp,m > CV,m gases Cp = Cv nR any amount ofan ideal gas I CP,m = CV,m R l one mole ofan ideal gas
Molar heat capacities
-?
-? •
-->
I Cp - �T - 6.T I -R-
Mf
and
but
Molar heat capacity relationship for an ideal gas (see derivation in text)
-?
-?
+
+
6 . 1 0 A Molecular I nterlude: The Origin of the Heat Capacities of Gases •
Contributions to the molar heat capacity for a monatomic ideal gas -?
-?
•
Consider only translational motion.
Um = l.. RT CV,m - 6.U6.Tm 2
6.Um = l.. R6.T R6.T = - R -6.T 3 "2
2
3 2
=
T) I C'm =�=%R I
(molar internal energy depends only on 1 2.47 1 77 J . K- 1 ·mor l
and
Contributions to the molar heat capacity for a linear-molecule ideal gas -?
-?
Consider translational motion and 2 degrees of freedom in rotational motion.
Um = 2RT CV,m - 6.U6.Tm 2
_
•
and
and
6.Um = 2R6.T R6.T =- R -6.T 5
"2
2
(molar internal energy depends only on
5
2
T)
and
Contributions to the molar heat capacity for a nonlinear-molecule ideal gas
-?
-? -->
Consider translational motion and 3 degrees of freedom in rotational motion. and (molar internal energy depends only on
Um = 3RT 6.Um = 3R6.T I CV,m = � = Tr = 3R I
Note:
�
24,943 53 JK ' mor'
and
T) -6.-T-m-=-4-R--'1 Ir--C-p,m-=-�
The contribution from molecular rotation is for room temperature. At lower temperatures, this contribution diminishes and the heat capacity approaches the value for a monatomic gas. At room temperature, a smal l contribution to the heat capacity from molecular vibrational motion leads to a value slightly larger than the ones given above.
SG-65
6.1 1 The Enthalpy of Physical Change •
Physical changes (phase transitions) at constant temperature and pressure �
� � �
�
Vaporization: Condensation:
liquid vapor
Fusion (melting): Freezing:
solid liquid
Sublimation: Deposition:
solid vapor
I1Hvap
=
-l1Hcond
� �
I1Hvap = Hvapor,m - �iquid,m
I1Hcond = �iquid,m - Hvapor,m
liquid solid
I1 Hfus
vapor solid
I1Hsub
� �
�
vapor liquid
�
I1Hfus
=
I1 Hfreez
�iquid,m - Hso1id,m
= Hso1id,m - �iquid,m
= Hvapor,m - Hso1id,m
I1Hdep = Hso1id,m - Hvapor,m
= -l1Hfi-eez
I1Hsub
=
> 0 (endothermic) < 0 (exothermic) > 0 (endothermic) < 0 (exothermic) > 0 (endothermic) < 0 (exothermic)
-l1Hdep
+ I1Hvap (for the same T and P ) Values of I1Hfuso and I1Hvap0 for selected substances are given in Table 6.3 in the text. Liquids may evaporate at any temperature, but they boil only at the temperature at which the vapor pressure of the liquid is equal to the external pressure of the atmosphere.
I1Hsub
=
I1Hfus
6.1 2 Heating C u rves •
Heating (cooling) curve
� �
The heating (cooling) curve is a graph showing the variation of the temperature, T, of a sample as heat is added (removed) at a constant rate. Two types of behavior are seen in a heating curve. The temperature increase of a single phase to which heat is supplied has a positive slope that depends on the value of the heat capacity. The larger the heat capacity, the more gentle the slope. At the temperature of a phase transition, two phases are present and the slope is 0 (no temperature change). The greater the heat associated with the phase change, the longer the length of the flat l ine. As more heat is added to the substance, the relative amounts of the two phases change. The temperature does not rise again until only one phase remains. In this schematic heating curve, l ine V with positive slope represents the heating of a solid phase. The flat line W shows z the phase transition between solid and l iquid (fusion). Line X y represents the heating of the liquid after the solid has completely x melted. Line Y shows the phase transition between liquid and w vapor (boiling at constant external pressure). F inally, l ine Z v represents the heating of the vapor after the l iquid disappears, Because the enthalpy of vaporization is greater than the enthalpy Heat Added of fusion, the length of l ine Y is longer than line W.
THE ENTHALPY OF CHEMICAL CHANGE (Sections 6.13-6.21) 6. 1 3 Reaction Enthalpies •
Thermochemical equation �
A chemical equation with a corresponding enthalpy change, 11H, expressed in kJ for the of each reactant and product
stoichiometric number of moles
,
SG-66
� •
Applying the principles of thermodynamics to chemical change
Enthalpy of reaction, flHr
�
The enthalpy change in a thermochemical reaction expressed in kJ ' mor l , where "mol" refers to the number of moles of each substance as determined by the stoichiometric coefficient in the balanced equation � Combustion is reaction with oxygen, 02 (g).
6. 1 4 The Relation Between I:!..H and I:!.. U •
Reactions with liquids and solids only
�
Recall that W = flU + fl(PV)
�
Only a small change in the volume of reactants compared to products at constant P
� W r::; flU
•
qP r::; qv
Reactions with ideal gases only
� W = flU + fl(PV) = flU + fl(nRT ) = flU + (6.n )RT
� •
�
constant temperature
The equation is exact for ideal gases only.
Reactions with liq uids, solids, and gases
� W r::; flU + (flngas )RT
� flngas �
constant temperature
= nfinal - ninitial = change in the number of moles of gas between products and reactants The equation is a very good approximation. The value of (flngas )RT is usually much smaller than fl U and flH.
6.1 5 Sta n d a rd Reaction Enthalpies •
•
Standard conditions
�
Values of reaction enthalpy, Wr , depend on pressure, temperature, and the physical state of each reactant and product. � A standard set of conditions is used to report reaction enthalpies. Standard state
Standard state of a substance is its pure form at a pressure of exactly 1 bar. � For a solute in a liquid solution, the standard state is a solute concentration of 1 mol· L- 1 at a pressure of exactly 1 bar. � Standard state of l iquid water is pure water at 1 bar. Standard state of ice is pure ice at 1 bar. Standard state of water vapor is pure water vapor at 1 bar. �
Note:
•
Compare the definition of standard state in thermodynamics to the definition of standard temperature and pressure, STP, for gases (Section 4.5). For gases, standard temperature is O°C (273 . 1 5 K) and standard pressure is 1 atm ( 1 .0 1 3 25 bar).
Standard reaction enthalpy, flHro
Reaction enthalpy when reactants in their standard states form products in their standard states � Degree symbol, 0 , is added to the reaction enthalpy. Reaction enthalpy has a weak pressure dependence, so it is a very good approximation to use standard enthalpy values even when the pressure is not exactly I bar. �
•
Temperature convention � �
Most thermochemical data are reported for 298. 1 5 K (25°C ) . Temperature convention is not part of the definition of the standard state.
SG-67
6. 1 6 Combining Reaction Enthalpies: Hess's Law •
Hess's law
�
�
�
�
•
Overall reaction enthalpy is the sum of the reaction enthalpies of the steps into which a reaction can be divided. Consequence of the fact that the reaction enthalpy is path independent and depends only on the initial reactants and final products Used to calculate enthalpy changes for reactions that are difficult to carry out in the l aboratory Used to calculate reaction enthalpies for unknown reactions
Using H ess's law
�
�
�
Write the thermochemical equation for the reaction whose reaction enthalpy is unknown. Write thermochemical equations for intermediate reaction steps whose reaction enthalpies are known. These steps must sum up to the overal l reaction whose enthalpy is unknown. Sum the reaction enthalpies of the intermediate reaction steps to obtain the unknown reaction enthalpy (see Toolbox 6. 1 in the text).
6. 1 7 The Heat Output of Reactions •
H eat output in a reaction
�
�
•
Heat is treated as a reactant or product in a stoichiometric relation. In an endothermic reaction, heat is treated as a reactant. In an exothermic reaction, heat is treated as a product.
Standard enthalpy of combustion, 6.He0
� �
Change in enthalpy per mole of a substance that is burned in a combustion reaction under standard conditions (see values i n Table 6.4 in the text) Combustion of hydrocarbons is exothermic, and heat is treated as a product.
6 . 1 8 Standard Enthalpies of Formation •
Treatment of chemical reactions �
�
�
•
Possible thermochemical equations are nearly innumerable. Method to handle this problem is the use of a common reference for each substance. The reference is the most stable form of the elements. Standard enthalpies of formation of substances from their elements are tabulated and used to calculate standard reaction enthalpies for other types of reactions.
Standard enthalpy of formation, 6.Ht
�
� �
�
Enthalpy change for the fonnation of one mole of a substance from the most stable form of its elements under standard conditions tJ! 0 = tJ!t Elements (most stable form) - substance (one mole) 6.Hfo = -23 8.86 Hmor i For example, C(s) + 2 H 2 (g) + � 02(g) - CH 3 0H(I)
Values of tJ!t at 298. 1 5 K for many substances are l isted in Appendix 2A. Most stable form of the elements at I bar (standard state) and 298. 1 5 K (thermodynamic temperature convention): Metals and semi metals are solids with atoms arranged in a lattice (see Chapter 5). The exception is Hg(l). Nonmetal solids are 8(s), C(s, graphite), pes, white), S(s, rhombic), Sees, black), and 1 2 (s).
SG-68
Monatomic gases are He(g), Ne(g), Ar(g), Kr(g), Xe(g), and Rn(g). Group 1 8 Diatomic gases are H 2 (g), N 2 (g), 02(g), F 2 (g), and Clz(g). Nonmetal l iquid is Br2 (1). Note: pes, red) is actual ly more stable than pes, white), but the white form is chosen because it is easier to obtain pure. Also, some of the nonmetal solids have molecular forms, for example, P4 and S 8 . •
!1Hro Use the most stable form of the elements as a reference state to obtain enthalpy changes for any reaction if Wfo values are known for each reactant and product.
Standard enthalpy of reaction,
�
�
Wr ° = I nWr O(products) - I nWr O(reactants) In this expression, values of n are the stoichiometric coefficients and the symbol I (sigma) means a summation. The procedure is to convert the reactants into the most stable form of their elements [-In!1HfO(reactants)] and then recombine the elements into products [+InWfO(products)]. This is an application of Hess's law. Use Wt values in Appendix 2A to obtain Wro.
6.1 9 The Born-Haber Cycle •
I on ic solids
� �
•
Cations and anions are arranged in a three-dimensional lattice (see Section 5 . 1 3) and held in place mainly by coulombic interactions (see Section 2.4). Enthalpy required to separate the ions in the solid into gas phase ions is the lattice enthalpy, WL H.n(ions, g) - H.n(ions, s) > O. See values in Table 6.6 in the text.
=
Born-Haber cycle
�
�
Thermodynamic cycle constructed to evaluate the lattice enthalpy from other reactions in which the enthalpy changes are known Steps in the Born-Haber cycle that start and end with the most stable form of the elements in amounts appropriate to form one mole of the ionic compound As an example, consider CaF 2(s). Start with one mole of Ca(s) and one mole of F 2 (g). Two moles of F atoms are required. ( I ) Atomize: Atomization of the metal element that will form the cation and of the nonmetal element that will form the anion !1H(atomization) > 0 !1Hro = WfO[Ca(g)] = 1 78 kJ . mor l Ca(s) Ca(g) !1Hro = 2Wt[F(g)] 2(79 kJ·mor l ) = 1 58 kJ·mor l F 2(g) 2 F(g) W(atomization) = ( 1 78 + 1 58) kJ·mor l +336 kJ . mor l (2) Ionize (cation) : I onize the gaseous metal atom to form the gaseous cation. Several ionization steps may be required. Because electrons created in the formation of cations in the ionization steps are consumed in the formation of anions in the electron-gain steps (3), energy and enthalpy are considered to be interchangeable when treating ionization and electron gain. !1H (ionization, cation) > O. Ca(g) Ca+(g) + e-(g) !1Hro = 11 [Ca] 5 90 kJ . mor l 2 Ca+ (g) Ca +(g) + e-(g) Wro = h [Ca] = 1 1 45 klmor l Ca(g) Ca2+(g) + 2 e-(g) !1H,0 = 11 + h = (5 90 + 1 1 45) kJ . mor l = 1 735 kJ . mor l W(ionization, cation) = + 1 735 kJ . mor l �
�
=
=
�
=
�
�
SG-69
Attach electron(s) to the gaseous nonmetal atom to form the gaseous anion. Several electron affinity values may be required. Recall that for electron gain, !1H = -Eea for each electron added. W(ionization, anion) < or > 0
(3)
Ionize (anion) :
F(g) + e-(g) F -(g) Wro = -Eea [F] = -328 kJ·morl 2 F(g) + 2 e-(g) 2 F -(g) Wro = -2Eea = -656 kJ ·mor l W(ionization, anion) = -656 kJ·mor l �
�
Form the lattice of ions in the solid from the gaseous ions. The enthalpy of lattice formation is the negative of the lattice enthalpy, WL . W(lattice formation) < 0
(4)
Latticize:
ci+(g) + 2 F -(g) CaF2(s) W(lattice fonnation) = -WL �
(5)
Wro = -!1HL
value unknown
Form the most stable form ofthe elements from the ionic solid. W(element formation) < 0
Elementize:
W(element formation) = + 1 220 kJ·mor l (6)
The sum of all the enthalpy changes for the complete cycle is 0 = W(atomization) + W(ionization, cation) + W(ionization, anion) + !1H (lattice formation) + !1H(element formation) 0 = 336 + 1 735 + (-656) + (-Wd + 1 220 kJ ·mor l = 2635 kJ . mor l - WL WL = +2635 kJ . mor l for CaF2 (s) Ca2+(g) + 2 F -(g) Lattice enthalpy:
o.
�
•
Summary
�
Strength of interactions between ions in a solid is determined by the lattice enthalpy, which is obtained from a Born-Haber cycle.
6.20 Bond Enthalpies •
Bond enthalpy, !1Ho
�
�
�
�
•
Enthalpy change accompanying the breaking of a chemical bond in the gas phase Difference between the standard molar enthalpy of the fragments of a molecule and the molecule itself in the gas phase Reactants and products in their standard states (pure substance at I bar) at 298. 1 5 K Bond enthalpies for diatomic molecules are given in Table 6.7 in the text.
Mean (average) bond enthalpies
�
In polyatomic molecules, the bond strength between a pair of atoms varies from molecule to molecule. � Variations are not very large, and the average values of bond enthalpies are a guide to the strength of a bond in any molecule containing the bond (see Table 6.8 in the text).
•
Using mean bond enthalpies
�
Atoms in the gas phase are the reference states used to estimate enthalpy changes for any gaseous reaction. A value of !1HB is required for each bond in the reactant and product molecules.
SG-70
�
�
f:.J!/
"'"
L nf:.J! B (bonds broken) - L nf:.J! B (bonds fonned)
reactants
products
I n this expression (not in the text), values of n are the stoichiometric coefficients and the symbol I (sigma) means a summation. The procedure is to convert the reactants into gaseous atoms [+UuVfB(bonds broken)] and then recombine the atoms into products [-Inf:.J!B(bonds formed)]. This is an application of Hess's law. Compare bond enthalpy to dissociation energy in Sections 2. 1 4 and 2. 1 5 .
6.2 1 The Variation of Reaction Enthalpy with Tem peratu re •
Temperature dependence of reaction enthalpy �
� •
f:.J!ro needs to be measured at the temperature of interest. Approximation method is possible from the value of 11H,0 measured at one temperature if heat capacity data on the reactants and products are available.
Kirchhoff's law
�
�
�
�
The difference in molar heat capacities of the products and reactants in a chemical reaction is t£p
=
L nCp,m (products) - L nCp,m (reactants)
An estimation of the reaction enthalpy at a temperature of T2 if the value at a temperature of TI and the difference in molar heat capacities are known is
I f:.J!r,2 °
=
f:.J!r, 1 ° + t£P(T2 - 71 )
I
A major assumption is that the molar heat capacities are independent of temperature. The law in this form does not account for any phase changes. Separate steps must be added.
SG-7 l
Chapter 7 THERMODYNAMICS: THE SECOND AND THIRD LAWS ENTROPY (Sections 7.1 -7.8) 7. 1 Spontaneous Change •
The big q uestion: What is the cause of spontaneous change?
•
Spontaneous (or natural) change � Occurs without an external i nfluence, can be fast or slow
•
Examples: Diamond converts to graphite (infinitesimally slow rate). I ron rusts in air (very slow, but noticeable change). 2 H 2 (g) + 02 (g) � 2 H2 0(g) (very fast reaction or explosion).
Nonspontaneous change �
Can be effected by using an external influence (using energy from the surroundings to do work on the system)
E xamples: Liquid water can be electrolyzed to form hydrogen and oxygen gases. Graphite
can be converted to diamond under extremely high pressures.
7.2 En tropy and D isorder �
•
Spontaneous changes
•
Entropy (S) � A measure of disorder; increase in disorder leads to an increase in S.
•
A ny spontaneous change is accompanied by an increase in the disorder of the universe (system plus the surroundings).
Second law of thermodynamics � For a spontaneous change, the entropy of an isolated system increases. �
For a spontaneous change, the entropy of the universe increases. [The universe is considered to be a (somewhat large) isolated system. ]
•
Isolated system � No exchange of energy or matter with the surroundings
•
Macroscopic defin;';on of entropy, infinitesimal change
•
Finite change, isothermal p rocess
I1S =
qrev
T '
I
dS
d =
�"
I
(See Section 7.3)
for constant T
� Heat transfer processes are carried out reversibly to evaluate 11S. (Temperature of surroundings equals that of the system at al l points along the path.)
� Reversible path: The system is only infinitesimally removed from equil ibrium at all points along
the path.
•
Entropy is an extensive p roperty � Proportional to the amount of sample
•
Entropy is a state function � Changing the path does not change !J.S. Note:
According to the definition of 11S, we must calculate I1S by using a reversible path. The result then applies to any path because I1S is independent of path.
SG - 73
7.3 Changes i n E ntropy •
E ntropy i ncreases � I f a substance is heated (increase in thermal disorder)
� I f the volume of a given amount of matter increases (increase i n positional disorder)
•
Temperature d ependence of entropy ( any subst ance)
� Let n = number of moles of a substance, C p = heat capacity (lX- ') at constant P, C p,m = molar heat
capacity (lX- "mor') at constant P, C v= heat capacity at constant V, C V•m = molar heat capacity at constant V, T , initial temperature, and T2 final temperature. =
=
� Isobaric heating of a substance (constant P)
� I sochoric heating of a substance (constant
V)
T T 6.S = C v In-2 = nC vmln-2 .
1]
1]
� Above relationships assume that the heat capacity is const ant over the range of T , to T2 .
Example: I f two moles of Fe are cooled from 300 K to 200 K at constant P, the entropy change is
200 K T 6.S = nC P.m In 2 =(2 mol)(25 .08 J . K- J . mol- J) In =-20.3 J · K- J ; 6.S
6.S of the system decreases as expected from the decrease in temperature. •
Volume dependence of entropy (ide al g as) � �
n number of moles, R = gas constant (8.31 4 5 1 J . K-' ·mor'), V , = initial volume, and V2 = final volume =
Isothermal volume change of an ide al g as (constant T )
6.S =
qrev = T
_
Wrev = nR In V 2 T V;
� Recall: 6. U = 0 because the energy of an ide al gas depends only on its temperature. •
Pressu re dependence of entropy (ide al g as)
�
n = number of moles, P ,
=
initial pressure, and P2 = fi nal pressure
� Isothermal pressure change (ide al g as, constant 1), 6. U = 0, (Boyle's Law)
and
6.S
=
V P nR In--1. = nR In--1 P2 VI
Example: Three moles of an ideal gas expand from 20 L to 80 L at constant
p, = --.!.. VI P2
V
--1.
T.
V 80 L 6.S = nR In 2 = (3 mol)( 8.3 I 4 5 I J . K -J. m 0 1-J) In =34 .6 J . K - J 20 L VI
6.S of the system increases as expected from the volume increase.
SG -74
Example: Four moles of an ideal gas undergo a pressure i ncrease from 0.500 atm to 1 .75 atm at constant T.
PI 0.500 atm -I -I I t:hl=nR l n-=(4 mol)(8.3 1 4 5 1 J · K ·mol ) In = - 4 1 .7 J·KP2 1 .75 atm
t:hl of the system decreases as expected from the volume decrease.
7.4 •
Entropy Changes Accompanyi ng Changes in Physical State
Changes of physical state
�
Fusion == fus (solid to l iquid); vaporization == vap (liquid to vapor); subl imation Sn(white)
� Solid-to-solid phase changes; for example, Sn(gray) •
==
sub (solid to vapor)
�
At the transition temperature � Temperature of substance remains co nst ant during change of physical state.
� Transfer of heat is reversible.
� Heat supplied is identified with e nth alpy ch ange because pressure is constant (for these
reversible processes at constant P,
•
qrev = qP = W ).
Entropy of vaporization � Normal boi ling point, h T at which l iquid boi ls when P = I atm .
'
I';. va . . I I';.Svap = H p t:hlvap IS the e ntropy of v apor izat IOn (Units: J . K_I · mol--
Tb
� •
I or kJ·K - · mol - I ).
/'),S vapo = st andard e ntropy of v apori zatio n (l iquid and vapor both pure and both at
I bar).
Trouton ' s rule
85 J . K- I ·mor l � Rationale: Approximately the same increase in positio nal disorder occurs when any liquid is vaporized (gas molecules are far apart and moving rapidly). � Exceptions: Liquids with very weak or very strong intermolecular interactions Examples: Liquid helium, He (20 J . K- I · mor l ); hydroge n-bo nded liquids such as water, H 2 0 I ( 1 09 J ·K- I · morl ), and methanol, CH 3 0H ( l OS J . K- · mor l ) � For many liquids,
•
/'),S vapo (fiq)
;::;
E ntropy of fusion (melting) �
t:hlfus °
�
/'),Sfllso /'),Srllso
�
=
Wfus ° , where Tr is the melti ng poi nt. Tf
is the st andard e ntropy offusio n.
is smaller than /'),S yap° because a liquid is only sl ightly more disordered than its sol id, whi le a gas is considerably more disordered than its liquid. Example: Methane melts reversibly at I bar and - 1 82.5°C : CH4 (s)
C H 4 (I). Under these l conditions, the enthalpy of fusion is + 0.936 klmor . The standard ( 0 ) molar entropy I';.H ° 936 J·mol-I . . . = 1 0.3 J · K - I· mol-l• of fusI On I S given by I';.Sfus o = __r_Ils_= 90.65 K Tr �
Because t:hl> 0 for melti ng , the final state is more disordered than the initial state, as expected.
SG-75
7.5 A Molecular Interpretation of Entropy •
� If entropy S is a measure of disorder, then a perfectly ordered
A bsolute value of the entropy
state of matter (perfect crystal) should have zero entropy.
•
� Entropies of perfect crystals are the same at T = 0 K .
Th ird law of thermodynamics
(Thermal motion almost ceases a t T = 0 K , and by convention S = 0 for perfect crystals at 0 K.)
•
Boltzmann approach
� Entropy increases as the number of ways that molecules or atoms can
be arranged in a sample ( at the s ame tot al energy) increases.
•
�
Boltzmann formula
IS
=
kIn W
i ; k = Boltzmann constant = 1 .3 80 658
x
3 1 0-2 lK-1
[ Note: R = NA k J . K- I . mor l = the gas constant.]
W = number of microst ates available to the system at a certain energy
•
Boltzmann entropy �
•
M icrostate
Also called statistical entropy
� One permissible arrangement of atoms or molecules in a sample with a given total energy � W = total number of permissible arrangements corresponding to the same total energy, also called an ensemble � I f only one arrangement is possible, W = I and S = 0 •
Boltzmann interpretation � Spontaneous change occurs toward more probable states.
•
Scaling S
�
For N molecules or atoms, the number of microst ates is related to the number of molecular orient ations permissible: W = (orient ations)N and S
=
kIn W
=
k In( orient ations)N
=
N k In( orient ations )
N
� For NA molecules or atoms, W = (orient ations) A and
S •
=
k In(orient ations)
Residual entropy
NA
=
N Ak In(orient ations )
=
R ln(orient ations)
� For some solids, S > 0 at T = O. (Such solids retain some disorder.)
Note: Units of S are usual ly J · K- I · mor l .
E xample: The residual entropy of CO(s) at 0 K is 4.6 J·K-I · mol-I . The number of orient ations of
CO molecules in the crystal is calculated as S
= RIn(orientations); (orientations)
=
eS / R
=
e(4.6/8314 51)
=
e(O.553 2 5) = 1 .7 .
This is less than 2, the number expected for a random orientation (disorder) of CO molecules at 0 K (see text Fig. 7.7), suggesting some ordering at 0 K, possibly caused by al ignment of the smal l permanent dipoles of neighboring CO molecules.
SG -76
7.6 The Equ ivalen ce of Statistical and Therm odyn amic Entro pies •
Boltzmann ' s molecular interpretation compared with the thermodynamic approach
•
Qualitative comparison Volume increase in an ideal gas:
V thermodynamic approach � I'!.S = nR In 2 ;
VI
if V2 > VI , then I'!.S > 0
molecular approach � Consider the gas container to be a box. From particle- in-a-box theory, as V
increases, energy levels accessible to gas molecules pack more closely together and W increases (see text Fig. 7.9). W
So, D.S = Sv., - Sv.1 = k In- 2
WJ
Temperature increase in an ideal gas: thermodynamic approach
�
I'!.S = nC (V
or
P,m )
>
T �;
In
I
O.
if T2 > TI , then D.S > 0
molecular approach � From particle- in-a-box theory, at low T, gas molecules occupy a small number of energy levels (W = small); at higher T, more energy levels are available (W = larger) (see text Fig. 7.10). W2
So, D.S = ST, - ST1 = kln- > 0. WI
Note: For liquids and solids, simi lar reasoning for the T dependence of S applies. •
Quantitative comparison � Assume that the number of microstates avai lable to any molecule is proportional to the volume
available to it: W
=
� For N molecules, W �
constant x V. =
(constant
x
V)N
. c: VI to V2, F or an expansion 0 fNA mo I ecu I es trom
(
AC' = NAkln constantxV2 co nstant x VI
LW
which is identic al to the thermodynamic expression. S ummary
�
) ( ) =
V2 R ln VI
Macroscopic and microscopic approaches give the same predictions for D.S. The Boltzmann approach provides deep insight into entropy on the molecular level in terms of the energy states avai lable to a system.
7.7 Standard Mola r En tropies, SmO(T)
•
Use of different formalisms � Use Boltzmann formalism to c alcul ate entropy (sometimes difficult to do) .
� •
Use thermodynamic formalism to me asure entropy.
Standard (P = 1 bar) molar entropies ofpure substances
�
Determined experi mentally from he at c ap acity data (C I',m), enth alpy data (f.Jf) for phase changes, and the third law of thermodynamics
SG-77
�
For a gas at final temperature T (assuming only one solid phase)
iT, CPm (solid)dT Sm (T) = S m ( 0) + o
'
O
/lHruS
+
T
__
Tf
T, Cpm (I iquid)dT f + T,
'
T
/lHvap
+ -- +
Tb
fT Cpm (gas)dT T,
' ---,--
T
� I f we assume that S li l ( 0) = 0 according to the third law, then
Sm 0 Note:
iT, Cpm (solid)dT + /lH, (T) = '
O
s + ,u_
_ _
Tf
T
Cpm (gas)dT fT. Cpm (I iquid)dT + /lHvap + fTT ----,,,-,' , T T, Tb T '
-'--
--
For solids, only the first term in the equation is used if only one solid phase exists. For liquids, the first three tenns are used. For gases, all five terms are required.
E xample: Plots of CI',1ll as a function of T and CI',m IT as a function of T for sol id
copper, Cu(s), are shown below. In the second graph, the area under the curve (T 0 to T = 298. 1 5 K) yields the experimental value of the entropy for Cu(s): For the pure substance at I bar, S ma (298. 1 5 K) = 3 3 . 1 5 lK-' · mor ' . =
II)
L 0 E N , �
;:---N L o
E I' �
a
N
II)
�
:2.
a
I-
:j,�
o •
100
T (K)
200
300
ci
N
II)
ci
a
ci
II) a
ci E Q a a (,) ci
0
100
T(K)
200
300
Appendix 2 A �
� �
Lists of val ues of experiment al standard molar entropies at 25°C for many substances (st andard == pure substance at 1 bar) Entropies of gases tend to be larger than those of solids or liquids (Sgas> S'iquid or S so'id), as expected from the association of entropy with r andomness and disorder. Other things equal, molar entropy increases with molar mass. Heavier species have more vibrational energy levels available to them than lighter ones, so Wand S are larger. (See text Fig. 7. 1 3 for use of particle-i n-a-box model to understand this better.)
7 . 8 Standard Reaction En tropies, IJ..S rO •
Standard reaction entropies, /').Sro �
•
For a ny chemical reaction �
/').S r °
=
Determine similarly to calculation of standard enthalpy of reaction, /lH0, , for a chemical reaction. InS m ° (products )- InSm ° (reactants )
+ C Vg) - 2 LiCI(s). Using Sm ° values from Appendix 2A, we obtain
Example: Calculate /':,.Sro for the reaction 2 Li(s)
/').S r ° = InSm o(product s)- InSm ° (reactants )= 2 S mO (LiCI) - [2 SmO (Li) =
2(59.33) - [2(29. 1 2) + 222. 96] = -1 62.54 lK-' · mor ' .
+
S m O (Cl 2 )]
/':,.S,o< 0, as expected for a net decrease in the number of moles of gas when reactants form products
SG-78
•
G eneralizations � Entropy of gases dominates change in reaction entropy.
� !1Sro is positive if there is a net production of gas in a reaction.
� !1Sro is neg ative if there is a net consumption of gas in a reaction.
• •
Properties of Sand !1S � S is an extensive property of state, like enthalpy. For any process or change in state � Reverse the process, ch ange the sign of !1S.
� Ch ange the amounts of all materials in a process; make a proportion al ch ange in the val ue of!1S.
� Add two reactions together to get a third one; add the !1S values for the first two reactions to get !1S for the third reaction.
GLOBAL CHANGES IN ENTROPY (Sections 7.9-7.11) 7.9 The Su rroundings • •
Second law � For spont aneous change, the entropy of an isol ated system incre ases.
A ny spontaneous change � Entropy of system plus surroundings incre ases.
•
� Criterion for spontaneity includes the surroundings; !1Stot = !1S + !1Ssurr
•
Criterion � A process is spontaneous as written if !1Stot > O.
� A process is nonspont aneous as written if!1Stot <
O.
In this instance, the reverse process is spontaneous.
� For a system at equilibrium, !1Stot = 0 (see Section 7. 1 1 ). •
Calculating !1Su s rr for a p rocess at constant T and P � Note:
!1S
qsurr surr = T
=
_
!:!J! T
!:!J! is the enthalpy change for the system. At constant pressure, the he at associated with the process is transferred reversibly to the surroundings. The heat capacity of the surroundings is vast, so its temperature remains constant. The process occurring in the system itsel f may be revers ible or irrever sible.
Example: The entropy change of the surroundings when I mol of H 20(l) vaporizes at 25°C is
q urr !1SSUIT = s
T
_
(j}{yap
T
1 40700 J · mol1 = - 1 3 7 J . mol- I . K - • ( 2 73 . 1 5 + 2 5 ) K
The entropy o f the surroundings decreases whi le that o f the system increases.
SG-79
7 . 1 0 Overall Change in Entropy •
Spontaneity depends on M and �Ssurr
� Four cases arise because M and �Ssurr can e ach be
either positive (+) or negative ( ). -
Case
I�Stot
�S
+
+
2
?
3
?
+
�Ssurr I
Spontaneity
+
alw ays spontaneous
+
spontaneous, if IMsurr I > I�S I spontaneous, if I M I > I�Ssurr I
+
never spontaneous
4 •
Application to chemical reactions � Exothermic reactions ( �SSUIT> 0) correspond to Cases if I Msurrl > I�S Iwhen M < o.
I and 2, that i s, spont aneous if �S> 0 or
� Endothermic reactions ( �SSUIT < 0) correspond to Cases 3 and 4, that is, spont aneous only if
•
�S> 0 and IM I> I�SSUITI.
For a given change in state (same �, the path can affect Msurn �Stot' and spontaneity .
� Comparison of a reversible and an irreversible isothermal expansion of an ide al g as for the same initial and final states (see text Example 7 . 1 2) :
(I) �S (irreversible) = M (reversible) (entropy is a state function)
(2) �U = q + w= 0 and t.Ji = 0 because energy and enthalpy of an ideal gas depend only on T, and �T = 0 for this process. (3) q and ware different for the two processes (q and ware p ath functions). H eat given off to the surroundings (-q = qsun) is different for the two processes. (4) A reversible process does the m aximum work w(see Chapter 6). Because q + w= 0, a reversible process del ivers less heat to the surroundings. (5) Because MSUIT = qSUlT/ T, �Ssun· (irreversible) > �Sslln (reversible). (6) Thus, �Stot (irreversible) > Mtot (reversible) = O.
(7) The reversible process corresponds to one in which the system is only i nfinitesim ally removed from equil ibrium as the process proceeds, whereas the irreversible process [�Stot(irreversible) > 0 ] is spont aneous. 7. 1 1 Equ ilibri u m • •
System a t eq uilibrium � N o tendency t o change in forward o r reverse direction without a n external
influence
Types of eq uilibrium thermal:
No tendency for heat to flow in or out of the system Example: An aluminum rod at room temperature
mechanical: No tendency for any part of a system to move Example: An undefonned spring with no tendency to stretch or compress physical:
Two phases of a substance at the transition temperature with no tendency for either phase to increase in mass Example: Steam and water at the normal boi ling poi nt ( 1 00°C)
SG-80
chemical:
E xample: A mixture of reactants and products with no net tendency for the formation
of either more reactants or of more products
Example: A saturated sol ution of sucrose in water in contact with sol id sucrose •
Universal thermodynamic criterion for equilibrium �
1'hl101 = 0 for any system at equil ibri um. I f this were not true, I'hlIOI would be greater than 0 in either the forward or the reverse direction and spont aneous ch ange would occur until l'hlo, , = o.
� Total entropy change may be calculated to determine whether a system is at equil ibrium.
GIBBS FREE ENERGY (Sections 7.12-7.16) 7. 1 2 Focusing on the System •
6.G, an alternative criterion for spontaneity ( I ) 6.S101 = I'hl + I'hlSUIT (always true) (2) If P and T are constant, /'-,SslIrr (3) I f P and T are constant, /'-,SIOI
till = -T till -
=
/'-,S
-
(function of system only).
T
(4) Definition of Gibbs free energy, G == H - TS (5) 1 f T is constant,
6.G = W - 6.(TS) = W T6.S. (6) According to (5) and (3), 6.G = -T 6.SIOI• •
Summary
•
I f a process at constant T and P is
-
� spont aneous �
� •
(1'hl101 > 0), then 6.G < o. nonspont aneous (1'hl101 < 0), then 6.G > O. at equilibrium (6.SI01 0), then 6.G = 0 as well. =
6.H of a system at constant T and P Four cases arise because both 6.S and W can be either positive (+) or negat ive (-).
Dependence of spontaneity on 6.Sand �
Case
I6.G
6.H
T�SI
+ 2
?
3
?
+
4
+
+
Spontaneity
alw ays spontaneous spontaneous, ifI6.H I > T I 6.S I
+
spontaneous, if TI6.SI> I WI never spontaneous
SG -SI
Case 1 �
� �
Exothermic process or reaction (Mf < 0) with 6.S > 0 Enthalpy and entropy favor spontaneity. Enthalpy- and entropy-driven process or reaction Temperature dependence: If 6.H and 6.S are T independent, the reaction is spontaneous at all
Case 2 �
� �
Case 3 �
� �
Case 4 �
�
�
•
Exothermic process or reaction (Mf < 0) with 6.S < 0 Only enthalpy favors spontaneity. If the process is spontaneous, it is enthalpy-driven. Temperature dependence: If Mf and 6.S are T independent, the reaction is spontaneous at low T (enthalpy "wins") but nonspontaneous at high T. A crossover temperature exists at which Mf =T 6.S and the system is at equil ibrium. Endothermic process or reaction (Mf> 0) with 6.S > 0 Only entropy favors spontaneity. I f the process is spontaneous, it is driven by entropy (entropy-driven process). Temperature dependence: If Mf and 6.S are independent of T, the reaction is nonspontaneous at low T and spontaneous at high T (entropy "wins"). A crossover temperature exists at which 6.H =T 6.S and the system is at equilibrium. Endothermic process or reaction (Mf> 0) with 6.S < 0 Nonspontaneous process or reaction (6.G > 0) Temperature dependence: If 6.H and 6.S are independent of T, then the reaction is nonspontaneous at all T
Temperature dependence of G for a pure su bstan ce �
G =H-TS
� The free energy of a substance decreases with increasing T � But, for a given substance, Sm,solid < Sm,liq < S,n,gas '
� So Grn,solid decreases more slowly than Grn,liq which decreases more slowly than G,n,gas'
� The above forms the basis for a thermodynamic understanding of melting, vaporization, and
subl imation (see text Figs. 7.25 and 7.26).
7. 1 3 Gibbs Free Energy of Reaction, 6.Gr •
•
Chemical reactions � Definitions (n
=
6.Gr and 6.Gro are defined in a manner simi lar to the reaction enthalpy, Mfr, and standard reaction enthalpy, Mfro.
stoiciometric coefficient)
� If G", is the molar Gi bbs free energy of a reactant or product, then the Gibbs free energy
of reaction is
SG-82
T
� I f Gnl° is the
standard molar Gibbs free energy, the standard Gibbs free energy of reaction is
� Values o fGm or Gm° cannot b e determined directly. So the equations given above
cannot be used to
� Values of D.Gr and D.Gro are determined from the Gibbs free energies of formation,
D.Gr and D.Gro.
determine D.Gr and D.Gro.
•
Standard G ibbs free energy of formation, D.Gro
� Standard Gibbs free energy of formation of a compound or element
� Gibbs free energy change for the formation of one mole of a compound from the most stable form
of its elements under standard conditions (I bar)
� •
D.Gt == 0 for all elements in their most stable form (same convention as for enthalpy).
Calculating D.Gro for a given compound 1.
Write and balance the formation reaction. One mole of compound on the product side and the most stable form of the elements with appropriate coefficients on the reactant side
2. Calculate Wro and D.Sro for the reaction by using the data in Appendix 2A. The value of D.Sro is obtained from the standard molar entropy values as follows: /j"Sro = Sma (compound) InSmo (reactants). -
3.
Solve for the formation of one mole of compound, using the expression D.Gro Wro- TD.Sro =
=
•
W r O(compound) T [Sm O(compound) L,:nSm O(reactants) ] -
-
Thermodynamically sta ble compound � Has a
�
negative standard free energy of formation (D.Gro < 0) Has a thermodynamic tendency to form from its elements
� Examples are ores such as Ah03(S) and Fe2 03(S). •
Thermodynamically unsta ble compound � Has a �
positive standard free energy of formation (D.Gt> 0) Has a thermodynamic tendency to decompose into its elements
� Examples are some hydrocarbons such as acetylene (C 2 H 2 ) and benzene (C6 H 6).
•
Properties of thermodynamically unsta ble compounds � �
•
Many decompose into their elements over a long time span. Thermodynamically unstable substances (l ike liquid octane and diamond) that decompose into their elements slowly are said to be kinetically stable.
Classification of thermodynamically unsta ble compounds labile:
Decompose or react readily, for example, TNT and NO
non labile:
Decompose or react slowly, for example, liquid octane
inert:
Exhibit virtually no reactivity or decomposition, for example, diamond
SG-83
•
Calculating the standard G ibbs free energy of reaction !1Gro for a given chemical reaction 1.
Write the balanced chemical equation for the reaction. 2. Calculate !1H,o and !1S,O for the reaction by using the data in Appendix 2A. 3. Solve for !1G,o, using the expression !1Gr o= !1Hr 0_ T !1 Sr °
= [I n!1Hr O(products) - In!1Hr O(reactants)J- T [ InSm O(products) - I nSm O(reactants) ]
4.
Alternatively, calculate !1Gro from the !1Gro values listed in Appendix 2A. !1Gr ° = In!1GrO(products) - In!1GrO(reactants) The equation directly above yields a faster result, but does not allow for an estimation of the temperature dependence of !1Gr 0. E xample: Calculate !1Gro at 25°C for the reaction
2 Ah03(S) + 3 C(graphite,s)
�
3 CO2 (g) + 4 AI(s).
!1Gro= ilz!1GrO(products) - ilz!1GrO(reactants) = 3 !1Gt[C02 (g)] + 4 !1GnAI(s)] - { 2 !1GrO[AI 20 3(s)] + 3 !1GrO[C(graphite,s)] } = 3(-394.36 kJ . mor l ) - 2(-1 582.3 kJ . mor l ) = 1 98 1 .5 kJ·mor ' Since !1Gro is positive, the reaction is not spontaneous at 25°C under standard conditions. Note: !1GnAI(s) and !1G?[C(graphite,s)] are zero (elements in their standard states). 7 . 1 4 The Gibbs Free Energy and Nonexpansion Work •
Processes at constant T and P E xpansion work � dw �
=
-PdV (infinitesimal change in volume, see Chapter 6)
W = -P!1V (jinite change in volume, constant opposing P)
� Expansion work is performed against an opposing pressure. � The sign of w is defined in terms of the
system:
I f work is done on the system by the surroundings, w> 0 (the system gains energy). I f work is done by the system on the surroundings, w < 0 (the system loses energy).
Nonexpansion work �
�
dWe We
(infinitesimal change, subscript "e" stands for extra) (jinite change)
� Any other type of work, including electrical work, mechanical work, work of
muscular contraction, work involved in neuronal signaling, and that of chemical synthesis (making chemical bonds)
•
Relationship between nonexpansion work and (finite) G ibbs free energy changes � For a reversible process, !1G = Wrev,e �
W,ev,e
(constant P and 1) = maximum nonexpansion work obta inable from a constant P, T process
SG-84
� �
Reversible process : maximum amount of work is done by system on the surroundings
(wrev,e is negative).
All real systems are irreversible and the maximum nonexpansion work is never obtained. E xample: The maximum electrical work obtainable at I bar and 25°C by burning one mole of
propane in a fuel cel l with excess oxygen according to the equation C 3 H 8(g) + 5 02(g) - 3 CO2(g) + 4 H 20(l) is given by Wrev, e = !!.Gro = In!!.GrO(products) - In!!.GrO(reactants)
= 3!!'Gt[C02(g)] + 4!!.GrO[H20(l)] - { !!' Gt[C3 H 8(g)] + 5 !!. G t[02(g)]} = 3(-394.36 klmor l ) + 4(-237.13 klmorl) - (-23.49 kJ·morl)
= -2108.11 kJ for I mol of C 3 H 8(g) The negative value for Wrev, e indicates work done on the surroundings. 7.1 5 The Effect of Temperature •
For a ny chemical reaction at constant P and T
�
Cases 1- 4 from the table i n Section 7.12 of this Study Guide apply to any chemical reaction, assuming that I1Hro and !!.Sro are independent of temperature.
� I n this case, the temperature dependence of !!.Gro arises from the (
-
T !!.SrO ) term.
E xample: To calculate the temperature range over which the reaction H 2 (g) +
+ 0ig) - H 20(g) is
spontaneous, use !!.Gro = !!.Hro T !!.Sro. Set !!.G ro = 0 to determine the crossover temperature, where neither the forward or reverse reaction is favored. -
Tcrossover -
I1H r ° !!.S
r
°
Sm ° [ H 2 0(g)]
-
{Sm ° [H2(g)] + (1/2)Sm ° [0 2 (g)]} -
241.82 kJ . mol-I
0.188 83 kJ · K- mol-I -{0.130 68 + (1 /2)(0.205 1 4 )} kJ · K- 1. mol-I 1.
-241.82kJ·mol-1
-0.044 42 kJ · K-1 · mol-1
=5444K
I n this case, enthalpy favors spontaneity; entropy does not. At 298.15 K, the reaction is spontaneous [!!.Gro = I1Hro T !!'sro = -241.82 - (298.15)(-0.044 42) = -228.58 kJ mor l]. I t is spontaneous unti l T = 5444 K. Above 5444 K, H 20(g) is expected to decompose spontaneously into its elements. -
7 . 1 6 Impact o n Biology : Gibbs F re e E ne rg y Chan ges i n Bio lo g i cal Systems •
Metabolism � I n metabol ic processes, reaction steps may have a
positive (unfavorable) reaction free
energy. They can be coupled to spontaneous reactions to make the overal l or net reaction spontaneous.
SG-85
•
Concept of coupled chemical reactions
--*
--*
--*
A way to make nonspontaneous reactions occur without changing temperature or pressure A prominent process in biological systems, but also found in nonbiological ones I f reaction (I) has a positive reaction free energy [�GrO (I) > 0] (nonspontaneous), it can be coupled to reaction (2) with a more negative reaction free energy [�Gro (2) < 0], such that reaction (3), the sum of the two reactions, has a negative reaction free energy [�GrO (3) < 0]: �GrO (3) = �Gr°(l) + �GrO (2) < 0 (overal l reaction is spontaneous). Example: The sugar glucose (a food) is converted to pyruvate i n a series of steps. The overall
process may be represented as: glucose + other reactants
�
2 pyruvate + other products; �Gro
=
- 80 .6 kJ·mor l .
The relatively large negative value of �Gro indicates that the overall reaction occurs spontaneously under biochemicql standard conditions. The biochemical standard state corresponds more nearly to typical conditions in a cel lular environment. The first step in the metabolic process is the conversion of glucose into glucose-6-phosphate: glucose + phosphate
�
gl ucose-6-phosphate + H 2 0; (I) �Gro (I) = + 14.3 klmorl . o
II
o-p-o
� � OH
HO
HO
Glucose
OH
HO
HO
_
OH
OH
G lucose-6-phosphate
The positive standard free energy implies nonspontaneity for reaction ( 1 ). To generate glucose-6-phosphate, reaction (I) is coupled with (2), the hydrolysis of adenosine triphosphate (A TP) to yield adenosine diphosphate (ADP). Reaction (2) has a favorable �Gro < 0 value (spontaneous): A TP + H 20 ADP + inorganic phosphate; (2) �Gro (2) -31.0 kJ ·morl . �
=
Adding (coupl ing) reactions (I) and (2) gives a net spontaneous reaction (3): glucose + A TP
glucose-6-phosphate + ADP (3) �GrO(3) = �Gr°(I) + �GrO (2) = -1 6.7 kJ . mor l Note:
�
Reactions in metabolic pathways are catalyzed by enzymes. Step (I) is catalyzed by the enzyme hexokinase. Biological catalysts are discussed in Chapter 13 and sugars, a class of carbohydrates, are discussed in Chapter 19.
SG-86
Chapter 8 PHYSICAL EQUILIBRIA PHASES AND PHASE TRANSITIONS (Sections 8.1-8.7) 8.1 Vapor Pressu re •
Liquids �
Evaporate to form a gas or vapor Puddle of rainwater evaporates. The reaction H 2 0(l) - H 2 0(g) goes to completion and the puddle disappears. Phase equilibrium is not attained here.
•
Gases �
•
Equilibrium � I n a closed system, both l iquid and gas phases exist in
Condense to form a liquid or solid at sufficiently low temperature Water vapor condenses to form rain. Steam condenses on cold surfaces.
equilibrium.
Represent equil ibrium system using double-headed arrow: H 2 0(l)
•
Liquid-gas phase equilibrium � Example: I n a
=
H 20(g)
Dynamic equilibrium Rate of evaporation equals rate of condensation.
covered j ar half-fi l led with water at room temperature, the liquid and gas
phases are in equil ibrium, H 20(l) = H 20(g). Water vapor exhibits a characteristic equilibrium pressure, its vapor pressure.
•
Vapor pressu re, P
� Characteristic pressure of a vapor above a
equilibrium (closed system)
confined l iquid or sol id when they are in dynamic
� Depends on temperature, increasing rapidly with increasing temperature
� Different vapor pressure for different liquids and solids
Examples: S everal solids with measurable vapor pressures are camphor, naphthalene,
paradichlorobenzene (mothbal ls), and dry ice [C02 (s)). Virtually all l iquids have measurable vapor pressures, but some [e.g., Hg(l)] may be very sma l l at room temperature.
•
Sublimation � Evaporation of solid to form a gas [C0 2(s) - CO2(g)]
•
Volatility � Related to the abil ity to evaporate
� Liquids with high vapor pressures at a given temperature are volatile substances.
•
Characteristics of eq uilibrium between phases � ->
Dynamic equilibrium occurs when molecules enter and leave the individual phases at the same rate, and the total amount in each phase remains unchanged. Molar free energies of the individual phases are equal. substance (phase I) = substance (phase 2), t-.Gm = 0
8.2 Volatility and Intermolecular Forces •
Increasing the strength of intermolecular forces in a liquid �
�
�
Decreases the volatility Decreases the vapor pressure at a given temperature Increases the normal boil ing point, Tb (defined in Section 8.4)
SG-87
•
•
London forces
�
�
More massive molecules are less volatile. M ore electrons in a molecule produce stronger forces.
Dipolar forces
� For molecules with the same number of electrons, dipolar molecules are associated with less volatile
•
•
�
substances than are molecules with only London forces. A greater dipole moment yields a lower volati lity.
Hyd rogen bonding � Ionic forces Note:
•
�
Molecules that form H-bonds may produce even less volatile substances than dipolar molecules.
Salts are essentially nonvolatile.
As with many rules in chemistry, the above qual itative guidelines must be applied cautiously.
Quantitative approach to vapor pressure (see derivation in text)
RT
•
Insight from the above equation
�
�
�
Because t..Svapo is about the same for all liquids (Trouton's rule), the vapor pressure of a liquid depends mainly on t..Hvapo, which is always a positive quantity. Stronger intermolecular forces result in a larger t..Hvapo and a decrease in vapor pressure. For a given l iquid, an increase in temperature results in an increase in vapor pressure.
8.3 The Variation of Vapor Pressure with Tem peratu re •
Solve the vapor pressure equation for two temperatures
(P2 at T2 and PI at TI)
Clausius-Clapeyron equation •
Uses of the Clausius-Clapeyron equation
�
� �
Measure the vapor pressure of a liquid at two temperatures (P2 at T2 and PI at TI) to estimate the standard enthalpy of vaporization t..Hvapo.
Measure t..Hvap° and the vapor pressure of a liquid at one temperature (PI at TI) to estimate the vapor pressure at any different temperature (P2 at T2)' M easure t..Hvapo and the vapor pressure at one temperature (PI at TI)' Estimate the normal boil ing point of the liquid (Tb = h when P2 = 1 atm) or the standard boil ing point of the l iquid (T2, when P2 = 1 bar).
SG -88
� Determine Wvapo and the vapor pressure at the normal boiling temperature (PI at
Tb)'
Estimate the vapor pressure at a different temperature (P2 at T2)' Note:
In all cases, the term estimate is used because the assumption that Wvap° is independent of temperature is not strictly true; it is an approx imation.
8.4 Boiling •
Liquid in an open container �
•
Boiling �
• •
• •
Rate of vaporization is greater than the rate of condensation, so the l iquid evaporates.
Vapor pressure of l iquid equals atmospheric pressure. Rapid vaporization occurs throughout the entire liquid.
Boiling point � Temperature at which the l iquid begins to boi l Normal boiling point �
Boiling point at 1 atm,
Tb
Within experimental error,
�
Tb = 99.974°e 1 aaoe for water.
Standard boiling point � Boiling point at 1 bar Effect of pressure on the boiling point � An
increase of pressure on a liquid leads to an increase in the boi l ing point. The pressure cooker makes use of this fact.
� A
decrease of pressure on a liquid leads to a decrease in the boi l ing point.
This helps explain why water boi ls at a lower temperature on a mountaintop. It also accounts for vacuum distillation at low temperatures to purifY l iquids that decompose at higher temperatures.
8.5 Freezing and Meltin g •
Freezing and melting � Two common phase transitions, one the reverse of the other
•
F reezing (melting) � Solidification of a liquid (liquefaction of a solid)
•
F reezing (melting) point
� Temperature at which liquid freezes (solid melts)
For a given substance, freezing and melting points are identical .
•
Normal freezing point, Tr � Temperature at which solid begins to freeze at I atm
Within experimental error,
•
•
Standard freezing point
Tr= ooe and I atm for water.
� Temperature at which sol id begins to freeze at
I bar
Difference between Tr and the standard freezing point is very small.
Su percooled liquid � A pure l iquid may exist below its freezing point if it is cooled extremely slowly. � Thermodynamically unstable with respect to formation of sol id
� If heat is withdrawn slowly from pure water, the temperature may drop below ooe, indicating
supercool ing. At some point, a tiny crystal (nucleus) of ice forms, and the entire sample crystal l izes rapidly to form ice whose temperature rises from the heat released by sudden freezing.
•
Pressure dependence of the freezing point: Most substances � Solid phase is more dense than the liquid phase (smal ler molar volume),
so the sol id sinks as it forms. Increasing pressure favors the phase with
SG -89
the smallest density; thus, the freezing point increases with increasing pressure. Few substances
Note:
---+ Solid phase is less dense than the liquid phase (larger molar volume) and the solid floats as it forms. An increase in pressure favors the phase with the greatest density; thus, the freezing point decreases with increasing pressure. Examples include water and bismuth.
The anomalous behavior of water is critical for the survival of certai n species. If ice sank, bodies of water would freeze solid in winter, ki lling aquatic life.
8.6 Phase Diagrams •
Component ---+ A single substance, for example, aluminum, octane, water, or sodium chloride ---+ A chemically independent species
•
Single-component phase diagram
•
---+
Map showing the most stable phase of a substance at different pressures and temperatures
Phase boundary
---+
---+ •
Lines separating regions on a phase diagram Represents a set of P and T values for which two phases coex ist in dynamic equilibrium
Triple point ---+
---+
Point where three phase boundaries intersect Corresponds to a single value of P and equilibrium ---+
T for which three phases coexist
in dynamic
•
Critical point
•
Critical temperature, Tc ---+ The temperature above which a gas cannot condense into a l iquid
High-temperature terminus of the liquid-vapor phase boundary Only one phase is observed above
Note:
Tc.
Critical point and critical temperature are i ntroduced in Section 8.7 in the text. The definitions given above are repeated later.
Phase Diagra m of Water
Note the regions (bounded areas) labeled ice (solid), l iquid, and vapor (gas). I n each area, only a single phase (ice, water, or water vapor) i s stable. Within each region, pressure and/or temperature may b e varied independently with no accompanying phase transition. Two independent variables, or two degrees of freedom, characterize such a region. The three lines that separate the regions define the phase boundaries of solid-liquid, l iquid-vapor, and solid-vapor. At a boundary, only P or T may be varied if the two phases are to remain in equilibrium. Only one independent variable, or one degree of freedom, exists for water in states corresponding to the boundary. All three phases coexist at the triple point with its specific P and T, or zero degrees of freedom.
t: o
t:. � ::3 '" '" OJ
ct
100
Temperature ( C) o
SG-90
•
Slope of the solid-liq uid phase boundary line � Positive for most substances:
Solid sinks because it is more dense than liquid. At constant temperature, a pressure increase yields no phase change for the solid. At constant temperature, a pressure increase may cause the l iquid to solidify.
� Negative for H 20:
Because the solid is less dense than the l iquid, ice floats. At constant temperature, a pressure increase may cause ice to melt. At constant temperature, a pressure i ncrease yields no phase change for the l iquid.
•
Increasing the p ressure on graphite to make diamond and liquid carbon � What occurs when the pressure on graphite at 3000 K is increased from 1 bar to 500 bar?
600 500 ro
400
Diamond Liquid
�300 Q
200 100
Graphite
O L-����--���� o
1000
Note:
2000
3000
T(K)
4000
5000
I n the initial state of graphite (point A), one phase exists. Temperature is held constant (3000 K), and the pressure is allowed to vary freely. The CCgraphite)CCdiamond) phase boundary is reached at approximately 230 kbar. At this point, CCdiamond) forms and exists in equilibrium with CCgraphite). A ll the CCgraphite) is eventually converted to CCdiamond). (This process may be quite slow.) In the diamond region, one phase exists. Additional increases in the pressure on CCdiamond) result in the equilibrium of CCdiamond) and liquid carbon at about 490 kbar. When all the diamond has melted, the pressure on CCl) is free to increase to attain the final state (point B).
There are three allotropes of carbon: graphite, diamond, and buckminsterfullerene (C 60); the l atter, discovered in 1985, is composed of soccer-ball-shaped molecules. The thermodynamic stability of buckminsterfullerene has not yet been determined. The validity of its inclusion on the C phase diagram is, therefore, uncertain. (Metastable phases, such as supercooled water, do not appear on phase diagrams.) The crystal structure is face centered cubic with C60 molecules at the corners and faces of a cubic unit cell . The unit cel l is shown below.
SG-9 1
8.7 Critical Properties � Terminus of the l iquid-gas phase boundary at high temperature
•
C ritical point
•
Critical temperature
� The temperature above which a vapor (gas) cannot condense into a
liquid. Only one phase is observed above
•
Supercritical fluid � Substance above its critical temperature,
Tc.
Tc
Phase Diagram of Water
Liquid 200 E :::l III III Ql
100
............. Ice
...
CI....
·· ······
· · ·· ·
Critical
( ii8 �;:J�� "
·
··
374"C
Vapor
This discontinuous diagram shows both low and high pressure regions. Point A represents typical "room temperature" conditions. Liquid water has a vapor pressure of23.76 Torr (0.03 1 26 atm) at 25°C (298. 1 5 K). At 200°C (473 . 1 5 K), the vapor pressure is 1 5 atm (point B). As the temperature is increased further, the densities of the liquid and vapor i n equilibrium approach one another and become nearly equal at the critical point (point C). Above the critical temperature, only one phase with the properties of a very dense vapor remains. The critical pressure, Pc, is the vapor pressure measured at the critical temperature. The temperature, pressure, and density values for water at the critical point are Tc = 374. 1 °C (647 K), 3 Pc 2 1 8.3 atm, and de 0.32 g . cm- , respectively. =
200 o 100 Temperature •
=
300
(0C)
I ntermolecular forces � Both
Tc and Tb have a tendency to increase with the increasing strength of intermolecular forces.
Substance
Tc (K)
Tb (K)
5.2
4.3
H e (heli um)
Note:
Ar (argon)
1 50
88
Xe (xenon)
290
1 66
N H 3 (ammonia)
405
240
H 2 0 (water)
647
373
The strength of the London forces increases with increasing molar mass (number of electrons) of a noble gas atom. With stronger intermolecular forces, higher critical and normal boil ing temperatures are expected. Ammonia and water form strong hydrogen bonds. Water forms more hydrogen bonds per molecule than does ammonia. Therefore, water has critical and normal boiling temperatures higher than those of ammonia. I n making these comparisons, care must be taken.
SG - 92
SOLUBILITY (Sections 8.8-8.13) 8.8 The Limits of Solu bility �
•
Two component solution
•
I nteractions � Solvent-solvent, solute-solute, and solute-solvent
•
•
Contains one solvent and one solute species
U nsaturated solution
�
�
All solute added to solvent dissolves. Amount of sol ute dissolved in the solvent is less than the equilibrium amount.
Saturated solution � Solubi lity limit of solute has been reached, with any additional solute present as a precipitate.
� The � •
equilibrium amount of solute has been dissolved in solvent. Dissolved and undissolved solute molecules are in dynamic equi librium; in other words, the rate of dissolution equal the rate of precipitation.
Supersaturated solution
�
Under certain conditions, an amount of solute greater than the equi librium amount (solubi lity limit) can be dissolved in a solvent. � Because more than the equilibrium amount is dissolved, the system is thermodynamically unstable. A slight disturbance (seed crystal) causes precipitation and rapid return to equilibrium. �
•
Solubility limit
•
Molar solubility � Molar concentration of a saturated solution of a substance
•
G ram solubility � Mass concentration of a saturated solution of a substance
•
Molal solubility � Molal ity of a saturated solution of a substance
Depends on the nature of both the solute and the solvent Units : (moles of sol ute)/(I iter of solution) or mol-L- 1 == M F or example, a saturated aqueous solution of AgCI has a molar concentration of 1 .33 x 1 0-5 mol -L- 1 or 1 .33 x 1 0-5 M at 25°C. Units : (grams of solute)/(liter of solution) or g· L- 1 For example, a saturated aqueous sol ution of AgCI has a mass concentration of l .9 1 x 1 0-3 g· L- 1 at 25°C. Units : (moles of solute)l(kg of solvent) or mol· kg- 1 For example, a saturated aqueous solution of AgCI has a molal ity of 1 . 33 1 0-5 mol ·kg- 1 at 25°C.
Note:
x
For dilute solutions, the molar and molal solubi lities wi ll be essentially equal as in the example above. This is because I L of water has a mass very nearly equal to I kg at 25°C and the mass of the solute is very small compared with that of the solvent. In concentrated solutions, the mass of the solute becomes an appreciable fraction of the mass of the sol ution, and a liter of solution contains substantial ly less than I kg of water. In such a solution, molality and molarity can differ considerably.
SG-93
8.9 •
The Like-Dissolves-Like Rule
Rule � I f solute-solute and solvent-solvent intermolecular forces (London, dipole, hydrogen-bonding,
or ionic) are simi lar, larger solubil ities are predicted. Lesser solubi l ities are expected if these forces are dissimi lar.
•
Use
� A
qualitative guide to predict and understand the solubil ity of various solute species in different solvents
E xample: Oil is composed of long chain hydrocarbons. That oil and water do not mix is an
observation of everyday life. Oil molecules are bound together by London forces, whereas water associates primarily by hydrogen bonds. The like-dissolves-like rule suggests a solvent with only cohesive London forces is needed to dissolve oil. Gasol ine, a mixture of shorter chain hydrocarbons such as heptane and octane, is a possible solvent, as is benzene, an unsaturated cyclic hydrocarbon.
E xample: Gl ucose, C 6H 1 2 06, has five -OH groups capable of forming hydrogen bonds. It is
expected to be soluble in hydrogen-bonding solvents such as water. Conversely, it should be insoluble in nonpolar solvents such as hexane.
Example: Potassium iodide, KI, is an ionic compound. Because both water and ammonia are
highly polar molecules, they are expected to be effective in solvating both K+ and l ions. Ethyl alcohol molecules are less polar than water, so KI is expected to be less sol uble in ethanol than in water. A simi lar argument can be used to explain the slight solubi l ity ofK I i n acetone. � Water-attracting
•
Hydrophilic
•
H yd rophobic � Water-repel ling
•
Soaps
� Long chain molecules, with hydrophobic and hydrophilic ends, which are soluble
in both polar and nonpolar solvents
•
S urfactant
� Surface-active molecule with a hydrophilic head and a hydrophobic tail
•
M icelle
� Spherical aggregation of surfactant molecules with hydrophobic ends in the interior
and hydrophi lic ends on the surface
Note: The term hydrophobic is somewhat of a misnomer. I n actuality, an
attraction between solute and solvent molecules always exists. I n the case of water, the sol ute-solvent interaction disrupts the local structure of water, which is dominated by hydrogen bonding. The stronger the sol ute-H 20 interaction, the greater the likelihood that the local solvent structure wi l l be disrupted by hydrated solute molecules, increasing the sol ubi l ity of the solute.
8. 1 0 Pressu re and Gas Solu bility : Hen ry's Law •
Pressure dependence of gas solubility: Qualitative features � For a gas and liquid in a container, an increase in gas pressure leads to an increase in
its solubil ity in the liquid.
� Gas molecules strike the liquid surface and some dissolve. An increase in gas pressure leads to an
increase in the number of impacts per unit time, thereby increasing solubil ity.
SG-94
� I n a gas mixture, the solubil ity of each component depends on its partial pressure because molecules
str ike the surface independently of one another.
Quantitative features � Henry's law �
r-I -S = k-H-P---'
P,
The sol ubility, s, of a gas in a l iquid is directly proportional to the partial pressure, of the gas above the l iquid. � s == molar solubil ity (see Section 8 . 14) � == Henry ' s law constant, a function of T (see Section 8. 1 1 ), the gas, and the solvent I � Unjts: s (mol· L- ), (mol·L- I · atm- I ), and (atm)
kH
kH
P
E xample: Estimate the molar solubility and gram solubility of O2 in dry air dissolved in a l iter of
water open to the atmosphere at 20°e. Assume that air is 20.95% O 2 by volume. The pressure of the atmosphere is I atm, thus the partial pressure of O 2 is 0.2095 atm. S (02) = ( 1 .3 x 1 0-3 mol-L- I ·atm- I )(0.2095 atm) = 2. 7 x 1 0-4 mol· L- 1 4 s x M = (2.7 X 1 0- mol - L- I )(32.0 g· mor l ) = 8.6 x 1 0-3 g-L- I
8. 1 1 Tem perature and Solubility •
Dependence of molar solubility on temperature � For solid and l iquid sol utes, solubil ity usu ally increases with increasing temperature, but �
�
for some salts, for example Li 2 C03 , sol ubi l ity decreases with increasing temperature (see text Fig. 8.22). For gaseous solutes, solubility usu ally decreases with increasing temperature. Despite these trends, solubil ity behavior can sometimes appear complex, as with sodium sulfate, Na2 S04 , whose solubi lity in water increases then decreases (text Fig. 8.22). I n this instance, complex solubility behavior arises because different hydrated forms of sodium sulfate precipitate at different temperatures. E xamples:
1.
The solute that precipitates from a saturated solution may be different fr om that of the solid initial ly dissolved. At room temperature, anhydrous potassium hydroxide, KOH(s), readily dissolves in water. The solid that precipitates from a saturated solution of KOH is the dihydrate, KOH · 2 H 2 0. Therefore, the chemical equilibr i um in the saturated solution is KOH · 2 H 20(s)
2.
=
K +(aq, saturated)
+
OH -(aq, saturated).
From the diagram on the next page, notice that between -50 and 1 60°C, the solubi lity of LiCl displays three discontinuities, corr esponding to temperatures at which one solid hydrated form is transformed into another. The four regions between the three discontinuities correspond, from low to high temperature, to precipitation of LiCl· 3 H20, LiCl· 2 H20, LiCl· H20, and LiCl, respectively. Note also that saturated aqueous solutions freeze at lower temperatures and boi l at higher temperatures than pure water (see Section 8. 1 6), accounting for the large temperature range of these solutions.
SG - 95
'D
o
_
Liel
o V)
oH,O "
o
-
o2H,O
o3H,O
-40
o
8.12 The E nthalpy of Solut i o n • •
80
40
120
160
T, ·C
Solutions of ionic substances � AmBn(s) - mA"\aq)
+
nBm-(aq)
Enthalpy of solution, /:,.Hso1
� Enthalpy change per mole of substance dissolved
� Depends on •
concentration of solute
Limiting enthalpy of solution � Refers to the fonnation of a very dilute solution
Val ues are given in Table 8.6 of the text. � Use of the limiting enthalpy of solution avoids complications arising from interionic interactions that occur in more concentrated solutions because ions are far apart in very dilute solutions. Note: All topics that fol low refer to the limiting enthalpy condition. •
Nature of the formation of solutions of ionic substances � Conceptualized as a two step process: subl imation of an ionic solid to fonn gaseous ions followed
by the solvation of the gaseous ions to form an ionic solution
� The enthalpy change for the subl imation step is designated the lattice enthalpy, MiL (see Table 6.6
in the text).
� The enthalpy change for the second step (formation of hydrated ions from the gas phase ions) is
designated as the enthalpy of hydration, /:,.Hhyd (see Table 8.7 in the text). � MiL always has a positive value, whereas Mihyd always has a negative value. � Misol = /:,.HL + Mihyd is, then, the difference of two numbers, both of which are typically quite large. •
Endothermic process:
•
Exothermic process:
/:,.HL > I /:,.Hhyd I
/:,.HL < I /:,.Hhyd I
Note: For smal l, highly charged ions, both MiL and
I Mihyd I have large values.
8. 1 3 The Gibbs F re e E n e rgy of Solutio n •
Sol utions of ionic substances
->
Am BI1(s) - mA 11+(aq)
SG - 96
+
nBm-(aq)
•
Free energy of solution, � Gsol
---+
---+ •
---+
Free energy change per mole of substance dissolved Depends on concentration of solute �Gsol = Msol - T �Ssol at constant T
Nature of solubility
---+ A substance wi l l dissolve if �Gsol < 0 and will continue to dissolve until a saturated solution, for which �Gsol 0, is obtained. =
---+ Both Msol and LlSso1 change with increasing concentration to make �Gsol more positive. ---+ I f the solute is consumed before �Gsol reaches 0, an unsaturated solution results, with the potential to dissolve additional sol ute. ---+
---+
---+
With excess solute present, �Gsol reaches 0 and a saturated solution results. For endothermic enthalpies of solution, the increase in entropy of solution drives the solubi lity process. A substance with a large endothermic enthalpy of solution is usually insoluble.
COLLIGATIVE PROPERTIES (Sections 8.14-8.17) 8. 1 4 Molality •
Molality, molarity, mole fraction
•
Molality, m
---+
Different concentration units used for quantitative treatment of coli igative properties
---+ Moles of solute per kilogram of solvent
---+
Independent of temperature (solute and solvent given as masses or mass equival ent) ---+ Units of moles per kilogram (mol . kg- I ) ---+ Used when relative number of molecules of components is to be emphasized
Note: •
nSOIUle
msolute = . --"=""'--kllogramssolvent The letter m is used to designate mass and molality. Care should be taken to avoid confusing the two. · · F or a b mary soIutlon,
Mole fraction, x
---+ Ratio of moles of sol ute to the total number of moles of all species in a mixture ---+ Independent of temperature ---+ Dimensionless quantity ---+ Used when relative number of molecules of components is to be emphasized
For a binary sol ution, •
Molarity
---+
nSOIUle
xsolule = ---""'-"-"'---nSOlule + nsolvent
Moles of solute divided by total volume of solution
SG-9 7
�
�
Temperature dependent (volume of solution changes with temperature) Units of moles per l iter (moiL- I ) Molar ity solute
For a binary solution,
•
=
n solute Vsolution
Converting from one concentration unit to another � The molar mass of one component is required to convert
molality to mole fraction or mole fraction to molality. � The density of the solution is required to convert molarity to molality or vice versa. Note:
When converting from one concentration unit to another, it is convenient to assume one of the fol lowing: (I) the solution has a volume of one l iter (molarity <=:> molality), (2) the total amount of solvent and solute is one mole (mole fraction <=:> molality), or (3) the mass of the solvent in the solution is one kilogram (molality <=:> molefraction).
8. 1 5 Vapor-Pressu re Lowering •
Qualitative features
�
� •
The vapor pressure of a solvent in equilibr i um with a solution containing a nonvolatile solute (for example, sucrose or sodium chloride in water) is lower than that of the pure solvent. For an ideal solution or a sufficiently dilute real solution, the vapor pressure of any volatile component is proportional to its mole fraction in solution.
Quantitative features �
Raoult's law
I
P
=
Xsolvent Ppure
I
� The vapor pressure P of a solvent is equal to the product of its mole fraction in sol ution,
Xsolvent , and its vapor pressure when pure, same temperature.
•
Ppure.
The quantities P and
Ppu,e
are measured at the
Ideal solution
�
An ideal solution is a hypothetical solution that obeys Raoult's law exactly for all concentrations of solute. � Solute-solvent interactions are the same as solvent-solvent interactions; therefore, the enthalpy of solution, f.J{sol , is zero. � Entropy of solution, L\Ssol > ° � Free energy of solution, i1Gsol < 0, leading to a lowering of vapor pressure of the solvent (see Section 8.2). � The process of solution formation is entropy driven. � Solutions approximating ideality are typically formed by similar solute and solvent species, such as hexane and heptane. � Real sol utions do not obey Raoult's law at all concentrations, but they do follow Raoult's law in the limit of low solute concentration (dilute solution). •
Non ideal (real) solution � Solution that does not obey Raoult's law at a certain concentration
SG -98
� Solute-solvent interactions differ from solvent-solvent interactions.
� f".JlsoJ is not equal to zero.
� Real solutions approximate ideal behavior at concentrations below I O- J mol·kg- J for �
nonelectrolyte solutions and below 10-2 mol· kg- J for electrolyte solutions.
Real solutions tend to behave ideally as the solute concentration approaches zero.
8. 1 6 Boili ng-Point Elevation and Freezi ng-Point Depression •
Boiling-point elevation
�
A nonvolatile sol ute lowers the vapor pressure of the solvent. As a result, a solution contai ning a nonvolati le sol ute wil l not boi l at the normal boiling point of the pure solvent. The temperature must be increased above that value to bring the vapor pressure of the solution to atmospheric pressure. Therefore, boil ing is achieved at a higher temperature (the boil ing point is elevated). � Arises from the influence of the solute on the entropy of the solvent � Quantitatively,
I boi ling-point elevation
=
kb
x
molality
I
� The boiling-point constant, kb, depends on the solvent and has units of K -kg·mor J (Table 8.8). � The boi ling-point elevation equation holds for nonvolatile solutes in dilute solutions that are
approximately ideal.
•
F reezing-point depression
�
Vapor-pressure lowering of a solution decreases the triple-point temperature, the intersection of the liquid-vapor (vapor pressure) and solid-vapor phase boundaries. The solid-liquid phase boundary originating at the triple point is moved slightly to the left on the phase diagram (the solid-vapor boundary is unchanged). The freezing temperature of the solution is thereby lowered (the freezing-point is depressed). � Arises from the influence of the solute on the entropy of the solvent � Proportional to the molal ity of the sol ute � Quantitatively,
freezing-point depression
=
kf
x
molal ity
� The freezing-point constant, k r, depends on the solvent and has units of K·kg·mor J (Table 8.8).
� •
The freezing-point depression equation holds for nonvolatile solutes in dil ute solutions that are approximately ideal.
Freezing-point depression corrected for ionization or aggregation of the solute �
freezing-point depression
=
i kf x molality
� The van 't Hoff factor,
i, determined experimental ly, is an adj ustment used to treat electrolytes (e.g., ionic solids, strong acids and bases) that dissociate and molecular solutes that aggregate (e.g.,
acetic acid dimers). � For electrolytes,
i is the number of moles of ions formed by each mole of solute dissolved in I kg of solvent if al l ions behave independently. In aqueous solutions, this only occurs in very dilute solutions; with increasing concentration, interionic effects cause i to be smal ler than the value expected for complete ionization.
� For molecular aggregates,
i may be used to estimate the average size of the aggregates.
SG-99
8. 1 7 Osm osis •
Qualitative features
�
Osmosis: The tendency of a solvent to flow through a membrane into a more concentrated solution. It is used to determine an unknown molar mass, particularly for large molecules such as polymers and proteins.
� Osmometry is a technique used to determine the molar mass of a solute if the mass concentration is
measured.
-7
-7
-7
-7
•
A semipermeable membrane allows only certain types of molecules to pass through. Typical membranes allow passage of water and small molecules, but not large molecules or ions. I n osmosis, the free energy of the solution is lower than that of the solvent; hence, dilution is a spont aneous process . The free energy of the solution can be increased by applying pressure on the solution. The i ncreased pressure at equilibrium is called the osmotic pressure, n. I n a static apparatus open to the atmosphere, pressure is applied by the increased height of the raised column of solution caused by the flow of solvent through the membrane i nto the solution. I n a dynamic apparatus i n a closed system, pressure i s applied by the increased force of a piston confining the solution (preventing solvent flow).
I f the external pressure P < n, osmosis (dilution) is spont aneous. I f P = n, the system is at equilibrium (no net flow). I f P > n, reverse osmosis occurs (flow of solvent in solution to the pure solvent). Reverse osmosis is used to purifY seawater.
Quantitative features
In = i RTM I
-7
van 't Hoff equation
-7
. . moles of solute M = molanty of the solutI On = -----liters of solution i = van' t H off factor, R = gas constant, T = temperature in kelvin Use units of n in atm and of R in L·atm·K- J ·mor J ; i is dimensionless.
BINARY LIQUID M IXTURES (Sections 8.18-8.20) 8. 1 8 The Vapor P ressure of a Bi nary Liq u id M ixtu re •
Ideal solution with two volatile components A and B; liq uid and vapor phases in eq uilibrium
-7
-7
-7
Let PA,pure = vapor pressure of pure A and PB,pure = vapor pressure of pure B . I n an ideal sol ution, each component obeys Raoult's law at all concentrations. PA XA,Jiquid PA,pure , where XA,Jiquid = mole fraction of A in the I iquid mixture =
� PB = XB,Jiquid PB,pure , where XB,Jiquid = mole fraction of B in the l iquid mixture
-7
-7
For a bi nary mixture, XA,Jiquid + XB,Jiquid = 1.
PA and PB are the partial pressures of A and B, respectively, in the vapor above the solution. PlolaJ is the total pressure above the solution. Dalton's law of partial pressures for ideal gases is PlolaJ = PA + PB (see Section 4. 8).
SG- I OO
�
�
•
Combination of Raoult's law and Dalton's law produces the expression
Nearly ideal sol utions with two volatile components are formed when the components are very similar (hexane/octane and benzene/toluene), and both components are completely soluble in each other (miscible). The composition may range from 0 < x < I; the l abels "solute" and "solvent" are not usual ly used.
Composition of the vapor above a binary ideal solution �
�
�
� �
Each vapor component obeys Dalton's law. PA
XA,vapor P, = XS,va r P, po =
where XA,vapor = mole fraction of A in the vapor mixture and P = Proral
where XS,vapor = mole fraction of B in the vapor mixture mixture, XA,vapor + XS,vapor = I. binary For a
Ps
Combination of Raoult's law and Dalton' s law yields an expression for XA,vapor x A,vapor
�
-
-
PA PA P - PA + RB _
XA,liquid PA,pure XA,liquid PA,pure + XB,liquid P,B,pure
I f PA,pure :t= PS,pure , then XA,vapor :t= XA,li qui d and XS,vapor :t= XS,liquid . The l iquid and vapor compositions differ and the vapor is always richer in the more volatile component.
8.19 Distillation •
Temperature-composition phase d iagram (ideal solution)
�
�
� •
�
Plot of the temperature of an equil ibrium mixture vs. composition (mole fraction of one component) Two curves are plotted on the same diagram, T vs. Xliquid and T vs. xvapor ' At a given temperature, the composition of each phase at equilibrium is given. See Fig. 8.37 in the text for a graphical ill ustration of a temperature-composition diagram.
Distillation
�
�
Purification of a l iquid by evaporation and condensation
•
Distillate
•
Fractional d istillation
� � � �
�
Vapor produced during distillation that is condensed and col lected in the final stage
Continuous separation (purification) of two or more liquids by repeated evaporation and condensation steps on a vertical column Each step takes place on a fractionating column in smal l increments. Liquid and vapor are in equi librium at each point in the column, but their compositions vary with height as does the temperature. At the top of the column, condensed vapor (distillate) is collected in a series of samples or fractions; the most volatile fraction is collected first. Progress may be displayed on a temperature-composition phase diagram.
SG- \ O \
8.20 Azeotropes •
Nonideal solutions with volatile components � �
•
For non ideal solutions with volatile components: Raoult's law is not obeyed, the enthalpy of mixing, /)Jf.nix :t:- 0, and solute-solvent interactions are different from solvent-solvent interactions.
Azeotrope �
•
Most mixtures of liquids are not ideal.
�
A solution that, like a pure liquid, distills at a constant temperature without a change in composition. At the azeotrope temperature, XJ iquid
�
�
�
�
greater than the value predicted by Raoult's law.
Enthalpy of mixing is endothermic, /)Jf.nix > O.
Solute-solvent interactions are weaker than solute-solute interactions.
See F ig. 8.40a for an illustration of the vapor-pressure behavior of a mixture of ethanol and benzene. To make an ethanol-benzene solution, strong hydrogen bonds in ethanol are broken and replaced by weaker ethanol-benzene London interactions. Weaker interactions increase the vapor pressure and decrease the boi l ing temperature. Fonns if the mini mum boi ling temperature is less than that of each pure liquid Distillation yields azeotrope as the distillate (see Fig. 8.4 1 in text).
Negative deviation from Raoult ' s law �
Vapor pressure of the mixture is smaller than the value predicted by Raoult's law.
�
Solute-solvent interactions are stronger than solute-solute interactions
� �
•
for each component
M inimum-boiling azeotrope �
•
xv"por
Positive deviation from Raoult's law � Vapor pressure of the mixture is
•
=
Enthalpy of mixing is exothermic, /'..H,nix < O. See F ig. 8.40b for an i l l ustration of the vapor-pressure behavior of a mixture of acetone and chloroform. In the acetone-chloroform solution, strong interactions, similar to hydrogen bonding, occur between a lone pair of electrons on the 0 atom of acetone and the H atom of chloroform. Stronger sol ute-solvent interactions decrease the vapor pressure and increase the boiling temperature.
Maximum-boiling azeotrope
�
�
Fonns if the maximum boiling temperature is greater than each of the pure liquids Distillation yields one pure component as the distillate (see F ig. 8.42 in text).
SG- I 02
IMPACT ON BIOLOGY AND MATERIALS (Sections 8.21 -8.22) 8.2 1 Colloids •
Colloids
� �
� �
� �
�
•
�
Particles (dispersed phase) with lengths or diameters between I nm and I j.1m dispersed or suspended in a dispersion medium (gas, liquid, or sol id solvent)
Classification of colloids is given in Table 8.9 in the text. Aerosols are sol ids dispersed in a gas (smoke) or liquids dispersed in a gas (hairspray, mist, fog).
Sols or gels are sol ids dispersed in a l iquid (printing ink, paint). Emulsions are l iquids dispersed in a liquid (mi lk, mayonnaise). Solid emulsions are l iquids dispersed in a solid (ice cream). Foams are gases dispersed in a liquid (soapsuds). Solidfoams are gases dispersed in a solid (Styrofoam).
Solid dispersions are sol ids dispersed in a solid (ruby glass, some al loys).
Aq ueous colloids �
Hydrophilic colloids contain molecules with polar groups that are strongly attracted to water.
�
Hydrophobic colloids contain molecules with nonpolar groups that are only weakly attracted to water. Examples incl ude fats that form emulsions (mi lk and mayonnaise).
�
Examples incl ude proteins that form gels and pUddi ngs.
Rapid mixing of silver nitrate and sodium bromide may produce a hydrophobic colloidal suspension rather than a precipitate of si lver bromide. The tiny silver bromide particles are kept from further aggregation by Brownian motion, the motion of small particles resulting from constant collisions with solvent molecules. The sol is further stabil ized by adsorption of ions on the surfaces of the particles. The adsorbed ions are hydrated by surrounding water molecules and help prevent further aggregation.
8.22 Bio-based Materials and Biom imetic Materials •
Bio-based materials
� Materials taken from or made from natural materials in l iving things �
�
•
Packing pel lets made from corn and soybeans, polylactic acid (a polymer used to make plastic packaging), and various kinds of pharmaceuticals Example mentioned in the text is hyaluronic acid (Margin Fig. 5). It is a major component of the fl uid that l ubricates j oi nts and plays a role in the repair of tissues especially the skin. With its many -OH groups, hyaluronic acid forms hydrogen bonds with water, and as a result, as it moves through the body, large numbers of water molecules move with it. This material is used in sport medicine to heal injuries by reducing inflammation.
Biomimetic materials
�
Materials that are modeled after natural ly occurring materials Gels or fl exible polymers modeled after natural membranes and tissues � Surfactant compounds cal led pospholipids are found in fats and form the membranes of living cel ls. Membranes of l i ving cel ls are double layers of phospholipid molecules that I ine up with their hydrocarbon tails pointing into the membrane and their polar head groups forming the membrane surface. Liposomes are artificial membranes shaped like tiny bags, which can be used to encapsulate drug molecules for delivery to different organs and regions of the body. �
SG - J 03
Chapter 9 CHEMICAL EQUILIBRIA REACTIONS AT EQUILIBRIUM (Sections 9.1-9.5) 9.1 The Reversibility of Reactions •
Chemical reaction � Reaction mixture approaches a state of dynamic equi librium. �
At equilibrium, the rates of the forward and reverse reactions are equal. H 2 (g) + 12 (g)
E xample: Forward reaction:
Reverse reaction:
•
2 H I (g)
�
�
2 HI (g)
H 2 (g) + h (g)
Dynamic equil ibrium: H 2 (g) + h (g) = 2 H I (g) o r 2 H I (g) = H 2 (g) + 12 (g) At chemical equil ibrium, the reaction may be represented either way. Note: The approach to equil ibrium may be too slow to measure. S uch a system may appear to be at equilibrium, but is not. Example: The sugar glucose, exposed to the air, is not in equi librium with its combustion products, CO2 and H 20.
9.2 Equilibri u m and the Law of Mass Action •
Law of mass action � The composition of a reaction mixture can be expressed in tenns of an equilibrium constant K, which is unitless. � The equi librium constant K has the fonn:
K •
=
{
activities of products activ iti es of reactants
}
A ctivity of substance J, aJ
equili brium
� Partial pressure or concentration of a substance relative to its standard value � Pure number that is unitless � I deal ized systems
aJ
p,J P, J
0
for an i deal gas
- _ -
partial pressure of the gas (bar) divided by the pressure of the gas in its standard state ( I bar) GJ
=
�
for a solute in a dilute solution
[J]O
molarity of substance J divided its molarity in the standard state ( 1 mol· L-1) aJ
=
I for a pure solid o r liquid
� Because the standard state values of activity are unity, they are omitted in the expression for activity
and numerical ly. aJ
=
;0 ; J =
=
11
or
I
aJ
=
l
11 for an ideal gas
SG- \ 05
I
Simi larly, aJ
•
I aj = 1
=
[J ] for a solute in a dilute solution
for a pure solid or liquid
I
and
A ctivity coefficient, YJ �
Pure number that is unitless
�
Accounts for intermolecular interactions
�
�
Equal to 1 for ideal gases and solutes in dilute sol utions Used for concentrated solutions and gases that do not behave ideally aj = yjPj
for a real gas
aj = yj[J] for any concentration of solute in a solution aj = I •
for a pure solid or liquid
Form of the equilibrium constant �
For a general reaction: a A + b B - c C + d D law of mass action, K is unitless
� �
�
K is the ratio of the activities of the products raised to powers equal to their stoichiometric coeffi cient to the corresponding expression for the reactants. Use activities given in the boxed equations above to write expressions for K. Distinguish between homogeneous and heterogeneous equil ibria.
•
Homogeneous equilibria
�
•
Heterogeneous equilibria
�
•
Note: For reactions in solution, expressions for equil ibrium constants should be obtained by using
Reactants and products al l in the same phase Reactants and products with different phases Activities of pure solids and liquids are set equal to l .
net ionic equations. 9.3 The Therm odynamic Origin of Eq u ilibri u m Constan ts •
t.Gr as a function of the composition of the reaction mixture �
�
Gm(J) is the molarfree energy of the reactant or product J .
Gm 0(J) i s the molarfree energy of the reactant o r product J in the standard state.
�
I Gm (J)=GmO (J) + RT ln aJ I for any substance in any state I Gm (J)=Gm° (J) + RT l n ?, I for ideal gases (from Section 8.3) I Gm (J)=GmO (J) + R T ln [J] I for sol utes in dilute solution I Gm (J)=Gm° (J) I for pure solids and liquids (a I)
�
In dilute aqueous solutions, the activity of water may be approximated as a(H20) = I.
�
�
�
=
SG- I 06
•
6Gm
for changes in pressure or concentration in idealized systems
� Standard state of a substance is its pure form at a pressure of I bar. � For a solute, the standard state is for a concentration of I mol· L- I . •
for chemical reactions (see Section 7. 13)
6 Gr
� Generalized chemical reaction :
�
aA + b B
-
cC + d O
the standard reaction free energy, i s the difference in molar free energies o f the products and reactants in their standard states.
6Gro,
� The stoichiometric coefficients,
n, are used as pure numbers (molar convention) to obtain l in klmor . The same convention is used for 6Gr below.
6Gro
� Standard free energies of formation, 6Gt, for a variety of substances are given in Appendix 2A. �
1
�
6Gr
�
•
�
6Gr
=
InGm (products) - InGm (reactants)
I
is the reaction free energy at any definite, fixed composition of the reaction mixture.
1 6Gr
Evaluated by using
=
6Gr ° +
RT In Q
I
Q is cal led the reaction q uotient.
Expressions for Q for a general reaction: a A + b B
-
cC + dD
> Real systems: � For ideal gases or dilute solutions, Q reduces to :
Q
> I deal gases:
= Pccp; o b pA pB
> Dilute sol utions: •
Q is unitless.
Q is unitless.
Chemical Equilibrium � At equilibrium, 6Gr = 0 and Q = K (equilibrium constant)
� K has the same form as Q, but it is determined uniquely by the equilibrium composition of
the reaction system.
� Real systems:
K
=
[ C d)
aC aD aAo aBb equili. .br.ium
law of mass action, K is unitless.
SG- \ 07
•
�
A ctivities can be expressed in concentration units multiplied by an activity coefficient.
�
For a gas,
aJ
=
Y
J � yJPJ
po
=
numerically. For a solute,
I 6.Gr 0
Evaluation of K from �Gr 0 and vice versa �
=
9.4 The Exten t of Reactio n •
Thermodynamically, !).Gr 0
•
0
-
T t0r 0
=
-
In K
RT I n K ;
=
_
Wr 0 RT
+
=
YJ
[J ]
[ J]O
=
[ J] numerically .
RT In K
!).Sr 0
Solve for K
R
I n a chemical reaction, products are favored if
�
�
•
= Wr
-
aJ
�
K is large. The reaction is exothermic in the standard state (negative !1Hr o , the more negative the better). The reaction has a positive standard entropy of reaction ( � Sr0
>
0 , final state more random).
G u idelines for attaining equilibrium
�
�
�
3 If K > 1 0 , products are favored. 3 3 I f 1 0- < K < 1 0 , neither reactants nor products are strongly favored. 3 If K < 1 0- , reactants are favored.
Example: At a certain temperature, suppose that the equil ibrium concentrations of nitrogen
and ammonia in the reaction
N 2 (g) + 3 H 2 (g) = 2 NH 3 (g)
were found to be 0. 1 I M and 1 . 5 M, respectively. What is the equilibrium concentration of H 2 if Kc 4.4 x 1 04 ? =
The value of Kc is in the range that favors products. Solving the equilibrium constant expression requires the evaluation of the cube root of the concentration of H 2 . Use the y' key on your calculator. Kc
=
(
[ NH3
P"
[ N 2 ] [ H 2 l'
l
eq
and [ H 2 ]3 =
[ N H3f
Kd N 2 ]
Note that the product ammonia is favored as expected. 9.5 The D i rection of Reaction •
For the general reaction a A (g) + b B (g)
and
Q
=
c C (g) + d O (g)
[ dJ C
=
aC a D
aA Oab B
not at
SG- I 0 8
. equil. .ibrium
•
G u idelines for approach to equilibrium � Compare Q with K.
K, the amount of products is too high or the amount of reactants is too low.
� If Q> � �
Reaction proceeds in the reverse direction, toward reactants ( - ).
I f Q < K, the amount of reactants is too high or the amount of products is too low. Reaction proceeds in the forward direction to form products ( - ).
= K, reactants and products are at equi librium and no observable change occurs ( = )
If Q
.
EQUILIBRIUM CALCULATIONS (Sections 9.6-9.8) 9.6 The E q uilibri u m Constant in Terms of Molar Con centrations of Gases •
Gas-phase eq uilibria (ideal gas reaction mixture) � a
A (g) + b B (g) = c C (g) + d O (g)
[ )
CPt K = PpCap, b
ideal gases (activity
A B equili . . brium .
K
c
�
� � � •
=
(general gas phase reaction at equi librium)
[ [[CtAt[D]d)
= partial pressure)
ideal gases (activity
[B] b equil ibrium
=
molar concentration)
K and Kc are unitless; only values of K are reported in Appendix 2A. n n J RT RT J RT [J] . I n general, � = -
Both
v-
Relationship between 6.n = (c + d)
=
V
=
K and Kc as derived in the text: I K = ( RT ) t.n Kc I
- (a + b )
=
I(coefficients of products) - L(coefficients of reactants)
Mixed-phase eq uilibria � a
A (g) + b B (aq) = c C (g) + d D (aq)
[
�
(some gases, some solution species at equil ibrium)
]
K = PCC [ D] bd P: [ B] equilibrium /).ng = c - a I(coefficients of gaseous products) - I(coefficients of gaseous reactants) =
� Note: 6.ng can be positive, negative, or zero.
SG- I 09
9.7 Alternative Forms of the Eq uilibri u m Constant •
Dependence of K on the di rection of reaction a nd the balancing coefficients �
a A (g) + b 8 (g) = c C (g) + d D (g)
�
c C (g) + d D (g) = a A (g) + b 8 (g)
KI
ac a o
aAa a Bb
. . rium . eqUili b _ -
� na A (g) + nb 8 (g) = nc C (g) + nd D (g) •
- ( C d]
�
K3 =
( aS:�:] aA a B
-
1 - K -I I KI
= Kln equil ibrium
Combining chemical reactions
�
Adding reaction 1 with equi librium constant KI to reaction 2 with equilibrium constant K2 yields a combined reaction with equil ibrium constant K that is the product of KI and K2 . (l\G 1 o + l\G20) leads to K = KI
X
K2
� Subtracting reaction 2 with i ndividual equilibrium constant K2 from reaction 1 with equi librium
constant KI yields a combined reaction with equi librium constant K that is the quotient of KI and K2 ••
•
KI (l\G 1 o - l\G20) leads to K = _ K2 Summary : Add two reactions ( 1 + 2), multiply equi libri um constants (KI x K2)' Subtract two reactions ( 1 2), divide equi librium constants (KI / K2)'
9.8 Using Equilibri u m Constants �
-
•
Equilibrium table
•
Format and use of the eq uilibrium table: Step
I
Reactant and product species taking part in the reaction are identified.
Step 2
The initial composition of the system is listed.
Step 3
Changes needed to reach equil ibrium are listed in terms of one unknown quantity. The equi librium composition in terms of one unknown is entered on the last line.
Step 4
•
Very helpful in solving all types of equilibrium problems
Step 5
The equilibrium composition is entered into the equil ibrium constant expression and the unknown quantity (equil ibrium constant or reaction species concentration) is determined.
Step 6
Approximations may sometimes be used to simpl ify the calculation of the unknown quantity in step 5.
Calculating equilibrium constants from pressures or concentrations Example: A sample of H I (g) is placed in a flask at a pressure of 0. 1 00 bar. After equilibrium is attained, the partial pressure of HI (g) is 0.050 bar. Eval uate K for the reaction
2 H I (g) = H2 (g) + 12 (g) (a) Set up an equil ibrium table. In the initial system, line I , there are no products so some must form to reach equilibrium.
SG - I I O
(b) On l ine 2, let x be the change in pressure of 1 2 required to reach equi librium. From the stoichiometric coefficients, the change in the pressure of H 2 is x whi le the change in the pressure of H I is -2x ( H I disappears twice as fast as iodine appears). (c) Lines I and 2 of the table are added to give the equil ibrium pressures of the reactants and products in terms of x. (d) The value of x is determined from the change i n pressure of H I (-2x) from the initial to the equilibrium value. (e) Using the calculated value of x, the equilibrium pressures from line 3 of the equilibrium table are calculated and entered into the equilibrium constant expression to obtain K. Species HI
0.100 -2x 3 . Equil ibrium pressure 0.100 - 2x
I. Initial pressure
2. Change in pressure
and
PHI = 0.100 - 2x = 0.050
K •
=
�
PH 1 l l PHI
=
x
=
�
(0.025) (0.050 )
H2
12
0 +x x
0 +x x
0.025 = RH ,
=
=
RI ,
0.25
Calculating the eq uilibrium composition Example: A sample of ethane, C 2 H 6 (g), is placed in a flask at a pressure of 2.00 bar at 900 K.
The equil ibrium constant K for the gas phase dehydrogenation of ethane to form ethylene C 2 H 6 (g) = C 2 H 4 (g) + H 2 (g)
is 0.050 at 900 K. Determine the composition of the equilibrium mixture and the percent decomposition of C 2 H 6• (a) Set up an equilibrium table and let x be the unknown equil ibrium pressure of each product, C 2 H 4 and Hz . The equil ibrium pressure of C 2 H 6 is then 2.00 - x. (b) Solve for x by rearranging the equi librium constant expression . (c) The new form is a quadratic equation, which is solved exactly by using the quadratic formula. (d) The equil ibrium pressures can then be obtained . (e) The percent decomposition is the loss of reactant divided by the initial pressure times 100%, or
(_x_)(IOO)%. 2.00
C2 H6
2.00 -x 3 . Equil ibrium pressure 2.00 - x
I. I nitial pressure
2. Change in pressure
SG - l l l
Species C2H4
H2
0 +x x
0 +x x
K=
pC,H" PH,
(x) 2
PC,H"
(2.00 - x)
= 0.050
Rearranging yields a quadratic equation: x2 + (O,050)x - 0. 1 0 = 0 Recall: ax
2
+
bx + c = 0
=>
x=
�
-b ± b2 - 4ac
(exact solution)
2a
Then, a = 1 , b = 0.050, and c = -0. 1 0. x= =
-(0.050) ± (0.050) 2 - 4(1 )(-0. 1 0)
�
2(1 )
-(0.050) ± (0.6 3 44) 2
-(0.050) ± JOA025 2
=
-(0.050) + (0.63 44)
=
2
0. 5844 2
=
0.29
The negative root is rejected because it is unphysical . PC,H , = PH , = X = 0.29 bar and PC,H " = 2.00 - x = 1 .7 1 bar 0.29 . . change in pressure x l 00% = x 1 00% Percent decomposition = 2.00 initial pressure --
=
1 4 .5%
3
Note: If K is less than 1 0- in a problem ofthis type, the adj ustment to equil ibri um is small.
The pressures of reactants then change by only a smal l amount. I n such a case, an approximate solution that avoids the quadratic equation is possible.
THE RESPONSE OF EQUILIBRIA TO CHANGES IN CONDITIONS (Sections 9. 9-9.13) 9.9 Add ing and Removing Reagents •
Le Chatelier's principle �
•
Qualitative features
When a stress is appl ied to a system in dynamic equilibri um, the equilibrium tends to adjust to minimize the effect of the stress.
� Use Le Chatel ier's principle to predict the qual itative adj ustments needed to obtain a new
equilibrium composition that rel ieves the stress.
� Assessing the effects of the applied stress on Q relative to K is an alternative approach.
Example: Use Le Chatelier's principle and reaction quotient arguments to describe the effect on
the equil ibrium
Q=
[
Ps 6,
Ps 6, Po,
1
of (a) addition of S02 , (b) removal of O2 , (c) addition of S03 , and (d) removal of S03 • Both reactant and product pressures appear in the equil ibrium constant expression, and in Q. Assuming no change in volume or temperature, adding reactant (or removing product) shifts the equil ibrium composition to produce more product molecules, thereby helping to relieve the stress . Adding product (or removing reactant) shifts the
SG - 1 1 2
equilibrium composition to produce more reactant molecules, thereby helping to relieve the stress. (a) Adding reactant S02 results in Q < K. The system adj usts by producing more product S03 (consumi ng S02 and O2 ) untiI Q = K ( - ). (b) Removing reactant O2 results in Q > K. The system adj usts by consuming product S03 (producing more S02 and O2 ) until Q = K ( - ). (c) Adding S03 results in Q > K. The system adjusts by producing more S0 2 and O2 (consuming S03 ) unti l Q = K (- ). (d) Removing S03 results in Q < K. The system adj usts by consuming S02 and O 2 (producing more S03 ) until Q = K (- ). Note: I n this example, the stress changes only the value of Q. The composition adjusts to
return the value of Q to the original value of K. The equi librium constant does not change. The equilibrium constant has, however, a very weak pressure dependence that is usually safe to ignore.
•
Quantitative features
�
Use the equi librium table method to calculate the new composition of an equil ibrium system after a stress is appl ied.
Example: Assume that the reaction 2 H I (g) = H 2 (g) + 1 2 (g) has the equi libri um composition PH = 0.050 bar, PH , = PI , = 0.025 bar and K = 0.25. I
I f HI, H 2 , and 1 2 are added to the system in amounts equivalent to 0.008 bar for H I , 0 .008 bar for H 2 , and 0.008 bar for 1 2 , predict the direction o f change and calculate the new equil ibrium compos ition. After adding 0.008 bar to each reactant and product, the nonequlibrium pressures are PH = 0.058 bar, PH , = 0.033 bar, and PI , = 0.033 bar. I Q=
(0.033)(0.033)
(0 . 058) 2
=
0.324 > K and the adjustment to the new equi librium is in the
direction toward the reactants. Quantitative treatment: Set up an equilibrium table, letting -x be the unknown adjustment to the initial nonequifibrium pressures of each product H 2 and 1 2 . The value of +2"( is then the adj ustment to the initial nonequilibrium pressure of H I . A negative value of x i s expected from our qual itative analysis, which suggests that reactants should form from products to relieve the stress on the system. Species HI 0.05 8 l . I nitial pressure +2x 2. Change in pressure 3. Equilibrium pressure 0.058 + 2x P P
K=� 2 = PH I
b
H2 0.033 -x 0.033 - x
(0 . 0.).... .....) - x ) 2
(0.058 + 2x)
2
0.033 -x 0.033 - x = 0 ·2 5
This expression is relatively easy to solve for x by taking the square-root of both sides:
JK
=
(0.0 33 - x)
(0.058 + 2x)
SG- J 1 3
=
';0.25
=
0.50
0.033 PH ,
=
-
x = 0.029 + x; 2x = 0.004; x = 0.002
P I, = 0.033 - 0.002 = 0.03 1 bar and P ] = 0.058 H
Equi l ibrium expression check: K
=
(0 . 0"J 1 )2 (0.062) 2
=
+
0.004 = 0.062 bar
0.25 as expected
9 . 1 0 Com p ressing a Reaction Mixture •
Decrease in volume of a reaction mixture at constant temperature � The partial pressure of each component J of the reaction mixture increases;
nJ RT (see Section 4. 8). V � To minimize the increase in pressure, the equilibrium shifts to form fewer gas molecules, if possible. recall �
•
=
Increase in volume of a reaction mixture at constant temperature
nJ RT . V � To compensate for the decrease in pressure, the equilibrium shifts to form more gas molecules, if possible.
� The partial pressure of each member J of the reaction mixture decreases; recal l
•
�
=
Increase in pressure of a reaction mixture by introducing an inert gas at constant T and V, as in a steel vessel. � The partial pressure of each member J of the reaction mixture does not change;
nJ RT V � The equilibrium composition is unaffected. The equilibrium constant has a weak pressure dependence that is safe to ignore unless the pressure change is large. recall �
•
= --
I ncrease i n volume of a reaction mixture by introducing an inert gas at constant T and total p ressure P, as in a piston.
� I ncreases the volume of the system without changing the amount of reactants or products �
The partial pressure of each member J of the reaction mixture decreases; recall �
nJ RT
. V � The return to equilibrium is equivalent to that which occurs by increasing the volume or decreasing the pressure of the system . The equilibrium shifts to form more gas molecules, i f possible. =
9 . 1 1 Temperat u re and Equ ilibrium •
Qualitative features �
•
For endothermic reactions, heat is treated as a reactant.
Use Le Chatel ier's principle to predict the qual itative adj ustments needed to obtain a new equilibrium composition that rel ieves the stress .
� Increasing the temperature supplies heat to the reaction mixture and the composition of the mixture
adjusts to form more product molecules.
� The value of the equil ibrium constant, K, increases.
� Decreasing the temperature removes heat from the reaction mixture and the composition of the
mixture adjusts to form more reactant molecules. � The value of the equil ibrium constant, K, decreases.
SG - 1 1 4
•
For exothermic reactions, heat is treated as a product. � �
Increasing the temperature suppl ies heat to the reaction mixture and the composition of the mixture adjusts to form more reactant molecules.
The value of the eq uilibrium constant, K, decreases. � Decreasing the temperature removes heat from the reaction mixture and the composition of the mixture adjusts to form more product molecules. � The value of the equilibrium constant, K, increases. •
Quantitative features � van 't Hoff equation
(see derivation in the text)
� Gives the qual itative results above (The equilibrium constant is K, not K, .)
� A maj or assumption in the derivation i s that both f1Hro and f1Sro are independent of temperature between Tl and T2 •
9.12 Catalysts and Haber's Achievement •
Catalyst increases the rate at which a reaction approaches equilibrium. � Rate of forward reaction is i ncreased. � Rate of reverse reaction is i ncreased. � Dynamic equil ibri um is unaffected
�
The equilibrium constant is not changed.
� The identity of the catalyst is unchanged.
•
� The catalyst is neither a reactant nor a product. It does not appear i n the chemical equation.
Haber's achievement � Production of ammonia: 3 H 2 (g) + N 2 (g) -= 2NH 3 (g) At 298 . 1 5 K, f1Hro = -92.22 kJ . mor 1 , f1Gro = -32 .90 kJ . mor 1 , and K = 5.80 x 1 0 5
� The reaction is very slow at room temperature and pressure. � Compression of the gas mixture favors product ammonia.
�
Removal of ammonia as it is formed encourages more to be formed.
� Increasi ng the temperature is required to increase the slow rate of the reaction at 298 . 1 5 K, but it
adversely affects the position of the equil ibrium.
� To circumvent the problem, an appropriate catalyst (a mixture of Fe203 and Fe3 04) was discovered
at about the time of World War I . It is still used today.
9. 1 3 The Im pact on Biology : Hom eostasis •
H omeostasis � Mechanism simil ar to chemical equil ibrium, governed by Le Chatel ier's principle � Helps l iving organisms to maintain constant internal conditions
� Allows living organisms to control biological processes at a constant level
� An example is the equi l ibrium involving oxygen and hemoglobin, Hb.
Hb(aq) + 02(aq) -= Hb02 (aq) [See text discussion of the ways in which hemoglobin and myoglobin provide needed oxygen to our bodies under varying physiological conditions.]
SG- 1 1 5
Chapter
10
ACIDS AND BASES
THE NATURE OF ACIDS AND BASES (Sections 1 0.1 -1 0.6) 1 0 . 1 Bnmsted-Lowry Acids a n d Bases •
B ronsted-Lowry definitions � A cid: proton donor; base: proton acceptor �
General theory for any solvent (for example, H 20 or N H 3) or no solvent at all � Acid (proton donor) is deprotonated when reacting with a base. � Base (proton acceptor) is protonated when reacting with an acid. � Acids and bases react with each other i n proton transfer reactions. �
A strong acid is completely deprotonated.
E xample: H Br(aq) + H 2 0(l) - H 30+(aq) + Br-(aq) (strong acid: reaction goes essenti ally to
completion; proton is transferred from H Br(aq) to H 2 0(l)).
� A strong base is completely protonated.
E xample: N H 2-(aq) + H 2 0(l) - N H3 (aq) + OW(aq) (strong base: reaction goes essentially to
�
completion; N H 2-(aq) accepts a proton from H 20).
A weak acid is partially deprotonated. =
E xample: H F(aq) + H 2 0(l)
is established).
H 3 0+(aq) + F-(aq) (weak acid reaction, a chemical equilibrium
� A weak base is partially protonated.
= NH/(aq) + OH-(aq) (weak base reaction , a chemical equilibrium is established).
E xample: N H 3(aq) + H 2 0(l) •
Conjugate base � Formed from acid after the proton is donated.
deprotonation => conjugate base Examples: The conj ugate base of HBr is Br-; the conjugate base of H 3 P04 is H 2 P04-.
� Acid
•
=>
Conj ugate acid � Formed from base after the proton is accepted. �
•
•
Base
=>
protonation
=>
conj ugate acid
Examples: The conj ugate acid ofNH 3 is N H/; the conj ugate acid of CH 3 COO- is CH3 COOH.
Conjugate acid-base pair
� Differ only by the presence of an H+ species:
H CN, CN- ; H 3 0+, H 2 0 ; H 2 0, OW ; H 2 S, H S-
Solvent � Need not be water. Other solvents incl ude HF(I), H 2 S04(1), CH 3 COOH(I), C2HsOH(I), N H3(1). �
Acid-base reactions can occur with no solvent as in: HCI(g) + N H3 (g) - N H4CI(s).
1 0.2 Lewis Acids and Bases •
Lewis definitions � A Lewis acid is an electron pair acceptor.
� A Lewis base is an electron pair donor. �
2
Examples: BF3 (g), Fe +(aq), H 2 0(l)
2
Examples: N H 3 (aq), 0 -(aq)
Most general theory (holds for any solvent and includes metal cations)
SG- 1 1 7
•
Comparison of Lewis and B rensted-Lowry Acids and Bases
�
�
Proton (rt) transfer (Bmnsted-Lowry theory) is a special type of Lewis acid-base reaction.
In a proton transfer reaction, H+ acts as a Lewis acid, accepting an electron pair from a Lewis base.
� Bmnsted acid: supplier of one particular Lewis acid, a proton
�
Bmnsted base: a particular type of Lewis base that can use a lone pair to bond a proton 2 E xample: The reaction 0 -(aq) + H 20(l) 2 OH -(aq) can be viewed as a Bmnsted-Lowry 2 reaction (H2 0 is the proton donor, 0 - is the proton acceptor) or as a Lewis acid-base 2 reaction (0 - is the electron pair donor (Lewis base), H+ on water is the electron pair acceptor (Lewis acid), as shown in the diagram below. �
•• 2. 0 ' /"\ H • • + H '-... / •• .0. •
•
�
. 2 (: 0--. . -- H )
1 0.3 Acid ic, Basic, and A m p hoteric Oxides • •
Oxides � React with water as Lewis acids or bases, or both Oxides of nonmetals � Called acidic oxides
� Tend to act as Lewis acids E xamples: CO2 , S03 , C h07 , N 2 0S � React with water to form a Bmnsted acid E xamples: CO2(g) + H 2 0(l)
H 2 C03(aq) H 2 S04(aq)
S03(g) + H 2 0(l)
�
Br@nsted acid
Lewis acid
React with bases to form a salt and water E xample: H 2C03(aq) + 2 NaOH(aq) Na2 C03(aq) + 2 H20(l) �
•
Oxides of metals � Called basic oxides
� Tend to act as Lewis bases E xamples: Na2 0, MgO, CaO
� React with water to form Bmnsted bases Example: CaO(s) + H 20(l) Lewis base
•
�
� React with acids to form a salt and water E xample: CaO(s) + 2 HBr(aq)
Amphoteric oxides �
Ca(OHMaq)
Br@nsted base
�
CaBr2(aq) + H 2 0(l)
Exhibit both acid and base character Examples: Ah03 , S n02
� React with both acids and bases
Example: See text for reactions of Ah03 as an acid and as a base.
•
Trends in acidity for the oxides of main-group elements
Increasing acidity
�
1
2
1 3/111
1 4/IV
l S/v
1 6/vI
1 7/vII
LhO
BeO
B203
CO2
N20s
(02)
OF2
Na2 0
MgO
Ah03
Si02
S03
Ch07
K20
CaO
G a203
Ge02
P4 O J O
As20s
Se03
Br207
Rb2 0
SrO
In203
S002
Sb20 s
Te03
h07
Cs20
BaO
TI 203
Pb02
B h Os
Po03
At2 07
SG - 1 1 8
Note:
Acidity of oxides in water tends to increase as shown in the preceding table. Amphoteric oxides are in bold type, basic oxides are to their left and acidic oxides to their right. Exceptions are O2 , which is neutral, and OF2 , which is only weakly acidic.
1 0.4 Proton Exchange Between Water Molecules
•
� A molecule or ion that can act as either a Br0nsted acid (H+ donor)
Amphiprotic species
or a Br0nsted base (H+ acceptor).
Examples: H20: H+ donor ( - OW) or H+ acceptor ( - H30+) •
NH 3 : H+ donor ( - NHn or W acceptor ( - NH/)
Autoprotolysis � A reaction in which one molecule transfers a proton to another molecule of
the same kind
Examples: Autoprotolysis of water: 2 H20(I)
=
H30\aq) + OH-(aq) =
Autoprotolysis of ammonia: 2 N H3(l) •
Equilibrium constant
2 H 2 0(l)
=
� Autoprotolysis of water:
NH/(am) + N H2-(am)
Kw = [H30+] [OH -] = I
H30+(aq) + OW(aq);
x
1 0- 1 4 at 25°C
1 0-7 mol· L - I
•
Pure water at 25°C
� [H30+] = [OW] = 1 .0
•
Neutral solution at 25°C
� [H30+] = [OH-] = 1 . 0 x 1 0-7 mol· L - 1
•
Acidic solution 25°C
� [H30+] > 1 .0
X
1 0-7 mol·L- 1 and [OH-] < 1 .0
x
1 0-7 moIL- I
•
Basic solution 25°C
� [H30+] < 1 .0
X
1 0-7 mol·L- 1 and [OH-] > 1 .0
x
1 0-7 mol · L- 1
1 0.5 The pH Scale •
Concentration variation of H30+
•
Logarithmic scale
� pH
�
x
4 Values higher than 1 moIL- I and l ess than 1 0- 1 mol· L- 1 are not commonly encountered.
-log [H30+] and [H30+] = 1 0-pH pH values range between I and 1 4 in most cases. =
� Pure water at 25°C: � Acidic solution:
Note: •
=>
pH = 7.00; pH < 7.00;
=>
Neutral solution: pH = 7.00 Basic solution: pH > 7.00
pH has no units, but [H30+] has units of molarity.
Measurement of p H � Universal indicator paper turns different colors at different p H values.
� pH meter measures a difference in electrical potential (proportional to pH) across electrodes that dip into the solution (see Chapter 1 2).
Examples: The pH of the sol utions for which the [H30+] = 5 .4
pH = -log [H30+]
=
-
( 3 27 ) -
.
=
X
1 0-4 mol· L- 1 is:
3 .2 7 (acidic solution)
The [H 3 0+] for a solution that has a pH value of 1 .70 is: [H30+] = 1 0-pH = 1 0-170 = 2.0 X 1 0 -2 mol · L- I . 1 0.6 T h e p O H o f Solutions •
pX as a generalization of pH � pX
=
-log X (X = anything)
SG- 1 1 9
� pOH
•
=
-log [OW] and [OW]
� pKw = -log Kv
=
-log ( 1 .0
x
=
1 0-pOH 14 1 0- ) 1 4.00 at 25°C =
Relationship between pH and pOH of a solution � [H 30+] [OW]
=
Kw = 1 .0
� log ([H 3 0+][OWD
�
=
+
1 0- 1 4 at 25°C
log [ H3 0+]
-log [ H30+] - log [OW]
� pH
x
=
+
log [OW]
=
log Kw
-log Kw
pOH = 1 4.00 at 25°C
WEAK ACIDS AND BASES (Sections 10.7-10.10) 1 0.7 Acidity and Basicity Constants •
Weak acid �
•
General weak acid H A
pH is larger than that of a strong acid with the same molarity. I ncomplete deprotonation
� HA(aq) + H 20(l) ,.: H 30+(aq)
But, a H o , � Thus, Ka
=
=
I ,
+
a H o+ aA,- A-(aq); Ka = ---,'a HA a H ,O
a HA "" [ HA], a H o+ "" [ H 30+], and aA - ,," [ A - ] . ,
[ H 3O+ ] [ A - ] [ HA]
Ka is the acidity constant; pKa = -log Ka is commonly used.
Some Weak Acids and Their Acidity Constants Species Neutral acids
x
9.3 1
x
CH 3COOH
1 .8
4.75
Hydrocyanic acid
HCN
4.9
x
1 0- 1 0
Pyridinium
C 5 H 5+
5.6
x
1 0-
5 .24
NH/
5.6
x
1 0- 1 0
9.25
Ethylammonium
C2H5N H 3+
1 .5
x
1 0.82
Hydrogen sulfate
HS04H 2 P04 -
1 .2
x
10 1 1 0-2
6.2
x
1 0-8
7.2 1
Dihydrogen phosphate •
3 .45
3.5
Ammonium Anion acids
pK.
1 0-4 1 0- 5
HF
Hydrofluoric acid Acetic acid
Cation acids
K.
Formula
I n the preced ing table � Hydrogen sulfate is the strongest acid (l argest Ko, smallest pKo). � Ethylammonium is the weakest acid (smallest Ka, largest pKo).
SG- 1 20
6
- 1
1 .92
•
Weak base �
•
General weak base B
p H is smaller than that of a strong base with the same molarity. I ncomplete protonation
� B(aq) + H 2 0(l)
� Thus, Kb
=
=
B H+(aq) + OH-(aq); Kb
[ B H + ] [O H - ] [ B]
=
aBH + aOW a B a H,O
Kb i s the basicity conslant;
pKb = -log Kb is commonly used.
Weak Bases and Thei r Basicity Constants Species Neutral bases
Ethylamine
C2 HsNH 2
6.5
x
Methylamine
CH 3NH 2
3.6
x
NH 3
1 .8
x
NH 20H
1.1
Ammonia Hydroxylamine A nion bases
P04 3F-
Phosphate Fl uoride
•
•
I n the preced ing table
� Phosphate, P04 3- , i s the strongest base (largest Kb, smallest pKb).
�
F l uoride, F -, is the weakest base (smallest Kh, largest pKb).
Acid-base strength � The larger the value of K, the stronger the acid or base.
�
The smaller the value of pK, the stronger the acid or base.
1 0.8 The Conj u gate Seesaw •
Reciprocity of conjugate acid-base pairs
•
Conjugate acid-base pairs of weak acids � The conj ugate base of a weak aci d is a weak base. �
� •
The conj ugate acid of a weak base is a weak acid.
" Weak " is defined by pK values between I and 1 4.
Conj ugate seesaw (qualitative) � The stronger an acid, the weaker its conj ugate base.
•
�
The stronger a base, the weaker its conj ugate acid.
Conjugate seesaw (q uantitative) �
Kb
Formula
Consider a general weak acid HA and its conjugate base A-. [ H 3 0+ ][A - ] HA(aq) + H 20(l) = H 30+(aq) + A-(aq) Ka [ HA] _ _
SG - 1 2 1
1 0-4 1 0-4
pKb
3. 1 9 3 .44 4.75
x
I O-s
1 0-8
7.9 7
x
1 0 -2
1 .32
2.9 x 1 0- 1 1
1 0.54
4.8
A- ( aq) + H 20(l) = HA(aq) + OW(aq) Ka
x
K
b
=
[ H P+ ] [A - ] [ HA]
x
[ H A ] [OH- ] [A - ]
� Or, for any conjugate pair, IKa � I f K. > I
� If Kb > I
x x
x
Kb
=
=
Kb _
[ H A ] [OH - ] [A-]
[ H 3 O+ ] [OH - ]
=
Kw
Kw I and pKa + pKb = pKw.
1 0-7 (pK. < 7), then the acid is stronger than its conjugate base. 1 0-7 ( pKb < 7), then the base is stronger than its conjugate acid.
See Table 1 0.3 in the text. Example: Sulfurous acid, H 2 S03 , is a stronger acid than nitrous acid, HN02 . Consequently, nitrite, N02-, is a stronger base than hydrogen sulfite, HS03-. �
An acid is strong if it is a stronger proton donor than the conjugate acid of the solvent.
•
Solvent leveling
•
Strong acid in water �
Stronger proton donor than H 3 0+ (Ka > I ; pK. < 0)
� H 3 0+(aq) + H 2 0(l) = H 20(l) + H 30+(aq)
Ka
� A l l acids with Ka > I appear equally strong in water. �
=
1
Strong acids are leveled to the strength of the acid H30+.
Example: HCI, HBr, HI, HN03 , and HCI04 are strong acids in water. When dissolved in
water, they behave as though they were solutions of the acid H 3 0+.
1 0.9 •
Molecular Stru ctu re and Acid Strength
Binary acids, HnX � For elements in the same period, the more polar the H -A bond, the stronger the acid.
E xample: H F is a stronger acid than H 2 0, which is stronger than NH3 .
� For elements in the same group, the weaker the H-A bond, the stronger the acid.
Example: HBr is a stronger acid than HCI, which is stronger than H F .
•
Summary
� from left to right across a period (polarity of H-A bond increases), from top to bottom down a group ( H-A bond enthalpy decreases).
� For binary acids, acid strength increases
� The more polar or the weaker the H -A bond, the stronger the acid.
1 0 . 1 0 The Strengths of Oxoacids and Ca rboxylic Acids •
Oxoacids (also called oxyacids) � The greater the number of oxygen atoms attached to the same central atom, the stronger the
acid.
� This trend also corresponds to the increasing oxidation number of the same central atom.
Example: HCI03 is a stronger acid than HC102 , which is stronger than HCIO.
� For the same number of 0 atoms attached to the central atom, the greater the electronegativity of
the central atom, the stronger the acid. Example: HCIO is a stronger acid than HBrO, which is stronger than H IO. The halogen is considered the central atom for comparison to other oxoacids.
SG- 1 22
•
�
The greater the electronegativities of the groups attached to the carboxyl (COO H) group, the stronger the acid. Example: CH8r2 COOH is a stronger acid than CH 2 8rCOOH, which is stronger than C H 3 COOH .
Carboxylic acids
Note: The preceding rules are summarized in text Tables 1 0.4, 1 0.5, and 1 0.6.
THE pH OF SOLUTIONS OF WEAK ACIDS AND BASES (Sections 10.11-10.13) 1 0. 1 1 Solutions of Weak Acids �
I nitial concentration of an acid ([HA] i nitial) or base ( [ 8 ] i ni tial) as prepared, assuming no proton transfer has occurred .
•
I nitial concentration
•
General result for a dilute solution of the weak acid H A
Main equilibrium
HA(aq) +
Initial molarity Change in molarity Equilibrium molarity
[HA]initial -x [HA ] initial - X
H 2O(l)
=
H 3O+Caq) +
A-(aq)
0 +x
0 +x
X
X
. [ H 30 + ] [ A - ] ---x2 . . = A CId Ity constant expression: Ka = [ HA ] [HA ]init ial - X x2 + (Ka)x - (Ka[HA ] initial) = 0 and x > 0 (positive root)
Quadratic equation:
- Ka + 'Ka 2 + 4 Ka [ HA] · n . t. al I i i = [ H3 0+] = [ A-] ;::: 1 0 -6 M " 2 [ HA] = [ HA] ini li al - X [ O H -] = Kw / [ H30+] from the secondary equil ibrium x x 1 00% (2) Percentage deprotonated = [ HA ] i n i tial
Positive root:
•
x
=
(I)
Approximation to the general result
Assume [HA]inilial
-X�
[HA]initial , then Ka =
x2
[HA ] initial
and x =
Assumption is j ustified if percentage deprotonated < 5%.
�Ka [ HA]initial .
( 3)
Example: For acetic acid, C H 3 COOH, Ka = 1 . 8 x 1 0-5. The value of [ H 3 0+] in a 0 . 1 00 M
sol ution of C H 3COOH is given by the preceding equation labeled ( I ). With [HA] i nitial = 0. 1 , the result is x = [ H3 0+] = 1.33 x 1 0-3 M. Using the 5% approximation method in the equation labeled (3) yields x = [ H 3 0+] = x = Ka [ HA]i n i tial = [ ( 1 . 8 x 1 0-5)(0. 1 00)]1/2 = I .34 x 1 0-3 M .
�
The approximation is excellent in this case.
SG- 1 2 3
1 0. 1 2 Solutions of Weak Bases •
General result for a weak base B at dilute concentrations
Main equil ibrium
B(aq)
I nitial molarity Change in molarity Equil ibrium molarity
[B] initi.1 -x [B] initi.1 - X
Basicity constant expression:
Kb
x
=
[B] = [B] ;"iti.1 - X
+
BH+(aq) 0 +x
OW(aq) 0 +x
X
[ BH + ] [OH - ] [B]
' -Kb + V Kb2 + 4Kb [B]-mtllal --
2 [H30+] = Kw / [OW]
Percentage protonated = •
H 2 O(l) -
=
X
x2
[ B ]initial - x
x2 + (Kb)x - (Kb[B]initial ) = 0 and x > 0 (positive root)
Quadratic equation: Positive root:
=
+
x
[ B ]initial
x
= [BH+] = [OH-]
�
1 0-6 M
(4)
secondary equilibrium
1 00% (5)
Approximation to the general result
Assume [B]initi.1 - x ;::; [B] initial , then Kb
=
x2 [ B ]initial
and x
Assumption is justified if percentage protonated < 5%.
=
i
�Kb [ B] nitial
(6)
4 1 0- • In a 0.200 M solution of methylamine, the [OH-] is given by the equation labeled (4), with [B] initi.1 0.200 M . S ubstituting the values of Kb and [B]initial yields x = [OW] = [CH3NH/] = 8.3 1 x 1 0-3 M. Using the approximate solution in the equation labeled (6), x = [OH -] 3 = [CH3N H 2+] = 8 .49 x 1 0- M . The percent difference between the exact and approximate answers is less than 5%, so the approximation is j ustified.
Example: Kb for the weak base methyamine, CH3N H2 , is 3.6
X
=
1 0 . 1 3 The pH of Salt Solutions •
Acidic cations
The conjugate acids of weak bases E xamples: Ammonium, N H/; anilinium, C6HSNH3+ Certain small , highly charged metal cations Examples: I ron(1 I 1), Fe3+; copper(I I ), C u2+ Neutral cations � Group I and 2 cations; cations with + 1 charge E xamples: Li\ Mg2+, Ag+
� -->
• •
Basic anions
�
Conjugate bases of weak acids
2
Examples: Acetate, CH3COO-; fluoride, F -; sulfide, S -
•
Neutral a nions
•
Acidic anions
•
Acid salt
�
Conjugate bases of strong acids Examples: C l -, CI04-; N03-.
� Only a few such as HS04-, H 2P04-
� Typically, a salt with an acidic cation
E xamples: FeCb, N H4N03 (in these examples, the anions are neutral - see above)
� Dissolves in water to yield an acidic solution
Example: N H4N03(aq) - NH/(aq) + N03-(aq) (dissociation of salt in water)
NH/(aq)
+
H20(l)
=
H30+(aq)
+
N H3(aq) (acidic cation reacts with water)
SG- 1 2 4
•
Basic salt � A salt with a basic anion
Examples: Sodium acetate, Na02 CCH3 ; potassium cyanide, KCN
(In these examples, the cations are neutral - see above.)
� Dissolves in water to yield a basic solution
Example: Na02 CCH3(aq) - Na+(aq) + CH3COO- (aq) (dissociation of salt in water) CH 3 COO-(aq) + H 2 0(l) = OH -(aq) + CH3COOH (aq)
•
(basic cation reacts with water to yield a basic solution)
Neutral salt � A salt with neutral cations and anions
E xamples: Sodium chloride, NaC l ; potassium nitrate, KN03 ; magnesium iodide, Mgh
� Dissolves in water to yield a neutral solution •
Method of solution (main equilibrium) � For acid salts, use the same method as for weak acids, but replace HA with B W.
� For basic salts, use the same method used for weak bases, but replace B with A . (Al l are anions. ) -
Example: I ron( I I I) chloride, FeCI" is the salt of an acid cation and a neutral anion. I t dissolves in
water to give an acidic solution:
3 FeCI3(s) + 6 H 2 0(l) - Fe(H 20)6 +(aq) + CI -(aq) (dissolution of iron( I I I) chloride) 3 Fe(H 2 0)6 +(aq) + H 2 0(l) = Fe(H 20)s(OH) 2+(aq) + H30+(aq); (reaction of a weak acid with water) 3 For the above equation, Ka X 1 0- (see text Table 1 0.7). The cation is a weak acid and the equation above is the main equilibrium; use the methodology in section 1 0. 1 1 above to calculate concentration. For example, the [H30+] of a 0.0 1 0 M sol ution of FeCI3(s) is given by equation ( 1 ): =
x=
-Ka +
3.5
�Ka 2 + 4Ka [HA]i nitial 2
-(3.5
x
1 0-3 ) +
�(3.5
x
3
= [ H 0+ ] 3
1 0-3 ) 2 + 2
4(3.5
x
1 0-3 )(0.0 1 0)
3 [H 0+ ] = 4.42 x 1 0-3 pH -Iog(4.42 x 1 0- ) 2. 5 � 2.4 3 The sol ution is quite acidic and would be considered corrosive. x =
=
=
3
POLYPROTIC ACIDS AND BASES (Sections 1 0.1 4-10.1 7) 1 0. 1 4 The pH of a Poly p rotic Acid Solution •
Successive deprotonations of a polyprotic acid � A polyprotic acid is a species that can donate more than one proton.
3 Stepwise equilibria must be considered if successive Ka values differ by a factor of ;::: 1 0 , which is normally the case. V al ues of Ka are given in Table 1 0.9 in the text. � Polyprotic bases (species that can accept more than one proton) also exist. �
SG- 1 2S
•
Polyprotic acids a nd pH
�
Use the main equ ilibrium for a dilute solution of a polyprotic acid to calculate the pH. Neglect any secondary equilibria except for H 2 S04 , whose first deprotonation is complete. � The calculation is identical to that for any weak monoproti c acid.
Example: The pH of a 0.025 M solution of diprotic carbonic acid, H2C03 is calculated using the
first dissociation step and equation ( I ) in Section 1 0. 1 1 : H 2C03(aq) x=
-Ka J +
+
H 2 0(l)
=
H30+(aq) + HC03-(aq) Kal = 4.3 x 1 0 -7 (main equi librium)
)Ka J 2 + 4 Ka [HA] iniliaJ 2
=
+
[ H 30 ]
7 7 -( 4 .3 x I 0-7 ) + (4 .3 x 1 0- ) 2 + 4(4 .3 x 1 0- )(0.025)
)
2
4 x == [H30+ ] == 1 .03 x I 0-4 pH = -log( 1 .03 x 1 0- ) = 3 .9935 � 4.0 1 0 . 1 5 Solutions of Salts of Poly p rotic Acids •
pH of a solution of an amphiprotic anion H A � Given by pH =
•
± (pKa
,
+
pKa2 )
� Independent of the concentration of the anion Solution of basic anion of polyprotic acid : A2- or A3- Examples: Solutions of Na2 S or K3P04
� Main equi librium is hydrolysis of the basic anion: A2-(aq) + H 20(aq) OH -(aq) + HA- (aq) � Treat using methodology for solutions for basic anions: [Section 1 0: 1 2, Eqs. (4 ) - (6), this guide ]
=
1 0 . 1 6 The Concentrations of Solute Species
•
Determination of the concentration of all species for a solution of a weak diprotic acid, H2A
�
Solve the main equil ibrium (Ka,) as for any weak acid.
�
Solve the secondary equil ibrium (Ka2 ) for [A2-] using the [H30+] and [HA -] values determined above.
� This yields [ H2A], [H30+], and [HA-].
•
+
� Determine [OW] using Kw and [H 30 ] .
Determination of the concentration of all species for a solution of a weak triprotic acid, H 3 A
�
Solve the main equil ibrium (Kal) as for any weak acid.
� This yields [H3A], [H30+ ] , and [ H2A-].
2 Solve the secondary equil ibrium (Kd for [HA -] using the [H30+] and [H2A -] values from above. 2 3 � Solve the tertiary equilibrium (Ka., ) for [A - ] using the [H30+] and [ HA - ] values from above. �
�
Determine [OH - ] using Kw and [H30+ ].
1 0 . 1 7 Com position a n d pH •
Polyprotic acid � Concentrations of the several solution species vary with the pH of the solution.
•
Composition and pH � Solution to the stepwise equilibrium expressions for a polyprotic
acid as a function of pH
SG- 1 26
•
Diprotic acid (H zA)
-
qualitative considerations (See text Fig. 1 0.20.)
Ka J _
Ka2 _
�
[ H 30 + ] [ HA ] -
[ H 2A ]
[ H 30+ ] [ A 2 - ] [ HA - ]
(7)
(8)
At low pH, equilibria represented by Equations (7) and (8) shi ft to the left (Le Chatelier) and the 2 solution contains a high concentration of H2A and very little A -.
� At high pH, equilibria represented by Equations (7) and (8) shift to the right and the solution � •
contains a high concentration of A2- and very l ittle H 2 A.
At intennediate pH values, the solution contains a relatively high concentration of the amphiprotic anion HA-.
Diprotic acid (HzA) �
�
q ua ntitative considerations
2 Fraction (ex) of H 2A, HA -, and A - in solution as a function of [H30+]
AUTOPROTOLYSIS AND pH (Sections 10.18-10.19) 1 0. 1 8 Very Dilute Solutions of Strong Acids and Bases •
Very dilute solution � Concentration of a strong acid or base is less than 1 0-6 M.
•
General result for a strong acid H A at very dilute concentrations
Three equations in three unknowns ([H30+], [OW], and [A-]) Kw
Autoprotolysis equilibrium:
=
[H30+][OW]
=
1 .0
x
1 0- 14
Charge balance: [A-]
Material balance: Solution:
[OH-]
Quadratic equation:
=
=
[HA ]i nitial
(complete deprotonation)
[ H30+] - [HA] initial (from charge and material balance)
Kw
=
Kw
=
x2
[H30+]([H30+] - [HA]init ial ) [H30+f - [HA] initial [H30+],
- [ HA]initial X - Kw
=
0 and x
SG- 1 27
>
let x = [ H 3 0+]
0 (positive root)
Positive root:
x ==
+
[ HA] initial
�[ HA ]tni ial + 4 Kw t
and [OW] = Kw /[H30+] •
[H30+]
=
2
(9)
( 1 0)
General result for a strong base M O H at very dilute concentrations
Three equations in three unknowns Kw
A utoprotolysis equilibrium:
+
=
[H30+][OW] = 1 .0
[M+]
Material balance:
[M+] = [MOH] ini tial [OW] = [ H30+]
Solution:
x
=
Kw
=
=
==
-
(complete dissociation)
[MOH]initial
[H30+] { [H30+]
+
[MOH]initial }
[H30+] 2 + [MOH]illitial[H30+], let x = [ H30+]
x2 + [MOH]ini tial X
Quadratic equation: Positive root:
Kw
+
1 0- 14
[OH - ]
Charge balance:
[H30+]
x
-
Kw = 0 and x > 0 (positive root)
�
[ MOH ] i n ilia l + [ MOH ] nitial + 4 Kw = [H30+] ( I I ) 2
t
and [OW] = Kw /[H30+]
( 1 0)
Note: Strong bases such as 02- and CH3- produce aqueous solutions that are equivalent to MOH .
1 0-8 M sol ution of the strong base KOH can be calculated from equations ( 1 0) and ( I I ).
E xample: The pH of a 1 .5
x
+
From ( I I ), [H30 ] ==
-[ MOH] in i tial
+
2
From ( 1 0), pH -log [H30+] -log (9.28 (sl ightly basic solution as expected) =
=
t
�[M OH]tni ial
+
4 K w ==
9. 2 8 x I 0
-
8
M.
1 0-8) = 7.03.
X
1 0. 1 9 Very Dilute Solutions of Weak Acids •
Very dilute solution � Concentration of a weak acid or base is less than about 1 0- 3 M.
•
General result for a weak acid HA at very dilute concentrations
Four equations infour unknowns I ) Weak acid equilibrium: HA(aq) + H 20(l)
=
H30+(aq) + A- (aq);
2) Autoprotolysis equil ibrium:
Kw
3) Charge balance:
[H30+] = [OW]
=
[H30+] [OW]
SG- 1 2 8
+
=
1 .0
[A-]
x
1 0-1 4
[ H P+ ] [ A - ] Ka == [ HA]
[HA] inilial
4) Material balance: Solution:
[A-]
=
[ H30+] - [OH-]
= =
[ HA] = [ HA] ini,ial - [A-]
[ HA] + [A-] (incomplete deprotonation) K [ H30+] (from charge balance) w [ H P+ ]
=
[HA] inilial - [ H30+] +
K
w [ H P+ ]
S ubstitute expressions for [HA] and [A -] into Ka .
Ka =
( �w )
x x-
Rearrange to a cubic form.
K
[HAlini'iaJ - x + � x
Cubic equation: Solve by using a graphing calculator, trial and error, or the mathemati cal software present on text web site: www.whfreeman.comlchemicalprinciples/. Only one of the three roots will be physically meaningful. Alternatively, consider the fol lowing approximations: I f [ H30+]
>
1 0-
6
M,
Further, i f [ H30+]
then
>
[ H 30 + ]
< 1 0-8, and this term can be neglected in ( 1 2) to yield
6 1 0- M and [H30+]
Use Equations ( 1 2), ( 1 3), Note:
Kw
or
«
[ HA] inilial , then ( 1 3) reduces to
( 1 4) as appropriate to solve problems.
Many software programs do not work when numbers in the polynomial equation differ by many orders of magnitude.
SG - 1 29
Chapter
11
AQUEOUS EQUILIBRIA
M IXED SOLUTIONS AND BUFFERS (Sections 11. 1-11.3) •
A com mon theme in Chapter 1 1 � A variety of equilibria can be treated in the following way: I . IdentifY the sol ute species in solution. 2 . I dentifY the equilibrium relations among the solute species. 3. Use the relations in number 2 above to determine the concentrations of the solute species.
1 1 . 1 Bu ffer Action •
M ixed solution: � A solution containing a weak acid or base and one of its salts �
•
Example: a solution of hydrofl uoric acid and sodium fluoride
Buffer � A mixed solution containing weak conj ugate acid-base pairs that stabilizes the p H of a solution
� Provides a source and a sink for protons •
•
Acid buffer
�
�
Consists of a weak acid and its conj ugate base provided as a salt E xamples: Acetic acid (CH 3 COOH) and sodium acetate (NaCH 3 COO);
phosphorous acid (H 3 P03) and potassium dihydrogenphosphite (KH2P03)
Buffers a solution on the acid side of neutral ity (pH < 7), if pKa < 7
Base buffer � Consists of a weak base and its conj ugate acid provided as a salt
Examples: Ammonia (NH 3) and ammonium chloride (NH 4CI);
�
sodium dihydrogenphosphate (NaH2P04) and sodium hydrogen phosphate (Na2H P04 )
Buffers a solution on the base side of neutral ity (pH > 7) if pKb < 7
1 1 .2 Designing a Bu ffer •
pH of a buffer � The pH of a buffer solution is given approximately by the Henderson-Hasselbalch equation pH
""
pKa
+
log
(
. . IJ
[base] Jnllla . [acid] initi a l
� The adjustment from initial concentrations to the equilibrium ones of the conj ugate acid-base pair is
usually negligible. I f it is not, the pH can be adjusted to the desired value by adding additional acid or base.
� Use of this equation is equivalent to setting up a table of concentrations and solving an equilibrium
problem if adj ustment from initial concentrations is negligible.
SG- 1 3 1
•
Preparation of an acid buffer � Use an acid (HA) with a pKa value close to the desired pH.
� Add the acid and its conj ugate base (A- ) in appropriate amounts to satisfy the Henderson
Hasselbalch equation.
� Measure the pH and make any necessary adjustments to the acid or base concentrations to
achieve the desired pH.
•
Preparation of a base buffer � Use a base (B) whose conj ugate acid (BH+) has a pKa value close to the desired pH. �
Add the base and its conjugate acid in appropriate amounts to satisfy the Henderson Hasselbalch equation.
� Measure the pH and make any necessary adjustments to the acid or base concentrations to
achieve the desired pH.
1 1 .3 Bu ffer Capacity •
Buffer capacity � The amount of acid or base that can be added before the buffer loses its ability to resist a
change in pH
� Determined by its concentration and pH
� •
The greater the concentrations of the conj ugate acid-base pair in the buffer system, the greater the buffer capacity
Exceed ing the capacity of a buffer � Buffers function by providing a weak acid to react with added base and a weak base to react with
added acid. � Adding sufficient base to react with all of the weak acid or adding sufficient acid to react with all of the weak base exhausts (completely destroys) the buffer. •
Limits of buffer action
�
A buffer has a practical limit to its buffering abi lity that is reached before the buffer is completely exhausted.
� As a rule of thumb, a buffer acts most effectively in the range
10
>
[base] initial [acid] initial
>
0. 1
� The molarity ratios above yield the following expression for the effective
pH range of a buffer
I
pH
=
pKa ± I
SG- J 32
TITRA TIONS (Sections 11.4-11.7) •
When working with titrations, the full chemical equation must be kept in mind to ensure the correct stoichiometry. � An example is the neutralization of calcium hydroxide with hydrochloric acid requiring two moles
of acid for each for each mole of base neutral ized: 2 HCI(aq) + Ca(OH)2 --> CaClbq) + 2 H 20(l)
1 1 .4 Strong Acid-Strong Base Titrations •
pH curve � Plot of the pH of the analyte solution as a function of the vol ume of the titrant added
during a titration
•
For all strong base-strong acid or strong acid-strong base titrations
�
�
•
The stoichiometric point has a pH value of 7. The net ionic equation for the reaction is H 3 0+(aq) + OH (aq) --> 2 H 2 0(l) for al l strong base strong acid or strong acid-strong base titrations .
Titration of a strong base (analyte) with a strong acid (titrant) (see text Fig. 1 1 .4) �
�
The pH changes (drops) slowly with added titrant unti l the stoichiometric point is approached. The pH then changes (drops) rapidly near the stoichiometric point.
� At the stoichiometric point, the base is neutralized, at which point the titration is normal ly ended. �
•
In the region wel l beyond the stoichiometric point, the pH approaches the value of the titrant solution.
Titration of a strong acid (ana lyte) with a strong base (titrant) (see Fig. 1 1 .3 in the text)
�
The pH changes (rises) slowly with added titrant unti l the stoichiometric point is approached.
�
At the stoichiometric point, all the acid is neutral ized.
�
The pH then changes (rises) rapidly near the stoichiometric point.
� The pH approaches the value of the titrant solution in the region well beyond the stoichiometric
point.
•
pH curve calculations � As a titration proceeds, both the volume of the analyte solution and the amount of acid or base �
present change.
To calculate [H30+] in the analyte solution, the amount of acid or base remaining should be determined and divided by the total volume of the solution.
� Before the stoichiometric poi nt is reached, follow the procedure outlined in Example 1 1 .4 in the
text.
� For a strong acid-strong base titration, the pH at the stoichiometric point is 7.
� After the stoichiometric poi nt, fol low the procedure outli ned in Toolbox 1 1 . 1 in the text. � Always assume that the volumes of the titrant and analyte sol utions are additive: �
( 11;0131 = V'malyte + lI;itrant added)
If a solid is added to a solution, its volume is usually neglected.
SG- J 3 3
1 1 .5 Strong Acid-Weak Base and Weak Acid-Strong Base Titrations •
•
C ha racteristics of the titrations � The titrant is the strong acid or base, the analyte is the weak base or acid. � Stoichiometric poi nt has a pH value different from 7. Strong base-weak acid titration
�
A strong base dominates the weak acid and the pH at the stoichiometric point is greater than 7.
� A strong base converts a weak acid, HA, into its conj ugate base form, A-.
� At the stoichiometric point, the pH is that of the resulting basic salt solution.
•
Strong acid-weak base titration
�
A strong acid dominates the weak base and the pH at the stoichiometric point is less than 7.
� A strong acid converts a weak base, B, into its conj ugate acid form, B H+.
� At the stoichiometric point, the pH is that of the resulting acidic salt solution. •
Shape of the p H curve � Similar to that of a strong acid-strong base titration
� Differences include:
> A buffer region appears between the initial analyte solution and the stoichiometric point. > The pH of the stoichiometric point is different from 7. > There is a less abrupt pH change in the region of the stoichiometric point.
•
pH curve calculations � The pH is governed by the main species in solution.
� Calculate amounts and solution volume separately; divide to obtain concentrations. �
Before the titrant is added, the analyte is a solution of a weak acid or base in water. Determine the pH of the solution by fol lowing the procedures in Chapter l O i n the text.
� Before the stoichiometric point is reached, the solution is a buffer with differing amounts of
acid/conjugate base. To calculate the pH, follow the procedure outlined in Example 1 1 .6 in the text.
� At the stoichiometric point, the analyte solution consists of the salt of a weak acid or of a weak base.
To calculate the pH, fol low the procedure outlined in Example 1 1 .5 in the text.
� Well after the stoichiometric point is passed, the pH of the solution approaches the value of the
titrant, appropriately dil uted.
� Assume that the volumes of the titrant solution added and the analyte sol ution are additive:
( 1-';olal = Vanalyte + 1-';itranl added) I f the sample is a solid, its volume is usually neglected. 1 1 .6 Acid-Base Indicators •
Acid-base ind icator
�
�
A water-soluble dye with different characteristic colors associated with its acid and base forms Exhibits a color change over a narrow pH range
� Added in low concentration to the analyte so that a titration is primarily that of the analyte, not the
indicator
� Selected to monitor the stoichiometric or end point of a titration
SG- 1 34
� An indicator is a weak acid. It takes part in the following proton-transfer equil ibrium:
Values of pKl n are given in Table 1 1 .3 in the text.
�
The color change is usually most apparent when [ Hln] = [1n-] and pH = pKl n . The color change begins typically within one pH unit before pK11l and is essentially complete about I pH unit after pKl n . The pKln value of an indicator is usually chosen to be within I pH unit of the stoichiometric point.
I pKln "" pH(stoichiometric point) ± I I •
End point of a titration
� Stage of a titration where the color of the i ndicator in the analyte is
midway between its acid and base colors
•
An ind icator should be chosen such that its end point is close to the stoichiometric point.
1 1 . 7 Stoichiom etry of Polyp rotic Acid Titrations •
Polyprotic acids
�
Review the l ist of the common polyprotic acids in Table 1 0.9 of the text.
� All the acids in Table 1 0. 9 are diprotic, except for phosphoric acid, which is triprotic. �
Phosphorous acid, H3P03 , is diprotic. The major Lewis structures are :0 :0: II I H -O -H � H -O . .- P-O . .- H . .-P-O . .
I
I
H
H
The proton bonded to phosphorus is not acidic. •
Titration curve of a polyprotic acid � Has stoichiometric points corresponding to the removal of each acidic hydrogen atom � Has buffer regions between successive stoichiometric points
� At each stoichiometri c point, the analyte is the salt of a conjugate base of the acid or one of its � •
hydrogen-bearing anions. See the examples of pH curves for H3P04 , phosphoric acid, in Fig. 1 1 . 1 3 and H 2 C204, oxalic acid, in Fig. 1 1 . 1 4 of the text.
pH calculations �
The pH curve can be estimated at any point by considering the primary species in solution and the main proton-transfer equil ibrium that determines the pH.
SG - J 35
SOLUBILITY EQUILIBRIA (Sections 11.8-11.14) 1 1 .8 The Solubility Product •
Solubility product, Ks p
�
Equilibrium constant for the equ i librium between a solid and its dissolved form (saturated solution)
� The activity of a pure solid is I , and, for
K
2
3
aSb 3+ as,
-sp = ---"'-"'-------"-aS b,S,(s)
dilute solutions (sparingly soluble salts), the activity of
a solute species can be replaced by its molarity.
� To a fair approximation, •
Ksp
=
[Sb3+ f [S2- ]3 .
(saturated solution)
Molar solubility, s � Maximum amount of solute that dissolves in enough water to make one
l iter of solution
•
Relationship between s and Ksp
� From the example above, the stoichiometric relationships are
3 I mol Sb2 S3 "" 2 mol Sb + and 1 mol Sb2 S3
""
3 mol S 2-
� The molar sol ubil ity, s, of the salt Sb2 S3(s) is related to the ion concentrations at
�
�
equi l ibrium as fol lows: 3 [Sb +] = 2s and [S 2-] = 3s.
The relationship between s and
The relationship s
=
[ Ksp )115 1 08
Ksp
is then
holds for any salt whose general formula is X 2 Y3 .
� Similar relationships can be generated for other type of salts; for example, for an AB salt such
as AgCl, s
1 1 .9 •
=
Ksp
1 12 .
The Com mon-Ion Effect
Common-ion effect
� �
A decrease in the sol ubil ity of a salt caused by the presence of one of the ions of the salt in solution Follows from Le Chatel ier's principle
� Used to help remove unwanted ions from solution
3 2 Sb +caq) + 3 S 2-(aq), increasing the concentration of 3 Sb + ions by addition of Sb(N03)3 shifts the equil ibrium to the left, causing some precipitation of Sb2 S3(s). Thus, less Sb2 S3(s) is dissolved in the presence of an external 3 source of Sb + (or S 2-) than in pure water.
E xample: For the eq uil ibrium Sb2 S3(s)
=
SG- 1 3 6
•
Quantitative features � Use the solubility product expression to estimate the reduction in solubil ity resulting �
from the addition of a common ion.
For the equilibrium Sb2 S 3(s) = 2 Sb3+(aq) + 3 S 2-(aq), Ksp [Sb3+] 2 [S 2-1 ' . I f Sb2 S 3 (s) is added to a solution already containing Sb3+ ions with a concentration [Sb3 +], then dissolution of thi s sparingly soluble salt will add a negligible amount to the cation concentration. The solubil ity, s, and its relationship to Ksp is then =
K"
=
[Sb " l' (3,) ' and s
=
[ �l' )' " [S
1 1 . 1 0 Predicting Preci pitation • •
A salt precipitates if QSJl is greater than Kspo To predict whether precipitation occurs
�
Calculate Qsp using actual or predicted ion concentrations.
�
I f Qsp > Ksp, the salt will preci pitate from solution.
� Compare Qsp to Ksp .
�
If Qsp = Ksp, the solution is saturated with respect to the ions in the expression for Ksp Any additional amount of these ions will cause precipitation.
� If Qsp < Ksp , the solution is unsaturated and no precipitate will form.
1 1 . 1 1 Selective Precipitation •
•
Ions in a mixture can be separated by adding a soluble salt containing an oppositely charged ion that forms salts having very different solubilities with the original ions. Separation of different cations in solution
�
Add a soluble salt containing an anion with which the cations form insoluble salts.
� If the i nsol uble salts have sufficiently different solubil ities, they will precipitate at
different anion concentrations and can be collected separately.
1 1 . 1 2 Dissolving Precipita tes •
Ion removal � Removal from solution of one of the ions in the solubil ity equil ibrium increases the
solubil ity of the solid.
� A precipitate will dissolve completely if a sufficient number of such ions are removed . � I n this section, we focus on the removal of anions. •
Removal of basic a nions
�
Addition of a strong acid, such as HCI, neutral izes a basic anion. react to form water:
SG - 1 3 7
For example, hydroxides
---,}
•
•
I n some cases, gases are evolved. For example, carbonates react with acid to form carbon dioxide gas: 2 H 3 0+(aq) + C032-(aq) � CO2(g) + 3 H 20(l)
Removal by oxidation of anions
---,}
In some cases, gases are evolved. For example, S2-(aq) from a metal sulfide is oxidized to S(s) with nitric acid : 3 S 2-(aq) + S H 30+(aq) + 2N03-(aq) � 3 S(s) + 2 NO(g) + 1 2 H2 0(l)
Summary
---,}
The solubility of a salt may be increased by removing one of its ions from solution. An acid can be used to dissolve a hydroxide, sulfide, sulfite, or carbonate salt. Nitric acid can be used to oxidize metal sulfides to sulfur and a soluble salt.
1 1 . 1 3 Com plex Ion Form ation •
Complex ions
---,}
---,} ---,}
Formed by the reaction of a metal cation, acting as a Lewis acid, with a Lewis base M ay disturb a solubil ity equilibrium by reducing the concentration of metal ions Net result is an increase in the solubility of salts. Example: Fe2+(aq) + 6 CN -(aq) = Fe(CN)64-(aq)
In the above reaction, Fe2+ acts as a Lewis acid and CN - acts as a Lewis base. Complexation reduces the concentration of Fe2+ in solution thereby causing a shift in any solubil ity equilibrium in which Fe2+ is involved.
---,} ---,} ---,}
A number of fonnation reactions are listed in Table 1 1 .5 in the text.
3 Electron pair acceptors (Lewis acids) that form complex ions incl ude metal cations such as Fe +(aq), 2 Cu +(aq), and Ag+(aq). Electron pair donors (Lewis bases) that form complex ions include basic anions or neutral bases such as Cqaq), N H3 (aq), CN -(aq), and the thiosulfate ion S 20 32-(aq).
Quantitative features ---,}
The equil ibrium for a/ormation reaction is characterized by the/ormation constant, Kr . [ Fe(CN) 4- ] Example: Fe2+(aq) + 6 CN-(aq) = Fe(CN)64-(aq) Kf 6 [Fe2+ ] [CN- ]6 _
---,}
---,}
Values of Kr at 25°C for several complex ion equil ibria are given in Table 1 1 .5 in the text. Formation and solubil ity reactions may be combined to predict the solubility of a salt in the presence of a species that forms a complex with the metal ion. Example: FeS(s) = Fe2+(aq) + S2-(aq)
Fe2+(aq)
+
6 CN -(aq)
=
Ksp = [Fe2+] [S2- ]
F e(CN )t (aq )
Kr
_
-
[ Fe(CN) 6 4- ] [Fe2 + J [CN - ]6
Combining the above two reactions produces the following: 4 2 FeS(s) + 6 CN-(aq) = Fe(CN)6 -(aq) + S -(aq)
solubility reaction formation reaction
K = K x K = [Fe(CN )64- J [ s2 - ] = ( 6.3 x l O- 18 )( 7.7 x l 036 ) = 4.9 X I O I 9 » 1 sp r [CN - t
SG - 1 3 8
The equi libri um lies far to the right indicating a large solubil ity for FeS(s) in a cyanide solution. 2 Because 1 mol FeS ec: 1 mol S -, the molar solubil ity, s , is equal numerically to the sulfide ion concentration. 1 1 . 1 4 Qualitative Analysis •
Qualitative analysis � �
•
Method used to identify and separate ions present in an unknown solution (for example, seawater) Utilizes complex formation, selective precipitation, and pH control
� Uses solubil ity equil ibria to remove and identify ions selectively
Outline of the method � The method is outlined in F ig. 1 1 . 20 in the text and is summarized below.
Partial Qualitative Analysis Scheme
Ksp
Possible precipitate
Step I ) Add HCI(aq)
2) Add H 2 S(aq) (in acid solution, there is a 2 low S - concentration)
3) Add base to H 2 S(aq) (in basic solution, there is a 2 higher S - concentration)
AgCI Hg2 C b PbCI 2
1 .6 1 .3 1 .6
x
Bi 2 S3 CdS CuS HgS Sb2 S3
1 .0 4.0 7.9 1 .6 1 .6
X
FeS MnS NiS ZnS
6.3 1 .3 1 .3
x
1 .6
x
x X
x X X X
x x
1 0- 1 0 1 0- 18 5 1 01 0-97 2 1 0- 9 1 0-45 1 0-52 3 1 0-9 1 0- 18 1 0- 1 5 1 0-24 2 1 0- 4
2
Note: I n solubi lity calculations, the fairly strong basic character of S - must be considered.
2
S -(aq) �
•
+
H 2 0(l)
=
HS -(aq) + O W(aq)
Kb
=
1 .4
A solution may contain any or all of the cations given in the preceding table. We wish to determine which ones are present and which are absent.
Procedure
I:
Add Hel to the solution.
� Most chlorides are sol uble and wi l l not precipitate as the [Cn increases. �
�
For AgCl and Hg2C1 2, we expect their Qsp values to be exceeded and for them to precipitate. PbCh is sl ightly more sol uble, but it should also precipitate.
SG- 1 3 9
� •
The precipitate obtained, if any, using Procedure I is collected and the remaining solution tested for additional ions using Procedure 2.
Procedure 2: Add H 2 S to the solution. � Because the solution is highly acidic after Procedure I , the concentration of sulfide wi ll be very low:
H 2 S(aq) + 2 H2 0(l)
=
2 2 H30+Caq) + S -(aq)
� Thus, only sulfide salts with very low solubi l ity products, such as CuS or HgS, will precipitate.
� The resulting precipitate, if any, is collected and the remaining solution tested for additional ions
using Procedure 3 .
•
Procedure 3 : Add base to the solution.
2 � Base reduces [ H30+] and, as seen from the equi librium above, increases [S - ] .
� A t higher sulfide concentration, more soluble sulfide salts such a s FeS and ZnS precipitate. •
•
�
The resulting precipitate, if any, is collected.
Precipitates obtained in Procedures 1 , 2, and 3 are analyzed sepa rately for the presence of each cation in the group. Determining the presence or absence of Ag+ , Hg/+ ' and Pb2+ in Procedure 1
� Of the three possible chlorides, PbCh is the most soluble and its solubi lity increases with increasing
temperature.
2 Rinse the precipitate with hot water, collect the l iquid, and test for Pb + by adding chromate. 2 � A precipitate of insoluble yellow lead(l I) chromate ind icates that Pb + is present in the initial solution: 2 P b +(aq) + CrO/-(aq) - PbCr04(s) �
� Add aqueous ammonia to the remaining precipitate.
�
Silver(1) wil l dissolve forming the diammine complex : Ag+(aq) + 2 NH3(aq) - Ag(NH3)2+(aq)
� In aqueous ammonia, mercury(1) wil l disproportionate to form a gray mixture of l iquid mercury and
solid HgNH 2 CI :
�
Hg2CI 2 (s) + 2 NH3(aq) - Hg(l) + HgN H2CI(s) + N H/(aq) + Cnaq)
Formation of the gray mixture confirms the presence of mercury(l ) in the original solution.
� Silver(1) is confirmed by adding HCI to the ammine solution to precipitate white AgCI(s):
Ag(N H3)/(aq) + C naq) + 2 H30+Caq) - AgCI(s) + 2 N H/(aq) + 2 H 2 0(l)
SG- 1 40
Chapter
12
ELECTROCHEMISTRY
REPRESENTING REDOX REACTIONS (Sections 12.1-12.2) 12.1 Half-Reactions •
•
Half-reaction
�
�
I mportant for understanding oxidation-reduction (redox) reactions and electrochemical cells
Oxidation �
� �
•
Conceptual way of reporting an oxidation or reduction process
Removal of electrons from a species Increase in oxidation number of a species Represented by an oxidation half-reaction Example: Zn(s) - Zn2+caq) + 2 e-. Zinc metal loses two electrons to form zinc( I I ) ions.
Reduction � Gain of electrons by a species �
�
Decrease in oxidation number of a species
Represented by a reduction half-reaction E xample: Cu2 +Caq) + 2 e- - Cu(s). Copper( I l ) ions gain two electrons each to form copper metal .
•
Redox couple �
•
H alf-reactions
Ox/Red. Oxidized (Ox) and reduced (Red) species i n a half-reaction with the oxidized form l i sted first by convention Example: The Zn2+/Zn redox couple
� Express separately the oxidation and reduction contributions to a redox reaction
�
May be added to obtain a redox reaction if the electrons cancel
Example: From the half reactions above, the following redox reaction is obtained :
Oxidation half-reaction: Reduction half-reaction: Redox reaction:
Zn(s) 2 Cu +Caq) + 2 e-
Zn(s) + Cu2+(aq)
Zn2+(aq) + 2 e-
C u(s)
Zn2+(aq) + Cu(s)
12.2 Balancing Redox Equations •
Balancing methods � Two methods are commonly used to balance redox equations: the half-reaction method and the �
oxidation number method.
We will use the half-reaction method.
� I n either method, mass and charge are balanced separately.
� I n redox reactions, it is conventional to represent hydronium ions, H 3 0\aq), as H+(aq). •
Half-reaction method � A six-step procedure (see also Toolbox 1 2. I in the text)
I . Identify all species being oxidized and reduced from changes in their oxidation numbers. 2. Write unbalanced skeletal equations for each hal f-reaction.
SG - 1 4 1
3 . Balance all elements in each half-reaction except 0 and H . 4 . a) For an acidic solution, balance 0 by using H 20, then balance H by adding H+. b) For a basic solution, balance 0 by using H 20, then balance H by adding H 2 0 to the side of each half-reaction that needs H and adding OW to the other side. Note:
Adding H 2 0 to one side and OW to the other side has the net effect of adding H atoms to balance hydrogen if charge is ignored. Charge is balanced in the next step.
5. Balance charge by adding electrons to the appropriate side of each half-reaction. 6. a) M ultiply the half-reactions by factors that give an equal number of electrons in each reaction. b) Add the two half-reactions to obtain the balanced equation. � An alternative procedure for basic sol utions is to balance the redox equation for acid
solutions first. Then, add
n H\aq) + n OW(aq) - n H 20(I) or
n H 2 0(l) - n H+(aq) + n OW(aq)
to the redox equation to eliminate n H+ from the product or reactant side, whichever is necessary.
GALVANIC CELLS (Sections 12.3-12.10) 1 2.3 The Structu re of Galvanic Cells •
•
Galvanic cell �
Electrochemical cel l in which a spontaneous reaction produces an electric current
Galvanic cell components � Two electrodes (metal lic conductors and/or conducting solids) make electrical contact
with the cell contents, the conducting medium.
� An electrolyte (an ionic solution, paste or crystal) is the conducting medium.
�
Anode, labeled (-) or e : the electrode at which oxidation occurs.
� Cathode, labeled (+) or EB : the electrode at which reduction occurs. �
•
I f the anode and cathode do not share a common electrolyte, the oxidation reaction and reduction reaction compartments are separated to prevent direct mixing of the electrolyte solutions. Electrical contact between these sol utions is maintained by a salt bridge, a gel containing an electrolyte such as KCl(aq), or an equivalent device.
Hyd rogen electrode or half cell � Inert Pt wire, used to conduct electricity, immersed in an acid solution, through which
hydrogen gas is bubbled
� Half reaction: 2W(aq) + 2 e- - H 2(g) (reduction) or H 2 (g) - 2H\aq) + 2 e- (oxidation) •
The Daniell cell, an example of a galvanic cell
�
Zinc metal reacts spontaneously with Cu( l I) ions in aqueous solution.
� Copper metal precipitates and Zn( l I ) ions enter the solution.
SG- 1 42
� The reaction is exothermic, as suggested by the standard enthalpy of the cell reaction:
Zn(s) + Cu2+(aq)
Zn2+(aq) + Cu(s)
�
bJirQ = -225.56 kJ . mor l
The galvanic cell to the left contains anode and cathode compartments l inked by a salt bridge to Electron flow prevent the mixing of ions. Salt bridge The anode compartment contains a zinc electrode i n a n electrolyte solution such a s K Cl o r ZnC b . The cathode compartment contains an electrode (not necessarily Cu) in an electrolyte solution containing Cu2+(aq) ions. The salt bridge contains an electrolyte such as KCl in an aqueous gelatinous medium to control the flow of K+(aq) and Cnaq) ions. When current is drawn from the cell, the electrodes are linked by an external wire to a load (examples include a l ight bulb or motor). Current (electrons) flows through the wire Reduction Oxidation Zn(s) --7 Zn2 +(aq) + 2 e- Cu2 +(aq) + 2 e- --7 Cu(s) from anode to cathode as indicated and the spontaneous chemical cell reaction Zn(s) + Cu 2+(aq) Cu(s) + Zn2+(aq) occurs. As the reaction proceeds, Zn(s) is oxidized to Zn2+(aq) in the anode compartment (oxidation) and Cu2+(aq) is reduced to Cu(s) in the cathode compartment (reduction). The Cu(s) deposits on the �
cathode, and the Zn(s) anode slowly disappears. The closed line labeled circuit traces the direction that current flows in the cel l. Current is carried by electrons in the electrodes and external connections, and by ions in the conducting medium.
12.4 Cell Potential and Reaction Free Energy •
Cell potential and emf � Cell potential, E: a measure of a cell reaction's ability to move electrons through the
external circuit
� Electromotive force (emf): cell potential when a cel l is operated reversibly �
For our purposes, cell potential E will be taken to mean emf unless otherwise noted.
� The amount of usefu l work a cel l can perform depends on the cell potential and the
amount of the l i miting reactant present.
•
Units and definitions � Current: � Charge:
� Energy:
ampere, A, the SI base unit of electric current joule, J
� Potential : volt, V
== ==
==
A-s, the S I derived unit of charge kg·m2 ·s-2 , the SI derived unit of energy
coulomb, C
J . e l , S I derived unit of the cel l potential
� A charge of one coulomb fal l ing through a potential difference of one volt produces one
joule of energy.
� Cel l potential : Measured using a voltmeter connected to the two electrodes of a cell (see
Fig. 1 2.4 in the text)
� Convention: Write galvanic cel ls with anode on the left and cathode on the right. This arrangement
gives a positive cell potential.
SG- 1 43
•
Examples of galvanic cell reactions with measured cell potentials � Mg(s) + Zn2+(aq, I
M)
Zn(s)
-
+
Mg2\aq, 1 M)
EO = 1 .60 V EO = 0.76 V
Note: For the preceding cells, all reactants and products are i n their standard states and the potentials
measured are called standard potentials, EO. •
Relation between reaction free energy, 8.Gn and cell potential, E � Reaction free energy, 8.Gr : maxim um nonexpansion work, constant T and P : 8.Gr = We
We ,
obtainable from a reaction at
� When n electrons move through a potential difference, E, the work done, is the total charge times
the potential difference:
We
= Q V.
� The charge o n one mole o f electrons is NA times the charge o f a single electron,
-e
=
- 1 .602 1 77 x 1 0- 19 C: Q = -eNA •
� The Faraday constant is the magnitude of the charge of one mole of electrons : F = -eNA = 9.648 53 1 x 1 04 C mor l .
� The work done and reaction free energy are given by the product of the total charge, -nF, and the cel l potential, E: I'lGr = We = -nFE. � In the express ion above, n is the stoichiometric coefficient of the electrons in the oxidation and
�
reduction half-reactions that are combined to determine the cell reaction.
Maximum nonexpansion work can only be obtained if the cel l operates reversibly.
� Reversibility is only realized when the applied potential equals the potential generated by the cel l .
� All working cells operate irreversibly and produce a lower potentials than the emf.
� In sum, the relationship between the free energy and the cel l potential for reversible
electrochemical cel ls is one of the most important concepts in this chapter and is given by
I 8.Gr •
=
- nFE
I
Standard cell potential, E O � The standard state of a substance (s, I, or g) i s the pure substance at a pressure of 1 bar.
�
For solutes in solution, we take a molar concentration of 1 mol · L-1 as the standard-state activity.
� The standard cel l potential, E O, is the cell potential measured when all reactants and products are in
their standard states.
� The standard reaction free energy can be obtained from the standard cel l potential (emf) and vice
versa:
12.5 The Notation fo r Cells •
Cell diagram �
A symbolic representation of the cell components.
� A single vertical line I represents a phase boundary. � A double vertical line IIrepresents a salt bridge. �
Reactants and products are represented by chemical symbols.
SG - 1 44
---*
---* ---*
•
---*
Components in the same phase are separated by commas. The anode components are written first, followed by the salt bridge, if present, and then the cathode components. Phase symbols (s, I, g) are normal ly used. Concentrations of solutes and pressures of gases may also be given.
Cel l diagrams with concentrations of reactants and products ---*
Often, the concentrations of reactants and products are incl uded in the cell diagram to give a more complete description of the cell contents. 2 2 cell reaction Zn(s) + Mg +(aq, 0.5 M) E xample: Mg(s) + Zn +(aq, 1 M) 2 2 Mg(s) IMg +(aq, 0 .5 M) IIZn +(aq, J M) I Zn(s) cel l diagram �
12.6 Standard Potentials •
Standard cell potential ---*
---*
Difference between the two standard electrodes of an electrochemical cell The standard cell potential, E O, is the difference between the standard potentials of the electrode on the right side of the diagram and the electrode on the left side of the diagram. EO = EO(eJectrode on right of cell diagram ) - EO(electode on l eft of cel l diagram)
---*
---*
---*
•
>
0, the cell reaction is spontaneous (Keq > J )under standard conditions as written.
I f EO < 0, the reverse cell reaction is spontaneous (Keq < 1 ) under standard conditions as written. The standard potential of an electrode is sometimes cal led the standard electrode potential or the standard reduction potential.
Cell potential
---*
---*
•
I f EO
---*
measured with an electronic voltmeter voltmeter draws negligible current measured voltage is positive (+) when the + terminal of the meter is connected to the cathode
Sta ndard electrode potentials, E O
---*
---*
---*
---*
The potential of a single electrode cannot be measured. A relative scale is required. By convention, the standard potential of the hydrogen electrode is ass igned a value of O at all temperatures: E O(H+IH2) == o. For the standard hydrogen electrode (SHE), the half-reaction for reduction is H 2(g, J bar), and the half-cell diagram is 2 W(aq, J M ) + 2 e�
---*
H+(aq, J M ) IH 2 (g, I bar) IPt(s)
E O(H+/H 2) == °
I f an electrode is found to be the anode in combination with the S H E, it is ass igned a negative potential. I f it is the cathode, its standard potential is positive.
-; Val ues of standard electrode potentials measured at 25°C are given in Table 1 2. I and Appendix 2B in the text.
SG- 1 45
•
Standard electrode potentials a nd free energy � The relation between free energy and potential also applies to standard electrode potentials:
•
�
t,Gro values i n Appendix 2A can be used to determine an unknown standard potential.
Calculating the standard potential of a redox couple from those of two related couples � If two redox couples are added or subtracted to give a third redox couple:
> The standard potentials in general cannot be added to give the unknown potential. > This is because potential is an intensive property. > Free energies are extensive and can be added. � To calculate the unknown potential, we convert potential values to free energy values. •
Su mmary of the process of combining half-reactions � Combining two half-reactions (A and 8) to obtain a third half-reaction (C) is accomplished by
addition of the associated free energy changes.
� The equation,
•
t,G,o = nFE o , applies to each half-reaction and leads to the following equation: -
Standard electrode potentials of ha lf-reactions
� Consider the half-reaction M 2+(aq, I M ) + 2 e- - M(s)
E = E O(M 2+/M). 2 � Metals with negative standard potentials (E E O(M +/M) < 0) have a thermodynamic tendency to reduce H30+ (aq) to H 2(g) under standard conditions ( I M acid). � Metals with positive standard potentials ((E = E O(M 2+ 1M) » can not reduce hydrogen ions under =
standard conditions; in this instance, hydrogen gas is the stronger reducing agent. � In general :
> The more negative the standard electrode potential, the greater the tendency of a metal to reduce H+.
> The more positive the standard electrode potential, the greater the tendency of a metal ion to oxidize H 2 . 1 2.7 The Electrochem ical Series •
Electrochemical series � Standard half-reactions arranged in order of decreasing standard electrode potential (see both
Table 1 2. 1 and text Appendix 28)
� Table of relative strengths of oxidizing and reducing agents (see table on the following page)
SG- 1 46
� I n going up the table, the oxidizing strength of reactants increases.
E xample: F 2(g) is an exceptional ly strong oxidizing agent with a strong tendency to gain electrons
and be reduced.
� F 2 (g) will oxidize any of the product species on the right side of the table in the reactions listed
below it.
� In going down the table, the reducing strength ofproducts increases.
Example: Li(s) is an exceptionally strong reducing agent with a large tendency to lose
electrons and be oxidized.
� Li(s) wi l l reduce any of the reactant species on the left side of the table in the reactions listed
above it.
� Standard electrode potentials have values ranging from about +3
v.
V to -3 V, a difference of about 6
� No single standard galvanic cell may have a potential larger than about
half-reactions that can give a larger potential.
E O (Y)
Reduction half-reaction
F 2 (g) + 2 e- - 2 F -(aq) 3 M n +(aq) + e- - M n2+(aq)
Oxidizing strength •
6 V because there are no
Reducing strength
+2.87 + 1 .5 1
h(s) + 2 e- - 2 naq)
+0.54
2 H+(aq) + 2 e- - H2(g) 2 F e +(aq) + 2 e- - Fe(s) AI3+(aq) + 3 e- - AI(s)
o
-0.44
Na+(aq) + e- - Na(s)
-2.7 1
Li+(aq) + e- - Li(s)
- 1 .66 -3 .05
Viewing half-reactions as conjugate pairs � For a Bf0nsted-Lowry acid-base conj ugate pair, the stronger the conj ugate acid, the weaker the
conjugate base.
� For a given electrochemical half-reaction, the stronger the conjugate oxidizing agent, the weaker the
conjugate reducing agent. F2 (g) is an extremely powerful oxidizing agent, whereas F -(aq) is an extremely weak reducing agent.
•
Uses of the electrochemical series
�
�
�
1 2.8 •
Predicting relative reducing and oxidizing strengths Predicting which reactants may react spontaneously in a redox reaction Calculating standard cel l potentials Standard Potentials and Equilibrium Constan ts
Equ ilibrium constants can be obtained from standard cell potentials for � Redox, acid-base, dissolution/precipitation, and dilution (change in concentration) reactions
SG- 1 47
•
Quantitative aspects
�
Combining !J.G,o = -RT l n K (from Chapter 9) with !J.G,o = -nFE o yields InK
�
�
=
nFEo RT
=
nEo 0.025 693 V
at T
=
298. 1 5 K
Standard electrode potentials may be used to calculate E O values. Because !J.G,o applies to all reactions, the equation above may also be used for acid-base reactions, precipitation reactions, and dilution reactions.
� For non-redox reactions, the overall "cell" reaction will not show the electron transfer that
occurs in the half-reactions explicitly.
1 2.9 •
•
The Nernst Equation
Properties of a galvanic cell
�
�
As a galvanic cell discharges, the cell potential, E, decreases and reactants form products. When E = 0, the cell is completely discharged and the cel l reaction is at equilibrium.
Quantitative aspects � From Chapter 9, !J.G, = !J.G,o + RT ln Q, where Q is the reaction quotient.
(Note that the identical symbol Q is also used for the quantity of electricity.)
� From this chapter, !J.G,
=
-nFE and !J.G,o = -nFE o.
� Combining the above equations leads to the Nernst equation:
E •
=
EO
_
RT InQ nF
=
EO
_
(0.025 693 V ) In Q at 298. 1 5 K n
The Nernst equation �
Describes the quantitative relationship between cel l potential and the chemical composition of the cell
� I s used to estimate the potential for the following:
> Cel l from its chemical composition > Half-cell not in its standard state
1 2 . 1 0 Ion-Selective Electrodes •
Ion-selective electrode � An electrode sensitive to the concentration of a particular ion �
•
Example: a metal wire in a solution containing the metal ion. The electrode potential for E 2 (Ag+, Ag) or E (Cu +, Cu) is sensitive to the concentration of Ag+ or C u2+, respectively.
Measurement of pH � An important appl ication of the Nernst equation
� Util izes a galvanic cel l containing an electrode sensitive to the concentration of the H+ ion
� A calomel electrode connected by a salt bridge to a hydrogen electrode:
SG- 1 48
Calomel electrode: Hg2Ch(s) + 2 e- - 2 Hg(l) + 2 Cqaq)
EO
Hydrogen electrode: H 2(g) - 2 H+(aq) + 2e
EO
Cell reaction: Hg2C12(s) + H 2 (g)
-
2 H+(aq) + 2 Hg(l) + 2Cqaq)
= =
+0.27 V 0.00 V
E O = +0.27 V
� Q in the Nemst equation for the cell reaction above will depend only on the unknown
concentration of H+ if the C I - concentration in the calomel electrode is fixed at a value determined for a saturated solution of KCI and the pressure of H2 gas is kept constant at I bar.
•
G lass electrode �
A thin-walled glass bulb containing an electrolyte with a potential proportional to pH
� Used to replace the (messy) hydrogen electrode in modern pH meters �
•
Modern pH meters use glass and calomel electrodes in a single unit (probe) for convenience. The probe contacts the test solution (unknown pH) through a small salt bridge.
pX meters �
�
Devices sensitive to other ions such as Na+ and CN Used in industrial appl ications and in pollution control
ELECTROLYTIC CELLS (Sections 1 2.11 -12.12) 1 2. 1 1 Electrolysis •
Electrolysis � Process of driving a reaction in a nonspontaneous direction by using an electric current
•
�
Conducted in an electrolytic cel l
Electrolytic cell � An electrochemical cell in which electrolysis occurs
� Different from a galvanic cell in design
� Both electrodes normally share the same compartment.
� As in a galvanic cel l , oxi dation occurs at the anode and reduction occurs at the cathode.
•
�
Unlike a galvanic cel l , the anode has a positive charge and the cathode is negative.
Example of an electrolytic cel l � Sodium metal is produced by the electrolysis of molten rock salt (impure sodium chloride) and
calcium chloride. Calcium chloride lowers the melting point of sodium chloride (recall freezing point depression), increasing the energy efficiency of the process.
� A schematic diagram of an electrolytic cel l for the production ofNa(l) and CI2(g) is shown on the
next page (also see text F ig. 1 2. 1 5).
� The cell has one compartment that contains both the anode and cathode. The electrode material is
often an inert metal such as platinum. Electrical current is supplied by an external source, such as a battery or power supply to drive this nonspontaneous reaction. This current/orees reduction to occur at the cathode and oxidation to occur at the anode. The cathode is placed on the right side
S G- 1 49
as in a galvanic cel l. Sodium ions are reduced at the cathode and chloride ions are oxidized at the anode: Na+Cl) + e- - Na(l) cathode: 2 Cr(1) - Ch(g) + 2 eanode: Note that liquid sodium metal i s formed in the electrolysis because the cell temperature of 600°C is above the melting point of sodium (98°C). Notes: In the molten salt mjxture, Na+ is reduced in preference to
2 Cql) •
�
Ca2+. I n aqueous salt solutions, care must be taken i n predicting the products of electrolysis because of the possibility of oxidizing or reducing water instead of either the anion or cation, respectively.
Clig) + 2 e
Na\l) + e- � Na(l)
Overpotential and the prod ucts of electrolysis ---+ ---+ ---+
---+ ---+ ---+
For electrolysis to occur, an external potential at least as great as that of the spontaneous cel l reaction must be applied to an electrolytic cell . The actual potential required i s often greater than this minimum value. The additional voltage required is called overpotential. Overpotential depends on the structure of a solution near an electrode, which differs from that in the bulk solution. Near an electrode, electrons are transferred across an electrode-solution interface that depends on the solution and the condition of the electrode. Because of overpotential, the observed electrolysis products may differ from those predicted using standard electrode potentials and the Nernst equation. In a solution with more than one reducible species, the reactant with the most positive reduction potential will be reduced. In a solution with more than one oxidizable species, the product with the most negative reduction potential will be oxidized.
1 2 . 1 2 The Products of Electrolysis •
Faraday's law of electrolysis ---+ ---+
•
The number of moles of product formed is stoichiometrically equivalent to the number of moles of electrons supplied. The amount of product depends on the current, the time it is applied, and the number of moles of electrons in the half-reaction.
Quantitative aspects of Faraday ' s law
---+
---+
Determine the stoichiometry of a half-reaction; for example, Mn+ + n e
-
- M(s).
The quantity of electricity, Q, passing through the electrolysis cell is determjned by the electric current, I, and the time, t, of current flow. Recall that I ampere (A) I coulomb (C) S- I . =
Charge supplied (C)
=
current (A) x time (s)
SG- J 50
or
Q
=
It
�
Because Q
=
nF, where n is the number of moles of electrons and F is the Faraday constant, then
M oI es 0f e I ectrons =
charge suppl ied (C) F
=
current (1) x time (t) F
It
or n ( e ) = F -
� The mass of product is determined from n(e-), the half-reaction, and the molar mass, M.
Moles of product = n(p) = Mass of Product = m (p) =
It ---
n(e- )F
[ l It
n (e )F
M(P)
-
where M(p) is the molar mass of the product p.
THE IMPACT ON MATERIALS (Sections 12.13-12.15) 1 2 . 1 3 Applications of Electrolysis •
Uses of electrolysis � Extracting metals from their salts. Examples include AI(s), Na (see text Fig. 1 2. I 5), and Mg. � Preparation of fluorine, chlorine, and sodium hydroxide
� •
Refining (purifyi ng) metals such as copper � Electroplating metals Electroplating
�
Electrolytic deposition of a thin metal fi lm on an object
� Object to be plated is made the cathode of an electrolysis cel l. � Cell electrolyte i s a salt of the metal to be plated.
� Cations are supplied either by the added salt, or from oxidation of the anode, which is then made of
the plating metal . For example, silver, gold, and chromium plating (see text Fig. 1 2 . 1 6)
1 2 . 1 4 Corrosion •
Corrosion �
•
Unwanted oxidation of a metal
� An electrochemical process that is destructive and costly
Reta rding corrosion
�
Coating the metal with paint (painting) or plastic. Non-uniform coverage or uneven bonding leads to deterioration of the coating and rust.
� Passivation, the formation of a nonreactive surface layer (Example: Ah03 on al uminum metal) �
G alvanization, coating the metal with an unbroken layer of zinc. Zinc is preferentially oxidized
and the zinc oxide that forms provides a protective coating (passivation).
� Use of a sacrificial anode, a metal more easily oxidized than the one to be protected, attached to it.
Used for large structures. A disadvantage is that sacrificial anodes must be replaced when completely oxidized (think of an underground pipeline).
SG-1 S l
� Cationic electrodeposition coatings. A sacrificial metal (yttrium) is used as the primer for
automobil e and truck bodies to reduce corrosion.
•
Mechanism of corrosion � Exposed metal surfaces can act as anodes or cathodes, points where oxidation or reduction can
occur.
� I f the metal is wet (from dew or groundwater), a film of water containing dissolved ions, may l i nk
the anode and cathode, acting as a salt bridge.
� The circuit is completed by electrons flowing through the metal, which acts as a wire in the external
circuit of a galvanic cel l .
•
Mechanism of rust formation on iron metal � At the anode, iron is oxidized to iron(l I) ions.
(I)
2 Fe(s) - Fe +(aq) + 2 e-
_E O
=
-(-0.44 V) = +0.44 V
� At the cathode, oxygen is reduced to water.
(2)
�
02(g) + 4 H +(aq) + 4 e- - 2 H 20(I)
[ron(1[) ions migrate to the cathode where they may be oxidized by oxygen to iron(I I l). 2 3 _E O = -(+0.77 V) = -0.77 V (3) F e +(aq) - Fe +(aq) + e-
� The overall redox equation is obtained by adding reaction ( I ) four times, reaction (3) four times, and reaction (2) three times. The resulting cell reaction has n = 1 2.
(4)
3 4 Fe(s) + 3 02 (g) + 1 2 W(aq) - 4 Fe +(aq) + 6 H 20(l)
E O = + 1 .27 V
� Next, the iron(1 l I) ions in reaction (4) precipitate as a hydrated iron( l I I) oxide or rust. The
coefficient, x, of the waters of hydration in the oxide is not well defined. 3 (5) 4 Fe +(aq) + 2(3 + x) H 20(l) - 2 Fe203 ·xH20(s) + 1 2 H+(aq)
� F i nal ly, the overal l reaction for the formation of rust is obtained by adding reaction (4) and (5):
(6)
4 Fe(s) + 3 02 (g) + 2xH20(l) - 2 Fe2 03 ·xH 20(s)
� From this mechanism, summarized in reaction (6), we expect corrosion to be accelerated by
moisture, oxygen, and salts, which enhance the rate of ion transport.
1 2 . 1 5 Practical Cells •
Practical galvanic cells � Commonly cal led batteries
� Should be inexpensive, portable, safe, and environmentally benign
� Should have a high specific energy (reaction energy per kilogram) and stable current •
Primary cells � Cannot be recharged
� Some types of primary cel ls:
>
Dry cells
-
(A, AA, D, etc. , batteries), used in flashlights, toys, remote controls, etc.
Cell diagram: Zn(s) IZnCI 2(aq), N H4CI(aq) IMnO(OH)(s) IMn02(s) IC(gr), 1 5 V .
2 Cell reaction : Zn(s) + 2 NH/(aq) + 2 Mn02(s) - Zn +(aq) + 2 MnO(OH)(s) + 2 NH3(g) >
simi lar to dry cells but with an alkal ine electrolyte that increases battery life Used in backup (emergency) power supplies and smoke detectors
Alkaline cells
-
SG- J S 2
Cell d iagram: Zn(s) IZnO(s) IOW(aq) I Mn(OH)2 (S) I Mn02 (s) I C(gr), 1 . 5 V Cell reaction : Zn(s) + Mn02 (s) + H 2 0(l) - ZnO(s) + M n(OHMs)
>
S ilver cells - have only solid reactants and products, high emf, long l i fe; used i n pacemakers, hearing aids, and cameras Cell d iagram: Zn(s) I ZnO(s) IKOH(aq) IAg2 0(s) IAg(s) I steel, 1 .6 V Cell reaction : Zn(s) + Ag2 0(s) - ZnO(s) + 2 Ag(s)
•
Secondary cells � Can be recharged
� M ust be charged before initial use � Some types of secondary cel ls:
>
Lead-acid cell (automobile battery) - has low specific energy but generates high current over smal l time period to start vehicles Cell d iagram: Pb(s) I PbS04 (s) I H\aq), HS04-(aq) I Pb02 (s) I PbS04(s) I Pb(s), 2 V Cell reaction : Pb(s) + Pb02(s) + 2 H 2 S04(aq) - 2 PbS04(s) + 2 H 2 0(l)
> >
>
•
Lithium-ion cell - high energy density, high emf, can be recharged many times; used in laptop
computers.
Sodium-sulfur cell - used to power electric vehicles, has all liquid reactants and products, a
solid electrolyte, and relatively-high voltage. Cell diagram: Na(l) I Na+(solution) IIS 2-(solution) IS 8(1), 2.2 V Cell reaction : 1 6 Na(l) + S 8(1) - 1 6 Na +(solution) + 8 S 2-(solution) Nickel-metal hyd ride (NiMH) cell - used in hybrid vehicles as a supplemental energy source; utilizes hydrogen storage in metal alloys, has a low mass, high energy density, high current load capabil ity, and is nontoxic (environmentally friendly)
Fuel cells � Galvanic cell s designed for continuous reaction
� Require a continuous supply of reactants
� Alkali-fuel cel l - used in the space shuttle
Cell d iagram: N i(s) I H 2(g) IKOH(aq) I02(g) INi(s), 1 .23 V
Cell reaction : 2 H 2 (g) + 02 (g)
-
2 H 2 0(l)
S G- J 5 3
Chapter 13
CHEMICAL KINETICS
REACTION RATES (Sections 13.1-13.3) 13.1 Concentration a n d Reaction Rate •
Definition of rate General
-+
Rate is the change in a property divided by the time required for the change to occur.
Chemical kinetics
•
•
•
Reaction rate is the change in molar concentration of a reactant or product divided by the time required for the change to occur.
Average vs. instantaneous rates: an analogy to travel -+
The average speed of an automobile trip is the length of the journey divided by the total time for the journey.
-+
The instantaneous speed is obtained if the car is timed over a very short distance at some point in the journey.
Average reaction rate -+
For a general chemical reaction, aA + b B cC + dO, the average reaction rate is the change in molar concentration of a reactant or product divided by the time interval required for the change to occur.
-+
The average reaction rate may be determined for any reactant or product.
-+
Rates for different reactants and products may have different numerical values.
-
U nique average reaction rate -+
For a general chemical reaction, aA + b B cC + dO, the unique average reaction rate is the average reaction rate of reactant or product divided by its stoichiometric coefficient used as a p ure number.
-+
The unique average reaction rate may be determined from any reactant or product.
-+
The unique average reaction rate is the same for any reactant or product.
-
Note:
•
->
Two critical underlying assumptions for generating a unique reaction rate are that the overal l reaction time is slow with respect t o the bui ldup and decay o f any intermediate and that the stoichiometry of the reaction is maintained throughout.
Reaction rates -+
The time requi red to measure the concentration changes of reactants and products varies considerably according to the reaction.
-+
Spectroscopic techniques are often used to monitor concentration, particul arly for fast reactions. An important example is the stopped-flow technique shown in Fig. 1 3 .3 in the text. 5 The fastest reactions occur on a time scale of femtoseconds ( 1 0- 1 s).
13.2 The Insta n taneous Rate of Reaction •
Reaction rates -+
Most reactions slow down as they proceed and reactants are depleted.
-+
If equilibri um is reached, the forward and reverse reaction rates are equal .
-+
The reaction rate at a specific time is cal led the instantaneo us reaction rate.
SG- J 55
•
•
I nstantaneous reaction rate -t
For a product, the slope of a tangent line of a graph of concentration vs. time
-t
For a reactant, the negative of the slope of a tangent line on a graph of concentration vs. time
Mathematical form of the instantaneous rate -t -t
The slope of the concentration vs. time curve is the derivative of the concentration with respect to time. For the instantaneous disappearance of a reactant, R, Reaction rate
-t
d [ R] dt
For the instantaneous appearance of a product, P, Reaction rate
-t
= _
=
d [ P] dt
For the general reaction, aA
+
+
b8 - c C
. . . Umque Instantaneous reactIOn rate -t
dO,
=
I d [ A] - - --
a dt
=
I d [8] ----
b
=
dt
I d [C]
1 d [O]
- -- = ---
c dt
d dt
In the remainder of the chapter, the term reaction rate is used specifically to mean the unique instantaneous reaction rate written here. The two conditions relating to unique average rates described in Section 1 3 . 1 of the Study Guide also apply here.
13.3 Rate Laws and Reaction Order •
•
•
I nitial reaction rate -t
Instantaneous rate at t = 0
-t
In what is called the method of initial rates, analysis is simpler at t = 0 because products that may affect reaction rates are not present.
-t
The method of initial rates is often used to determine the rate law of a reaction.
Rate law of a reaction -t
Expression for the instantaneous reaction rate in terms of the concentrations of species that affect the rate such as reactants or products
-t
Always determined by experiment and usually difficult to predict
-t
May have relatively simple or complex mathematical form
-t
The exponents in a rate expression are not necessarily the stoichiometric coefficients of a reaction; therefore, we use m and n in place of a and b in rate equations.
Simple rate laws
I Rate
k[ A] 11I
I
-t
Form of a simple rate law:
-t
k = rate constant, which is the reaction rate when all the concentrations appearing in the rate law are 1 M (recall that the symbol M stands for the units of mol·L-1).
-t
k is independent of concentration(s).
-t
k is dependent on temperature, k (T).
=
SG- J S 6
�
A refers to the reactant, and m, the power of [A], is called the order in reactant A and is determined experimental ly. Order is not related to the reaction stoichiometry. Exceptions include a special class of reactions, which are cal led elementary (see Section 1 3 . 7).
�
Typical orders for reactions with one reactant are 0, I, and 2. Fractional values are also common. Zero-order reaction
(m = 0):
F irst-order reaction
(m
=
Rate = k
Rate = k[A] k[A]2
I):
Second-order reaction (m = 2): •
Rate
=
More complex rate laws �
Many reactions have experimental rate laws containing concentrations of more than one species.
�
Form of a complex rate law: The overall order is the sum of the powers m
+
n
+ ....
�
The powers may also be negative values, and forms of even greater complexity with no overall order are found (see Section 1 3 . 1 5 for an example).
�
More complex rate laws are treated in Section 1 3 .8.
�
The fol lowing table lists experimental rate laws for a variety of reactions:
Reaction
Rate law
2N H3(g) - N2(g) + 3 H2Cg)
Rate
k
Zero
2 N20S(g) - 4N02(g) + 02(g)
Rate = k[N20S]
First
2N02(g) - NO(g) + N03(g)
Rate
2NO(g)
+
02(g) - 2 N02(g)
13-(aq) + 2N3-(aq) - 3 I -(aq)
+
=
=
Order
k[N02]
2
2 Rate = k[N0] [02]
Second First in O2 Second in NO Third overall First in CS2 First in N3Second overall
3 N2(g)
in the presence of CS2(aq) Rate = k = �
U nits of k
[03f [02]
k [03]2[02r1
Second in 03 Minus one in O2 First overal l
In general, the reaction order does not fol low from the stoichiometry of the chemical equation. I n the case o f the catalytic decomposition of ammonia on a hot platinum wire, the reaction order is zero initial ly because the reaction occurs on the surface of the wire, and the surface coverage is independent of concentration. The rate of a zero-order reaction is independent of concentration until the reactant is nearly exhausted or until equi librium is reached.
SG-IS7
•
•
�
Reaction rates may depend on species that do not appear in the overal l chemical equation. The reaction of triodide ion with nitride ion, for example, is accelerated by the presence of carbon disulfide, neither a reactant nor product. The concentration of carbon disulfide, a catalyst, appears in the rate law. Catalysis i s discussed in Section 1 3 . 1 4.
�
I n the last example above, the order with respect to the product O2 is -I. Negative orders often arise from reverse reactions in which products reform reactants, thereby slowing the overal l reaction rate. Oxygen i s present at the beginning o f this reaction as commonly studied because ozone is made by an electrical discharge in oxygen gas. It is possible to study the kinetics of the decomposition of pure ozone. In the absence of oxygen, the initial rate law is q uite different from the one listed in the above table.
Procedure for determining the reaction orders and rate law �
The reaction order for a given species is determined by varying its initial concentration and measuring the initial reaction rate, while keeping the concentrations of al l other species constant.
�
The procedure is repeated for the other species until all the reaction orders are obtained.
�
Once the fonn of the rate law is known, a value of the rate constant is calculated for each set of initial concentrations. The reported rate constant is typically the average value for all the measurements. See Table 1 3 . 1 in the text for examples of rate laws and rate constants.
Pseudo-order reactions �
If a species in the rate-law equation is present in great excess, its concentration is effectively constant as the reaction proceeds. The constant concentration may be incorporated into the rate constant and the apparent order of the reaction changes.
�
Reactions in which the solvent appears in the rate law often show pseudo-order behavior because the sol vent is usually present in large excess.
�
A technique called the isolation method is often used to determine the order in a particular reactant by keeping the concentration of other reactants in excess. The method is repeated for all reactants and the complete rate law is obtained. A hidden danger is that occasional ly the rate law is not the same in the extreme ranges of the concentrations of reactants. Different reaction mechanisms that yield different rate laws may occur under the extreme conditions.
CONCENTRATION AND TIME (Sections 13.4-13.6) 13.4 First-Order I ntegrated Rate Laws •
I ntegration of a first-order rate law �
First-order rate law:
�
I ntegration with limits :
_
_
d[ A] = k [A] dt
erA), d [A] J[A ] o [ A]
=
r' kdt Jo
SG-JS8
( ] [A],
= -kl
Result in logarithmic fonn :
�
Result in exponential fonn :
�
I f a plot of In [A], vs. I gives a straight l ine, the rate law is first-order in [A].
�
The rearrang ed logarithmic expression, In [A], y=
� •
In
�
intercept
+
(slope
x
[A]a
I [A], = [A]ae-k' I =
In [A]a
x ), of a plot of In [A], vs. I.
Plot of In [A], vs. I: slope = -k and intercept
=
-
kl , shows the straight-line b ehavior,
In [A]o
I ntegrated first-order rate law �
[A] decays exponentially with time.
�
Used to confinn that a reaction is first order and to measure its rate constant
�
Known k and [A]o values can b e used to predict the value of [A] at any time I.
13.5 Half-Lives for First-Ord e r Reactions •
Half-life expression �
The half-life, initial value.
�
At'
�
Solve for
= '1/2,
11/2,
[ AJr '12
1112
of a substance is the time required for its concentration to fal l to one-half its
=
b y rearranging the logarithmic form of the integrated rate law to give
( ].
[A] o .!. 11 1 2 = . - ln k [AJI", -->
•
[ A]o [ AJo 1 = =2. -2 [A]o. The ratio of concentrations is [ AJ I", '2I [ AJ o
Half-life express; on,
Then substitute
1
/112 =
In 2 k
1
[A] o =2. [A] I",
Half-life of a first-order reaction �
I ndependent of the initial concentration
�
Characteristic of the reaction
�
I nversely proportional to the rate constant, k
�
The logarithmic form of the rate law may b e modified as fol lows:
In
( ]
( ]
' [A]o = kl= (In2) [AJr 11 1 2
SG- J S 9
13.6 Second-Order I n tegrated Rate Laws •
•
•
I n tegration of a second-order rate law _ d[A] = k[A f dt
�
Second-order rate law:
�
I ntegration with limits:
�
I ntegrated form:
I I --=kt [A] I [A]o
�
Alternative form:
[AJr =
�
I f a plot of I/ [A]I vs. t gives a straight l ine, the rate law is second-order in [A].
�
The rearranged expression, I/[AL = I/[A]o + kt, reveals the linear b ehavior of a plot of I/[AL vs. t.
�
Plot of I/[AL vs. t:
_
i
[AL d [ A] = 2 [AJo [A]
i
kd t
1
0
--
[A]o I + [A]Okt
slope = +k and intercept
=
II [A]o
I n tegrated second-order rate law �
For the same initial rates, [A] decays more slowly with time for a second-order process than for a first-order one (see Fig. 1 3 . 1 4 in the text).
�
Used to confirm that a reaction is second order and to measure its rate constant
�
If k and [A]o are known, the value of [A] at any time t can b e predicted.
Summary of rate laws
�
See Tab le 1 3 . 2 in the text.
REACTION MECHANISMS (Sections 13.7-13.10) 13.7 Elementary Reactions •
Reactions � �
•
A net chemical reaction is assumed to occur at the molecular level in a series of separab le steps, called elementary reactions. A reaction mechanism is a proposed group of elementary reactions that accounts for the reaction's overal l stoichiometry and experimental rate law.
Elementary reaction �
A single-step reaction, written without state symb ols, that describ es the b ehavior of individual atoms and molecules taking part in a chemical reaction
�
The molecularity of an elementary reaction specifies the numb er of reactant molecules that take part in it.
SG-160
---+
Only one reactant species is written (molecularity = 1 ).
Unimolecular:
C3H6 (cyclopropane) CH2=CH-CH3 (propene) Cyclopropane molecule decomposes spontaneously to form a propene molecule. �
---+
Bimolecular :
Two reactant species combine to form products (molecularity = 2).
Termolecular:
H 2 + Br HBr + H A hydrogen molecule and a bromine atom collide to form a hydrogen bromide molecule and a hydrogen atom. Three reactant species combine to form products (molecularity = 3). I + I + H2 HI + H I Two iodine atoms and a hydrogen molecule collide simultaneously to form two hydrogen iodide molecules. �
---+
�
•
Reaction mechanism ---+
Sequence of elementary reactions which, when added together, gives the net chemical reaction and reproduces its rate law
---+
Reaction intermediate, not a reactant or a product, is a species that appears in one or more of the elementary reactions in a proposed mechanism. It usually has a small concentration and does not appear in the overal l rate law.
---+
The plausibility of a reaction mechanism may be tested, but it cannot be proved.
---+
The presence and behavior of a reaction intermediate are sometimes testable features of a proposed reaction mechanism.
1 3.8 The Rate Laws of Elementary Reactions •
Forms of rate laws for elementary reactions ---+
Order fol lows from the stoichiometry of the reactants in an elementary reaction. Note that products do not appear in the rate l aw of an elementary reaction. See Table 1 3 . 3 in the text.
---+
Unimolecular elementary reactions are first-order. Reaction: A
---+
Rate = k [A]
products
Bimolecular elementary reactions are second-order overal l . If there are two different species present, the reaction is first-order in each one. Reaction: A Reaction: A
---+
�
+
+
A B
�
�
products
Rate = k [A f
products
Rate = k [A] [B]
Termolecular elementary reactions are third-order overall . I f there are two different species, the reaction is second-order in one of them and first-order in the other. I f there are three different species, the reaction is first-order in each one. 3 Rate = k [A] Reaction : A + A + A products Rate = k [A f [B] Reaction : A + A + B products �
�
Reaction: A + B + C ---+
�
products
Rate = k [A] [B] [C]
The molecularity of an elementary reaction that is written in the reverse direction also follows from its stoichiometry. The products have become reactants. Forward reaction:
A+B
Reverse reaction:
P
+
Q
�
+
P+Q+R
R
�
A+B
Forward rate = k [A] [B] Reverse rate = k' [P] [Q] [ R]
SG-161
Note:
•
•
The prime symbol, " is used here and in the following sections to denote a rate constant written for a reverse elementary reaction. The same symbol is used in Section 1 3 . 1 0 to denote a rate constant for the same reaction (elementary or otherwise) measured at a different temperature. Do not confuse these two very different meanings.
Rate laws from reaction mechanisms �
The time evolution of each reactant, intermediate, and product may be determined by integrating the rate laws for the elementary reactions in a proposed reaction mechanism. Recall that a successful mechanism must account for the stoichiometry of the overal l reaction as wel l as the observed rate law.
�
Mathematical solution of kinetic equations often requires numerical methods for solving simultaneous differential equations. In a few cases, exact analytical sol utions exist.
�
For multistep mechanisms, approximate methods are often used to determine a rate law consistent with a proposed reaction mechanism. If the derived rate law is in agreement with the experimental one, the plausibil ity of the mechanism is tested by further experimentation, if possible.
Approximate methods for determining rate laws �
Characteristics of reactions commonly encountered allow for the determination of rate laws with the use of three types of approximations.
�
The rate-determining step approximation is made to determ ine a rate law for a mechanism i n which one step occurs at a rate substantially slower than any others. The slow step is a bottleneck, and the overal l reaction rate cannot be larger than the slow step. The rate law for the rate-determining step is written first. If a reaction intennediate appears as a reactant in this step, its concentration term must be el iminated from the rate law. The final rate law only has concentration terms for reactants and products.
�
The pre-equilibrium condition describes situations in which reaction intermediates are formed and removed in steps prior to the rate-determining one. If the formation and removal of the intermediate in prior steps is rapid, an equilibrium concentration is established. If the intermediate appears in the rate-determining step, its relatively slow reaction in that step does not change its equilibrium concentration. The pre-equi librium condition is sometimes called afast equilibrium.
�
The steady-state approximation is a more general method for solving reaction mechanisms. The net rate of formation of any intermediate in the reaction mechanism i s set equal to O. An intermediate is assumed to attain its steady-state concentration instantaneously, decay ing slowly as reactants are consumed. An expression is obtained for the steady-state concentration of each intermediate in terms of the rate constants of elementary reactions and the concentrations of reactants and products. The rate law for an elementary step that leads directly to product formation is usually chosen. The concentrations of all intermediates are removed from the chosen rate law, and a final rate law for the formation of product that reflects the concentrations of reactants and products is obtained.
�
Often, a predicted rate law does not quite agree with the experimental one. Reaction conditions that modify the form of the rate law predicted by the mechanism may lead to final agreement (see Section 1 3 . 1 5 for an example). I f complete exploration of conditions fai Is to reproduce the experimental result, the proposed reaction mechanism is rejected.
SG-1 62
1 3.9 Chain Reactions •
Chain reaction �
Series of linked elementary reactions that prop agate in chain cycles
�
A reaction intermediate is produced in an initiation step.
�
Reaction intermediates are the ch ain carriers, which are often radicals.
�
In one propagation cycle, a reaction intermediate typically reacts to form a different intermediate, which in turn reacts to regenerate the original one.
�
In each cycle, some product is usually formed.
�
The cycle is eventual ly broken by a termination step that consumes the intennediate.
�
In ch ain branching, a chain carrier reacts to fonn two or more carriers in a single step.
�
Chain branching often leads to chemical explosions.
1 3 . 1 0 Rates and Equ i librium •
Elementary reaction �
At equilibrium , the forward and reverse reaction rates for an elementary reaction are equal. For a given elementary reaction at equil ibrium, A + B Forward rate = k[A][B] = reverse rate = k' [P][Q]
�
=
P + Q.
The equilibrium constant for an elementary reaction is equal to the ratio of the rate constants for the forward and reverse reactions. [ P][Q] (reverse rate)Ik' = k I n th e ab ove examp Ie, K = [A][B] (forward rate)lk k' ---
=
-
for any elementary reaction
•
Equilibrium constants from reaction mechanisms �
Consider an overal l reaction with a multistep mechanism in which the rate constants for the elementary reactions are kJ , k2 , k3 , . . . in the forward direction and kJ', k2" k3', . . . in the reverse direction.
�
If each elementary step is in equilibrium, the overal l reaction is obtained by summing the elementary reactions, and the eq uili brium constant for the overall reaction is obtained by mUltiplying the equilibrium constants for the individual steps. For the reaction at equi l ibrium, the above procedure gives
�
K=
•
kJ
k2
k3
kJ'
k 2'
k)
-x-x-x
Temperature dependence of the equilibrium constant of an elementary reaction �
Follows from the temperature dependence of the ratio of the rate constant for the forward reaction to that of the reverse reaction, with both expressed in Arrhenius form
�
If the reaction is exothermic, the activation energy in the forward direction is smal ler than in the reverse one (see the two figures in the Study Guide in Section 1 3 . 1 3 ).
SG-J63
�
�
The rate constant with the larger activation energy increases more rapidly as temperature increases than the one with smal ler energy.
�
As temperature is increased, k' is predicted to increase faster than k and the equilibrium constant decreases.
�
Reactants are favored.
I f the reaction is endothermic, the activation energy is larger in the forward direction than in the reverse one. � �
The rate constant with the larger activation energy increases more rapidly as temperature increases than the one with smaller energy. As temperature is increased, k is predicted to increase faster than k' and the equilibrium constant
increases. �
Products are favored.
MODELS OF REACTIONS (Sections 13.11-13.13) 13.11 The Effect of Tem perature •
•
Qualitative aspects �
A change in temperature results in a change of the rate constant of a reaction, and, therefore, its rate.
�
For most reactions, increasing the temperature results in a larger value of the rate constant.
�
For many reactions in organic chemistry, reaction rates approximately double for a I aoc increase in temperature.
Quantitative aspects �
The temperature dependence of the rate constant of many reactions is given by the Arrhenius equation, shown here in two forms:
Exponential form:
� �
Logarithmic form:
In k =lnA-
E a
-
RT
I R is the universal gas constant (8.3 1 4 5 1 J·K- ·mor l ).
T is the absolute temperature (K). �
The two constants, A and Ea, cal led the Arrhenius parameters, are nearly independent of temperature.
�
The parameter A is the pre-exponential factor. The parameter Ea is the activation energy. The two parameters depend on the reaction being studied.
�
The Arrhenius equation applies to all types of reactions, gas phase or in solution.
SG-164
•
Obtaining A rrhenius parameters ---*
Plot In k VS. liT. If the plot displays straight-l ine behavior, the slope of the l ine is -E.lR and the intercept is InA. Note that the intercept at ( liT = 0) impl ies T = 00.
---*
The parameter A always has a positive value. The larger the val ue ofA, the larger is the value of k.
---*
The parameter Ea almost always has a positive value. The larger the value of Ea, the smaller the value of k. If Ea > 0, the value of k increases as temperature increases.
---*
The rate constant increases more quickly with increasing temperature if Ea is l arge.
---*
If Ea and k are known at a given temperature, T, the logarithmic fonn of the Arrhenius equation can be used to calculate the rate constant, k', at another temperature, T'. Or alternatively, a value of Ea can be estimated from values of k' at T' and k at T . In � k
•
Ea
R
(�_J...) T
T'
Units of the Arrhenius parameters ---*
A has the same units as the rate constant of the reaction.
---*
Ea has units of energy, and values are usual ly reported in klmor .
'
Note:
•
=
3 An equivalent S I unit for the l iter is dm . Rate constants and Arrhenius pre-exponential factors 3 3 i ncorporating volume units are sometimes reported using dm or cm instead of l iters. Those i ncorporating amount units are sometimes reported in units of molecules i nstead of moles. For ' example, a second-order rate constant of 1 . 2 x 1 0 " L· mor ·s- ' is equivalent to 3 3 ' ' 10 ' " ' 1 .2 x 1 0 dm ·mor · s- or 2.0 x 1 0- cm ·molecule- · s- .
Selected Arrhenius Parameters
---*
See Table 1 3 .4 in the text.
1 3 . 1 2 Collision Theory •
•
•
Collision theory ---*
Model for how reactions occur at the molecular level
---*
Applies to gases
---*
Accounts for rate constants and the exponential form of the Arrhenius equation
---*
Reveal s the significance of the Arrheni us parameters A and Ea
Assumptions of collision theory ---*
Molecules must collide in order to react.
---*
Only collisions with kinetic energy in excess of some minimum value, Emili, lead to reaction.
---*
Only molecules with the correct orientation with respect to each other may react.
Total rate of collisions in a gas mixture ---*
Determined quantitatively given from the kinetic model of a gas
SG-J65
�
Rate of coll ision = total number of collisions per unit volume per second
I Rate of col l ision a=
=
avreIN/[A][B]
I
collision cross-section (area a molecule presents as a target during col lision) Larger molecules are more likely to coll ide with other molecules than are small er ones.
vrel = mean speed at which molecules approach each other in a gas �
For a gas mixture of A and B with molar masses MA and MB, respectively,
vrel -
_
_
�8RT 7tf.!
Molecules at high temperature col lide more often than at low temperature.
NA = Avogadro constant [A] and [B] are the molar concentrations of A and B, respectively. •
Total rate of reaction in a gas mixture �
Rate of reaction = number of coll isions that lead to reaction per unit volume per second Rate of reaction
=
Pavrel N/[A][B]
x
e -Em;,' RT
=
minimum kinetic energy required for a col lision to lead to reaction Collisions with energy less than the minimum do not lead to reaction.
Emili
T e -Em;,'R = fraction of collisions with at least the energy Emili Derived from the Boltzmann distribution (see Fig. 1 3 .26 in the text)
P
= steric factor < 1 Reactive collisions often require preferred directions of approach.
•
Reaction rate constant �
Rate constant
=
k
rate of reaction
[A ][B] 2e ; T k - Pa Vrel N -Em ,' R
=
------
A
�
Comparison with the Arrhenius equation in exponential form, k Ea"'" Emin and A "'"
Pa vrel N/
=
Ae
-.!...... RT
T 1 2,
the model predicts that the pre-exponential factor and Because the speed is proportional to activation energy both have a slight temperature dependence. •
S ummary �
According to collision theory, reaction occurs only if reactant molecules collide with a kinetic energy equal to or greater than the Arrhenius activation energy.
SG- J 66
The reaction rate constant i ncreases with the size and speed of the colliding species (molecules). �
The steric factor accounts for collisions that lead to no reaction because the orientation of the molecules during such a collision is unfavorable. See Table 1 3 . 5 in the text for some P values.
13.13 Transition State Theory •
•
Transition state theory (also often called Activated Complex Theory) �
Applies to gas phase and solution reactions
�
Reacting molecules col lide and distort.
�
During an encounter, the kinetic energy, EK, of reactants decreases, the potential energy increases, the original chemical bonds lengthen, and new bonds begin to form.
�
Products may form if EK for the collision is equal to or greater than Ea and if the reactants have the proper orientation.
�
Reactants reform if EK is less than Ea, regardless of orientation.
�
If EK is equal to or greater than Ea, the molecules form either reactants or products. The lowest kinetic energy that may produce products is equal to Ea. An activated complex is a combination of the two reacting molecules that are at the transition point between reactants and products.
Reaction profile �
Shows how energy changes as col l iding reactants form a transition state or an activated complex and then products along the minimum energy pathway. (An analogous process is hiking through a mountain range and choosing the pathway of lowest elevation through it.)
�
Progress of reaction refers to the relevant spatial coordinates of the molecules that show bond breaking as reactants approach the energy maximum and bond formation as products form.
�
Reaction profiles for exothermic and endothermic reactions are shown in the schematic drawings on the next page. Exothermic reaction Endothermi c reaction
I
Activated complex Activation energy.
Ea
/ Activation barrier
Reactants
I
"
/ Activation
'
Activation energy, Ea
� Q)
i
.
barrier
Products
c: Q)
-;:; .� 13
Q)
Products
0..
Progress 0 f reaction •
Activated complex
Reactants
Progress of reaction
Potential energy surface �
Three-dimensional plot in which potential energy is plotted on the z-axi s, and the other two axes correspond to interatomic distances. See Fig. 13.31 in the text for a potential energy surface.
�
Similar to a mountain pass, the saddle point corresponds to an activated complex with the minimum activation energy, Ea, required for reactants to form products.
SG- 1 67
IMPACT ON MATERIALS AND BIOLOGY: ACCELERATING REACTIONS (Sections 13.14-13.15) 1 3 . 1 4 Catalysis •
Nature of a catalyst -t
Accelerates a reaction rate without being consumed
-t
Does not appear in the overal l reaction
-t
Lowers the activation energy of a reaction by changing the pathway (mechanism)
-t
Accelerates both the forward and reverse reaction rates
-t
Does not change the final equil ibrium composition of the reaction mixture
-t
May appear in the rate equation, but usually does not
•
H omogeneous catalyst
-t
Present in the same phase as the reactants
•
Heterogeneous catalyst
-t
Present in a different phase from the reactants
•
Poisoning a catalyst -t
Catalysts can be poisoned or inactivated.
-t
For a heterogeneous catalyst, irreversible adsorption of a substance on the catalyst' S surface may prevent reactants from reaching the surface and reacting on it.
-t
For a homogeneous catalyst, a reactive site on a molecule may be similarly "poisoned" by the binding of a substance directly to the site or nearby either in a permanent or reversible way. The properties of the site are significantly altered to prevent catalytic action.
1 3 . 1 5 Livin g Catalysts: Enzymes •
Enzyme, E -t
A biological catalyst
-t
Typically, a large molecule (protein) with crevices, pockets, and ridges on its surface (see Fig. 1 3 .39 in the text)
-t
Catalyzes the reaction of a reactant known as the substrate, S, that produces a product, P : E+S - E+P
-t •
•
Capable of increasing the rate of specific reactions enormously
I nduced -fit mechanism -t
Enzymes are highly specific catalysts that function by an induced-fit mechanism in which the substrate approaches a correctly-shaped pocket of the enzyme, known as the active site.
-t
On binding, the enzyme distorts to accommodate the substrate, allowing the reaction to occur.
-t
The shape of the pocket determines which substrate can react.
-t
The induced-fit mechanism is a more real istic version of the lock and key mechanism, which assumes that the substrate and enzyme fit together like a lock and key.
M ichaelis-Menten mechanism of enzyme reaction -t
A two-step mechanism that accounts for the observed dependence of the reaction rate of many enzyme-substrate reactions
SG- 1 6 8
�
For an enzyme, E, and a substrate, S, the overal l reaction and the M ichaelis-Menten mechanism are E
+
S - E
Step I E
+
S
Overal l reaction: Mechanism:
+
P
ES
=
Step 2 ES - E + P �
J n step I , the enzyme and substrate form an enzyme-substrate complex, E S , that can dissociate to reform substrate (reverse of step I) or decay to form product P (step 2). The rate of formation of product is given by step 2.
Rate of formation of P �
=
k2[ES]
The steady-state approximation may be applied to the intermediate, ES, and its concentration i s kl[ E ] [ S] [ES] k if + k 2 =
�
Because the free enzyme concentration may be substantially diminished during the reaction, it is customary to express the rate in terms of the total enzyme concentration, [E]o [E] + [ES]. Then, [E] is replaced by [E]o - [ES] in the steady-state result, leading to =
KM �
=
kif + k 2 . ' hae I'IS constant. IS known as th e MlC kl
At low concentration of substrate, [S]
«
KM,
and the rate law is given by
k2 [ E1o [S] Rate of formation of P = KM
The rate varies linearly with the concentration of substrate in this region.
Physically, in this region, most enzyme sites lack substrate molecules, so doubling the substrate concentration doubles the number of occupied sites and the rate as well. �
At high concentration of substrate, [S]
Rate of formation of P
=
»
k 2 [ E] o [S] [S]
K M and
=
the rate law becomes
k2 [ E]0
In this region, the rate is independent of the substrate concentration (see Fig. 1 3 .4 1 in the text).
Physically, all the enzyme sites are occupied with substrate molecules, so adding substrate has no effect on the rate.
SG-J69
•
Summary �
Enzymes are large proteins that function as biological catalysts.
�
Substrate molecules (reactants) bind to active sites on enzyme molecules.
�
The model of enzyme binding and action is cal led the induced-fit mechanism.
�
S ubstrate molecules undergo reaction at the active site to form product molecules.
�
Product molecules are released and the active site is restored.
SG-170
Chapter 14
THE ELEMENTS:
THE FIRST FOUR MAIN GROUPS
PERIODIC TRENDS (Sections 14.1-14.2) 1 4 . 1 Atomic P roperties •
Atomic p roperties mainly responsible for the properties of an element: ---+
•
Atomic and ionic radii ---+
•
Typically decrease from left to right across a period and increase down a group
First ionization energy ---+
•
Atomic radius; first ionization energy; electron affinity; electronegativity; polarizability
Typically increases from left to right across a peri od and decreases down a group
---+
Increase across a period: Caused by increasing attraction of nuclear charge to electrons in the same valence shel l
---+
Trend down a group: Concentric shells of electrons are progressively more distant from the nucleus.
E lectron affinity (Ee.) Measure of the energy released when an anion is formed. The greater the Eea, the more stable the anion formed.
---+
---+ ---+
•
E lectronegativity (X) ---+ Measure of the tendency of an atom to attract bonding electrons when part of a compound
---+
Typically increases from left to right across a period and decreases down a group
---+
A useful guide to the type of bond an element may form A predictor of the polarity of chemical bonds between nonmetallic elements
---+
•
Polarizability ---+ A measure of the ease of distorting an atom's electron cloud ---+
---+
---+
---+
•
Highest values of Eea are found at the top right of the periodic table. Note that CI has a higher electron affinity than F, but F has a higher electronegativity. Elements with high values of Eea are present as anions in compounds with metallic elements and commonly have negative oxidation states in covalent compounds with other nonmetallic elements. For example, F has an oxidation state of - I in sulfur hexafluoride, SF 6.
Typically decreases across a period (left to right) and increases i n a group (top to bottom) Greatest for the more massive atoms in a group Anions are more polarizable than parent atoms. Diagonal neighbors in the periodic table have similar polarizability values and a similar amount of covalent character in the bonds they form.
Polarizing power 2 3 Associated with ions of small size and high charge, such as A1 + or Be +
---+
---+
Cations with high polarizing power tend to have covalent character in bonds formed with highly polarizable anions. Example: Beb
1 4.2 Bon d i n g Trends •
Elements bonded to other elements ---+
Period 2 elements: Valence is determined by the number of valence-shell electrons and the octet rule.
SG- 1 7 1
•
�
Period 3 or higher (Z;:::: 1 4) : The valence of an element may be increased through octet expansion allowed by access to empty d-orbitals. Two examples are PCls and SF 6 .
�
Inert-pair effect: Elements at the bottom of the p block may display an oxidation number two less than the group number suggests. For example, lead has an oxidation number of +2 in PbO and +4 in Pb02 .
Pure elements With low ionization energies tend to have metallic bonds, for example, Li(s) and Mg(s)
� �
With high ionization energies are typically molecular and form covalent bonds, for example, F2(g) and 02(g)
�
In or near the diagonal band of metal l oids (intennediate ionization energies and three, four, or five valence electrons), they tend to form network structures in which each atom is bonded to three or more other atoms. Example: The network structure of black phosphorus, in which each P atom is bonded to three others
Chemical Properties: Hydrides •
Binary hyd rides �
B inary compounds with hydrogen (hydrides)
�
Formed by al l main-group elements-except the noble gases, and possibly In and Tl-and several d-block elements Properties show periodic behavior.
�
�
•
Formulas directly related to the group number Examples: S i H4, N H3, H2S, and HCl in Groups 1 4, 1 5, 1 6, and 1 7, respectively
Saline (or salt-like) hydrides �
Formed by all members of the s block except Be
�
Ionic salts of strongly electropositive metals with hydrogen, present as the hydride ion, H -
� �
White, high-melting solids with crystal structures resembling those of the corresponding halides Made by heating the metal in hydrogen, for example, 2 Na(s) + H2(g) 2 NaH(s)
�
React readily with water (H - is a strong base) forming a basic sol ution and releasing H2(g):
�
H -(aq) + H20(l) •
�
H2(g) + OH -(aq)
Metallic hyd rides �
�
Black, powdery, electrically conducting solids formed by certai n d-block elements Release hydrogen when heated or when treated with acids. A typical reaction in acid is: CuH(s) + H30+(aq) Cu\aq) + H2(g) + H20(l) �
•
Molecular hydrides � Formed by nonmetals and consist of discrete molecules �
� �
Volatile, many are Br0nsted acids or bases. Gaseous compounds include the base NH3(g), the hydrogen hal ides HX(g), X= F, C I , Sr, and I, and the lighter hydrocarbons, such as CH4 and C2H4 (ethene). Liquids include water and heavier hydrocarbons, such as benzene, C6H6 , and octane, C sH ls .
Chemical Properties: Oxides •
Binary oxides �
�
Formed by all the mai n-group elements except the noble gases Main-group binary oxides of metals are basic. Example: MgO
SG - I 72
� �
Main-group binary oxides of nonmetals are acidic. E xample: S 03 Exception: AI203 is amphoteric. Other amphoteric oxides incl ude BeO, Ga203 , S n02 , and Pb02 . Soluble ionic oxides: formed from elements on the left side of the periodic table
�
Oxides with low melting points, often gaseous: formed from elements on the right of the p block
�
•
Metallic elements and metalloids �
I onic oxides: formed from metal l ic elements with low ionization energies E xample: K20
�
Amphoteric oxides: formed from metal lic elements with intermediate ionization energies (Be, B, and AI) and from metalloids Amphoteric oxides: do not react with water but dissolve in both acidic and basic solutions
�
•
I nsoluble oxides with high melting points: formed from elements in the left of the p block
Non metallic elements �
�
Many oxides of nonmetals are gaseous molecular compounds. Most can act as Lewis acids. They form acidic solutions in water and are cal led acid anhydrides. Examples: The acids HN03 and H2S04 are derived from the oxides N20S and S03 , respectively. Oxides that do not react with water are calledformal anhydrides of acids. The formal anhydride is obtained by removing the elements of water (H, H, and 0) from the molecular H .. , formula of the acid, for example, the formal anhydride of acetic acid, CH3COOH, is c=c=o: / CH2CO (ketene), whose Lewis structure is given on the right. H
HYDROGEN (Sections 14.3-14.4) 14.3 The Element •
Hyd rogen � � �
•
One valence electron but few simi larities to the alkal i metals A nonmetal that resembles the halogens but has some different, unique properties Does not fit clearly into any group and is therefore not assigned to any group
Commercial p rod uction of hydrogen � Formed I) as a by-product of the petroleum refining and 2) by the electrolysis of water
�
I n the refining process, a Ni-catalyzed re-forming reaction, produces products called synthesis gas: CH4(g) + H20(g)
�
CO(g) + 3 H2(g)
In a second step, a shift reaction employing an Fe/Cu catalyst yields hydrogen gas: CO(g) + H20(g) � •
�
CO2(g) + H2(g)
Electrolysis afwater uti lizes a salt solution to improve conductivity: 2 H20(l)
�
2 H2(g) + 02(g)
Laboratory production and uses of hyd rogen Produced by reaction of an active metal such as Zn with a strong acid such as HCI: 2 Zn +(aq) + H2(g) + 2 H20(I) Zn(s) + 2 H30+(aq)
�
�
�
Produced by reaction of a hydride with water: 2 CaH2(s) + 2 H20(l) � Ca +(aq) + 2 0 H -(aq) + 2 H2(g)
SG-1 73
1 4.4 Compounds of Hydrogen •
U nusual properties of hyd rogen It can form a cation, H +, or an anion, hydride, H -.
� � •
It can form covalent bonds with many other elements because of its intermediate X value of 2.2.
The hyd ride ion, H � Large, with a radius ( 1 54 pm) between those of F- ( 1 3 3 pm) and cr ( 1 8 1 pm) ions �
->
Highly polarizable, which adds covalent character in its bonds to cations The two electrons in H - are weakly attached to the single proton, and an electron is easily lost, making W a good reducing agent.
�
Saline hydrides are very strong reducing agents. The reduction half-reaction and standard reduction potential are H2(g) + 2 e- - 2 H -(aq) and EO -2.25 V, respectively. The alkali metals are also strong reducing agents with reduction potentials simil ar to the hydride value. =
� � •
Hydride ions in saline hydrides reduce water upon contact: W(aq) + H20(I) - H2(g) + OH -(aq).
Hydrogen bonds � �
Relatively strong intermolecular bonds Fonned by hydrides of N, 0, and F
�
Can be understood in terms of the coulombic attraction between the partial positive charge on a hydrogen atom and the partial negative charge of another atom
�
The H atom is covalently bonded to a very electronegative N, 0, or F atom, giving it a partial positive charge.
�
The partial negative charge is often the lone-pair electrons on a different N, 0, or F atom.
�
A hydrogen bond is represented as three dots between atoms not covalently bonded to each other: [ O-H . . · : O].
�
A hydrogen bond is typically about 5% as strong as a covalent bond between the same types of atoms. Example: Hydrogen bonding in HF . . . H-F: . . . H-F: ... H-F: .
.
.
' 0
' 0
' 0
GROUP 1: THE ALKALI METALS (Sections 14.5-14.7) 14.5 The G ro u p 1 Elements •
Alkali metals �
•
f Have the valence electron configuration ns , where n is the period number
Preparation Pure alkali metals are obtained by electrolysis of the molten salts. � Exception: K, prepared by reaction of molten KCI and Na vapor at 750°C :
�
KCI(I) + Na(g) - NaCI(s) + K(g) This reaction is driven to form products by the condensation of K(g) because the equilibrium constant is unfavorable.
SG-174
•
•
Properties of alkali metals �
Are dominated by the ease with which their single valence electron is lost
�
Soft, silver-gray metals with low melting points, boiling points, and densities
�
Melting and boi ling points decrease down the group. Cs (melting point 28°C) is very reactive and is transported in sealed ampoules. Fr is very radioactive and l ittle is known about its properties.
�
All alkali metals are highly reactive.
Applications �
Li metal is used in the rechargeable lithium-ion battery, which holds a charge for a long time.
�
Li is also used in thermonuclear weapons.
1 4.6 Chem ical P roperties of the Alkali Metals •
Alkali metals �
Are strong reducing agents (low first ionization energies lead to their ease of oxidation) Example: Reduction of Ti (IV) : TiCl4 + 4 Na(l) 4 NaCl(s) + Ti(s) �
�
Reduce water to form basic solutions, releasing hydrogen gas E xample: 2 N a(s) + 2 H20(l) 2NaOH(aq) + H2(g) �
�
Dissolve in NH3(1) to yield solvated electrons, which are used to reduce organic compounds React directly with almost all nonmetals, noble gases excluded
�
However, only one alkali metal , Li, reacts with nitrogen to form the nitride (LbN).
�
•
Formation of compounds containing oxygen �
�
�
Li forms mainly the oxide Li20. Na forms a pale yel l ow peroxide Na202 . K, Rb, and Cs form mainly a superoxide, for example, K02 .
1 4. 7 Compounds of Lit h i u m , Sod i u m , and Potass i u m •
Lithium � �
�
�
•
Differs signjficantly from the other Group I elements, a common occurrence for elements at the head of a group. For Group I, the differences originate in part from the sma l l size of Li +. Cations have strong polarizing power and a tendency toward covalency in their bonding. Has a diagonal relationship with Mg and is found in the minerals of Mg Is found in compounds with oxidation number + I ; its compounds are used in ceramics, l ubricants, and medicine. Examples: Lithium carbonate is an effective drug for manic-depressive (bipolar) behavior. Lithium soaps, which have higher melting points than sodium and potassium soaps, are used as thickeners in l ubricating greases for high-temperature applications.
Sod ium compounds �
Are important in part because they are plentiful, inexpensive, and water soluble
�
NaCl is mined as rock salt and is used in the electrolytic production of C h (g) and NaOH.
�
�
NaOH is an inexpensive starting material for the production of other sodium salts. Na2 C03· 1 0 H20 was once used as washing soda.
�
NaH C03 is also known as sodium bicarbonate, bicarbonate ofsoda, or baking soda.
�
Weak acids [lactic acid (milk), citric acid (lemons), acetic acid (vinegar)] react with the hydrogen carbonate anion to release carbon dioxide gas: HC03-(aq) + HA(aq) A-(aq) + H20(l) + CO2(g). �
SG- 1 7S
� •
Double-acting baking powder contains a solid weak acid as well as the hydrogen carbonate ion.
Potassium compounds �
Are usually more expensive than the corresponding sodium compounds
�
Use may be j ustified because potassium compounds are generally less hygroscopic than sodium compounds. The K+ cation is larger and is less strongly hydrated by H20 molecules.
�
KN03 releases 02(g) when heated and is used to help matches ignite. KN03 is also the oxidizing agent in black gunpowder (75% KN03, 1 5% charcoal , and 1 0% sulfur). KCI is used directly in some ferti lizers as a source of potassi um, but KN03 is required for some crops that cannot tolerate high chloride ion concentrations.
� �
M i neral sources of potassi um are sylvite, KCI, and carnallite, KCI· M gCh·6 H20.
GROUP 2: THE ALKALINE EARTH METALS (Sections 14.8-14.10) 1 4 .8 •
Alka line earth metals �
•
The G ro u p 2 Elements
Have the valence electron configuration ni, where n is the period number
Preparation �
Pure metals: obtained by electrolysis of the molten salts (all) or by chemical reduction (except Be)
�
Ca(s), Sr(s), and Ba(s) may be obtai ned by reduction with AI(s) in a variation of the thermite reaction, for example, 3 CaO(s) + 2 AI(s) AI203(s) + 3 Ca(s). Be occurs in nature as beryl, 3 BeO· Ah03·6 Si02 . 2 M g i s found i n seawater as M g +(aq) and i n the mineral dolomite, CaC03 · M gC03 . �
�
� •
Properties � � �
S i l ver-gray metal s with much higher melting points, boil i ng points, and densities than those of the preceding alkali metals in the same period Melting points decrease down the group with the exception of Mg, which has the l owest value. The trend in boiling point i s irregular. A l l Group 2 elements except Be reduce water. 2 E xample: Sr(s) + 2 H20(l) Sr \aq) + 2 0H -(aq) + Hig) �
�
�
�
->
•
Be does not react with water. Mg reacts with hot water, and Ca reacts with cold water. Be and Mg do not dissolve in HN03 because they develop a protective oxide fi lm. Mg burns vigorously in air because it reacts with 02(g) , N2(g) , and CO2(g). Reactivity of the Group 2 metals with water and oxygen increases down the group, j ust like the Group I metals.
Applications of beryllium �
�
� �
Has a low density, making it useful for missile and satellite fabrication Used as a "window" for x-ray tubes Added in sma l l amounts to Cu to increase its rigidity Be/C u alloys are used in nonsparking tools (oil refineries and grain elevators) and for nonmagnetic parts that resist defonnation and corrosion (electronics industry).
SG- 1 76
•
Applications of magnesium �
•
�
Light and very soft, but its alloys have great strength and are used in applications where lightness and toughness are desired (airplanes)
�
Limitation in widespread usage is related to its expense, low melting temperature, and difficulty in machining.
Applications of calcium, strontium, and barium �
1 4.9 •
Group 2 metal s can be detected in burning compounds by the colors they give to flames. Ca burns orange-red, Sr crimson, and Ba yellow-green. F ireworks are made from their salts. Com pou nds of Berylli u m a n d Magnesiu m
Bery llium � � � �
�
•
A silver-white metal protected from air oxidation by a white oxide fi l m, making it appear dul l gray
Differs significantly from the other Group 2 elements, as is typical of an element at the head of its group 2 The difference originates in part from the small size of the Be + cation, which has strong polarizing power and a tendency toward covalency in its bonding. Has a diagonal relationship with AI. It is the only member of the group that, like AI, reacts in 2 aqueous NaOH. In strongly basic solution, it forms the belyllate ion, [Be(OH)4] -. Compounds are extremely toxic. 2 The high polarizing power of Be + produces moderately covalent compounds, whi le its small size allows no more than four groups to be attached, accounting for the prominence of the tetrahedral BeX4 unit.
�
BeH2(g) and BeCh(g) condense to form solids with chains of tetrahedral BeH4 and BeCl4 units, respectively.
�
In BeC12, the Be atoms act as Lewis acids and accept el ectron pairs from the CI atoms, forming a chain of tetrahedral BeCl4 units in the solid.
Magnesium � �
� � �
�
�
�
More metallic properties than Be Compounds are primarily ionic with some covalent character. MgO and Mg3N2 are formed when magnesium burns in air. MgO dissolves very slowly and only slightly in water. It can withstand high temperatures (refractory) because it melts at 2800°C. MgO conducts heat wel l, but electricity only poorly, and is used as an insulator in electric heaters. Mg(OH)2 is a base. It is not very sol uble in water but forms instead a white coll oi dal suspension, which is used as a stomach antacid cal led milk of magnesia. A side effect of stomach acid neutralization is the formation of MgCI2, which acts as a purgative. MgS04, (Epsom Salts) is also a common purgative. Chlorophyll is the most important compound of Mg. Magnesium also plays an important role in energy generation in living cel ls. It is involved in the contraction of muscles.
1 4 . 1 0 Com p o u n ds o f Calci u m •
Carbonate, oxide, and hydroxide �
�
Ca is more metallic in character than Mg, but its compounds share some sim i lar properties. CaC03 occurs naturally as chalk and limestone. Marble is a very dense form with colored impurities, most commonly, Fe cations.
SG- I 77
�
The most common fonns of pure calcium carbonate are calcite and aragonite. A l l carbonates are the fossilized remains of marine life.
�
CaO is formed by heating CaC03: CaC03(s)
�
CaO(s) + COz(g).
CaO is called quicklime because it reacts rapidly and exothermically with water: CaO(s) + HzO(l) � �
•
�
Ca
Z
\aq) + 2 0H - (aq)
Ca(OH)z , the product of the equation above, is known as slaked lime because in this fonn the thirst for water has been quenched, or slaked. An aqueous solution of Ca(OH)z, which is only slightly soluble in water, is called lime water, which is used as a test for carbon dioxide: Ca(OH)z(aq) + COz(g) � CaC03(s) + HzO(I).
Carbide, sulfate, and phosphate �
CaO may be converted into calcium carbide: CaO(s) + 3 C(s)
�
Ca i s found in the rigid structural components of living organi sms, either as CaC03 (shellfish shel l s) or as Ca3(P04)Z (bone).
�
Tooth enamel is a hydroxyapatite, CaS(P04)30H, which is subject to attack by acids produced when bacteria act on food residues. A more resistant coating is fonned when the OH - ions are replaced by F - ions. The resulting mineral is calledjluoroapatite:
�
�
CaCz(s) + CO(g).
Calcium sulfate dihydrate (gypsum), CaS04·2 HzO, is used in construction materials.
CaS(P04)30H(S) + F -(aq)
�
CaS(P04)3F(S) + OH -(aq)
Tooth enamel is strengthened by the addition of fluorides (fluoridation) to drinking water, and by the use of fluoridated toothpaste containing tin(I!) fluoride, SnFz , or sodium monofluorophosphate (MFP), NazFP03 .
GROUP 13 / 111 : THE BORON FAMILY (Sections 14.11-14.14) 1 4. 1 1 The G roup 1 3 / 111 E lements � •
Have the valence electron configuration ni np '
Preparation of boron �
Mined as the hydrates borax and kernite, NazB407 • xHzO, with x = 1 0 and 4, respectively
�
The ore is converted to the oxide, B Z03, and extracted in its amorphous elemental form by reaction of the oxide with Mg(s): B Z 03(S) + 3 Mg(s) 2 B(s) + 3 MgO(s). A purer fonn of B is obtained by reduction of volatile BBr3(g) or BCb(g) with Hz(g).
�
�
•
Preparation of aluminum �
�
�
M i ned as bauxite, AhO} · xH20, where x ranges to a maximum of 3 The ore, which contains considerable iron oxide impurity, is processed to obtain alumina, AbO) . Al is then extracted by an electrolytic process (Hall process). A lzO) is mixed with cryolite, Na3AIF6 , to lower the melting point from 2050°C (alumina) to 950°C (mixture). The half-cel l and cell reactions are: 3 Cathode: AI +(melt) + 3 e AI(l) 2 COz(g) + 4 e2 0 -(melt) + C(s,gr) Anode: z 3 4 AI(l) + 3 C02(g) 4AI +(melt) + 6 0 -(melt) + 3 C(s,gr) Cel l : -
�
�
�
SG- I 78
•
Properties of boron �
A nonmetal in most of its chemical properties
�
Has acidic oxides and forms a fascinating array of binary molecular hydrides It exists in several allotropic forms in which B atoms attempt to share eight electrons, despite the sma l l size of B and its ability to contribute only three electrons. In this sense, B is regarded as electron deficient.
�
One common form of B is a gray-black, nonmetallic, high-melting solid.
�
�
•
The element is attacked by only the strongest oxidizing agents.
�
Another common form is a dark brown powder with a structure (icosahedral with 20 faces) that has clusters of 1 2 atoms as shown on the right. The bonds create a very hard, three-dimensional structure.
Properties of aluminum �
�
�
�
Has a l ow density, but is a strong metal with excellent electrical conductivity Easily oxidized, but its surface is passivated by a protective oxide film If the thickness of the oxide layer is increased electrolytically, anodized aluminum results. AI is amphoteric, reacting with both nonoxidizing acids and hot aqueous bases. The reactions are: 3 2 AI(s) + 6 W(aq) - 2 A I \aq) + 3 H2(g) 2 AI(s) + 2 0H -(aq) + 6 H20(l)- 2 AI(OH)4-(aq) + 3 H2(g)
•
Applications � �
•
•
B fibers are incorporated into plastics, forming resilient material stiffer than steel and lighter than A I(s). Boron carbide is similar in hardness to diamond, and boron nitride is similar in structure and mechanical properties to graphite, but, unlike graphite, boron nitride does not conduct electricity.
�
AI has widespread use in construction and aerospace industries. Because it is a soft metal , its strength is improved by alloy formation with Cu and Si.
�
Because of its high electrical conductivity, A I is used in overhead power lines. Its high negative electrode potential has l ed to its use in fuel cells.
Gallium �
Produced as a by-product in the production of AI
�
Common doping agent for semiconductors
�
GaAs, gallium arsenide, is used in light emitting diodes (LEDs).
Thallium �
Poisonous heavy metal sometimes used as a rat poison
1 4 . 1 2 G ro u p 1 3 /111 Oxides •
Boron oxide �
Boron, like most nonmetals, has acidic oxides. Hydration of B 203 yields boric acid, H3B03 or B(OHh in which B has an incomplete octet and can therefore act as a Lewis acid: (OH)3B + : OH2 - (OH)3B-OH2
�
The complex formed is a weak monoprotic acid (pKa = 9. 1 4). H3B03 is a mild antiseptic and pesticide. It is also Llsed as a fire retardant in home insulation and clothing.
SG-179
�
The main use of H}BO}, is as a source of boron oxide, the anhydride of boric acid: 2 H3B03(S)
heat
) B203(S) + 3 H20(l)
B203, which melts at 450°C and dissolves many metal oxides, is used as a fl ux for soldering or welding. It is also used in the manufacture of fiberglass and borosilicate glass, such as Pyrex. •
Aluminu m oxide �
Ah03 , known as alumina, exists in several crystal forms, each of which is used in science and i n commerce.
�
a-Alumina is a very hard substance, corundum, used as an abrasive known as emery.
�
y-Alumina is absorbent and is used as the stationary phase in chromatography. It is produced by heating AI(OH)3 and is moderately reactive and amphoteric: 3 Ah03(S) + 6 H30+(aq) + 3 H20(l) � 2 [AI(H20)6 +](aq) Ah03(S) + 2 0W(aq) + 3 H20(l) � 2 AI(OH)daq) 3 The strong polarizing effect of the small , highly charged A 1 + ion on the water molecules surrounding 3 it gives the [AI(H20)6 +] ion acidic properties.
�
A luminum sulfate is prepared by reaction of Ah03 with sulfuric acid:
A h(S04)3 (papermaker 's alum) is used in the preparation of paper. �
Sodium aluminate, NaAI(OH)4 , is used along with Ah(S04)3 in water purification.
1 4 . 1 3 Nitrides a n d Halides •
Nitrides �
When B is heated in N H3, it forms boron nitride: 2 B(s) + 2NH3(g) � 2 BN(s) + 3 H2(g). The structure BN is simi lar to that of graphite, but with the planes of hexagons of C atoms replaced by hexagons of alternating B and N atoms.
�
BN is a fl uffy, slippery powder that can be pressed into a solid rod that is easy to machine. Unlike graphite, BN is a good insulator.
�
At high pressure, BN is converted to a diamond-l ike crystal l ine material cal led Borazon.
�
�
BN nanotubes, simi lar to those formed by C, are semiconductors. Boron trifluoride reacts with NH3 to produce the Lewis acid-base complex, F3BNH3: F3B(g) + : N H3(g) � F3B-NH3(S) Solid F 3BNH3 is stable up to about 1 25°C, where it decomposes: 4 F3BNH3(S) � BN(s) + 3NH4BF4(s)
•
H a lides � � �
Boron halides are made either by direct reaction of the elements or from the oxide. Example: B203(S) + 3 CaF2(s) + 3 H2S04(1)
heat
) 2 BF3(g) + 3 CaS04(s) + 3 H20(l)
All boron hal ides are trigonal-planar, covalently bonded, and electron-deficient, with an incomplete octet on B, as a consequence of which, they are strong Lewis acids. AICb is formed as an ionic compound by direct reaction of the elements, or of al umina with chlorine in the presence of carbon: 2 AI(s) + 3 CI2(g) � 2 AICb(s) Ah03(S) + 3 Ch(g) + 3 C(s, gr) � 2 AIC I3(s) + 3 CO(g)
SG-J80
I onic A ICb melts at 1 92°C to form a molecular liquid, AIzCI6. This dimer of AICh, formed by donation of an unshared electron pair on CI to a vacant orbital on an adjacent AI atom, contains AICI4 tetrahedral units and two bridging CI atoms, as shown on the right. --+
Aluminum halides react with water in a highly exothermic reaction.
--+
The white solid, lithium aluminum hydride, prepared from AICh is an important reducing agent in organic chemistry: 4 L i H + AICI3 LiAI� + 3 LiCI.
�
1 4 . 1 4 Boranes, Borohyd ri des, and Borides •
Boranes --+ --+
·
Compounds of boron and hydrogen, such as B2H6 and B I OH 1 4 Electron-deficient compounds with three-center, two-electron B-H-B bonds
--+
Valid Lewis structures cannot be written.
--+
Molecular Orbital theory provides a framework for understanding the delocalized behavior of an electron pair associated with three atoms, as shown j ust above.
--+
Diborane, B2H6 , is highly reactive. When heated, it decomposes to hydrogen and boron: B2H6(g)
2 B(s) + 3 H2(g)
�
It reacts with water, reducing hydrogen and forming boric acid: B2H6(g) + 6 H20(l) •
�
2 B(OHMaq) + 6 H2(g)
Borohyd rides --+
Anionic versions of boranes. The most important borohydride is B�-, as in sodium borohydride, NaBH4' which is produced by the reaction of NaH with BCh: 4 NaH + BCh
�
NaB� + 3 NaCI
The borohydride anion B H4- i s an effective and important reducing agent. --+
Diborane is produced by the reaction of sodium borohydride with boron tri fluoride: 3 BH4- + 4 BF3
•
�
3 BF4- + 4 B2H6
Borides --+
Many borides of the metals and nonmetals are known.
--+
Formulas of borides are often unrelated to the position of the two elements in the periodic table. E xamples : A 1 B2 , CaB6 , B l 3C2, B 1 2S2 , Ti3B4 ' TiB, and TiB2 .
--+
B atoms i n borides commonly form extended structures such as zigzag chains, branched chains, or networks of hexagonal rings.
GROUP 14 / IV: THE CARBON FAMILY (Sections 14.15-14.20) 1 4 . 1 5 The G ro u p 1 4 / IV Elemen ts --+
Have the valence electron configuration ni n/
--+
The lighter elements have a l l four valence electrons available for bonding.
SG-1 8 1
•
�
Heavier elements (Sn, Pb) display the inert-pair effect and exhibit oxidation numbers 2 and 4.
�
For Pb, the most common oxidation number is 2.
�
Metal l i c character increases from C to Pb.
General properties �
Carbon forms covalent compounds with nonmetals and ionic compounds with metals.
�
The oxides of C and Si are acidic.
�
Ge, a metal loid, exhibits metal lic or nonmetal lic properties, depending on the compound.
�
Sn and Pb are classified as metals, but tin has some intermediate properties.
�
Sn is amphoteric and reacts with hot, concentrated hydrochloric acid and hot alkal i : 2 Sn(s) + 2 H 30+(aq) - Sn \aq) + H2(g) + 2 H20(l) Sn(s) + 2 0H -(aq) + 2 H20(l) - Sn(OH)/-(aq) + H2(g)
�
Carbon, at the top of the group, differs greatly in its properties from the other members.
�
Carbon tends to form multiple bonds, whereas silicon does not.
�
CO2 is a molecular gas, while S iOis) (the mineral sil ica) contains networks of -O-Si-O- groups.
�
Si compounds can act as Lewis acids, whereas C compounds usually cannot. Si may expand its octet, whereas C does not.
1 4 . 1 6 The Different Forms of Carbon •
G raphite �
The thermodynamically stable allotrope of carbon (Soot contains smal l crystals of graphite.)
�
�
Produced pure commercially by heating C rods in an electric furnace for several days 2 Contains Sp -hybridized C atoms
�
Consists structural ly of large sheets of fused benzene-like hexagonal units. A n-bonding network of delocalized electrons accounts for the high electrical conductivity of graphite.
�
When certain impurities are present, graphitic sheets can slip past one another, and graphite becomes an excel lent dry l ubricant.
�
Only sol uble in a few l iquid metals
�
•
Soot and carbon black have commercial applications in rubber and inks. Activated charcoal is an important, versatile purifier. Unwanted compounds are adsorbed onto its microcrystalline surface.
Diamond �
It is the hardest substance known and an excellent conductor of heat.
�
It is an excellent abrasive, and the heat generated by friction is rapidly conducted away.
�
� �
�
Natural d iamonds are uncommon. Synthetic diamonds are produced from graphite at high pressures (> 80 kbar) and temperatures (> I 500°C) and by thermal decomposition of methane. It is soluble in liquid metal s, such as Cr and Fe, but less soluble than graphite. This solubility difference is ut ilized in synthesizing diamond at high pressure and high temperature. Carbon in diamond is sp3-hybridized, and each C atom in a diamond crystal is bonded directly to four other C atoms and indirectly i nterconnected to all of the other C atoms in the crystal through C-C single a-bonds. The a-bonding network of localized electrons accounts for the electrical insulating properties of diamond.
SG- J 82
•
Fullerenes �
Buckminsterfullerene, C60 , is a soccer-ball-like molecule first identified in 1 985. N umerous fullerenes, containing 44 to 84 carbon atoms, were discovered quickly thereafter.
�
Fullerenes are molecular electrical insulators and are soluble in organic solvents.
�
They have relatively low ionization energies and large electron affinities, so they readily l ose or gain electrons to form ions.
�
Fullerenes contain an even number of carbon atoms, differ by a C2 unit, and have pentagonal and hexagonal ring structures. Solid samples of ful l erenes are called jitllerites.
�
C60 has the atomic arrangement of a truncated icosahedron (soccer ball), with 32 faces, 1 2 pentagons, and 20 hexagons. The i nterior of the molecule is large, and other atoms may be inselted to produce compounds with unusual properties. The crystal structure of C60 is face-centered cubic. The anion C603- is very stable, and K 3C60 is a superconductor below 1 8 K .
�
Buckminsterfullerene, C60
1 4 . 1 7 Silicon , Germ an i u m , Tin , and Lead •
Silicon �
4 Occurs naturally as silicon dioxide Si02(s) and as silicates, which contain S i04 - units
�
Forms of Si02 include q uartz, quartzite, and sand.
�
Pure Si, which is widely used in semiconductors, is obtained from quartzite in a three-step process. S ilicon produced in the first step is crude and requires further purification: S i02(s) + 2 C(s) S iCs) + 2 CI2(g)
�
�
S iCI4(1) + 2 H2(g) •
•
S iCI4(1) �
SiCs) + 4 HCI(g)
Germanium �
Mendeleev predicted the existence of Ge, "eka-silicon," before its discovery.
�
Occurs as an impurity in Zn ores
�
Used mainly in the semiconductor industry
Tin �
Reduced from its ore, cassiterite (Sn02), through reaction with C at 1 200°C: Sn02(s) + C(s)
� •
SiCs) + 2 C02(g)
�
Sn(l) + CO2(g)
Used to plate other materials and for the production of al loys
Lead �
The principal ore is galena, PbS, which is converted to the oxide in air. The oxide in turn is reduced with coke to produce the metal : 2 Pb S(s) + 3 02(g) PbO(s) + C(s)
�
�
�
2 PbO(s) + 2 S02(g)
P b(s) + CO(g)
It is very dense, mal leable, and chemically inert. It is used as an electrode material in car batteries.
SG- 1 83
1 4 . 1 8 Oxides of Carbon •
Carbon dioxide, CO2 �
•
A nonmetal oxide and the acid anhydride of carbonic acid, H2C03 (The equilibrium between CO2 and H 2C03 favors CO2 . )
�
CO2(aq) is a n equilibrium mixture o f CO2 , H2C03 , HC03-, and a very sma l l amount of CO/-.
�
Solid CO2 (dry ice) sublimes to the gas phase, making it useful as a refrigerant and cold pack.
Carbon monoxide, CO � �
A colorless, odorless, flammable gas that is nearly insoluble in water and highly toxi c H a s a n extremely large bond enthalpy ( 1 074 Hmor l ).
�
CO is produced when an organic compound or C itself is burned in a limited amount of oxygen. It is commercially produced as synthesis gas:
�
CO is theformal anhydride of formic acid and is produced in the laboratory by dehydration of formic acid with hot, concentrated sulfuric acid:
C H4(g) + H20(g)
HCOOH(I)
�
�
CO(g) + 3 Hig)
CO(g) + H20(l)
�
CO is a moderately good Lewis base. One example is the reaction of CO with a d-block metal or ion: Ni(s) + 4 CO(g) � Ni(CO)4(1).
�
Complex formation is responsible for the toxicity of CO, which attaches more strongly than O2 to iron in hemoglobin, thereby blocking the acceptance of O2 by red blood cel l s.
�
CO can act as a reducing agent, yielding CO2.
1 4 . 1 9 Oxides of Silico n : The Silicates •
Silica or silicon d ioxide, Si0 2 �
Derives its strength from its covalently bonded network structure
�
•
Occurs natural ly in pure form only as quartz and sand 4 Silicates, compounds containing Si04 - units 4 � Orthosilicates, the simplest silicates, contain Si04 - ions. Examples: Na4Si04 and ZrS i 04 , zircon 4 � Pyroxenes contain chains of tetrahedral Si04 - units in which one 0 atom bridges two S i atoms (Si-O-S i) from adjacent tetrahedra, such that the "average" unit is SiO/-. Cations placed along the chains provide electrical neutrality. Examples: AI(Si03)2 and CaMg(Si03)2 �
�
•
Amphiboles contai n chains of silicate ions that fonn a cross-linked l adder-l i ke structure containing 6 S i40 1 1 - units. One example is the fibrous mineral tremolite, Ca2M gs(Si40 1 I MOH)2, one form of asbestos, a material that withstands extreme heat but is a known carcinogen. A lmost all amphiboles contain hydroxide ions attached to the metal. 2 Other types of silicates have sheets containing Si20s - units, with cations lying between the sheets and linking them together. One example is talc, Mg3(Si20sMOH)2 . Additional structural types of silicon oxides exist.
Aluminosilicates and cements �
3 4 A luminosilicates form when a S i + ion is replaced by A1 + plus additional cations for charge balance. One example is mica, KMg3(Si3AI0 I OMOH)2 . The extra cations hold the sheets of tetrahedra together, accounting for the hardness of these materials. Feldspar is a silicate material in which more than hal f the sil icon is replaced by al uminum.
SG- 1 84
�
•
Cements are produced by melting aluminosilicates with l imestone and other materials and allowing them to solidify. Cement is one of the materials in concrete.
Silicones �
Contai n long -O-Si-O- chains with the remaining two silicon bonding positions occupied by organic groups, such as the methyl group, -CH 3
�
Possess both hydrophobic and hydrophilic properties, which makes them useful in waterproofing fabrics and in biological appl ications, such as surgical and cosmetic implants
1 4.20 Other Im portan t G ro u p 1 4 / IV Compou n d s •
Carbides �
Can be ionic (saline), molecular, or interstitial
�
Saline carbides are most commonly formed from Group I and 2 metals, AI, and a few other metals. s-Block metals form saline carbides when their oxides are heated with C. 4 Anions present in carbides are normally methanide, C -, or acetylide, C /- , both very strong Bmnsted bases. Methanides react with water to produce a basic solution and methane gas:
�
AI4C3(s) + 1 2 H20(l)
�
4 AI(OH)3(s) + 3 CH4(g)
Acetylides release acetylene, HC=CH, upon reaction with water: CaC2(s) + 2 H20(I)
�
Ca(OHMs) + C2H2(g)
�
Covalent carbides are formed by metal s with
X
�
S i licon carbide, S iC (carborundum), a n extremely hard covalent carbide and a n excel lent abrasive, is produced by the reaction of si l icon dioxide with carbon at 2000°C :
values similar to that of C, such as S i and B .
S i02(s) + 3 C(s) - SiC(s) + 2 CO(g) �
•
Interstitial carbides are formed by direct reaction of a d-block metal with C at temperatures above 2000°C. The C atoms lie in holes in close-packed arrays of metal atoms. Bonds between the C and metal atoms stabil ize the lattice. Examples: WC, Cr3C, and Fe3C, a component of steel
Tetrachlorides �
All Group 1 4/1V elements form liquid molecular tetrachlorides, of which PbCI4 is the least stable. Carbon tetrachloride, a carcinogenic liquid, is formed by the reaction of chlorine and methane: C Hig) +4 Ch(g) - CCI4(1) + 4 HCI(g)
�
Si reacts directly with chlorine to form silicon tetrachloride. Unlike CCI4 , S iC I4 reacts with water as a Lewis acid, accepting a lone pair of electrons from H20: S iCI4(l) + 2 H20(l) - Si02(s) + 4HC I(aq)
•
Cyanides �
Cyanides are strong Lewis bases that form a range of complexes with d-block metal ions. They are extremely poisonous, combining with cytochromes involved with the transfer of electrons and the supply of energy in cel ls.
�
The cyanide ion, CN -, is the conj ugate base of the acid HCN, which is made by heating methane, ammonia, and air in the presence of a Pt catalyst at I 1 00°C : 2 CH4(g) + 2NH3(g) + 3 02(g) - 2 HCN(g) + 6 H20(g)
•
Methane and silane �
C bonds readily with itself to fonn chains and ri ngs and forms a multitude of compounds with hydrogen (hydrocarbons). See chapter 1 8 .
SG-185
�
S i fonns a much smaller number of compounds with hydrogen; the simplest is the analogue of methane, silane, S i H4, formed by the reaction of the reducing agent LiAlH4 with si licon hali des: SiCI4 + LiAIH4 - S i H4 + LiCI
�
+
AICh
SiH4 is more reactive than CH4 and ignites spontaneously in air. Higher si lanes are unstable and decompose on standing.
THE IMPACT ON MATERIALS (Sections 14.21-14.22) 1 4.2 1 Glasses •
Preparation and properties � �
A glass is an ionic solid with an amorphous structure simi lar to a l iquid.
�
Glass is characterized by a network structure based on a nonmetal oxide, commonly silica, Si02 , melted with metal oxides that function as network modifiers.
�
Heating S i02 ruptures S i-O bonds, enabling metal ions to form ionic bonds with some of the 0 atoms. The nature of the glass depends on the metal added. 2 Soda-lime glass, used for windows and bottles, is made by adding Na+ ( 1 2% Na20) and Ca + ( 1 2% CaO). Heating Na2C03 (soda) produces the Na20; heating CaC03 (lime) produces the CaO.
� �
•
The development of optical fibers led to considerable recent advances in glassmaking.
Borosilicate glass, such as Pyrex, is produced using less soda and l ime and adding 1 6% B203 . These glasses expand very little on heating and are used for ovenware and laboratory glassware.
Reaction with acids and bases �
Glass is resistant to attack by most chemicals.
�
B ut, the silica in glass can react with HF:
�
S i lica also reacts with the Lewis base OH - in molten NaOH and the carbonate anion in molten Na2C03 at 1 400°C:
2 Si02(s) + 6 H F(aq) - S iF6 -(aq) + 2 H30+(aq)
Si02(s) �
+
2Na2C03(1) - Na4Si04(1) + 2 C02(g)
Removal of silica from glass by the ions F - (from H F), OH -, and CO/- is cal led etching.
1 4.22 Cera m ics •
Preparation and properties Ceramics �
I norganic materials, such as clays, hardened by heating to a high temperature
�
Typically very hard, insol uble in water, and stable to corrosion and high temperatures
�
Can be used at high temperatures without fail ing, and resist defonnation
�
Tend to be brittle, however
�
Often are oxides of elements on the border between metals and nonmetals
SG - 1 86
�
Used in many automobile parts, including spark plugs, pressure and vibration sensors, brake l injngs, and catalytic converters
�
Some d-metal oxides and compounds of B and S i with C and N are also ceramic materials.
�
Most ceramics are electrical insulators, but some are semiconductors and superconductors.
A luminosilicate ceramics �
Formed i n many cases by heating clays to expel water trapped between sheets of tetrahedral aluminosil icate units
�
The result is a rigid heterogeneous mass of small interlocking crystals bound together by glassy sil ica.
�
One form is white china clay, used to make porcelain and china. It is free of iron impurities that tend to make clays reddish brown.
�
Methods developed to make ceramics less brittle are the sol-gel process and the composite material technique.
SG- 1 87
Chapter 1 5 THE ELEMENTS: THE LAST FOUR MAIN GROUPS
GROUP 15 / V: THE NITROGEN FAMILY (Sections 15.1-15.4) 1 5 . 1 The G ro u p 1 5 / V Elements •
Nitrogen -+
Nitrogen differs greatl y from other group members owing to its high electronegativity (X = 3 .0), sma l l size (radius of 74 pm), abi lity to form m ultiple bonds, and lack of available d-orbitals. It is found with oxidation numbers from -3 to +5.
-+
Elemental nitrogen, N 2 , is found in air (78. 1 % by volume) and is prepared by fractional l distillati on of l i quid air. The strong triple bond in N2 (I'lHs = 944 kJ . mor ) makes it very stable. Nitrogenfixation is the process by which N2 is converted into compounds usable by plants. The Haber synthesis of ammonia is the maj or industrial method for fixing nitrogen, but it requires high temperatures and pressures and is, therefore, expens ive. Lightning produces some oxi des of nitrogen, which are washed into soi l by rain. Bacteria found in the root nodules of l egumes also fix N2 . An important area of current research is the search for catalysts that can mimic bacteria and fix nitrogen under ordi nary temperatures.
-+
•
Phosphorus -+
Phosphorus differs from nitrogen in its lower electronegativity (X 2.2), larger size (radius of 1 1 0 pm), and the presence of avai lable d-orbitals. Phosphorus may form as many as six bonds, whereas nitrogen can form a max imum of four.
-+
Phosphorus is prepared from the apatites, mineral forms of calcium phosphate, Ca3(P04)2 . The rocks are heated in an electric furnace with carbon and sand. The phosphorus vapor formed condenses as white phosphorus, a soft, white, poisonous molecular solid consisting of tetrahedral P4 molecules. It bursts into flame upon exposure to air and is normally stored under water.
-+
Red phosphorus is less reactive than white phosphorus and is used in match heads because it can be ignited by friction. It is a network soli d consisting, most l ikely, of l i nked P4 tetrahedral units. Redphosphorus is the most thermodynamically stable form of phosphorus, yet white phosphorus crystal lizes from the vapor or from the liquid phase. It is the form chosen to have an enthalpy and free energy of formation equal to O. For red phosphorus, I'lHfo = - 1 7.6 kJ . mor l and I'lGt = - 1 2. 1 kJ ·mor l at 25°C.
-+
-+
•
•
=
Black phosphorus is a third form of the element; it has a network structure in which each P atom is bonded to three others at approximately right angles.
Arsenic and antimony -+
Arsenic and antimony are metalloids. They are el ements known since ancient times because they are easily reduced from their ores.
-+
As elem ents, arsenic and antimony are used in lead alloys in the electrodes of storage batteries and in the semiconductor industry. Gallium arsenide is used as a l ight-detecti ng material with near infrared response and in lasers, including ones used for CD players.
Bismuth -+
Bismuth is a metal with large, weakly bonded atoms. Bismuth has a low melti ng poi nt, and it is used in alloys that serve as fire detectors in sprinkler systems.
SG- 1 8 9
�
Bismuth is also used to make l ow-temperature castings. Like ice, solid bi smuth is less dense than 2 the l iquid. Bismuth (Z = 83) is the last element in the periodic tabl e with a stable isotope, 09 Bi.
1 5 .2 Com pou n d s with Hydrogen and the Halogens •
Ammon ia, N H 3 , and phosphine, PH3 � Ammonia is prepared in large amounts by the Haber process. Ammonia is a pungent, toxic gas that condenses to a colorless liquid at -3 3°C. I t acts as a solvent for a wide range of substances. Autoprotolysis occurs to a much smaller extent in liquid ammonia (am) than in pure water. 33 2 NH3(am) = NH/(am) + NH2-(am) Kam = I X 1 0- at -35°C Very strong bases that are protonated by water survive in l i quid ammonia.
•
�
Ammonia is a weak Bmnsted base in water; it is also a fai rly strong Lewis base, particularly toward 2 d-block elements. One reaction is Cu +(aq) + 4NH3(aq) - Cu(NH3)/\aq). 2 In this reaction, unshared electron pairs on nitrogen interact with the L ewis acid Cu + to form four Cu-NH3 bonds.
�
Ammonium salts decompose when heated; if the salt contains a nonox id izing anion, NH3 is produced. The reaction is (NH4)2S04(S) - 2 NH3(g) + H2S04(1). Decomposing ammonium carbonate has a pungent odor and is used as a "smell ing salt" to revive people who have fainted.
�
Phosphine is a poisonous gas that smel l s faintly of garlic and bursts into flame in air if it is impure. It is much less stable than ammonia. Because it does not form hydrogen bonds, it is not very sol uble in water and is a very weak base (pKb = 27.4). Its aqueous solution is nearly neutral .
Hydrazine, N 2 H 4 �
Hydrazine is an oily, colorless li quid. I t is prepared by gentle oxidation of ammonia with alkaline hypochlorite solution. 2 NHJCaq) + CIO-(aq) - N2H4(aq) + Cr(aq) + H20(l)
�
•
•
•
I ts physical properties are similar to water; its melting point i s 1 . 5°C and its boi l i ng point is 1 1 3°C. I t i s dangerously explosive and is stored in aqueous solution. A mixture of methylhydrazine and l i quid N204 is used as a rocket fuel because these two l iquids ignite on contact and produce a large vol ume of gas.
4 CH3N HNH2CI) + 5 N204(1) - 9 N2(g) + 1 2 H20(g) + 4 C02(g) 3 N itrides, N -, and phosphides, p 3
�
Nitrides are solids that contain the nitride ion. Nitrides are stable only for sma l l cations such as l ithium, magnesium, or al uminum. Nitrides dissolve in water to form ammonia and the corresponding hydroxide. Zn3N2Cs) + 6 H20(l) - 2 NH3(g) + 3 Zn(OHMaq)
�
Phosphides are solids that contain the phosphide ion. Because PH3 is a very weak parent acid of 3 the strong Bmnsted base p -, water is a sufficiently strong proton donor to react with phosphides 3 to form phosphine. 2 p -(s) + 6 H20(I) - 2 PH3(g) + 6 0 H- (aq)
Azides, N3�
The azide ion is a highly reactive polyatomic anion of nitrogen. Sodium azide is prepared by the reaction of din itrogen oxide with molten sodium amide, NaNH2 .
�
Some azides, such as AgN3 , Cu(N3)2 , and Pb(N3)2 , are shock sensitive. The azide i on is a weak base, and its conj ugate acid, hydrazoic acid, HN3 , is a weak acid simi lar in strength to acetic acid.
Halides �
Nitrogen has an oxidation number of +3 in the nitrogen halides. N itrogen trifluoride, NF3 , is the most stable and does not react with water. Nitrogen trichloride, however, reacts with water to
SG - J 90
form ammonia and hypochlorous acid, HCIO. Nitrogen triodide is so unstable that it decomposes explosively upon contact with light. �
Phosphorus forms chlorides with an oxidation number of +3 in PCh and +5 in PCls . Phosphorus trichloride, a l iquid, is formed by direct chlorination of phosphorus. It is a maj or intermediate in the production of pesticides, oi l additives, and flame retardants. Phosphorus pentachloride, a solid, is made by al lowing the trichloride to react with additional chlorine. It exists in the solid as tetrahedral PCI/ cations and octahedral PCI6- anions, but it vaporizes to a gas of trigonal bipyramidal PCls molecules. Phosphorus pentabromide is also molecul ar in the vapor and ionic as a solid; but in the solid, the anions are Br- anions, presumably because of the difficulty in fi tting six large Br atoms around a central P atom.
�
Both phosphorus trichloride and phosphorus pentachloride react with water in a hydrolysis reaction, a reaction with water in which new element-oxygen bonds are formed and there is no change in oxidation state. An oxoacid and hydrogen chloride gas are formed i n the reacti on. PCh(l) + 3 H20(l) - H3P03(S) + 3 HCI(g) PC ls(s) + 4 H20(I) - H3P04(l) + 5 HCI(g)
1 5.3 Nitrogen Oxides a n d Oxoacids •
•
Dinitrogen oxide, NzO �
Dinitrogen oxide is commonly cal led nitrous oxide (laughing gas), and it is the oxi de of nitrogen with the lowest oxidation number for nitrogen (+ I ). I t is formed by gentl e heating of ammonium nitrate. NH4N03(s) - N20(g) + 2 H20(g)
�
N2 0 is fairly unreactive, but it is toxic if inhaled in large amounts.
Nitrogen oxide, NO �
�
�
•
Nitrogen oxide is commonly cal led nitric oxide; the nitrogen atom has an oxidation number of +2. Nitrogen oxide, a colorless gas, is produced industrially by the catalyti c oxi dation of ammonia at I OOO°C. 4 NH3(g) + 5 02(g) - 4 NO(g) + 6 H20(g) The endotherm ic formation of NO from the oxidation ofNz occurs readi ly at the high temperatures that exist i n automobile engines and turbine engine exhausts. N2(g) + 02(g) - 2NO(g) NO is further oxidized to N02 upon exposure to air. I n this form, it contributes to smog, acid rain, and the destruction of the stratospheric ozone layer.
Nitrogen dioxide, NOz �
Nitrogen dioxide is a choking, poisonous, brown gas that contributes to the color and odor of smog. The oxidation number of nitrogen in N02 is +4. N02 , l ike NO, is paramagnetic. It disproportionates in water to form nitric acid and nitrogen oxide: 3 NOlCg) + H20( I) - 2 HN03(aq) + NO(g)
�
Nitrogen dioxide is prepared in the laboratory by heating lead( l l ) nitrate: 2 Pb(N03Ms) - 4NOz(g) + 2 PbO(s) + Oz(g) And it exists in equi l i brium with its colorless dimer, dinitrogen tetroxi de: 2NOz(g) = N Z 04(g) Only the dimer ex ists in the sol id, so the brown gas condenses to a colorless solid.
•
Nitrous acid, HNOz Nitrous acid can be produced in aqueous solution by mixing its anhydride, dinitrogen trioxide, with water. The oxidation number of nitrogen in HN02 is +3 . N203(g) + HzO( I) - 2 HN02(aq)
�
SG - 1 9 1
�
Nitrous acid has not been isolated as a pure compound, but it has many uses in aqueous solution. It i s a weak acid with pKa 3 .4. The conjugate base, nitrite ion, N02-, forms ionic solids that are soluble in water and m ildly toxic. =
•
Nitric acid, H N03 �
Nitric acid, a widely used industrial and laboratory acid, is produced by the three-step Ostwald process . The oxidation number of nitrogen in HN03 is +5. 4NH3(g) +2 02(g) � 4NO(g) + 6 H2(g) 2NO(g) + 02(g) 2N02(g) 3 HN03(aq) + NO(g) 3 N02(g) + H20( 1) �
�
�
Nitric acid has been isolated as a pure compound, and it is a colorless l i quid with a boi l i ng point of 83°C. It is usual ly used in aqueous solution. It is an excel lent oxi dizing agent as wel l as a strong acid.
1 5.4 Phosp h o rus Oxides and Oxoacids •
•
Phosphorus(III) oxide, P406 , and ph osphorous acid, H3 P03 �
Phosphorus(IlJ) oxide is made by heating white phosphorus in a l imited supply of air. P406 molecules consist of P4 tetrahedral units, with each oxygen atom lying between two phosphorus corner atoms. Six oxygen atoms lie along the six edges of the tetrahedron, one per edge. The edge bonds are represented as -P-O-P-.
�
P406 is the anhydride of phosphorous acid: Phosphorous acid is a diprotic acid.
�
P406(S) + 6 H20(l)
Phos phorus(V) oxide, P40IO , and phosphoric acid, H3P04 � Phosphorus(V) oxide is made by heating white phosphorus in an excess supply of air. P40 l O molecul es are similar to P406 but have an additional oxygen attached to each phosphorus at the apices of the tetrahedron. �
•
4H3P03(aq)
P 40 1 0 is the anhydride of phosphoric acid, H3P04 • It traps and reacts with water very effici ently and is widely used as a drying agent. Phosphoric acid is a triprotic acid. It is only mildly oxidizing, despite the fact that phosphorus is in a high oxidation state. Phosphoric acid i s the parent acid of phosphate salts that contain the tetrahedral phosphate ion, pol-. Phosphates are generally not very soluble, which makes them suitable structural material for bones and teeth.
Polyphos phates �
�
3 Polyphosphates are compounds made up of l inked P04 - tetrahedral 4 units. The simplest structure is the pyrophosphate ion, P207 -, in which two phosphate ions are l i nked by an oxygen atom through -O-P-O- bonds.
[
.
.
. 14-
'0 '0 II II : o- p- o - p- o: .. I I :0: :0: .. . • .
.
• •
.
• •
• .
.
More complicated structures that form with longer chains or rings also exist. The bi ochemically most important polyphosphate is adenosine triphosphate, ATP, which contai ns three phosphorus tetrahedral units linked by -O-P-O- bonds. The hydrolysis of ATP to adenosine diphosphate, ADP, by the rupture of an O-P bond releases energy that is used by cel ls to drive biochemical reactions within the cel l . ATP + H20 ADP + HPO/- 6.H = - 4 I kJ ATP is often cal led the "fuel of l ife" because it provides the energy for many of the biochemical reacti ons occurring in cel ls. �
SG- I 92
GROUP 16 / VI: THE OXYGEN FAMILY (Sections 15.5--15.8) 1 5.5 The G ro u p 1 6 / VI E lements •
•
Oxygen �
Elemental oxygen, O2 , is found in air (2 1 .0% by volume) and is prepared by fractional disti l l ation of l iquid air. It is a colorless, odorless, paramagnetic gas.
�
A satisfactory Lewis structure for oxygen cannot be drawn. The Molecular Orbital theory accounts for the paramagnetic properties (see Chapter 3).
�
Elemental oxygen exists i n two al lotropic forms, O2 and 03 (ozone). I norganic chemists often use the term dioxygen to refer to molecular oxygen. Ozone is produced by an electric discharge in oxygen gas; its pungent odor is apparent near sparking electrical equipment and l i ghtning strikes.
Sulfu r �
Sulfur i s found i n ores and as elemental sulfur. I t i s widely distributed as sulfide ores, which i nclude galena, PbS ; cinnabar, HgS; iron pyrite, FeS2 (fool 's gold), and spahlerite, ZnS. Sulfur is found in elemental form as deposits, cal led brimstone, which are created by bacterial action on H2S .
�
Elemental sulfur i s a yell ow, tasteless, almost odorless, insol uble, nonmetal l i c molecular solid of crownl ike S8 rings. The more stable allotrope, rhombic sulfur, forms beautiful yel low crystals, whereas the less stable allotrope, monoclinic sulfur, forms needlelike crystals. The two all otropes differ in the manner in which the S8 rings are stacked together. For simplicity, the elemental form of sulfur is often represented as S(s) rather than S8(S).
�
One method of recovering sulfur is the Claus process, in which some of the H2S that occurs in oil and natural gas is first oxid ized to sulfur diox ide. 2 H2S(g) + 3 02(g) 2 S02(g) + 2 H20( I) Sulfur dioxide oxidizes the remaining hydrogen sulfide and both are converted to elemental sulfur at 3 00°C in the presence of an alumina catalyst. 2 H2S(g) + S02(g) 3 S(s) + 2 H20(J) �
�
•
•
�
Sulfur is used primarily to produce sulfuric acid and to vulcanize rubber. V ulcani zation increases the toughness of rubber by introducing cross- links between the polymer chains of natural rubber.
�
An i mportant property of sulfur is its ability to form chains of atoms, catenation. The -S-S l inks that connect different parts of the chains of amino acids in protei ns are an important example. These disulfide links contribute to the shapes of proteins (see Chapter 1 9).
Seleniu m and tel l urium �
Selenium and tellurium are found in sulfide ores; they are also recovered from the anode s l udge formed during the electrolytic refining of copper.
�
Both elements have several allotropic forms; the most stable one consists of long zigzag chains of atoms. The appearance of the all otropes is that of a si lver-white metal, but electrical conductivity is poor. Exposing selenium to light increases its electrical conductivity, so it is used in solar cells.
Polonium � Polonium is a radioactive, low-melting metalloid. It was identi fied in uranium ores in 1 898 by the Curies. I ts most stable isotope, 209pO, has a half-life of 1 03 y. 4 2 � Polonium decays by emission of an alpha particle, He +. It is used i n antistatic devices in text i l e mills to counteract the buildup o f negative charges o n fast moving fabric.
SG-J93
1 5.6 Com po u n d s with Hyd rogen •
Water, H 20, and hyd rogen s u lfide, H 2 S �
Water is a remarkable compound possessi ng a unique set of physical and chemical properties. Water is purified in domestic water suppl ies in a mu lti-step procedure that includes aeration, addition of slaked lime, removal of particles by coagulation and flocculation, addition of CO2 , filtration, reduction of pH, and addition of chlorine disinfectant to kill bacteria. H igh-purity water is obtained by distillation.
�
Water has a higher boi ling point than expected on the basis of its molar mass because of the extensive hydrogen bonding between H 20 molecules. Hydrogen bonding also causes a more open structure and, therefore, a lower density for the sol i d than for the l i quid. Because of its high polarity, water is an excellent solvent for ionic compounds.
�
Chemically, water is amphiprotic, and it can both donate and accept protons; so it is both a Brgnsted acid and a Bnonsted base. H20(I , acid) + N H3(aq)
�
NH4 +(aq) + OH -(aq)
C H3COOH(aq) + H20(l, base) �
�
H30+(aq) + CH3COO-(aq)
Water can act as a Lewis base by donating its unshared pair of electrons on the oxygen atom. 3 3 Fe(H20)6 +(aq) Fe +(aq) + 6 H20(l ) �
�
Water can act as an oxidizing agent and a reducing agent. 2Na(s) + 2 H20(l, ox. agent) 2 H20(l, red. agent) + 2 F2(g)
�
�
�
2NaOH(aq) + H2(g) 4 HF(aq) + 02(g)
A hydrolysis reaction is one in which water as a reactant is used to form a bond between oxygen and another element. Hydrolysis can occur with or without a change in oxidation number. PC l5(s) + 4 H20( I) H3P04(aq) + 5 HCI(aq) �
C h(g) + H20(I)
�
CIOH(aq) + HCI(aq)
The first exampl e given above has no change in oxidation number. The second is an examp le of a disproportionation reaction, in which the oxidation number of chlorine changes from 0 in CI2 to + I in HCIO and - 1 in HC! . �
Hydrogen sulfide is an examp Ie of a Group 1 6/ V I binary compound with hydrogen. I t is a toxic gas with an offensive odor (rotten eggs). Egg proteins contain sulfur and release H2S when they decompose. !ronO I ) sulfi de sometimes appears as a pale green discoloration at the boundary between the white and the yolk.
�
Hydrogen sulfide is prepared by the direct reaction of hydrogen and sulfur at 600°C or by protonation of the sulfide ion, which is a Bf0nsted base. FeS(s)
�
•
+
2 HCI(aq)
�
FeCb(aq) + H2S(g)
Hydrogen sulfide dissolves in water and is slowly oxidized by dissolved air to form a coll oidal dispersion of particles of sulfur. Hydrogen sul fi de is a weak diprotic aci, and the parent acid of the 2 hydrogen sulfides, H S-, and the sulfides, S -. Sulfides of s-block elements are moderately sol uble in water, whereas the sulfides of the heavy p- and d-block metals are generally very insol uble.
Hydrogen peroxide, H 2 02 , and polys u lfanes, H S-SII-SH �
Hydrogen peroxide is a highly reactive, pale blue liquid that is appreciably more dense ( 1 .44 g· mL- 1 ) than water. I ts melting point is - O AoC and its boiling point is 1 52°C.
SG-194
�
Chemical ly, hydrogen peroxide is more acidic than water (pKa = 1 1 . 75). It is a strong ox idizing agent, b ut it can also function as a reducing agent. 3 2 2 Fe +(aq) + H202(aq, ox. agent) + 2 H+(aq) - 2 Fe +(aq) + 2 H20(l) Ch(g) + H202(aq, red. agent) + 2 0H -(aq) - 2Cl -(aq)
� �
+
2 H20(l) + 02(g)
The 0-0 bond in hydrogen peroxide is very weak. The oxidation state of oxygen in hydrogen peroxide is - 1 . A polysulfane i s a catenated molecular compound of composition H S-Sn-SH, where n can take values from 0 through 6. The sulfur analogue of hydrogen peroxide occurs with n = o. Two polysulfide ions obtained from polysulfanes are found to occur in the mineral lapis lazuli (see Fig. 1 5 . 1 5 in the text); its color derives from impurities of S 2-(bl ue) and S3-(hint of green). The S2- ion is an analogue of the superoxide ion, O2-; and S3- is an analogue of the ozonide ion, 03-.
1 5.7 Sulfu r Oxides and Oxoacids •
Su lfu r dioxide, SOz , sulfurous acid, HZS03 , and sulfite, SO/�
Suljilr dioxide, a poisonous gas, is made by burning sulfur in air. Volcanic activity, the combustion of fuels contam inated with sulfur, and the oxidation of hydrogen sulfide are the maj or sources of S02 in the atmosphere. The chemical equation for the oxidation of hydrogen sulfide in the atmosphere i s : 2 H2S(g) + 3 02(g) - 2 S02(g) + 2 H20(g)
�
Sulfurous acid is formed by the reaction of its aci d anhydride, S02 , and water. S 02(g) + H20(l) - H2S03(aq) Sulfurous acid is an equ i l ibrium mixture of two molecules, (HO)S H 02 and (HO)SO(OH). These molecules are also in equ i l i brium with molecules of S02 , each of which is sUITounded by a cage of water molecules. When the solut ion is cooled, crystals with a composition of roughl y S02 · 7 H20 form. This substance is an examp le of a clathrate, in which a molecule sits in a cage of other molecules. Methane, carbon dioxi de, and the noble gases also form clathrates with water.
�
I n sulfite ions and suljilr dioxide, the oxi dation number of the sulfur atom is +4. Because this value i s intermediate in sulfur's range of -2 to +6, these compounds can act as either oxi dizing agents or reducing agents. The most important reaction of sulfur diox ide is its slow oxi dation to sulfur trioxide, in which the oxidation number of the sulfur atom is +6. 2 S02(g) + 02(g) - 2 S 0}(g) This reaction is catalyzed by the presence of metal cations in droplets of water or by certain surfaces. I ndirect pathways also exist in the atmosphere for the conversion of S02 to S O} .
•
Sulfu r trioxide, S03 , sulfu ric acid, HzS04, and sulfate, SO/�
At normal temperatures, sulfur trioxide is a volatile l iquid with a boiling point of 45°C. I ts shape is trigonal planar with bond angles of 1 20°. I n the sol id, and to some extent in the l iquid, the molecu les aggregate to form trimers, S309, as wel l as larger groupings.
�
Sulfuric acid is produced by the contact process, in which sulfur is first burned in oxygen at I OOO°C to form sulfur diox i de, which then reacts with oxygen to form sulfur trioxide over a V20S catalyst at 5 00°C. Because sulfur trioxide forms a corrosive acid with water, it is absorbed in 98% concentrated sulfuric acid to give a dense, oily liquid cal led oleum, H2S207 . S03(g) + H2S04(l) - H2S207(1)
�
Sulfuric acid is a colorless, corrosive, oily l i quid that boils (decomposes) at 300°C. It is a strong Br@nsted acid, a powerfu l dehydrating agent, and a mild oxidizing agent. In water, its first
SG- J 95
ionization is complete, but HS04- is a weak acid with pKa = 1 . 92. As a dehydrating agent, sulfuric acid i s able to extract water from a variety of compounds, including sucrose, C ' 2H220" , and formic acid, HCOOH. C ' 2H220 1 l (S) � 1 2 C(s) + 1 1 H20(l) HCOOH(l) � CO(g) + H20(l) Because of the large exothermicity associated with dilution, m ixing sulfuric acid with water can cause v iolent splashing, so for reasons of safety, sulfuric aci d (which i s more dense than water) is always carefu l l y added to water, not the water to acid. The sulfate ion i s a mild oxidizing agent. 4 H+(aq) + SO/-(aq) + 2 e- � S02(g) + 2 H20(l) EO = 0. 1 7 V 1 5. 8 Sulfu r Halides •
•
Fluorides �
S ulfur ignites in fluorine to produce sulfur hexafluoride, SF6, a dense, colorless, odorless, nontoxic, thermally stable (�Gfo = - 1 1 05 .3 kJ . mor ' ), i nsolubl e gas.
�
The oxidation number of sulfur in sulfur hexafluoride attains its maximum value of +6, but the large number of strongly electronegative fl uorine atoms around the sulfur atom protects the compound from attack. Because it has a high ionization energy, sulfur hexafluoride is a good gas phase electrical insulator.
�
The predicted shape of SF 6 is octahedral and the molecule is nonpolar, 11 = 0 (see Section 3 .3).
Chlo rides �
Disulfur dichloride, S2C h , is one of the products of the reaction of sulfur with chlorine. A yel l ow, toxic liquid with a nauseating odor, disulfur dichloride is used mainly for the vulcanization of rubber.
�
The reaction of disulfur dichloride with excess chlorine in the presence of a catalyst, iron( I I I ) chloride, produces a fou l smel l i ng, red l iquid o f sulfur dichloride, SCh . I t reacts with ethene, C2H4 , to produce mustard gas, S(CH2CH2CI)2 , which has been used in chemical warfare. Mustard gas causes symptoms of severe discomfort, skin blisters, nasal discharges, and vom iting. It also destroys the cornea of the eye, leading to bli ndness.
GROUP 17 / VII: THE HALOGENS (Sections 15.9-15.10) 1 5.9 The Group 1 7 / VII Elements •
Fluorine �
Fluorine is a reactive, almost colorl ess gas of F2 molecules. Its properti es fol l ow from its high electronegativity, small size, and lack of avai lable d-orbitals for bonding. It is the most electronegative element and has an oxidation number of - I in all its compounds. Elements combined with fl uorine are often found with their highest oxidation numbers, such as +7 for r in I F 7 . The small size of the fl uoride ion results in high lattice enthalpies for fluorides, a property that makes them less sol uble than other hali des.
�
Fluorine occurs widely in many minerals includingfluorspar, CaF2 ; cryolite, N a3AIF6 ; and the fluoroapatites, CaSF( P04)3 .
SG- J 96
•
�
Elementaljluorine is obtained by the el ectrolysis of an anhydrous, molten KF-HF m ixture at 75°C, using a carbon electrode. F l uorine is highly reactive and highly ox idizing; it is used in the production of SF6 for electrical equipment, Teflon (polytetrafluoroethylene), and UF6 for isotope enrichment.
�
The strong bonds formed by jluorine make many of its compounds relatively inert. Fluorine's abil ity to form hydrogen bonds results in relatively high melting points, boil i ng points, and enthalpies of vaporization for many of its compounds.
C hlorine �
Chlorine is a reactive, pale yellow-green gas of Ch molecules that condenses at -34°C. It reacts directly with all the elements except carbon, nitrogen, oxygen, and the noble gases. I t is used in a large number of industrial processes, including the manufacture of plastics, solvents, and pesticides. It is also used as a disinfectant to treat water supplies.
�
Elemental chlorine is obtained by the electrolysis of molten or aqueous NaCI . I t is a strong oxidizing agent and oxidizes metals to high oxidation states. 2 Fe(s) + 3 Cb(g)
•
�
2 FeCI3(s)
Bromine �
Bromine is a corrosive, reddish brown l i quid of B r2 molecules. It has a penetrating odor. Organic bromides are used in texti les as fire retardants and in pesticides. I norganic bromides, particularly si lver bromide, are used in photographic emulsions.
�
Elemental bromine is produced from brine wells by the oxidation of Br- by C12 . 2 Br-(aq) + Cb(g) 8 r2(1) + 2 C r(aq) Iodine � Iodine is a bl ue-black, l ustrous sol id Of l2 molecules that readily subl imes to form a purple vapor. It occurs as 1 - in seawater and as an impurity in Ch ile saltpeter, KN03 . The best source is the brine from oil wel ls. Elemental iodine is produced from brine wells by the oxidation of l - by Cb . � Elemental iodine is sl ightly sol uble in water, but it dissolves well in iodide solutions because it reacts with aqueous 1 - to form the triodide ion, 1 3 -. � Iodine is an essential trace element in human nutrition, and iodides are often added to table salt. This "iodized salt" prevents iodine deficiency, which leads to the enlargement of the thyroid gland in the neck (a condition cal led goiter). �
•
•
Astatine �
Astatine is a radioactive element that occurs in uranium ores, but only to a tiny extent. Its most 2 stable isotope, 1 °At, has a half-life of 8.3 h. The isotopes formed in uranium ores have much shorter lifetimes. The properties of astatine are surmised from spectroscop ic measurements.
�
Astatine is created by bombarding bismuth with alpha particles in a cyclotron, which accelerates particles to high speed.
1 5 . 1 0 Compounds of the Halogens •
I n terh a logens �
Interhalogen compounds consist of a heavier halogen atom at the center of the compound and an odd number of l i ghter ones bonded to it on the periphery. An exception is l 2C 1 6 , which contains two (-I-C l-I-) bridges between the two central iodine atoms. The two bridging chlorine atoms act as Lewis bases (el ectron pair donors), in a manner similar to A12C16.
�
Interhalogens are typical ly formed by direct reaction between stoichiometric amounts of the elements. l 2(g) + 7 F2(g) 21 F 7(g) �
SG- J 97
•
�
As the size of the central atom in an interhalogen compound increases, the bond enthal pies also increase, usually resulting in lower reacti vity of the compound. (X-F): CIF s ( 1 54 kJ ·mor 1 ) < BrF s ( 1 87 klmor l ) < l Fs (269 kJ·mor 1 )
�
Table 1 5. 5 in the text provides a compilation of the known interhalogens.
Hydrogen halides and metal ha lides � Hydrogen halides are prepared by direct reaction of the elements or by the action of nonvolat i le acids on metal halides. Ch(g) + H2(g)
�
2 HCl(g)
CaF2(s) + H2S04(aq, conc.)
�
Ca(HS04Maq) + 2 HF(g)
Because Br - and 1 - are oxid ized by sulfuric acid, phosphoric acid is used in the preparation of HBr and H I . �
Hydrogen halides are pungent, colorless gases, but hydrogen fluoride i s a l i qu i d at temperatures below 20°e. I ts low volatility is a sign of extensive hydrogen bonding, and short zigzag chains up to about (HF)s persist in the vapor.
�
Hydrogen halides dissolve readily in water to produce acidic solutions. Hydrogenfluoride has the unusual property of attacking and dissolving si lica glasses, and it is used for glass etching (see Section 1 4.2 1 ).
�
Anhydrous metal halides may be formed by di rect reaction. 2 Fe(s) + 3 Ch(g)
2FeCb(s)
�
Halides of metals tend to be ionic unless the metal has an oxidation number greater than +2. For example, sodium chloride and copper( 1 1 ) chloride are ionic compounds with high melting points, whereas titanium( 1 V) chloride and iron( 1 l 1 ) chloride sublime as molecules. •
H alogen oxides and oxoacids �
Hypohalous acids, HXO or HOX, consist of the halogen atom as a terminal atom with an oxidation number of + 1 . They are prepared by the direct reaction of a halogen with water; the corresponding hypohalite (XO-) salts are prepared by the reaction of a halogen with aqueous alkali solution. CI2(g) + H20( l) HOCl(aq) + HCl(aq) CI2(g) + NaOH(aq) NaOCl(aq) + HCl(aq) �
�
�
Hypochlorites cause the oxidation of organ ic material in water by producing oxygen in aqueous solution, which readily oxidizes organic materials. The production of O2 occurs in two steps; the net result is 2 OCl -(aq) 2 C l -(aq) + 02(g). �
�
Chlorates, Cl03-, contain a central chlorine atom with an oxidation number of +5 . They are prepared by the reaction of chlorine with hot aqueous alkal i. 3 Ch(g) + 6 0 W(aq)
�
Cl03-(aq) + 5 Cl -(aq) + 3 H20(l)
Chlorates decompose upon heating; the identity of the final product is determi ned by the presence of a catalyst. 4 KC103(1) 2 KC103(1)
�
�
3 KCI04(s) + KCI(s) 2 KCl(s) + 3 02(g)
no catalyst Mn02 catalyst
Chlorates are good oxidizing agents and are also used to produce the important compound, Cl02 . Sulfur dioxide is a convenient reducing agent for this reaction: 2NaCl03(l) + S02(g) + H2S04(aq, dilute)
SG - 1 9 8
�
2NaHS04(aq) + 2 C l 02(g)
�
Chlorine dioxide, CI02 , contains a central chlorine atom with an oxidation number of +4. I t has an odd number of electrons and is a paramagnetic yell ow gas that is used to bleach paper pulp. It is highly reactive and may explode violently under the right conditions.
�
Perchlorates, CI04-, contain a central chlorine atom with an oxidation number of +7. They are prepared by the electrolysis of aqueous chlorates. The half-reaction is: CI03-(aq) + H20(l ) CI04-(aq) + 2 H+ (aq) + 2 e�
Perchlorates and perchloric acid, HCI04 , are powerful oxidizing agents; perchloric acid in contact with small amounts of organic materials may explode. �
The oxidizing strength and acidity of oxoacids both increase as the oxidation number of the halogen i ncreases. For oxoacids with the same number of oxygen atoms, the oxidizing and acid strength both increase as the electronegativity value of the halogen increases. The general rules for acid strength of oxoacids are summarized in Chapter 1 0.
GROUP 18 / VIII: THE NOBLE GASES (Sections 15.11-15.12) 1 5. 1 1 The G roup 1 8 / VIII E lements •
Helium �
�
•
•
Helium derives its name from the Greek word helios, sun. The fi rst evidence for hel ium was discovered in the solar spectrum taken during a solar eclipse in 1 868. Helium is the second most abundant element in the universe after hydrogen. Helium atoms are light and travel at velocities suffici ent to escape from the earth' s atmosphere. 4 2 Alpha particles, H e +, are released by nucl ear decay of naturally occurring uranium and thorium ores. They are h igh-energy nuclei of hel i um-4, which capture two electrons apiece to become helium atoms. The low density and lack of flammabil ity of helium make it an appropriate gas for lighter-than-air airships such as blimps. Helium is found in the atmosphere with a concentration of 5 .24 ppm (parts per mill ion by vol ume).
Neon �
Neon derives its name from the Greek word neos, new. It is widely used in advertising display signs because it emits an orange-red glow when an electric current passes through it. Of all the noble gases, the discharge of neon is the most i ntense at ordinary voltages and currents. The neon atom is the source of l aser light in the helium-neon laser.
�
Neon has over 40 times more refrigerating capacity per unit volume than l iquid hel ium and more than three times that of liquid hydrogen. It is compact, inert, and less expensive than helium when used as a cryogenic coolant. Neon is found in the atmosphere with a concentration of 1 8 .2 ppm.
Argon �
Argon derives its name from the Greek word argos, inactive. Its main use is used to fi l l electric l ight bulbs at a pressure of about 400 Pa, where it conducts heat away from the fi lament. It is also used to fi 11 fl uorescent tu bes.
�
A rgon is also used as an inert gas shield (to prevent oxidation) for arc welding and cutting, and as a protective atmosphere for growi ng silicon and germanium crystals. Argon is the most abundant noble gas and is found in the atmosphere with a concentration of about 0.93% (or about I part per hundred by vol ume).
SG- J 99
•
Krypton � Krypton derives its name from the Greek word kryptos, hidden. It gives an intense white l i ght when an electric discharge is passed through it, and it is used in airport runway l i ghting. �
•
•
Krypton is produced in nuclear fission, and its atmospheric abundance is a measure of worldwide nuclear activity. Krypton is found in the atmosphere with a concentration of about I ppm.
Xenon �
Xenon derives its name from the Greek word xenos, stranger. It is used in halogen lamps for automobile headl ights and in high-speed photographic flash tubes.
�
Xenon is used in the nuclear energy area in bubble chambers, probes, and other applications where a high atomic mass is desirable. Currently, the anesthetic properties of xenon are bei ng explored. Xenon is found in the atmosphere with a concentration of about 0.087 ppm.
Radon Radon derives its name from the element radium. The gas is radioactive and is formed by radioactive decay processes deep in the earth. Uranium-238 decays very sl owly to radium-226, which further decays by alpha particle emission to radon-222 (see Chapter 1 7). The half-life of radon-222 is 3 . 825 days. Other shorter-lived isotopes are formed from the decay of thori um-232 and uranium-235 . Every square mile of soi l to a depth of 6 inches is estimated to contain about 1 g of radium, which releases radon in tiny amounts into the atmosphere.
�
�
Radon is used in implant seeds for the therapeutic treatment of local ized tumors. Radon is found 2 in the atmosphere with a concentration estimated to be 1 in 1 0 1 parts of air.
1 5 . 12 Compounds of the Noble Gases •
•
•
Helium, neon, and a rgon �
The ionization energies of helium, neon, and argon are very high. They form no stable neutral compounds.
�
If electrons are removed from noble gas atoms in ionization processes, the resulting noble gas ions are radicals, and they may form stable molecular ions such as NeAr+, ArH+, and HeNe+ i n the gas phase. I f a n electron is added t o them, these ions rapidly dissociate into neutral atoms.
Krypton �
Krypton forms a thermodynamically unstable neutral compound, KrF2 (recal l that diamond is also thermodynamically unstable). It is a volatile white sol id, which decomposes slowly at room temperature.
�
In 1 988, a compound with a Kr-N bond was discovered, but it is stable only at temperatures below -50°C.
Xenon �
�
Xenon is the noble gas element with the richest chemistry. It forms several compounds with fl uorine and oxygen, and even compounds with Xe-N and Xe-C bonds. The compound, XeF2 , is thermodynamically stable. Xenon is found in compounds with oxi dation numbers of +2, +4, +6, and +8. Direct reaction ofjluorine with xenon at high temperature results in the formation of compounds with oxidation numbers of +2 (XeF2), +4 (XeF4), and +6 (XeF6). I n the case of XeF 6 , high pressure is required. A l l three fluorides are crystall ine sol ids. In the gas phase, all are molecular compounds. Solid xenon hexajluoride is an ionic compound with a complex structure of XeF 5+ cations bridged by F - anions.
SG-200
�
Xenonjluorides are powerfu l jluorinating agents, reagents that attach fl uorine to other substances. Xenon trioxide is synthesized by the hydrolysis of xenon tetrajluoride. 6 X eF4(s) + 1 2 H20(1) - 2 Xe03(aq) + 4Xe(g) + 3 02(g) + 24 HF(aq) The triox ide is the anhydride ofxenic acid, H2Xe04 . In aqueous basi c solution, the acid forms the hydrogen xenate ion, HXe04-, which further di sproportionates to the perxenate 4 ion, Xe06 -, in which xenon attains its highest oxidation number of +8. 4 2 H X e04-(aq) + 2 0W(aq) - Xe06 -(aq) + Xe(g) + 02(g) + 2 H20( I)
•
Radon �
Radon chemistry is diffi cult to study because all of its isotopes are radioactive.
�
A radonjluoride of unknown composition is found to form readily, but it decomposes rapidly during attempts to vaporize it for further study.
THE IMPACT ON MATERIALS (Sections 15.13-15.14) 1 5 . 1 3 Lum inescent Materials •
•
Light emission from materials �
Incandescence is light emission from a heated obj ect, such as the fi l ament in a l amp or the particles of hot soot i n a candle fl ame. See the discussion of black-body radiation in Section 1 . 2 .
�
Luminescence i s light emission from materials caused by other processes, such a s light absorption, chemical reaction, impact with electrons, radioactivity, or mechanical shock.
�
Chemiluminescence is light emiss ion from materials caused by chemi cal reaction. The products of reaction are formed in energetical ly excited states that decay by light emission. The highly exothermic reaction of hydrogen atoms and fl uorine molecules provides an exampl e of infrared chemiluminescence. Hydrogen fluoride is produced with considerable vibrational energy, and it decays by em ission in the infrared region of the spectrum (see Major Technique I for a discussion of infrared absorption). Visible chemilum inescence is usually produced by electronic excitation of product molecules.
�
Bioluminescence is a form of chemiluminescence produced by living organisms, such as firefl i es and certain bacteria.
Light emission from molecu les �
Fluorescence is the prompt emission of light from molecules excited by radiation of higher frequency. Normally, absorption of ultraviolet radiation by a molecule l eads to emiss ion in the visible region; higher frequency absorption leads to lower frequency emission. [Note: With high-power l asers, simultaneous absorption of two or more photons by a molecu le may lead to emission of a single photon with a greater frequency than that of the individual photons absorbed.]
�
Phosphorescence is the slow or delayed emiss ion of light from molecules excited by radiation of higher frequency. In this case, the initially excited molecule undergoes a transition to a state that decays more slowly.
SG-20 1
�
•
Triboluminescence (from the Greek word, tribos, a rubbing) is luminescence that is produced by a mechanical shock to a crystal . It is readily observed in striking or grinding sugar crystals. Trapped nitrogen gas escapes whi le in an excited state and produces the radiation.
Phosphors �
Phosphorescent materials that glow when activated by the impact of fast electrons, as wel l as high frequency radiation such as ultraviolet or x-ray
�
Clusters of three phosphors are used for each dot in a color television or computer display screen. Commonly used phosphors for this purpose are europium-activated yttrium orthovanadate, YV04 , for the red color, silver-activated zinc sulfide for blue, and copper-activated zinc sulfide for green .
�
Fluorescent lamps make use of fl uorescent materials that are activated b y ultraviolet light. Mercury atoms in the lamp are excited by electrons in a discharge. The atoms emit l ight at 254 and 1 85 nm, which is absorbed by a phosphor thi nly coated on the surface of the lamp. A commonly used phosphor is calcium halophosphate, Ca5(P04)3F I xCix , doped with manganese(l I ) and antimony(l I l ) ions. An antimony(I I I)-activated phosphor emits blue l i ght and a manganese(l l )-activated phosphor emits yel low l ight. The net result is l ight with a spectral range that is approximately white. Fluorescent materials have important applications in medical research. Dyes such asjluorescein are attached to protein molecules to probe biological reactions. Fluorescent materials, such as sodium iodide and zinc sulfide, can be activated by radioactivity and are used in scintillation counters to measure radiation (see Chapter 1 7).
�
1 5 . 1 4 Nanomaterials •
•
Nanoparticles �
Particles ranging in size from I to 1 00 nm
�
Manufactured and manipulated at the molecular level
Nano materials �
Materials composed of nanoparticles or regular arrays of molecules or atoms such as nanotubes
�
Properties differ from the properties of atoms and from those of bulk materials
�
•
Used to create miniature circuits and drug del ivery systems
Nanotechnology �
Physical methods such as lithography, "top-down" approaches
�
Chemical methods such as the solution and vapor-phase synthesis, "bottom-up" approaches
�
See Box 1 5 . 1 in the text on Self-Assembling Materials .
SG-202
Chapter 16 THE ELEMENTS : THE d BLOCK
THE d-BLOCK ELEMENTS AND THEIR COMPOUNDS (Sections 16.1-16.2) 1 6. 1 Trends in Physical Properties •
d-Block elements -4
All are metals, most are good electrical conductors, particularly Ag, Cu, and Au.
-4
Most are malleable, ductile, l ustrous, and silver-white in color. Exceptions are Cu (red-brown) and Au (yellow).
-4
•
•
Most have higher melting and boiling points than main group elements. A maj or exception is Hg, which is a l iquid at room temperature.
Shapes of the d-orbitals -4
Some properties of the d-block elements fol low from the shapes of the d-orbitals.
-4
d-Orbital lobes are relatively far apart from each other, so electrons i n different d-orbitals on the same atom repel each other weakly.
-4
Electron density in d-orbitals is low near the nucleus. Because d-electrons are relatively far from the nucleus, they are not very effective in shielding other electrons from the positive charge of the nucleus.
Trends in atomic radii A cross a Period -4
Nuclear charge and the number of d-electrons both increase across a row from left to right.
-4
The first five electrons added to a d-subshel l are placed in different orbitals (Hund's rule).
-4
Since repulsion between d-electrons in different orbita ls is small, increasing effective nuclear charge is the dominant factor, so atoms tend to become smaller as Z increases in the first half of the d block. Exception: M n
-4
Further across the block, radii begi n to increase slightly with Z because electron-electron repulsion in doubly-fi l led orbitals outweighs the effect of increasing effective nuclear charge.
-4
Attractions and repulsions are finely balanced, and the range of atomic d-metal radii is small.
-4
d-Metals form many al loys because atoms of one metal can easily replace atoms of another metal .
Down a group - Periods 4 anti 5 -4
d-Metals in Period 5 are typical ly larger than those in Period 4.
-4
The effective nuclear charge on the outer electrons is roughly the same, whereas the n quantum number increases down the group. The usual pattern is an increase in size for group members from top to bottom. This pattern is followed in Periods 4 and 5 for the d-metals. Exception: M n
A cross Period 6 - the lanthanide contraction. -4
d-Metal radii in Period 6 are approximately the same as those in Period 4. They are smal ler than expected because of the lanthanide contraction, the decrease in radius along the first row of the fblock.
-4
There are 1 4 J-block elements before the first 6d-block element in Period 6, Lu. At Lu, the atomic radius has fal len from 224 pm for Ba to 1 73 pm for Lu.
SG-203
�
Period 6 d-block elements are substantially more dense than those in Period 5 owing to the lanthanide contraction. The radii are about equal, but the masses are almost twice as large. I r and 3 Os are the two most dense elements (approximately 22.6 g· cm- ).
�
A second effect of the contraction is the low reactivity of Pt and Au, whose valence electrons are so tightly bound that they are not readily available for chemical reaction.
1 6.2 T rends in Chemical Properties •
Oxidation numbers (states) �
We wil l use oxidation state and oxidation number interchangeably.
�
M ost d-block elements have more than one common oxidation number and one or more less common oxidation numbers (see text Figure 1 6.6). 2 Oxidation n umbers range from negative values to +8. Example: In [Fe(CO)4] -, Fe has an oxidation number of -2, whereas in OS04 , Os has an oxidation number of +8.
�
•
�
Oxidation numbers show greatest variability for elements in the middle of the d block. Example: Mn, at the center of its row, has seven oxidation numbers, whereas Sc, at the beginning of the same row, has only one (in addition to zero).
�
Period 5 and Period 6 d-block elements tend to have higher oxidation numbers than those in Period 4. Example: The highest oxidation number of Ni (Group 1 0, Period 4) is +4, whereas that of Pt (Group 1 0, Period 6) is +6.
Chemical properties and oxidation states �
Species containing d-block elements with high oxidation numbers tend to be good oxidizing agents. Examples: Mn04- with Mn +7 and CrO/- with Cr +6
�
S pecies containing d-block elements with high oxidation numbers tend to exhibit covalent bonding. E xample: M n207 with Mn +7 is a covalent l iquid at room temperature.
�
S pecies containing d-block elements with high oxidation numbers tend to have oxides that are acid anhydrides.
�
S pecies containing d-block elements with low oxidation numbers tend to be good reducing agents. Examples: CrO with Cr +2 and FeCIz with Fe +2 S pecies containing d-block elements with low oxidation numbers tend to exhibit ionic bonding. Example: M n304 , with Mn+2 and Mn +3 is an ionic sol id.
� �
Species containing d-block elements with low oxidation numbers tend to have oxides that are basic anhydrides.
SELECTED ELEMENTS: A SURVEY (Sections 16.3-16.4) 1 6.3 Sca n d i u m Through Nickel •
Properties of the metals (see text Table 1 6. 1 ) 3 3 � Densities increase from S c (2. 99 g· cm- ) to N i (8.9 1 g·cm- ). �
M elting points increase from Sc ( 1 540 K) to V ( 1 920 K), then decrease un iformly to Ni ( 1 45 5 K). Exception: Mn has an anomalously low mp, 1 25 0 K.
�
Boiling points increase from Sc (2800 K) to V (3400K), then show i rregular behavior to Ni (2 1 50 K).
SG -204
) 2 Valence electron configurations fol low the fill ing rule: 4s fi l ls before 3d from Sc, [Ar] 3d 4s , to 2 8 ) s Ni, [Ar] 3d 4s , with the exception of Cr, [Ar] 3d 4s . 2 Scandium, Sc, I Ar1 3d ) 4s �
•
�
A highly reactive metal that reacts with water as vigorously as Ca
�
Has only a few commercial uses and is thought not to be essential to l ife
�
•
Has one oxidation state, +3, i n compounds 3 � Small, highly charged S c + is strongly hydrated in water, forming the Sc(H20) /+ ion, which is a weak acid (about as strong as acetic acid). 2 2 Titanium, Ti, I Ar ) 3d 4s �
A light, strong metal passi vated by an oxide coating, which masks its inherent reactivity
�
Commonly found with the +4 oxidation number in its ores: rutile, Ti02 , and ilmenite, FeTi02
�
Requires strong reducing agents to extract from its ores
�
The metal is used in j et engines, dental appliances; the oxide, T i02 , is used as a paint pigment
�
Forms a series of oxides cal led titanates
� •
Barium titanate, BaT i 03 , is a piezoelectric material (develops an electrical signal when stressed). 3 2 Vanad ium, V, I Ar ) 3d 4s
�
A soft, silver-gray metal
�
V metal i s used inferroalloys (Fe, V, C) to make tough steels for automotive springs
�
Exhibit positive oxidation numbers ranging from + I to +5
�
Common oxidation numbers are +4 and +5.
�
Many colored compounds containing vanadium species are used as ceramic glazes.
�
•
Orange-yellow Vanadium(V) oxide, V20S, is used as an oxidizing agent/catalyst in the contact process for the manufacture of sui furic acid. Chromium, C r, I Ar) 3d s 4s J
�
•
A bright, corrosion-resistant metal
�
Used to make stainless steel and for chromium plating
�
Obtained from chromite, FeCr204 , by reduction with C in an electric furnace:
�
FeCr204(s) + 4 C(s) � Fe(l) + 2 Cr(l) + 4 CO(g) 2 An exception to the normal fi l l ing order of electrons (3d s 4s ) not 3d 4 4s ,
Compounds of C r �
Positive oxidation numbers range from + I t o +6.
�
Common oxidation numbers are +3 and +6.
�
Chromium(IV) oxide, Cr02 , is used as a ferromagnetic coating for "chrome" magnetic tapes.
�
Sodium chromate, Na2Cr04(s), is used to prepare many other Cr compounds. l n acid solution, CrO/- forms the dichromate ion, Cr20/-; in both ions, the oxidation number of Cr is +6. 2 CrO/-(aq) + 2 H30+(aq)
�
Cr20/-(aq) +3 H20(l)
� •
)
Cr compounds are used as pigments, corrosion inhibitors, fungicides, and ceramic glazes. 2 Manganese, M n, I A rl 3d s 4s
�
A gray metal that corrodes easily and is rarely used alone
�
Important in alloys such as steel and bronze
SG-205
� •
Obtained from the ore, pyrolusite, mostly Mn02
Compounds of M n � Positive oxidation numbers range from + 1 to +7. �
The most stable oxidation number i s +2, but +4, +7, and, to some extent, +3 are also common.
�
The most important compound of Mn, Mn02 or manganese dioxide, is a brown-black solid used as a decolorizer in glass, to prepare other Mn compounds, and in dry cells.
�
•
•
The permanganate ion, Mn04-, with M n i n the +7 oxidation state, i s an important oxidizing agent and a mild disinfectant. 6 2 I ron, Fe, [Ar1 3d 45
�
The most abundant element on earth and the most widely used d-metal
�
Obtained from the principal ores hematite, Fe203 , and magnetite, Fe304
�
Reactive metal that readily corrodes in moist air, forming rust
�
Forms corrosion-resistant al loys and steels (see text Table 1 6.2)
�
Aferromagnetic metal, as are its oxides and many of its alloys
Compounds of Fe �
Positive oxidation numbers range from +2 to +6.
�
Common oxidation numbers are +2 and +3.
�
S alts of iron vary in color from pale yellow to dark green. 2 Color in aqueous solution is dominated by [FeOH(H20)s] +, the conj ugate base of F e(H20)6] 3+.
� �
•
•
Fe(!!) is readily oxidized to Fe(JJ!); the reaction is slow i n acid solution and rapid in base. 2 Cobalt, Co, JAr) 3d ? 45 �
A silver-gray metal used mainly in Fe alloys
�
Permanent magnets such as those used in loudspeakers are made of alnico steel, an alloy of Fe, Ni, Co, and AI.
�
Co steels are hard and are used for dril l bits and surgical tools.
Compounds of Co �
Positive oxidation numbers range from + 1 to +4.
�
Common oxidation numbers are +2 and +3.
� •
Co(!!) oxide, CoO(s), i s a deep blue salt used to color glass and ceramic glazes. 2 Nickel, Ni, JAr) 3d B 45
�
A hard, si lver-white metal used mainly in stainless steel
�
Used as a catalyst for the hydrogenation of organic molecules
�
Obtained as a pure metal by the Mond process (the fi rst and third steps require heat): NiO(s,ore) + H2(g)
�
Ni(s, impure) + H20(g)
N i(s, i mpure) + 4 CO(g) Ni(COMI, pure) •
�
�
Ni(COMI, pure)
Ni(s, pure) + 4 CO(g)
Compounds of Ni �
Positive oxidation numbers range from + 1 to +4.
�
Common oxidation numbers are +2 and +3 ; +2 is the most stable. 2 The green color of aqueous solutions of nickel salts arises from [N i(H20)6] + ions.
�
SG-206
� •
I n nickel-cadmium (nicad) batteries, Ni(IJI) is reduced to Ni(JJ) .
Carbonyl compounds �
Many transition metal s form compounds with CO.
�
In carbonyl compounds, both the metal and carbonyl (CO) groups are regarded as nearly neutral species, and the metal is assigned an oxidation state of o.
�
When Fe is heated in CO, it reacts to form trigonal-bipyramidal iron pentacarbonyl, Fe(CO)5 , a yellow molecular l iquid that melts at -20°C, boi ls at 1 03°C, and decomposes in visible light.
�
Other examples of transition metal carbonyls are nickel tetracarbonyl, Ni(CO)4' a colorless, toxic, flammable l iquid that boils at 43°C and chromium hexacarbonyl, Cr(CO)6, a colorless crystal that sublimes readily. Ni(CO)4 is tetrahedral; and Cr(CO)6 is octahedral.
�
Some metal carbonyl compounds can be reduced to produce anions in which the metal has a negative 2 oxidation number. E xample: In the [Fe(CO)4] - anion, produced by reduction of Fe(CO)5, Fe has an oxidation number of -2 : Fe(CO)5
Na ---:----,--=--� ) tetrahydro furan
-
2 [Fe(CO)4] - + CO
1 6.4 Gro u ps 1 1 a n d 1 2 •
Properties of the metals (see Text Table 1 6.3) �
A l l the elements of Groups I I and 1 2 are metals with completely fi l led d-subshel ls.
�
The Group I I metal s, copper, silver, and gold, are cal led the coinage metals and have valence electron configurations of (n I )d IOns 1 .
�
The low reactivity of the coinage metals derives partly from the poor shielding abil ities of the d electrons, and hence the strong attraction of the nucleus on the outenTIost electrons. The effect is enhanced in Period 6 by lanthanide contraction, accounting for the inertness of gold.
-
�
The Group 1 2 metal s, zinc, cadmium, and mercury, have valence electron configurations of 2 (n I )d IOns . Zinc and, to a lesser extent, cadmium resembl e beryllium or magnesium in their chemistry. Copper, C u, IAr) 3d 1 0 4s 1 -
•
�
Found as the metal and in sulfide ores such as chalcopyrite, CuFeS2
�
Used as an electrical conductor i n wires
�
Forms alloys such as brass and bronze, which are used in plumbing and casting
�
Corrodes in moist air forming green basic copper carbonate: 2 Cu(s) + H20(l) + Oig) + CO2 (g) - CU2(OH)2C03(S)
� •
I s oxidized by oxidizing acids such as HN03
Compounds of Cu � � � �
Common oxidation numbers are + I and +2; +2 is more stable.
Curl) disproportionates in water to form metallic copper and copper( l I ).
Cu(JJ) forms a pale-blue hydrated ion in water, [Cu(H20)6f+.
Cu is essential in animal metabolism. In mammals, Cu-conta ining enzymes are required for healthy nerve and connective tissue. In species such as octopi and lobsters, Cu (not Fe) is used to transport oxygen in the blood.
SG-207
•
Silver, Ag, [Kr[ 4([ 1 0 5s �
Rarely found as the free metal and obtained as a byproduct of the refining of Cu and Pb
�
Reacts with sulfur to produce a black tarnish on silver dishes and cutlery
�
Like Cu, Ag i s oxidized by oxidizing acids. 3 Ag(s)
•
1
+
4 H30\aq) + N03-(aq) - 3 Ag+(aq) + NO(g) + 6 H20(l )
Compounds o f Ag �
Oxidation numbers are + 1 , +2, and +3 , but + 1 is most stable.
�
Ag(J) does not disproportionate in water.
�
Except for AgN03 and AgF, Ag salts are insoluble in water.
�
•
Silver halides are used in photographic film. Gold, Au, [ Xe) 4f14 5([ 1 0 6s 1 �
So inert that it is usual ly found as the free metal
�
Highly malleable, easily made into thin foi l (gold leaf)
�
Cannot be oxidized by nitric acid, but it dissolves i n aqua regia, a mixture of sulfuri c and hydrochloric acids: Au(s) + 6 H \aq) + 3 N03-(aq) + 4Cr(aq) - AuCI4-(aq) + 3 N02(g) + 3 H20(l)
�
A carat is 1 /24; 1 0- and 1 4-carat gold contains 1 0/24 and 1 4/24 parts of gold(s) by mass, respectively.
Positive oxidation numbers are + I, +2, and +3, but + 1 and +3 are most common. 2 Zinc, Zn, [Ar) 3 ([ 1 0 4s
• •
•
�
A silvery, reactive metal , found mainly in the sulfide ore, sphalerite, ZnS
�
Obtained from the ore by froth flotation, fol lowed by smelting with coke
�
Used primari ly for galvanizing Fe
�
Like Cu, Zn i s protected by a hard film of basic carbonate, Zn2(OH)2C03 .
Compounds of Zn �
Common positive oxidation number is +2.
�
An amphoteriC metal , Zn dissolves in both acidic and basic solutions: 2 Acid: Zn(s) + 2 H30+(aq) - Zn +(aq) + H2(g) + 2 H20(l) Base: Zn(s) + 2 0H-(aq) +2 H20(l) - [Zn(OH)4f- + H2(g) Zn(OH)/- is called the zincate ion. The reactions above suggest that galvanized containers should not be used to transport either acids or alkalis.
�
Zn is an essential element for human health; it occurs in many enzymes.
� •
Zn is toxic only in very large amounts. Cadmium, Cd, [ Kr) 4([ 1 0 5s 2
� •
A silvery, reactive metal
Compounds of Cd �
Common positive oxidation number is +2.
�
Cadmiate ions (analogous to zincate ions) are known, but Cd does not react with strong bases. Like Zn, Cd reacts with nonoxidizing acids.
�
Unl ike Zn, Cd salts are deadly poisons.
SG-208
�
Cd disrupts human metabolism by replacing other essential metals in the body, such as Zn and Ca, leading to soft bones, and to kidney and lung disorders.
� •
•
Cd and Zn are alike chemical ly in many ways, but both are quite different from Hg. 10 4 Mercu ry, Hg, I Xel 4f1 5£1 6s 2
�
A volatile, si lvery metal , and a l iquid at room temperature (Ga and Cs are l iquid on warm days)
�
Found mainly in the sulfide ore, cinnebar, HgS
�
Obtained from the ore by froth flotation, followed by roasting in air; vapor is poisonous.
�
Combines with many metals to form mercury alloys called amalgams
�
The liquid temperature range (from -39°C to 357°C) is unusual ly wide and makes Hg well suited for use in thermometers, si lent electrical switches, and high-vacuum pumps.
Compounds of Hg � �
Positive oxidation numbers are + I , +2 . 2 The Hg(J) cation, [Hg-Hg] +, is diatomic, with a covalent bond between the nuclei. It is usual ly . 2 wri tten as H g2 +.
�
Hg does not react with nonoxidizing acids or with bases, but does react with oxidizing acids in a reaction simi lar to those for Cu and Ag: 2 3 Hg(l) + S H +(aq) + 2 N03-(aq) - 3 Hg +(aq) + 2 NO(g) + 4 H20(l)
�
Hg compounds, particularly organic ones, are acutely poisonous. Frequent exposure to low levels of Hg(g) results in the accumulation of Hg in the body. Damaging effects after such chronic exposure include impaired neurological function and hearing loss.
COORDINATION COMPOUNDS (Sections 16.5-16.7) 1 6.5 Coord ination Com plexes •
U nderstanding complexes (terminology) �
Coordination number is the number of l inks formed by a central metal atom or ion, where a link corresponds to a single or multi ple bond . .
�
Ligand i s a molecule o r ion attached to a metal atom or ion by one or more coordinate-covalent bonds.
�
A ligand coordinates to the metal when it attaches.
�
For ionic complexes, l igands attached directly to a central atom are enclosed in brackets. Example: [ZnC14f
�
The charge of the ion is displayed in the upper right corner as usual .
�
Ligands attached to a metal define the coordination "sphere " of the central atom or ion.
�
Common coordination spheres include octahedral (almost all 6-coordinate complexes), tetrahedral (4-coordinate complexes) or square planar (also 4-coordinate).
�
The isocyano l igand is attached to the metal by the electron pair on the nitrogen end.
SG-209
•
•
N am ing d-metal complexes and coordination compounds �
Use the formulas and names for the common ligands given in text Table 1 6.4.
�
Follow the procedure outlined in text Toolbox 1 6. 1 .
Ligand substitution reaction �
A reaction in which one Lewis base takes the place of another
�
Sometimes all the l igands are replaced, but often the substitution is i ncomplete. 2 E xample: [N i(H20)6f+(aq) + 6 NH3(g) [Ni(NH3)6] +(aq) + 6 H20(l) �
1 6.6 The Shapes of Com plexes �
Exhibit a variety of shapes (coordination geometries) depending on the coordination number
•
Complexes
•
Coordination number 6 � �
•
•
Complexes are usually octahedral, with ligands at the vertices of the octahedron and the metal at the center. 3 Some examples are hexamminecobalt( l l l), [Co(NH3)6] +, and hexacyanoferrate(ll), [Fe(CN)6t-.
l°
3+
3
_ _ NH3 H3N _ _ H_
H3N
/
1
I
/
NH3
NH3
Coordination number 4 �
Two shapes are common, tetrahedral and square planar. 2 Examples: [Zn(OH)4] - and TiCl4 are tetrahedral complexes.
�
Square planar complexes are most common for metal species with a d 8 electronic configuration, 2 2 3 such as N i +, Pt +, and Au + . 2 2 Examples : [PtCI4] - and [Ni(CN)4] - are square planar complexes.
Coordination number 2 �
Complexes are linear. Examples: Dimethyhnercury(O), Hg(CH3)2 , and diamminesilver(l ), [Ag(NH3)2 r
•
Metallocenes
�
"Sandwich" compounds with aromatic ri ngs bonded by 2pn:-electrons to a metal atom or ion. The aromatic rings act as ligands.
Examples: D ibenzenechromium, Cr(C6H6)2 , and ferrocene, [Fe(CsHs)2], both shown on the right. •
•
Other coordination geometries �
Complexes with coordination number 5 (trigonal-bipyramidal or square- pyramidal)
�
Coordination number 8 (square-anti prism or dodecahedral) is also known (see text).
Monodentate and polydentate ligands �
A monodentate ligand can bind to a metal at only one site. Examples: Chloro, C l -; cyano, CN -; carbonyl, CO; and methyl , CH3 .
�
A polydentate ligand can bind to a metal at more than one site simultaneously.
�
A bidentate ligand can bind to a metal at two sites.
SG-2 1 0
Fe
�
Tridentate l igands may bind at three sites.
�
Chelates
-
polydentate l igands that can form rings with a metal ion.
Examples: The ox and en l igands above can form five-membered rings with metal ions. 1 6.7 Isomers •
Isomers �
•
Isomers are species that contain the same numbers of the same atoms but in different arrangements. Two major classes of isomers:
•
�
I ) Structural isomers, in which some atoms have different partners (different connectivity)
�
2) Stereoisomers, in which all atoms have the same partners (same connectivity) but some atoms are arranged differently in space.
�
Types of structural isomers : ionization, hydrates, linkage, and coordination.
�
Types of stereoisomers : geometrical and opticals.
Ionization isomers � �
Differ by the exchange of a ligand with an anion or molecule olltside the coordination sphere Form different ions in sol ution Example: [CoBr(N H3)S] S04 and [CoS04(NH3)s]Br are ionization isomers.
•
Hyd rate isomers �
Special type of ionization isomer
�
Differ by the exchange of a H20 molecule and another l igand in the coordination sphere E xample: The sol id hexahydrate of chromium(II1) chloride, CrCI3 · 6 H20, is known in three forms: [Cr(H20)6]C I3, [CrCI(H20)s]Ch ' H20, and [CrCI2(H20)4] CI · 2 H20.
•
Linkage isomers �
Differ in the identity of the atom in a given l igand attached to the metal
�
A l igand may have more than one atom with a lone pair that can bond to the metal, but because of its size or shape, only one atom from the l igand may bond to the metal at one time. E xamples: N02- (M-N02 nitro and M-ONO nitrito); CN - (M-CN cyano and M -NC isocyano) The Lewis structure for CN - clearly shows the lone pairs on C and N that can bond to a metal ion. [:C = N : r
•
Coordination isomers �
Occur when one or more ligands is exchanged between a complex that is a cation and one that is an anion in a coordination compound Example: [Cu(N H3)4] [PtC I4] and [Pt(NH 3 )4] [CuCI4]
SG-2 1 1
•
Geometrical isomers �
Atoms bonded to the same neighbors, but with different orientations
�
Geometrical isomers are also stereoisomers. E xample: Cis (adjacent) and trans (across) isomers of the
NH 3 I CI - �t - NH 3 CI Cis
NH 3 I CI - �t - CI NH 3 Trans
square-planar complexl [Pt(NH)3CI2] •
Optical isomers �
Special type of stereoisomers: They are nonsuperimposable mirror images of each other. Example: C H BrCI F has a tetrahedral shape and two optical isomers:
�
•
Sr I
C C I ..... �/H F (a)
(b)
' Molecules (a) and (b) are not superimposable; therefore, they are different compounds. They are related to each other as the left hand is related to the right hand (mirror images).
Chirality and optical activity �
A chiral species is not identical to its mirror image. All optical isomers are chiral.
�
An achiral species is identical to its mirror image.
�
A chiral species and its mirror image form a pair of enantiomers. M olecules (a) and (b) shown above are enantiomers.
�
Chiral molecules display optical activity, the abil ity to rotate the plane of polarization of l ight. For a given pair, one enantiomer rotates the plane clockwise, whereas its mirror i mage rotates it by the same amount counterclockwise.
�
A racemic mixture is a mixture of enantiomers in equal molar proportions.
�
Because enantiomers rotate light equally in opposite directions, a racemic sample is not optically active.
THE ELECTRONIC STRUCTURES OF COMPLEXES (Sections 16.8-16.12) 1 6. 8 Crystal Field Theory
•
�
Accounts for the optical and magnetic properties of complexes
�
Ligands are considered to be point negative charges.
d-Orbital splitting in an octahedral complex, M L 6 � Ligands approach the positively charged metal ion along the X-, y-, and z-axes, and the complex is stabilized by coulombic attraction. But, the ligands, represented by negative charges, also repel electrons in the d-orbitals. �
This repulsion causes the d-orbitals to be split into two sets: t2g (the three orbitals whose lobes lie between the ligands (dXY ' dxz , and dyz ) , and eg (the two orbitals whose lobes point essentially towards the ligands (dx, -y2 and dz, ) .
SG-2 l 2
•
�
Both the t2g and eg orbitals are raised i n energy, but the eg orbitals are raised more than the t2g orbitals because they are closer to the point negative charges.
�
With respect to the average energy of the d-orbitals, the three t2g-orbitals are lowered in energy by (2/5)L10 and the two eg-orbitals and are raised in energy by (3/5)L1o .
�
The separation in energy between the t2g- and eg-orbitals is the octahedral ligandfield splitting, L1o, where the 0 denotes octahedral .
d-Orbital splitting in a tetrahedral complex, M L 4 �
Ligands approach the positively charged metal ion along the corner directions o f a tetrahedron.
�
The d-orbitals are split into two sets: t2 (three orbitals, dxy , dxz , and dyz ) and e (two orbitals, dX , -y, and dZ 1 ).
�
I n this instance, with respect to the average energy of the d-orbitals, the three t2-orbitals (dry ' dx: ' and dyJ are raised in energy by (2/5)L1T and the two e-orbitals (dx' _y' and dZl ) are lowered in energy by (3/5)L1T . The separation in energy between the t2- and e-orbitals is the tetrahedral ligandfield splitting, L1T , where the T denotes tetrahedral .
---*
•
Comparison o f octahedral and tetrahedral complexes �
Tetrahedral ligand-field splitting values tends to be smaller than corresponding octahedral splitting values, in part because there are fewer l igands to repel electrons in the d-orbitals in the tetrahedral case.
�
The relative energies of the t2 and e sets of orbitals are reversed in octahedral and tetrahedral complexes.
�
I n tetrahedral complexes, the subscript g is not used to describe the orbitals.
1 6.9 The Spectrochemical Series •
•
•
•
Relative strength of the l igand field splitting, L1 �
For a given d-metal ion in a given oxidation state, L1 depends on the ligand.
�
Some ligands produce larger values of L1 than others.
�
The relative values of L1 are approximately in the same order for all d-metal ions.
Spectrochemical series �
Ligands are arranged according to the magnitude of the ligand-field splitting they produce.
�
Weakfield ligands produce smal l values of L1: 1 - < Br- < C I - ;:: SCN- < F - < OH - < (ox) < H 2 0.
�
Strongfield ligands produce large values of L1: NH3 < (en) < N 02- < CN- ;:: CO.
Electron configurations in isolated atoms �
In an isolated atom, the five d-orbitals in a subshell have the same energy.
�
According to Hund's rule, electrons occupy each degenerate orbital separately with spins parallel until each has one electron.
Octahedral complexes �
Similar rules apply to d-metal complexes; however, the five d-orb itals in a subshel l are divided into two groups.
�
The lower energy 12g-orbitals fill first; degenerate orbitals (dxy dy: , d:x ) fill with spins parallel } until the 12g-orbitals are half full (d complex). ,
SG-2 1 3
�
7 The electron configurations of d 4 through d complexes depend on the magnitude of �o compared with the energy required to pair an electron in a t2g-orbital.
�
I f � o is large compared with the spin-pairing energy (strong-jield ligands), electrons wil l enter the lower-energy t2g-orbitals with spins paired before fi ll ing the eg-orbitals.
�
If � o is small compared with the spin-pairing energy (weak-jield ligands), electrons will enter the higher-energy eg-orbitals with spins parallel before pairing begins i n the t2g-orbitals. 3 For d I through d and d 8 through d 10 complexes, the orbitals fi l led are the same for strong
�
•
field and weak-field ligands.
�
A similar procedure is fol lowed for tetrahedral complexes, but the two groups of d-orbitals are reversed in energy, and the e-orbitals fi l l first.
�
Electron configurations of dn complexes are given in text Table 1 6. 5 .
H igh- and low-spin complexes �
A dn complex with the maximum number of unpaired electrons is called a high-spin complex. High-spin complexes are expected for weak-jield ligands.
�
A d" complex with the minimum number of unpaired electrons is call ed a low-spin complex. Low-spin complexes are expected for strong-field ligands. 4 7 The d through d octahedral complexes may be high- or low-spin.
� �
Because �T is smal l with respect to the energy required to pai r an electron i n the e-orbitals, t2orbitals are always accessible and tetrahedral complexes are almost always high-spin.
1 6 . 1 0 T h e Colors of Com plexes •
•
Absorption of l ight and transmission or reflection �
Visible light contains all wavelengths from about 400 nm (blue) to 800 nm (red).
�
Transition metal complexes often absorb some visible light and transmit or reflect the rest.
�
Transmitted or reflected l ight (the complementary color of the l ight absorbed) is the l ight we see and determines the color of an object.
�
The color wheel (text F ig. 1 6.30) provides a simple way of relating color absorbed to color seen in some systems.
E lectronic transitions in complexes �
Light absorption in complexes is often associated with d-d transitions. An electron is excited from a t2g- to an eg-orbital in an octahedral complex or from an e- to a t2-orbital in a tetrahedral complex.
�
Charge-transfer transitions may also occur in which an electron is excited from a ligand centered orbital to a metal-centered orbital or vice versa. These transitions are often q uite intense. Example: the deep purple color of the pennanganate ion, Mn04-
�
Color arises when a substance absorbs l ight from a portion of the visible spectrum and transmits or reflects the rest. Both d- d and charge-transfer transitions can occur in the visible region and either or both processes can contribute to the color of a transition metal complex.
SG-2 l 4
•
Color, d- d transitions, and the spectrochemical series �
Weak-jield complexes have small � values and absorb low-energy (red or infrared ) radiation.
�
Absent strong charge-transfer transitions, solutions of these complexes appear green (the complementary color of red) or colorless (if the absorption is in the infrared).
�
Strong-jield complexes have large � values and absorb high-energy (blue, violet, or ultraviolet) radiation.
�
In the absence of complicating charge-transfer transitions, solutions of these complexes appear orange or yellow (the complementary colors of blue and violet, respectively) or colorless (if absorption is in the ultraviolet).
�
The ligand field splitting � can be determined from the electronic spectrum of a given complex, but, in most cases, it is not related simply to the color of its solution.
1 6. 1 1 Magnetic Properties of Com plexes •
•
Complexes may be diamagnetic or paramagnetic �
Diamagnetic complexes have no unpaired electrons and are repel l ed by a magnetic field.
�
Paramagnetic complexes have one or more unpaired electrons and are attracted into a magnetic field. The greater the number of unpaired electrons, the greater the attraction.
Magnetic properties of complexes �
Determined from the d-electron configuration of the complex
�
Strong-field ligands produce weakly paramagnetic low-spin complexes.
�
Weak-field ligands produce strongly paramagnetic high-spin complexes.
1 6 . 1 2 Ligand Field Theory •
Ligand field theory �
Molecular Orbitals (MOs) are generated from the available Atomic Orbitals (AOs) in a complex.
�
Only one Atomic Orbital (AO) is used for each monodentate l igand (two for bidentate, etc.). Examples: For a chloride l igand, a CI 3p-orbital, directed toward the metal, is used. For an ammonia I igand, the sp 3 lone-pair orbital is chosen. For bidentate ethylenediamine(en), two nitrogen Sp 3 lone-pair orbitals are chosen.
•
�
The treatment of n-bonding in complexes provides further insight into the spectrochemical series.
�
Ligand field theory explains why uncharged species such as CO are strong-field l igands, whereas negatively charged species such as C I - are weak-field ligands.
�
Based on electrostatic considerations, we would expect the opposite to be true.
Octahedral complexes �
For first-row transition elements, the 4s-, 4p-, and 3d-orbitals (nine total orbitals) of the metal i on are chosen because they have similar energies.
�
The nine metal and six l igand orbitals (one from each ligand) yield a total of I S AOs, which overlap to form I S MOs (see text Fig. 1 6.37).
�
The final result is that six MOs are bonding, six are antibonding, and three are nonbonding.
�
The 1 2 ligand electrons and the available d-valence electrons are placed in the MOs in accordance with the bui lding-up principle to determine the ground-state electron configuration of the complex.
SG-2 1 S
•
•
�
I n transition metal ions, the valence s- and p-orbitals are typically vacant, whereas the d-orbitals are partially occupied.
�
The first 1 2 electrons form six metal-ligand sigma bonds. The d-electrons of the metal enter the t2g- and eg-orbitals in the same way they did in the crystal field theory. Ligand field theory, however, identifies the t2g-orbitals as nonbonding and the eg-orbitals as antibonding.
�
n-Bonding in octahedral complexes arises from overlap of the metal t2g-orbitals with p- or n-ligand orb itals to form bonding and antibonding MOs, n and n*, respectively.
�
The details of this interaction are shown in text F ig. 1 6.38.
Weak-field ligand (text Fig. 1 6.38a) �
The antibonding l igand n:*-orbital is too high in energy to take part in bonding.
�
The bonding ligand n:-orbital combines with a metal 12g-orbital and the combination raises the energy of the MO that the metal d-electron occupies.
�
The electron from the metal enters an antibonding MO. The energy of the eg-orbitals is unchanged, but the spl itting between the t2g- and eg-orbitals is reduced.
�
The l igand field spl itting,
�
Some examples of weak-field ligands are C I - and Br-, which have a valence shell p-orbital, with two electrons in it, oriented perpendicular to the metal-ligand sigma bond.
60,
is decreased in this case.
Strong-field ligand (text Fig. 1 6.38b) �
The bonding n:-orbital of the ligand is too low in energy to take part in the bonding. The empty antibonding n:*-orbital of the ligand combines with a t2g-orbital of the metal , and the combination lowers the energy of the MO that the metal d-electron occupies. The electron from the metal enters a bonding M0.
�
The energy of the eg-orbitals is unchanged, and the spl itting between the t2g- and eg-orbitals is increased.
�
The l igand field splitting,
�
Some examples of strong-field ligands are CO and eN - which each have an unoccupied valence shell n*-orbital oriented perpendicular to the metal-ligand sigma bond.
60,
is increased in this case. ,
THE IMPACT ON MATERIALS (Sections 16.13-16.15) 1 6 . 1 3 Steel •
I ron �
Pig iron, which contains 3-5% e, -2% Si, and lesser amounts of other impurities, is produced by reduction of iron ores in a blast furnace.
�
Cast iron is simi lar to pig iron, but with some impurities removed.
�
Both forms are hard and brittle, whereas pure iron is malleable and more flexi ble.
� •
I n the blast furnace, l imestone is used to remove impurities, such as Si02, Ah03, and P40 1 O . Steels
�
Steels are homogeneous al loys (solid solutions) made from pig iron. First, pig iron is puri fied to lower the carbon content and to remove other impurities. Then, appropriate metals are added to form the desired steel.
SG-2 1 6
�
Steels typical ly contain 2% or less carbon; their hardness, tensile strength, and ducti l ity depend on the carbon content. Higher carbon content produces harder steel. The strength of steel can be increased greatly by heat treatment.
�
The corrosion resistance of steel is significantly improved by alloying with other metals. H ighly corrosion-resistant stainless steel is an alloy contai ning about 1 5% Cr by mass.
1 6 . 1 4 Nonferrous Alloys •
•
Alloys �
Alloys are solid mixtures of metals and are homogeneous or heterogeneous solutions.
�
Homogeneous alloys include brass, bronze, and coins containing gold or silver.
�
Heterogeneous alloys include tin-lead solder and mercury amalgams for dental fil l ings.
Substitutional alloys �
Homogeneous al loys with metals of similar radius. Example: Brass, consisting of Cu and Zn
�
Typically harder than the pure metals but with lower electrical and thermal conductivity
1 6. 1 5 Magnetic Materials •
Magnetism �
Arises from unpaired electrons in substances
�
Three general types: paramagnetic, ferromagnetic, and anti ferromagnetic
�
Paramagnetic: Unpaired electron spins are oriented randomly. Paramagnetic materials behave as weak magnets in a magnetic field, but the magnetism dissipates rapidly when the magnetic field is removed.
�
Ferromagnetic: Unpaired electron spins are all al igned parallel in domains. Ferromagnetic materials are used to make permanent magnets .
�
Antiferromagnetic: Unpaired electron spins are al igned antiparallel, as i n manganese, and their magnetic moments cancel.
�
Many d-block elements have unpaired d-electrons and are paramagnetic or ferromagnetic. �
•
Ferromagnets
•
Ceramic magnets
•
•
Permanent magnets used for coating cassette tapes and computer disks
�
Made from barium ferrite, BaO· n Fe203 , or strontium ferrite, SrO· n Fe203
�
Used as refrigerator magnets; brittle, relatively lightweight, and inexpensive
Ferrofluids �
Magnetic liquids; suspensions of powdered magnetite, Fe304, in a viscous oil/detergent mixture
�
The Fe304 particles are attracted to the polar ends of the detergent molecules, which form micel les that are distributed throughout the oil. When a magnet approaches the fl uid, the particles try to al ign, but are inhibited from doing so by the viscous oil . As a result, it is possible to control the flow and position of theferrojluid by means of an applied magnetic field.
Single- molecule magnets �
Contain d-block metal atoms bonded to groups of nonmetals
�
Behave like nano-size compass needles in a magnetic field
�
Can be made with shapes such as rings, tubes, or spheres
�
Have potential uses in el ectronic media storage devices such as computer chips
SG-2 1 7
Chapter 17 NUCLEAR CHEMISTRY
NUCLEAR DECAY (Sections 17.1-17.5) 1 7 . 1 The Evidence for Spontaneous N u clea r Decay •
Radioactivity �
•
Spontaneous emission of radiation or particles by a nucleus undergoing decay
�
Radioactivity includes the emission of ex particles, � particles, and y radiation.
�
These three types, the most common forms of radioactivity, were the first ones to be discovered.
N uclear radiation (see Table 1 7. 1 in the text): ex
particles
i H e 2 + , traveling at speeds equal to about
�
Helium-4 nuclei,
�
Deflected by electric and magnetic fields
�
When ex particles interact with matter, their speed is reduced and they are neutral ized.
�
Denoted by
i ex
or
1 0% of the speed of light, c
ex
� particles �
Rapidly moving electrons, e-, emitted by nuclei at speeds less than 90% of the speed of light
�
Deflected by electric and magnetic fields, sometimes called negatrons
�
When traveling between positively and negatively charged plates, opposite directions. Denoted by _� e , �- , or �
�
ex
and � particles are deflected in
y radiation �
High-energy photons (electromagnetic radiation) travel ing at the speed of light
�
Uncharged and not deflected by electric and magnetic fields
�
Denoted by y or � y
�+ particles �
Called positrons
�
Have the mass of an electron but carry a positive charge
�
When an electron and a positron, its antiparticle, meet, they are annihilated and are completely transformed i nto energy, mostly y radiation.
�
Denoted by
p particles � �
�e
+
or �+
Protons are positively charged subatomic particles found in nuclei . When emitted in nuclear decay, protons typically travel at speeds equal to about 1 0% of the speed of l ight. Denoted by : H + , : p , or p
n pa rticles �
Neutrons are uncharged subatomic particles found in nuclei . The mass of a neutron is sl ightly larger than the mass of a proton (see Table B. I in the text). When emitted in nuclear decay, neutrons typically travel at speeds less than about 1 0% of the speed of l ight.
�
Denoted by
bn
or n
SG-2 1 9
•
Penetrating power of nuclear radiation ---t
Penetrating power into matter decreases with increased charge and mass of the particle.
---t
Uncharged y radiation and neutrons are more penetrating than charged positrons or alpha particles.
---t
•
Singly charged � particles (electrons) are more penetrating than the more massive, doubly charged particles, which are virtual ly nonpenetrating.
ex
Antiparticles ---t
S ubatomic particles with equal mass and opposite charge
---t
Annihilate each other upon encounter, producing energy
1 7.2 Nuclear Reactions •
N ucleons and nuclides ---t
A nucleon is a proton or a neutron in the nucleus of an atom.
---t
A nuclide is an atom characterized by its atomic number, mass number, and nuclear energy state.
---t
Nuclides are often denoted by
4 E , where Z is the atomic number, A
plus neutrons), and E is the symbol for the element. Note that A
-
is the mass number (protons
Z equals the number of neutrons
in the nuclide.
•
---t
A nuclide is a neutral atom and therefore is uncharged.
---t
Radioactive nuclei can change their structure spontaneously by nuclear decay, the partial breakup of a nucleus.
N uclear reactions ---t
Any transformations that a nucleus undergoes
---t
Differ from chemical reactions in three important ways: I ) I sotopes of a given element often have different nuclear properties but always have similar chemical properties. 4 2 Example: 1 C and 1 C are difficult to separate chemically because they have similar chemical properties, yet the two isotopes have different nuclear stabi l ity. Carbon- 1 2 is a stable nuclide, whereas carbon- 1 4 decays with a half-life of 5730 y (see Section 1 7. 7) . 2 ) Nuclear reactions often produce a different element. Example: Titani um-44 captures one of its electrons to produce the nuclide scandium-44. 3) Nuclear reactions involve enormous energies relative to chemical reactions.
•
Balancing nuclear reactions ---t ---t
Like chemical reactions, nuclear reactions must be balanced with respect to both charge and mass . Charge balance: For the several nucl ides
4E
in a nuclear reaction, the sum of the atomic numbers
Z of the reactants must equal the sum of the Z values of the products.
---t
Mass balance: The sum of the mass numbers A of the reactants must equal the sum of the mass numbers A of the products.
---t
The result of most nuclear decay reactions is nuclear transm utation, the conversion of one element into another.
---t
For nuclear decay reactions >
>
A reactant nuclide is called a parent nuclide. A product nucl ide is called a daughter nuclide.
SO-220
17.3 The Pattern of Nuclear Stability •
Characterization of nuclei: Even-even �
�
Nuclei with an even number of protons and neutrons, consequently A and Z are even numbers 157 even-even nuclides are stable. Examples: helium-4, iHe;carbon- 1 2, I�C;oxygen- 1 6, I�O;and neon-22, :gNe
Even-odd
� �
Nuclei with an even number of protons and an odd number of neutrons, consequently Z is even and A is odd 53 even-odd nuclides are stable. Examples: helium-3, �He;berylium-9, �Be;carbon-13, I�C; and neon-21 , fciNe
Odd-even �
�
Nuclei with an odd number of protons and an even number of neutrons, consequently both Z and A are odd numbers 50 odd-even nuclides are stable. Examples: hydrogen-I, : H; lithium-7, ;Li; boron-II, I�B; and nitrogen-IS, I�N
Odd-odd �
�
�
Nuclei with an odd number of protons and neutrons, consequently Z is odd and A is even Only 4 odd-odd nuclides are stable.
The stable odd-odd nuclides are hydrogen-2 (deuterium), �H;lithium-6, �Li ; boron-la, I�B; and nitrogen- 1 4, IjN .
Summary � •
�
Strong force
�
�
•
�
An attractive force that holds nucleons together in the nucleus Overcomes the coulomb repulsion of protons Acts only over a very short distance, approximately the diameter of the nucleus
Magic numbers
�
�
�
� •
Even-even nuclides are the most stable and most abundant nuclides. Odd-odd nuclides are the least stable and least abundant nuclides.
Are 2, 8, 20, 50, 82, 1 14, 1 26, and 184 for either protons or neutrons Nuclei with magic numbers of protons and/or neutrons are more likely to be stable. Analogous to the pattern of electronic stability in atoms associated with electrons in tilled subshells Doubly magic nuclides are very stable. Examples: helium-4, i He; oxygen-I6, I�O; calcium-40, i�Ca; and lead-208, 2�� Pb
Band of stability and sea of instability
�
�
�
�
�
Stable nuclides are found in a narrow band, the band of stability, that ends at Z = 83 (bismuth). All nuclides with Z> 83 are unstable. Unstable nuclides are found in the sea of instability, a region above and below the band of stability. Nuclides above the band of stability are neutron rich. Neutron rich nuclei are likely to emit � particles (neutron - proton + �). SG-221
�
� �
•
Nuclides below the band of stability are proton rich. Proton rich nuclei are likely to emit �+ particles (proton - neutron + �+) or to capture electrons (proton + � neutron). Heavier nuclides below the band of stability and those with Z> 83 may also decay by emitting ex particles. -
Models of nuclear structure
� �
Three common models of nuclear structure Description and order of sophistication: I) The liquid drop model, in which nucleons are considered to be packed together in the nucleus like molecules in a liquid 2) The independent particle model, in which nucleons are described by quantum numbers and assigned to shells. Nucleons in the outermost shells are most easily lost as a result of radioactive decay. 3) The collective model, in which nucleons are considered to occupy quantized energy levels and to interact with each other by the strong force and the electrostatic (coulomb) force
17.4 Predicting the Type of Nuclear Decay •
Massive nuclei with Z> 83
�
�
Lose protons to reduce their atomic number and generally lose neutrons as well Stepwise decay gives rise to a radioactive series, a characteristic sequence of nuclides. Examples of radioactive series:
I) The uranium-238 series (see Fig. 1 7. 1 6 in the text) is one in which ex (alpha) and � (beta) particles are successively ejected. The final step is the formation of a stable isotope of lead (magic number 82), lead-206. 2) The uranium-235 series proceeds similarly and ends at lead-207. This series is sometimes called the actinium series. 3) The thorium-232 series ends at lead-208. •
Neutron rich nuclides
�
�
•
�
Nuclides that have high neutron to proton (nip) ratios with respect to the band of stability Tend to decay by reducing the number of neutrons Commonly undergo � decay, which decreases the nip ratio
Proton rich nuclides
�
�
�
Nuclides that have high proton to neutron (pin) ratios with respect to the band of stability Tend to decay by reducing the number of protons Commonly undergo electron capture, positron emission, or proton emission (least likely), which decreases the pin ratio
17.5 Nucleosy nthesis •
Formation of the elements
�
�
Elements are formed by nucleosynthesis. Nucleosynthesis occurs when particles collide vigorously, as in stars.
SG-222
�
�
Transmutation is the conversion of one element into another.
The first artificial transmutation was observed in 1 919 by Rutherford, who converted IjN to I�O by bombarding the parent nuclide with high-speed.- , particles: IjN
�
�
•
iex
-
I�O
+
:p
The reactants must collide with enough energy to overcome the repulsive coulomb force between positively charged nuclei. Nucleosynthesis reactions are commonly written in shorthand as: target (incoming species, ejected species) product In the Rutherford example above, the shorthand notation is IjN (ex,p) I�O . Nucleosynthesis can be effected using a variety of projectile species including y radiation, protons, neutrons, deuterium nuclides, ex particles, and heavier nuclides. Because they are uncharged, neutrons can travel relatively slowly and still react with nuclides.
Transuranium elements �
�
•
+
Elements following uranium (Z = 92) in the periodic table Elements from rutherfordium (Rf, Z = 104) to meitnerium (Mt, Z 1 09) were given official names in 1997. =
Transmeitnerium elements �
�
Elements following meitnerium; six have been reported as of this writing. Temporary names are based on shorthand numbers derived from Latin (see Table 1 7.2 in the text).
NUCLEAR RADIATION (Sections 17.6-17.8) 17.6 The Biological Effects of Radiation •
Penetrating power
�
� � �
� �
ex, p, and y particles are forms of ionizing radiation with different penetration power. Different amount of shielding is required to prevent harmful effects (see Table 17.3 in the text). ex particles have the least penetrating power and are absorbed by paper and the outer layers of skin. They are extremely dangerous if inhaled or ingested. p particles are 100 times more penetrating than ex particles and are absorbed by 1 cm of flesh or 3 mm of aluminum, for example. y radiation is 100 times more penetrating than � particles. It can pass through buildings and bodies, leaving a trail of ionized or damaged molecules, and is absorbed by lead bricks or thick concrete. The order of increasing penetrating power is ex < p < y.
SG-223
•
Radiation damage to tissue
�
� •
Absorbed dose of radiation
�
�
•
Energy deposited in a sample, such as a human body, when exposed to radiation Units of the absorbed dose of radiation are: I rad = amount of radiation that deposits 0.0 I J of energy per kg of tissue I gray (Gy) = an energy deposit of I J·kg-I (SI unit) I Gy = 100 rad Note: The name rad stands for "radiation absorbed dose."
Dose equivalent
�
� � �
�
•
Depends on the type and strength of the radiation, the length of exposure, and the extent to which the radiation can reach sensitive tissue For example, Pu 4+ is an a emitter and, if ingested, replaces Fe3+ in the body, resulting in inhibition of the production of red blood cells.
Dose modified to account for different destructive powers of radiation Relative destructive power is called the relative biological effectiveness, Q. For �- andy radiation, Q I . For a radiation, Q 20. The natural unit of dose equivalent is the rem (roentgen equivalent man). Definition: Dose equivalent in rem = Q absorbed dose in rad Dose equivalent in sievert (Sv) = Q absorbed dose in gray (Gy) Sl Units: I Sv = 100 rem ==
�
x
x
Common radiation dose and harmful effects
�
�
�
�
Typical annual human dose equivalent from natural sources is 0.2 rem-y-I, a number with a wide range depending upon habitat and lifestyle. A typical chest x-ray gives a dose equivalent of about 7 millirem. 30 rad (30 rem) ofy radiation may cause a reduction in white blood cell count. 30 rad (600 rem) of a radiation causes death.
17.7 Measuring the Rate of Nuclear Decay •
Activity of a sample
�
�
�
�
•
Number of nuclear disintegrations per second One nuclear disintegration per second is called I becquerel, Bq (SI unit). The curie, Ci, an older unit for activity, is equal to 3.7 1010 Bq, which is equal to the radioactive output of I g of radium-226. Because the curie represents a large value of decay, the activity of typical samples is given in millicuries (mCi) or microcuries (�Ci). Note: See Table 17.4 in the text for a list of the radiation units. x
Nuclear decay
�
�
The decay of a nucleus can be written as: Parent nucleus - daughter nucleus + radiation This form is the same as a unimolecular elementary reaction, with an unstable nucleus taking the place of an excited molecule. SG-224
� •
Law of radioactive decay
�
�
•
The rate of nuclear decay depends only on the identity of the nucleus (isotope), not on its chemical form or temperature. The rate of nuclear decay is proportional to the number of radioactive nuclei N: Activity rate of decay k N
I
=
x
=
k is the decay constant (or rate constant for the reaction).
I
�
The units of the decay constant are (timerl, the same as for any first-order rate constant.
�
Noe -kl The number of radioactive nuclei N decays exponentially with Exponential form: time. Large values of k correspond to more rapid decay.
Integrated rate law for nuclear decay and half-life
�
IN
Logarithmic form: In
I
=
(:0)
=
-kt
On the left is a plot of N vs. t for nuclear decay. No is the number of radioactive nuclei present at t = O. N is the number of radioactive nuclei present at a later time t. The time required for the nuclei to decay to half their initial number is equal to the half-life, tl/2 . At this time, I
N = "iNo
o ... '" ..0
'
E
"
Z
Time, •
•
tll2
�
:
1 2
1
Half-lives of radioactive nuclides span an enormous range, from picoseconds el5Fr, 120 ps) to billions of years (238U, 4.5 109 y). See Table 17.5 in the text for a list of half-lives of common radioactive isotopes. x
t
Isotopic dating
� �
•
1
Substituting this value into the integrated rate law yields:
�
Used to determine the ages of rocks and of archeological artifacts Carried out by measuring the activity of a radioactive isotope in the sample Useful isotopes for dating are uranium-238, potassium-40, tritium, � H , and carbon-14.
Radiocarbon dating
�
�
Most important example of isotopic dating Uses the beta decay of carbon-14, for which the half-life is 5730 y
Radioactive 14C
�
-
Formed in the atmosphere by neutron bombardment of 14N IjN + 6n I � C + l p
SG-22S
�
�
�
� �
Enters living organisms as 14 C02 through photosynthesis and digestion Leaves living organisms by excretion and respiration Achieves a steady-state concentration in living organism with an activity of 15 disintegrations per minute per gram of total carbon Shows decreasing activity in dead organisms because they no longer ingest 14C from the atmosphere The time of death of an organism can be estimated by measuring the activity of the sample and using the integrated rate law.
17.8 Uses of Radioisotopes •
Radioisotopes
�
�
•
Used to cure disease, preserve food, and trace the mechanisms of chemical reactions Used to power spacecraft, locate sources of water, and determine the age of nonliving materials including wood C4 C dating), rocks (238U/206Pb ratio), and ground water (tritium dating using the lH/3H ratio)
Radioactive tracers
�
�
Radioactive isotopes that are used to track changes and locations Example: Phosphorus-32, a � emitter with tJ/ 2 = 14.28 d, can be incorporated into phosphate containing fertilizer to follow the mechanism of plant growth. In chemical reactions, tracers can help determine the mechanism of the reaction. Nonradioactive isotopes are also used for this purpose.
NUCLEAR ENERGY (Sections 17.9-17.12) 17.9 Mass-Energy Conversion •
Nuclear binding energy
�
� �
Energy released when protons and neutrons join together to form a nucleus: (Z) ] p + (A - Z) b n 4 EZ+ Since nuclides are neutral atoms, the energy released is also given approximately by: (Z)]H + (A - Z) b n - 4E The binding energy is then given by applying the Einstein mass-energy equation to each reactant and product species: Ebind I M I I IE (products) IE (reactants) I 2 2 = I E (product) - IE (reactants) I = I mc (product) - I mc (reactants) I = 1m (product) - Im (reactants) I c2 -
=
�
=
-
I Ebind = 16mlxc21
x
Binding energy is a measure of the stability of the nucleus.
SG-226
�
The binding energy per nucleon is defined as EbindlA and is a better measure of the stability of the nucleus.
A plot of the binding energy per nucleon vs. atomi c mass number is shown o n the left (see Fig. in the text). Nucleons are bound together most strongly in the elements near iron and nickel, which accounts for the high abundance of i ron and nickel in meteorites and on planets similar to the Earth. From the figure, it appears that light nuclei become more stable when they "fuse" together to form heavier ones. Heavy nuclei become more stable when they undergo "fission" and spl it i nto l ighter ones. Also, note the increased stabil ity of the nuclei with magic numbers, particularly oxygen- l which is doubly magic.
17.20
6Li
i c:: 0 ...
u
'" c � " 0. >. b.O .... ... c:: " b.O C
�
i:i5
•
o
6,
2H A
Units used in nuclear calculations �
�
Mass is usually reported in atomic mass units, u. One atomic mass unit is defined as exactly
I 1.660 x 10-27 I Iu
�
�
� the mass of one atom (nucl ide) of 12C :
=
54
kg
Binding energy is usually reported in electronvolts, eV, or mil lions (mega) of electronvolts, MeV. An electronvolt is the change in potential energy of an electron (charge of it is moved through a potential difference of I V:
IleV=1.60218xI0-19
1.602 18 x10-19
C) when
J
17.10 Nuclear Fission •
Nuclear fission �
Occurs when a nucleus breaks into two or more smaller nuclei
�
May be spontaneous or induced
� •
Releases a large amount of energy
Spontaneous nuclear fission �
�
Occurs when oscil lations in heavy nuclei lead them to break into two smaller nuclei of simi lar mass Yields a variety of products for a given nuclide (see Fig. Example:
07.
17.22
in the text)
Consider the spontaneous fission of americium-244. Two of the daughter nuclides formed are iodine-134 and molybdenum- I The reaction is : 3 3 1 95Am - I 5431 + + on
244
10427Mo
SG-227
•
Induced nuclear fission �
�
Caused by bombarding a heavy nucleus with neutrons Yields a variety of daughter nuclides for a given parent Example: Consider the induced fission of plutonium-239 by neutron bombardment. number of products are formed. Two important reactions are: 2§�PU + bn ��Mo + 1��Te + 4 bn 2§�Pu
•
1��Tc
+
l�fSb
+
5 bn
Energy released during fission is calculated by using Einstein's equation. For a balanced nuclear reaction,
1 6m
=
Im(products) - Im(reactants)
Fissionable and fissile nuclei �
�
I
and
I
M
=
x
6 m c2
Fissionable nuclei are nuclei that can undergo induced Fissile
fission. nuclei are fissionable nuclei that can undergo induced fission with slow-moving neutrons.
Examples of fissile nuclei:
•
bn
Energy changes in fission reactions
� �
•
+
A
2§�U, 2§� U , and 2��Pu , which are nuclear power plant fuels. The nuclide, 2��U, is fissionable by fast-moving neutrons but is not fissile.
Nuclear branched chain reactions �
�
Neutrons are chain carriers in a branched chain reaction (see Section 13.12). Chain branching occurs when an induced fission reaction produces two or more neutrons.
2��U + 6n - �6Zr + l�iTe + 2 6n , in which two neutrons are produced for each uranium-235 nuclide that reacts. The two neutrons can either escape into the surroundings or be captured by (react with) other uranium-235 nuclides. A critical mass is the mass of fissionable material above which so few neutrons escape from the sample that the fission chain reaction is sustained. The critical mass for pure plutonium of normal density is about 15 kg. A sample of plutonium with mass greater than IS kg is supercritical; the reaction is self sustaining and may result in an explosion. A subcritical sample is one that has less than the critical mass for its density. Example:
� �
�
� •
Controlled and uncontrolled (explosive) nuclear reactions �
�
�
� �
Explosive nuclear reactions occur when a subcritical amount offissile material is made supercritical so rapidly that the chain reaction occurs uniformly throughout the material. Controlled nuclear reactions occur in nuclear reactors. Controlled reactions are not explosive because there is a subcritical amount of fuel. The rate of reaction is controlled by moderators, such as graphite rods. Moderators slow the emitted neutrons so that a greater fraction can induce fission. SG-228
�
The difference between a fission explosion and a controlled fission is shown in the figure below. Each symbol Q9 represents a fission event. The neutrons from the fission are shown but products are not.
;r-@---7D ;r-@---7 0 :r-@-> 0 ---@-l> @ ---@-l> @ --@-> @ � "'--@->
/
O'� O'-@------7 0
Control rod and reactor geometry prevent some neutrons from sustaining fission,
Uncontrolled tission (explosion) Note:
•
Controlled fission
The number of neutrons released in an event is variable as is the identity of the products (see Fig. 17.24 in the text).
Breeder reactors �
�
Nuclear reactors used to synthesize fissile nuclides for fuel and weapon use. Breeder reactors run very hot and fast because no moderator is present. As a result, they are more dangerous than nuclear reactors used for power generation.
17. 1 1 Nuclear Fusion •
Nuclear fusion �
� � �
� �
•
�
Occurs when lighter nuclei fuse together to form a heavier nucleus Is accompanied by a large release of energy because heavier nuclei have larger binding energies per nucleon Occurs in stars and in the hydrogen bomb, in which the fusion reaction is uncontrolled Is difficult to achieve because charged nuclei must collide with tremendous kinetic energies to fuse Fusion is essentially free of radioactive waste. Fusion has not yet been sustained for an appreciable time in a controlled reaction on Earth. A safe, sustained fusion reaction could provide an almost unlimited source of energy.
Energy changes in fusion reactions
�
�
Energy released during a fusion reaction is calculated by using Einstein's equation. The procedure is the same as for fission reactions.
SG-229
17.12 The Chemistry of Nuclear Power •
Chemistry �
� � •
�
Uranium
�
� � •
The key to the safe use of nuclear power Used to prepare nuclear fuel Used to recover important fission products Used to dispose of nuclear waste safely The fuel of nuclear reactors Obtained primarily from the ore pitchblende, U02, by reducing the oxide to the metal and then enriching or increasing the fraction of the fissile nuclide, uranium-235 To be useful as a fuel, the percentage of 235U must be increased from its natural abundance of 0.7% to about 3%.
Enrichment � �
Exploits the mass difference between 2 35U and 238U Separation is accomplished by repeated effusion of 235UF6 and 238UF6 vapor (see Graham's law of effusion in Section 4.12).
238
Rate of effusion of 235U F6 Rate of effusion of U F6 Note:
•
= )MM238
235
=
Because the ratio is close to one, repeated effusion steps (hundreds) are required to get the desired separation.
Nuclear waste �
�
�
� �
Spent nuclear fuel called nuclear waste is still radioactive. Waste is a mixture of uranium and fission products. Nuclear waste must be stored safely for about 10 half-lives, and is generally buried underground. For underground burial, incorporation of the highly radioactive fission (HRF) products into a glass or ceramic material is better than placing it in metal storage drums. Storage drums can corrode, allowing the waste to seep into aquifers and/or contaminate large areas of soil.
SG-230
Chapter 18 ORGANI C CHEMISTRY I: THE HYDROCARBONS
ALIPHATIC HYDROCARBONS (Sections 18.1-18.6) 18. 1 Types of Aliphatic Hydrocarbon s •
•
Types of hydrocarbons (compounds containing only C and H) �
Aromatic hydrocarbon: hydrocarbon with one or more benzene rings. Examples: Benzene (C6H6), toluene (C6HsCH3), and naphthalene (C1oHg), all shown on the right � Aliphatic hydrocarbon: hydrocarbon without a benzene ring � Saturated hydrocarbon: an aliphatic hydrocarbon with no carbon-carbon multiple bonds Example: Butane (CH3CH2CH2CH3 ) � Unsaturated hydrocarbon: an aliphatic hydrocarbon with at least one carbon-carbon multiple bond. Examples: Butene (CH2=CHCH2CH3) and butyne (CH=CCH2CH3) Aliphatic hydrocarbons: structural formula � Shows chemical connectivity, how many atoms are attached to one another Example: Butane and methylpropane have the H H H H H CH3 H same empirical (C2HS) and molecular I I I t t i t H-C-C-C-H H-C-C-C-C-H (C4HIO) formulas, respectively, but t t t t t t t H H H H H H H different structural formulas. Methyipropane
Butane
In the structure of methylpropane, an abbreviated notation for the CH3 t methyl group (CH3 ) attached to the propane chain is used (the CH3 C-CH3 three C-H bonds are not shown). A structure with all three I H terminal methyl groups in abbreviated form is shown on the right. -
•
Condensed structural formula �
�
�
�
•
Simple way to represent structures of complicated organic molecules Shows in an abbreviated form how the atoms are grouped together The condensed structural formula for butane is CH3CH2CH2CH3 or CH3(CH2)2CH3: the latter is often written wheri the chain is lengthy. The condensed structural formula for methylpropane is CH3CH(CH3)CH3 or (CH3)3CH because the methyl groups are equivalent. The first formula is the one chosen to name the compound (methylpropane not trimethylmethane).
Line structure
� � � �
�
A line structure represents a chain of carbon atoms as a zigzag line. The end of each short line represents a carbon atom. C-H bonds and H atoms are not shown but are added "mentally" to the structure. Other atoms, such as 0 and N atoms, are written explicitly, along with any H atoms to which they CH3 are bonded. CH Example: The zigzag structural formula and the line structure of methyl propane are shown on the right. CH3/ 'CH3 A benzene ring is represented by a circle inside a hexagon or by a Kekule structure. The presence of a H atom attached to each C atom is implied. These forms of the benzene ring are also used in t
SG-231
structural formulas. The respective structural formulas and line structures of ethylbenzene, C6HsCH2CH3, are shown below.
6 0 Nomenclature of Hydrocarbon s •
Alkanes
�
�
� � � � � •
I f a hydrogen atom is removed from an alkane, the name of the alkyl group that results is derived by replacing the suffix -ane with -yl. The first three members of the alkyl family are the methyl (CHr), ethyl (CH3CH2-), and propyl (CH3CH2CHZ-) groups. Methylpropane (CH3CH(CH3)CH3) is a propane molecule in which a H atom on the middle C atom is replaced by a methyl (CH3) group. Cycloalkanes are alkanes that contain rings of -CHz- units. The first three members are cyclopropane (C3H6), cyclobutane (C4Hs), D and cyclopentane (CsH ) all shown on the right.
o
Properties of alkanes
�
� �
� �
Structures may be written as linear, but they are actually linked tetrahedral units. Rotation about C-C bonds is fairly unrestricted, so the chains often roll up into a ball in gases and liquids to maximize the attraction between different parts of the chain. Electronegativity values of C and H are similar, so hydrocarbon molecules can be regarded as nonpolar. The dominant interaction between the molecules is therefore the London interaction, whose strength increases with the number of electrons in the molecule. Thus, alkanes should become less volatile with increasing molar mass (see Text Fig. 18.4). Many alkane molecules appear in forms with the same atoms bonded together in different arrangements, such as the isomers butane and methylpropane shown on the previous page. In general, isomers are different compounds with the same molecular formula.
Alkenes
�
� �
�
•
(CH3CH2CH3)
10 ,
�
•
Alkanes are saturated hydrocarbons. In naming single-chain alkanes, add a prefix denoting the number of C atoms in the chain to the suffix -ane (see Table 18.1 in the text). The first three members of the alkane family are methane (CH4), ethane (CH3CH3), and propane
Alkenes are a series of compounds with formulas derived from H2C=CH2 by inserting CH2 groups. The next member of the family is propene, H-CHz-CH=CH2 (CH3CH=CH2). Ethene or ethylene, C2H4 or H2C=CH2, is the simplest unsaturated hydrocarbon. Alkenes are named in a manner analogous to that of alkanes except the ending is -ene. The double bond is located by numbering the C atoms in the chain and writing the lower of the two numbers of the two C atoms joined by the double bond (see Text Toolbox 18.1).
Alkynes �
� �
Alkynes are aliphatic hydrocarbons with at least one carbon-carbon triple bond. Ethyne or acetylene, HC=CH, is the simplest alkyne. Alkynes are named like the alkenes but with the suffix -yne.
SG-232
•
•
Nomenclature rules
Alkane nomenclature
�
� � �
•
for aliphatic hydrocarbons are given in Text Toolbox 18. 1 .
If the C atoms in an alkane are bonded in a row with a formula of the form CH3(CH2)mCH3, the compound, called an unbranched alkane, is named according to the number of C atoms present. All are named with a stem name and the suffix -ane. A cyclic (ring) alkane has the general formula (CH2)m and is designated by the prefix cyclo-. Examples: Pentane (CH3(CH2)3CH3) and cyclopentane (CsHlo) A branched alkane has one or more C atoms not in a single row, and these carbons constitute side chains. Butane is an unbranched alkane, and methylpropane is a branched alkane with a CH3 side chain. Side chains are treated as substituents, species that have been substituted for a H atom. A systematic procedure for naming alkanes is: Step I. Identify the longest chain of C atoms in the molecule. Step 2. Number the C atoms on the longest chain so that the carbon atoms with substituents have the lowest possible numbers. Step 3. Identify each substituent and the number of the C atom in the chain on which it is located. Step 4. Name the compound by listing the substituents in alphabetical order, with the numbered location of each substituent preceding its name. The name(s) of the substituent(s) is/are followed by the parent name of the alkane, which is determined by the number of C atoms in the longest chain. Example: Heptane with a methyl group at the number 3 carbon is named 3methylheptane. Step 5. If two or more of the same substituents are present (such as two methyl groups), a Greek prefix such as di-, tri-, or tetra- is attached to the name of the group. Numbers of the C atoms to which groups are attached, separated by commas, are included in the name. Example: Heptane with methyl groups at C atoms 3 and 4 is named 3,4-dimethylheptane. Step 6. As in the example above, numbers in a name are always separated from letters by a hyphen, while numbers are separated from each other by commas.
0
Alkene nomenclature
�
Alkenes are named by using the stem name of the corresponding alkane with a number that specifies the location of the double bond. The number is obtained by numbering the longest carbon chain so that the double bond is associated with the lowest numbered carbon possible and assigning to the double bond the lower number of the two double-bonded carbons. 6
5
4
3
2
I
CH3CH2CH2CH=CHCH3
� � �
2-hexene (not 3-hexene)
Other substituents are named as for alkanes, with a number specifying the location of the group and an appropriate prefix denoting how many groups of each kind are present. Numbering the longest chain so that the double bond has the lowest number takes precedence over keeping the substituent numbers low. If more than one double bond is present, the number of these bonds is indicated by a Greek prefix. The suffix is -ene. Example: I,3-hexadiene (H2C=CHCH=CHCH2CH3) or �
SG-233
•
Alkyne nomenclature
--+
--+
Alkynes are named by using the stem name of the corresponding alkane with a number specifying the location of the triple bond. The numbering convention is the same as that used for double bonds. If more than one triple bond is present, the number of these bonds is indicated by a Greek prefix. The suffix is -yne. When numbering atoms in the chain, the lowest numbers are given preferentially to (a) functional groups named by suffixes (see Toolbox 1 9. 1 ), (b) double bonds, (c) triple bonds, and (d) groups named by prefixes. Example: 4-chloro-2-hexyne (CH3CH2CH(CI)C=CCH3)
� CI
18.2 Isomers •
Structural isomers
--+
--+
•
Stereoisomers
--+
--+ --+
•
Molecules with the same connectivity but with some of their atoms arranged differently in space Geometrical isomers are stereoisomers with different arrangements in space on either side of a double bond, or above and below the ring of a cycloalkane. Optical isomers are stereoisomers in which each isomer is the mirror image of the other, and the images are nonsuperimposable.
Geometrical isomers
--+
--+ --+ --+
•
Molecules constructed from the same atoms connected differently Have the same molecular formula, but different structural formulas Example: Butane (CH3(CH2)2CH3 ) and methylpropane (CH3CH(CH3)CH3) are structural isomers with the same molecular formula (C4H10)' The molecules have a different connectivity.
Compounds with the same molecular formula and with atoms bonded to the same neighbors but with a different arrangement of atoms in space For molecules with a double bond, effective lack of rotation about the double bond makes this type of isomer possible. Geometrical isomers of organic compounds are distinguished by the italicized prefixes: cis (from the Latin word for on this side) and trans (from the Latin word for across). An example is 2-butene, which exists as cis and trans geometrical isomers. In the trans isomer, the two methyl groups lie across the double bond from each other. In the cis isomer, they are on the same side of the double bond. Rotation about the double bond does not normally occur because the 1t-bond is rigid, and the two isomers are distinct compounds with different chemical and physical properties. If sufficient energy is added to either compound, a cis-trans isomerization reaction may occur in which the 1t-bond is partially broken and part of the sample is converted to the other isomer. cis·2· butene
trans-2-butene
Optical isomers
--+
--+ --+ --+
--+
Nonsuperimposable mirror images A chiral molecule has a mirror image that is nonsuperimposab le. Your hand is a chiral object; its mirror image is not superimposable. An organic molecule is chiral if at least one of its central C atoms has four different groups attached to it. Such a C atom is called a stereogenic center. A pair of enantiomers consists of a chiral molecule and its mirror image. Enantiomers have identical chemical properties, except when they react with other chiral species.
SG-234
�
•
Enantiomers differ in only one physical property; chiral molecules display optical activity, the ability to rotate the plane of polarization of light. � Mixtures of enantiomers in equal proportions are racemic mixtures, which are not optically active. � A molecule that has a superimposable mirror image is said to be achiral. Summary � The three different types of isomerism are displayed in Text Fig. 1 8.1.
18.3 Properties of Alkanes •
Physical properties
�
� �
�
•
Chemical properties
�
� �
•
Alkanes are best regarded as nonpolar molecules held together primarily by London forces, which increase with the number of electrons in a molecule. Consequently, for unbranched alkanes, melting and boiling points increase with chain length. Unbranched alkanes tend to have higher melting points, boiling points, and heats of vaporization than their branched structural isomers. Molecules with unbranched chains can get closer together than molecules with branched chains. As a result, molecules with branched chains have weaker intermolecular forces than their isomers with unbranched chains. Example: Unbranched butane has a higher boiling point (-0.5°C) than its branched-chain structural isomer, methylpropane (-11.6°C). Alkanes are not very reactive chemically. They were once called paraffins, derived from the Latin "little affinity." Alkanes are unaffected by concentrated sulfuric acid, boiling nitric acid, strong oxidizing agents such as KMn04, and by boiling aqueous NaOH. The C-C bond enthalpy (348 kJ·mor') and the C-H bond enthalpy (412 kJ·mor') are large, so there is little energy advantage in replacing them with most other bonds. Notable exceptions are C=O (743 kJ.mor'), C-OH (360 kJ·mor'), and C-F (484 kJ·mor').
Combustion (oxidation) reactions
�
�
Alkanes are used as fuels because their enthalpies of combustion are high. The products of combustion are carbon dioxide and water. Strong C-H bonds are replaced by even stronger O-H bonds in H20, and the 0=0 bonds are replaced by two strong C=O bonds in CO2• Example: Combustion of one mole of octane releases 5471 kJ of heat: CgH,g (l) + 12.5 02(g) 8 CO2(g) + 9 H20(1) f..Hro = -5471 kJ·mor' �
•
Substitution reactions
�
� � �
Alkanes are used as raw materials for the synthesis of many reactive organic compounds. Organic chemists introduce reactive groups into alkane molecules in a process called junctionalization. Functionalization of alkanes is achieved by a substitution reaction, in which an atom or group of atoms replaces an atom in the original molecule (hydrogen, in the case of alkanes). Reaction of methane (CH4 ) and chlorine (Ch) is an example of a substitution reaction. In the presence of ultraviolet light or temperatures above 300°C, the gases react explosively: CH4 (g) + CI2(g) lightor heat) CH3 CI(g) + HCI(g) Chloromethane (CH3C I) is only one of four products; the others are dichloromethane (CH2Ch), trichloromethane (CHCb), and tetrachloromethane (CCI4 ), which is carcinogenic.
SG-235
18.4 Alkane Substitution Reaction s •
Radical chain mechanism
�
� �
�
�
Kinetic studies suggest that alkane substitution reactions proceed by a radical chain mechanism. The initiation step is the dissociation of chlorine: Cb light or heat ) 2 Cl . Chlorine atoms proceed to attack methane molecules and abstract a hydrogen atom: Cl· + CH4 HCl + . CH3 Because one of the products is a radical, this reaction is a propagation step. In a second propagation step, the methyl radical may react with a chlorine molecule: Cl2 + . CH3 CH3Cl + Cl . The chlorine atom formed may take part in the other propagation step or attack a CH3Cl molecule to eventually form CH2CI2 . CHCh and CCl4 can also be formed by a continuation of this process. A termination step occurs when two radicals combine to form a nonradical product, as in the reaction: Cl . + CH3 CH3Cl The substitution reaction is not very clean and the product is usually a mixture of compounds. One may limit the production of the more highly substituted alkanes by using a large excess of the alkane. �
�
.
�
18.5 Properties of Alkenes •
Double bond �
�
�
�
� � •
The carbon-carbon double bond, C=C, consists of a a-bond and an-bond. Each carbon atom is si-hybridized and one of the hybrid orbitals is used to form the a-bond. The unhybridized p-orbitals on each atom overlap with each other to form an-bond. All four atoms attached to the C=C group lie in the same plane and are fixed in that arrangement by resistance to twisting of then-bond (see Text Fig. 18.7). Alkenes can not roll up into a compact arrangement as alkanes can, so alkenes have lower melting points. In C=C, then-bond is weaker than the a-bond. A consequence of this weakness is the reaction most common in alkenes: replacement of then-bond by two new a-bonds.
Formation of alkenes by elimination reactions
�
�
�
In the petrochemical industry, alkanes are converted to more reactive alkenes by a catalytic process called dehydrogenation: CH3CH3(g) Cr20a ) CH2=CH2(g) + H2(g) This is an example of an elimination reaction, one in which two groups on neighboring C atoms are removed from a molecule, leaving a multiple bond (see Text Fig. \8.8). In the laboratory, alkenes are produced by dehydrohalogenation of haloalkanes, the removal of a hydrogen atom and a halogen atom from neighboring carbon atoms: CH3CH2Br CH:1CHP- ill cthallol at 70°C ) CH2=CH2 + HBr This elimination reaction is carried out in hot ethanol containing sodium ethoxide, CH3CH20Na. Dehydrohalogenation occurs by attack of the ethoxide ion on a hydrogen atom of the methyl group. A H atom is removed as a proton and CH3CH20-H is formed. When the methyl C atom forms a second bond to its neighbor, the Br- ion departs. States of reactants and products in organic reactions are often not give because the reaction may take place on a catalyst surface or in a non-aqueous solvent, as in the reaction above.
SG-236
18.6 Electrophilic Addition •
Addition reaction
�
�
� � � •
Estimating the reaction enthalpy of an addition reaction
�
� •
Use bond enthalpies for the bonds that are broken and formed. Note that bond enthalpies strictly apply to gas-phase reactions. Recall:
I1Hr
0
"'"
I
reactants
nl1HBCbonds broken) - I
products
nl1HBCbonds formed)
Mechanism of addition reactions
� � � �
�
•
A characteristic reaction of alkenes, in which atoms supplied by the reactant form a-bonds to the two C atoms joined by the 1t-bond, which is broken (see Text Fig. 18.9). Almost all addition reactions are exothermic. A hydrogenation reaction is the addition of two hydrogen atoms at a double bond: CH3CH=CHCH3 + H2 - CH3CHrCH2CH3 A halogenation reaction is the addition of two halogen atoms at a double bond: CH3CH=CHCH3 + CI2 - CH3CHCI-CHCICH3 A hydrohalogenation reaction is the addition of a hydrogen atom and a halogen atom at a double bond: CH3CH=CHCH3 + HCI - CH3CH2-CHCICH3
Double bonds contain a high density of high energy electrons associated with the 1t-bond. This region of high electron density is attractive to positively charged reactants. An electrophile is a reactant that is attracted to a region of high electron density. It may be a positively charged species or one that has or can acquire a partial positive charge during the reaction. An example treated in the text is the bromination of ethene to form dibromoethane: H2C=CH2 + Br2 - CH2BrCH2Br Bromine molecules are polarizable; a partial positive charge builds up on the bromine atom closest to the double bond. Bromine acts as an electrophile. The partial charge becomes a full charge and a bromine cation attaches to the double bond, leaving a Br- ion behind. The cyclic intermediate is called a bromonium ion. A Br - ion is attracted by the positive charge of the bromonium ion. It forms a bond to one C atom, and the bromine atom in the cyclic ion forms another bond, giving 1,2-dibromoethane.
Hydrogenation reaction
�
�
Hydrogen can be added across a double bond with the help of a solid-state catalyst: catalyst CH CH • • • H2(g) + This reaction is used in the food industry, for example, to convert liquid oils to solids. Recall that alkenes have lower melting points than alkanes, other things equal. ••• = ...
-----J.�
SG-237
•• •
--
AROMATIC COMPOUNDS (Sections 18.7-18.8) 18.7 Nomen clature of Arenes •
Definition �
�
�
Aromatic hydrocarbons are called arenes. Benzene (C6H6) is the parent compound. The benzene ring is also called the phenyl group, as in 2-phenyl-trans-2-butene (CH3C(C6H5)=CHCH3) shown on the right. Aromatic compounds include those with fused benzene rings, such as naphthalene (CIO Hg ), anthracene (CI4H ) and phenanthrene (CI4HIO). Naphthalene Anthracene \0
•
,
Phenanthrene
Benzene ring numbering �
�
60
Benzene ring substituents are designated by numbering the C atoms from I to 6 around the ring. Compounds are named by counting around the ring in the direction that gives the smallest numbers to the substituents. The prefixes ortho-, meta-, and para- are commonly used to denote substituents at C atoms 2, 3, and 4, respectively, relative to another substituent at C atom I. Example: The three possible dichlorobenzene CI isomers are shown in the diagram on the right. Numbering carbon atoms in fused ring systems is somewhat more complicated and is not covered in the text
&el
1.2-Dichlorobenzene Mho-Dichlorobenzene
1,3-Dichlorobenzene meta-Dichlorobenzene
r(ji 1
S
5
�
2(MhO)
3(meta)
4 (para)
CI
¢ CI
1,4-Dichlorobenzene para-Dichlorobenzene
18.8 Electrophilic Substitution •
• •
Substitution reactions �
Arenes have delocalized n:-electrons; but, unlike alkenes, arenes undergo predominantly substitution reactions, with the n:-bonds of the ring unaffected. Example: The reaction of benzene (C6H6) with chlorine produces chlorobenzene when one chlorine atom substitutes for a hydrogen atom: C6H6 + CI2 Fe ) C6H5CI + HCI Electrophilic substitution � If the mechanism of substitution of a benzene ring involves electrophilic attack, the reaction is called electrophilic substitution.
Halogenation
� �
� •
In the halogenation of benzene, an iron catalyst is used. The iron is converted to iron( I I I) halide (FeX3 ). Iron(lll) halide reacts further to polarize the X2 molecule. The commonly accepted mechanism for aromatic halogenation, X2, is then FeX3 + X-X X3Fe O-·· ·x-X'S+ fast equilibrium o O slow X3Fe -···X-X + + C6H6 - FeX4- + C6H6X+ =
fast C6H6X+ + FeX4- - C6H5X + HX + FeX3 In the last step, the hydrogen atom is easily removed from the ring, for in that way the n:-electron delocalization, lost in the slow step, is regained.
Nitration
�
A mixture of HN03 and concentrated H2S0 4 slowly converts benzene into nitrobenzene.
SG-238
�
�
•
�
The nitronium ion, NOt, is the electrophile and the nitrating agent. The commonly accepted mechanism for aromatic nitration is: HN03 + 2H2S04 NOt + H30+ + 2HS04slow NOt + C6H6 - C6H6NOt C6H6N02+ + HS04- - C6HSN02 + H2S04 fast In the last step, the hydrogen atom is removed from the ring, restoring aromaticity. =
Ortho- and para-directing activators
�
Electrons are donated into the delocalized Molecular Orbitals of the ring (resonance effect). The reaction rate is much faster than in unsubstituted benzene, and the substituent is called an activator. � Products of substitution reactions favor the ortho and para positions of the ring because these locations have more electron density than the meta positions. � Examples of ortho- and para-directing activators include -OH, -NH2' and substituted amines. � The atom bonded to the benzene ring has a nonbonding pair of electrons. Note: Alkyl groups have no nonbonding electron pair, yet they are also ortho- and para-directing activators. The mechanism in this case is a different one (see any organic chemistry text). �
•
Ortho- and para-directing deactivators
� �
� �
•
�
Electrons are donated into the delocalized Molecular Orbitals of the ring (resonance effect), but the substituent is very electronegative. The reaction rate is slightly slower than in unsubstituted benzene, and the substituent is called a deactivator. Deactivation occurs when the substituent is highly electronegative and withdraws some electron density from the ring. Products of a substitution reaction stiII favor the ortho and para positions of the ring because these positions have relatively more electron density than the meta positions. The only examples are -F, -CI, -8r, and -I. The atom bonded to the benzene ring has a nonbonding pair of electrons.
Meta-directing deactivators
� � � � �
Highly electronegative substances that can withdraw electrons partially and/or a substituent that removes electrons by resonance The reaction rate is much slower than in unsubstituted benzene, and the substituent is called a deactivator. Products of a substitution reaction favor the meta position because the ortho and para positions have greatly decreased electron density. Examples include -COOH, -N02 , -CF3 , and -C=N. All electron pairs on the atom bonded to the benzene ring are bonding pairs.
SG-239
IMPACT ON MATERIALS: FUELS (Sections 18.9-18.1 0) 1 8.9 Gasoline •
•
Gasoline
�
Derived from petroleum by fractional distillation and further processing � Contains primarily Cs to C I I hydrocarbons � Petroleum is refined to increase the quantity and quality of gasoline. Quantity � The quantity of gasoline is increased by cracking (breaking down long hydrocarbon chains) and by alkylation (combining small molecules to make larger ones). Example: Cracking fuel oil to obtain an octane/octane isomeric mixture Example: Alkylation converts a butane/butene mixture to octane. ca a l ys
t
•
t
..
Quality
�
�
�
�
Octane rating: used to measure the quality of gasoline Other things equal, branched hydrocarbons have higher octane ratings than unbranched ones. Isomerization and aromatization are used to improve octane rating of gasoline. Isomerization is used to convert straight chain hydrocarbons to branched ones: AlCl3
�
Aromatization converts alkanes to arenes, as in the conversion of heptane to methylbenzene:
�
Ethanol, a renewable fuel, also increases the octane rating of gasoline.
A lCI3, Cr203
1 8. 1 0 Coal �
�
�
�
�
End product of anaerobic decay of vegetable matter Less environmentally friendly than gasoline, in part because it releases ash when burned Contains many aromatic rings and is primarily aromatic in nature Coal fragments, when heated in the absence of oxygen, yield coal tar, which contains many aromatic hydrocarbons and their derivatives. Consequently, coal is used as a raw material for other chemicals. Many pharmaceuticals, fertilizers, and dyes are derived from coal tar. Examples: Naphthalene for making indigo dyes (for blue jeans) and benzene, used to make nylon, detergents, and pesticides
SG-240
Chapter 19 ORGANIC CHEMISTRY II: POL YMERS AND BIOLOGICAL COMPOUNDS
COMMON FUNCTIONAL GROUPS (Sections 19.1-19.8) 19.1 Haloalkanes •
Haloalkanes
-+
-+
-+
-+ -+
-+ -+
Alkanes in which a halogen atom, X, replaces one or more H atoms Insoluble in water Some are highly toxic and environmentally unfriendly. Example: The chlorofluorocarbon (CFC) I ,2-dichloro- I-fluoroethane, HFCIC-CCIH2 , is partly responsible for depletion of the ozone layer. The C-X bonds in haloalkanes are polar; C carries a partial positive charge and the halogen, X, a partial negative one. C-X bond polarity governs much of the chemistry of haloalkanes. Haloalkanes are susceptible to nucleophilic substitution, in which a reactant that seeks out centers of positive charge in a molecule (a nUcleophile) replaces a halogen atom. NUcleophiles include anions such as OH - and Lewis bases with lone pairs such as NH3. Example: Hydroxide acts as a nucleophile in the following hydrolysis reaction: CH3Br + OW - CH30H + Br-
19. 2 Alcohols •
Definitions
-+
-+
Hydroxyl group: An -OH group covalently bonded to a C atom Alcohol: An organic compound that contains a hydroxyl group not directly bonded to an aromatic ring or to a carbonyl group, C=O /
-+
Alcohols are named by adding the suffix -01 to the stem of the parent hydrocarbon. Examples: Methanol for CH30H; ethanol for CH3CH20H -+ To identify the location of the -OH group, the number of the C atom attached to it is given. Examples: I-Propanol for CH3CH2CH20H; 2-propanol for CHJCH(OH)CHJ -+ A diol is an organic compound with two hydroxyl groups. Example: I ,2-Ethanediol for HOCH2CH20H (ethylene glycol) •
Classes of alcohols
-+
-+
-+
Primary alcohol: RCH2-OH, where R can be any group Examples: Methanol and I -propanol Secondary alcohol: R2CH-OH, where the R groups can be the same or different Examples: 2-Propanol, (CH3)2CH-OH (same R groups); 2-butanol, CH3CH2CH(-OH)CH3 (different R groups) Tertiary alcohol: R 3C-OH, where the R groups can be the same or different Examples:
2-Methyl-2-propanol, (CH3)3C-OH (same R groups, CHr); 3methyl-3-pentanol, CH3CH2C(CH3)(-OH)CH2CH3 (different R groups, CH3CHr and CH}-). A structure for the latter is:
SG-24 1
•
Properties of alcohols
� �
�
Alcohols are polar molecules that can lose the -OH proton in certain solvents, but typically not in water. Alcohols have relatively high boiling points and low volatility because of hydrogen bond formation through the -OH group. In this way, they are similar to water. Compare the normal boiling point of ethanol (7S.2°C) with a molar mass of 46.07 g·morl to that of pentane (36.0°C) with a molar mass of 72.14 g.morl.
19.3 Ethers •
Ethers � � �
�
�
•
Organic compounds of the form R-O-R, where R is any alkyl group and the two R groups may be the same or different More volatile than alcohols with the same molar mass because ethers do not form hydrogen bonds with each other Act, however, as hydrogen-bond acceptors by using the lone pairs of electrons on the 0 atom Useful solvents for other organic compounds because they have low polarity and low reactivity Quite flammable and must be handled with care Examples: Diethyl ether (CH3CH2-O-CH2CH3) and I-butyl methyl ether (CH3CH2CH2CHrO-CH3)
Crown ethers �
�
�
Cyclic polyethers of formula -fCH2CH2- O tn Name reflects the crownlike shape of the molecules. Bind strongly to alkali metal ions such as Na+ and K+, allowing inorganic salts to be dissolved in organic solvents Example: The common oxidizing agent potassium permanganate (KMn04 ) is insoluble in nonpolar solvents such as benzene (C6H6). In the presence of [I S]-crown-6, it dissolves in benzene according to the [ 1 8J-Crown-6 reaction on the right:
19.4 Phenols •
Phenols
�
�
Organic compounds with a hydroxyl group attached directly to an aromatic ring The parent compound, phenol, is a white, crystalline, molecular solid_ Phenol, C,H,OH,
�
6 6 oc
Melting point: 40.9°C
CH3 ,"' Ht5H CH3 \
Substituted phenols occur naturally, and some are responsible for the fragrances of plants. Examples: Thymol is the active ingredient in oil of thyme, and eugenol provides the scent and flavor in oil of cloves.
SG-242
Thymol
Eugenol
•
Acid-base properties of phenols
�
�
�
Phenols are generally weak acids in contrast to alcohols, which typically are not acidic. The acidity of phenols can be understood on the basis of resonance stabilization of the negative charge of the conjugate base of phenol (C6H50-). Because of resonance stabilization, the phenoxide (C6H 5 0-) anion is a weaker conjugate base than the corresponding conjugate bases of typical alcohols. Example: The ethoxide (CH3CH20-) anion is a stronger base than the phenoxide anion shown above.
19.5 Aldehydes and Ketones •
Aldehydes �
� � � � �
�
Organic compounds of the form shown on the right, where R is a H atom, an aliphatic group, or an aromatic group The group characteristic of aldehydes is written as -CHO, as in formaldehyde (HCHO), the first member of the family. Named systematically by replacing the ending -e by -ai, but many common names exist, such as formaldehyde for methanal and acetaldehyde for ethanal The C atom on the carbonyl group is included in the count of C atoms when determining the alkane from which the aldehyde is derived. II II A few common aldehydes are shown to the right. o
Ethanal
Benzaldehyde
Methanal
Aldehydes generally contribute to the flavor of fruits and nuts, and to the odors of plants. Example: Benzaldehyde provides part of the aroma of almonds and cherries. Aldehydes can be prepared by the mild oxidation of primary alcohols. Example: Formaldehyde is prepared industrially by oxidizing methanol with a Ag catalyst. The overall reaction is: 600°C,
•
o
Ag
> 2 HCHO(g) + 2 H20(g)
Ketones
� �
� � �
Organic compounds of the form shown on the right, where the R groups (alkyl or aryl) may be the same or different The carbonyl group, -C=O, characteristic of ketones, is written -CO, as in propanone, CH3COCH3 (acetone), the first member of the family. Named systematically by replacing the ending -e by -one, but many common names exist such as acetone or dimethyl ketone for propanone, and methyl ethyl ketone for butanone The C atom on the carbonyl group is included in the count of C atoms when determining the alkane from which the ketone is derived. A number is used to locate the C atom in the carbonyl group to avoid ambiguity.
SG-243
�
A few common ketones are shown on the right: Propanone
� �
�
Benzophenone
2-Pentanone
Ketones contribute to flavors and fragrances. Example: Carvone, shown on the right, is the essential oil in spearmint. Ketones can be prepared by the oxidation of secondary alcohols. There is less risk of further oxidation than with aldehydes, so stronger oxidizing agents are used. Dichromate oxidation of secondary alcohols produces ketones with little excess oxidation and in good yields. Example: The oxidation of 2-propanol to propanone is shown schematically on the right:
1 9.6 Carboxylic Acids •
Carboxylic acids
�
� �
� � �
Organic compounds of the form shown on the right, where R is a H atom, an alkyl, or an aryl group The carboxyl group characteristic of carboxylic acids is written -COOH as in formic acid (HCOOH), the first member of the family. The carboxyl group contains a hydroxyl group, -OH, attached to a carbonyl group, -C=O. Named systematically by replacing the ending -e by -oic acid, but many common names are used such as formic acid for methanoic acid and acetic acid for ethanoic acid The carbonyl C atom is included in the C atom count to determine the parent hydrocarbon molecule. o Some common carboxylic acids II II II are shown on the right: H/
�
""" O H
Methanoic acid
Ethanoic acid
Benzoic acid
HO
/ /
c,
0
/c, C H2 'OH
Malonic acid
Malonic acid contains two -COOH groups; it is called a diacid. Carboxylic acids contain hydroxyl groups, -OH, which can take part in hydrogen bonding. Carboxylic acids can be prepared by oxidation of aldehydes or of primary alcohols in an acidified solution containing a strong oxidizing agent such as KMn04, or Na2Cr 207:
Note: �
C
o
R/
o II C
......
H
Aldehyde
Primary alcohol
SG-244
�
�
In some cases, alkyl groups can be oxidized directly to carboxyl groups.
+
An industrial example is the oxidation of the methyl groups on p-xylene by a cobalt(lI I ) catalyst to form terephthalic acid, which is used in the production of artificial fibers.
Co(llJ)
3 02 �
COOH Q + 2H'O COOH
19.7 Esters •
Esters
�
�
� �
�
o
An ester is the product of a condensation II reaction between a carboxylic acid and an /c" H O R' alcohol. A water molecule is also Carboxy lic acid Alcohol produced. Esterification reactions can be acid catalyzed as in the example above. R
+
0H
H30+
Condensation reaction : one in which two molecules combine to form a large one, and a small molecule is eliminated. Many esters have fragrant aromas and contribute o to flavors of fru its. II / c , /CH2CH2CH2CH2CH3 Examples: The esters n-amyl acetate and n-octyl 0 CH3 acetate are responsible for the aromas n-Octyl acetate n-Amyl acetate of bananas and oranges, respectively. Naturally occurring esters also include fats and oils such as tristearin, a component of beef fat. o
o
o
Tristearin 19.8 Amines, Amino Acids, an d Amides •
Amines
�
�
�
� �
Derivatives ofNH 3, formed by replacing one or more H atoms with organic groups, R The amino group (-NH2 ) is the functional group of amines. Named by specifYing the groups attached to the nitrogen atom (N) alphabetically, followed by the suffix amine Des ignated as primary, secondary, or tertiary, depending on the number of R groups attached to the N atom H I Ammonia and several representative N am ines are shown on the right: H/ CH3 '
Ammonia
SG-245
Methylam i n e primary
Ethylmethylamine secondary
Dimethylethylamine tertiary
•
Properties of amines
� �
� •
Amines are widespread in nature. They often have disagreeable odors (for example, ammonia itself, are weak bases.
�
Tetrahedral ions of formula R4N+
�
Negl igible aci d or base properties and l i ttle effect on pH
�
Four groups bonded to the central nitrogen atom, which may be the same or different I solated as quaternary ammonium salts Example:
Dimethylethylpropylammonium chloride, with four R groups, is a quaternary ammonium salt.
Amino acids
�
�
�
amino group and the carboxyl group separated by one C atom �H 2 A more precise name is an aminocarboxylic acid. Carboxyl ic acids with an
Have the general formula R(NH2)CH(COOH) Examples: Glycine, methionine, and
phenylalanine Note:
•
putrescine, N H2(CH2)4NH2 ) and, simi lar to
Quaternary ammonium ions �
•
Characterized by four s/ hybrid orbital s on the N atom; three participate in three single bonds and the fourth has a lone pair of electrons
� H2 HOOC- C\ " H H
�H 2 HOOC- C\" H CH2 CH2SCH3
Glycine
I n amino acids, the central C atom is
HOOC- C\" H
6
Phenylalanine
Methionine
stereogenic and the molecule is chiral except for glycine.
Properties of amino acids
�
�
�
� � � � �
B ui lding blocks of proteins (see S ection 1 9. 1 3) Essential for human health Form hydrogen
bonds with both the amino and carboxyl groups Form double ions (zwitterions) by transferring a proton from the carboxyl to the amino group as shown on the right: I n aqueous solution, amino acids exist in four forms whose concentrations are pH dependent.
� H2 C HOOC- �"H R
Am i n o acid
HOOC
The concentrations of these species can be determi ned as with polyprotic acids (see Section.
NH + I 3 C -OOC- �"H R
Double ion form
-; '\ V
"" H
CH3
1 0. 1 6).
Form 1 predominates at low pH while form 4 predomi nates at high pH. At neutral pH, form 2 predominates. The neutral form 3 exists only in trace amounts in aqueous solutions.
SG-246
•
Amides
o II ..... C ,
�
Molecu les with the general formula:
�
Formed by the condensation reaction of carboxyl ic acids with amines, eliminating water
� �
R
N I H
/R
Near room temperature, the reaction m ixture forms an ammonium salt that reacts further to form the amide at higher tem perature. The general reaction is: R
Note:
o II "'" C
1,--9_H Acid
H
+
I N
heat
H--,1 R Amine
_ _
"'
R
o II ..... C,
N I
/R
Amide H
The amide shown above has an N -H group that may participate in intermolecular hydrogen bonding. The atoms that form the product water molecule are shown in a box.
THE IMPACT ON MATERIALS (Sections 1 9.9-1 9 . 1 2) 19.9 Addition Polymerization •
Add ition polymers
�
�
�
�
Form when alkene monomers react with themselves with no net loss of atoms to form polymers Prepared by radical polymerization, a radical chain reaction Radical polymerization reactions are started by using an initiator such as an organic peroxide, R -O-O-R, which decomposes when heated to form two free radicals. A radical polymerization mechanism is shown for ROOR and CH2=CHX: R-O-O-R
I n itiation :
heat
R-O'
+
' O-R
Propagation : /
H
R-O'
Initiator
+
Monomer
Terminati on: �
�
�
\
H2C = C
(
_
X
2
/
H
)
\
R - O - CH2 - C '
X
/ ,
R - O- CH2- C
X
-
Radical polymerization reactions include initiation and
R-O-CH2CHXCHXCH2-0·R
propagation steps as shown above.
The reaction termi nates when all the monomer is consumed or when two radical chains of any length react to form a single diamagnetic (nonrad ical) species. One possible termination step is shown in the scheme above.
SG-247
•
•
•
•
Stereoregular polymer �
Each unit or pair of repeating units in the polymer has the same relative orientation.
�
These polymers pack wel l together, are relatively strong, dense, and impact-res istant.
�
A stereoregular polymer with all substituents on the same side of the extended carbon chain
Isotactic polymer
Example:
�
fH3 H .fH3
�.
Syndiotactic polymer
H �H3 H f H3 H fH3 H ,f H3 H fH3 �
Isotactic configuration - same side
A stereoregular polymer with substituents alternating regularly on either side of the extended carbon chain H H3 CH3 Example: A porti on of a syndiotactic polypropylene chain with . /, alternating methyl groups:
f
Syndiotactic configuration - alternating sides
�
A polymer with substituents randomly oriented with respect to the extended carbon chain
Atactic polymers are not stereoregular.
They tend to be amorphous and less well suited for many
applications. Example:
9H3 C H3 !1
An example of an atactic polypropylene chain with randomly oriented methyl side groups:
f
Atactic configuration - random sides
!
j,
Ziegler-Natta catalysts
�
�
�
Consist of aluminum- and titanium-containing compounds, such as titanium tetrachloride, TiCI4 , and triethyl aluminum, (CH3CH2)3AI Used in the production of stereoregular polymers, including synthetic rubber Chemists were unable to synthesize rubber with useful mechanical properties until the discovery of Ziegl er-Natta catalysts in 1 953.
19.10 Condensation Polymerization •
r
Atactic polymer
�
•
H
A portion of an isotactic polypropylene chai n :
Condensation polymers
�
Are formed
:> By a series of condensation reactions :>
� �
�
From reactants, each of which has two functional groups
:> From stoichi ometric amounts of the reactants
Encompass the classes of polymers known as polyesters and polyamides Are characterized by reacti ons proceeding without the need for an initiator Typically have shorter chain lengths than addition polymers because each monomer can initiate the reaction
SG-248
•
Polyesters � �
HO
Terep hthalic acid
For polyester polymerization, it is necessary to have two functional groups on each monomer and to m ix stoichiometric amounts of the reactants. Example:
•
O�'c---D-
Form from the condensation of a diacid and a dial
•
I ,4-B is(hydroxymethyl )-cyclohexane
Kodel polyester is formed from the esterification of terephthal i c acid and I ,4-bis(hydroxymethyI)-cycl ohexane as shown by the reactions above.
Polyamides �
�
Are formed from the condensation of a diacid with a diamine For polyamide polymerization, it is necessary to have two functional groups on each monomer and to mix stoichiometric amounts of the reactants . Example:
Nyl on-66 forms by condensation of 1 ,6-hexanedioic acid and 1 ,6hexanediamine. The reactants first form a salt, which then condenses to the pol yami de.
Ho
I
�
�
"O
� ::: :: II
1 6-He
e dio
id
o
H
+
1 ,6-Hexanediamine
19.11 Copolymers and Composites •
Copolymers
�
�
•
�
Polymers with more than one type of repeati ng unit Produced from more than one type of monomer Four different forms:
alternating, block, random, and graft copolymers (see Text Fig.
Alternating copolymers
�
Fol low the pattern -A-8-A-8-A-8-A-8- , where A and 8 are monomer units
SG -249
1 9. 1 4)
� •
Nylon-66 (see above) is an alternating copolymer. The index (66) indicates the number of C atoms (6) in each type of monomer.
Block copolymers
�
�
Follow the pattern -A-A-A-A-8-8- B-8-A-A-ALong segments of one monomer, A, are followed by long segments of monomer 8 . Example:
•
Random copolymers
�
3-Methylstyrene
I
I
I
B
B
I
B
I
B
I
I
B
I
I
B
I
�
I
B B I
Graft copolymer
I
B B
The different types of copolymers extend the range of physical properties obtainable for materials. Example:
t>
j
Soft polyurethane foams formed from diisocyanates and glycols are used for i nsulation and for furniture stuffing. The formation of a polyurethane is shown below:
CH3 1)N= C= O HOCH2CH20H I
�
+
t>�
-c
--
Ethylene glycol
OCH2CH20- C-N CH3 HN-Ct> OCH2CH20-t>C"'"
O I
�
n
A polyurethane
Composite materials
�
Consist of two or more materials solidified together
�
Can exh ibit properties superior to those of the component materials
�
Combine the advantages of the component materials Example:
Fiberglass, a material of great strength and flex ibility, consists of inorganic materials in a polymer matrix.
19.12 Physical Properties of Polymers •
I
B B
I
B
f
B B
I
B
B
B B
I
Properties of copolymers
TOluene-2 . 6-di isocyanate
•
I
-A-A-A-A-A-A-A-A-A-A-A-A-A-
Soft contact lenses are composed of a graft copolymer that has a backbone of nonpolar monomers but side groups of a different water-absorbing monomer.
O =C = N •
Styrene
Radical polymerization of the monomers styrene and 3-methylstyrene is expected to form a random copolymer:
Consist of long chains of one monomer, A, with pendant chains of the second monomer, 8 : Example:
•
1 .3·Butadiene
-A-A-8-A-8-A-8-8-A-8-A-8-8-8-A-8-A-A-8-8-8-
Graft copolymers �
Styrene
Fol low no particular pattern Example:
•
High-impact polystyrene is a block copolymer of styrene and butadiene:
Polymers can be designed to have specific properties for a specific application Synthetic polymers
�
Have no definite molar mass, only an
average value
SG-250
�
�
� •
Contain chains of various lengths mixed together For a given polymer, longer average chain length leads to higher viscosity and a higher softening point.
Properties of polymers
�
�
�
�
�
� •
Tend to soften gradually upon heating and have no defi nite melting point
Depend on the average chain l ength Depend on the polarity of the functional groups Polar groups are associated with stronger intermolecular forces and tend to increase both softening points and mechanical strength. Depend on the manner in which chains pack Long unbranched chains form crystalline regions that lead to strong, dense materials. Branched-chain polymers exh ibit more tangled arrangements; they are less li kely to form crystall ine regions and tend to be weaker, less dense materials.
Elasticity
�
�
�
� �
Abil ity of a polymer to return to its original shape after being stretched
Elastomers are materials that return easily to their original shape after stretching. Elasticity of natural rubber is improved by vulcanization (heating with S), which forms disulfide (-S -S-) links between chains, increasing the resil ience of the polymer (see Text Fig. 1 9. 1 7). Extensive cross-linking provides a rigid network of interl inked polymer chains and leads to very hard materials. Most polymers are electrical insulators, but conducting polymers are known (see Box 1 9. I ).
THE IMPACT ON BIOLOGY (Sections 19.13-19.15) 19.13 Proteins •
Proteins �
�
�
� •
Are condensation copolymers of up to 20 naturally occurring amino acids (see Text Table 1 9.4) N ine of the 20 am ino acids, known as human body.
essential amino acids,
cannot be produced by the
Perform highly specific functions in the human body Enzymes (globular proteins) act as specific and efficient catalysts. Example:
The enzyme
alcohol dehydrogenase, a globular protein, oxidizes ethanol to ethanal .
Peptides �
� �
�
Are molecules formed from two or more amino acids Named starting with the amino acid on the left (N-terminus)
peptide bond, and each amino aci d in a peptide is called a residue. Typical proteins contain polypeptide chains of more than a hundred residues joined through
The -CO-NH- l ink is cal led a
peptide bonds and arranged in a particular order.
SG-2 S 1
�
Oligopeptides are molecules with only a few amino acid residues. Example:
•
Reaction of the naturally occurring amino acids, aspartic acid (Asp) and phenylalanine (Phe), produces the artificial sweetener aspartame (Asp-Ph e), a dipeptide which contains two residues. Notice the peptide l ink, -CO-N H-, shown in the box.
Structure of proteins �
H2N
"-.
ID,(0 I C
CH
I
"""'-
N H
CH OH ....... / . C C-tenmnus
0
\
CH2
/
C=O
HO
Aspartame, a dipeptide
Primary structure
� Sequence of res idues in the peptide chain
Example: �
N-terminus
The primary structure of aspartame is Asp-Phe. The dipeptide with the N- and C termini interchanged is Phe-Asp, a different oligopeptide.
Secondary structure
� Describes the shape of the polypeptide chain
� Controlled by intramolecular interactions between different parts of the peptide chain
� Common secondary structures include the
and
� The
1 9.20).
a.
helix and the I) sheet (see Text Figs.
19. 1 9
a.
helix is a helical portion of a polypeptide held together by hydrogen bonding. � The I)-pleated sheet is a portion of a polypeptide in which the segments of the chain l ie side by side to form nearly flat sheets. �
Tertiary structure
� Tertiary structure describes the overall three-dimensional shape of the polypeptide,
including the way in which macromolecule.
a.
helix and I) sheet regions fol d together to shape the
� Folding is a consequence of the hydrophobic and hydrophilic interacti ons between
residues lying in different parts of the primary structure. One i mportant link responsible for tertiary structure is the disulfide link (-S -S -) between amino acids containi ng sulfur.
� Proteins are classified as globular (ball-shaped) or fibrous (hair-shaped with long chai ns of
polypeptides that occur in bundles).
� Globular proteins are soluble in water; fibrous proteins are not. Essential ly, all enzymes
are globular proteins.
Example:
�
Hemoglobin (respons ible for oxygen transport in blood) and cytochrome c (a component of electron transport chains in mitochondria and bacteria) are globular proteins; fibroin (the protein of silk) is a fibrous protein.
Quaternary structure
> Describes the arrangement of subunits in proteins with more than one polypeptide chain
> Not al l proteins contain more than one subunit, so not all have a quaternary structure.
Example: •
Hemoglobin (see Text Fig. quaternary structure.
1 9.20) contains four polypeptide units and has a
Denaturation
�
Loss of structure of proteins
SG-2S2
---,) Occurs when a protein loses quaternary, tertiary, or secondary structure (disruption of noncovalent interactions)
---,) Denaturation is caused by heating and other means, is often irreversible, and is accompanied by loss of function of the protei n. 19.14 Carbohydrates •
Carbohydrates
---,) Most abundant c lass of naturally occurring organic compounds ---,) Constitute more than 50% of the Earth ' s biomass ---,) Members include starches, cellulose, and sugars. ---,) Often have the empirical formula CH20 (hence the name carbohydrates) •
D-Glucose
---,) Most abundant carbohydrate ---,) Forms starch and cellulose by polymerization reactions ---,) Has the molecular formula C6H 1206 ---,) I s classified as both an alcohol and an aldehyde (pentahydroxyl aldehyde) ---,) Exists in an acyclic (open-chain) and in two cyclic (cl osed-chain) forms cal led anomers ---,) I n water, the cyclic structures, also called glucopyranoses are favored. •
Structure of D-glucose
R ":
---,) The straight-chain structure of D-glucose and the interconversion to the two cyclic anomers are shown on the right.
HO-C
=l�
H -C-OH
I
---,) The percentages indicate proportions of the several species present in water solution.
H-C
I
---,) The anomers, a-D-glucose and �D-glucose, differ in the stereochemistry at the carbon atom derived from the aldehyde C atom.
a-D-G Jucose 36%
---,) More real istic representations of the cycl ic forms of D-gi ucose are shown on the right: Notes:
�
HO
I ) L-G lucose, not found in nature, is the
CH,OH
rl-D-GJ ucose 64%
D-G Jucose 0.02%
CH'OH
HO
0
OH
a - D-G J ucose
OH
HO
CH,OH
O
� \ � HO
OH
I3-D-GJ ucose
mirror image of D-glucose, and together they form a pair of enantiomers, which differ in the direction of rotation of polarized l ight.
2) •
H
The symbols D and L refer to a configuration relationship to the compound D-glyceraldehyde and not necessari ly to the sign of rotation of light.
Polysaccharides
---,) Polymers of gl ucose ---,) I ncl ude starch (digestible by humans) and cellulose (not digestible by humans)
SG-253
OH
Example:
Cellulose, the structural material of plants and wood, is a condensation polymer of P-D-glucose. A three subunit chain of cellulose is shown below. The polymer i s formed formall y by elimination of H20 from glucose monomers.
19.15 Nucleic Acids •
DNA �
�
Abbreviation for deoxyribonucleic acid DNA molecules > Carry genetic information from generation to generation
> Control the production of proteins •
> Serve as the template for the synthesis of RNA (see RNA subsection)
Structure of DNA � �
�
� •
DNA molecules are condensation copolymers of enormous size.
�
�
�
o-p -o-
I
0-
base
base
0
II I sugar-O-P-O-
0
�
�
�
su ar-o- - o- -
I
0-
0-
Nucleotide
A base-sugar-phosphate grouping is called a nucleotide.
54� CH,OH O H OH
In DNA, the sugar is deoxyribose.
NH, H OH H H Cl H,N:en N ex"> '�'CJ:
There are four possible bases: cytosine (C), guanine (G), adenine (A), and thymine (T). The sugar and four bases are shown on the right :
3
N ucleosides
�
II
Nucleoside
2
NH,
0
1
0
�
Deoxyribose
•
o
I sugar
A base-sugar unit is called a nucleoside.
Sugar and bases �
-�
DNA is composed of a sugar phosphate backbone and pendent bases as shown on the right:
base
�
N
Guanine (G)
Cytosine (C)
Formed in DNA by condensation involving the OH group at carbon an appropriate amine hydrogen atom of a base
0
�
Adenine (A)
1
0
�
Thymine (T)
i n deoxyribose and
"''CJ: �l:i �CC,L � yt � ---l
Cytosine, guanine, adenine, and thym ine nuc leosides are shown HO here: H
H
OH
NH,
o
A
N�O
H
H
H
Cytosine nucleoside
N
HO
H
H
OH
H
H
H
Guanine nucleoside
SG-2S4
NH,
HO
H
H
�
0
<,N
H
H
0
'
),
N.r" HO
H
Adenine nucleoside
H
H
OH
0
N
H
H
0
H
Thymine nucleoside
•
Nucleotides � �
•
Condense at carbon 3 and carbon 5, el im inating water, to form the DNA molecul e (nucleic acid), a polynucleotide A trinucleotide segment of a polynucleotide is shown on the right:
�
� �
N
0
N
l
�i-r!
NH
--l A
':!�o� . � � f � \
o= -o 0-
Association between strands of two polynucleotides occurs when hydrogen bonds between bases on different strands are formed.
H
H
H
0
N
H
H o I o=p-o ' I � 0- I
0
H
H�N-H ----- -O N CH3 O-----H-N H N nN-�NN � ----H , , /-\ - - H -N� /--\ � N N )= - -- - - H - N
In DNA, only two types of base pairs (G with C and A with T) occur.
These are shown below with the hydrogen bonds indicated by dashed l i nes : /
Note:
sugar
0-
Cytosine
\
H
Guanine
sugar
0
Thymine
Adenine
sugar
CG and AT base pairs have approximately the same size and shape, which reduces dist0l1ions in the double helix structure.
RNA
�
The abbreviation for ribonucleic acid
�
Is shorter than DNA, and generally single-stranded
�
I s a polynucleotide, l ike DNA
Three types of RNA �
�
•
o = -o � o � 0-
Polynucleotide (nucleic acid) strands l ink to each other in pairs to form the wel l-known double helix structure (see Text Fig. 1 9.30).
sugar
•
rl· --l ) I �
.
DNA double helix
�
•
-'-i
�
Messenger RNA (mRNA) carries geneti c information from a cell nucleus to the cytoplasm where translation to protein takes place. Ribosomal RNA (rRNA) appears to act as a catalyst in the biosynthesis of proteins. Transfer RNA (tRNA) is used to transport amino acids during protein synthesis.
Structure of R N A
�
�
RNA contains four base pairs as in DNA, but uracil (U) substitutes for thymine (T).
Uracil lacks the methyl group of thymine at the carbon-5 position. The structures of thymine and uracil are shown on the right.
Note:
Thymine (T)
Uracil (U)
The position of the methyl group on thymine does not interfere with its hydrogen bonding with adenine (A) (see figure above of an AT base pair).
SG-2 5 5
�
�
� �
0H O H OHI :C� H20H 0 O� 45CH2� H OH OH H H OH H H
The sugar in RNA is ribose (deoxyribose is shown for comparison): A lthough RNA molecules are generally si ngle stranded, they contain regions of a double helix produced by the formation of hairpin loops. I n these regions, the usual base pairing is A with U and G with C.
3
2
Ribose
However, in RNA, imperfections i n base pairing are common.
SG-256
3
2
Deoxyribose
SOLUTIONS MANUAL
SM-l
FUNDAMENTALS
A.I
A.3
A.5.
A.7
A.9
(a) chemical; physical; (c) physical The temperature of the injured camper and the evaporation and condensation of water are physical properties. The ignition of propane is a chemical change. (a) physical; chemical; (c) chemical (a) intensive; intensive; (c) extensive; (d) extensive (a) 2xl0-6 km 2xl03 j.!m; 2xIQ5 j.!m; (c) 2xl07 nm 2xl04 j.!m. Therefore, is the largest pint )x ( 1 q�art ( 0.946 L ( 1000 mL 236 mL 1.00 cup x ( 21 cups 2 pmts 1 quart 1 L (b)
(b)
(b)
=
=
(b)
(b)
A. l l
A.13
)
x
d= m V = =
112. 3 2 g 1 mL3 (19.29.3 2g7. cm-mL3 23.45 mL J ( 1 cm J -
SM-3
)x
)
=
A. 1 7
rearrange gIves 20. 0 g 2.70 g .cm3 7. 4 1 cm3 Since the area is 1 centimeter, the thickness must be 7. 4 1 cm. The result should have 3 significant figures because the number 3.2 5 has the least significant figures. The results should be 0. 989 x 10 3 pm (a) 4.82 nm x ( 1000pm 4. 8 2 ) 1nm cm3 x ( 1000mm3 3 )x ( 1 min ) 30. 5 mm3Is (b) ( 1.8mm3.mL ) x ( 11mL 1cm 60s 1kg (c) 1. 88 ng x ( � )x ( 3 109 ng 10 g 1. 88 x 10- 1 2 kg 3 2. cm 10 6 6g � X X 6 (d) ( cm3 ) ( 1000g ) ( 1m3 ) 2.66 x 103 kg/m3 1L 3 X 0. 0 44 mg/cm ( (e) ( 0.0L44g )X ( 1000mg ) 3 19 1000cm v
A.19
A.21
.
m d =-, V =
V=
m d
=
=
)
=
)
=
=
)
A.23
(a)
d
=
m V
=
- ( 1.100 cm x 0.0.523113 cmg x 0.2 12 cm ) 0.2 13 g 3 ) - ( 0.1238 cm 1 . 72 g . cm =
-3
SMA
This detennination is more precise because the volume is not limited to two significant figures as it is in part (b). (b) = m ( 41. 003 g - 39. 7 53 g ) 20.37 mL - 19.65 mL ( 1.2 50 g )(�)3 0. 72. mL 3 1 cm = 1. 7 g cm.!.2. mv2 2 2 m 1000 h 1 2 J ( J ( . = .!.2. (4.2 kg)(14 km h-') 3600 s 1 km = 32 kg· m2 . S-2 = 32 J m = 2. 8 metric tons, = 100 km· hr-I, vr = 50 km· hr-' 1 2 =-mv 2 . ( 103 .kg 1= -(2. 8 metnctons) 1 metnc ton2 J 2 [( 1 001 hrkm ) ( 601 mihrn. ) ( 601 misec.n. )] =4. 3 2 kg·km2 ·S-2 =4.32 x 106 kg·m2 ·S-2 =4,320 kJ 10.3 .kg . ( 1 metnc = -(21 2. 8 metnctons) ton2 J [( 501 hrkm )( 1 min. ) ( 1 misec.n. )] = 0.27 kg· km2 . S-2 = 0.27 x 1 06 kg· m2 . S-2 = 270 kJ = (4,320 - 270) kJ 4,050 kJ 4.0 x I 03 kJ (2 SF) d
V
=
=
A.2S
A.27
E
K
=
Vi
EK
EK(inil)
EK(final)
60
EK(inil) - EK (final)
hr
60
=
SM-5
=
This amount of energy could have been recovered, neglecting friction and other losses, or used to drive the vehicle up a hill. Ep = 9.8 1 ms 2 Setting potential energy equal to 4,050 kJ 4.05 kg m2 s-2 and solving for height gives Ep ( 4.05 X106 kg · m2 ' S-22 ) m 150 m (2 SF). (2800 kg)(9. 8 1 ms- ) Ep =(40.0 g)(9. 8 1 m . s-2 )(0.50 m) ( 10001 kgg ) = 0.20 kg· m2 . S-2 for one raise of a fork. For 30 raises, (30)(0.20 kg·m2 'S-2 ) 6.0 We need to use the expansion given in Exercise A.22 to help solve this problem. We also need to recognize that Ep = for the small difference in distance, can be represented by subtracting Ep at distance between the proton and electron from Ep at distance = mgh
g
-
=
h=
A.29
mg
= 1 47
=
= mgh
=
A.3 1
=
J.
egh
h,
r + h.
r
A.33
The relationship between distance of separation and potential energy for charged particles is given in section A.2, Equation = E -(1.62022 0-'1 9 C)'2 x1 '2 m- )(53 0- m) (8.85419 10 SF) 352 X 0- ( 1. 602x10- 19 J ) SM-6 4.
p
= VCr) =
!lrlL. 4nco r
( - e)( +e)
4ncor l x = ----------���--��------�-4n x - ' c · rl l eV = -27 . 1 7 eV= - 27 eV (2 1 18 J = -4 .
Considering the proton and electron beginning at rest and at infinite separation sets the initial total energy to Since the electron is not at rest in a hydrogen atom, its total energy is represented by Equation 5: We have only calculated the potential energy. The discrepancy between the calculated value of the potential energy, -27.7 eV, and the measured amount released, -13. 6 eV, is the kinetic energy of the electron, 13. 6 eV. O.
E = EK + Ep
B.l
mass of sample number of beryllI. um atoms mass of one atom - ( 1. 50 1�:�i�:. atom- ' ) ( 1�:�g ) 1.40 1 022 atoms (a) 5p, 6n, 5e; (b) 5p, 5n, 5e; (c) 15p, 16n, 15e; (d) 92p, 146n, 92e (a) 1 94 Ir; (b) 2 Ne; (c) S IV =
=
B.3
B.5
B.7
B.9
B. 1 1
Element Chlorine Zinc Calcium Lanthanum
---=---'
-
x
x
Symbol Protons Neutrons Electrons Mass number 19 17 3°Cl 36 17 O)Zn 65 35 30 30 4UCa 20 20 40 20 IT/La 57 137 80 57
protons (a) They all have the same mass. (b) They have differing numbers of protons, neutrons, and electrons SM-7
B.13
B.IS
B.17
B.19
B.2 1
C.l
C .3
C .7
(a) 0.5 519; (b) 0.4479; (c) 2.459 10-4; (d) 551.9 kg (a) Scandium is a Group 3 metal. (b) Strontium is a Group 2 metal. (c) Sulfur is a Group 16 nonmetal. (d) Antimony is a Group 15 metalloid. (a) Sr, metal; (b) Xe, nonmetal; (c) Si, metalloid Fluorine, F, Z 9, gas; Chlorine, CI, Z 17, gas; Bromine, Br, Z 35, liquid; Iodine, I, Z 53, solid (a) block; (b) block; (c) block; (d) block; (e) block; (t) block Container (a) holds a mixture (one is single compound and another is single element); container (b) holds a single element. The chemical formula of xanthophyll is C4oH5602. x
=
d
=
=
=
p
s
d
d
(a) Cesium is a metal in Group 1; it wil form Cs+ ions. (b) Iodine is a nonmetal in Group 17/VII and wil form ions. (c) Selenium is a Group 16NI nonmetal and2 will form Se ions. (d) Calcium is a Group metal and wil form Ca + ions. (a) l oBe2+ has 4 protons, 6 neutrons, and el8e0ctrons. (b) 1 702- has 8 protons, 9 neutrons, and 10 electrons.7 3 (c) Br-has 35 protons, 45 neutrons, and 36 electrons. (d) 5As -has 33 protons, 42 neutrons, and 36 electrons. r-
2-
2
C .9
p
2
SM 8 -
C.13
(a) Aluminum forms Ae+ ions; tellurium forms Te2- ions. Two aluminum atoms produce a charge of Three tellurium atoms The formula for aluminum telluride is produce a charge of Al2Te3 (b) Magnesium forms Mg2+ ions and oxygen forms 02- ions. A magnesium ion produces a charge of whi c h is requi r ed to bal a nce the charge on one 02- ion. The formula for magnesium oxide is MgO. (c) Sodium forms ions; sulfur forms ions. The formula for sodium sulfide is Na2 S. (d) Rubidium forms ions and iodine forms ions. One iodide ion is required to balance the charge of one rubidium ion, so the formula is RbI. (a) HCI, molecular compound (in the gas phase); (b) S8, element (molecular substance); (c) CoS, ionic compound; (d) Ar, element; (e) CS2, molecular compound; (f) SrBr2, ionic compound (a) Group III; (b) aluminum, Al The formula is Ah03 . 2
3
x
-2
=
x
+3
=
+6.
-6.
•
+2,
-2
+1
+1
C.lS
C.l7
C.l9
-1
( cm cmg cm ) g . cm-3
d = m
V
-
2.5
= 3 .4
D.l
x
1 02
3.0
x
4.0
(a) MnCh. Mn forms ions and chlorine forms ions. (b) Ca3(P04h. Calcium forms ions and the phosphate ion is pol-. (c) Ah(S03h Aluminum forms ions and the sulfite ion is 3SO/-. (d) Mg3N2 . Magnesium forms ions and the nitride ion is N -. 2+
-1
+2
+3
2+
SM-9
D .3
D.S
D .7
D.9
D.13
D.IS
D.17
D.19
D.21
(a) calcium phosphate; (b) tin(IV) sulfide, stannic sulfide; (c) vanadium(V) oxide; (d) copper(l) oxide, cuprous oxide (a) Ti02 ; (b) SiCI4 ; (c) CS2 ; (d) SF4 ; (e) Li2 S; (f) SbFs; (g) N2 0S ; (h) IF? (a) sulfur hexafluoride; (b) dinitrogen pentoxide; (c) nitrogen triiodide; (d) xenon tetrafluoride; (e) arsenic tribromide; (f) chlorine dioxide (a) hydrochloric acid; (b) sulfuric acid; (c) nitric acid; (d) acetic acid; (e) sulfurous acid; (f) phosphoric acid (a) sodium sulfite; (b) iron(III) oxide or ferric oxide; (c) iron(lI) oxide or ferrous oxide; (d) magnesium hydroxide; (e) nickel(lI) sulfate hexahydrate; (f) phosphorus pentachloride; (g) chromium(III) dihydrogen phosphate; (h) diarsenic trioxide; (i) ruthenium(lI) chloride (a) CUC03 Copper (II) carbonate; (b) K2S03 potassium sulfite; (c) LiCI, lithium chloride (a) heptane; (b) propane; (c) pentane; (d) butane (a) cobalt(III) oxide monohydrate; C02 03 . H 2 0; (b) cobalt(lI) hydroxide; Co(OH)2 Si; SiH4, silicon tetrahydride; N�Si, sodium silicide E
=
SM-IO
D.23
D .25
D.27
E.1
(a) lithium aluminum hydride, ionic (with a molecular anion); (b) sodium hydride, ionic (a) selenic acid; (b) sodium arsenate; (c) calcium tellurite; (d) barium arsenate; (e) antimonic acid; (f) nickel(III) selenate (a) alcohol; (b) carboxylic acid; (c) haloalkane 23 6. 0 22 10 1 .0 mo e 1.0 moleAgAgatom 1 Ag144pmatom 1011m2 pm 8.67 x 101 3 m 8.67 x 101 0 You wil to add 18 Ca atoms to balance the 9 Br atoms on the left pan because a Br atom is twice as massive as a Ca atom. people (a) moles of people 6.0226.01021093 people· mor 4 1.0 10-1 mol . 1 mol- 1 peas -I I .0 1014 s (b) hme 1.0 10 4 mol· S (1. 0 101 4 s) ( 36001 h s ] ( 124dayh ] ( 365 days ] 3.2 106 years (a) mass of average Li atom ( 7.10042 ) (9.988 10-24 g) ( 92100. 5 8 ) (1.165 10-23 g) 1.153 10-23 g . atom-I molar mass (1.153 10-23 g . atom-I ) (6.022 102 atoms · morl) 6.94 g . mor l (b) mass of average Li atom x
x
I
=
=
E .3
E .5
x
km
=
=
=
x ----,,---,-
x
x
x
1
x
=
X
1 yr
x
E.7
x
=
=
=
+
X
x
X
=
=
x
x
SM l -
1
3
=
(�.�� ) (9.988 X 10-24 g) + C001��·67 ) (1.165 10-23 g) x
1.1556 10-2323g . atom-I molar mass (1.1556 10- g . atom-I) (6.022 1023 atoms· morl) 6.96 g . mol-I (a) MgS04 • H2 0 formula mass 246.48 g . mol-I mol 0 atoms atoms f 0 ( 246.45.158 g .gmOrl J ( mol11 MgSO 4 · 7 H2 0 J (6.022 1023 atoms· mor l ) 1.3 8 023 ( (b) formula units 246.45.158 g .gmol- I J (6.022X 1023 atoms . mol- I ) 1 .26 1022 (c) moles of H2 0 7 ( 246.45.158 g .gmol- I J 0.146 mol The percentage lO B 100 - percentage l i B (0; ; molarmass 100%10 B J (mass 10 B)+ (0;;100%II B J (mass l i B) I I B (mass l i B) ( 100%100%- % II B (mass lO B) + (-100% J J Rearranging gives B l O 100· mass % l i B 100 · molmassar mass l i B - mass lO B . mor l 100(10.8 1 g . morl) 100(10. 0 13 g ) I 11.093 g . mol-I - 10.0 13 g . mol73. 8 % % l O B 26.2 % =
x
x
=
x
=
E.9
=
7
=
0
=
x
xl
=
=
x
=
E.1 1
=
=
=
0
0
%
=
=
=
=
=
SM- 12
E.13
E.15
(a) 114.8275gg. mol-, = 0.65 mol In 80 g. ---, = 0.63 mol Te ----'--127.60 g mol 75 g of indium contains more moles of atoms than 80 g oftellurium. g =0.484 mol P (b) 30.915.7 g0. mor' 15. 0. g = 0.468 mol S 32. 07 g mor' 15. 0 g ofP has slightly more atoms than 15.0 g ofS. (c) Because the two samples have the same number of atoms, they wil have the same number of moles, which is given by 2.49 x?103 22 atoms = 0.0413 mol 6. 022 x 10- atoms · mor' (a) ( 69;7236g g. molGa_' Ga J x ( 102.1 mol90 gRhRh ) = 53 g Rh (b) = ( 114.8236g gIn. mol-' In J x ( 102.1 mol90 gRhRh ) 32 g Rh m Rh
=
mRh
E.17
=
(a) molar mass of Al2 03 =101.96 g · mor' = 101. 910.6 g0 .gmol-' = 0.0981 mol = (0.0981 mol)(6.022 x 1023 atoms · mor') 5. 9 1 1022 molecules (b) molar mass ofHF 20.0 1 g . mol-' = 20.25.0921 gx .10-mor'3 g = 1 . 30 x l 0-3 mo1 = (1.30 10-32 mol)(6.022 x 1023 atoms· mol-I) = 7. 83 10 0 molecules (c) molar mass of hydrogen peroxide 34.02 g . mol-' n
A ' lO)
NA ' O l )
=
=
n
HF
NHF
X
X
=
SM-13
X
3 g 4.56 x 10 mol 1. 105 5 = " 34.02 g . mol-I (4.56 10-1 5 mol)(6.022 x 1023 atoms· mol-I ) 2.75 10 9 molecules (d) molar mass of glucose 180.15 g . morl 125 0. g l 6.94 mol 180.15 g mor 23 = (6.94 mol)(6.022 x 10 atoms· morl) = 4.18 x 1024 molecules (e) molar mass ofN atoms 14. 0 1 g ·mor l 4.3 7 g I = 0.3 12 mol 14. 0 1 g·mol- 23 = (0.3 12 mol)(6.022 xl 0 atoms· mol-I ) 1. 88 x 1023 atoms molar mass of N2 molecules = 28.02 g . morl 4.37. g l 0.156 mol 28.02 g mor 23 = (0.156 mol)(6.022 x 10 atoms· mol-I ) 9.39 x 1022 molecules (a) molar mass of CuBr2 = 223. 3 5 g . mol-I Cu2+ 0.0 1 34mol Cu2+ ( 223. 33.50g0.gmorlCuBrCuBr 2 J x ( 11 molmolCuBr 2J 2 (b) molar mass of S03 = 80.06 g . morl = ( 80.00.67g00. morg S0l 3S03 J 8.74 10-3 mol SO3 (c) molar mass ofUF6 352.03 g . mol-I = 25.2 kg x ( 10001 kgg J x ( 352.1 mol03 gUFUF6 6 J x ( 16molmolUFF-6 J = 430. mol FnH
X
°
NH ° , ,
=
-5
=
X
=
X
=
n lucose g
=
=
N luCOSe g
=
=
nN
=
NN
=
nN ,
=
=
NN ,
E.19
n
Cu
,+
=
=
=
nso,
=
n
. F
SM-14
X
(d) molar mass ofNa2C03 · 10H20 286. 2 1 g-mol- 1 ( 286.22.10g·morl 0 gNa2 C03 ·10H20 ] x ( 10 mol H2 0 ] Na2C03 ·10H2 0 1 mol Na2C03 ·10H 20 0. 0699 mol H 20 (a) number of formula units (0.750 mol)( 6. 022 x 1023 formula units · morl) 4.52 x 1023 formula units (b) molar mass of Ag2 S04 = 311. 80 g . mOrl 20 formula units (311.80 g . morl) 1000 mg x 10 2. 3 9 ( 1g ] ( 6. 022 x 1023 formula units· mol-I ] = 124 mg (c) molar mass of NaHC02 = 68. 0 1 g . mOrl 29 g 1 ] (6.022X 1023 formula units· mol-I ) ( 68. 03.14g·mor = 3.036 x l 022 formula units (a) molarmass of H 2 0 = 18.02g·mol- 1 2 3 g·molecule- I x l0=2. 9 92 ( 6.022 x 18.10203 2molg·morl ] ecules · morI g. ] (6.022 X 1023 molecules· mol-I) (b) ( 18. 01000 2 g mol = 3. 3 4 X 1025 molecules (a) molar mass ofCuCb-4H20 206.53 gmol- 1 ( 206.58.3 6g1. mor g CU:12 ·4H20 ] = 0.041 7 mol CuCl2 . 4H2 0 CuCI2 · 4H 20 =
n
H 20
=
=
E.21
=
=
E .23
NH 2 0
E.25
=
_I
=
n
=
SM-15
(b) Because there are 2 mol CI- per mole of compound, the number of moles will be twice the amount in part (a), 0.0834 mol CI(c) There are 4 mole of H20 for each mole compound. Therefore, 23 number of water molecules (0.203417 x 4) mole H20 x (6.022 10 molecules ' moC') 1.00 x 10 H20 molecules 00g·mol -l) 0.3 099 (d) fractI. on of mass due to 4(16. . 206. 53 g mol(a) [ 1 mol10hydrated mol H 2 0 [ 18. 0 1 g H2 0 [ 1 mol hydrated cpd x I 00 cpd J 1 mol H 2 J 264. 166 g hydrated cpd J 68.2% H 2 0 Therefore, 6.82 kg out of 10 kg was water. $ - $72. 00 J [ 10 kg hydrated CPd [ 1 kg H2 0 L H 2 0 [ 10 kg hydrated cpd 6.82 kg H2 0 J 1 L H2 0 J $10.60 per L H 2 0 (b) Since the anhydrous compound costs $80.00 for 10 kg, or $8.00/kg, and only 10-6.82 3.18 kg is NaHC03 , a fair price would have been $8.00 x 3 . 18 kg $25.44. kg Zr atoms/mole (90.0000 g 90Zr) [ 1.49291 atom x 1 0-22 g J 6.028535 x 1023 atoms 1 mole Since there are more atoms per mole, each mole wil weigh more and all of the molar masses wil be proportionately higher by 23 - 6.02214 x l023 X l00 0.10619% [ 6.028535XI0 J 6.02214 x 1023 V
=
=
°
E .27
=
I
°
=
?
=
=
=
E.29
=
=
=
=
SM- 16
=
(a) 6 � 12. 0000 .... g'mol-I I1.0010619 12. 0 13 g'mol-I I (b) 1 9 .97Au � 196.97 .... g'mol- 1.0010619 197.18 g'molSolve by factor label (dimensional analysis). 2 3 H atoms ] ( 1. 0 74 g )( 2. 5 0XI0 3 ?. moIe H atoms 28 .0 cm NaBH 1 cm3 3.93 g x ( 6.022 101 mol23 H atoms ) 3.18 moles H atoms For example, Brhas m l 80. 9 163 u, m 78.9 183 u 12C
=
x
x
E.31
=
4
x
=
E .33
=
=
2
=
. :7""f� � . . ./ ·-_··. ·. . ·. .·+V . . // �� .. . . �.+. . . . . . . . . . . . . .. .. .
81
80 . 5 180
.
..
. ... .......... ... .................. +
-
.
+·........·..· ··.. .
_... ..........
79 . 5
E
.................................
..
.. .. . .. ··....·..............
.. .. .......................... ... ....-
/
,........ ..... ................................._+. ........... .... .. .... ... ........
79
78 . 5
f + . .... ................................ ........................................................
o
20
40
60
percent m 1
E.35
(a) (b)
79; 158, 81 160, Bromine-79 and bromine-81 occur in nature. Bromine-79 is more abundant. (Brisotopel + Brisotopel )
=
(Brisotopel + Brisotope2 ) =
Brisotopel =
Br(isotope2) =
SM-I7
.+ . ..... ...........
80
.
............-
1 00
F.l
C I OH I 6 0 10 x 12. 0 1 g 120.1 g x ( 152.1002264 ) 78.90% 16 1.0079 g 16.1264 g x ( 152.1002264 ) 10. 5 9% 1 16.00 g 16. 00 g ( 152.1002264 ) - 10. 5 1% =
x
x
=
=
=
=
x
84.07 g x ( 161.10020 g J = 52. 1 5% 15.1185 g x ( 161.10020 g J = 9.3 787% 14. 0 1 g x ( 161.10020 g J =8.691% 48.00 g x ( 161.10020 g J =29.78% 161.20 g 100.00% (a) M 2 0 88. 8% M For 100 g of compound, 88. 8 g is M, 11.2 g is 0. g M 2 0 ( 100 g M2 0 ( 16.00 g 0 ) ( 1 mol mole M2 0 11.2 g ) J 1 mol 1 mol M2 0 J 143 g·mor M20 Therefore, 143- 16 127 glmol are due to M. Since there are 2 moles of M per mole ofM2 0, the molar mass ofM 63.4 glmol. That molar mass matches Cu. (b) copper(I) oxide (a) For 100 g of compound, 79 g 1.426 mol moles of Na = 22.932.9 g·mor moles of Al 26.913.8 g02. molg -) 0.4826 mol SM 18 7 x 12. 0 1 g 15 x 1.0079 g 1 x 14. 0 1 g 3x16.00 g
=
=
=
=
F.5
?
°
_
=
°
°
=
=
F.7
)
=
=
=
-
g = 2.852 mol moles of F = 19.054.19 0 g ·mor Dividing each number by 0.4826 gives a ratio of 1 Al : 2.95 Na : 5. 9 1 F. The formula is Na3 AIF6 . (b) For 100 g of compound, 31.9 1 g ' =0.8 161 mol moles of K = 39.10g·molmoles of CI = 35.428.5 g9.3molg - = 0.8 161 mol mass of is obtained by difference: moIes f O = 100 g 16.-31.00 9g1. gmor- 28.l 93g = 2.448 moI Dividing each number by 0. 8 161 gives a ratio of 1.00 K : 1 CI : 3.00 O. The formula is KCI03 • (c) For 100 g of compound, moIes f N = 14. 012.1 g2. molg - I = 0. 8 71 mol moles f H = 1.00795.2g6g. mor = 5.22 mol g = 0. 869 mol moIes f P = 30.926.7 g9. mol moles of 0 = 16.055.0 g6. molg - = 3.475 mol Dividing each number by 0.869 gives a ratio of 1.00 N : 6. 0 1 H : 1.00 P : 4.00 O. The formula is NH6 P04 or [NH4 ] [H2 P04 ], ammonium dihydrogen phosphate. g I =0.134mol (a)molesof P = 30.94.14 . 7 g molmoIes f CI = 27.35.845g g-. 4.14g morl 0.667 mol SM-19 I
I
°
0
0
0
0
I
I
I
F.9
0
=
F.I l
Dividing each number by 0.134 mol gives a ratio of 4.98 CI : 1 P. The formula is PCIs ' (b) Name is Phosphorus pentachloride. For 100 g of compound, 49g 5.6 19 mol Moles ofC ( 12. 067.1g·mor J g J 4.563 mol Moles of H ( 1.008g4.60· mor Moles ofCI ( 35.412.5 g4.5gmol- J 0.3512 mol 84g 0.7024 mol Moles ofN ( 14. 09.1g·mol -J Moles of 0 ( 16.05.0g62g· mol- J 0.3512 mol Dividing each number by 0.3512 mol gives a ratio of 16.00 C : 13.00 H : 1.00 CI : 2.00N : 1.00 O. The formula is C I6H I 3CIN20. =
=
,
=
=
=
,
=
=
,
=
F.13
=
,
=
,
For 100 g of the osmium carbonyl compound, 89 g 1.323 mol moles of C 12. 015.1 g·mol moles of 16.021.18g 0g·mor 1.324 mol 93 g 0.3309 mol moles of Os 190.62.2 g·mor Dividing each number by 0.3 309 mol gives a ratio of 4.00 C : 4.00 0 : 1.00 Os. (a) The empirical formula is OSC4 04 • (b) The formula mass of OSC4 04 is 302.24 g . mor' . The molar mass is 907 g . mor' which is 3 times the formula mass, so the molecular formula is OS3 C' 2 0' 2 ' SM-20 =
°
=
=
=
,
=
,
,
=
F. l S
For 100 g of caffeine, moles of = 12. 049.1 g4.8molg - = 4.12 mol moles f = 1.00795.19g g . mol- I = 5.15 mol moles of = 14. 028.1 g8.5molg - I = 2.059 mol moles of = 16. 016.0 g48. molg -I = 1.03 mol Dividing each number by 1.03 mol gives a ratio of 4.00 : 5. 00 : 2.00 1.00 0. The formula is with a molar formula mass of 97.10 g . mOr l . Because the molecular molar mass is twice this value, the C
0
I
H
N °
C
N:
F. 1 7
H
C4 H5N20
Glucose has amolarmassof180.15 g·morl and wil have the following composition: 0 1 g .. mor ll ) = 40.00% = 6(12. 180.15 g mor 0079 g . mol-I) = 6. 7 1 % %H = 12(1.180.15 g·mol-I 00 g .. mol-Il ) = 53.29% %0 = 6(16. 180.15 g mor Sucrose has a molar mass of 342.29 g . morl and will have the following composition: 0 1 g .· mol-Il ) = 42.10% %C= 12(12. 342.29 g molg. . mol-I)I = 6.48% %H = 22(1.342.0079 29 g mol(C6 H I 2 06 )
%C
(CI 2 H 22 0 1 1 )
SM-21
00 g .. mor') 51.42% %0 11(16. 342.29 g mol-' While the values for glucose and sucrose are too close to allow us to distinguish between them by this value alone, %C (40.00 versus 42.10%) and %0 (53.29 versus 51.42%) values are sufficient that, when taken together, can give us a reasonable amount of confidence in distinguishing between them. Calculate the mass percent carbon for each fuel from its formula. 2(-'-1_2--- _.0-1).: x1 00 85. 63% C 2(12.0 1) + 4(1.0079) 3..(_1..:.. 2_.0_1..) .:.... __ _ _ 3(12. 0 1) + 8(1.0079) + 16.00 x 100 59.96% 7(12.0 1) x100=83. 9 1% 7(12. 0 1) + 16(1.0079) ethene (85.63%) heptane (83. 9 1 %) propanol (59.96%) (a) empirical formula: molecular formula: (b) empirical formula: molecular formula: C2HSN2 This problem requires that we relate unknowns to each other appropriately by writing a balanced chemical equation and using other information in the problem. NaNO) + y Na2S04 � (x + 2y) Na + x NO) +y S0241.6 1 g Na+ 0.07003 mol Na+ x + 2y 22.99 g·mor' Na+ 5. 3 7 gtotal - 1.6 1 gNa+ =3.76 g =(62. 0 1 g·mor')x + (96.07 g ' mol-' )y =
=
%H
F.19
_ _ _
=
_ _
_ _ _
C
=
C
>
F.2 1
>
C2 H3CI,
C4H6Cb
CH4N,
F.23
+
X
=
-
=
SM-22
Rearrange and substitute: 3. 76 62. 0 1(0.07003 -2y) 96.07 0.06065 0.07003 - 2y 1. 548y 0.009385 0.4516y y 0.02078 moles sulfate, 0.02847 moles nitrate Therefore, the mass of sodium nitrate in the mixture was 0.02847 mOI ( 8S.001 gmolNaN03 ) 2.42 g NaN03 2.42 g NaN03 xl 00 45.1 % NaN03 . 5.3 7 g total (a) density; (b) the abilities of the components to adsorb; (c) boiling points (a) heterogeneous, decanting; (b) heterogeneous, dissolving followed by filtration and distillation; (c) homogeneous, distillation mass of AgN03 (0.179 mol· L-' )(0.5000 L)(169.88 g . mol- I) 15.2 g 2.111 g (a) moI anty· f Na2 C03 (l05.99g·mor I )(0.2500L) 0.079 67 =
Y
+
=
+
=
=
x =
=
=
G.!
G.3
G.5
G .7
=
=
=
=
0
M
(2.15 10-3 mol Na+(l mol Na2 C02 ) 1.3 5 x 1 0-2 L or 13. 5 mL (0.079 67 mol· L-' Na2 C03 )(2 mol Na+ ) (4.98 10-3 mol cot)(l mol Na2 C022-) (b) (0.079 672mol· L-' Na2 C03 )(1 mol C03 ) 6.25 x 10- L or 62. 5 mL 3 (50. 0 x lO2 C0L-'3 ) Na C0 ) (c) (105.99 g . mol-I )(0.079g67Namol· 2 3 3 5.92 10- Lor 5.92 mL V
X
=
V
=
X
=
=
V
= =
X
SM-23
G.9
(a) Weigh 1.6 g (0.0 10 mol, molar mass of KMn04 158.04 g . mol- J) into a 1.0-L volumetric flask and add water to give a total volume of 1.0 L. Smaller (or larger) volumes could also be prepared by using a proportionally smaller (or larger) mass of KMnO (b) Starting with 0.050 mol· L-J KMn04 ' add four volumes of water to one volume of starting solution because the concentration desired is one fifth of the starting solution. This relation can be derived from the expressIOn � molarity; Vf molarityf where i represents the initial solution and f the final solution. But Vf � + Vd, where Vd represents the volume of solvent that must be added to dilute the initial solution. Rearranging the first equation gives � molaritYf molarity; � molarityf So ifthe ratio of final molarity to initial molarity is 1 : 5, we can write 1 � +� 5 5� � + Vd 4� Vd • For example, to prepare 50 mL of solution, you would add 40 mL of water to 10mLof 0.050mol·L-J KMn04 . (a) V(0.778 mol · L-J) (0.1500 L)(0.0234 mol · L-J) 4. 5 1 10-3 L or 4.5 1 mL (b) The concentration desired is one-fifth of the starting NaOH solution, so the stockroom attendant wil need to add four volumes of water to one volume of the 2.5 mol· L-1 solution. To prepare 60.0 mL of solution, =
4'
x
=
x
=
=
----'---- =
---=-=-
-
V
----'-,- = =
=
G. l l
=
v
=
X
SM-24
divide 60.0 by 5; so 12. 0 mL of 2. 5 mol· L-I NaOH solution are added to 48.0 mL of water. (See the solution to G.9 .) (a) mass of CuS04 (0.20 mol· L-I )(0.250 L)(159.60 g . mol- I) 8. 0 g (b) mass of CuS04 ·5 H 2 0 (0.20 mol· L-I )(0.250 L)(249.68 g . morl ) 12 g (a) Total moles ofK+: ( 74.0.5500gKCI ( 0.500gK. ?S l J ( 2molK+ J + 5g·morl 110.26g m�r ImolK2S J 3molK+ ( 212.0.500gK3P0 4 J ( l 27g·mor ImolK3P04 J = 6.7 1 10-3 mol + 9.07 10-3 mol + 7.07 10-3 mol = 2.29 10--2 mol Molarity of K+ ( 2.29 0. 50-00L2 mol K+ J = 4.58 10-2 M (b) molarity of S2( 110.0.52060gg .Kmol2S- I J ( 1lmolmoISK2-S J ( 0.5100 L J = 9.07 10-3 M 2 (a) mass of K2 S04 (0.125 mol· L-I )(1.00 L)(174.26 g . mol- I ) 21. 8 g (b) mass of NaF (0.0 15 mol·L-I)(0.375 L)(41.99 g·mor l ) 0.24 g C1 2 H22 01 1 (0. 3 5 mol· L-I )(0.500 L)(342.29 g . mol-I ) (c) massof 60. g The total molar concentration of CI- : [( 0.50 g .NaCI I J + ( 0.30 g. KCI J] ( 1 = 0.13 M CI58.44 g mol- 74.5 5 g mol-I 0.100 L ) =
G. 1 3
=
=
=
G. 1 S
+
x
xl
=
x
x
x
x
:
x
=
G. 1 7
=
=
=
=
=
G. 1 9
x
SM-25
G.2 1
We can show that fewer than 1 molecule of X would be left after only 70 doublings. 23 0.10mol 6. 0 2214xl0 10 mL x 1000 mL x 1 molmolecules = 6.0 x 1 020 mo1ecu1es are in the first 10 mL aliquot of the solution. In order to find the number of times the volume must be doubled to get to one molecule, we can solve for the number of times this amount of molecules must be cut in half until it equals 1. 6.0 xl 02°molecules x ("21 )1 = 1 molecule log(6.0 x 1020 ) + nIOg ( �) 10g(1) 20.8 + n(-0.3 01) = 0 20.8 = 0.3 01n n = 69 The other 21 additional doublings involve solutions with no X remaining. There can be no health benefits ifthere are no molecules of the active substance, X, left in the solution. Find the volume of concentrated HCI solution that is equivalent to 10.0 L of the dilute HCI solution with respect to the number of moles of solute present in each volume. mL con. HCI(aq) = 10.0 L dilute HCI(aq) ( 1 0.L7di436lutemoHCIl HC(aq)I J ( 361 .mol46 gHCIHCl ) 3 HCI ( aq) ( 1 cm x ( 10037.g con. 5 0 g HC1 J 1.205 g J 600. mL con. HCI Therefore, 600. mL of concentrated HCI(aq) must be diluted up to a final volume of 10.0 L by adding water in order to form 0. 7436 M HCl(aq). =
G.23
?
=
SM-26
H.t
(a) You cannot add a different compound or element to the chemical equation that is not produced (or involved) in the chemical reaction. In this case, atom is not produced by the defined reaction. (b) 2 Cu + S02 � 2 CuO + S °
H.5
(a) NaBH4(s) + 2 H20(l) � NaB02(aq) 4 H2(g) (b) LiN3(s) + H20(l) � LiOH(aq) + HN3(aq) (c) 2 NaCl(aq) + S03(g) H20(l) � Na2S04(aq) + 2 HCl(aq) (d) 4 Fe2 P(s) + 18 S(s) P4 S,o (s) + 8 FeS(s) (a) Ca(s) + 2 H20(1) H2(g) + Ca(OH)2(aq) (b) Na2 0(s) + H2 0(l) � 2 NaOH(aq) (c) 3 Mg(s) + N2(g) � Mg3N2(S) (d) 4NH3(g) 7 02(g) � 6 H20(g) + 4 N02(g) (I) 3 Fe2 0) (s) + CO(g) 2 Fe)04 (s) + CO2 (g) (II) Fe)04 (s) + 4 CO(g) � 3 Fe(s) + 4 CO2 (g) (I) N2 (g) + 02 (g) � 2 NO(g) (II) 2 NO(g) + 02 (g) � 2 N02 (g) +
+
�
H.7
�
+
H.9
H. l l
�
SM-27
H.19
(I) H2 S(g) + 2 NaOH(s) Na2 S(aq) + 2 H 2 0(l) (II) 4 H2 S(g) + Na2 S(alc) Na2 SS (alc) + 4 H 2 (g) (III) 2 Na2Ss(alc) + 9 °2 (g) + 10 H20(l) 2 Na2S2 03 ·5 H 2 0(s) + 6 S02 (g) (a) We can find the empirical formulas from the percent compositions. First oxide: P 43.64 g 30.97 g/mol 1.409 mol ° 56.36 g 7 16.00 g/mol 3.523 mol 3.523 mol 2. 500 1:2 . 5 or 2:5 P2 0S 1.409 mol Second oxide: P 56.34 g 7 30.97 g/mol 1.8 19 mol ° 43.66 g 16.00 g/mol 2.729 mol 2.729 mol 1. 500 1. 8 19 mol These empirical formulas could be named diphosphorus pentoxide and diphosphorus trioxide. The names according to the Stock system are given below with the formulas of the actual compounds. (b) P40 IO (phosphorus (V) oxide), P406 (phosphorus (III) oxide). Since the molar masses of the empirical formulas are 142 gimol and 110 gimol, respectively, and these masses are both half as big as the molar masses of the compounds, both molecular formulas are twice the empirical formulas. �
�
�
H.2 1
7
=
=
=
7
=
=
=
1.1
1.3
The picture would show a precipitate, CaS04(s), at the bottom of the flask composed of red and pink particles in a 1: 1 ratio . Sodium (black) and chloride (blue) ions, NaCl(aq), would remain throughout the solution. You should use the reagent to only react with one of the ions and form SM-28
precipitate, so that two ions can be separated. (a) Dil2u+ted H2S04 solution wil be used as reagent. Pb (aq) SO/-(aq) PbS04(s) (b) H2S2 solution wi2l also be used as reagent. Mg \aq) S - (aq) MgS(s) (a) CH3 0H, nonelectrolyte; (b) BaCb, strong electrolyte; (c) KF, strong electrolyte (a) soluble; (b) slightly soluble; (c) insoluble; (d) insoluble +
+
1.5
I. 7
1.11
1.13
�
�
The very small amount that does go into solution wil be present as Ag+ and CO/- ions. (c) NH/(aq) and PO/-(aq);(d) Fe2+ (aq) and SO/-(aq) (a) Fe(OH)3 , precipitate; (b) Ag2 C03 ,precipitateforms; (c) No precipitate wil form because all possible products are soluble in water. (a) net ionic equation: Fe2+ (aq) S2-(aq) FeS(s) ; spectator ions: Na+ , CI(b) net ionic equation: Pb2+ (aq) + 2 (aq) PbI2 (s); spectator ions: K+ , N0'l(c) net ionic equation: Ca2+(aq) + SO/-(aq) CaS04(s) ; spectator ions: N03-, K+ (d) net ionic equation: Pb2+ (aq) + CrO/- (aq) PbCr04 (s); spectator ions: Na+ , N0'l�
+
r
�
�
�
SM-29
1.15
spectator ions: K+, N03(a) overall equation: (NH4)2Cr04(aq) + BaCI2 (aq) BaCr04(s) + 2 NH4CI(aq) complete ionic equation: 2 NH/(aq) + CrO/-(aq) + Ba2+(aq) + 2 CI-(aq) BaCr04(s) + 2 NH/ (aq) + 2 Cr(aq) net ionic equation: Ba2+(aq) + CrO/-(aq) BaCr04(s) ; spectator ions: NH4+ , Clcomplete ionic equation: Cu2+ (aq) + SO /-(aq) + 2 Na+ (aq) + S2-(aq) CuS(s) + 2 Na+(aq) + SO/-(aq) net ionic equation: Cu2+ (aq) + S2- (aq) CuS(s) ; spectator ions: Na+, SO/(c) 3 FeCI2 (aq) + 2 (NH4)3P04(aq) Fe3 (P04)2 (s) + 6 NH4Cl(aq) complete ionic equation: 3 Fe2+ (aq) + 6 Cl- (aq) + 6 NH/ (aq) + 2 PO/-(aq) Fe3 (P04)2 (s) + 6 NH/ (aq) + 6 Cl- (aq) net ionic equation: 3 Fe2+ (aq) + 2 pot (aq) Fe3 (PO4)2 (s) ; spectator ions: NH4+ complete ionic equation: �
�
�
�
�
�
�
�
cr ,
SM 30 -
2 K+(aq) C2 0/-(aq) Ca2+(aq) 2 N03-(aq) CaC2 04 (s) 2 K+ (aq) 2 N03- (aq) net ionic equation: Ca2+(aq) C2 0/-(aq) CaC2 04(s) ; spectator ions: K+ , N03complete ionic equation: 2 Ni2+ (aq) 2 SO/-(aq) Ba + (aq) + 2 N03-(aq) Ni + (aq) 2 N03-(aq) + BaS04(s) net ionic equation: Ba2+(aq) SO/-(aq) BaS04 (s) ; spectator ions: Ni2+ , N03+
+
+
�
+
+
�
+
+
+
�
+
�
+
1.17
(a) Ba(CH3 C02 )2 (aq) + Li2 C03 (aq) BaC03 (s) 2 LiCH3 C02 (aq) Ba2+ (aq) + 2 CH3 C02- (aq) + 2 Li+ (aq) CO/-(aq) BaC03 (s) + 2 Li+(aq) 2CH3 C02- (aq) net ionic equation: Ba2+ (aq) CO/-(aq) BaC03 (s) (b) 2 NH4Cl(aq) Hg2 (N03 )2 (aq) 2 NH4N03 (aq) + Hg2 C12 (s) 2 NH/(aq) 2 Cl-(aq) + Hg22 + (aq) + 2 N03-(aq) Hg2 C12 (s) 2 NH/(aq) 2 N03-(aq) �
+
�
+
+
�
+
�
+
�
+
+
+
(c) Cu(N03 )2 (aq) + Ba(OH)2 (aq) CU(OH)2 (S) Ba(N03 )2 (aq) Cu2+(aq) 2 N03-(aq) Ba 2+(aq) 2 OH-(aq) CU(OH)2 (S) Ba2+ (aq) 2 N03-(aq) net ionic equation: Cu2+ (aq) + 2 OH-(aq) CU(OH)2 (s) �
+
+
+
+
�
+
+
�
1.19
(a) AgN03 and Na2 Cr04 (b) CaC12 and Na2 C03
SM-31
1.21
(a) 2 Ag+ (aq) + SO/- (aq) Ag2 S04 (s) (b) Hg2+(aq) + S2-(aq) HgS(s) (c) 3 Ca2+ (aq) + 2 PO/-(aq) Ca3 (P04)2 (s) �
�
�
1 .23
white ppt.2+ = AgCI(s), Ag+; no ppt. with H2S04, no Ca2+; black ppt. ZnS, Zn
complete ionic equation: 2 Na+ (aq) 2 OH-(aq) + Cu2+(aq) + 2 NO;- (aq) CU(OH)2 (S) 2 Na+(aq) + 2 NO;- (aq) net ionic equation: (b) ( 0.0 mL50.0 mL0.100 M ) = 0.0800M (a) Find the number of moles of potassium chromate per liter of solution. mol K2 Cr04 = ( 3.5 0 g K 2Cr04 ) ( 1 mol K 2 Cr04 ( 1000 mL ) L solution 75. 0 mL solution 194.20 g K2 Cr04 J 1 L = 0.240 M K2 Cr04 (aq) (b) Find the number of grams of potassium in the amount of potassium chromate that was dissolved. �
+
4 + M Na =
1.27
=
+
x
?
SM-32
4 ( 2 mol K+ g K+ = 3. 5 0 g K 2 Cr04 ( 194.1 mol20 gKK2 Cr0 2 Cr04 J 1 mol K 2 Cr04 J x ( 39.1 0mol983 K+g K+ J =1.4 1 g K+ (c) MgCr04 ?
J.1
J.3
(a) base; (b) acid; (c) base; (d) acid; (e) base (a) overall equation: HF(aq) + NaOH(aq) NaF(aq) + H 2 0(l) complete iDnic equation: net ionic equation: HF(aq) + OH- (aq) F- (aq) + H 2 0(l) (b) overall equation: (CH3 )3N(aq) + HN03 (aq) (CH3 )3 NHN03 (aq) 3N(aq) + H30+(aq) + N03- (aq) complete ionic equation: (CH3 )(CH 3 )3NH+ (aq) + N03- (aq) + H2 0(l) net ionic equation: (CH3 )3N(aq) + H30+ (aq) (CH3 )3 NH+ (aq) + H 20(l) (c) overall equation: LiOH(aq) + HI(aq) LiI(aq) + H 2 0(l) complete ionic equation: Li+ (aq) + OH- (aq) + H3 0+ (aq) + (aq) Li+ (aq) + (aq) + 2 H2 0(l) net ionic equation: OH-(aq) + H30+ (aq) 2 H20(l) (a) HBr(aq) + KOH(aq) KBr(aq) + H 2 0(l) (b) Zn(OH)2 (aq) + 2 HN02 (aq) Zn(N02 )2 (aq) + 2 H 2 0(l) (c) Ca(OH)2 (aq) + 2 HCN(aq) Ca(CN)2 (aq) + 2 H 2 0(l) (d) 3 KOH(aq) + H3 P04 (aq) K3 P04 (aq) + 3 H 2 0(l) ---+
---+
---+
---+
---+
---+
r
---+
---+
J.S
---+
---+
---+
---+
SM-33
r
J.7
J.9
(a) acid : H)O+ (aq); base: CH)NH2 (aq) ; (b) acid: HCI(aq); base:C2 HS NH2 (aq) ; (c) acid: HI(aq); base: CaO(s) Since X turns litmus red and conducts electricity poorly, it is a weak acid . We can find the empirical formula from the percent composition. C 26.68 g 12. 0 1 g/mol 2.221 mol H 2.239 g 71.0079 g/mol =2.221 mol o 71. 0 81 g 7 16. 0 0 g/mol = 4. 443 mol So the subscripts are 1: 1 :2 on the empirical formula. (a) CH02 (b) Since the molar mass of the empirical formula is 45. 0 g·moC 1 while the molar mass of X is 90.0 g·mol-1 , the molecular formula is twice the empirical formula or C2H204. (c) The weak acid whose formula matches the one given in part (b) is oxalic acid. (COOHh(aq) + 2 NaOH(aq) Na2C204(aq) + 2 H20(l) net ionic equation: (COOHh(aq) + 2 OH- C20/- (aq) + 2 H20(l) (a) C6HSO- (aq) + H20(l) C6HSOH (aq) + OH-(aq) (b) CIO- (aq)+ + H20(l) HCIO(aq) + OH-(aq) + (c) CsHsNH (aq) + H20(I) CsHsN(aq) ++ H30 (aq) (d) NH/ (aq) + H20(I) NH3(aq) + H30 (aq) (a) AS043- (aq) + H20(l) HAsO/- (aq) + OH-(aq) HAsO/- (aq) + H20(l) H2As04- (aq) + OH-(aq) H2As04- (aq) + H20(l) H3As04(aq) + OH- (aq) In each equation, H20 is the acid. 7
=
--+
--+
J. l t
--+
--+
--+
--+
J.13
--+
--+
--+
SM-34
(b)
) x ( 1 mol3 molNaNa+ 3 As04 0.505 mol Na+ (a) 2 NO2 (g) + 03 (g) N2 ° (g) + °2 (g) (b) Ss(s) + 1 6 Na(s) 8 Na2S(s) (c) 2 Cr2+ (aq) + Sn4+ (aq) 2 Cr3+ (aq) + Sn2+ (aq) (d) 2 As(s) + 3 CI2 (g) 2 AsCI3 (l) (a) Mg(s) + Cu2+ (aq) Mg2+ (aq) + Cu(s) (b) Fe2+ (aq) + Ce4+ (aq) Fe3+ (aq) + Ce3+ (aq) (c) H2 (g) + CI2 (g) 2 HCI(g) (d) 4 Fe(s) + 3 °2 (g) 2 Fe2 03 (s) (a) +4; (b) +4; (c) -2; (d) +5 ; (e) + 1 ; (f) 0 (a) +2; (b) +2; (c) +6; (d) +4; (e) + 1 (a) Methanol CH3 0H (aq) is oxidized to formic acid (the carbon atom goes from an oxidation number of +2 to +4). The 02 (g) is reduced to 02present in water. (b) Mo is reduced from +5 to +4, while sulfur (that which ends up as S(s)) is oxidized from -2 to The sulfur present in MoS2 (s) remains in the -2 oxidation state. (c) TI+ is both oxidized and reduced. The product TI(s) is a reduction of TI+ (from + 1 to 0) while the T13+ is produced via an oxidation of Tt . A =
K. l
---+
5
---+
---+
K.3
---+
---+
---+
---+
---+
K.5
K.7
K.9
O.
SM-35
some
reaction in which a single substance is both oxidized and reduced is known as a (a) Cl2 wil be reduced more easily and is therefore a stronger oxidizing agent than (b) N2 0S wil be a stronger oxidizing agent because it will be readily reduced. NS+ wil accept e- more readily than wil N+. (a) oxidizing agent: H+ in HCI(aq); reducing agent: Zn(s) (b) oxidizing agent: S02 (g); reducing agent: H2 S(g) (c) oxidizing agent: B2 03 (s); reducing agent: Mg(s) (a) CI04- CI02, CI goes from +7 to +4; need a reducing agent (b) SO/- S02, S goes from +6 to +4; need a reducing agent CO2{g) + 4 H2(g) CH4(g) + 2 H20(l) Oxidation-reduction reaction; C02 is oxidizing reagent; H2 is reducing reagent (a) oxidizing agent: W03 (s) ; reducing agent: H2 (g) (b) oxidizing agent: HCI reducing agent: Mg(s) (c) oxidizing agent: Sn02 (s); reducing agent: C(s) (d) oxidizing agent: N2 4 (g) ; reducing agent: N2 H 4 (g) (a) 3 N2H4(l) 4 NH3(g) + N2(g); (b) -2 in N2H4; -3 in NH3, 0 in N2; (c) N2H4 is both oxidizing and reducing agent; (d) Factor label (dimensional analysis) can be used to find the volume of nitrogen. disproportionation reaction.
K. l l
cr .
K. 1 3
K. l S
--+ --+
K. 1 7
K. 1 9
--+
°
K.2 1
--+
SM-36
L N2 (g) 1.0 L N2 H4 (1) ( 10001 Lcm3 J ( 1.10cm04 3g )( 32.1 mol0 g J x ( 3 �:���. J ( �8�o�) ( ���J =2. 5 xl 02 LN2 (g) (a) 2 Cr2+ + Cu2+ 2 Cr3+ + Cu(s); (b) 2 e-transferred; (c) Use factor label (dimensional analysis): mol NO:;- ( 50.5 g Cr(N03 )2 )( 1 mol Cr(N03 )2 L solution 250.0 mL solution 176.0 g Cr(N03 )2 J NO:;- J ( 1000 mL ) x ( 1 mol2 molCr(N0 3 )2 1 L 2.27 M NO:;mol SO�- - ( 60.0 g CuS04 )( 1 mol CuS04 L solution 250.0 mL solution 159.62 g CuS04 J SO!- J ( 1000 mL ) x ( 11 molmolCuS0 4 1L 1 .5 0 M SO�4+ o ri d e; Si (a) SiCI4(l) + 2 H2(g) SiCs) + 4 HC1(g); silicon tetrachl (b) Sn02(S) + C(s) Sn(l) + C02; tin(IY) oxide; Sn4+ S+ y (c) Y20S(s) + 5 Ca(l) 2 Yes) 5 CaO(s); vanadium(Y) oxide; (d) B203(S) + 3 Mg(s) 2 B(s) + 3 MgO(s); boron oxide; B3+ ?
K.23
=
�
?
_
=
?
=
K.2S
�
�
�
+
�
(a) moles of Na2 S2 03 needed to dissolve 1.0 mg AgBr 2 mol Na2 S2 03 ( = 1.0 mg AgBr ( 10001 gmgAgBrAgBr J ( 187.1 mol78 gAgBr AgBr J 1 mol AgBr J = 1.1 x 10-5 mol Na 2 S2 03 SM-37
(b) mass of AgBr to produce 0.033 mol Na3 [Ag(S2 03 )J 187. 7 8 g AgBr ] 0.033 mol Na3 [Ag(S?0- 3 )J ( 1 mol 1Namol3 [AAgBr ] ( g(S2 °3 )2 ] 1 mol AgBr 6.2 g AgBr 6 NH4 CI04 (s) + 10 AI(s) 5 Al2 03 (s) + 3 N2 (g) + 6 HCI(g) + 9 H2 0(g) 4 CI04 ] ( 1000 g ] (a) (1.325 kg NH4CI04) ( 117.1 mol49 gNHNH4CI 04 1 kg ( 6 mol10 molNH4AlCIO4 ] ( 26.1 mol98 gAlAl ] 507.1 g Al (b) 3500kg AI ( 10001 kggAlAI ] ( 26.1 mol98 gAlAl ] ( 5 10molmolA1Al2 03 ] ( 101.1 mol96 gAlAl2 02 03 3 J 6.6 13 106 g Al2 03 or 6.6 13 x 103 kg Al2 03 2 CS7HJJ0 06 (S) + 163 °2 (g) 114 CO2 (g) 110 H 2 0(l) 1 mol fat ] ( 110 mol H2 0 ] ( 18. 02 g H 2 0 ] (454 g fat) ( (a) 891.44 g fat 2 mol fat 1 mol H2 0 505 g H2 0 1 mol fat ] ( 163 mol O2 ] ( 32.00 g O2 ] (454 g fat) ( 891. 4 4 g 2 mol fat 1 mol 0 (b) 2 3 1.3 3 10 g 02 =
=
L.3
�
=
x
X
=
L.5
�
+
=
=
x
0. 79 g . mL-J , density of gasoline (3.785 L gas) ( 10001 LmL ] ( 0.719mLg gas ] ( 114.1 mol22 ggasgas ] ( 218molmolCgHH 2 0J g ] ( 18.1 mol02 =--gH-H-2-=.02_.O J 4.246 103 g H2 0 or 4.2 kg H 2 0 SM-38
d
=
_ _
=
X
L.9
L.l l
(a) HCI + NaOH NaCI H2 0 34 mol HCI ] ( 1 mol NaOH ] = 0. 00407 mol 17.40 mL ( 0.21000 mL 1 mol HCI mol concentratI. on of NaOH = 15.0.000407 0 x l0-3 L = 0.271 mol· L-' (b) (0.271 mol · L-')(0.0 1500 L) (40.00 g·mol-' ) = 0.163 g NaOH (a) Ba(OH)2 (aq) + 2 HN03 (aq) Ba(N03)2 (aq) + 2 H 2 0(l) mol HN03 = ( 11.5 6 mL Ba(OH)2 (aq) ) ( 9.670 g Ba(OH)2 ] 250. mL Ba(OH)2 (aq) L HN03 (aq) 2 mol HN03 ] ( 171.1 mol36 gBa(OH) ]( 2 mol--Ba(OH)2 ------Ba(OH) -- 2 1 -(25.0 mL HN03 (aq)) ( 10001 LmL ) = 0.209 mol· L-' (b) mass of HN03 in solution: HN03 ] = 0.329 g ( 01000.209 mLmOl ] (25.0 mL) ( 63.1 mol02 gHN0 3 HX(aq) + NaOH(aq) NaX(aq) + H2 0(l) 750 mol NaOH ] = 0.0516 mol NaOH (68. 8 mL) ( 0.1000 mL NaOH 3.25 g HX corresponds to 0.0516 mol NaOH used 3.25 g = 63.0 g . mol- ' molar mass of acid 0.0516 mol NaI(aq) + CuN03(aq) CuI(s) + NaN03(aq) 1 ( ( 190.15.4755 ggCuI ] . mOrl ] ( lmoI1 molCuN03 CuI 0.0500 L ) = 1 . 65 M SM-39 �
+
�
?
--��
x�
L.13
�
=
L.IS
�
M('uN03
=
--��
(b) First find the concentration of the diluted acid. g Na2 Ca3 J ( 1 mol Na2 Ca3 M HCl(aq) d ' lute ( 0.1000.832L base solution 105.99 g Na2 Ca3 J x ( 0.0.03102525LLbaseacidsolsoluutitionon J x ( 1 mol2 molNaHCI2 Ca3 J = 0.126 M HCI(aq) dilute The original HCI solution is 100 times more concentrated than the solution used for titration (diluted 10.00 mL to 1000 mL), so the original concentration of the HCI solution is 12.6 mol· L-I . 1;- + SnCll" + x)CI- 3 + SnCly The information given can be used to find the molar mass of the reactant in order to identify it. 25. 00 mL ( 0.1201000molmL1;- J 3.00xl0-3 mol 1.;g tin chloride ) = 0. 5 70 g tin chloride 30.00 mL ( 19.0 1000 mL If the reaction is 1: 1 then the moles of 1;- is the same as the number of moles of SnCll" ' In that case, the molar mass of the tin chloride reactant is 0.5703g = 190. g . This molar mass matches that ofSnCI-? , 3.00xl0- mol 189. 6 1 g . mOr l . Tin (II) chloride also has the correct mass percent tin. 118. 7 1 g . mo�- I Sn xl00=62.6% 189. 6 1 g·mol- SnCl2 Since the product compound is oxidized relative to the reactant, we can expect it to be Sn(IV). The net ionic equation for the reaction is ?
L.19
=
1
(y
-
�
C
=
#
mOr l .
SM-40
1;- + Sn 2+ 3 + Sn4+ . Another way to write a balanced reaction would �
L.2l
C
(a) S20/- is both oxidized and reduced. (b) Find the number of grams of thiosulfate ion in 10.1 mL of solution. HSO�(aq) ( 55.0 g HS_O;g S20�- = 10. 1 mL HSO;- (aq) ( 11.4mL5 g HS0 3 (aq) J 100 g HS03 (aq) J [ 81.1 mol0 g HSO � [ 1 mol S2 0�= [ 112. 0 g S2 0r HS03 J 1 mol HS03 J 1 mol S20� J = 11.1 g S20rpresent initial y ?
x
The reactant and product that contain X are in a 1: 1 ratio, so 3. 5 71 g of the reactant is equivalent to 3.180 g of the product. The molar mass of the reactant is x + 4(35.453 g/mol) while that of the product is x + 2(35.453 glmol)+2(l4. 0 1 glmol) + 6(1.0079 glmol). Therefore, we can set up the following proportion in order to solve for x in glmol: x+141. 8 = 3.5 71 x + 1 04.97 3.180 1.1230x - x 23.923 x=194.6 g·mol-', orPt The number of moles of product is 2.27 g -;- 208.23 g . mor' 0.0 109 moles BaCl2 • An equivalent number of moles is represented by 3.25 g ofBaBrx, so its molar mass is 3.25 0.0 109 moles 298 BaBrx• Since 137. 3 3 g is attributable to Ba, 161 g must be Br. Each Br has a mass of 80.4 glmol, so there must be 2 moles ofBr for each mole ofBa in the reactant. =
L.2S
=
g
-;-
=
g · mol-
SM-41
'
L.27
First equation must be balanced as following: Fe + Br2 FeBr2 3 FeBr2 Br2 Fe3BrS Fe3BrS 4 Na2C03 8 NaBr + 4 C02 + Fe304 The masses of Fe are needed to produce 2. 5 t ofNaBr: 1 mol Fe3Brg ) 2. 5 0 t x ( 10001 t kg ) ( 10001 kgg J ( 102.1 mol9 NaBr ( NaBr J 8 mol NaBr x ( 13 molmolFeFeB3Brr2g J ( 1 1molmolFeBrFe 2 J ( 55.1 mol84 gFeFe ) = 5.09 105 g Fe 509 kg Fe (a) is used for calculation of dilution problems: v, = ( 1.00 L x 0. 5 0 M ) = 0.031 L = 31 mL 16.0 M Pipette 31 mL of 16 M HN03 into a 1.00 L volumetric flask which contains about 800 mL of H20. Dilute to the mark with H20. Shake the flask to mix the solution thoroughly. (b) = ( 100 mL0.20 M0.50 M ) = 2.5 102 mL �
�
+
+
�
g
x
=
L.29
MI VI
=
M2 V2
I
V,
x
NaOH
L.3 1
x
Total mass of tin oxide is: 28.35g - 26.45 g = 1 .90 g (a) 1.5 0 Sn ( 118.1 mol7 1 gSnSn J = 1.264 x 10-2 mol Sn moles of 0: (1.90 - 1.50)g x ( 16.1 mol00 00 J 0.025 mol 0 mole ratio of Sn:O is 1:2 empirical formula: Sn02 (b) tin(lV) oxide g
x
g
SM-42
=
theoretical yield: ( 1 mol CO? J ( 44.0 1 g CO?- J 3 (42.73 g CaC03 ) ( 100.1 mol09 gCaC0 CaC03 J 1 mol CaC03 1 mol CO2 = 18.79 g CO2 actual yield: 17.5 g x 100% = 93.1 % yield 18.79 g C H CL + (x + Y2 )02 CO2 + Y H?- 0+ �2 CI2 1. 52 .g =4.2 1x10-3 moI ArochIoryl.elds 2.224. g , 360.88 g mol2 44.0 g mol=5. 0 55x10- mol CO2 2. 5 3 .g ' = 7.01 x 10-3 moIArochIor Yl.elds 18.0.0 12530g g . mol- ' 360.88 g mol2 = 1.405 x 1 0- mol H2 0 2 = 12.0 and y = 1.405x10-2 = 2.00 Therefore, x = 5.4.0255x101xl0-3 7. 0 1x10-3 12.0 11x + 1.0079Y + 35.453z = 360.88 12. 0 11(12. 0) + 1.0079(2.00) + 35.453z = 360.88 35.453z = 214.7 z = 6.06 Since the number or CI atoms per Arochlor 1254 molecule must be a whole number, the number of chlorine atoms is 6. (a) P4 (s) + 3 02 (g) P4 06 (s) P4 06 (s) + 2 02 (g) P4 0, o (s) In the first reaction, 5.77 g P4 uses °2 ( 32.00 g O?- = 4.47 g °2 (g) (5.77 g P4) ( 123.1 mol88 gP4P4 J ( 31 mol mol P4 J 1 mol 02 J SM-43 -
M.3
,
y
� X
.
--=-,
-
-=------,--
M.5
�
�
excess 02 5. 77 g - 4.47 g 02 1.3 0 g 02 In the second reaction, 5.77 g P4 uses ( 123. 85.87g7. gP4mol-I P4 J ( 1 1molmolP4P406 J ( 1 2molmolP4°026 J ( 32.1 mol00 g0°2 2 J 2.98 g 02 limiting reagent: 02 =
=
=
0 1. 3 0 g 0 219. 8 8 g 1 molP40 P4 2 6 6 ( ( ( l J J . (c) 32.00 g mOr 02 2 mol 02 1 mol P4 06 J 4.47 gP406 used In the first reaction, 5.77 g P4 produces =
excess reagent: 10.2 g - 4.47 g 5. 7 gP406 (a) Cu2+(aq) 2 OH-(aq) Cu(OHh(s) NaOH J = 0.0500 mol NaOH (b) 2.00 g NaOH x ( 40.1 mol0 gNaOH (0.0800 L)(0. 5 00 M) 0.0400 mol CU(N03)2 Moles ofNaOH required to react with 0.04 mol Cu(N03h: 2 mol NaOH J = 0.0800 mol NaOH 0.0400 mol CU(N03)2 x ( 1 mol Cu(N03 h 0.0800 mol > 0.0500 mol; therefore, NaOH is limiting reagent. The mass ofCu(OHh(s): CU(OH)2 J 2. 44 g H)2 ) x [ 97.1 mol5 7 gCu(OHh 0.0500 mol NaOH x [1 2molmolCU(O NaOH =
M.7
�
+
=
=
SM-44
M.9
(0.682 (0.0155 (0.174 (0.0193 (0.110 (0.00785 0.061 16.00g0 : 1.
( 44.1 01 J (1 1 J 0.0155 C)(12.01 0.186 (18.1 02 J (12 J 0.0193 0.0195 (1.0079 (28.02 J (12 J 0.00785 0.110 0.376 - (0.186 + 0.0193 0.110 0.061 0.0038 0.0038 4.1 : 5.1 : 2.1 194 2 mol CO 2
g CO 2 ) mol
g CO 2
1
mol N
mol N 2 g N2 g
mol H
gH
=
mol N 2 mol N) (14.01 g · mol - ' N) =
=
=
mol H 2 0
g . mor' H) =
g N2 )
mol C
gC
mol H
mol H 2 0 g H2 0
mol H)
mass of 0 =
=
mol CO 2
g · mor' C) =
g H 2 0)
g°
mol C
mol N
gN
g
g
+
mol °
gives C : H : N : ° ratios
Dividing each amount by
The empirical formula is C 4 H s N 2 0.
=
g . mor ' . Its empirical mass is
The molecular mass of caffeine is
97.10
g°
g) =
g ·mol-' .
molecular formula =
x
empirical formula
=
Cg H N 4 0 2 IO
M.11 Calculate the mass percentage composition; doing so eases the comparison
of data from multiple analyses.
2.20 ( 44.1 01 Jx(1 1 Jx(12.1 01 ) 0.600 %C 1.350.600 x 100% 44.5% g CO2
=
mol CO2 g CO2
x
gC
-=--
---
g unknown
mol C
mol CO2
=
SM-45
C
gC
mol C
=
gC
0.901 x (18.1 02 ) (1 2 )x(1.10079 ) 0.101 %H 1.350.101 x 100% 7.48% 0.130 x ( 28.1 02 )x(12 )x(14.1 01 ) 0.130 0.130gN 0.500 ---=-- x 100% 26.0% 78.0% % 100 -78.0 22% 100 44.5 x (12.1 01 ) 3.71 x (1.381 ) 2.69 x 3 8.07 7.48 x (1.10079 ) 7.42 x (1.381 ) 5.38 x 3 16.1 26.0 x (14.1 01 ) 1.86 x (1.381 ) 1.35 x 3 4.05 22.0 x (16.1 00 ) 1.38 x (1.381 ) 1.00 x 3 3.00 mol H20
g H20
gH
=
=
mol N2
gN
mol N2 =
--
g unknown
=
gH
H
mol N
g N2
---
mol H
mol H20
=
g unknown
g N2
%N
g H20
gH
mol H
X
gN
=
mol N
N
Oxygen must be present in the compound because the percentages of C, H, and N only account for
of its composition. Combustion analysis
does not generate data directly for oxygen; we calculate it by difference: 0=
=
°
To find the empirical formula, assume a sample size of
g and find the
mole ratios. gC
gH gN
g°
mol C
mol H
gH
mol N
gN
mol 0 g
mol
=
gC
°
=
mol
=
mol
=
mol
mol
=
=
=
=
=
=
=
mol
mol
mol
(a) The solid is calcium phosphate, Ca (PO 4 ) 2 . 3 (b)
(206 - (164.1 10 ( 97.1 99 ) 82.01
mol Ca(NO) 2 g Ca(NO) 2
g Ca(NO)? )
g H)P04
mol H)P04
=
) ( 32
g H)P04
Therefore, Ca(NO) 2 is the limiting reagent.
SM-46
=
mol H )P04
mol Ca(NO) 2
)
M.lS
If the 2-naphthol (144.16 g . mor') were pure, it would give the following combustion analysis : 01g·mor') xl00% =83.31%C %C= (144.1O(12. 16 g'mol-'naphthol) ') x100% =5.59% H %H= (144.8(1.160g079. mol-g .'molnaphthol) The observed percentages are low as is expected for a sample contaminated with a substance that contains no C or H. Because the sample does not contain C or H, the percent purity can be easily obtained by _Do_ u_ id_c_al 83.77.31%48%puremixture %purity (based on C)= %_otYoh_ eoret naphthol x l 00% =93.00% Yo_Do_ u_ d_ _ = 5.20% mixture x100% %purity (based on H) %__otheoreti cal 5.59% pure naphthol =93.0% (a) CrHyOz +(x + � -�) 02 CO2 + y H20 2.492 g. CO2 5.664xl0-2 mol CO2= mol C 44.0 g mol-' 0.6495 g H2� =3.608xl0-2 mol H20 7.216xl0-2mol H 18.01 g·mor (5.664xl 0-2mol C)(12.mol011 g) 0.6803 g C (7.216xl 0-2mol H)(1.0mol079 g) 0.07273 g H SM-47 =
=
M.l7
�X
=
=
=
=
1.000 g compound -(0.6803 g 0.07273 g) 0.2470 g 0.2470 g 0-;.-16.00 g·mol-I =1.544xl0-2 mol 0 5.664x10-22 = 3.67 7.216xl0-22 =4.67 1.544x101.544x10The mole ratio ofC:H:O is 3.67:4.67:1 or 11:14:3, so the empirical formula is CllH1403. (b) The molar mass of the empirical formula is 194 glmol, which is half of 388.46 glmol. Therefore, the molecular formula of the compound is C22H2S06. Determine the number of moles of each element present in the compound and then find their ratios to get the subscripts for the empirical formula. 0.055 g CI =1.55x10-3 mol Cl 35.453 g·morl 0.0682 g CO� =1.55x10-3 mol CO2 mol C 44.0 g'mol0.0140 g. H2� =7.78x10-4 mol H20 1.56x10-3mol H 18.01 g mol0.100 g compound _ (0.055 g CI 1.55x10-3 mol.12.mol0 g C 1.55xI 0-3 mol .1.0079molg H ) = 0.0247( gO ] x 16.1 mol 00 g =1.55xl0-3mol 0 The mole ratio is 1:1:1:1, so the empirical formula is CHOCl. As 0.100 g of the compound contains 1.55xl0-3 moles of each element, its molar mass is 1.550.x1lO-003gmol 64.52 g·mol-1• The molar mass of the empirical formula is 64.47 g'mol-1, so the molecular formula is CHOCl. +
M.19
=
=
=
+
+
=
SM-48
M.2l
HA + XOH Moles of HA:
(
Moles of XOH:
�
H20 + XA
2 .45 g HA
23 1 g · mol- I
(
J
=
J
1 .50 g XOH
1 25 g · mol- I
Molar mass o f XA
=
0.0 1 06 mol
=
0.0 1 20 mol
(23 1 - 1 .0) + ( 1 25 - 1 7)
Mole a f XA produced:
(
2 . 9 1 g XA 3 3 8 g . mol- I
J
=
=
3 3 8 g ·moC 1
8.61
x
-3
10
mol
Theoretical yield: (based on the limiting reagent, HA) 0.0 1 06 mol HA
x
. Percentage Yield:
( (
1 mol XA 1 mole HA
8.61
x
)
=
0.0 1 06 mol XA
1 0-3 mol XA
0.0 1 06 mole XA
SM-49
J
x
1 00 % = 8 1 .2 %
CHAPTER
1
ATOMS: THE QUANTUM WORLD
1.1
(a) Radiation may pass through a metal foi l .
(b) All light
(electromagnetic radiation) travels at the same speed; the slower speed supports the particle model . model .
( c) This observation supports the radiation
(d) This observation supports the particle model;
electromagnetic radiation has no mass and no charge.
1.3
microwaves < visible light < ultraviolet light < x-rays < y-rays
1.5
All of these can be determined using
E=hv
and
vA,=c.
For example, in
the first entry frequency is given, so :
A, =
C u
=
2.998x108 m · s· l 8.7x10l4 S - l
=
. m 34xlO·7
=
340 nm .,
Event
Frequency
Wavelength
Energy of photon
(2 s.f.)
(2 s.f.)
(2 s.f.)
8.7 x 1014 Hz
340 nm
5.8 x 1 0 - 19 J
Suntan
5.0 x 1014 Hz
600 nm
3.3 x 1 0 - 19 J
Reading
300 MHz
1 m
2 x 1 0-2 5 J
Microwave popcorn
1.2 x
1 1 0 7 Hz
2.S nm
7.9 x 1 0- 17 J
SM-SO
Dental X-ray
1.7
Wien ' s law states that If TI K = 1540°C
+
TAmax = constant = 2.88 x 10 -3 K· m.
273°C = 1813 K, then AInaX
=
2 .88 x 10-3 K· m
-----
1813 K
Amax = 1.59 x 10-6 m, or 1590 nm
1.9
(a) From e = nA and E = hn, we can write E = heA-I
= (6.626 08 x 10-34 J . s) (2.997 92 x 108) (589 X 10-9 mrl
(
= 3 .3 7 x 10-19 J E=
(b)
5.0 0 x l 0-3 g Na (6.022 x 10 23 atoms · mor l ) 22.99 g . mol-I Na
J
x (3 . 3 7 X 10-19 J . atom-I )
= 44.1 J
2 (c) E = (6.022 X 10 3 atoms · mol-I )(3 . 3 7 x 10-19 J . atom-I ) = 2 .03 x 105 J or 203 kJ
1.11
The energy is first converted from eY to joules: E = (140. 511 x l 03 eY) (1.6022 x 10-19 J · ey- I ) = 2 .2513 x l 0-14 J From E = hv and e = vA we can write A=
he E
(6.626 08 x 10-34 J . s) (2 .997 92 x 108 m · S-I ) = �------------�------------� 2.2513 X 10-1 4 J = 8.8237 x 10-1 2 m or 8.823 7 pm
1.13
(a) false. The total intensity is proportional to Law)
(b) true;
T4 . (Stefan-Boltzmann
(c) false. Photons of radio-frequency radiation are
lower in energy than photons of ultraviolet radiation.
-
SM 51
1.15
(a) Use the de Broglie relationship,A =
hp-I
=
h(mvfl .
= (9. 1 09 3 9 x 1 0-28 g) ( l kg/ 1 000 g) = 9. 1 09 3 9 x 1 0-3 1 kg (3 . 6 x 1 0 3 km · s-I) ( l 000 m · km- I ) =3 .6 x 1 06 m · s - I
me
A = h(mvfl
6.626 08 x 1 0-34 J . S = -----(9. 1 09 3 9 x 1 0-31 kg) (3.6 x 1 0 6 m · S- I ) = 2 . 0 x 1 0-1 0 m
(b) E
=
=
=
hv (6.626 08 x 1 0-34 J. s) (2.50 x 1 01 6 S - I )
1 .66 X 1 0-1 7 J
(c) The photon needs to contain enough energy to eject the electron from the surface as well as to cause it to move at 3 .6 x 1 03 km· S - I . The energy involved is the kinetic energy of the electron, which equals EphOlon
+ mv2
•
1 = 1 .66 x 1 0-17 J + - mv 2 2 = 1 .66 x 1 0- 1 7 J + .!.. (9. 1 09 39 x 1 0-3 1 kg) (3 .6 x 1 06 m · s- I ) 2 2 = 1 .66 X 1 0-17 J + 5.9 X 1 0- 1 8 J =
2.25 X 1 0-17 J
But we are asked for the wavelength of the photon, which we can get from E
=
hv
and
C =
VA or E = hCA- I •
kg·m2 . S-2 = (6.62608 x 1 0-34 kg·m2 . S-I ) x (2 . 99792 x 1 08 m·s-I )A-I
2.25 x l 0-1 7
A = 8 .8 x l O-9 = 8 .8
m
nm
(d) 8 . 6 nm is in the x-ray/gamma ray region.
1.17
To answer this question, we need to convert the quantities to a consistent set of units, in this case, SI units.
SM-52
(5. 1 5 ounce) (28 . 3 g . ounce- I ) ( 1 kg/1 000 g)= 0. 1 46 kg
( )( )( 92 mi
1h
1 km
h
3 600 s
0.62 1 4 mi
)(
1 000 m 1 km
Use the de Broglie relationship.
)
4 1 m . s- 1
=
-
A= hp I= h(mvfl h(mvfl =
6.626 08 x 1 0-34 J . S
= ------
(0. 1 46 kg) (0.041 km · S-I) 6.626 08 x 1 0-34 kg· m 2 S-I
.
= --------=-----,--
= 1.19
(0. 1 46 kg) (4 1 m · S - I) 1 . 1 X 1 0- 34 m
From the de Broglie relationship, p
= hA-l
or
h = mVA,
we can calculate
the velocity of the neutron:
h V=mA = =
1.21
(6.626 08 x 1 0-34 kg · m 2 · s-l) (rememb er th at 1 J= 1 kg . m 2 . s -2 ) 2 2 7 ( 1 .674 93 x 1 0- kg) ( 1 00 x 1 0-1 m)
3 .96 x 1 03 m · s- I
Yes there are degenerate levels. The first three cases of degenerate levels are: n, = l,n2 =
2 is degenerate with
=
3 is degenerate with
nl= 1, n 2= 3 nl= 2, n2
n, =
is degenerate with n l
nl
=
=
2
,
n2
1
=
3, n2= 1 3, n2
=
2
__
1.23
Given the expression for the cubic box is E
xyz
h2 8m
l( � n
� n;
+n
+
L2
where Exyz represents the energy of a given level having n values corresponding to x, y and z we can describe each new level by SM-53
J
incrementing each n by one; any levels with the same energy will be degenerate. The three lowest energy levels will therefore be E1 1 1 , 1:-'2 1 1 , and E22 1 . From this we can arrive at expressions for the energy of each l evel : E1 1 1
=
energy levels are determined to be E2 1 1
3h
2
--
4 mL
2
and E22 1
=
9h
2
8 mL
2
The 2 1 1 and the 22 1 levels are degenerate; that is E2 1 1 =E1 2 1 =El 1 2 and
1.25
(a) Refer to the plot below for parts (a) thru (d); nodes are where the wavefunction is zero : 1.42
-----,-----.-----"-:>- \------,
.----r-� ,...,� �,,-, ,, .
\,,
,.
\{'(x)/m-1I2
0.00
--------------
\\ n=2
\
'\
\�
-------
\.
------
."
\ '.
-1.42
\
--------------
I I I I I I I I I I
\1
\{"
I · '--------'----4 ,, '"-'-"'--__-----' �---"'-'''''''____,---''
0.00
(b) for n
=
(c) for n
=
0.333
0.667
0.500
x (m)
1.00
2 there is one node at x = 0.500 m.
3 there are two nodes, one at x
(d) the number of nodes is equal to
n
-
=
1
Refer to the plot below for parts (e) and (f):
SM-54
0 . 3 3 3 and 0.667 m .
2.00 ,-----....-r,..".... ----." ....------r,,..,---p-<;:;----,
\ .0 = 2
\
(e) for m.
n2 =
(f) for n = 3
\''-, .1 . .. / __l..L.----' ... _---'__ ---"" 0.00 "'-_ _-'--_'----"-"-__----"-'"'-0.00 0.17 0.25 0.50 0.75 0.83 1.00 x
a particle is most likely to be found at a particle is most likely to be found at
m.
1.27
(m)
(
x x
=
=
0.25 m and
2L [( 2nn nnx . nnx X)f] n [� ( L�2 ) _ + � t � t( n�x ) ' dx 2L [( 2nn nnx . nnx x ) ] = Lonon nn . nn sin
=
-1 . cos ·S n T I T + "2
given
is an integer:
0
�
(b) Integrate over the "left third of the box" or from 0 to 1/3 L: '1"
�
=
-
-
1
sin
1 . t cos n + ToSI T "2 0
I
COS - oSI n - + 3 3 3
SM-55
=
0.75
0. 1 7, 0.50 and 0.83
(a) Integrate over the "left half of the box" or from 0 to 12 L:
f �2 = � f n�x r dx
x
if n is a multiple of 3 , the first term in this sum is zero and the
probability of finding an electron in the left third of the box is 1 1 3 . Also, as n becomes large the probability of finding the
electron in the left third of the box approaches 1 /3 .
1.29
(a) The Rydberg equation gives
v
one can calculate A from the relationship V =� and
( � �) n1
-
n2
c = VA = 2.99792 x 1 08
c=�
( � �) n,
-
n2
when � = 3 .29 X 1 015
c = VA.
-I S ,
from which
m · s-'
A
2.99792 x 1 08 m · s-' = (3 .29 x 1 0'5
S
-'
)
(� ) l
4
__
16
A
A = 4.86 x 1 0-7 m = 486 nm (b) Balmer series ( c) visible, blue
1.31
For hydrogen-like one-electron ions, we use the Z-dependent Rydberg relation with the relationship
c = AV
to determine the transition
wavelength. For He+ , Z = 2. V = Z 2� 1 /l,
1.33
�2 ) = 9.87 (\.�-�1) = (22 X3 .29 x 1 0' 5 s- , {� �1 n,
n2
X
1 0'5 Hz
2.99 7 92 x 1 08 m . s-' -- = 3 . 04 x l 0-8 m = 3 0 . 4 nm = � = -----V 9.87 x 1 0' 5 s-'
(a) This problem is the same as that solved in Example 1 .5, but the electron is moving between different energy levels. For movement between energy levels separated by a difference of 1 in principal quantum number, the expression is
SM-56
For n
= 2 and n + 1 1
Then /"3 2 .
=
he E
=
=
Sh 2 3 , M = -8mL 2
8mheL 2 Sh 2
=
8meL 2 Sh
For an electron in a ISO-pm box, the expression becomes
�
.2
=
=
8(9. 1 09 3 9 x l 0- 3 1 kg) (2.997 92 x 1 08 m · S-I ) ( lS0 x l 0- 1 2 m) 2 S(6.626 08 x 1 0-34 J . s) 1 .48 x 1 0-8 m
(b) W e need to remember that the equation for �E was originally determined for energy separations between successive energy levels, so the expression needs to be altered to make it general for energy levels two units apart: M
A
=
=
he
M
For n A
=
En + =
= 2,
2
_
En
=
(n + 2) 2 h 2 8mL 2
he(8mL 2 ) h 2 [4n + 4]
=
_
n 2 h2 8mL 2
8meL 2 h 2 [4n + 4]
=
(n 2 + 4n + 4 - n 2 )h 2 8mL 2
=
(4n + 4)h 2
the expression becomes
8meL2 h[(4 x 2) + 4]
=
8meL2
----
1 2h
8(9. 1 0939 x l o - 31 kg) (2.99792 x I 0 8 m's -l) (lS0 x 1 0 -12 m)2
= �------------��--------------�--------�12(6.62608 x 1 0- 34 kg · m2 'S-I ) =
1.35
6 . 1 8 x 1 0 -9 m
In each of these series, the principal quantum number for the lower energy level involved is the same for each absorption line. Thus, for the Lyman series, the lower energy level is n Paschen series, n
=
=
1; for the Balmer series, n
3; and for the Brackett series, n
SM-S7
=
4.
=
2; for
1.37
2p
(a) Is
3d
(b) A node is a region in space where the wavefunction l.f/ passes through O.
(c) The simplest
s
-orbital has 0 nodes, the simplest p -orbital has 1
nodal plane, and the simplest d-orbital has 2 nodal planes.
(d) Given the
increase in number of nodes, an f-orbital would be expected to have 3 nodal planes.
1.39
The Px orbital will have its lobes oriented along the x axis, the Py orbital will have its lobes oriented along the y axis, and the Pz orbital will have its lobes oriented along the z axis .
1.41
The equation derived in Illustration 1 .4 can be used:
1.43
To show that three p orbitals taken together are spherically symmetric, sum the three probability distributions (the wavefunctions squared) and show that the magnitude of the sum is not a function of e or ¢ . Px
=
R(r)C sin e cos¢
pz
=
R(r)C cos e
Py
=
R(r)C sin e sin¢
where C =
I
( �J 4
S quaring the three wavefunctions and summing them:
SM-58
R(r) 2 C 2 sin 2
e cos 2 ¢ + R(r) 2 C2 sin 2 e sin 2 ¢ + R(r) 2 C2 cos 2 e = R(r) 2 C2 ( sin 2 e cos 2 ¢ + sin 2 e sin 2 ¢ + cos 2 e) = R(r) 2 c 2 ( sin 2 e( cos 2 ¢ sin 2 ¢) + cos 2 e) +
U sing the identity cos 2 R(r) 2 C2 sin 2
(
x+
sin 2
x
= 1 this becomes
e + cos2 e) = R (r) 2 c 2
With one electron in each p orbital, the electron distribution is not a fuction of e or
1.45
¢ and is, therefore, spherically symmetric.
(a) The probability (P) of finding an electron within a sphere of radius ao may be determined by integrating the appropriate wavefunction squared from 0 to ao :
ro r 2 exp
4 P =� ao
( }
2r -- r ao
This integral is easier to evaluate if we allow the following change of variables: 2r Z =-
. . . z = 2 when r = ao' z = 0 when r = 0, and dr =
r
2 1 1 2 P = - z 2 exp(-z)dz = --(z 2 + 2z + 2) exp(-z) 2 2 0 1 ( 4 + 4 + 2) exp( -2) - 2 = 2 = 0.323 or 3 2 . 3 %
)
--[(
]
1
(� )
dz
(b) Following the answer developed in (a) changing the integration limits to 0 to 2 ao: z = 4 when r = 2ao' z = 0 when r = 0, and dr =
( ;) 14 a
1 1 P = - 1j1 Z 2 exp(-z)dz = --(z 2 + 2z + 2) exp(-z) 2 2 0 =-
� [(
)
26 eXp(-4) - 4
= 0.76 1 or 76. 1 %
]
SM -59
dz
orbital ;
1.47
(a)
1.49
(a) 7 values: 0,
1
values:
(b)
orbitals;
5
(c) (b)
1 , 2, 3, 4, 5, 6;
orbital s ;
3
5 values; -2, -1, 0, 1, 2;
-1, 0, 1; (d) 4 subshells: 4s, 4p, 4d, (b)
1.51
(a)
n=6;1=1;
1.53
(a)
- 1, 0, +1;
(d)
- 3,-2, -1, 0, +1,+2, + 3.
(b)
n=3;1=2;
(d) 7 orbitals
(c)
-2, -1, 0, +1, +2;
3
and 4/
n=2;1=1; (c)
(c)
(d)
n=5;1=3
- 1, 0, +1;
1.55
(a) 6 electrons;
1.57
(a) 5d, five;
1.59
(a) six;
1.61
(a) cannot exist;
1.63
(a) The total Coulomb potential energy VCr) is the sum of the individual
(b)
10 electrons;
(b) Is, one;
(b) two;
(c)
(c) eight; (b) exists;
6 /,
(c)
2 electrons;
seven;
(d)
2 p,
(d) 1 4 electrons three
(d) two (c) cannot exist;
(d) exists
coulombic attractions and repulsions. There will be one attraction between the nucleus and each electron plus a repulsive term to represent the interaction between each pair of electrons. For lithium, there are three protons in the nucleus and three electrons. Each attractive Coulomb potential will be equal to (-e)(+3e)
2 = -3e
where
is the charge on the electron and
-e
nucleus,
Co
+3e
is the charge on the
is the vacuum permittivity, and r is the distance from the
electron to the nucl eus. The total attractive potential will thus be
SM-60
The repulsive terms will have the form ( -e)( -e)
e
=
2
where rab represents the distance between two electrons a and b. The total repulsive term will thus be 2 e
47rcO'i2
+
e
2
47rcO'i3
2 e
+ 47rc r 3
This gives
O2
=
e
2
471B
o
(
1
1
1
� + �+ r 3 2
J
(b) The first term represents the coulombic attractions between the nucleus and each el ectron, and the second term represents the coulombic repulsions between each pair of electrons.
1.65
(a) false. Zeff is considerably affected by the total number of electrons present in the atom because the electrons in the lower energy orbitals will "shield" the electrons in the higher energy orbital s from the nucl eus. This effect arises because the e-e repulsions tend to offset the attraction of the electron to the nucl eus.
(b) true;
(c) false. The electrons are
increasingly less able to penetrate to the nucl eus as I increases. 1.67
(d) true.
Only (d) is the configuration expected for a ground-state atom; the others all represent excited-state configurations.
1.69
(a) This configuration is possible. because 1=0 here, s o
m)
(b) This configuration is not possibl e
must also equal O.
(c) This configuration is
not possible because the maximum value I can have is n
- 1;
n
= 4, so lmax= 3. SM-6 1
1.71
(a) silver
[Kr]4d105s1
(b) beryllium
[He]2s2
(c) antimony
[Kr]4dlo 5s2 5 p3
(d) gallium
[Ar]3d10 4s2 4i
(e) tungsten
[Xe ]4[14 5d4 6s2
(f)
iodine
[Kr]4dlo 5s2 5 p5
1.73
(a) tellurium;
1.75
(a) 4p;
1.77
(a) 5;
(b) 1 1 ;
1.79
(a) 3;
(b) 2;
(b) vanadium;
(b) 4s;
(c) 3;
(d) thorium
(d) 6s
(c) 6s;
(c) 5;
(c) carbon;
(d) 20 (d) 2
1.81 Unpaired Element
Electron Configuration
electrons
Ga
[Ar] 3dlo 4i 4i
1
Ge
[Ar] 3dlO 4i4p2
2
As
[Ar] 3dlo 4s2 4p3
3
Se
[Ar] 3dlO 4i4p4
2
Br
[Ar] 3dlO 4i4p5
1
(b) ns2ni;
(c) (n-1)d5ns2 ;
1.83
(a) nsl;
1.85
(a) oxygen (1310 kJ mor l ) .
>
I
(d) (n-1)dlo nsl
selenium (941 kJ mol- I ) .
>
tellurium
(870 kJ mor l ) ; ionization energies generally decrease as one goes down .
SM-62
a group .
(b) gold (890 kJ . mor l )
>
osmium (840 kJ . mor l )
>
tantalum (76 1 kJ· mor l ) ; ionization energies generally decrease as one goes from right to left in the periodic table. barium (502 kJ . mol- I )
>
(c) lead (7 1 6 kJ . mor l )
cesium (3 76 kJ . mol- I ) ; ionization energies
>
generally decrease as one goes from right to left in the periodic table.
1.87
The atomic radii (in pm) are Sc
161
Fe
1 24
Ti
1 45
Co
1 25
V
1 32
Ni
1 25
Cr
1 25
Cu
1 28
Mn
1 37
Zn
1 33
The major trend is for decreasing radius as the nuclear charge increases, with the exception that Cu and Zn begin to show the effects of electron electron repulsions and become larger as the d-subshell becomes filled. Mn is also an exception as found for other properties; this may be attributed to having the d-shell half-filled. 1.89
(a) Sb3 +, Sb5+; (c) Tt, TI 3 +; (b) and (d) only form one positive ion each.
1.93
(a) fluorine;
1.95
(a) A diagonal relationship is a similarity in chemical properties between
(b) carbon;
(c) chlorine;
(d) lithium.
an element in the periodic table and one lying one period lower and one group to the right. (b) It is caused by the similarity in size of the ions. The lower-right element in the pair would generally be larger because it lies in a higher period, but it also will have a higher oxidation state, which will
SM-63
cause the ion to be smaller. (c) For example, A1 3 + and Ge4+ compounds show the diagonal rel ationship, as do Li+ and Mg 2 + .
1.97
Only (b) Li and Mg exhibit a diagonal relationship.
1.99
The ionization energies of the s -block metals are considerably lower, thus making it easier for them to lose electrons in chemical reactions.
1.101 (a) metal;
(f)
1.103
(b) nonmetal ;
(c) metal;
(d) metalloid;
(e) metalloid;
metal
(a)
v
- = 3 600 cm- I c
v =
c(3600 cm - I )
V =
(2.997 92 x 1 08 m · S - I ) (3600 cm- I )
(2.997 92 x 1 010 cm · S -I ) (3600 cm- I ) = 1 . 1 X 1 0 1 4 S- I
v = v
(b) From =
E = hv: E = (6.626 08 x 1 0-34
7.2 X 1 0-20 J.
(c) 1 .00 mol of molecules
=
J . s)( 1 .079 X 1 0 1 4 S - I )
6.022 x 1 02 3 molecules, so the energy
absorbed by 1 .00 mol will be (7 . 15 1 x 1 0-20 J. molecule-I )(6.022 x 1 02 3 molecules · morl ) =
4.3 x 1 04 J . mor l or 43 kJ . mOr l .
1.105 Cu = [Ar] 3 io 4s1 and Cr = [Ar]3d5 4s1 ; In copper it i s energetically
favorable for an electron to be promoted from the 4s orbital to a 3d orbital, giving a completely filled 3d subshell . In the case of Cr, it is energetically favorable for an electron to be promoted from the 4s orbital to a 3d orbital to exactly 12 fill the 3d subshell.
SM-64
1.107 This trend is attributed to the inert-pair effect, which states that the s
electrons are less available for bonding i n the heavier elements. Thus, there is an increasing trend as we descend the periodic table for the preferred oxidation number to be 2 units lower than the maximum one. As one descends the periodic table, ionization energies tend to decrease. For Tl, however, the values are slightly higher than those of its lighter analogues. 1.109 (a) The relation is derived as follows : the energy of the photon entering, Etatai'
must be equal to the energy to eject the electron, Eejectian' plus the
energy that ends up as kinetic energy, Ekinetic' in the movement of the electron: Etata'
= Eejectian
+ Ekinetic
= h v and
But Etata' for the photon
Eki elic n
=
(�)
m v2 where m is the mass
of the object and v is its velocity. Eejectian corresponds to the ionization energy, (b) EIOl l a
I , so we arrive at the final relationship desired.
= hv= hC).-1 = (6.62608
X
1 0-34 J · s) (2 .99792
= 3 .40 1 x 1 0- 18 J = EejeCliOn + Ekil1elic Ekinelic
=
(� ) = ( �) m v2
=(�)
(9. 1 093 9
x
(9. 1 093 9
x
X
1 08 m · s-') (58.4
1 0-28 g) (2450 km · s-I )2
1 0-3 1 kg) (2.450
X
1 06 m · s-I)2
= 2.734 x 1 0- 18 kg · m2 . S -2 = 2.734 x 1 0- 18 J
3 .40 1
X
1 0- 18 J = Eejection + 2. 734 x 1 0- 18 J
ec n=
E je ti o
x
6.67 x 1 0 -1 9 J
SM-65
1 0-9 mrl
1.111 B y the time we get to the lanthanides and actinides- the two series off
orbital filling elements-the energy levels become very close together and minor changes in environment Cause the different types of orbitals to switch in energy-l evel ordering. For the elements mentioned, the electronic configurations are
6s25dl 2 [ Rn] 7S 6dl
La Ac
6s24114 5d 1 2 [Rn] 7 S 5/4 6dl
Lu
[Xe]
[Xe]
Lr
As can be seen, all these elements have one electron in a d-orbital, so placement in the third column of the periodic table could be considered appropriate for either, depending on what aspects of the chemistry of these elements we are comparing. The choice is not without argument, and it i s
J. Chern. Ed.
discussed b y W . B . Jensen (1982),
59,634.
1.113 (a)-(c) We can use the hydrogen 2s wavefunction found in Table 1.2.
Remember that the probability of locating an electron at a small region in
lfI2,not lfI. 1 2 r -2- - e 2a" ao 2 r -2 - - e a, a"
space is proportional to
( ) ( ) lfI2s -4 -2nao3 ( ( J ) 2 lfI2s -_ 32nao3 ( ) ( )' . ao lfI2/ (r,e,¢) 32nao3 ( ) lfI2/ (O,e,¢) 32nao3 (2-J - . =
1
1
-
r
r
1
1
2-
1
Because
e a,
0
=
=
, --
�
22 e -;-'
e
, G,
4
r will be equal to some fraction
simplify further:
SM-66
x
times
ao'
the expression will
[ J o xa_ 2__ a°
2
:
e --
.
III s 2 ( e , If' A.) 't' 2 '-:-'-'---. .::. _ --'--'- = --0..-_....:._ ...:. _ =
r,
2 !fI 2 s (0, e, r/J )
(2 _ X )
4
4
2
e-x
Carrying out this calculation for the other points, we obtain: x
relative probability
0. 1
0.82
0.2
0.66
0.3
0.54
0.4
0.43
0.5
0.34
0.6
0.27
0.7
0.2 1
0.8
0. 1 6
0.9
0. 1 2 0.092
1.1
0 .067
1 .2
0 . 048
1 .3
0.033
1 .4
0 . 022
1 .5
0.01 4
l .6
0 . 008 1
l .7
0 . 004 1
l .8
0 . 00 1 7
1 .9
0.0004
2
0.0000
2. 1
0.0003 1
2.2
0.00 1 1
2.3
0. 0023
2 .4
0 . 0036
2.5
0 . 005 1
2.6
0 . 0067
SM-67
2.7
0. 0082
2.8
0.0097
2 .9
0.0 1 1
3
0.0 1 2
This can be most easily carried out graphically by simply plotting the function f (x)
=
(2 - X) 4
2 -x e
from 0 to 3 .
::: �J.�· �i,· · -r-tt � tt� ITt �
!
.
-Lo
i!
.
_1+_'\ ! I i I. � I I
I
- -
0.30
I i
-t+J -
0.10
I
\J,
-
I
.
I I I
.!
I!
0 20 .
I!
I
li B -
-
! I
i
iH-+h-I t�_I --t\_, -I f-r r-+t-1+-::���I ::-fr- r; t- !+- +;i+-+f--+ --
0. 0 4
0.00
!
i\
!
0.50
�[
!
i�
0.60
.�
I I
I
I!
I
.
'
-f-f-
1 --t-T-r--f-fI
i ! ! I --+--h-f 1 1
-I i- +� -
��I- +- +rH-' Ii .
I
c-
-
--
I I � 1 �!�����I�I��� · i� i�����* i* I ��
0.0
0.5
1.0
I
1.5 x
The node occurs when x = 2, or when
2.0
r =
2.5
3.0
2a . This is exactly what is o
obtained by setting the radial part of the equation equal to O.
1.115 The approach to showing that this is true involves integrating the
probability function over all space. The probability function is given by the square of the wave function, so that for the particle in the box we have If/ =
( ) ( L) 1
2 2
L
.
sm
nnx
and the probability function will be given by 2
If/ =
() ( ) 2
-
L
. 2 sm
nnx
--
L
SM-68
Because
x
can range from 0 to
L (the length of the box), we can write the
integration as
1'? X lj/-dx I =
lJ
() ( ) 2
-
L
nn x
.
sm 2
--
L
dx
for the entire box, we write
probability of finding the particle somewhere in the box =
( �) ( � ) ( �) r ( � ) t 'P' �� t ( � J [( ) L] ) ] [(2 � [� ]
t
sin 2
n x
dX
sin 2
probability =
Sin
2
n
x
n
x
dx
dx
=
L
-1 . nn x . nn x x + · sm cos T T "2 2nn
=
�
. cos nn . sin nn +
L
2nn
L 2
0
-0
if n is an integer, sin nn will always be zero and
probability
=
=
1
0 1.117 (a) 4.8 x 1 0-1 esu; (b) 1 4 electrons l 1.119 Given that the energy needed to break a C - C bond is 248 kJ ·mor , 348 kJ mol C-C bonds
X
1 mol C-C bonds 1 000 J = 5.7 8 x 1 0.1 9 J x 6.022 x 1 0 23 C-C bonds 1 kJ
Substituting this value into A
=
he/ M
,
this energy corresponds to a
wavelength of 3 .44 x 1 0-7 m or 344 nm. This is in the ultraviolet region of the electromagnetic spectrum.
SM-69
1.121 Given that E
M
E
=
n
h9i
=
- E
final
=
2. 1 8
1 0- 1 8 J and 2 n
x
initial
(a) for an electron to fall from the 4d to I s level, the energy of the photon .
IS
M
=
EI -E4 = -2. 1 8
x
10
-18
(
l"i21
J
-
1 1 2 4
)
=
-
2 . 04
x
1 0-1 8 J (the
negative sign means that energy is released or emitted); (b) Similarly, an electron moving from the 4d to 2p level will emit a photon of 4.09
x
1 0-19 J ;
( c) same a s (b); an electron moving from the 4 d t o 2 s emits the same amount of energy as it would if it were moving to the 2p orbital; this is due to the fact that all orbitals having the same principal quantum number n are degenerate (i.e. they have the same energy). (d) In a hydrogen atom, no photon would be emitted on moving between orbitals possessing the same n (due to degeneracy of the 4d and the 4s
orbitals in hydrogen).
(e) Since potassium has both more electrons and protons than hydrogen, the individual orbitals within a given shell will have different energies (arising from the attractions and repulsions of electrons with the nucleus and other electrons in the atom). As a result we would expect to see emission lines for all the transitions; thus potassium should show four lines while hydrogen only shows two.
1.123 A
=
1 064 nm = 1 .064
is E
=
-A € he
=
x
1 0-6 m; The energy of a photon of this wavelength
.626 x 1 0- 34 J · sec 1 . 064
x
lO
)
-6
.998
x
x
=
m
the energy of the ejected electron is 0. 1 3 7 eY 2. 1 95
1 08 m
x
( 1 .602
x
1 . 867 x 1 0
-19
1 0-19 J·ey-I)
=
J;
1 0-20 J . The difference between these two values is 1 .65 x 1 0-19 J
SM-70
22
or 1 .65 x 1 0-
kJ per atom of thulium. Multiplication by Avogadro's
number gives an electron affinity of 99.2 kJ per mol of thulium. 1.125 The radial distribution of a 3s orbital is given by curve (b), while curve (a)
shows the same for a 3p orbital ; this can be determined by examining the electron density near the origin (which is the nucleus); the plot with the most electron density closest to (0, 0) arises from the s orbital. 1.127 (a) The electron configuration of atomic chlorine is [Ne]3s 2 3p5 ; it has
one unpaired electron. The electron configuration of a chloride ion is [Ne]3i 3p6 ; this configuration is identical to neutral argon. (b) Assuming a one quantum level jump, an excited chlorine should have an electron configuration of [Ne]3i 3 p4 4s1 (c) The energy of a given level by E
n
?
Z;ffh91 = 2
=
Z;ff
n
n
•
in a non-hydrogen atom can be estimated
)
t; . 1 8 x 1 0- 1 8 J � n
2
. For chlorine, Zeff is
approximately equal to 6 (Fig. 1 .45). For an electron to jump from the 3 to
-
�
n =
fr- 2
)
4 quantum level the energy needed is: M = E - E = 4 3 .18 x 10
-1 8
(1 J 2 4
l
1 1 -32") = 3 . 82 x 1 0 -1 8 J .
n =
This energy
corresponds to a wavelength of 52.0 nm (the X-ray region). (d) This amount of energy corresponds to 2 . 3 0x 1 0 3 kJ·mor 1 or 2 3 . 8 eV per chlorine atom ( e) If the proportion of 3 7 CI in a sample of chlorine atoms is reduced to 3 7.89 % (half of its typical value), the proportion of 35 CI will be increased to 62 .1 1 %. Based on this, the average mass of a chlorine atom will be: Clave mass
=
=
(
3 7. 8 9% 1 00
)
6. 1 3 9
X
1 0-2 3 g
5 .933 x 1 0-2:1g/atom
=
+
(
62 . 1 1 % 1 00
3 5 .73 g/mol
SM-7 1
)
5 . 807
X
1 0-23 g
(f) thru (h): refer to the table below: Compound
Chlorine
Name
oxidation number
ClO2
+4
Chlorine dioxide
NaClO
+1
Sodium hypochlorite
KClO3
+5
Potassium chlorate
NaClO4
+7
Sodium perchlorate
SM-72
CHAPTER 2 CHEMICAL BONDS
2.1
The coulombic attraction is directly proportional to the charge on each ion (Equation 1 ) so the ions with the higher charges will give the greater 3 2 coulombic attraction. The answer is therefore (b) Ga + , 0 -.
2.3
The Li+ ion is smaller than the Rb + ion (58 vs 1 49 pm). Because the l attice energy is related to the coulombic attraction between the ions, it will be inversely proportional to the distance between the ions (see Equation 2). Hence the larger rubidium ion will have the lower lattice energy for a given anion.
2.5
(a) 5 ;
2.7
(a) [Ar] ;
2.11
2.17
(b) 4;
(c) 7;
(d) 3
(b) [Ar] 3io 4/ ;
(c) [Kr] 4d 5 ;
o
(a) [Kr ] 4i 5/ ; same; In+ and Sn2+ lose (b) none (c) [Kr] 4io ; Pd
(a) 4s;
(b)
3p;
(c)
3p;
(d) 4s
SM 73 -
(d) [Ar] 3io 4/
5p valence electrons.
2.19
(a) - 1 ;
(b) -2 ;
(d) +3 (+ 1 sometimes observed);
(c) + 1 ;
(e) +2
2.21
2.23
(a) 3 ;
(c) 6;
(b) 6;
(d) 2
(a) [Kr ] 4d ' °S/; no unpaired electrons; (b) [Kr] 4 dl O ; no unpaired 4 electrons; (c) [Xe]4/4 S d ; four unpaired electrons; (d) [Kr] ; no unpaired electrons;
2.25
(a) 3p;
(b) Ss;
2.27
(a) +7;
(b) - 1 ;
(e) [Ar]3d8 ; two unpaired electrons (c) Sp;
(d) 4d
(c) [Ne] for +7, [Ar] for - I ;
(d) electrons are lost
or added to give noble-gas configuration.
2.33
(a)
:CI: I : CI C - CI:
(b)
I
'00 • 0 II : CI - C-CI:
: CI:
(c)
.
.
O = N -F :
(d)
SM-74
o
f 0
I F : -N o
0
-
F:
2.35
H (a)
H
-
I
B- H
I
H
(c)
2.37
H
: 0:
(a)
II
(b)
H - C -H
I
H - C - O-H
I
H H (c)
I
II
H - N - C - C - O -H
I
H
2.39
: 0:
I
H
The structure has a total o f 3 2 electrons; of these, 2 1 are accounted for by the chlorines (3 CI ' s
x
7 valence electrons each); of the 1 1 electrons
remaining, 6 come from the oxygen. This leaves 5 electrons unaccounted for; these must come from E. Therefore, E must be a member of the nitrogen family and since it is a third period element E must be phosphorous (P)
SM-75
2.41
(a)
[H-Z-H] H
+
I
:C..1:
(b)
�: [RJ3-
(c) Na+
K+
[�l-RJ -
2.43 H
H
H
H
H
H
I
H
/c
......
C
�C ......
C
�C ......
I
I
H
C
I
H
H
�C"
I
I
H
I
I
I
2.45
.. : Cl :
I
b· === N-O :
2.47 o
(a) : N _I
(d) : C
+1] + 0:
_I
C:
]-2
0 (b) : N -I
(e) : C
SM-76
-I
0 N: 0
( c) : C
]
N:
-
+1
O:
2.49
(a)
O=CI0: .. o .
n
II (I
:0: n
(b)
()
..
0= C
(I
r
H
(l
:O-CJ-o:
Illwa t:'tIL;t�)
=
..
I)
(" )
0
·1
..
I:Q-C:::::S . •
S (I
H--C=N:
(c)
., -I
..
1
.,.2 I
:0: ..
H
(I
... 1
J-\-C=N:
(I
(J
·1
+
I
2.51
(a)
H
1 -
.. 0 I 0 :O - S - O :
1
-
H
I
-1 0 . . +1 :O - S - O.:
I
I
: 0: 0
: 0:
lower energy
(b)
-
1
0 : 0:
II +1
I
:O - S - O: 1 -
I
1-
H 0
"
'. 0'
�2
.
H
I
I 0 : O- S - O: 1 -
: 0:
-
lower energy
2.53
-I
-1
1
I
1-
1
-
:0:
(a) In the first structure, the formal charges at Xe and F are 0, whereas, in the second structure, Xe is
-1, one F is 0, and the other F is + 1. The first
structure is favored on the basis of formal charges.
(b) In the first
structure, all of the atoms have formal charges of 0, whereas, in the second SM-77
structure, one 0 atom has a formal charge of + 1 and the other 0 has a
formal charge of -1. The first structure is thus preferred. 2.55
(a) The sulfite ion has one Lewis structure that obeys the octet rule: -2
:o-s-o:
I
:0 : and three with an expanded octet: -2
-2
-2
:o-s-o:
o=s-o:
.
:o-s=o
.
I
II
I
:0:
:0 :
:0 :
The structures with expanded octets have lower formal charges. (b) There is one Lewis stucture that obeys the octet rule: :o-s-o:
I
H - O: The formal charge at sulfur can be reduced to 0 by including one double bond contribution. This change gives rise to two expanded octet structures. :o-s=o
••
o=s-o:
I
I
H- O :
H- O:
Notice that, unlike the sulfite ion, which has three resonance forms, the presence of the hydrogen ion restricts the electrons to the oxygen atom to which it is attached. Because H is electropositive, its placement near an oxygen atom makes it less likely for that oxygen atom to donate a lone pair to an adjacent atom.
SM-78
(c) The perchlorate ion has one Lewis structure that obeys the octet rule: :0: . .
I
:O - Cl-O:
..
I
: 0:
..
The fonnal charge at Cl can be reduced to 0 by including three double bond contributions, thereby giving rise to four resonance fonns. :0:
: 0:
O=Cl=O
: O=Cl=O
II
II
..
II
I
: 0:
:0:
:0 :
:0:
..
. .
II
I
O=Cl-O:
O=Cl=O
: 0:
: 0:
..
. .
II
II
..
For the nitrite ion, there are two resonance fonns, both of which obey the octet rule:
]
:O-N=O
O=N-O: .
2.57
.
.
.
.
.
.
.
]-
Radicals are species with an unpaired electron, therefore only (b) and (c) are radicals since they have an odd number of electrons whil e (a) and (d) have an even number of electrons allowing Lewis structures to be drawn with all electrons paired.
SM-79
2.59
The Lewis Structures are
(a)
(b)
: C I-Q:
: C..I-O-O ..- C I:
. O· .
(c)
I
.
O===N-O-Cl:
Radicals are species with an unpaired electron, therefore (a) and (c) are radicals. 2.61
(a)
(b)
[: ¢i-X-�l=r : C I: : C...... I .. I . "/I
I has 2 bonding pairs and 2 lone pairs I has 4 bonding pairs and 2 lone pairs
•
��l:
: C I:
(c)
I has
: C I:
lone pairs
I .
: C I0 I :0 o
3 bonding pairs and 2
I
: �! : (d)
I has
: C I: : ci00...... 1I 00 00 / \ - C00:I : C I: : C :I
5 bonding pairs and
lone pair
SM-80
1
2.63
•• .. F ·. •• F· • � I /•••
(a)
.F ••
(b)
I
.
•
·. F ·
•
.
:F:
.
.
As
I
.
(d)
�.
.. : CI : : CI .. �
I
Te
: CI � I : Cl :
. F ·.
:F:
I
10 electrons
•• .. F ·. F· � 1/··· . . F. /
•
Xe :
:F:
.
1 2 electrons
.F ...
I
•
/S�
·. F ·.
(c)
: F:
.
10 electrons
1 2 electrons
2.65
(a)
(b)
: 0:
II
: F --Xe-F : . .
.
.
F : : F."" .. /.. Xe /" . . � .. .F : .F : . .
.
2 lone pairs
.
2 lone pairs
.
(c)
. . : 0: . . · F F· · . ."" II /- . Xe / . � .. · F· F . .
.
.
.
1 lone pair
2.67
1 (2.7) < Br (3 .0) < CI (3 .2) < F (4.0)
2.69
In (1.8) < Sn (2.0) < Sb (2.1 ) < Se (2.6)
2.71
(a) The bond in HCI would be more ionic. The electronegativity difference is greater between H and CI then between H and I, making the H ---C 1 bond more ionic. (b) The bonds in CF4 would be more ionic. The electronegativity difference is greater between C and F than between C and H, making the C-F bonds more ionic.
SM-81
.
.
.
(c) C and S have nearly identical electronegativities, so the C -S bonds would be expected to be almost completely covalent, whereas the C - O bonds would be more ionic.
2.73
2 2 Rb+ < Sr + < Be +; smaller, more highly charged cations have greater polarizing power. The ionic radii are 1 49 pm, 1 1 6 pm, 27 pm, respectively.
2.75
2 0 -
<
3 N -
<
C I-
<
Br -; the polarizability increases as the ion gets larger
and less electronegative. The ionic radii for these species are 140 pm, 1 7 1 pm, 1 8 1 pm, 1 96 pm, respectively.
2.77
/
(a) C O - > C O 2 > C O 2 C O 3 - will have the longest c-o bond l ength. In C O there is a triple bond and in C O 2 the C -O bonds are double bonds. In carbonate, the bond is an average of three Lewis structures in which the bond is double in one form and single in two of the forms. We would thus expect the bond order to be approximately
1.3. Because the bond l ength is inversely related
to the number of bonds between the atoms, we expect the bond length to be longest in carbonate.
/
(b) S O - > S 0 2
�
S0 3
Similar arguments can be used for these molecules as in part (a). In S02 and S 0 3 ' the Lewis structures with the lowest formal charge at S have double bonds between S and each O. In the sulfite ion, however, there are three Lewis structures that have a zero formal charge at S . Each has one s-o double
bond and two S-O single bonds. Because these s-o
bonds would have a substantial amount of single bond character, they would be expected to be longer than those in S0 2 or S0 3 . This is consistent with the experimental data that show the s-o bond lengths in
SM-82
/
S 0 2 and S 0 to be 143 pm, whereas those in SO - range from about 145 3 pm to 152 pm depending on the compound. (c) C H N H 2 > C H 2N H > HC N 3 The C -N bond in HC N is a triple bond, in C H 2N H it is a double bond, and in C H N H 2 it is a single bond. The C -N bond in the last molecule 3 would, therefore, be expected to be the longest.
2.79
(a) The covalent radiu s ofN is 75 pm, so the N -N single bond in hydrazine would be expected to be ca. 150 pm. The experimental valu e is
145 pm. (b) The c-o bonds in carbon dioxide are double bonds. The
covalent radius for doubly bonded carbon is 67 pm and that of 0 is 60 pm. Thu s we predict the C RO in C O 2 to be ca. 127 pm. The experimental bond length is 116.3 pm.
(c) The C -O bond is a double bond so it
would be expected to be the same as in (b), 127 ppm. This is the experimentally found valu e. The C -N bonds are single bonds and so one might expect the bond distance to be the sum of the single bond C radiu s and the single bond N radiu s (77 plu s 75 pm) which i s 152 pm. However, becau se the C atom is involved in a multiple bond, its radiu s is actu ally smaller. The sum of that radiu s (67 pm) and the N single bond radiu s gives
142 pm, which is close to the experimental valu e of 133 pm.
(d) The
N -N bond is a double bond so we expect the bond distance to be two times the double bond covalent radiu s ofN , which is 2 pm. The experimental valu e is 123.0 pm. 2.S1
(60 pm) or 120
= 149 pm (b) 111 pm + 72 pm 183 pm 141 pm + 72 pm = 213 pm. Bond distance increases with size going
(a) 77 pm + 72 pm (c)
x
=
down Group 14/IV.
SM-83
2.83 -I
+1
-I
+1
+1
+1
+1
+1
-1
. .
N - N = N = .. N
:N
2.85
(a)
o :0:
II
-I
/C ,
: 0/ •
•
-2
0
'
0
c
II
II
-I
/ ""-
: 0:
-2
: 0: ../ C �
0: ..
0
.
.0
I
�C
(b)
:�:===�]
-2
�
:0:
0
I
: 0:
c �o: � ,' ::/" :o� c
-I
o:
:0: -2
: 0:
� c""-
:0
..
C
II
/0 ..
:
: 0:
-I
(C):C
_1
C:
]
-2
2.87
(a)
:0:
I
CH 3 -C
SM-84
=
O •
•
(b) H
H
I
I
H-C-C=O
H-C�C-O :
I
I
CH 3
CH 3
(c)
:0:
II
:0:
H
I
I
P and
I
CH 3 - C = N
CH 3 - C - N:
2.89
H
. .
•
S are larger atoms that are less able to form multiple bonds to
themselves, unlike the small N and 0 atoms. All bonds in P4 and S8 are single bonds, whereas N 2 has a triple bond and O 2 a double bond.
2.91
(a) H - C :::;:: C - H
H - Si = Sj - H
H - C - Si - H H-C= N:
SM-85
(b) H , "" N ,:'_' / H � C C
!I
H
I
",.. c , � N . . "' c "
I
H H
H
....... ...
C II
/
N
,/ C ,_,
'c I
H
.. .. �
"N
'C/
II
. N ",
-:;:- C ,,_
"H
·
H
H
"-' C, /
II
N
N
�' C / 'i'
c
I
N. .
H H
'� .N ' ,",
"-,�
C
I
N
/
N ' ,; ',
�
N
N
I
+
C �7 " "
H
N/ II
..
N
" �N ;
C
I
�� C .... ......
II
, N,
H
·
N, �"""" N ' •
. c ?' N. /
I
II
+
'
•
" ,I'N ,/'
I
"
N ............
H
H
suitable resonance form
SM-86
H
2.93
(a) In H202, hydrogen peroxide, the oxidation number of oxygen is - 1 ; the oxygen in water has an oxidation number of -2. Since oxygen i s gaining electrons on going from H202 to H20 it is undergoing reduction. Ascorbic acid (C 6Hg06), on the other hand i s undergoing oxidation: the oxidation number of carbon changes from +2/3 to + 1 on conversion from ascorbic acid to C 6Hg0 6 which means it is losing elecctrons. (b) The Lewi s structures for hydrogen peroxide and water, respectively, are o
o
H -O
-
o
O- H
H -O- H
N ote that all atom have formal charges of zero, suggesting that there i s no driving force for this conversion based solely on formal charges; however, since we know that hydrogen peroxide does react to form water in the lung, and that thi s conversion i s borne by the change in oxidation states, oxidation numbers are far more useful in determini ng whether a material has been oxidized or reduced.
2.95
(a)
( 1 - d*/d)/d
MI
d(M-I), pm
Li
274
3.19
N aI
Lattice Energy, kJ · mor l
1 0-3
759
294
x
3 .00 x 1 0-3
700
KI
329
2.72
1 0- 3
645
RbI
345
2.6 1
x
1 0-3
632
CsI
361
2.5 1
1 0-3
60 1
x
x
2 A high correlation (R = 0.973 1 ) exists between latti ce energy and d(M-I). 2 A better fit (R = 0.9847) is obtained between latti ce energy and ( 1 - d */d)/d. =
2 1 83 3 1 ( 1 - d*/d)/d) + 54.887 and the Ag-I distance (309 pm), the estimated AgI lattice energy is 683 kJ · mor l . (b) Using the equation L.E.
SM-87
(c) There is not close agreement between the estimated (683 kJ · mor 1 ) and experimental (886 kJ · mor 1 ) lattice energies. A possible explanation is that the Ag+ ion is more polarizable than the alkali metal cations of similar size and therefore the bonding in AgI is more covalent. 2 .97
(a) . .
: 0: H
H
H
H : 0:
(b) All the atoms have formal charge 0 except the two oxygen atoms, which are - 1 . The negative charge is most likely to be concentrated at the oxygen atoms. (c) The protons will bond to the oxygen atoms . Oxygen atoms are the most negative sites in the molecule and act as Lewis bases due to their lone pairs of electrons. The resulting compound is named hydroquinone. :O-H H
H
H
H : O- H
SM-88
2.99
Number of Z
-
Configuration
unpaired e
Element
Charge
Energy state
4
Fe
2+
ground
2
Te
2-
excited
0
S
2-
ground
26 [Ar] 3cf
J0 5 1 52 [Kr] 4d 5i5p 6s 16 [Ne]3i3p6 39
[Kr]4i
1
Y
2+
ground
30
[Ar] 3cf4i
2
Zn
2+
excited
2.101
(a)
+1
H-O-N-C-O:
H- O-N=C=O
H-O=N-C=O -I
+1
(b)
H H
(c)
H H
(d)
"
/'
"
/'
H
C=S=O
H
C=N=N +1
+I
-I
: I.
O=N=C=N
SM-89
-I
"-
/'
C = S - O .' +1
-I
2.103
R H H H R' I I I I I H -C=C-C-C=C-H
~
R H H H R' I I I I I H-C-C=C-C=C-H
/
R H H H R' I I I I I H-C=C-C=C-C-H
ThO}; II: TbO; (b) 3+; 1 +; (c) [Xe]4/4sio; [Xe] 4/ 4 sio6i; (d)
2.105 (a) I :
Because compound II has a lower melting point, it is probably more covalent which is consistent with the fact that the 3+ ion is more polarizingo 2.107 (a)
l
00
:0 II
00
:O - CI =O 00 1 00 -I
+1 : 0 : -1
l r r l -I r r l r � lO 1 : O -�I O; I 1 l 1 :0
00
II
00
O0 0 = CI - O: 1 1 00 + : 0 : -1
O:
00
II
_1
00
:O - CI : 00 00 11+ - O
� l O
-I
O = I=O 00 00 I
:0:
-I
:0:
1
O:
00
II
00
O0 0 = CI - O0 0 : II
_1
:0:
O � I=O 00 00 II
:0:
l lO 00
1
00
:O - CI 00 00 1 2= O -I
+ :0 : -1
·0
00
-I
+2
1 0 00 0 0 0 0_1
00
SM-90
O l : () � l = () ll r 1 l 1 IO l_ 1 l r -l I l:O 1 O l i . 1 l _I O 1 00
00
I
-I : 0 :
00
O0 0 = C\ - O: 11 0 0 +1 :0:
_1
l :O:
00
I
I
-I
-
00
:0:
00
O0 0 = C\ 00 1 =O +1 : 0 : -1
O-
:O - � I = O0 0 00
O0 0 = � I = O 00 II
00
I
0 0-
:O - Cl - O0 0 00
00
0
1 +30
0 00 0 0 0 _1
1
0
0 0- 1
O0 0 = CI : 00 1 2- O + :0 : - 1
II
-I
:0:
:0:
1
00
0
I
0
0 :1
- CI -O: :O 00 11+ 0 0
-I
2 :0 :
The four structures with three double bonds (third row) and the one with four double bonds are the most plausible Lewis structures according to formal charge arguments because these five structures minimize the formal charges. (b) The structure with four double bonds fits these observations best since its bond lengths would all be 1 40 pm, or only 4 pm shorter than the observed length. However, the four structures with three double bonds also fit because, if the double bonds are delocalized by resonance, we can estimate the average bond length to be
� (1 70 pm) + 1 (1 40 pm) = 1 47.5
pm, or just 3 . 5 pm longer than observed.
(c) 7 + ; The structure with all single bonds fits this criterion best. (d) Approaches (a) and (b) are consistent but approach ( c) is not. This result is reasonable because oxidation numbers are assigned by assuming ionic bonding. 2.109 The alkyne group has the stiffer C-H bond because a large force
constant, k, results in a higher-frequency absorption. 2.111 Look at the Lewis structures for the molecules : -I
(a) : C
+1
0:
(b)
.- I. +1 (d) :O - S = Q
+1 .-1. (c) : O- O = Q
�= Q -I
+1
(e) : N = N = Q
(f)
..
: Ar:
Of these, a, c, d, and e can all function as greenhouse gases
2.113 Since the halogen all have an odd number of valence electrons (7),
interhalogen compounds of the type XX ' n will be radicals and therefore
SM-9 1
extremely reactive unless the total number of halogens is an even number. This can only be achieved i fn is odd. Look at ICh vs. ICh as examples:
. . . . ---- 'I -...... . . : Cl .. .. .Cl .:
2.115
'.
.
I - Cl : I : Cl:
: Cl
-
'
(a) Compare the length of a S-S bond to that of a CI-CI bond. From Table 2.5, the CI-CI bond length is 1 99 pm; here the S-S bond l ength in
S2F2 is reported to be 1 90 pm. From Table 2.5 it can be determined that, on average, an X-Y bond is approximately 20 pm longer that a X=Y bond; this suggests that the S-S bond (which is 9 pm shorter than a CI-C1 single bond) has some double bond character. (b) and (c) The Lewis structures for the two possible S2F2 are:
-I +1 : F - S = .� - F :
: F - S - .� - F : favored
: .F." . . .. S = .S. / ' F .' ..
-I : F." +. 1. .. S - .S. : / F '. ·
isomer 2
,
·
·
isomer 1
.
favored
If resonance is occurring then one would expect that the S-S bond l ength is indeed between a single and a double bond in length.
2.117
(a) The Lewis structures of NO and N02 are: .
N =Q
(best possible structure)
+1 -I . . : O- N =Q .
-1 +1 . O = N - Q:
(equivalent resonance structures)
SM-92
From Table 2 .4, the average bond dissociation energy of a N= O bond is 630 kJ mor l , which i s right in line with the Lewis structure of NO. The bond dissociation energy of each NO bond in N02 is 469 kJ mor l , which is about half-way between a N-O double and an N-O single bond, suggesting that the resonance picture of N02 is a reasonable one. (b) An N-O single bond should have a bond length of 1 49 pm while an NO double bond should have a bond length of 1 20 pm (values are obtained by summing the respective covalent radii values from Figure 2 .2 1 ). From Table 2 . 5 the length of a N-O tripl e bond can be estimated to be between 1 05 and 1 1 0 pm. Since NO itself has a bond length of 1 1 5 pm, this suggests that its actual bond order i s somewhere between that of a double and a triple bond. o
(c)
N=O
I
..
o
0
(d)
: 0: � +I
� = +1N - R: -I o
0
+1 1'
N-O-N I I
:0:
0
: 0:
:0: -I
-I
(e) N20S(g) + H20(l) -7 2 HN03 (aq); Nitric acid is produced
(t) Molarity
=
7.50
x
1 0-
2
mol HN0 3
1 .00 L
=
7.50
x
2 1 0-
mol · LI HN03
(g) The oxidation number of nitrogen for the various nitrogen oxides are as follows: NO: +2 ; N02 : +4; N203 : +3 ; and N20S : +5 . An oxidizing agent is a species that wants to gain electrons; based on oxidation number, N20S should be the most potent oxidizing agent of the nitrogen oxides, as it possesses the most positive oxidation number for N of the group.
SM-93
CHAPTER 3 MOLECULAR SHAPE AND STRUCTURE
3.1
(a) Must have lone pairs; (b) May have lone pairs.
3.3
(a) The shape of the thionyl chloride molecule is trigonal pyramidal. (b) The O-S-CI angles are identical. The lone electron pair repels the bonded electron pairs equally and thus, all O-S-CI bond angles are compressed equally. (c) The expected bond angle is slightly less than 1 09.5°.
3.5
(a) angular, the electron pair on the central atom results in a trigonal planar arrangement . (b) The bond angle will be slightly less than 1 20° .
3.7
(a) linear;
(b) slightly less than 1 80°
3.9 (a) :Cl- S'�Cl: .. I : CI :
'0
: CI :
(c)
[ ]:
of °
�: p f. � .
:F : ..
:
(d)
..
..
:0: I
: O -Xe : I
:0:
(a) The sulfur atom will have five pairs of electrons about it: one nonbonding pair and four bonding pairs to chlorine atoms. The arrangement of electron pairs will be trigonal bipyramidal; the nonbonding pair of electrons will prefer to lie in an equatorial position, because in that location the e-e repulsions will be lowest. The actual structure is described as a seesaw. AX 4 E (b) Like the sulfur atom in (a), the iodine in iodine trichloride has five pairs of electrons about it, but here there are two lone pairs and three bonding pairs. The arrangement of electron pairs will be the same as in
SM 94 -
(a), and again the lone pairs will occupy the equatorial positions. Because the name of the molecule ignores the lone pairs, it will be classified as T shaped. AX E 3 2 ( c) There are six pairs of electrons about the central iodine atom in IF4 - . Of these, two are lone pairs and four are bonding pairs. The pairs will be placed about the central atom in an octahedral arrangement with the lone pairs opposite each other. This will minimize repulsions between them. The name given to the structure is square planar. AX 4 E 2 (d) In determining the shape of a molecule, double bonds count the same as single bonds. The Xe0 3 structure has four "obj ects" about the central Xe atom: three bonds and one lone pair. These will be placed in a tetrahedral arrangement. Because the lone pair is i gnored in naming the molecule, it will be classified as trigonal pyramidal. AX3 E
3.11
(a) The 1 - molecule is predicted to be linear, so the < I -1-1 should 3 equal 1 80°. AX 2 E 3 (b) The SbCIs molecule is trigonal bipyramidal. There should be three CI-Sb-CI angles of 1 20° , and two of 90°. AXs ( c) The structure of 10 4 - will be tetrahedral, so the 0-1-0 bond angles should be 1 09.5°. AX 4 (d) The structure of NO; is angular with a bond angle slightly l ess than
3.13
The Lewis structures are
. : F: : C I- C - F : : F: .
(a)
••
••
I I
••
••
(b)
.. : C, I:. .. : <;.l -Te��l : : Cl:
(c)
: F- C =O: : F: ' "
SM-95
d)
�H - c"H - H�-
(a) The shape of CF3 CI is tetrahedral; all halogen-C-halogen angles should be approximately 1 09.5°. AX 4 ; (b) TeCl 4 molecules will be see saw shaped with CI-Te-CI bond angles of approximately 90° and 1 20°. AX 4 E; (c) COF molecules will be trigonal planar with F-C-F and O-C-F 2 angles of 1 20°. AX 3 ; (d) CH 3- ions will be trigonal pyramidal with H-C-H angles of slightly less than 1 09.5°. AX3 E
3.15
(a) a and b are expected to be about 1 20°, c is expected to be about
1 09.5° in 2, 4-pentanedione. All of the angles are expected to be about 1 20° in the acetylacetonate ion. (b) The major difference arises at the C of the original Sp3 - hybridized CH group, which upon deprotonation goes to Sp 2 hybridization with 2 only three groups attached.
3.17
(a) slightly less than 1 20° ;
(b) 1 80° ;
(c) 1 80° ;
(d) slightly less
then 1 09.5°
3.19
The Lewis structures are ( a)
H I
CI C· ·l : : CI:
H-
••
-
: CI : : C I- CI - CI: : CI: ••
(b)
••
I
••
••
· ··s = c = s ·
( c) ··
. .
(d)
..
I · ··
:F :
: F - S ":' F : ••
I
••
:F :
Molecules (a) and (d) are polar; (b) and (c) are nonpolar.
3.21
(a) pyridine: polar (b) ethane: nonpolar (c) trichloromethane: polar
SM 96 -
3.23
(a) Of the three forms, 1 and 2 are polar; only 3 is nonpolar. This is because the C-CI bond dipoles are pointing in exactly opposite directions in 3. (b) The dipole moment for 1 would be the largest because the C-CI bond vectors are pointing most nearly in the same direction in 1 ( 60° apart) whereas in 2 the C-CI vectors point more away from each other ( 1 20° ), giving a larger cancellation of dipole.
3.25
H
H
'\ /
H I
C = C - C := N :
2
The first two carbons (CH2 and CH) are Sp hybridized with H-C-H and C-C-H angles of 1 20°. The third carbon (bonded to N) is sp hybridized
3.27
with a C-C-N angle of 1 80°.
(a) tetrahedral, bond angle of 1 09.5° (b) Tetrahedral about the carbon atoms ( 1 09.5° ) C-Be-C angle of 1 80°. (c) angular, H-B-H angle slightly less than 1 20° (d) angular, CI-Sn-CI angle slightly less than 1 20°
3.29
H
(a) H-C-H and H-C-C angles of 1 20°. (b) linear, 1 80°.
H
'\ /
/ c = c "-
H
H
C] - C = N : • •
(c) Tetrahedral, 1 09.5° .
:Cl: I
: O-P-Cl: • •
I
: �l :
• •
• •
:Cl: I
O=P-Cl: • •
• •
I
:9:
• •
• •
(Note: Both are acceptable Lewis structures; the presence of the double bond makes the second molecule the more stable Lewis structure.) (d) The arrangement of atoms about each N is trigonal pyramidal giving H-N-H and H-N-N bond angles of approximately 1 07°.
SM-97
H-N-N-H I
H
I
H
..
3.31
(a) tetrahedral :
••
: Cl: I
(b) tetrahedral :
••
: O - Sb-Cl: ••
I
••
••
[: Oj�·F· : l••
I
:0:
t
••
••
: O - S -C1:
: Cl:
(c) seesaw:
:Cl: I
:0:
••
Note: While all are acceptable Lewis structures, those with double bonds are the more stable structures.
3.33
(a)
Sp3 ,
orbitals oriented toward comers of a tetrahedron ( 1 09.5° apart);
(b)
sp, orbitals oriented directly opposite to each other ( 1 800
apart);
(c)
Sp3 d 2 , orbitals oriented toward the comers of an octahedron (interorbital angles of 900 and 1 800 ) ;
(d)
Sp 2 , orbitals oriented toward the comers
of an equilateral triangle trigonal planar array (angles
=
1 200 ); trigonal
planar.
3.41
The Lewis structure of tetrahedral
P4 is:
;
P
:P
:::: \P ;: :
\
P
(a) Each phosphorous is attached to three other phosphorous atoms and a single lone pair, therefore its hybridization is
SM-98
sp3.
(b) While each P within this structure is polar (due to the presence of a single lone pair on each phosphorous), P4 is nonpolar due to the 3 D orientations o f those lone pairs.
3.43
As the s-character of a hybrid orbital increases, the bond angle increases.
3.45
Atomic orbitals a and b are mutually orthogonal if Ja . b
dr
=
0 (assuming a
::f::.
where the integration is over all space.
b)
Furthermore, an orbital, a, is normalized if
Ja 2 dr = 1 .
In this problem, the two hybrid orbitals are : hi
=
S + Px + Py + Pz and h2
=
S - Px + Py - Pz . Therefore, to show
these two orbitals are orthogonal we must show
J� h2 dr
=
O.
fhl h2 dr = J( s + Px + Py + pJ( s - p, + Py - pJ dr =
J(S 2 - sPx + SPy - spz + sp, - p� + PxPy - PxPz + SPy -
P, Py + P� - pypz + spz - pz p, + pz Py - p� ) dr Of course, this integral of a sum may be written as a sum of integrals:
Js 2 dr - JsPx dr + fspy dr - Jspz dr + . . . Because the hydrogen wavefunctions are mutually orthogonal, the members of this sum which are integrals of a product of two different wavefunctions are zero. Therefore, this sum of integrals simplifies to:
Js 2 dr - Jp�dr + fp� dr - fp�dr
=
1-1+1-1
=
0
(recall that the integral of the square of a normalized wavefunction is one.)
3.47
We are given: A =
cos B
.
In the H 2 0 molecule, the bond angle I S cos 2 (li B) .
1 04.5°. Therefore, A = 0.67 and the hybridization is Sp 0 67 .
SM-99
3.49
Li
(a)
2 BO - 2 (2 2 - 2) -- 1 2+ BO = t (2 2 - 2 - 1) = t - 1.
+
diamagnetic, no unpaired electrons +
Li
(b)
paramagnetic, one unpaired electron Li
(c)
2- BO = t (2 + 2 - 2 - 1) = t
paramagnetic, one unpaired electron
3.51
((725 ) ((72/ J ((72P ) 02P, J 02Py J (7r2 P, * J (7r2 Py * J ((72 /) (2) ((725 ) ((72 5 * J ((72P ) 02P, J 02Py J (7r2 P, * J (7r2 Py * ) ((725) ((725 * ) ((72P ) 02P, J 02Py J (7r2 P, * J 02PY * J ((72 P * )
(a) ( 1 )
(3)
(b) ( 1 ) 0 . 5 ;
(2) 1 .5 ;
(c) ( 1 ) and
(2) are paramagnetic, with one unpaired electron each
(d)
3.53
(7
for ( 1 ) and (3),
7r
(3) 0
for (2).
2
(a) The energy level diagram for N is as follows:
(b) The oxygen atom is more electronegative, which will make its orbitals lower in energy than those ofN. The revised energy-level diagram
SM- I OO
is shown below. This will make all of the bonding orbitals closer to 0 than to N in energy and will make all the anti bonding orbitals closer to N than to 0 in energy. Energy level diagram for NO +
,
2s
ft-- -
-
_ - -
-
-
-
,�,
- - - -
,,
+r
-
-
-
-
-
cr;s -
-
-
-
- ---- ....1.L T, - - - - -cr 2s -
Orbitals on N
-
-
� - :-.. I , 2s
- - -
Orbitals on 0
(c) The electrons in the bonding orbitals will have a higher probability of being at 0 because 0 is more electronegative and its orbitals are lower in energy.
bond order
=
1.
2 (b) Be 2 (4 valence electrons) : ( 0"2J ( 0" s* ) 2 , bond order = o . 2 (c) F2 (1 4 valence electrons) :
SM- l O I
3 .57
(0"2S ) (0"2S * ) 02Px J 02Py J (0"2P ) CO + (0"2S ) (0"2S * ) 02Px J 02P J (0"2P ) Y ;
CO
bond order = 3 ;
bond order = 2.5
Due to the higher bond order for CO, it should form a stronger bond and therefore have the higher bond enthalpy
3 .5 9
(a) - ( c) All of these molecules possess unpaired electrons and therefore are paramagnetic.
B2
-
and
+
B2
have an odd number of electrons and must,
therefore, have at least one unpaired electron; indeed, both have only one unpaired electron.
B2 has an even number of electrons, but in its
molecular orbital energy level diagram, the HOMO is a degenerate set of
n
2p
orbitals that are each singly occupied, giving this molecule two
unpaired electrons. For
B2 , one more electron will be placed in this -
degenerate set of orbitals, causing one of the original unpaired electrons to now be paired.
will therefore have one unpaired electron. Likewise,
B2
-
B/ will have one less electron than B2; thus one of the originally unpaired electrons will be removed, leaving one unpaired electron in this molecule as well .
3
.61
(a) F with 14 valence electrons has a valence electron configuration of
2 (0"2s )2 ( 0"2s* f (0"2P ) \ n2pJ2(n2P)2(n2P/ )2(n2py * )2 with a bond order of 1 . After forming F2 from F , an electron is added into a 0"2P' orbital. The 2 -
addition of an electron to this antibonding orbital will result in a reduction of the bond order to 1 12 (See 5 1 ). F will have the stronger bond.
2
with a bond order of 1 . Removing one electron to form
SM- I 02
(b)
B2+ will eliminate
one electron in the bonding orbitals, creating a bond order of 1 /2 .
B 2 will
have the stronger bond.
3.63
C 2+ (()2S J (()2S * J 02 P, J 02 Py ); C 2 (()2 S J (()2 S * ) 02 P, J 02 Py J;
C2 - (()2S J (()2S * ) 02P, J 02 Py J (()2 P ) ; C2 is expected to have the lowest ionization energy because its electron is lost from a higher energy MO ( ()2 P ) than either C 2+ or C 2 ( Py ) -
7[2
3.65
.
The conductivity of a semiconductor increases with temperature as increasing numbers of electrons are promoted into the conduction band, whereas the conductivity of a metal will decrease as the motion of the atoms will slow down the migration of electrons. (a) In and Ga;
3.69
Given the overlap integral S \f'
=
\f' A i s + \f' B is
(b)
P and Sb
3.67
'
=
f\f' A i s \f' B ls dr , the bonding orbital
and the fact that the individual atomic orbitals are
normalized, we are asked to find the normalization constant N which will normalize the bonding orbital \f' such that :
fN2 \f' 2 dr =N2 f(\f' Ai s +\f'BIJ2 dr = 1 N 2 f(\f' A i s +\f' B IJ 2 dr N 2 f(\f' �" + 2\f' Ai s \f' B i s + \f'�Jdr =
Given the definition of the overlap integral above and the fact that the individual orbitals are nonnalized, this expression simplifies to :
SM- I 03
N-
Therefore,
3.71
/1 Ij�
The antibonding molecular orbital is obtained by taking the difference between two atomic orbitals that are proportional to e- r/a" . Halfway
between the two nuclei, the distance from the first nucleus, l'j , is equal to the distance to the second nucleus, If'
proportional to:
e-r/a"
ex
-
r2 ,
e-r/a"
=
and the antibonding orbital is 0
3.73 (a)
Shape: tetrahedral
: Cl :
I
Hybridization:
I Cl :
Bond angles: 1 09.5°
: Cl - Ga - Cl : :
•
•
•
•
Shape: see-saw
:F:
I
: F ,,--
Hybridization:
.. Te - F :
/
:F:
.
Bond angles: > 90° and > 1 20° Shape: see-saw
..
: Cl : : Cl •
•
_____
I
: Cl :
.
: Cl :
I
: Cl - Si - Cl :
I Cl :
Hybridization: •
dSp 3
•
Sb.- Cl :
/
:
dSp 3
Polar
(c)
(d)
Non-polar
..
(b)
Sp 3
Bond angles: > 90° and > 1 20° Polar Shape: tetrahedral Hybridization:
Sp 3
Bond angles: 1 09.5° Non-polar
SM - I 04
3.75
(a)
SiF4 : SiF4 i s nonpolar (tetrahedral AX 4 structure) but PF is polar 3
(trigonal pyramidal AX E structure); 3 b)
SF6 : SF6 is nonpolar (octahedral AX 6 structure) whereas SF4 is polar
(seesaw, AX4 E structure); (c) AsFs : IFs is polar (square pyramidal, AXs E structure) whereas AsFs i s nonpolar (AXs ' trigonal bipyramidal structure).
3.77
(a) The elemental composition gives an empirical formula of CH 4 0, which agrees with the molar mass. There is only one reasonable Lewis structure; this corresponds to the compound methanol . All of the bond angles about carbon should be 1 09.5° . The bond angles about oxygen should be close to 1 09 . 5° but will be somewhat l ess, due to the repulsions by the lone pairs. (b) Both carbon and oxygen are Sp3 hybridized. (c) The molecule is polar. Bonding
3.79
Anti bonding
(a)
(b)
(c) The bonding and antibonding orbitals for HF appear different due to the fact that a p-orbital from the F atom is used to construct bonding and antibonding orbitals whereas in the H2 molecule s orbitals on each atom are used to construct bonding and antibonding orbitals.
3.81
(a) The expected molecular orbital diagram for CF is
SM- I 05
C2p
-t- +
#+
F2p
C2s F2s
The bond order for the neutral species is 2 .5 because one electron occupies a
7r2p
* orbital. Adding an electron to fonn CF- will reduce the bond
order by 1 12 to 2, while removing an electron from form CF+ will increase the bond order to 3. Because bond lengths increase as bond order decreases, the C-F bond length varies in these molecules in the fol lowing manner: CF+ < CF < CF-
.
(b) The CF+ ion will be diamagnetic but both
CF and CF- will have unpaired electrons (one in the case of CF and two in the case of CF- ).
SM- I 06
3.83
The Lewis structure ofborazine is nearly identical to that of benzene. It is obtained by replacing alternating C atoms in the benzene structure with B
and N, as shown. The orbitals at each B and N atom will be s/ hybridized.
3.85
(a) The Lewis structures are:
(b) All of these species are expected to be diamagnetic. None are radicals. (c) The predicted bond angles in each species based upon the Lewis structure and VSEPR theory will be CH + 3
AX 3
trigonal planar 1 20°
CH 4
AX 4
tetrahedral
1 09.5°
CH 3 -
AX 3 E
pyramidal
slightly less than 1 09.5°
CH 2
AX 2 E
angular
slightly less than 1 20°
2 CH 2 +
AX 2
linear
1 80°
2 CH 2 -
AX 2 E 2
angular
less than 1 09.5° , more so than CH 3 - due to the presence of two lone pairs
The order of increasing H-C-H bond angle will be CH/-
<
CH 3 -
<
CH 4
<
CH 2
<
CH 3 +
SM- I 07
<
CH/+
3.87
Acetylene : H - C=C - H
Polymer:
[
H I
=C-
H I
Z Z =C-
H I
1
=C-
Polyacetylene retains multiple bonds along the chain. It is through the series of orbitals that electrons can be conducted. A resonance form of the Lewis structure can be drawn showing that the electrons may be delocalized along the polyacetylene chain. No such resonance form is possible for polyethylene. Note: The dark color of the material results from the formation of a large number of molecular orbitals that are not very different in energy. These molecular orbitals are made up of combinations of the p-orbitals on the carbons that make up the double bonds. Because the orbitals are closely spaced in energy, electrons in them can readily absorb visible light to be promoted to a higher energy orbital. See section 3.13. 3.89
The energy of an electron in the box is given by: E
=
nth n2 h2 /(8mL2)
quantum state of a one-dimensional
, where h is Planck ' s constant,
mass of the electron and L is the length of the box (recall
n
m
is the
must be an
integer). The lowest energy transition for this system will be from the highest occupied quantum state, which we will identify with the quantum number state,
nHO, nw.
to the next highest state or the lowest unoccupied quantum
Since each quantum state can hold 2 electrons and each carbon
atom contributes one electron, case
N
is odd,
N/2
and
=
=
nHO N/2 nw (N/2)
+ 1.
(In the
must be rounded up to the nearest integer.) Also, the
length of the box is given by L
=
NR
where
R
is the average C-C bond
length. Therefore, the lowest energy transition is given by:
2] ) ) ( ( [ 2 _ _ 2 n nwh _ oh2 h2 N N +l H HO - 8mN2 R2 8mN2 R2 - 8mN2 R2 2 2 h2(N +l ) 8mN2 R2 N,
/)'£ -E - E W =
_
To shift the wavelength of the absorption to longer wavelengths (lower energies) the length of the carbon chain, SM- I 08
must increase.
3.91
(a)
(b) The overlap between the orbitals will decrease as one goes from (T
3.93
to Jr to 0 , so we expect the bond strengths to decline in the same order.
The effect of changes (a) and (b) will be similar. The overall bond order will change. In the first case, electrons will be removed from Jr orbitals so that the net Jr-bond order will drop from three to two. The same thing will happen in (b), but because two electrons are added to antibonding orbitals, a net total of one Jr -bond will be broken. Based on this simple model, the ions formed should be paramagnetic because the electrons are added to or taken from doubly degenerate orbitals.
3.95
(a) The Lewis structure of benzyne is H H
H ) sp� Sp2 C sp
III
C sp s/........... ... sp�
,
H
(b) B enzyne would be highly reactive because the two carbon atoms that are sp hybridized are constrained to have a very strained structure compared to what their hybridization would like to adopt-namely a linear arrangement. Instead of 1 800 angles at these carbon atoms, the angles by necessity of being in a six-membered ring are constrained to be close to 1 200 A possibility that allows the carbon atoms to adopt more reasonable •
angles is the formation of a diradical :
SM- I 09
H H �
c·
H � H 3.97
II
C·
3 (a) The carbon atoms are all Sp hybridized.
(b) The C-C-C,
H-C-H and H-C-C bond angles should be 1 09.5° based upon the answer to (a).
(c) B ecause of the ring structure, however, the C-C-C
bond angles must be 60° .
(d) The cr-bond will have the electron
density of the bond located on a line between the two atoms that it joins. ( e) If the C atoms are truly s/ hybridized, then the bonding orbitals will not necessarily point directly between the C atoms .
(f) The Sp
3
hybridized orbitals can still overlap even if they do not point directly between the atoms as shown. Such bonds are sometimes called "bent" bonds, or "banana" bonds. As a result of the situation in the C-C-C bond angles, the H-C-H bond angles are also distorted from
3.99
(a) and (d) have the possibility of
n- to-lr
* transitions because these
molecules possess both an atom with a lone pair of electrons (on 0 in
H COOH and on N in H CN) and a lr-bond to that atom. The other molecules have either a lone pair or a lr -bond , but not both.
SM- I I 0
3.101 (a-b)
/
o(C2s/,C2s/)
o(Hls,C2sp2)
....H
�
..
Os/
./
..--
- I //0 . H - C -\ C - C\
o(C2Sp2,02s/),
, 7i(C2p,02p)
'
7i(C2p,C2p)
All atoms in this molecule have a formal charge of zero.
[
JS based on the information
3.103 (a) A possible Lewis structure for Sb F 2 7
provided is as follows: •
•
F
•
: F·� \Sb •
•
•
•
•
+1
•
/'
I -I
• •
•
F
·
•
y:� Sb/;, f : -I I
•
•
:F :
: F:
..
..
Each Sb has a formal charge of - 1 while the bridiging F has a formal charge of + 1 . (b) Each Sb atom in this structure has a hybridization of s/d. (NOTE: the crystal structure of this ion has been done for both the potassium and the cesium salt; to read more about this controversial ion,
[
see "Crystal Structure of KSb Fr On the Existence of the Sb F7 2 2 S . H. Mastin and R.R. Ryan,
3.105
In organ ic Chemistry, 10,
A possible Lewis structure for
[Bi2C14 t
r
1 757 ( 1 97 1 ) .
is as follows:
..
+1
2-
Cl
. . / _ . """ . .
: Cl - Bi. B i - Cl : . "" . . / . . 2 -2 Cl +1
Formal charges for the bridging Cl atoms and the two Bi are shown.
SM- I I I
ion",
3.107
A possible Lewis structure for 15- that fits the conditions given in the problem is:
:
:- _ ;
C
-
;
b
. _ ./
I
. � • I
.... .
'
": i-
a
."
I: .
. . ;--....... .. . . ' I • d •• • e
In this structure, two atoms (Ib and Id) have a formal charge of - 1 , one
atom (Ie) has a formal charge of + 1 , and two atoms (Ia and Ie) have no formal charge; the ion has an overall charge of - 1 . To explain the shape, Ib and Id are s/d hybridized (trigonal bipyramid) with the attached atoms occupying axial positions, which results in a bond angle of 1 800 around each axial atom. Ie has a VSEPR formula of AX2E2; its four electron pairs adopt a tetrahedral electron geometry, which is consistent with s/
hybridization.
3.109
Al 2 Cl 6 is nonpolar; all dipoles in the molecule (due to the AI-CI bonds) will cancel : these cancel
t
SM-1 1 2
...
--.
these cancel
CHAPTER
4
THE PROPERTIES OF GASES
4.1
(a) 8 x 1 09 Pa; (d) 1
4.3
x
(b) 80 kbar;
1 0 6 lb · in-
2
(c) 6 x 1 0 7 Torr;
(a) The difference in column height will be equal to the difference in pressure between atmospheric pressure and pressure in the gas bulb. If the pressures were equal, the height of the mercury column on the air side and on the apparatus side would be the same. The pressure in the gas bulb is 0.890 atm, or 0 . 890 x 760 Torr · atm- I
=
676 Torr. The difference would
be 762 Torr - 676 Torr = 86 Torr = 86 mm Hg.
(b) The side attached to
the bulb will be higher because the neon pressure is less than the pressure of the atmosphere.
(c) If the student had recorded the level in the
atmosphere arm to be higher than the level in the bulb arm by 86 mm Hg, then the pressure in the bulb would have been reported as 762 Torr + 86 Torr = 848 Torr.
7 3 . 5 cm x
4.7
1 3 . 6 g . cm-3 3 1 . 1 0 g . cm-
= 909 cm or 9.09 m
(20. in)(1 O. in)( l 4. 7 lb · in -2 ) = 2.9 x l 03 lb
SM- 1 1 3
4.9
P;
1 2 (a) In Boyle's l aw, V oc - V = nr Therefore,
h = P. h2 �
I _
2 , _
h; h
oc
V
since the atmospheric pressure is 29.85 inHg.
PJ (29.85 + 1 2.0) inHg 4 1 .85 inHg P2 = (29.85 30.0) inHg = 59.85 inHg 1 .85 inHg = 22.4 in h2 = hl � = 32.0 in59.854inHg � =
+
=
x
(b) If the atmospheric pressure is 29.85 inHg, the pressure of gas in tube ( 1 ) is 29.85 inHg + 1 2.0 inHg = 4 1 .85 inHg.
The pressure of gas in tube (2) is 29.85 inHg + 30.0 inHg = 59.85 inHg.
4.11
nR
(a) Volume, L
- atm · KV '
0.01
8.2 1
0.02
4. 1 0
0.03
2 .74
0 . 04
2.05
0.05
1 .64
(b) The slope is equal to
I
nR .
V
(c) The intercept is equal to 0.00 for all the plots.
4.13
(a) From
� V;
= � V2 we have '
,
(2.0 X 1 05 kPa) (7.50 mL) = (� ) (1 000 mL); solving for � we get
1 .5 x 1 03 kPa.
�V;
4.15
(b) Similar to (a),
= � V2 or (643 Torr) (54.2 cm 3 ) '
=
3 ( � )(7.8 cm ), �
=
4.5
X
3 1 0 Torr .
Using
� �
- =
1 . 1 0 atm . · KeI vms, · and expressmg T m 298 K T2
�
-
SM- 1 1 4
=
;
� p.2 898 K
--
=
3 . 3 1 atm.
4.17
1 . 5 atm . � . D � � . - 1 . 6 atm. = ,[ ' Ke vms, Usmg � = -- an d expressmg T m 283 K 303 K 2 r;
4.19
�� � If P and T are constant, then - = , or n, n2 0. 1 00 mol
I'
V2 O . l l O mol
---= '-- = -
n � 0. 1 1 0 � Solving for � in terms of � , we obtain V2 = 2 = = 1.10 �. n, 0. 1 0 So the volume must be increased by 1 0.% to keep P and T constant.
4.21
S lope =
(�)
= 2.88 x 1 0-4 LX- I . For a given T, we can find a V value,
vice versa . Assuming at T = 25°C = 298 K, V = 0.0858 L P= 0.90 bar x
[
1 atm 1 .0 1 325 bar
.
J
= 0.89 atm;
Based on the equatlOn PV = nRT, n=
Mass =
(
P V(MOl. mass) RT
)
= 0.050 g. 4.23
=
[
[
mass Mol. mass
J
0.89 atm x 0.0858 L x 1 6.04 gxmol-' 0.08206 Lx atmxmor ' x K - ' x 298K
_
J
P. V _ P2 V2 . '_I = _ (a) Because P, V, and T all change, we use the relatlOn � .
r;
Substituting for the appropriate values, we get (0.255 atm)(3 5 . 5 mL) (1 . 00 atm)(V2 ) -'-------'----' --'-- = ; V? = 1 1 . 8 mL . 228 K 298 K (b) The same relation holds as in (a), but here the final temperature and volume are known: (0.255 atm) (3 5 . 5 mL) 228 K
=
( � )(l 2.0 mL) . P2 293 K
SM- 1 1 5
=
0.969 atm.
(c) Similarly, we can use the same expression, with P and V known and wanted. .2 5_ (3_5_ (0_ a_ tm .5_m L) _--'-. _)_ _5_ _ = 228 K
4.25
(
(500 Torr) 760 Torr · atm- I T2
J
(1 2.0 mL)
T
. ' T2 = 1 99 K
(a) Using the ideal gas law with the gas constant R expressed in kPa: P(0.3500 L) = (0. 1 500 mol) (8.3 1 4 5 1 L · kPa · K-I . mol-I ) (297 K); P = 1 .06 x 1 03 kPa. (b) BrF3 has a molar mass of 1 36.91 g . mOr l . We then substitute into the ideal gas equation:
(
1 0.0 Torr 760 Torr · atm-I =
(
J
V
23.9 X 1 0-3 g _ 1 3 6.9 1 g . mol 1
J
(0.0 82 06 L · atm · K-I . mol - I ) (373 K)
V = 4.06 X 1 0 2 mL
(
(0.77 atm)(O. l 000 L).
(c)
m
=
m
_ 64.06 g · mol I
= 0.20 g
(d)
J
(0.082 06 L . atm . K - I . mol - I ) (3 03 K)
(1 29 kPa)(6.00 x 1 03 m 3 )
(
1 x 1 06 cm3 m3
J
(
1L 1 000 cm 3
J
= n (8 . 3 1 4 5 1 L · kPa · K- I · morl )(287 K) n
= 3 .24 x 1 05 mol CH 4
(e) The number of He atoms is the Avogadro constant
NA
multiplied by
the number of moles. The number of moles is obtained from the ideal gas equation: PV = nRT PV n = -· n = RT '
N NA
SM- 1 1 6
so the number of atoms N will be given by
(PV)
N=N
A
RT
= (6 . 022 x 1 0 23 atoms · mo 1 - 1 ) = 9.2 x 1 0 1 4 atoms.
4.27
(a)
(b)
4.29
=
nRT
=
nRT
=
V P V P P, V,
=
Because
(
(2.0 0 kPa) (1 .0 x l O -6 L) (8.3 1 4 5 1 L · kPa · K- 1 · mo r l ) (1 5 8 K)
( 1 mol)(0.082 06 L · atm · K-I . mol- I )(773 K) 1 atm (1 mol)(0.082 06 L · atm · K- 1 · mol-I )(77 K) 1 atm
and
T
)
= 63 .4 L
= 6.3 L
are state functions, the intermediate conditions are
irrelevant to the final states. We can simply use the ideal gas law in the form
;:v;
[
T;
=
��
�
759 Torr
)
[
() V
252 Torr _ 2) 760 Torr · atm I 760 Torr . atm -I -- ----'------' -----'----- - = --'----253 K 1 523 K (1 .00 L)
� = 1 8 . 1 L.
4.31
Because
� V;
T
is constant, we can use
= ��
(1 .00 atm) (1 .00 L) = � (0.239 L) � = 4. 1 8 atm.
4.33
The volume of each tank is 3980 cm 3 The pressure of each tank,
P
=
=
3 .980 L, V;wo
5 860 kPa x
SM- 1 1 7
(
1 atm
tanks=
1 0 1 .325kPa
)
=
7.960 L. 5 7 . 83 atm.
Based on the equation PV = nRT, n = (Mass of O2 in two tanks) =
4.35
(
P V(MOl. maSS) RT
)( =
PV = nRT
(
24.5 kPa 1 0 1 .325 kPa · atm-I =
)
(
mass Mol. mass
�
) )
57.83 atm X 7.960 L 32. O g . morl � = 62 1 g. 0.08206 L · atm · mol I · K I x 289K
(0.2500 L)
n(0.082 06 L · atm · K- I . morl )(292.7 K)
n = 2.52 x 1 0-3 mol
4.37
(a)
E;� T;
=
� V2
T;
(1 04 kPa)(2.0 m 3 ) 294.3 K
=
(52 kPa) V2 268.2 K
=
(0.880 kPa) V2 22 1 .2 K
V2 = 3 .6 m 3 ( b)
E;� T;
=
� V2 T2
(1 04 kPa)(2 .0 m 3 ) 294.3 K 2 V2 = 1 . 8 x 1 0 m 3
4.39
Density is proportional to the molar mass of the gas as seen from the ideal gas law: PV = nRT PV = � RT M m
MP . = mass per umt. voIume = - = .
d enslty
V
SM- 1 1 8
RT
The molar masses of the gases in question are 2 8 .0 1 g . mol- I for CO(g), 44. 0 1 g . mor l for CO? (g), and 34.0 1 g . morl for H 2 S(g) . The densest will be the one with the highest molar mass, which in this case is CO 2 . The order of increasing density will be CO
4.41
<
H2 S
<
CO 2 •
The pressure of the Ar sample will be given by
P=V Ar
nRT
(
=
(
2.0 0 x 1 O- ' g 39.95 g . mol ,
�
)
}
0.0 82 06 L . atm . K - ' . ma r ' )(2 93 K) 0.050 0 L
2.0 0 x 1 0-3 (0. 082 06 L · atm · K-I . mo l- I )(T ) 2 8 3 . 8 0 g . mol I
p = -'-------=-----"-----------Kr
0.050 0 L
B ecause we want the pressure to be the same, we can set these two equal to each other. Because volume, mass of the gases, and the gas constant
R
are the same on both sides of the equation, they will cancel .
(
83.80
:
. ma l - '
Solving for
4.43
)(1; ) ( �
39.9 5
:
. mar'
r; , we obtain temperature
)
(29 3 K)
= 6 1 5 K, or 342°C.
(a) Density is proportional to the molar mass of the gas as seen from the ideal gas law. See S ection 4.9. (1 1 9. 3 7 g . mol- I )
d=
(
200. Torr
)=
760 Torr · atm - I (0.082 06 L · atm · K- I . mol-I )(298 K)
(b) d =
1 .28 g . L- I
(1 1 9. 3 7 g . mol- I )(1 .00 atm) (0.082 06 L · atm · K - I . mol- I ) (373 K)
SM- 1 1 9
=
3 .90 g . L- I
4.45
(a)
(8.0 g . L-' )(0.082 06 L · atm · K-' . mol- I )(300. K) = M = dRT P 2.8 l atm = 70. g . mol- I
(b) The compound is most likely
CH F3, for which
M = 70 g . mOr ' .
(c) You can use the relationship in (a) to calculate the new density, or you can apply the proportionality changes expected from the change in pressure and temperature to the original density:
d2 = (8 . 0 g . L- I ) 4.47
(
l .00 atm 2 . 8 1 atm
(J J 3 00. K 298 K
= 2.9 g · L- ' .
From the analytical data, an empirical formula of
CHCI is calculated. The
empirical formula mass is 48 .47 g . mol- I . The problem may be solved using the ideal gas law:
PV = nRT PV = .!!!:. M .. R T M = mRT PV · atm · K- ' . morl )(273 K) = 95 .9 g . mol-I M = (3 .557 g)(0.082( l .061 0 Latm)(0.755 L) n in the formula (CHCI)1l is equal to 95.9g · morl -;- 48.47g · mol- I = 1 .98. The formula
4.49
Density is proportional to the molar mass of the gas as seen from the ideal gas law:
PV = nRT PV = .!!!:. M .. R T
.
I
. denslty = mass per umt vo ume
MP = -V = RT m
SM- 1 20
0.943 g . C I M
4.51
=
(
5 3 . 1 kPa 1 0 1 .325 kPa · atm- 1 (0.082 06 L · atm · K -I . mol- I )(298 K) M
- - J-
= � -
44.0 g mOr l . •
� --,-
-
-
(a) The number of moles of H2 needed will be 1 .5 times the amount of NH produced, as seen from the balanced equation: 3 or Once the number of moles is known, the volume can be obtained from the ideal gas law:
V
=
nH RT 2
p
=
U· nNH J R T p
=
[( --'--'-
] -'--
3 mol H2 (1 .0 X I 03 kg)( 1 03 g . kg-I ) RT 1 7.03 g . morl 2 mol NH3
----'-J
---
--------p
(l)2 ( 1 7.031 0g6· gmol- I J (0.082 06 L · atm ·
K -I . mol- I )( 623 K)
-
= -��-�--�------------=
1 5 .00 atm
3 . 0 X 1 05 L.
(b) The ideal gas equation
_ _ P,v. 1__ 1
� R�
=
p.2_ V_ 2
� R�
. . sImplIfies to
P,v. _ 1_ 1
�
=
_ V
P.2 2 _
�
because R and n are constant for this problem. (1 5 . 00 atm)(3 .0 x I 05 L) (376 atm)V2 (523 K) 623 K V2 = 1 .0 x l 0 4 L =
4.53
We need to find the number of moles of CH4 (g) present in each case. Because the combustion reaction is the same in both cases, as are the temperature and pressure, the larger number of moles of CH 4 (g) should
SM- 1 2 1
produce the larger volume of CO (g). We will use the ideal gas equation 2 to solve for
n = PV RT =
n
in the first case:
(1 .00 atm)(2.00 L) (0.082 06 L · atm · K- 1 mol- I )(348 K) •
2.00 g of CH 4 will be
2 .00 g 1 6.04 g . mor I
=
=
0.0700 mol CH 4 .
0. 1 24 mol.
The latter case will have the greater number of moles of CH 4 and should produce the larger amount of CO (g). 2
4.55
The molar mass of glucose is 1 80. 1 5 g . mol-I . From this, we can calculate the number of moles of glucose formed and, using the reaction stoichiometry, determine the number of moles of CO needed. With that 2 information and the otherinformation provided in the problem, we can use the ideal gas law to calculate the volume of air that is needed:
PV= nRT V
=
x
= 4.57
[(
1 80. 1 5 g
����s!l::��,
(
glucose
0.26 Torr
760 Torr · atm- I
1 ( �:o�l���e l] 1 1
(0.082 06 L · atm · K- 1 mor l )(298 K) 2 .4 x 1 04 L. •
(a) This is a limiting reactant problem. Our first task is to determine the number of moles of NH and HCI that are present to start with. This can 3 be done from the ideal gas equation:
SM-1 22
PV
( (
= nRT
n NH}
=
n HCI =
RT
-
RT
) )
_ (0.0 1 50 L) 760 Torr · atm I = 7.94 x l 0-5 mol = (0.082 06 L · atm · K - I . mor l )(303 K)
PV
PV
1 00 Torr
=
1 50 Torr
76 0 Torr · atm - I
(0.0250 L)
(0.082 06 L · atm · K - I . mol- I )(298 K)
= 2.02 x l O -4 mo I .
The ammonia is the limiting reactant. The number of moles of NH 4 CI(s) that form will be equal to the number of moles ofNH that react. From the 3 molar mass of NH 4 CI (53 .49 g . morl ) and the number of moles, we can calculate the mass of NH 4 CI that forms : (7.94 x l 0-5 mol NH 4 CI(s))(53 .49 g . mol-I ) = 4.25 x 1 0 -3 g. (b) There will be (2.02 x 1 O-4 mol - 7.94 x 1 O-5mol) = 1 .23 x 1 0-4 mol HC I left after the reaction. This quantity will exist i n a total volume after mixing of 40.0 mL or 0.0400 L. Again, we use the ideal gas law to determine the final pressure: P V = nRT P=
4.59
(1 .23 x 1 0-4 mol)(0.082 06 L · atm · K - I · mol- I ) (300 K)
( a) XHCl _
(
0.0400 L
n HC I n HC
Xbenzene =
+ nbenzene I
)( _
-
= 0. 0757 atm.
)( )
# of HCl molecules _ 9 - 0 . 9 ,. 10 Total # molecules -
1 - 0.9 = 0. 1
-
(b) PHCl = 0.9 x 0 .8 atm = 0.72 atm; Pbenzene = 0. 1 x 0. 8 atm = 0.08 atm; 4.61
(a) The molar volume of an ideal gas is 22.4 L at 273 . 1 5 K. 1 .0 mol of ideal gas will exert a pressure of 1 .0 atm under those conditions. The partial pressure of N? (g) will be 1 .0 atm. Because there are 2.0 mol of H 2 (g) , the partial pressure of H2 (g) will be 2.0 atm. pressure will be 1 .0 atm
+ 2.0 atm = 3.0 atm.
SM - 1 23
(b) The total
4.63
n total = n N2 + n02
.n
'
N2
= -
=
n total =
(
0.02 1 mol
0.02 1 + 0. 1 0 = 0. 1 2 mol;
Xnitrogen =
I f 0.020 mol were released, then n left =
p
O2
=
0. 1 0 mol and nRT
n oxygen
=
( ) 0.02 1
--
0. 1 2
(0. 82) x (0. 1 0 mol)
=
=
0. 1 8;
Xoxygen =
0.82
0.082 mol
(0. 082mol) x (0.08206 L · atm · K- I . mOr l ) X (28 8K) 1 . 00 L
V = 1 .9 atom.
4.65
)
1 atm x (1 .00 L) (0.50 bar) x PV 1 .0 1 325 bar ----' '-----'--- ---= ---RT (0.08206 L . atm . K- l . mor l ) X (28 8 K)
(a) Of the 756.7 Torr measured, 1 7.54 Torr will be due to water vapor. The pressure due to H 2 (g) will, therefore, be
756.7 Torr - 1 7.54 Torr = 739.2 Torr. (b) H 2 0(l) � H 2 (g) + t 0 2 (g).
(c) To answer this question, we must determine the number of moles of H 2 produced in the reaction. Using the partial pressure of H 2 calculated in part (a) and the ideal gas equation, we can set up the following:
(
739.2 Torr 760 Torr · atm - I
]
(0.220 L) = n(0.082 06 L · atm · K - I . mol - I )(293 K)
Solving for n, we obtain n
=
0.008 90 mol. According to the
stoichiometry of the reaction, half as much oxygen as hydrogen should be produced, so the number of moles of 0 2 = 0.004 45 mol. The mass of 0 2 will be given by (0.004 45 mol)(32.00 g . morl ) = 0. 1 42 g.
4.67
No. The reason is that not all the molecules are traveling at the same vel Deity.
4.69
Force
�
( N,": 2 J v,
; if v, vari es, force will be di fTeren!.
Graham ' s law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:
SM- 1 24
rate of effusion
= .lM
Diffusion also follows this relationship. If we have two different gases whose rates of diffusion are measured under identical conditions, we can take the ratio 1
�-
rate, _ ,JM, _ -1 rate2
-
�M 2
M,
M ' I
If a compound takes 1 .24 times as long to diffuse as Kr gas, the rate of diffusion of Kr is 1 .24 times that of the unknown. We can now use the expression to calculate the molar mass of the unknown, given the mass of Kr:
= 83 . 8 0Mg .2 mol- I M2 = 1 29 g . mol - I l .24
A mass of 1 29 g . morl corresponds to a molecular formula of
4.71
C lo H lo '
The rate of effusion is inversely proportional to the square root of the molar mass. Using a ratio as follows allows us to calculate the time of effusion without knowing the exact conditions of pressure and temperature:
The rate will be equal to the number of molecules time interval . For the conditions given, for the second gas chosen.
SM- 1 25
N
N
that effuse in a given
will be the same for argon and
N N
time 1 47 s
=
time 1
39 .95 g . mo l-l
=
M
1 47 s
l
In order to calculate the time of effusion, we need to know only the molar mass of the gases. (a) For CO 2 with a molar mass of 1 44.0 1 g . mol-I :
timeco 2
=
1
1 47 s
3 9.95 g . mol-l 44.0 1 g . mor l
time = 1 54 s.
(b) For C 2 H 4 with a molar mass of 1 28 .05 g . morl
:
time C2H. 1
=
1 47 s
39.95 g . morl 28. 05 g . morl
time = 1 23 s.
(c) For H 2 with a molar mass of 2.0 1 g . mol- I :
time co 2 1
3 9.95 g . mol - l
=
2.0 1 g . mol-l
1 47 s time = 3 3 . 0 s.
(d) For S0 2 with a molar mass of
64.06 g . mol-I :
time co2 1
1 47 s
=
3 9.95 g . mol-l 64.06 g . mol-l
time = 1 86 s.
SM- 1 26
4.73
The fonnula mass of C H 3 is 27.04 g . mOr l . From the effusion data, we 2 can calculate the molar mass of the sample:
Because time is inversely proportional to rate, we can alternatively write: 1 349 s
3 9.95 g . morl
=
1
MI
210 s 210 349 M
I
3 9 .95 g . morl
=
=
M
I . 1 1 0 g mol- I
The molar mass is 4 . 1 times that of the empirical fonnula mass, so the molecular fonnula is C 8 H 1 • 2
4.75
(a) The average kinetic energy is obtained from the expression average kinetic energy
=
t RT.
The value is independent of the nature of the
monatomic ideal gas . The numerical values are: (a) 4 1 03.2 J . mor l ; (c) 4 1 03.2 J . mor l
4.77
(b) 4090 . 7 J . mol- I ; -
4090.7 J . mor l
=
1 2 .5 J . mol-I
The root mean square speed is calculated from the fol lowing equation: vnns
-
�
3RT
-
M
(a) methane, CH4 , M vnns = =
=
1 6.04 g . morl
3 x (8.3 1 4 kg · m 2 · S-2 · K-I · morl ) x (253 K) 627 m · S- I
1 .604 x l 0.2 kg · mol-I
SM - 1 27
(b) ethane, C H 6 , M = 3 0.07 g . mol - I 2
= 45 8 m · s- 1 (c) propane, C 3 Hp M = 44.09 g . marl yrillS =
3 x (8.3 1 4 kg · m2 ' S-2 · K-I · morl ) x (25 3 K) 4.409 X 1 0-2 kg . mol-I
= 378 m ' S - I 4.79
Bas ed on equation
VrlllS =
( )
1 . 3RT / 2 . ' If T IS constant, M
( VrIllS ) (M ) 1/ 2 = ( VrmJ (M ) 1 / 2 , (550. ffi' S - 1 )( 1 6. 06) 1 /2 = (vrms) K1' x I 2 2 1 (83 . 80) 1 12 ( Vrms) Kr = 24 1 m s - 1
Note: You do not need to change M into kg/mol in this calculation.
4.8 1
Use the expression for the root mean square speed to determine the temperature: yrms T
�
3RT M
y2rms . M 3R
=
(1 477 m · s- I )2 · (4.00 x 1 0 -3 kg . morl ) 3(8.3 1 4 J · K- I · morl )
= 349.9 K Use the Maxwell Distribution of Speeds (Equation 4 . 22) appropriately for both gases:
SM- 1 28
1
( 4.00 g . mol-I Y (1 477 m · s - I )2
=
(39.95 g · morl
= =
4.83
(
)1 2
( )
(46 7 m · s- I )2
X
-(4.00 x 1 0-3 kg · mol- ' ) x ( 1 477 m · S-, ) 2 e 2 x (8 3 1 4 J . K- ' mol ' ) x (349 9 K)
---,-:...,--..,.--
-(39.95 x 1 0- 3 kg . mol- ' ) x (467 m · s-' ) 2
-
e
2 x (8.3 1 4 J · K ' . mol-' ) x (349 9 K)
4.00 1 1 4 77 1 (0 . 99 77 ) ) 39.95 46 7 0.3 1 6
(a) The most probable speed is the one that corresponds to the maximum on the distribution curve. (b) The percentage of molecules having the most probable speed decreases as the temperature is raised (the distribution spreads out).
4.85
Hydrogen bonding is important in HF. At low temperatures, this hydrogen bonding causes the molecules of HF to be attracted to each other more strongly, thus lowering the pressure. As the temperature is increased, the hydrogen bonds are broken and the pressure rises more quickly than for an ideal gas. Dimers (2 HF molecules bonded to each other) and chains of HF molecules are known to form.
4.87
The pressures are calculated very simply from the ideal gas law: p=
nR T
V
=
(1 .00 mol)(0.082 06 L · atm · K- I . morl )(298 K)
V
Calculating for the volumes requested, we obtain 48.9 atm;
We can rearrange this to solve for P:
P
( V J ( VJ n RT -
=
(a) 1 .63 atm;
(b)
(c) 489 atm. The calculations can now be repeated using the
van der Waals equation:
=
P
nb
-
an
2
?
SM- 1 29
[ -[ =
( 1 .00 mol)(0.082 06 L · atm · K - I . mol-I )(298 K)
V
-
( 1 .00 mol) (0.04267 L · mol-I )
n;,�mol-' ) ( I .OO' )
(3 . 640 L' . at
J
J
Using the three values for V, we calculate for P = (a) 1 .62;
(b) 3 8 .9;
(c) 1 . 8 8 x 1 03 atm. Note that at low pressures, the ideal gas law gives
essentially the same values as the van der Waals equation, but at high pressures there is a very significant differences.
4.89 a
4.91
(atm · L2 . mor2)
Ammonia: Oxygen: Volume
1 7. 5 8
CH3CN
3 . 3 92
CO2
2.253
CH4
0.2 1 07
Ne
a =
a =
Substance
4.225 e . atm · mol-2 ; b = 0.037 07 L · atm- I
1 .3 7 8 e . atm · mor2 ; b
P, ammonia
=
0.03 1 8 8 3 L · atm-I
P, oxygen
P, ideal
3581
1 897
489
0. 1
81 1
497
245
0.2
256
1 80
1 22
0.3
1 40
1 06
82
0.4
94
75
61
0.5
70
58
49
0.6
55
47
41
0.7
46
39
35
0.8
39
34
31
0.9
34
30
27
1
30
27
24
0.05
SM- 1 3 0
Clearly , the greater deviation from the ideal gas law values occurs at low volumes or higher pressures. Ammonia deviates more strongly and its van der Waals constants are larger than those for oxygen. This may likely arise because ammonia is more polar and will have stronger intermolecular interactions.
4.93
(a) XNe =
( �) l
C�)=0.4 (;, ) (;, ) XAr
=0.6;
At constant T and V, PAr=
(
->
�
)
(0.4)(420. TOrr) =280. Torr 0.6
(b) Ptotal = (280. Torr) 4.95
=
+
(420. Torr)
=
700. Torr
(a) . .
.
. .
. .
O=N- O:
I ___ _
. .
.
: O -N= O
(b) Since the wavelength of the absorbed photons is 197 the energy per photon: Ephoton
=
run,
we can find
( )
he
(6.626XI0-34 J 'S)(2.998 xI08m . s-I) 109 run = 197 run m A =1.008 x l0-18 J
The number of photons in 1.07 m J must be equal to the number ofN02 molecules.
(
)(
1 photo� 1 J 1000 m J 1.008 xl0- 8 J =1.062 x10 15 photons =1.062 x10 15 N0 molecules 2 The pressure is created by all the molecules in the sample, so the ideal gas
? photons =1.07 m J
law can be used to find the total molecules.
SM-131
)
? molecules total N =
=
AV
=
x nlol N
AV
X
(
PV RT
_ _ 10 1_
(6.022 xI023molecules·mol-')
(0.85 atm)(2.5 L) (0.08206 L 'atm . K- ' . mor ')(293 K) = 5.32 x 1022molecules total x
J
Therefore, the proportion ofN02 molecules in the sample is 1.062 x10'5 N0 molecules 2 0.020 ppm. 5.32 x1022molecules total
=
4.97
Use the ideal gas law to calculate the number of moles ofHCI:
n-
.----
-
( 1 L) (690. Torr)(200. m L) PV . (1 atm) RT (0.082 06 L· atm· K-' . m L-' )(293 K) (760 Torr) (1000 m L)
=7.55 x 10-3 molHCI
Since the reaction betweenHCI andNa OH occurs in a 1:1 mole ratio , this number of moles ofNa OH is also present in the volume ofNa OH(aq) required to reach the stoichiometric point of the titration. Therefore, the molarity of theNa OH solution is molesNa OH 1 L of solution = 0.481 M.
(7.55 x10-3 molNa OH) 1000 m L 1 L ( 15.7 m L)
moles can be calculated from the ideal gas equation: p=
nRT V
(
J
43.78 g , (0.082 06 L· atm· K- ' . mor' )(298 K) 92.02 g. mol5.00 L = 2.33 atm (c) The only difference in the calculation between part (b) and part (c) is that the molar mass of N0 is half that of N 04• 2 2
SM- 132
p=
nRT V
( ---
43.78 g 46.01 g . mol
=--1
J
(0.082 06 L . atm· K-1 . mol- I)(298 K)
�-------------------- ------= �--------� =
5.00 L
4.65 atm
(d) Because both N204 and N02 are present , we need to determine some way of calculating the relative amounts of each present. This can be done by taking advantage of the gas law relationships. The total pressure at the end of the reaction will give us the total number of moles present: = I:otal 2.96 atm (2.96 atm)(5.00 L) = ntotal(0.082 06 L . atm· K-1 . mol-I )(298 ntotal = 0.605 mol = :. nN 04 + nNO 0.605 mol 2
K)
2
This gives us one equation, but we have two unknowns , so another relationship is needed. We can take advantage of knowing the stoichiometry of the reaction. If we assume that all of the gas begins at
N204 and we allow some to react, we can write the following: Initial amount of N204 0.476 mol Amount of N204 that reacts x(mol) Amount of N02 formed 2x (mol) When the reaction is completed , there will be 0.476 -x mole of N204 and 2x mole N02• The total number of moles will be given by: (0.476 - x) + 2x = ntotal 0.605 mol = 0.476 mol + x x = 0.129 mol =
2(0. 129 mol) = 0.258 mol
SM- 133
= 0.476 mol - x = 0.347 mol XN
02
XN20
4.101
.
0.2S8 mol = 0.426 0. 60S mol = 0.347 mol = 0.S74 0.60S mol
=
(a) The elemental analyses yield an empirical fonnula of NH2• The fonnula unit has a mass of 16.02 g . mol- I. The mass , volume , pressure, and temperature data will allow us to calculate the molar mass , using the ideal gas equation:
PV = nRT PV = �RT M (0.473 g)(0.082 06 L· atm· K-I . morl )(298 M = mRT = PV (1.81 atm)(0.200 L)
K)
= 31.9 g . mol-I
The molar mass divided by the mass of the empirical fonnula mass will give the value of n in the fonnula (NHJn·31.9 g . mol-I N 2H4,
7 16.02 g . mol-I =1.99 , so the molecular fonnula is
which corresponds to the molecule known as hydrazine.
(b) H-N-N-H I
H
I
H
(c)
ra__ te A_ r ateB
3.S x 10-4 mol IS.0 min X
-----
�
_M_B MA
. l = 32.0S g mor
2S.0 min X
� [
17. 03 g . morl
]
3.S x 1O-4 mol 17.03 g . morl (2S.0 min) . 32.0S g . morl IS.0 m m = 4.2 x 10-4 mol
=
SM-134
4.103
Using vrms = root mean s quare speed and E quation 4.22 (the Maxwell Distribution of Speeds): f( 1 0vrmJ f (vrmJ
1 M ( ) 47rN -2RT - 47rN ( � )1
2 ( 1 0vrms ) e
_
2RT
-M(IOv",,)2 j
j2RT
2 -M(V'''')/{ RT ( vrms) e
The ratio is not independent of temperature since the variable Tappears in the denominator of a negative exponent on e. It makes sense that the ratio should become bigger at higher temperatures as the distribution spreads out such that the number of molecules with higher speeds increases while the number with lower speeds decreases. 4.105
The two scents will dif fuse according to E quation 4.17a:
.
rate fru_It-,-Y rate. mlllty _ _
#J
MC,H,02 36 = _ = 0. 889 MCItIH2,,02 172
=
Let x = the distance traveled by ethyl octanoate (fruity) and y the =
distance traveled by p-anisaldehyde (minty) in the same amount of time. Then x + y =5 m
and
x 0 . 889 m = ,or x 1m y
==
O.889y.
=
Substituting for x gives 0 . 8 89 y + y 5 1.8 89y = 5 Y
=
2.65.
So the fruity smell will travel 5-2.65 2.35 m in the same time that the =
minty smell will travel 2.65 m. A person must stand more than 2.35 m
SM-135
away from the north end of the room where the fruity smell originates in order to smell the minty scent first. 4.107
The molar mass calculation follows from the ideal gas law:
PV = n RT PV = .!!!.- RT M 06 L· atm· K-1 morl)(473 � M - mRT (l.509 g)(0.082 PV 745 Torr ...-= -- - ...-'.-
(
J
K)
---------''----'--:-
760 Torr· atm-I
= 254 g . mol-I
•
(0.235 L)
If the molecular formula is Os O then the molar mass will be given by x,
=
190.2 g . mol-I + x( l 6 . 00 g . mol-I) 254 g . mol-I x = 3.99.
The formula is Os O
4.109
4.
In this problem, the volume, pressure , and molar mass of the substance stay constant. In order to calculate the new mass with the same conditions, we can resort to using the ideal gas e quation rearranged to group the constant terms on one side of the e quation:
PV = nRT PV = .!!!.- RT M But M, P, and Vare constants, so we can write MPV = mT. R Now we have two sets of conditions, 1 and 2, for which
MPV is constant, R
so we can set them e qual: (46.2 g)(300. K) = (m 2 )(600 . K), therefore m2 = 23.1 g The mass of gas released must therefore be 46.2 g -23.1 g = 23.1 g.
-
SM 13 6
4.111
(a) volume of one atom = molar volume -;- Avogadro's number: 2 . 370 X 1 0-2 L · mor l -;- 6 . 022 x1 0 23 atoms · mor l = 3 . 936 x1 0-26 L · atom-I 3 . 936 x1 0-2 6 L . atom-I x1 000 cm 3 . L-I = 3 . 936 X 1 0- 23 cm 3 . atm-I 3 . 936 x1 0-23 cm 3 · atm-I x ( 1 01 0 pm · cm-I ) 3 = 3 . 936 x1 07 pm 3 3 . 936 x 1 07 pm 3 = � Jrr3 3 r = 2 1 1 pm (b) The atomic radius of He is 1 28 pm (Appendix 2D). The volume of the He atom, based on this radius, is 4
V = -7rr 3 3 = � 7r(1 28 pm) 3 3 = 8 . 78 x1 0 6 pm 3 . ( c) The difference in these values illustrates that there is no easy definition for the boundaries of an atom . The van der Waals value obtained from the correction for molar volume is considerably larger than the atomic radius, owing perhaps to longer range and weak interactions between atoms. One should also bear in mind that the value for the van der Waals b is a parameter used to obtain a good fit to a curve, and its interpretation is more complicated than a simple molar volume . 4.113
(a) ClN02 (b) ClN0 2 (c) : Cl: .. I O= -O: N .. .. ..
:CI: I .. :O=O N .. .. ..
(d) trigonal planar
SM- 1 37
4.115
(a) Substitute the van der (vdW) parameters for ammonia as well as the given values of n, R, P, and T into the vdW equation and then solve for V. Since the equation is cubic, solve graphically for the three roots or use an appropriate program such as Math Cad . Only one of the three roots is physically possible: V3 + n
( RT; hP ) V2 + ( n:a
V 3 + (0 . 505 mol)
( (
J
;
v - n h �O
J
0 . 08206 L · atm · K-' · mor' )(298 K) + (3 .707 L · mor' )(95 . 0 atm) 2 V (95 . 0 atm) 0 . 505 mol) 2 (4 . 225 e . atm . mol-2 ) v + (95 . 0 atm) (0 . 505 mol) 3 (4 . 225 e · atm · mol-2 )(3.707 L · mol-' ) =0 (95 . 0 atm) V 3 + (2 . 002 L) V 2 + (0 . 0 1 1 34 e)V - 0 . 02 1 23 e = 0 V = -1 . 99 1 , or - 0. 1 089, or 0 . 09789 L but only the positive root is physically possible, so V= 0 . 0979 L x
_
J
(b) Compare the volume calculated in part (a) to that of an ideal gas under the same conditions: PV = nRT (0 505 mol)(0 . 08206 L · atm · K-I . mol-I )(298 K) V= . = 0 . 1 30 L 95 . 0 atm Videal 0. 1 30 L> VvdW, 0 . 0979 L. Attractive forces dominate because the van der Waals, or "real," gas occupies less volume than the "ideal" gas . If the molecules are attracted to one another, they will behave less independently, reducing the effective number of moles of gas .
=
=
SM - 1 38
CHAPTERS LIQUIDS AND SOLIDS
5.1
(a) London forces, dipole-dipole, hydrogen bonding; (b) London forces, dipole-dipole, hydrogen bonding; (c) London forces, dipole-dipole; (d) London forces
5.3
Only (b) CH 3 CI, (c) CH 2 CI 2 , and (d) CHCl 3 will have dipole dipole interactions . The molecules CH4 and CCl4 do not have dipole moments.
s.s
The interaction energies can be ordered based on the relationship the energy has to the distance separating the interacting species . Thus, ion- ion interactions are the strongest and are directly proportional to the distance separating the two interacting species . Ion-dipole energies are inversely proportional to d 2 , whereas dipole-dipole for constrained molecules (i.e . , solid state) is inversely proportional to d 3 • Dipole- dipole interactions where the molecules are free to rotate become comparable to induced dipole-induced dipole interactions, which are both inversely related to d 6 • The order thus derived is (b) dipole-induced dipole � (c) dipole-dipole in the gas phase < (e) dipole-dipole in the solid phase < (a) ion- dipole < (d) lOn-IOn .
5.7
Only molecules with H attached to the electronegative atoms F, N, and 0 can hydrogen bond . Additionally, there must be lone pairs available for the H's to bond to. This is true only of (c) H 2 S0 3 • SM- 1 39
5.9
I I because the dipoles are aligned with oppositely charged ends closest to each other thereby maximizing dipole-dipole attractions.
5.11
(a) NaCl (801°C vs. -114.8°C) because it is an ionic compound as opposed to a molecular compound; (b) butanol ( -90°C vs. -l16°C) due to hydrogen bonding in butanol that is not possible in diethyl ether; (c) triiodomethane because it will have much stronger London dispersion forces ( -82.2°C for tri fluoromethane vs. 219°C for triiodomethane); (d) Methanol ( -94 °C vs. - 169 °C) because of the hydrogen bonding in methanol but not in ethylene.
5.13
(a) PF 3 and P Br 3 are both trigonal pyramidal and should have similar intermolecular forces, but P Br 3 has the greater number of electrons and should have the higher boiling point. The boiling point of PF 3 is -101.5°C and that of P Br 3 is 173.2 °C. (b) S0 is bent and has a 2 dipole moment, whereas C O is linear and will be nonpolar. S0 should 2 2 have the higher boiling point. S0 boils at -10°C, whereas C O 2 2 sublimes at -78°C. (c) BF 3 and BCl 3 are both trigonal planar , so the choice of higher boiling point depends on the difference in total number of electrons. BCl 3 should have the higher boiling point (12.5°C vs. -99.9°C).
5.15
The ionic radius of A1 3+ is 53 pm and that of Be2+ is 27 pm. The ratio of energies will be given by
SM-140
E p8e2+
oc
(-I d2 1 Jl) = ( (27-+12100)1Jl 2 ) . z
The electric dipole moment of the water molecule (/-l) will cancel:
.
RatIo
2 ] -_ ( 3(127)2 -_ 1. 03 [EE--pAI2+)_- ( --112I3IJlJl//(53+100) (27+100) 2 2(153)2 J p8e
2+
The attraction of the A13+ ion will be greater than that of the Be2+ ion. Even though the A1 3+ ion has a larger radius, its charge is higher than that of Be2+, making the attraction greater over all. 5.17
(a) Xenon is larger, with more electrons, giving rise to larger London forces that increase the melting point. (b) Hydrogen bonding in water causes the molecules to be held together more tightly than in diethyl ether. (c) Both molecules have the same molar mass, but pentane is a linear molecule compared to dimethylpropane, which is a compact, spherical molecule. The compactness of the dimethylpropane gives it a lower surface area. That means that the intermolecular attractive forces, which are of the same type ( London forces) for both molecules, will have a larger effect for pentane.
5.19
5.21
F
=
-dE dr
p
=
(�) -(-�)
-d dr r
=
r
oc
� r
(a) cis-Dichloroethene is polar, whereas tr ans-dichloroethene, whose individual bond dipole moments cancel, is nonpolar. Therefore, cis dichloroethene has the greater intermolecular forces and the greater surface tension. (b) Surface tension of li quids decreases with increasing temperature as a result of thermal motion as temperature rises. Increased thermal motion allows the molecules to more easily break away from each other, which manifests itself as decreased surface tension.
SM-141
5.23
At S O°C all three compounds are li quids. C6H6 (nonpolar) (polar, but no hydrogen bonding)
<
<
C6H5S H
C6H50 H (polar and with hydrogen
bonding).The viscosity will show the same ordering as the boiling points, which are 80°C for C6H6,169°C for C6H5S H,and 182° for C6H 50 H. 5.25
C H4, - 162°C; C H3C H3, -88.SoC; (C H3)2C HC H2C H3, 28°C; C H3(C H2)3C H3,36°C; C H30 H, 64.SoC; C H3C H20 H, 78.3 °C; C H3C H O HC H3, 82.SoC; CS H90 H (cyclic, but not aromatic), 140°C; C6 HsC H30 H (aromatic ring), 20SoC; O HC H2C H O HC H20 H, 290°C
5.27
(a) hydrogen bonding; (b) London dispersion forces increase
5.29
Using
-
h
2 = y we can calculate the height. For water:
gdr
1 2
- [ J ( J[ � J
1 2
1m = 7.S x 10-5 m 1000 mm 6 3 1 kg 1 0 m 3 d = 0.997 g . cm= 9.97 X 102 kg· m- 3 m 1000 g 2(72.7S x 10- 3 N 'm-I) h= = 0.20 m or 200 mm (9.81 m· S-I)(9.97 x 102 kg· m- 3)(7.S x 10-5 m)
r = di ameter = ( O. l S mm)
Remember that 1 N = 1 kg· m-I . S-2 For ethanol:
[ J[
J
6 10 �m 3 =7.9 X102 kg . m- 3 m 2 (22.8x10-3 N . m- I ) = 0.078 m or 78 h= (9.81 m·s-I ) (7.9x 102 kg·m-3 )(7.S x 10-5 m)
d =0.79 g . cm-3
1 kg 1000 g
mm
Water will rise to a higher level than ethanol. There are two opposing effects to consider. While the greater density of water, as compared to ethanol, acts against it rising as high, it has a much higher surface tension.
SM-142
5.31
Glucose will be held in the solid by London forces, dipole-dipole interactions, and hydrogen bonds; benzophenone will be held in the solid by dipole-dipole interactions and London forces; methane will be held together by London forces only. London forces are strongest in benzophenone, but glucose can experience hydrogen bonding, which is a strong interaction and dominates intermolecular forces. Methane has few electrons so experiences only weak London forces. We would expect the melting points to increase in the order C H4 (m.p. <
5.33
=
=
- 182°C)
benzophenone (m.p. 48°C) < glucose (m.p. = 148 - 155°C).
(a) network; (b) IOmc; (c) molecular; (d) molecular; (e) network
5.35
(a) At center: 1 center 8 corners
x
x
1 atom· center-I =1 atom; at 8 comers,
t atom· corner-I
=
1 atom; total = 2 atoms. (b) There are
eight nearest neighbors, hence a coordination number of 8. (c) The direction along which atoms touch each other is the body diagonal of the unit cell. This body diagonal will be composed of four times the radius of the atom. In terms of the unit cell edge length a, the body diagonal will be
J3 a. The unit cell edge length will, therefore, be given by 4r
5.37
=
3 a or '"r;:;
a
=
4r
-
J3
=
4·( 124 pm)
J3
=
286 pm.
(a) a = length of side for a unit cell; for an fcc unit cell, a = 18 r, or 2.fi r = 404 pm. v
a 3 = (404 pm x10 - 2 m· pm-1) 3 = 6.59 x l0-29 m' = 6.59 x 10-23cm 3. Because for an fcc unit cell there are four atoms per unit cell, we have =
1
SM- 143
1 mol Al atoms 2 6.022 x 10 3 atoms· mol-
mass(g) = 4 Al atoms x
1
X
26.98 g mol Al atoms
-=--
---
=1.79 x 10-22 g d=
1.79 xl O-22 g =2.72 g· em-3 . 6.59 x10-23 em3
( b)
a=
v
4r
13
=
4 x 235 pm
13
= 543 pm
= (543 X 10- 12 m)3 =1.60 x 10-28 m3 =1.60 x 10-22 em3
There are two atoms per b e e unit eell: mass(g) = 2 K atoms x
1 mol K atoms 6.022 x 1023 atoms· mol- 1
X
39. 10 g mol K atoms
-=---
= 1.30 x 10-22 g 22 d = 1.30 x 10- g =0.8 13 g . em-3 1.60 x 10-22 em3 5.39
a = length of unit cell edge v
=
mass of unit cell d
(a)
(
)(
)(
)
195.09 g Pt Im �I Pt 4 atoms ( 1 unit eell) 6.022 x 10 -3 atoms Pt 1 unit eell -- mol Pt ----,------'---'------' -V = a3 = --------'-------'-----'21. 450 g . em3 a = 3.92 x 10-8 em r;; Be eause for an fee eell, a = v'8 r, r =
-J2a = -J2(3.92 x l0-8 em)
----- ---'4 4 = 1.39 X10-8 em =139 pm.
(b) V = a3
(
--
)(
)(
)
2 atoms 1 mol Ta 180.95 g Ta . ( 1 umt ee11) 1-- mol Ta 6.022 x 1023 atoms Ta 1 -----'--unit cell = --------'----�---'--------�----'16.654 g . em3 = 3.6 1 X 1 0-23 em3
SM- 144
a = 3.30 x 10-8 cm J3 a J3(3.30 x10-8 cm ) r =
5.41
-4
=
4
=(
=
1.43 X 10-8 cm = 143 pm
(a) The volume of the unit cell is Vunilcell
543 pm x
=
)
10-I ?- m 100 cm 3 = 1.601 x 10-22 cm 3 x pm m
mass in unit cell ( 1.601 x 10-22 cm 3) x(2.33 g . cm- 3) = 3.73 x10-22g. (b) The mass of a Si atom is 28.09 g . mol-I -7- (6.022 x1023) atoms·morl
=
4.665 x10-2 3 g . atom- I.
Therefore, there are 3.73 xI0-22g -7- 4.665 x 10-23 g . atom-I = 8 atoms per unit cell. 5.43
volume of a cylinder = base area x length
=
m2 1
volume of a triangular prism = base area x length
=
(bhl)12
=
-V3r21 for a
triangle inscribed in the centers of three touching cylinder bases Each of the inscribed triangle's 60° angles accounts for 1/6 of the volume of each of the three touching cylinders, or a total cylinder volume of (3/6) m2 I. So the percent of the space occupied by the cylinders is (0.5) m2 1 *100/ -V3r21 = 0.5n/-V3 90.7%. =
5.45
(a) There are eight chloride ions at the eight comers, giving a total of 8 comers x t atom ·corner-I =1 cr ion. There is one Cs that lies at the center of the unit cell. All of this ion +
belongs to the unit cell. The ratio is thus 1 : 1 for an empirical formula of CsCI, with one formula unit per unit cell. (b) The titanium atoms lie at the comers of the unit cell and at the body center: 8 comers x t atom·corner- I + 1 at body center = 2 atoms per unit cell.
SM-145
Four oxygen atoms lie on the faces of the unit cell and two lie completely within the unit cell, giving: 4 atoms in faces
x
1 atom·face-I + 2 atoms wholly within cell 4 atoms =
The ratio is thus twoTi per four 0, or an empirical fonnula of Ti0 with 2 two fonnula units per unit cell (c)TheTi atoms are 6-coordinate and the ° atoms are 3-coordinate. 5.47
Y:
x
8 atoms
t atom·comer-I = 1 Y atom
Ba: 8 atoms x ± atom·edge-I = 2 Ba atoms Cu: 3 Cu atoms completely inside unit cell 3 Cu atoms =
0: 10 atoms on faces cell
549 .
=
x
1 atom·face-I + 2 atoms completely inside unit
7 ° atoms
. 149 pm . ·h (8,8) = 1. 1 2, pred· (a) ratIO = lct ceSlUm-chlon·de structure WIt 133 pm coordination; however, rubidium fluoride actually adopts the rock-salt structure (b) ratio
=
72 pm 140 pm
=
0.5 1, predict rock-salt structure with (6,6)
=
0.520, predict rock-salt structure with (6,6)
coordination (c) ratio
=
102 pm 196 pm
coordination 5.51
(a) In the rock-salt structure, the unit cell edge length is e qual to two times the radius of the cation plus two times the radius of the anion.Thus, for Ca O,
a =
2( 100 pm) + 2( 140 pm) = 480 pm. The volume of the
SM-146
unit cell will be given by (converting to cm 3 because density is nonnally given in tenns of g . cm-3 )
=(
V 480 pm x
)=
10- 12 m 100 cm 3 x pm m
1 1 1 x 10-22 cm 3. .
There are four fonnula units in the unit cell, so the mass in the unit cell will be given by
(
)(
)
4 fonnula units 56.08 g cao x 1 unit cell 1 mol Ca O . . mass mumt cell = =3.725 x10-22 g. 6.022 x 1023 molecules·mol- I The density will be given by the mass in the unit cell divided by the volume of the unit cell: d
3.725 xl0-22 g = 3.36 g·cm-3 1.11 x 10-22 cm3
(b) For a cesium chloride-like structure, it is the body diagonal that represents two times the radius of the cation and plus two times the radius of the anion. Thus, the body diagonal for Cs Br is equal to 2 (170 pm) + 2 (196) = 732 pm. For a cubic cell, the body diagonal = J3 a = 732 pm a = 423 pm
)
(
10- 12 m l Oa cm 3 = 3 7.57 x 10-23 cm 3. x a = V = 423 pm x m pm There is one fonnula unit of Cs Br in the unit cell, so the mass in the unit cell will be given by
(
)(
)
1 fonnula units 212.82 g CS Br 1 unit cell 1 mol Cs Br = 3.534 x l0-22 g mass in unit cell = 6.022 x 1023 molecules·mol- I d=
5.53
3.534 X 10-22 g = 4.67 g . cm3. 7.57 X 10-23 cm3
(a) the smallest possible rectangular unit cell is as following:
SM-147
(b) There are four carbon atoms in each unit cell. (c) The coordination number is 3 (C3). 5.55
Each unit cell contains 4 Rb I; total of unit cells = (6.022 molecules/mole)/(4 Rb I molecules/unit cell) mole; volume of one unit cell (732.6 pm)3
=
=
x
1023 Rb I
1.505 1023 unit cells/ x
Assume that the edge length of a cubic crystal of Rb I that contains one mole Rb I is X, the volume of one mole Rb I
[ 1.505
X
1023 unit cell );
.xJ/mole
=
[
(732.6 pm)3 unit cell
)
x
mole
X = (732.6
5.57
=
pm)
x
\/1 . 505 1023 x
3.897
X
1010 pm
=
3.897 cm.
(a) The alloy is undoubtedly interstitial because the atomic radius of nitrogen is much smaller (74 pm vs. 124 pm) than that of iron. The rule of thumb is that the solute atom be less than 60% the solvent atom in radius in order for an interstitial alloy to fOlm. That criterion is met here. (b) We expect that nitriding will make iron harder and stronger, with a lower electrical conductivity.
5.59
(a) These problems are most easily solved by assuming substance. In
100 g of
100 g ofNi-Cu alloy there will be 25 gNi and 75 g Cu, SM- 148
corresponding to 0.43 molNi and 1.2 mol Cu. The atom ratio will be the same as the mole ratio: 1.2 mol Cu 0.43 molNi
=
2 8 C u per N·1 .
(b) Pewter, which is 7% Sb, 3% Cu, and 90% Sn, will contain 7 g Sb, 3 g Cu, and 90 g Sn per 100 g of alloy. This will correspond to 0.06 mol Sb, 0.05 mol Cu, and 0.76 mol Sn per 100 g. The atom ratio will be 15 Sn : 1.2 Sb : 1 Cu. 5.61
There are too many ways that these molecules can rotate and twist so that they do not remain rod-like. The molecular backbone when the molecule is stretched out is rod-like, but the molecules tend to curl up on themselves, destroying any possibility of long-range order with neighboring molecules. This is partly due to the fact that the molecules have only single bonds that allow rotation about the bonds, so that each molecule can adopt many configurations. If multiple bonds are present, the bonding is more rigid.
5.63
Use of a nonpolar solvent such as hexane or benzene (etc.) in place of water should give rise to the formation of inverse micelles.
5.65
Because benzene is an isotropic solvent and its viscosity is the same in every direction. However, a liquid crystal solvent is an anisotropic solvent and its viscosity is smaller in the direction parallel to the long axis of the molecule than the "sideway" direction. Since methylbenzene is a small sphere molecule, its dif fusion movement is not affected greatly by whether its solvent is isotropic or anisotropic.
5.67
(a) TheN I3 Lewis structure is
SM-149
N
: X/ I "':�: :I :
. The molecular shape is trigonal pyramidal and it is a
polar molecule. It can participate dipole-dipole interactions. (b) The Bh Lewis structure is . .. I . / ... .
:L�
B
I
:I :
. The molecular shape is trigonal planar and it is a
non-polar molecule. It can not participate dipole-dipole interactions. 5.69
(a) Pentane and 2,2-dimethylpropane are isomers; both have the chemical formula CsHl . We will assume that 2,2-dimethylpropane is roughly 2 spherical and that all the hydrogen atoms lie on this sphere. The surface area of a sphere is given by A = 4Jrr 2 . For this particular sphere, A = 4Jr(2S4 pm)2 = 8.1 1 x 105 pm 2. For pentane, the surface area of rectangular prism is 2 (295 pm x 766 pm) + 2 (295 pm x 254 pm) + 2 (254 pm x 766 pm) = 9.91 x 105 pm2. (b) Pentane has the larger surface area. (c) The pentane should have the higher boiling point. It has a signi ficantly larger surface area and should have stronger intermolecular forces between the molecules.
5.71
(a) anthracene, CI4 H 10 H I
H I
H I
SM-150
London forces (b) phosgene, C OCl2 :0:
.. II .. :C..l - C-Cl: .. Dipole-dipole forces, London forces (c) glutamic acid, CS H9N04 :0:
II
..
H2N-C H - C- O.. H
I
C H2
I
C H2
I
..
C= O""
I
:O H Hydrogen bonding, dipole-dipole forces, London forces 5.73
The unit cell for a cubic close-packed lattice is the fcc unit cell. For this cell, the relation between the radius of the atom r and the unit cell edge length a is
4r = 12a 4r
a=
12"
The volume of the unit cell is given by
(
)3
If r is given in pm, then a conversion factor to cm is required:
V = a3
=
12
4r x 10- m x 100
12
pm
em
m
Because there are four atoms per fcc unit cell, the mass in the unit cell will be given by SM-151
mass
=
(
4 atoms unit cell
(J
M 3 6.022 x102 atoms·mol-I
The density will be given by d
=
=
mass of unit cell volume of unit cell
=
(
( J
J
.
M 4 atoms x 3 2 6.022 x 10 atoms· mol- I unit cell 3 4 r 10- 12 m 100 cm - x x --J2 pm m
(
J
J
(2.936 x 105 )M
or r
=
(2.936 x l05 )M --3 --'--------'d
where Mis the atomic mass in g . morl and r is the radius in pm. For the different gases we calculate the results given in the following table: Gas
Neon
5.75
3 Density (g. em )
J Molar mass (g. mor )
Radius (pm)
1.20
20.18
170
Argon
1.40
39.95
203
Krypton
2. 16
83.80
225
Xenon
2.83
13 1.30
239
Radon
4.4
222
246
There are two approaches to this problem. The information given does not specify the radius of the tungsten atom. This value can be looked up in Appendix 2 D. We can calculate an answer, however, based simply on the fact that the density is19.3 g . cm-3 for the bcc cell, by taking the ratio between the expected densities, based on the assumption that the atomic radius of tungsten will be the same for both. The unit cell for a cubic
SM- 152
close-packed lattice is the fcc unit cell. For this cell, the relation between the radius of the atom r and the unit cell edge length a is 4r =.J2 a 4r a= . .J2 The volume of the unit cell is given by
If r is given in pm, then a conversion factor to cm is required:
J
(
10- 12 m 100 cm 3 � 3 x = = a V x .J2 pm m Because there are four atoms per fcc unit cell, the mass in the unit cell is given by mass =
(
4 atoms unit cell
J
(
J
M . 2 6.022 x 10 3 atoms·mol- I
The density is given by
d=
=
mass of unit cell volume of unit cell (2.936
or r= 3
x
105 )M
=
[
J (-
4 atoms unit cell
(
J ---J3
M 6.022 x 1023 atoms· mol-I 4 r 10- 12 m 100 cm x x .J2 m pm
---(2.936 x 105)M
d
where Mis the atomic mass in g . morl and r is the radius in pm. Likewise, for a body-centered cubic lattice there will be two atoms per unit cell. For this cell, the relationship between the radius of the atom r and the unit cell edge length a is derived from the body diagonal of the
SM-153
cell, which is e qual to four times the radius of the atom. The body diagonal is found from the Pythagorean theorem to be e qual to the 13a : 4r
=
a =
J3 a
4r -
J3
The volume of the unit cell is given by
( ) (
4r 3
V = a3 =
J3
]
If r is given in pm, then a conversion factor to cm is re quired:
V = a3
=
1 0- 1 2 m 1 00 cm 3 x m pm J3
4r
x
Because there are two atoms per bcc unit cell, the mass in the unit cell will be given by mass
=
(
2 atoms unit cell
]( x
M
6.022 xl023 atoms·mol-I
The density will be given by d
. ce11 mass 0f umt = volume of unit cell
=
(�n�:o:�] ( (
=
(2.698
.
�
6.022 x 1023 toms . mOl-I 4r 10-1 2 m 100 cm 3 x x m pm J3
( �n�:O:� ---- ) ( ----------� -- )
= �
]
---]
]
6 022 x 1 O" toms . mor' ---� (2.309 x 10-10 r)3
��
x
105 )
M
or
Setting these two expressions e qual and cubing both sides, we obtain
SM-154
The molar mass of tungsten Mis the same for both ratios and will cancel from the e quation:
dfee
dbee
Rearranging, we get d (2.936 x105 ) = fcc (2.698xl0 5) dbee =
1.088 dbee•
For
5.77
W,
dfee =1.088x19.6 g . cm3 =
21.3 g·cm-3 .
(a) The oxidation state of the titanium atoms must balance the charge on 'the oxide ions, 02-. The presence of 1.18 02- ions means that theTi present must have a charge to compensate the -2.36 charge on the oxide ions. The average oxidation state ofTi is thus +2.36.
(b) This is most
easily solved by setting up a set of two e quations in two unknowns. We know that the total charge on the titanium atoms present must e qual 2.36, so if we multiply the charge on each type of titanium by the fraction of titanium present in that oxidation state and sum the values, we should get 2.36:
=
let x fraction of Ti2+, y = fraction of Ti3+, then 2x + 3y = 2.36 Also, because we are assuming all the titanium is either +2 or +3, the fractions of each present must add up to 1: x+ y =1 Solving these two e quations simultaneously, we obtain y =0.36, x = 0.64. 5.79
(a) True. If this is not the case, the unit cell will not match with other unit cells of the same type when stacked to form the entire lattice. (b) False. Unit cells do not have to have atoms at the comers.
SM-I55
(c) True. In order for the unit cell to repeat properly, opposite faces must have the same composition. (d) False. If one face is centered, the opposing face must be centered, but the other faces do not necessarily have to be centered. 5.81
There are several ways to draw unit cells that will repeat to generate the entire lattice. Some examples are shown below. The choice of unit cell is determined by conventions that are beyond the scope of this text (the smallest unit cell that indicates all of the symmetry present in the lattice is typically the one of choice). (a)
(b)
I K:\:
•
8·� •
5.83
L/ .
•
(a) The face diagonal the unit cell e quals to 4r (r is the radius of the buckminster fullerene). Therefore, (4r i = 2(142pm l Solve the e quation: 4r
=
.fi (142) pm; r = 50.2 pm
(b) The radius of K+ is 133 pm. In close-packed cubic unit cell (face centered cubic unit cell), the tetrahedral holes are smaller than the octahedral holes. The sizes of tetrahedral holes can be calculated as follows: One unit cell can be divided into eight subunit cells and each subunit cell contains one tetrahedral hole. The face diagonal of each
SM-156
subunit cell J'
=
(;) + (�J '
(a is the edge !ength); the body
The radius of the tetrahedral hole is rtetrahedral + rbuckminsterfullerene rtetrahedral
=
2 1 3pm 2
-
50.20 pm
=
diagona!
= 2!2.
11.3 pm which is not large enough to K+
ion. Therefore, K+ must take octahedral holes. Each unit cell has four octahedral holes and three are occupied by the K+. The percentage of holes filled is 75%. 5.85
Fused silica, also known as fused quartz, is predominantly Si0 with very 2 few impurities. This glass is the most refractory, which means it can be used at the highest temperatures. Quartz vessels are routinely used for reactions that must be carried out at temperatures up to 1000°e. Vycor, which is 96% Si02 , can be used at up to ca. 900°C , and normal borosilicate glasses at up to approximately 550°C . This is due to the fact that the glasses melt at lower temperatures, as the amount of materials other than Si02 increases. The borosilicate glasses (Pyrex or Kimax) are commonly used because the lower softening points allow them to be more easily molded and shaped into different types of glass objects, such as reaction flasks, beakers, and other types of laboratory and technical glassware.
5.87
For a bcc unit cell (see Example 5.3), d = 33/2 (55.85 g/mol)/32(6.022x1023/mole)x(1.24xl 0-8cm)3 7.90 g/cm3 =
(1.5 cm i x 7.90 g/cm3 x (1 mole/55.85 g) 0.477 mole Fe =
(1.00 atm x15.5 L) / (0.08206
L atm /K mol) (298K) = 0.634
SM-157
mol O 2.
Fe is L R, so (0.5)(0.477 mole)(159.70 glmole) = 38.1 g Fe203. This is the theoretical yield, or maximum mass that can be produced. 5.89
For M, there is a total of one atom in the unit cell from the comers; x
cations: 8 comers
t atom·comer-I + 6 faces
+
+ atom · face-I
=
4 atoms ;
anions: 8 tetrahedral holes x1 atom·tetrahedral hole-I =8 atoms. The cation to anion ratio is thus 4 : 8 or 1 : 2; the empirical formula is
M�. 5.91
The cesium chloride lattice is a simple cubic lattice of CI- ions with a Cs + at the center of the unit cell (see 5.43). In the unit cell, there is a total of one
cr
ion and one Cs + ion. If the density is 3.988 g . cm-3 , then we
can determine the volume and unit cell edge length. The molar mass of CsCI is 168.36 g·mol-I.
3.988 g·cm-3
a
3
(
=
(
168.36 g . mol-I 6.022 x 10 23 formula units· mol-I a3
168.36 g·morl 6.022 x 1023 formula units· morl 3.988 g . cm -3
J
J
= �--------------�------�
4.12 x 10-8 em = 412 pm
a =
The volume of the unit cell is (412 pm)3 = 6.99 x 107 pm3 • We will determine the size of the Cs + and Cl- ions from ionic radii given in Appendix 2D, but we can check these values against the unit cell dimensions. For this type of unit cell, the body diagonal will be e qual to 2 r(Cs +) + 2 r(CI- ) = a J3 = 714 pm. The sum of the ionic radii gives us 2 (170 pm) + 2 (181 pm) 702 pm, which is in very good agreement. =
Note that we cannot calculate the size of these ions independently from the
SM-158
unit cell data without more information because this lattice is not c1osepacked. We will assume that the ions are spherical. The volume occupied in the unit cell will be 4 3
4 3
VCs =-nr3 =- n(170 pm)3 =2.06xl07 pm 3 +
The total occupied volume in the cell is 2.06 x107 pm3 + 2.48xl07 pm3 = 4.54 x107 pm3. The empty space is 6.99 x107 pm3 -4.54 x 107 pm3 =2.45x107 pm3. The percent empty space is 2.45x107 pm3 -;-6.99xl 07 pm3 xl 00 =35%. 5.93
(a) In a simple cubic unit cell there is a total of one atom. The volume of the atom is given by t nr 3. The volume of the unit cell is given by a3 and a = 2r, so the volume of the unit cell is 8r3. The fraction of occupied space in the unit cell is given by 4 3 4 -nr -n 3-- - _ 3 _ - 0.52. 3 8r 8 52% of the space is occupied, so 48% of this unit cell would be empty. (b) The percentage of empty space in an fcc unit cell is 26%, so the fcc cell is much more efficient at occupying the space available.
5.95
The Lewis structure of (HF) 3 chain is as follows: H-F: --- H-F: --- H-F:
The bond angle is 1800• �
5.97
(a) This oxidation will increase the electrical conductivity of graphite because of the mobile HS04- anions.
SM-159
(b) This oxidation will increase the spacing, d, due to the HS04- . Based on the Bragg equation, 2d sine
=
A,
the angle of the x-ray beam will
decrease (if the wavelength is the same). 5.99
The lattice layers from which constructive x-ray diffraction occurs are parallel . First draw perpendicular lines from the point of intersection of the top x-ray with the lattice plane to the lower x-ray for both the incident and diffracted rays. The x-rays are in phase and parallel at point A . If we want them to still be in phase and parallel when they exit the crystal, then they must still be in phase when they reach point C. In order for this to be true, the extra distance that the second beam travels with respect to the first must be equal to some integral number of wave-lengths . The total extra distance traveled, A
----jo
B
----jo
C, is equal to 2x. From the diagram, we can
see that the angle A-D-B must also be equal to e . The angles e and sum to 90° , as do the angles and x = d sin
e.
a
a
and A -D-B. We can then write sin e = �
The total distance traveled is 2 x
=
d
2 d sin e. So, for the
two x-rays to be in phase as they exit the crystal, 2 d sin e must be equal to an integral number of wavelengths. 5.101 The answer to this problem is obtained from Bragg's law
where
/L is the wavelength of radiation, d is the interplanar spacing, and e
is the angle of incidence of the x-ray beam . H ere d
/L
=
/L = 2 d sin e,
=
40 1 . 8 pm and
=
7 1 . 07 pm .
7 l . 07 pm 2(40 l . 8 pm) sin e e = 5 . 074° 5.103 3/2 kT = 1 12 mv
2 /L = hlmv 3kT = h2/m/L2 T= 633 K
SM-160
Chemistry Connections
5.105
(a) The Lewis structures of ethyl ammonium nitrate (EAN) and the formal charge of each atom are showing as follows: o
o
o
-1
+
H H H I I I I H- Csp-Cj-N H j j .8. sp�' .. I HI sp Isp H H (b) The hybridization schemes of C and N atoms are assigned in the Lewis structure of (a). (c) The anion (N03-). N5+ can be reduced to a lower oxidation state by a reducing agent. (d) CH3CH2NH2 + HN03 CH3CH2NH3+ N03- (acid-base) ( (0.960atm)(2. 00L) (e) nethylamine= (PV) RT (0. OS206L . amt . K-1. mol-1 )(296.2K) J 0. 0790 mole nnitric acid= (0. 2 50 L)(0 . 2 40M) 0 . 0 600 mole (limiting reagent) Theoretical yield= 0.0600 mol HN03 ( (1ImoleHNO} mole EAN) ( 1 OS.1 04 gEAN = 6.49 g EAN J ImoleEAN J ) Percentage yield= ( 4.10 6.49 100% = 63. 2% (f) The forces holding EAN together in the solid contain ion-ion, hydrogen-bonding, and London forces. NaCI and NaBr only have ion-ion and London forces to hold them together (No H-bonding involved). However, the ion-ion forces in NaCI and NaBr are much o
+1
0
0
:0:
0
.
o
/
N
+1
q
0
0
0
-7
=
=
=
x
x
SM-161
stronger that those ofEAN because of the smaller ion sizes ofNaCl and NaBr. (g) A s the ion size increases, the distance between ions will increase, the ion-ion forces will decrease, and the melting point will decrease .
SM- 1 62
CHAPTER 6 THERMODYNAMICS: THE FIRST LAW 6.1
(a) isolated;
(b) closed;
(c) isolated;
(d) open;
(e) closed;
(f) open
6.3
(a) Work is given by W = -?'xtfl V . The applied external pressure is known, but we must calculate the change in volume given the physical dimensions of the pump and the distance, d, the piston in the pump moves : � V = -Jrr2 d = Jr (1 . 5 cm) \ 20 . cm)
[
J
1L 3 = -0 . 1 4 L 1 000 cm
V is negative because the air in the pump is compressed to a smaller volume; work is then: ( 1 0 1 .325 ) = 28 J w = -(2 . 00 atm)( -0 . 1 4 fl
L
)
L
L · atm
(b) Work on the air is positive by convention as work is done on the air, it is compressed. 6.5
The change in internal energy flU is given simply by summing the two energy terms involved in this process . We must be careful, however, that the signs on the energy changes are appropriate. In this case, internal energy will be added to the gas sample by heating and through compression . Therefore the change in internal energy is: flU = 524 kJ + 340 kJ = 864 kJ
6.7
(a) The internal energy increased by more than the amount of heat added. Therefore, the extra energy must have come from work done on the system. (b) w = flU - q = 982J - 492J = +4 . 90 x 1 0 2 J.
SM-1 63
6.9
To get the entire internal energy change, we must sum the changes due to heat and work . In this problem, q = +5500 kJ . Work will be given by w=
-'?"xtL1V because it is an expansion against a constant opposing
(
pressure: w
=-
750 Torr 760 Torr · atm - I
(
](
]
1 846 mL - 345 mL = - 1 .4 8 L . atm 1 000 mL . L- I
K-I
To convert to J we use the equivalency of the ideal gas constants: w=
-(l .48 L . atm)
]
. mor l 8 .3 1 4 J . = - 1 . 50 x 1 0 2 J 0 . 082 06 L · atm · K- I . mol- I
L1U = q + w = 5500 kJ - 0. 1 50 kJ=5500 kJ
The energy change due to the work tenn turns out to be negligible in this problem .
6.11
Using 6 U
=
q
+ W
, where L1U = -2573 kJ and q
=
- 2573 kJ = -947 kJ + w. Therefore, w = - 1 626 kJ.
-947 kJ ,
1 626 kJ of work can be done by the system on its surroundings . 6.13
(a) true if no work is done; (b) always true; (c) always false; (d) true only. if w = 0 (in which case LtU q = 0); ( e) always true =
6.15
(a) During melting heat is absorbed and q is positive . Since the change is occurring at constant temperature L1£
=
o.
Therefore work is done by the system and w is negative . (b) During condensation heat is released and q is negative. Since the change is occurring at constant temperature L1£ Therefore work is done on the system and w is positive.
SM- 1 64
=
o.
6.17
(a) The heat change will be made up of two tenns: one tenn to raise the temperature of the copper and the other to raise the temperature of the water:
q = (75 0 . 0 g)( 4. 1 8 J . ( OC) -I . g- I )(1 OO.O°C - 23 . 0 °C) + (5 00 . 0 g)( 0 . 3 8J . (oC) - 1 . g- I )(1 00 . 0°C - 23 . 0 °C) 2 = 2 . 4 x 1 05 J + 1 . 5 x 1 04 J = 2 . 6 x 1 05 J = 2 . 6 x l 0 kJ (b) The percentage of heat attributable to raising the temperature of water will be
( )
24 1 kJ (1 00) = 94 . 3% 256 kJ
6.19
heat lost by metal
-
heat gained by water I (2 0 . 0 g) ( Tfina l - 1 00 . 0°C) ( 0 . 3 8 J . ( 0 C) - I . g- ) = - (5 0 . 7 g) ( 4. 1 8 J . ( OCr l . g- I ) ( Tfinal - 22 . 0 °C) ( Tfina l - 1 00 . 0°C) (7.6 J . ( O Cr l ) = - (2 1 2 J . ( O Cr l ) ( Tfina l - 22 . 0 °C) Tfina l - 1 00. 0°C = - 28( Tfina l - 22 . 0°C) Tfina l + 28 Tif na l = 1 00 . 0°C + 6 1 6 °C =
29 Tif lla l = 7 1 6°C Tfina l = 25°C
=
22 . 5 kJ = 14 . 8 kJ . COC)- I 0 23 . 97 0 C - 22 . 45 C
6.21
ca cl
6.23
(a) The in-eversible work of expansion against a constant opposing pressure is given by W=
w=
-P'xi1V
-(1. 00 atm) (6. 52 L - 4 . 29 L) -2.23 L . atm = -2 . 23 L · atm xl 0 1 .325 J . L- I . atm - I = -226 J =
SM- 1 65
(b) A n isothennal expansion will be given by V w=-nRT-2
�
n is calculated from the ideal gas law: PV
(1 .79 atm) (4.29 L) = 0 . 307 mol RT (0.082 06 L · atm· K- ' . mor ' ) (305 K) 6.52 w = - (0.307 mol) (8 . 3 1 4 J . K . mor ' ) (305 K) In 4 . 29 = - 326 J n=
=
-,
Note that the work done is greater when the process is carried out reversibly.
6.25
N0 2 . The heat capacity increases with molecular complexity-as more atoms are present in the molecule, there are more possible bond vibrations that can absorb added energy .
6.27
(a) The molar heat capacity of a monatomic ideal gas at constant pressure is C q=
P
(
.111
= -t R. The heat released will be given by
5 . 025 g 83 . 80 g . mol
_,
J
(25 . 0°C - 97.6°C) (20 . 8 J . mor ' . ( O Cr ' ) = -90 . 6 J
(b) Similarly, the molar heat capacity of a monatomic ideal gas at constant volume is CV.m = + R. The heat released will be given by q=
6.29
(
5.025 g 83 . 80 g . mol
_,
J
(25 . 0°C - 97.6 ° C) (1 2 . 5J . mor ' . ( O Cr ' ) = -54.4 J
(a) HCN is a linear molecule . The contribution from molecular motions will be 5/2 R. (b) C 2 H 6 is a polyatomic, nonlinear molecule . The contribution from molecular motions will be 3R.
SM- 1 66
( c) A r is a monatomic ideal gas. The contribution from molecular motions to the heat capacity will be 3/2 R. (d) HBr is a diatomic, linear molecule. The contribution from molecular motions will be 5/2 R . 6.31
The strategy here is to determine the amount of energy per photon and the amount of energy needed to heat the water. Dividing the latter by the former will give the number of photons needed. Energy per photon is given by: E
he
=- =
/L
(6.626 x l O-
34 J · s)(2 9979 x l 0 8 m · s- I ) . 4 . 50 x l 0 - 3 m
The energy needed to heat the water is: 3 50 g (4 . 1 84 J . g - I . 0 C- I ) (1 00.0 0 C - 25 .0 0 C)
=
=
-2 4 . 4 1 x l 0 3 J · p hoton - I
1 . 1 0 x 1 05 J
The number of photons needed is therefore: 1 . l O x 1 05 J = 2 . 49 x l 0- photons 4.4 1 x l 0-23 J . photon?7
1
6.33
(a) Using the estimation that 3R = C: 3R = C = (0.3 92 J .K - ' . g- I )(M) M=
3R = 63 . 6 g mol _ I 0 . 392 J .K - . g1
1
This molar mass indicates that the atomic solid is Cu(s) (b) From Example 5.3 we find that the density of a substance which forms
;M
a face-centered cubic unit cell is given by: d = 3 / 8
N/
3 . Therefore, we
expect the density of copper to be: 4 (63 .55 g · mol- ' )
d = 312 8 (6.022 x 1 023 mol- ' )(1 . 28 x l 0-8 cm) 3
SM- 1 67
=
8 . 90 g . cm -3
6.35
(a) !1Hyap (b) !1H
yap
6.37
4.76 kJ = 8.22 kJ . mol- I 0.579 mol
=
=
(
2 1 . 2 kJ 22 .45 g 46.07 g . mol- '
J
=
43 .5 kJ . mor '
This process is composed of two steps: melting the ice at
ooe
and then
raising the temperature of the liquid water from O°C to 25°C : Step 1 : !1H =
(
J
80.0 g _ (6.01 kJ · mor ' ) = 26.7 kJ 1 8 .02 g · mol ,
Step 2: !1H = (80.0 g) ( 4. 1 8 J . ( O Cr ' . g- ' ) (20.0°C - O.O°C) Total heat required
6.39
=
=
6.69 kJ
26.7 kJ + 6.69 kJ = 3 3 .4 kJ
The heat gained by the water in the ice cube will be equal to the heat lost by the initial sample of hot water. The enthalpy change for the water in the ice cube will be composed of two terms: the heat to melt the ice at O°C and the heat required to raise the ice from O°C to the final temperature. heat (ice cube) =
(
50.0 g 1 8 .02 g · mol
_,
J
(6.01 X 1 03 J · mor ' )
+ (50.0 g) ( 4. 1 84 J . ( O Cr ' . g- ' ) (Tr - 0°) = 1 .67 x 1 04 J + (209 J . ( O Cr ' ) ( Tr - 0°) heat (water)
=
(400 g) ( 4. 1 84 J . ( O Cr ' . g- ' ) (Tr - 45°) ' = (1 .67 x 1 03 J . (O Cr ) (Tr - 45°)
Setting these equal: - (1 .67 x 1 0 3 J . ( O Cr ' ) Tf + 7.5 x 1 0 4 J = 1 .67 xl 0 4 J + (209 J . ( O Cr ' )Tr Solving for Tr : 1:f
=
5 . 8 x 1 04 J = 3 1 0C 1 .8 8 X 1 03 J . eCr '
SM- 1 68
6.4 1
=
= =
First, we'll work out how much heat was required to raise the temperature of the water sample: q gC OT ( 1 50 g) (4. 1 8 l oCi .go l )(5 . 00°C) 3 1 3 5 J Since both the water and ice samples were at 0.00° C and the water sample took 0.5 h to get to 5.00° C, we can estimate that the ice took 1 0 . 0 h to melt . If 3 1 3 5 J of heat were transferred in 0. 5 h, then the amount of heat transferred in 1 0.0 h is, 1 0. 0 h 0.5 h
--
x
3 1 3 5 J = 62 . 7 kJ (for 1 50 g sample).
62 . 7 kJ 8 . 3 3 mol
1 50 g 1 8 .0 g. mol -
---=----:-1 = 8 . 3 3 mol H -? O
7 . 53 kJ . mor l
Which given the assumptions, is fairly close to the W ',s of 6 . 0 1 kJ. mol- I . ;
6.43
Based on the Wfus = 1 0.0 kJ · mol- ' and Wvap = 20.0 kJ · mor ' and a constant heating rate, melting should occur twice as fast as vaporization which is observed in (b), (c) and (d) . H eating curve (d) can be eliminated because the slope for the solid, liquid and gas are all the same . Curve (c) can also be eliminated because the heating slope for the solid and liquid are the same (yet they have different heat capacities) . This leaves (b) as the best match.
6.45
(a) W = (1 .25 mol) (+35 8 . 8 kJ · mor l ) = 448 kJ (b) W =
(
1 97 g C 1 2 . 0 1 g . mol- ' C
(c) W = 4 1 5 kJ = (nCS2 ) nCS2
=
(
J(
3 5 8 . 8 kJ 4 mol C
]
=
3 5 8 . 8 kJ . mol- I ) 4 mol CS 2
1 .47 x 1 0 3 kJ
]
4.63 mol CS 2 or (4.63 mol) (76. 1 3 g . mol- I )
SM- 1 69
=
352 g CS 2
6.47
40°F = 4A4°C, 78°F = 25. 5 5°C 6,.T = 25 . 55°C - 4A4°C = 21 . 1 °c The heat required is (3.26 X 107 cm3 )(0.001 52 g . cm-3 )(1. 0 1 J . (OCrl • mol-I)(21.1 °C) = 1. 0 56 X 106 J 1. 056 X 103 kJ The mass of octane required to produce this much heat will be given by ( -1. 0 56 X I03 kJ ) (1 14.22 g ·mol-I) = 22. 0465 g = 2. 2 x l01 g -5471 kJ . mol-I =
((1. 0 gal) (3 . 785 x 103 mL· gal-I) (0. 7 0 g . mCI ) ) = (b) 1 14.22 g . mol- I kJ ) x ( 2-mol10 942 octane = - 1. 3 x 105 kJ (a) (1250 kJhr )(l dayhr J(150 days year year J 1 . 9 x 105 � mL )(0. 7 02 ) ) ( 1000 (b) ( 150 triyearPs )( OAO gtnp�l. J ( 3. 785 � gal. L mL ( 114.1 mo23l g ) (5471 � year mol ) = 7. 6 XI 06 � MJ
6.49
=
�
SM-170
6.51
From I1H = I1U + PI1V at constant pressure, or I1U = I1H - PI1V. Because = - PI1V = +22 kJ, we get -IS kJ + 22 kJ = I1U = +7 kJ. w
6.53
To determine the enthalpy of the reaction we must start with a balanced chemical reaction and determine the limiting reagent: 2HCl(aq) + Zn(s) H2 (g) + ZnCl 2 (aq) . 0. 800 L . O. S OO M HCl = OAOO mol HCl 8.S g = 0.130 mol Zn 6S. 3 7g· molExamining the reaction stoichiometry and the initial quantities ofHCl and Zn, we note that Zn is the limiting reagent (0.260 mol ofHCl is needed to completely react with 0.130 moles of Zn). The enthalpy of reaction may be obtained using tabulated enthalpies of formation: kJ ) -2 ( -167.16kJ 2 ( -167.16kJ ) - 0 I1Hr = -IS3.89-+ mol mol mol = -IS3.89 � mol This is the enthalpy per mole of Zinc consumed. Therefore, the energy released by the reaction of 8.S g of Zinc is: (-IS3.89�)(0.130 mol ) = -20.0 kJ mol The change in the temperature of the water is then: -20000 J = ( -4.l84 -°CJ-g J (800 g) I1 I1T = S. 9 8°C and = 2SoC + S. 9 8°C = 31°C The enthalpy of reaction for the reaction 4 C7HSN306(S) + 21 02(g) 28 CO2(g) + 10 H20(g) + 6 N2(g) may be found using enthalpies of formation: kJ ) + 10( -241.82kJ ) -4( - 67kJ ) = -13168kJ 28( -393.S1mol mol mol mol �
---=)
T
Tr
6.55
�
SM- 171
This is the energy released per mole of reaction as written. One fourth of this amount of energy or 3292 � per mole of TNT mol will be consumed. The energy density in kJ per may be found by dividing this amount of energy with the mass of one mole of TNT and then by multiplying with the density of TNT: 10 3 cm 3 ] = +23.9 10 3 kJ _3_2_92----=.�:..=.�.:..: =_l.. _ ( 1.65 � ) ( cm 1 227.14 � mol (a) (b) Bomb calorimeter means volume is fixed. Therefore, Use CtJ.T Need to find First work out how much heat was released on burning 1.40 g of carbon monoxide. + 0 -C T - (3.00 kJ· COCrl)(22.799 °c - 22.113 °C) - 2. 0 58 kJ MW (CO) 12. 0 107 g.morl + 15.9994 g.morl 28. 0 101 g.morl moles (CO) 28. 0 1011.40g.gmor) 4.998 x 10-2 - 2. 0 58 kJ -For 1. 00 mol CO(g), ---4.998 x 10-2 mol - 41.176 kJ. mor l - 41.2 kJ. morl For the combustion of 1.00 mol CO(g), - 41.2 kJ. morl The combustion reaction of diamond is reversed and added to the combustion reaction of graphite to give the desired reaction: MiO = -393.51 kJ re leased
L
x
L
6.57
L
w = o.
tJ. U = q + w
q =
q.
qcalOlimeter =
qreact ion
qreaction = - qcalOlimeter =
tJ.
=
=
=
=
=
=
q=
=
=
tJ.U = q =
6.59
SM-l72
1::Jl0 +395. 4 1 kJ 1::Jl0 = +1.90 kJ
CO2 (g) � C(dia) + 02 (g) C(gr) � C(dia) 6.61
6.63
=
The first reaction is doubled, reversed, and added to the second to give the desired total reaction: 2[S02 (g) � S(s) + 02 (g)] (2)[+296 . 83 kJ] 2S(s)+ 3 02 (g) � 2 S03 (g) -791.44 kJ 2 S02 (g) + 02 (g) � 2 S03 (g) 1::Jl0 = (2)(+296.83 kJ . morl ) -(791.44 kJ . mol-I) = -197 . 7 8 kJ First, write the balanced equations for the reaction given: C2 H2 (g) + t 02 (g) � 2 CO2 (g) + H 2 0(1) 1::Jl0 -1300 kJ C2 H6 (g) + ! 02 (g) � 2 CO2 (g) + 3 H 2 0(1) 1::Jl0 = -1560 kJ H 2 (g) + t0 2 (g) � H 2 0(1) 1::Jl0 = -286 kJ The second equation is reversed and added to the first, plus two times the third: C2 H 2 (g) + t 02 (g) � 2 CO2 (g) + H 2 0(1) 1::Jl0 -1300 kJ 2CO (g) + 3 H 2 0(l) � C2 H 6 (g) + + 02 (g) 1::Jl0 + 1560 kJ 2[H2(g) + t02(g) � H 2 0(1)] 2 [1::Jl° - 286 kJ] C 2 H 2 (g) + 2 H 2 (g) � C 2 H 6 (g) 1::Jl0 -1300 kJ . mol -I + 1560 kJ . mol-I + 2(-286 kJ . mol -I ) = -312 kJ . mol-I =
=
-
2
=
=
6.65
=
The reaction enthalpy for this reaction is given by:
SM- l73
M-I 0
=
=
=
6.67
6.69
6.71
6.73
12 r(H 2 0,1)) - [4 (HN0 3 , 1)) + 5 r(N 2 H 4 , 1))] 12(-285. 83 kJ . mol-I) -[4(-174.10 kJ . mol-I) + 5( +50. 63 kJ . mol-I)] -2986 . 71 kJ . mol-I ( M-I 0
( M-I 0 f
( M-I 0
2 NH4Cl(s) --+ 2 NH3(g) + 2 HC1(g) N2(g) + 4 H2(g) + Cb(g) --+ 2 NH4Cl(s) 2 NH3(g) --+ N2(g) + 3 H2(g) H2(g) + Cb(g) --+ 2 HC1(g) A --+ 2 B 2B --+ 2C + 2D 2 D --+ E A --+ 2 C + E
= 2(176. 0) kJ M-JO = -628. 8 6 kJ M-JO = 92. 2 2 kJ M-JO = -184. 6 kJ
M-I0
2NH4Br(s) --+ 2NH3(g) + 2HBr(g) � = 2(+188. 3 2 kJ) 2 NH3(g) --+ N2(g) + 3 H2(g) � = +92.22 kJ N2(g) + 4 H2(g) + Br2(l) --+ 2 NH4Br(s) � = -541. 6 6 kJ �=-72. 8 0 kJ From Appendix 2A,M-I°r (NO) = +90. 2 5 kJ The reaction we want is Adding the first reaction to half of the second gives 2 N02(g) + t °2 (g) � N20s (g) 2 NO(g) + % 02 (g) � N20s (g) SM-174
-1 14.1 kJ -55.1 kJ -169.2 kJ
M-I 0
=
M-I 0
=
The enthalpy of this reaction equals the enthalpy of formation of N 2 °S (g) minus twice the enthalpy of formation of NO, so we can write -169.2 kJ = IJ1-fDr (N20S ) 2(+90.25 kJ) WOr (N20S ) = +1 1. 3 kJ -
6.75
The enthalpy of the reaction PCl3 (1) + Cl2 (g) � PCIs (s) WO = - 124 kJ is WO = L WOf (products) - L WOf (reactants) - 124 kJ = WOf (PCIs ' s) WOf (PCI3 , 1) Remember that the standard enthalpy of formation of C12 (g) will be 0 by definition because this is an element in its reference state. From the Appendix we find that WOf (PCI3 ,1) = -319.7 kJ ·morl - 124 kJ = WO f (PCIs 's) (-319 . 7 kJ) WOf (PCIs ,s) -444 kJ . morl r
-
-
=
6.77
(a) For H2 °(l) , we want to find the enthalpy of the reaction The enthalpy change can be estimated from bond enthalpies. We will need to put in (1 mol) (436 kJ . mol- I ) to break the H-H bonds inl moIH2(g), (tmol) (496 kJ ·mol-I) to break the 0-0 bonds in t mol 02 (g) ; we will get back (2 mol) (463 kJ . mol - I ) for the formation of 2 mol O-H bonds. This will give /jj{ -242 kJ . mol-I . This value, however, will be to produce water in the gas phase. In order to get the value for the liquid, we will need to take into account the amount of heat given off when the gaseous water condenses to the liquid phase. This is 44. 0 kJ . mol-I at 298 K: =
SM- 175
!J.H 0 f,water(l) = !J.H0 f,water(g) - !J.H0 yap =
= -286 kJ . mol-1
-242 kJ . mol-1 -44. 0 kJ . mol-1
(b) The calculation for methanol is done similarly: for individual bond contributions: atomize 1 mol C(gr) (1 mol)(717 kJ . mol-') break 2 mol H-H bonds (2 mol)(436 kJ . mor') break 1mol bonds (1mol)(496 kJ . mol- ') form 3 mol C-H bonds -(3 mol)(412 kJ . mol- ') form 1 mol bonds -(1 mol)(360 kJ . mol- ') form 1 molO-H bonds - (1 mol) (463 kJ . mol-') !J.H
02
c-o
Total
!J.H 0 f,methanol(l) - !J.H0 f,methanol(g) - W O yap
-222kJ
= -222 kJ .. mol-1 -35.3 kJ . mol-1 = -257 kJ mol-1
Without resonance, we do the calculation considering benzene to have three double and three single C-C bonds: atomize: 6 mol C(gr) (6 mol)(717 kJ ·mor') break: 3 mol H-H bonds (3 mol) (436 kJ . mol-') form: 3 mol CRC bonds -(3 mol)(612 kJ 'mor') form: 3 mol C-C bonds -(3 mol) (348 kJ . mol-') form: 6 mol C-H bonds -(6 mol)(412 kJ 'mol-') +258 kJ
Total SM-176
t:Ji 0 f ,benzene(l) = t:Ji 0 f,benzene( g ) =
- t:Ji0 = +258 kJ . mol- 1 -30.8 kJ . mol- 1 ya p
+227 kJ . mol-1
With resonance, we repeat the calculation considering benzene to have six resonance-stabilized C-C bonds: atomize: break: form: form:
6 mol C(gr) 3 mol H-H bonds 6 mol C-C bonds, resonance 6 mol C-H bonds
Total t:Ji 0 f,benzene(l ) = t:Ji 0 f, benzene( g ) = - l kJ · mol- 1
6.79
- t:Ji0
+30 kJ = +30 kJ . mol- 1 -30. 8 kJ . mol- 1
2 Na+ (g) + 02- (g) t:JiL = 2 t:Ji ° r (Na, g) + t:Ji ° r (O , g) + 2 I, (Na) ( 0) ( 0 ) - t:Jir (Na 2 0 (s) ' ' ' t:JiL = 2(107. 3 2 kJ · mor ) + 249 kJ · mol- + 2(494 kJ · mol - ) -141 kJ . mol- ' + 844 kJ . mol- ' + 409 kJ . mol- ' ' t:JiL = 2564 kJ . mol-
For the reaction Na 2 0 (s) � -Eea '
6.81
yap
(6 mol) (717 kJ . mol- ' ) (3 mol) (436 kJ · mol - i ) -(6 mol) (518 kJ . mor ' ) -(6 mol) (412 kJ . mar ' )
(a)
-
Eea2
+
+
t:JiL = t:Ji ° r (Na, g) t:Ji° r (CI, g) I, (Na) Ee. of CI - t:Jir (NaCI(s)) kJ . mol - ' = kJ . mol - ' kJ . mol- '
-
787 108 + 122 - 349 k J . mol- ' - t:Jir (NaCI(s))' t:Jir (NaCI(s)) = -412 kJ · mor (b)
t:JiL = t:Ji 0 r (K, g) + t:Ji 0 r (Br, g) + I, (K) - Eea (Br) - t:Jir (KBr(s))
SM- l77
+ 494 kJ . mar '
=
MfL
(c) MfL
89 kJ . mar ' + 97 kJ . mar ' + 4 1 8 kJ . mol- ' - 325 kJ . mol - ' + 394 kJ . mol - ' ' = 673 kJ . mol-
=
L'1H O r (Rb, g) + Mfor (F, g) + I, (Rb)
-
Eea (F)
- Mfr (RbF(s))
774 kJ · mol- ' = Mfor (Rb, g) + 79 kJ · mol- ' + 402 kJ . mol- ' - 328 kJ . mol - ' + 558 kJ . mar ' Mfo r (Rb, g) 63 kJ . mol - '
=
6.83
(a) break: form:
3 mol C
==
C bonds 3(837) kJ . mol- 1
6 mol C=C bonds -6(5 1 8) kJ · mar '
- 597 kJ . mol- 1
Total (b) break:
4 mol C-H bonds 4(4 1 2) kJ . mar ' 4 mol CI-CI bonds 4(242) kJ . mol - '
form:
4 mol C-CI bonds -4(33 8) kJ · mar ' 4 mol H-CI bonds -4(43 1 ) kJ . mol- ' -460 kJ . mol- '
Total
( c) The number and types of bonds on both sides of the equations are equal, so we expect the enthalpy of the reaction to be essentially O.
6.85
(a) break:
form:
1 mol N-N triple bonds
( 1 mol) (944 kJ . mar ' )
3 mol F-F bonds
(3 mol) ( 1 5 8 kJ . mol- ' )
6 mol N-F bonds
(6 mol) ( -1 95 kJ · mor ' ) + 248 kJ . mol- 1
Total (b) break:
1 mol C=C bonds
( 1 mol) (6 1 2 kJ · mol- ' )
1 mol O -H bonds
( 1 mol) (463 kJ . mol- ' )
SM- I 78
fonn :
1 mol C-C bonds
-( 1 mol) (348 kJ . mol- I )
1 mol C-O bonds
-(1 mol) (360 kJ . mol - I )
1 mol C-H bonds
-(1 mol) (41 2 kJ · mol- I )
- 45 kJ . mol - 1
Total (c) break:
fonn :
1 mol C- H bonds
(1 mol) (4 1 2 kJ · mor l )
1 mol Cl-Cl bonds
( 1 mol) (242 kJ . mol- I )
1 mol C-Cl bonds
- (1 mol)(338 kJ . mol - I )
1 mol H -Cl bonds
-(1 mol) (43 1 kJ . mor l )
- 1 1 5 kJ · mol - 1
Total
6.87
The value that we want is given simply by the difference between three isolated C=C bonds and three isolated C-C single bonds, versus six resonance-stabilized bonds: 3 C=C bonds
+
3 C-C bonds = 3 (348 kJ) + 3 (6 1 2 kJ) = 2880 kJ
6 resonance-stabilized bonds
=
6( 5 1 8 kJ) = 3 1 08 kJ
A s can be seen, the six resonance-stabilized bonds are more stable by ca . 228 kJ .
6.89
(a) The enthalpy of vaporization is the enthalpy change associated with the conversion C 6 H 6(1) � C 6 H 6 (g) at constant pressure . The value at 298 .2 K will be given by Moor vaporizalion a l 298 K
=
=
=
f.J! 0f (C 6 H 6 ' g) - f.J! 0 (C6 H 6 ' 1) f
82.93 kJ . mol- I - (49.0 kJ . mol- I ) 3 3 . 93 kJ . mol- I
(b) In order to take into account the difference in temperature , we need to use the heat capacities of the reactants and products in order to raise the
SM-1 79
temperature of the system t0 353.2 K. We can rewrite the reactions as follows, to emphasize temperature, and then combine them according to H ess's law:
Mfo
C 6 H 6 (l) a l 298 K � C 6 H 6 (1)31 353.2 K
=
( 1 mol) (3 53.2 K - 298.2 K) ( 1 36 . 1 J . mol - I . K- I )
= 7 48 k J .
C 6 H 6 (g) a l 298 K � C 6 H 6 (g)aI 35 3. 2 K Mfo = ( 1 mol) (3 5 3 .2 K - 298.2 K) (8 1 . 67 J · mor l . K- I ) = 4.49 kJ To add these together to get the overall equation at 3 5 3 .2 K, we must reverse the second equation:
C 6 H 6 (1) aI 35 3.2 K � C 6 H 6 ( g ) aI 353.2 MfO = 3 3 .93 kJ . mol- I - 7.48 kJ . mol- I + 4.49 kJ . mol- I = 30 . 94 kJ . mol- I K
(c) The value in the table is 30.8 kJ . mol - I for the enthalpy of vaporization of benzene. The value is close to that calculated as corrected by heat capacities . At least part of the error can be attributed to the fact that heat capacities are not strictly constant with temperature.
6.91
For the reaction: A + 2 B --+ 3C + D the molar enthalpy of reaction at temperature 2 is given by: Mf;2 = H I�,.2 (products) - H�.2 (reactants)
= 3 H,�.2 (C) + H,� .2 (D) - H,�.2 ( A ) - 2 H,�.2 ( B ) =
=
3[ H,� . 1 (C) + C .m (C)(T; - � )] + [ H,�. , (D) + C .m (D)(T; - � )] p p - [ H,�. , ( A ) + C .m (A)(T; - � )] - 2[ H,�., (B) + C .m ( B )(T; - � )]
p
3 HI�. 1 (C) + H,�. , (D) - HI�,. 1 ( A ) - 2 HI�.1 (B) SM- 1 80
p
+ [ 3 Cp.m (C) + Cp, m ( D) - Cp,m ( A ) - 2 Cp,m ( B)] ( T2
- 7; )
= !ill;I + [ 3 Cp,m (C) + Cp,m ( D) - Cp,m (A ) - 2 Cp, m (B)] ( I; - 7; ) Finally, !illr ,2 ° = !illr , 1 ° + � Cp ( T2 - TI ) , which i s Kirchhoff s l aw . 6.93
This process involves five separate steps: ( 1 ) raising the temperature of the ice from -5 . 042 °C to 0 . 00 0c. (2) melting the ice at O.OO°C, (3) raising the temperature of the liquid water from O. OO°C to 1 00 . OO°C , (4) vaporizing the water at 1 00.00°C, and (5) raising the temperature of the water vapor from 1 OO . OO°C to 1 50,35°C . Step 1 :
!ill = (42,30 g)(2.03 J . ( O Cr l . g- I )(0.00 ° c - (-5 . 042°C)) = 0.433 kJ Step 2: !ill =
(
]
42 . 30 g (6.0 1 kJ . mol- I ) = 1 4 . 1 kJ 1 . mol1 8 .02 g
Step 3 :
!ill = (42.30 g)( 4. 1 8 J . (0C)- I . g- I )(1 00 . 00 ° c - O . OO ° C) = 1 7.7 kJ Step 4: !ill =
(
]
42 . 30 g ( 40 . 7 kJ · mol- I ) = 95 . 5 kJ 1 . 1 8 . 02 g mol-
Step 5 :
!ill = (42,30 g)(2 . 0 1 J . (O C) - I . g - I )(1 50 . 3 5 °c - 1 OO.OO°C) = 4.3 kJ The total heat required
= 0 . 4 kJ + 1 4 . 1 kJ + 1 7 . 7 kJ + 95 .5 kJ + 4.3 kJ = 1 32 . 0 kJ 6.95
Appendix 2 A provides us with the heat of formation of 1 2 (g) at 298K ( + 62.44 kJ · mor l ) and the heat capacities of 1 2 (g) (3 6.90 J . K- I . mor l ) and 1 2 (s) (54.44 J . K- I . mol- J ). We can
calculate the !illsub 0 at 298K: !illsub
SM- 1 8 1
0
= +62.44 kJ . mol- J
We can calculate the enthalpy of fusion from the relationship M-Isub
°
= M-Ifus °
+ M-Iyap
°
but these values need to be at the same temperature. To correct the value for the fact that we want all the numbers for 298K, we need to alter the heat of vaporization, using the heat capacities for liquid and gaseous
=
iodine . M-Iyap
°
+ 4 1 .96 kJ . mor '
From Section 6.22, we find the following relationship
M-Iyap,298K O M-Iyap,2 98K
= M-Iyap, 475 5Ko
+ ( Cp,m ° ( l 2 , g) - Cp,m ° ( l 2 , 1 )) (� ° = + 4 1 .96 kJ . mor '
-
�)
+ (36.90 J . K- ' . mol- ' - 80.7 J . K- ' . mor ' ) (298K - 475.5K) ' = + 49.73 kJ ' molSo, at 298K: +62 . 44 kJ . mol- ' M-Ifus O
6.97
=
=
' M-IfusO + 49 . 73 kJ . mol-
+ 1 2 .7 1 kJ ' mor '
(a)
7 6 5
....,
4
'r 3
2 1 0
0
2
4
V/Vj
6
8
10
(b) The amount o f work done i s greater at the higher temperature. This can be seen from the equation:
SM- 1 82
V,
w = -nRT ln � V:nitial
The amount of work done is directly proportional to the temperature at which the expansion takes place. (c) The comparison requested is the comparison of the terms In Vfinal
�nitial
for the two processes . Even though in both cases the gas expands by 4 L , the relative amount of work done is different . We can get a numerical
( ) ( )
comparison by taking the ratio of this term for the two conditions: L 1 n 9.00 5 . 00 L 0. 5 8 8 --;..------:- = = 0 . 365 1 .61 5 . 00 L In 1 . 00 L --
The second expansion by 4.00 L produces only about one third the amount of work that the first expansion did. 6.99
First, we need to calculate how much energy from the sunshine will be hitting the surface of the ethanol, so we convert the rate kJ . cm -2 . S - I : 1 kJ . m -2 . S- I
( )= 1m 1 00 cm
2
1 X 1 0-4 kJ . cm -2 . S - I
(
�
)
6 S I = 3 kJ ( l x l O-4 kJ · cm -2 . S - ) (50 . 0 cm 2 ) 1 0 min x mm The enthalpy of vaporization of ethanol is 43 . 5 kJ · mol- I (see Table 6 .2). We will assume that the enthalpy of vaporization is approximately the same at ambient conditions as it would be at the boiling point of ethanol.
(
3 kJ ____ 43 . 5 kJ . mol-
1
_ _
)
(46 . 07 g . mol- I ) = 3 g
SM- 1 83
(b) m
eo
( ( (
0. 1 754 g aniline
2 = 93 . 1 2 g . mol- I anline
= 0.4873 g C O 2 (g) m
1 754 g
)( J( )(
aniline 0. H 20 = 93 . 1 2 g . mor l aniline
= 0. 1 1 88 g H 2 0(l) m
=
N2
0. 1 754 g ani line 93 . 1 2 g . mol- I aniline
= 0. 02 6 39 g N 2 (g) (c) n 02
=
(
0 . 1 754 g aniline 93 . 1 2 g · mol- I aniline
= 0 . 0 1 4 60 g 0 2 (g)
J J
6 mol C O 2 (28 . 0 1 g . mor l C O 2 ) 1 mol aniline 3 . 5 moI H 2 o (1 8 . 02 g . mo l- I H 2 0) 1 mol aniline
J
0.5 mo l N 2 (28.02 g . mol- I N 2 ) 1 mol aniline
J(
¥ mo 1 0 2 1 mol aniline
J
p = nRT = (0 . 0 1 4 60 moI 0 2 ) (0 . 082 06 L · atm · KV
I · mor l ) (296K)
0 . 355 L
= 0 . 999 atm
6.103 (a) The reaction enthalpy is obtained by H ess's law: I':-Jr r
= !::H ° r (C O , g) - !::H° r ( H 2 0, g) !::H ° r = (1) (- 1 1 0. 53 kJ · mor l ) - (1) (-24 1 . 82 kJ · mor l ) !::H ° r = + 1 3 1 .29 kJ · mor l
endothermic (b) The number of moles of H 2 produced is obtained from the ideal gas law:
n=
PV
-
RT
=
(
J
500 Torr (200 L) 760 Torr · atm- I (0.082 06 L · atm · K - I . mol- I ) (3 3 8 K)
= 4.74 mol
The enthalpy change accompanying the production of this amount of hydrogen will be given by !::H
= (4.74 mol) (1 3 1 .29 kJ · mol- I ) = 623 kJ
SM- 1 84
6.105 (a) The number of moles burned may be obtained by taking the difference
in the number of moles of gas present in the tank before and after the drive using the ideal gas equation:
n 1 - n2 = PRT1 V - PRT2 V = (� P2 )( RTV ) = ( 1 6 .0 atm - 4.0 atm) = 1 4.7 mol
(
-
L
3 0_ 0 ._
_ _ _ _ _ _
-
(0.0820574 L · atm · K - I . mol - I )(298 K)
J
(b) From a table of enthalpies of combustion, the enthalpy of combustion of H 2 is found to be - 286 kJ . mol- I . The energy change is, therefore, ( 1 4.7 mol)( -28 6 kJ . mol- I ) = -4.20 x 1 0 3 kJ . 6.107 (a) First we must balance the chemical reaction:
For 1 mol C 6 H 6 (1) burned, the change in the number of moles of gas is
tJ"n (tJ"nRT) = -PtJ"V = - P -----p- = - tJ" n RT
(6.00 - 7 . 50) mol = - 1 .50 mol = w w
= -( - 1 . 5 0 mol) (8 . 3 1 4 J . K- ' . mol- ' ) (298 K) = + 3 . 7 1 6 x 1 03 J = + 3 .72 kJ
(b)
Mfe = 6(-393 . 5 1 kJ · mol- I ) + 3(-285 . 8 3 kJ · mol - I ) - (+49 . 0 kJ · mol- ' )
= - 3267 . 5 kJ (c)
tJ"uo = Mfo + H 1
6.109
(a)
w
H H, / ' / C C(1) 1 1 H /c � C,H H C/ H H1
= ( -3267 . 5 + 3 . 72) kJ = - 3 263 . 8 kJ H H H ,cI H / H -\ / ,C-
H 1
H H, /' / H C C(2) 1 I H-C / C-H H /, C / \H H �
I H -C / , /C-H H / C \H H �
(3t ?
(b) From bond enthalpies, each step is identical, as the number and types of bonds broken and formed are the same: break:
1 mol C=C bonds
SM- 1 85
6 1 2 kJ
form:
1 mol H-H bonds
436 kJ
1 mol C-C bonds
-348 kJ
2 mol C-H bonds
2( -4 1 2 kJ)
Total :
-1 24 kJ
The total energy change should be equal to the sum of the three steps or 3 (-1 24 kJ) = -372 kJ. (c) The Hess ' s law calculation using standard enthalpies of formation is easily performed on the composite reaction: C 6 H 6 (1) + 3 H 2 (g) � C 6 H 2 (1) I Wa r = � WO r (products) - � WOr (reactants) = WOr (cyclohexane) - WOr (benzene) = -1 5 6.4 kJ . mol-I - ( + 49.0 kJ . morl ) -205 .4 kJ . mol-I
=
(d) The hydrogenation of benzene is much less exothermic than predicted by bond enthalpy estimations . Part of this difference can be due to the inherent inaccuracy of using average values, but the difference is so large that this cam10t be the complete explanation . As may be expected, the resonance energy of benzene makes it more stable than would be expected by treating it as a set of three isolated double and three isolated single bonds . The difference in these two values [-205 kJ - (-372 kJ) 1 67 kJ]
=
is a measure of how much more stable benzene is than the Kekule structure would predict . 6.111
(a) The combustion reaction is The enthalpy of formation of C 60 (s) will be given by WOe = 60 WOr (C0 2 ,g) - WOr (C 6o ' s) -25 937 kJ = 60 mol x ( -393 . 5 1 kJ ' mol-I ) - WO f (C 60 , S ) WO r (C 60 ' s) = +2326 kJ . morl SM- 1 86
(b) The bond enthalpy calculation is 60 C(gr) � 60 C(g)
(60) (+7 1 7 kJ · morl»
Form 60 mol C-C bonds
-60 (348 kJ . mor I )
Form 30 mol C=C bonds
-30 (6 1 2 kJ . mol-I )
C60 (g) � C60 (s)
-23 3 kJ
60 C(gr) � C60 (s)
+3 547 kJ
(c) From the experimental data, the enthalpy of formation of C60 shows that it is m ore stable by (3 547 kJ - 2326 kJ) = 1 22 1 kJ than predicted by
=
the isolated bond model . (d) 1 22 1 kJ ..;- 60 20 kJ per carbon atom (e) 1 50 kJ ..;- 6 = 25 kJ per carbon atom (f) Although the comparison of the stabilization of benzene with that of C60 should be treated with caution, it does appear that there is slightly less stabilization per carbon atom in C60 than in benzene . This fits with expectations, as the C60 molecule is forced by its geometry to be curved . This means that the overlap of the p-orbitals, which gives rise to the delocalization that results in resonance, will not be as favorable as in the planar benzene molecule . Another perspective on this is obtained by noting that the C atoms in C60 are forced to be partially Sp3 hybridized because they cannot be rigorously planar as required by Sp 2 hybridization . 6.113
The balanced combustion reactions are C6H 3 (N0 2 ) 3 (s) + ;I 0 2 (g) � 6 CO 2 (g) + -t H 2 0(l) + -t N 2 (g) C6H 3 (NH 2 ) 3 (s) + 31 0 2 (g) � 6 CO 2 (g) + t H 2 0(l) + t N 2 (g) Because the fundamental structures of the two molecules are the same, we need only look at the differences between the two, which in this case are concerned with the groups attached to nitrogen . From the combustion equations we can see that the differences are ( 1 ) the consumption of ¥ SM- 1 87
more moles of O 2 (g) and (2) the production of three more moles of H 2 0(I) for the combustion of aniline . Because the MfO f of 0 2 (g) is 0, the net difference will be the production of 3 more moles of H 2 0(1) or 3 x (-285 . 83 kJ . mol- I ) = -857049 k1 . nRT
6.115
= = (a) Vini! p = 1 .5 L
(0 . 060 mol)(0.0820578 L · atm · K - I . mol- I )(298. 1 5 K) 1 . 00 atm
(b) The combustion reaction is: 2 S02(g) + 02(g) -7 2 S03(g) . If equal molar amounts of S02 and 02 are mixed, as in this case, S02 is the limiting reagent . (c) The total number of moles remaining in the container will be: 0 . 030 mol S03(g) + 0.01 5 mol 02(g) 0.045 mol of gas at the end of the reaction. The final volume will, therefore, be:
=
nRT
Vf = = P = 1.1 L
(0 . 045 mol)(0 . 0820578 L · atm · K - I . mol- I )(298 . 1 5 K) 1 . 00 atm
=
(d) � V = 1 . I L - 1 . 5 L = -004 L w = -P� V (1 . 00 atm)(-Oo4 L)(1 0 1 . 325 J . L- J atm- J ) = 40 J of work done on the system (work is positive) •
(e) The enthalpy of reaction may be found using standard enthalpies of formation and the balanced equation given above: Mfr 2(-395 .72 kJ · mol- I ) - 2(-296.83 kJ · mol- I ) - 1 97 . 78 kJ · mol- I .
=
=
If 0 . 030 mol of S02 are consumed, then enthalpy change is: (0 . 030 mol S0 2 )
(
- 1 97 . 78 kJ 2 mol S0 2
]
=
-2966 . 7 kJ = -3 . 0 kJ (2 sf) = -3000 1.
(f) �U r = Q + w = -3000 J + 40 J = -2960 J
SM- 1 88
6.117
(a)
H
H
I H -C - H I H
=
H
H
H
\C - H I � / I : q. HI H
___ C
I I H
H
I I .. . H
H - C - C -O
___ H
Mf; (4 x 4 1 2) + (2 x 496) + (2 x -743) + (4 x -463)
=
+ 2640 - 3338
=
- 698 kJ.mol-1
H ) C - O-CH ) (g) + 30 2 (g) � 2C0 2 (g) + 3H 2 0 (g) Mf;
=
(2 x 360) + (6 x 4 1 2) + (3 x 496) + ( 4 x -743) + (6 x -463)
=
- 1 070 kJ . mol-1
CH)CH 2 0H(g) + 30 2 (g) � 2C0 2 (g) + 3H 2 0 (g) Mf;
==
(360) + (463) + (348) + (5 x 4 1 2) + (3 x 496) + ( 4 x -743) + (6 x -463) - 1 03 1 kJ . mol-1
=
=
For the burning of 1 mole of CH)CH 2 0H(l), Mf� 43 . 5 - 1 03 1 - 988 kJ.morl . The burning of 1 mole of dimethyl ether releases the most heat . -890 kJ . mol-1 1 6.01 g. mol-1 -1 368 kJ. mol-1 46.02 g. mol-1
(c)
-547 1 kJ . mol-1 1 1 4.08 g. mol-1
= =
=
_
- 55 . 6 kJ. g-1 29.73 kJ . g-1
- 47 . 96 kJ . g·1
Methane as it releases the most heat per gram . (d) mass of 1 0 . 00 L octane
=(
0 . 70 g. mol-I ) (1 0,000 mL)
SM - 1 89
=
7000g
heat released
=
( 7000 g) ( -55.6 kJ . g.l )
=
3 . 892 x 1 05 kJ moles of methane gas = 890 kJ . mol· I V=
=
nRT
--
P
( 437.3 mol H 8 .20 574
=-
-'--
-
-
=
3 . 892 x 1 05 kJ
437 . 3 mol
)(
)
1 0-2 L . atrn . K -1 . mol· l 298 K --'---1 0 .00 atm
x
-
1 069 L = 1 . 1 x 1 03 L
(e) methane gas, -890 kJ. morl CO 2 (less CO 2 ) ethanol liquid, -684 kJ. morl CO 2
(more CO)
octane liquid, -684 kJ. marl CO 2
(more CO 2 )
SM - 1 90
CHAPTER 7 THERMODYNAMIC S : THE SECOND AND THIRD LAWS
7.1
=
(a) rate of entropy generation
=
�
surr = . q rev· T tzme tzme rate of heat generation
tJ.
_
T -(1 00 . J · S - ) = = 0 . 34 1 J . K-' . S - I 293 K
I
(b)
tJ.Sday =
(0 . 341 J . K - 1 S -I ) (60 sec · min-' ) (60 min · hr-' ) (24 hr · day-I ) = 29 . 5 kJ · K-1 • day- ' •
(c) Less, because in the equation tJ.S =
7.3
(a)
tJ.S = q rev
(b)
tJ.S =
T
=
65 J 373 K
-M£
T
, if T is larger, tJ.S is smaller .
65 J = 0 . 22 J . K-' 298 K =
0. 1 7 J . K-'
( c) The entropy change is smaller at higher temperatures, because the matter is already more chaotic . The same amount of heat has a greater effect on entropy changes when transferred at lower temperatures .
7.5
(a) The relationship to use is dS =
dq . At constant pressure, we can T
substitute dq n Cp dT : =
dS =
n Cp dT T
Upon integration, this gives
tJ.S = n
Cp In � . The answer is calculated by
T; simply plugging in the known quantities. Remember that for an ideal monatomic gas
SM- 1 9 1
IlS = ( 1 .00 mol) ( +- x 8 . 3 1 4 J . K-I . mol-I ) In
43 1 .0 K = 6 . 80 J . K - I 3 1 0.8 K
(b) A similar analysis using Cy gives IlS = n Cv In 1; , where Cv for a 1; monatomic ideal gas is % R : 43 1 . 0 K IlS = (l . 00 mol) ( +- x 8 . 3 1 4 J . K-I . mol-I ) In 4.08 J . K - I 3 1 0. 8 K =
7.7
d Because the process is isothermal and reversible, the relationship dS = q T
can be used . Because the process is isothermal, IlU = 0 and hence q = -w, where w = -Pd V. Making this substitution, we obtain dS =
P dV T
=
nRT TV V2
dV =
nR dV V
:. IlS = nR In � Substituting the known quantities, we obtain IlS = (5.25 mo l) (8 . 3 1 4 J . K - I . mol-I ) In
34 .0 58 L 24 . 252 L
= 1 4. 8 J · K - 1 7.9
The change in entropy for each block (block 1 and block 2) are: IlS = !l..l and IlS2 = 2l energy is transferred, then the change in entropy I 1; 1; for the system is:
IlS = IlS, + IlS2 = 9..l + 92 T,
T2
If 1 J of energy is transferred from block 2 to block 1 , 1 1 q , = + l J, q 2 = -I J, and IlS = - - - . T,
T2
For the transfer of heat from block 2 to block 1 to be spontaneous, IlS must be positive and, therefore, T2 would have to be greater than T, . SM - I 92
7.11
(a)
fj,s o =
!L =
T
tili °
T
=
1 .00 mol x ( - 6 . 0 1 kJ . mor ' ) = -22 0 J . K-' . 273.2 K
50.0 g x 43 . 5 kJ . mol- ' tili 46 . 07 g · mol- I !L = + 1 34 J . K- ' = (b) fj,S = = 3 5 1 .5 K T T 7.13
(a) The boiling point of a liquid may be obtained from the relationship fj,Svap
=
tilivap
, or TB =
tilivap
. This relationship should be rigorously true fj,Svap TB if we have the actual enthalpy and entropy of vaporization . The data in the Appendix, however, are for 298 K . Thus, calculation of tili ° yap or fj,s o vap ' using the enthalpy and entropy differences between the gas and liquid forms at 298 K, give a good approximation of these quantities but the values are not exact . For ethanal(l) � ethanal(g), the data in the appendix give tiliva p � - 1 66 . 1 9 kJ . mol- ' - ( -1 92 . 30) kJ . mol-' = 26 . 1 1 kJ . mol- ' fj,Svap � 250 . 3 J . K-' . mol-' - 1 60 . 2 J . K -I . mol-' = 90 . 1 J . K- ' . mol- ' T. 26 . 1 1 x 1 03 J . mor l = 290 . K B= 90 . 1 J · K-' · mol- ' (b) The boiling point of ethanal is 20 . 8°C or 293.9 K. (c) These numbers are in very good agreement . (d) Differences arise partly because the enthalpy and entropy of vaporization are slightly different from the values calculated at 298 K, but the boiling point of ethanal is not 298 K . 7.15
(a) Trouton ' s rule indicates that the entropy of vaporization for a number of organic liquids is approximately 85 J . K . mol-I . Using this -I
information and the relationship tili° ya p 2 1 .5 1 X 1 03 J . morl ------- = 253 K . T.B = fj,s o yap 85 J . K-' . mol-'
SM - 1 93
.17
(b) The experimental boiling point of dimethyl ether is 248 K, which is in reasonably close agreement, given the nature of the approximation. (a) The value can be estimated from 11J{0 = T6.S0 = (35 3 K) (85 J · mol-I · K-I ) = +30. kJ . mol- I (b) 6.S0 = _ 11J{0T -l l = -I I J . K- 1 6.S0 = _(l 78 .l l10g .9mol-l 1) (l 30 kJ·mol 35 3 K ) l1J{
°
v ap
vap
vap
surr
syslem
surr
'.19
'.21
COF2. COF2 and BF3 are both trigonal planar molecules, but it would be possible for the molecule to be disordered with the fluorine and oxygen atoms occupying the same locations. Because all the groups attached to boron are identical, such disorder is not possible. There are six orientations of an S02 F2 molecule as shown below: o
0
\ "S -, ,F F '
The Boltzmann expression for one mole of S02 F2 molecules having six possible orientations is S = In66 I On = (1. 3 8 10-23 J . K ) ln66 02 S = 14 .9 J ·K-1 k
02 x
X
-I
SM-194
x
1 021
7.23
(a) HBr(g), because Br is more massive and contains more elementary particles than F in HF; (b) NH3(g), because it has greater complexity, being a molecule rather than a single atom; (c) 1 2 (1), because molecules in liquids are more randomly oriented than molecules in solids; (d) 1 . 0 mol Ar(g) at 1 .00 atm, because it will occupy a larger volume than 1 . 0 mol of Ar(g) at 2.00 atm .
7.25
It is easy to order H 2 0 in its various phases because entropy will increase when going from a solid to a liquid to a gas. The main question concerns where to place C(s, diamond) in this order, and that will essentially become a question of whether C(s, diamond) should have more or less entropy than H 2 0(s) , because we would automatically expect C(s, diamond) to have less entropy than any liquid . Because water is a molecular substance held together in the solid phase by weak hydrogen bonds, and in C(s, diamond) the carbon is more rigidly held in place and will have less entropy. In summary, C(s, diamond) < H20(s) < H20(1) < H20(g) .
7.27
(a) Iodine is expected to have higher entropy due to its larger mass and consequently a larger number of fundamental particles. (b) When we consider the two structures, it is clear that I -pentene will have more flexibility in its framework than cyclopentane, which will be comparatively rigid. Therefore, we predict I -pentene to have a higher entropy . (c) Ethene (or ethylene) is a gas and polyethylene is a solid, so we automatically expect ethene to have a higher entropy . Also, for the same mass, a sample of ethene will be composed of many small molecules, whereas polyethylene will be made up of fewer but larger molecules .
SM- 1 95
7.29
Diethyl ether, (CH3CH2)zO, has the greater standard molar entropy because its molecular structure consists of more atoms than dimethyl ether, (CH3)zO .
7.31
(a) Entropy should decrease because the number of moles of gas is less on the product side of the reaction . (b) Entropy should increase because the dissolution of the solid copper phosphate will increase the randomness of the copper and phosphate ions. (c) Entropy should decrease as the total number of moles decreases .
7.33
L'lSB
<
L'lSc < L'lSA.
The change in entropy for container A is greater than
that for container B or C due to the greater number of particles . The change in entropy in container C is greater than that of container B because of the disorder due to the vibrational motion of the molecules in container C.
- I s o products - I s o reactants = 69.9 1 J . K- ' . mol-' [1 30 . 68 J . K- ' . mor ' + + (205 . 1 4 J . K- ' . mor ' )] = -1 63 . 34 J . K- ' . mol-' The entropy change is negative because the number of moles of gas has decreased by 1 . 5 . N ote that the absolute entropies of the elements are not 0, and that the entropy change for the reaction in which a compound is formed from the elements is also not O. L'lSO r
-
-
(b) CO(g) ++ 0 2 (g) � CO2 (g) J . K- I . mol-I - [1 97.67 J K I . mol- I + + (205 . 1 4 J K I . mol-I )] = -86 . 5 J . K -I . mol-I
L'lSI� = 2 1 3 . 74
·
·
-
-
The entropy change is negative because the number of moles of gas has decreased by 0 . 5 .
SM- 1 96
(c) CaC0 3 (s) � CaO(s) + CO 2 (g) !1S0r = 39 . 75 J . K-' . morl + 2 1 3.74 J . K -' . mol - I - [+92 . 9 J . K - ' . mol-I ] = + 1 60 . 6 J . K-' . mol-I The entropy change is positive because the number of moles of gas has increased by 1 . (d) 4 KCIO ) (s) � 3 KCI04 (s) + KCI(s) !1S0r
=
3(1 5 1 .0 J . K-' · mol-I ) + 82 . 59 J . K-' · mol-I - [4(1 43. 1 J · K-' · mol-I )] = - 36 . 8 J . K-' · mol-I
It is not immediately obvious, but the four moles of solid products are more ordered than the four moles of solid reactants .
7.37
dT C and therefore, !1S = . � dT. , T T " If C = a + bT + ciT 2 , then ('2 a + bT + ciT 2 = !1S dT Jrl T
dq dS = � = T
C
P,III
i'
P, III
=
('2 (� + b + -;T )dT ( a ln(T) + bT _ � ) Jr, T 2T =
(
(J
J
T 2
TI
T 1 _ 1 a ln 2 + b(T2 - T, ) _ � _ - 2 2 I 2 T2 T, T. I I !1S(true) for heating graphite from 298 K to 400 K is =
=
(1 6.86 J . K - 1 mOI- 1 ) ln •
( J ( 400 K 298 K
+ (0 . 00477 J . K -2 . mo]-l )( 400 K - 298 K) X I 05 J . K . mol- 1 ) 1 1 _ (-8.54 _ 2 2 (400 K) (298 K) 2 = 3 . 3 1 J . K- 1 mo]-l
J
•
If we assume a constant heat capacity at the mean temperature of 350 K:
SM- 1 97
Cpm = (1 6.86 J · K-I ' mol-I ) + (0 . 00477 J · K-2 ' mol-I )(350 K) (-8 . 54 x 1 05 J · K · mol-l ) + �-------��--� (350 K)2 = I 1 . 6 J · K-I · mol -1 and
[ J
T 400 K = 3.4 1 J . K - I . mol-I �S(mean) = Cpm, In -2 = (1 1 .6 J . K -I . mol - I ) In TJ 298 K N ot integrating the heat capacity leads to roughly a 3.0% error in �S. 7.39
The entropy of vaporization of water at 85 °C may be carried out through a series of three reversible steps . N amely, reversibly heating the reactants to 1 00 °C , carrying out the phase change at this temperature, and finally cooling the products back to 85 °C . The sum of the �S' s for these three steps will be equivalent to vaporizing water at 85 °C in one irreversible step. Step 1 , heating the reactants to 1 00 °C :
[J
[ J
T 373 K = 3.09 J . K-I · mol-I �S = Cpm, In 2 = (75 . 3 J . K-I . mol-I )In I 358 K � Step 2, the entropy of vaporization of H2 0 at 1 00 °C is 1 09 . 0 J . K-I . mol-I . Step 3, cooling the products to 85 °C : f>S,
= CP m In
(i, J
= (33.6 J T' mol- ' )In
(!���J
= -1 .38 J K - , mol- '
Therefore, the molar entropy of vaporization is H20 at 85 °C is: �S v,m = �S I + �S 2 + �S3 = 1 1 1 J · K- 1 · mol- 1 • 7.41
To calculate the change in entropy for the hot and cold water, the amount of energy which flows from one to the other must first be calculated: = -qh = n c ' Cp,m ( H 2 0) . (Tf - T;,J = -n h . Cp.m ( H 2 0) . (Tf - T;,h ) qc
SM- 1 98
where n c and n h are the moles of cold and hot water, respectively, and � , c and � ,h are the initial temperatures of the cold and hot water,
respectively. Dividing both sides by Cp.m (H 2 0) we obtain: n c . (Tf - � ,c ) = -n h . (Tf - �,h )' The moles of hot and cold water are: 65 g 50 g 1 = 2 . 78 mol and n h = 1 nc = 1 8 . 0 1 5 g ' mol1 8.0 1 5 g · mol-
=
3.61 mol
and the final temperature Tf is therefore: (2. 78 mol · 293 . 1 5 K) + (3. 6 1 mol · 323 . 1 5 K ) = 3 1 0. 1 K. Tf = (2 . 78 mol + 3 .6 1 mol) With Tf, we can calculate I1S for the hot and cold water, and the total I1S for the entire system: I1Sc = n c ' Cp.m In
= + 1 1 . 9 J . K- 1
( J
I1Sh = n h CP,111 In •
(
)
(
J
(
)
(
J
Tf 0l = (2.78 mol) 75 . 3 J . K- 1 · mol - I In 3 1 . K 293 K � ,c
( J Tf �,h
= (3 . 6 1
3 1 0. 1 K mOl) 75 .3 J . K- 1 . mol- 1 In 323 K
J . K- 1 Therefore, I1Stot is : I1Stot = I1Sc + I1Sh = +0.8 J . K - I . = -1 1 . 1
7.43
(a) The change in entropy will be given by -1 .00 mol x 8 . 2 x 1 03 J . mol-I y I1S = I::J!s stem = = -73 J . K - 1 1 1 1 .7 K T 1 00 mol x 8 . 2 x 1 03 J . morl I1Ssystem = I::J!system = . = +73 J · K - I 1 1 1 .7 K T SUIT
(b) I1S
SUIT
I1S . = '>,Im.
=
-I::J!system
T
I::J!system
T
=
=
-1 .00 mol x 4 . 60 x l 03 J . mol-I = - 29 . 0 J . K-1 1 58 . 7 K
1 . 00 mol x 4.60 x 1 03 J . mol-I 1 58 . 7 K
SM - 1 99
=
+ 29 . 0 J . K - I
(c)
I1SSUIT =
-/).}{system
T
-(1 . 00 mol x -4 . 60 x l 0 = 1 58 . 7 K
3 . J morl )
= + 29 . 0 J · K-1 I1Ssystem =
7.45
/).}{system
T
=
1 . 00 mol x -4 . 60 x I 0 1 5 8 .7 K
(a) The total entropy change is given by
3 . J mol-I
=
I1Stot = I1Ss lT u
- 29 . 0 J . K-I
+ I1S. I1S for an
isothermal, reversible process is calculated from I1S = qrev = -wrev = T T
nR In
V2
To do the calculation we need the value of � n, which is obtained by use of the ideal gas law: •
(4 . 95 atm) ( 1 . 67 L) = n(0 . 082 06 L · atm · K -I . mol-I ) (323 K); n = 0 . 3 1 2 mol . 7 . 33 L = +3 . 84 J . K-I . Because the I1S = (0 . 3 1 2 mol) (8 . 3 1 4 J . K-I . mol-I ) In 1 . 67 L process is reversible, I1St01 = 0, so I1Ss lT = -I1S = -3 . 84 J . K-I . u
(b) For the irreversible process,
I1S
is the same, + 3 . 84 J . K-I . N o work is
done in free expansion (see Section 6 . 6) so
W = O.
Because
I1 U =
0 , it
follows that q = O. Therefore, no heat is transferred into the surroundings, and their entropy is unchanged: therefore 7.47
I1St01 =
+3 . 84 J . K-I .
I1Ss lT = u
O . The total change in entropy is
Spontaneous change means I1Stot > O. A vapor to liquid change is condensation and it releases heat . However as shown in the diagram, the temperature of the system does not change . Therefore the released heat must have left system to the surroundings . Since heat flows from hot to cold, the temperature of the surroundings ( Tsun) must be lower than the temperature of the system ( Tsys), and since q I1Ssys = rev Tsys
and I1Ssurr
=
q rev
Ts
urr
' then I1Stot > O.
SM -200
7.49
Exothermic reactions tend to be spontaneous because the result is an increase in the entropy of the surroundings . Using the mathematical relationship
f.,. G,. = M-l,.
negative compared to
- T f.,.S,. , it is clear that if M-lr is large and
f.,.Sr '
then the reaction wiII generaIIy be
spontaneous . 7.51
(a)
M-l°r
f.,.Sor
f.,.Go r
f.,.Gr
= 3(-824 . 2 kJ · mol-I ) - [2(-1 1 1 8 . 4 kJ · mol-I )] = -235 . 8 kJ . mol-I = 3(87 . 40 J . K-I . mol-I ) - [2(1 46 . 4 J . K - I . mol-I ) + + (205 . 1 4 J . K-I . mol-I )] = -1 33 . 1 7 J . K-I . morl = 3(-742 . 2 kJ . mol-I ) - [2(-1 0 1 5 . 4 kJ . morl )] = -1 95 . 8 kJ . mol-I
may also be calculated from
M-l°r
and
f.,.Sor
(the numbers calculated
differ slightly from the two methods due to rounding differences): f.,.Gor = M-l ° r - T f.,.Sor = -23 5 . 8 kJ · mol-I - (298 K) (-1 33 . 1 7 J . K-I . mol-I )/(1 00 J . krl ) = -1 96 . 1 kJ · morl (b)
M-l°r
f.,.Sor
f.,.Gor
= - 1 208 . 09 kJ · mol-1 - [- 1 2 1 9 . 6 kJ · morl ] = 1 1 . 5 kJ · mol-I = - 80 . 8 J . K-I . mol-I - [68 . 87 J . K-I . mol-I ] = - 1 49 . 7 J . K-I . morl = - 1 1 1 1 . 1 5 kJ . morl - [- 1 1 67 . 3 kJ . mol-I ] = + 56 . 2 kJ . mor l
or f.,.Go r = M-l 0r
- T f.,.So r = 1 1 . 5 kJ · mol - I - (298 K) (-1 49 . 7 J . K-I . morl )/(l OOO J . krl ) = 56 . 1 kJ . mol - I
SM-201
=
9 . 1 6 kJ . mol-I - [ 2(33 . 1 8 kJ · mol-I )] = -57 . 20 kJ . mol-I I1sor = 304 . 29 J . K-I . morl - [2(240 . 06 J . K-I . mol-I )] = -1 75 . 83 J . K-I . mol-I I1G or = 97 . 89 kJ . mol-I - [2(5 1 . 3 1 kJ . mol-I )] = -4 . 73 kJ . mol-I
(c) MfOr
or = -57 . 2 kJ . morl
- (298 K) ( -1 75 . 83 J . K . mol-I )/(1 000 J . krl ) -I
= -4 . 80 kJ . mol-I
7.53
(a) + N 2 (g) + 1 H 2 (g) � NH3 (g) Mf O r = MfO r (NH3 ) = -46. 1 1 kJ . mol-I
I1sor = S O m ( NH 3 ' g) - [+ So m ( N 2 ' g) + 1 Som ( H 2 ' g)]
J . K . mOrl - [+ (1 9 1 . 6 1 J . K -I . mol-I ) + 1 (1 30 . 68 J . K-I . mol-I )] = -99 . 38 J . K-I . mOrl = 1 92 . 45
-I
I1GOr = -46 . l 1 kJ · morl - (298 K) ( -99 . 38 J . K-I . mol-I )/(1 000 J · kJ-l ) = -1 6 . 49 kJ . mOrl
S O m (NH 3 ) = 1 92 . 45 J . K-I . mol-I I1So r (NH3 ) is negative because several gas molecules combine to form 1 NH3 molecule .
(b) H 2 (g) + + 0 2 (g) � H 2 0( g ) Mfor
=
Mfor (H 2 0, g) = -24 1 . 82 kJ · mol - I
I1So r = S O m (H 2 0 , g) - [ S O m ( H 2 ' g) + + S O m ( ° 2 , g)]
J . K-I . mor l - [1 30 . 68 J . K-I . mol-I + + (205 . 1 4 J . K-I . mol-I )] = -44 . 42 J . K-I . morl �G or -24 l . 82 kJ · mol-I - (298 K) ( -44 . 42 J . K-I . morl )/(l OOO J · krl ) = -228 . 58 kJ . mol-I S O m (H 2 0, g) 1 88 . 83 J . K-I . mol-I = 1 88 . 83
=
=
I1S 0 (H 2 0 , g) is a negative number because there is a reduction in the r
number of gas molecules in the reaction when S O m is positive . SM -202
(c) C(s, graphite) + + 0 2 (g) � CO(g) WO r = WO (CO, g) = -1 1 0.53 kJ . mol-I !1So r = Sam (CO, g) - [SO m (C, s) + 1 Sam ( ° 2 , g)] = 1 97 . 67 J . K-I . mol-I - [5 . 740 J . K-I . mol-I + 1 (205. 1 4 J . K-I . mol-I )] = + 89 . 36 J . K-I . mol-I !1Go = -1 1 0 . 53 kJ . mol-I - (298 K) (89.36 J . K-I . mol-I )/(1 000 J . krl ) = -1 37 . 2 kJ . morl S am (CO, g) = 1 97 . 67 J . K -I . morl r
r
The Sam (CO, g) is larger than !1So (CO, g) because in the formation r
reaction the number of moles of gas is reduced . WO = WO r ( N0 2 ) = +33 . 1 8 kJ . mol-I !1sor = Sam ( N0 2 , g) - [ 1 SO nJ N 2 , g) + S am ( ° 2 , g)] = 240 . 06 J . K-I . mol-I - [1 (1 9 1 . 6 1 J . K-I . mol-I ) + 205 . 1 4 J . K-I . mol-I ] = -60 . 89 J . K-I . mol-I !1Gor = 33 . 1 8 kJ · mol-I - (298 K) ( -60.89 J · K-I . morl )/(1 000 J · krl ) = +5 1 . 33 kJ . mol-I S am ( N0 2 , g) = 240.06 J . K -I . mol-I r
The !1SO r ( N0 2 , g) is somewhat negative due to the reduction in the number of gas molecules during the reaction . For all of these, the important point to gain is that the Sam value of a compound is not the same as the !1SO for the formation of that compound . !1SO r is often r
negative because one is bringing together a number of elements to form that compound . 7.55
Use the relationship !1Gor = L: !1Gor (products) - L: !1Gor (reactants) : (a) !1Gor = 2!1Gor (S03 ' g) - [2!1Gor (S0 2 ' g)] = 2(- 3 7 1 .06 kJ · morl ) - [2(- 300 . 1 9 kJ · mol-I )] = - 1 4 1 . 74 kJ · morl The reaction is spontaneous . SM -203
(b) /)'Gor
=
/)'Gof (CaO, s) + /),Gof (C0 2 , g) - /),Gof (CaC0 3 , s) = (- 604 . 03 kJ · moI- I ) + (- 394.36 kJ · moI-I ) - (- 1 1 28 . 8 kJ · moI- I ) = + 1 30 . 4 kJ . mol-I
The reaction is not spontaneous . (c) /)'Gor = 1 6/),Gof (C0 2 , g) + 1 8/)'Gof (H 2 0, 1) - [2/)'Gof (C8 H I 8 , 1)] = 1 6(-394.36 kJ . mol-I ) + 1 8(- 237 . 1 3 kJ . mol-I ) - [2(6 . 4 kJ . mol-I )] = -1 0 590 . 9 kJ . mol-I The reaction is spontaneous . 7.57
The standard free energies of formation of the compounds are: (a) PCI s (g), -305.0 kJ . mol-I ; (b) HCN(g), +1 24 . 7 kJ . mol-I ; (c) NO(g), +86 . 55 kJ . mor' ; (d) S0 2 (g), -300 . 1 9 kJ . mOr ' . Those compounds with a positive free energy of formation are unstable with respect to the elements . Thus (a) and (d) are thermodynamically stable.
7.59
To understand what happens to /)'Go r as temperature is raised, we use the relationship /)'Go
r
=
Mfor
-
T /)'So r .
From this it is clear that the free
energy of the reaction becomes less favorable (more positive) as temperature increases, only if /)'S O r is a negative number. Therefore, we need only to find out whether the standard entropy of formation of the compound is a negative number. This is calculated for each compound as follows: (a) pe s) + � CI 2 (g) � PCI s (g) /)'Sor = Sam (PCI s ' g) - [Sam (P, s) + � S a m (CI 2 , g)] = 364 . 6 J . K-' . morl - [ 4 1 . 09 J . K-I . mol-I + � (223.07 J . K - ' . mol-I )] = - 234 . 2 J . K-' . morl The compound is less stable at higher temperatures . (b) C(s), graphite + 1 N2 (g) + 1 H 2 (g) � HCN(g) SM-204
;j,S or = S a m (H eN, g) [ Sam (C, s) + 1 S a m (N2 , g) + 1 s o m (H 2 , g) ] = 201 . 78 J . K -I . mOrl [5 . 740 J . K-I . mOrl + 1 ( 1 9 1 . 6 1 J · K-I . mOrl ) + 1 ( 1 30 . 68 J · K-I . mol-I ) ] = +34 . 90 J . K-I . mOrl -
-
HCN(g) is more stable at higher T. (c) 1 N 2 (g) + 1 0 2 (g) � NO(g) ;j,S O r = S O m (NO , g) [1 S 0 m (N 2 , g) + 1 S 0 m (0 2 ' g) ] = 2 1 0 . 76 J · K-I . mOrl [ 1 ( 1 9 1 . 6 1 J · K-I . mol-I ) + 1 (205 . 1 4 J · K-I . mol-I )] +1 2.38 J . K-I . mol-I -
-
=
NO(g) is more stable as T increases . (d) S(s) + 0 2 (g ) � S0 2 (g) ;j,S O r = S O(S0 2 ' g) - [SO(S , s) + SO(0 2 ' g)] = 248 . 22 J . K-I . mOrl - [3 1 . 80 J . K-I . mol-I + 205 . 1 4 J · K-I . mOrl ] = +1 1 . 28 J . K-I . mOrl S 0 2 (g) is more stable as T increases . 7.61
(a) 2 H 2 0 2 ( I) � 2 H 2 0(l) + 0 2 (g) ;j,S or
=
2 s o n, CH 2 0 , 1) + S a m (°2 , g) 2 s o m (H 2 0 2 , I) = 2(69 . 9 1 J . K-I . mOrl ) + 205 . 1 4 J . K-I . mol-I 2( 1 09 . 6 J · K-I . mOrl ) = + 1 25 . 8 J . K-I . mOrl -
-
W ; = 2 W; (H 2 0, aq) - 2 W ; (H 2 0 2 , 1) = 2(-285.83 kJ · mol- I ) - 2(-1 87 . 78 kJ · mol- I ) = -1 96 . 1 0 kJ mol-I
=
·
;j,G I� 2;j,G; (H 2 0, aq) - 2;j,G; (H 2 0 2 , 1) = 2(-237. 1 3 kJ · mol-I ) - 2(-1 20 . 35 kJ · mol- I ) -233 .56 kJ . mol- I =
SM-205
6.GO can also be calculated from 6.sor and MiO r using the relationship: r
6.GO r
MiO r - T 6.So = -1 96 . 1 k J . mol-I - (298 K) (+1 25 . 8 J . K-I . mol-I )/(l OOO J . krl ) = -233 . 6 kJ . mol-I
=
r
(b) 2 F2 (g) + 2 H 2 0(l) � 4 HF (aq) + ° 2 (g) 6.S; = 4S � (HF, aq) + S � ( ° 2 , g) - [2S �1 (F2 ' g) + 2S �1 (H 2 °, 1) ] I = 4(88.7 J . K -1 . mol- ) + 205 . 1 4 J . K - 1 . mol- 1 - [2(202 . 78 J . K -1 . mol- I ) + 2(69 . 9 1 1 . K - 1 . mol- I )] = +1 4 . 6 J · K- 1 · mol- 1 Mi; = 4Mi; (HF, aq) - 2Mi; (H 2 0,1) I = 4( -330.08 kJ . mol- I ) - 2( -285 . 83 kJ . mol- ) = -748 . 66 kJ . mol- 1 6.G J� = 46.G ; (HF, aq) - 26.G ; (H 2 °, 1) = 4( -296.82 kJ . mol- I ) - 2( -237 . 1 3 kJ . mol- I ) = -7 1 3 . 02 kJ · mol- 1 6.GO r can also be calculated from 6.S0 r and MiO r using the relationship: =
-748 . 66 kJ . mol-I - (298 K) ( 1 4 . 6 J . K-I . mor l )/( 1 000 J . krl ) = -753 . 0 1 kJ . mol-I 7.63
In order to find 6.Gor at a temperature other than 298 K, we must first calculate
MiO r
and 6.So r and then use the relationship
(a) In order to determine the range over which the reaction will be spontaneous, we consider the relative signs of MiO r and 6.S0 r and their effect on 6.Go r '
SM -206
2 WOr (B F:" g) + 3 WOr (H 2 0, I) - [WO r (B 2 0 3 , s) + 6W Or (HF, g)] = 2( - 1 1 3 7 . 0 kJ . mol-I ) + 3( - 285 . 83 kJ . mol-I ) - [( -1 272 . 8 kJ . mor ' ) + 6( - 27 1 . 1 kJ . mol-I )] = -232 . 1 kJ . mol-I !1So = 2som (BF3 ' g) + 3som (H 2 0' I) - [S Om (B2 0 3 , S) + 6som (HF, g)] = 2(254 . 1 2 J · K-' · mol-I ) + 3(69.9 1 J · K-' · mol-I ) - [53 . 97 J . K-' . mol-I + 6( 1 73 .78 J . K-' . mol-I )] = -378 . 68 J . K-' . mol-I !1Gor = -232 . 1 J · K-' · mol-I - (3 53 K) ( - 3 78 . 68 J · K- ' · morl )/(l OOO J · kr' ) = -98 . 42 kJ . mol-I
WOr
=
r
Because W O r is negative and !1sor is also negative, we expect the reaction to be spontaneous at low temperatures, where the term T !1So r will be less than W O r• To find the temperature of the cutoff, we calculate the temperature at which !1Gor = O . For this reaction, that temperature is !1G O r = 0 = - 232 . 1 kJ · morl - (T) (-378 . 68 J . K-' · mol-' )/( 1 000 J . kr' ) T = 6 1 2.9 K The reaction should be spontaneous below 6 1 2 . 9 K . (b) WO r
= WO r (CaCI 2 ,
aq) + W O r (C 2 H 2 , g) - [WO r (CaC 2 , s) + 2W O r (HCI, aq)] = (- 877 . 1 kJ · mol-' ) + 226 . 73 kJ . mol-I - [( -59 . 8 kJ . mol-I ) + 2( - 1 67 . 1 6 kJ . mol-I )] = - 256 . 3 kJ . mol-I !1sor = SC m (CaCI 2 , aq) + SCm (C 2 H 2 , g) - [SCm (CaC 2 , s) + 2soll, (HCI, aq)] = 59 . 8 J . K-' . morl + 200 . 94 J . K-' . mol-I - [69 . 96 J . K -I . morl + 2(56.5 J . K-' . mol-I )] = + 77 . 8 J . K- ' . mol-I !1Gor = - 256 . 2 kJ . mol-I - (3 53 K) ( +77 . 8 J . K-' . mol-I )/( 1 000 J . kr' ) = - 283 . 7 kJ · mol-'
SM-207
Because
Mfo,
is negative and
is positive, the reaction will be
I1So,
spontaneous at all temperatures . (c) Mfo,
= Mfof (C (s), diamond) = + 1 . 895 kJ · mol-l
I1So, = SOm (C
(s), diamond) SOm (C (s), graphite) = + 2.3 77 J . K-l . mol-l - 5 .740 J . K-l . morl = - 3 . 363 J . K-l I1Go, = + 1 . 895 kJ . morl - (353 K) ( - 3 . 363 J . K- l . mol-l )/ ( 1 000 J . kr l ) = + 3 .0 82 kJ · morl Because
Mfo,
-
is positive and
I1So,
is negative, the reaction will be
nonspontaneous at all temperatures . N ote: This calculation is for atmospheric pressure . Diamond can be produced from graphite at elevated pressures and high temperatures . 7.65
Assuming standard state conditions, I1Gro for C2H4(g) + H20(g) � CH3CH20H(l) is: I1Go, = (- 1 74.8 kJ . mol-I
) - (68. 1 5 kJ . mol-I ) - ( - 228.57 kJ . mol-I )
= - 1 4.4 kJ . morl Negative I1Go, indicates a spontaneous reaction . I1Gro
for C2H6(g) + H20(g) � CH3CH20H(l)
)
I1Go, = (- 1 74.8 kJ . morl - ( -32.82 kJ . mol-I =
+
H2(g) is:
) - ( - 228.57 kJ . mol-I )
. mol-l P ositive I1Go, indicates a nonspontaneous reaction . Reaction A is spontaneous, but reaction B is not spontaneous at any temperature . 7.67
+ 86.6 kJ
The reactions ( 2) and (3) in this problem are used to drive the regeneration of ATP . For every three moles of NADH that undergo reaction: 3 NAD+(aq) + 3H+ (aq) + 6e I1G, = 3( - 1 5 8 . 2 kJ) 3 NADH(aq) �
f 02(g) + 6 H\ aq) + 6e
3 N ADH(aq) +
I1G, = 3 ( -6 1 .9 kJ) 3H20(I) f 02(g) + 3H +(aq) 3 H 20(l) + 3NAD +(aq) -
�
�
I1G, =
SM-208
-660 . 3 kJ
Therefore, for every three moles of N ADH, 660.3 kJ / 30.5 kJ = 2 1 . 6 moles of ATP would be regenerated if all of the free energy released were used to drive reaction ( 1 ) . 7.69
(a) I -propanol (C 3 HsO) and 2-propanone (C 3 H 6 0) have similar numbers of electrons so that we would expect the molar entropies to be similar. Because I -propanol exhibits hydrogen bonding, however, we might expect the liquid phase to be more ordered than for 2 -propanone . This is observed . The standard molar entropy for 2 -propanone is 200 J . K-I . mol-I while that of I -propanol is 1 93 J . K -I . mOr l . (b) In the gas phase, hydrogen bonding will not be important because the molecules are too far apart, so the standard molar entropies should be more similar.
7.71
7.73
(a) IJ. Go < 0 (b) MP unable to tell ( c) IJ.S unable to tell (d) IJ.Stotalr > 0 For the cis compound there will be 1 2 different orientations: For the trans compound there will only be 3 different orientations . Comparing the Boltzmann entropy calculations for the cis and trans forms: CI S :
S = k In 1 2 6 02 x l021 = ( l . 3 8 1 0-23 J . K - I ) In 1 2 6 02 x 1 021 S = 20.6 J . K-I trans: S = k In 3 6 0 2 x l o2) = ( l .38 1 0-2 3 J · K-I ) In 3 6 02 x l021 X
x
S = 9 . 1 3 J · K-I The cis form should have the higher residual entropy. 7.75
(a)
W=3
SM-209
(b) (c)
7.77
W = 12 Initially one ofthe three atom systems had two atoms in higher energy states. In part (b) the system will be at equilibrium when each three atom system has one quantum of energy . Therefore energy will flow from the system with two quanta to the system with none .
According to Trouton ' s rule, the entropy of vaporization of an organic liquid is a constant of approximately 85 J . mol- ' . K- ' . The relationship between entropy of fusion, enthalpy of fusion, and melting point is given by
f.,SO fus =
,').}[O fus 'T' J: fus
5 1 00 J = 8 . 50 J . K -1 600 K 2290 J = -I = 9.79 J . K 234 K 2640 J . -1 = 7. 1 2 J K = 371 K
=
For Pb : f.,S;us o
For Hg : f.,Sfus For Na : f.,S;us
These numbers are reasonably close but clearly much smaller than the value associated with Trouton ' s rule. 7.79
This is best answered by considering the reaction that interconverts the two compounds We calculate !1Gor f.,Gor
!1Gor
using data from Appendix 2A:
=
6f.,Gof (Fe 2 0 3 , s) - [4!1Gor (Fe3 04 , s)] = 6(-742 . 2 kJ · mol- ' ) - [4(- 1 0 1 5.4 kJ · mor' ) ] = - 3 9 l . 6 kJ · mor'
Because
f.,Gor
is negative, the process is spontaneous at 25°C.
Therefore, Fe2 0 3 is thermodynamically more stable .
SM-2 1 0
7.81
We can calculate the free energy changes associated with the conversions: (a) 2 FeS(s) � Fe(s) + FeS 2 (s) (b) FeS 2 (s) � S(s) + FeS (s) For (a), !::" G or = - 1 66 . 9 kJ · mor l - 2 ( -1 00 . 4 kJ · mor l ) = + 3 3 . 9 kJ . mol- I
.
This process is predicted to be nonspontaneous.
For (b), !::" G O r = - 1 00.4 kJ . mor l - ( - 1 66.9 kJ . mor l ) = + 66.5 kJ mol- I This process is predicted to be nonspontaneous . 7.83
H2 (g) + Br2(g)
(a)
�
2 HBr(g)
Use thermodynamic data at 25 °e from the appendix, and the relationship !::" G or
=
!::" G or
=
L:
!::" G or (products)
-L:
!::" G or (reactants) :
2!::" G or ( HBr, g) - [ !::" G or ( H 2 ' g) + !::" G o r ( B r2 ' g)] l I = 2(-53 . 45 kJ · mol- ) - [0 + (3 . 1 1 kJ · mor )] = -1 1 0.0 kJ . mol- I STP : ooe or 273 . 1 5K and I atm
(b)
moles of H2 gas: n=
p·v R.T
=
1 . atm · 0 . 1 20 L ( 0 . 0820574 L . atm . K - I . m 01-
I
) . 273. 1 5 K
= 5 . 3 5 x I 0 - 3 mol
A stoichiometric amount of bromine gas is 5 . 3 5 x 1 0-3 mol . Moles of HBr(g) formed = 2( 5 .35 x 1 0-3 mol )
=
2 l . 07 x 1 0- mol
Molar concentration of the resulting hydrobromic acid, !!... l .07 x 1 0-2 2 M= = = 7 . 1 4 x 1 0- mol · C l 0 . 1 50 L
V
7.85
(a) Because the enthalpy change for dissolution is positive, the entropy
[
)
change of the sUlToundings must be a negative number !::"S osurr
=
-
!::,.}f O
system
T
.
. Because spontaneous processes are accompamed
SM-2 1 1
by an increase in entropy, the change in enthalpy does not favor the dissolution process . (b) In order for the process to be spontaneous (because it occurs readily, we know it is spontaneous), the entropy change of the system must be positive . (c) P ositional disorder is dominant . (d) Because the surroundings participate in the solution process only as a source of heat, the entropy change of the surroundings is primarily a result of the dispersal of thermal motion . (e) The driving force for the dissolution is the dispersal of matter, resulting in an overall positive !1S . 7.87
(a) (i) !1G,o = L !1G; (products) - I !1G; (reactan ts) = - 1 20 . 35kJ . mol- ' Method (i) releases more energy per mole of O2.
7.89
= +
(ii)
!1G,O = 2 (- 1 20.3 5) - 2 (-237 . 1 3)
(b)
Method (i) has the more negative standard Gibbs free energy.
(c)
No .
233 . 56 klmor'
The entries all correspond to aqueous ions . The fact that they are negative is due to the reference point that has been established . Because ions cannot actually be separated and measured independently, a reference point that defines SOm (H+ , aq) = 0 has been established . This definition is then used to calculate the standard entropies for the other ions . The fact that they are negative will arise in part because the solvated ion M (H2 0 )x"+ will be more ordered than the isolated ion and solvent molecules ( Mn+
7.91
!1G
=
/jj{ -
T!1S, therefore,
SM-2 1 2
+ x
H 2 0).
f:,Jf - f:,G T Given the reaction H 2 (g) + ± 02 (g) � H 20(l) and 0.50 g of H 2 (g) consumed : f:,Sr
f:,Gr
r
=
=
r
g (- 237.25 kJ . mol- 1 {\ 2 . 0 1 0.50 6 g ' mol- 1
(
Therefore, f:,S
7.93
.
I
=
- 70.9 kJ - (- 5 8 . 8 kJ ) 298 K
][ J
J
= -5 8 . 8 kJ.
1 03 J = -40.6 J . K- 1 • 1 kJ
(a) In order to calculate the free energy at different temperatures, we need to know f:,Jf 0 and f:,SO for the process : H 2 0(l) � H 2 0(g) f:,Jf ° r = f:,Jf °r (H 2 0, g) - f:,Jf ° r ( H 2 0, 1) = ( - 24 l . 82 kJ . mor l ) - [ - 285 . 83 kJ . mol- I ] 44 . 0 1 kJ . mol- I f:,Sor = Sam ( H 2 0, g) - Sam ( H 2 0, 1) = 1 88 . 83 J · K- I . mol- I - [69.9 1 J · K- I . mor l ] = 1 1 8 . 92 J . K- I . mol- I =
= 44.01 kJ . mol- I - T ( 1 1 8.92 J . K- I . mol- I )/( 1 000 J . kr l ) T(K) f:,G o r (kJ) 298
8 . 57 kJ
3 73
- 0 . 3 5 kJ
423
- 6.29 kJ
The reaction goes from being nonspontaneous near room temperature to being spontaneous above 1 00°C . (b) The value at 1 00°C should be exactly 0, because this is the normal boiling point of water. (c) The discrepancy arises because the enthalpy and entropy values calculated from the tables are not rigorously constant with temperature. Better values would be obtained using the actual enthalpy and entropy of vaporization measured at the boiling point .
SM -2 1 3
7.95
(a) Since standard molar entropies increase with temperature (more translational, vibrational and rotational motion), the -T!1S; term becomes more negative at higher temperatures. (b) For gases with translational motion, their increase in standard molar entropy, on heating, is much larger than for solids and liquids. Therefore their - T !1S,� term is more negative .
7.97
The dehydrogenation of cyc1 0hexane to benzene follows the following equation:
We can confirm that this process is nonspontaneous by calculating the !1Gor for the process, using data in Appendix 2 A : !1Gor
=
!1Go r (C 6 H 6 , 1) - !1Gor (C 6 H I 2 , 1) l I = 1 24.3 kJ · mol- - 26 . 7 kJ · mor I = + 97 . 6 kJ . mol-
The reaction of ethane with hydrogen can be examined similarly: C 2 H 2 (g) + H 2 (g) � C 2 H 6 (g) !1Gor = !1Go r (C 2 H 6 , g) - !1Gor (C 2 H I 2 , g) = ( - 3 2.82 kJ . mor l ) - 68. 1 5 kJ . mor l = - 1 00.97 kJ . mol- I We can now combine these two reactions so that C 2 H 2 (g) accepts the hydrogen that is formed in the dehydrogenation reaction: C 6 H I 2 ( 1 ) � C 6 H 6 (l) + 3 H 2 (g)
+ 3 [C 2 H 2 (g) + H 2 (g) � C 2 H 6 (g)] !1Gor !1Gor
=
=
+ 97.6 kJ · mor l 3 ( - 1 00.97 kJ · mor l )
C 6 H I 2 (l) + 3 C 2 H 2 (g) � C 6 H 6 (l) + 3 C 2 H 6 (g) !1Gor = - 205 . 1 3 kJ · mol - I
SM-2 1 4
We can see that by combining these two reactions, the overall process becomes spontaneous. Essentially, we are using the energy of the favorable reaction to drive the nonfavorable process.
7.99
(a)
H3C
,
c=c
CH3
\
/
H
/
H
cis-2-Butene
(1)
H3C
,
c=c
/
H
/
H
\
CH3
trans-2-butene
(2)
H3C
,
c=c
/
H3C
/
H
\
H
2-Methylpropene
(3)
(b) For the three reactions, the calculation of !1Go,
Mr,
and !1So are as
follows: !1Go, = !1Gor (2) - !1Gor (1) = 62.97 kJ . mor l - 65. 86 kJ . mol- I I = -2.89 kJ . mol Mfo, = Mfor (2) - Mfor (l) ' ' = (-1 l . 1 7 kJ · mol- ) - (- 6. 99 kJ · mol- ) !1Go , - 2 . 89 kJ . mol - I !1So,
= -4. 1 8 kJ . mol- I = Mf O , - T !1So , l = -4. 1 8 kJ . mor - (298 K) (!1So, )/(1 000 J . kr ' ) ' l = -4.33 J . K - . mor
!1Go, = !1Gor (3) - !1Gor (l) I I = 5 8 .07 kJ . mol - - 65 . 86 kJ . mol= -7 . 79 kJ · mor l Mfor = Mfo r (3) - Mfor (l) = (-1 6.90 kJ . mor ' ) - (-6.99 kJ . mol- I ) = -9.9 1 kJ . mor l
!1Go , = Mfo, - T !1So, -7.79 kJ · mol- I = -9 . 9 1 kJ · mor l (298 K) (!1S0, )/( l 000 J . kr ' ) !1So , = -7 . 1 1 J · K- ' · mor ' -
SM-2 1 5
tJ.GOr = tJ.Gof (3) - tJ.Gof (2) =
5 8 . 07 kJ . mol- ' - 62.97 kJ . mor ' ' = -4 . 90 kJ . mor bHo r = bHo f (3) - bHo f (2) ' ' = (-1 6 . 90 kJ · mor ) - (- 1 1 . 1 7 kJ · mor ) = -5 . 73 kJ · mor ' tJ.Gor = bHor - TtJ.sor =
-5.73 kJ · mor ' - (298 K) (tJ.S O r )/ ( l OOO J · kr ' ) tJ.So r = -2 78 J · K- ' · mor '
-4.90 kJ · mor '
•
(c) The most stable of the three compounds is 2-methylpropene . (d) B ecause tJ.SO is also equal to the difference in the S am values for the compounds, we can examine those values to place the three compounds in order of their relative absolute entropies . The ordering is S O m (I) > Sam (2) > Sam (3). 7.101 (a) 2NaN 3 { S) � 3N 2 {g) + 2Na{g)
(b) tJ.S� is positive because a gas is produced (from a solid). o
+
(d) tJ.S� = I S,�, ( products)
N
-
==
0
N
Overall the nitrogen is oxidized .
I S�, ( reactants)
= [ 3 ( 1 9 1 .6 1 ) + 2 ( 1 53 . 7 1 ) ] - [ 2 ( 96.9») ] =
[ 574.83 + 307 . 42 ] - [ 1 93 . 8 ]
= + 688.45J.K- ' .mot ' = + 688.4J.K- ' . mol- ' ( e) bH,o = I tJ. H� ( products) - I tJ. H� ( reactants) = [ 3 ( 0) + 2 ( 1 07 . 32 ) ] - [ 2 ( 2 1 . 7) ] = 2 1 4.64 - 43 . 4 = 1 7 1 .24 = 1 7 1 .2 kJ.mol- '
)
(
(
' ' tJ.G,O = bH� - TtJ.S� = 1 7 1 .2 x 1 0 3 J.mol· ' - 298K 688 . 4 JY . mol-
)
= - 3 . 39 x 1 0 4 J . mol - ' (g) bH,o and tJ.S� are both positive . Therefore lowered . (f) yes
SM-2 1 6
CHAPTER 8 PHYSICAL EQUILIBRIA
8.1
In a 1 . 0 L vessel at 20oe, there will be 1 7 . 5 Torr of water vapor . The ideal gas law can be used to calculate the mass of water present:
PV
let
(
=
m =
nRT
mass of water
(
)
)
m 1 7.5 Torr (1 0 L) = (0.082 06 L . atm . K- 1 . rnol- I ) (293 K) . ' 1 1 8 .02 g · mol760 Torr · atm(1'-7 . 5 Torr) ( 1 .0 L)( 18 . 02 g . mol- I ) m = --:-(760 Torr · atm- ' ) (0.082 06 L · atm · K- ' · mol- ' ) (293 K ) -
m
8.5
=
-
-
-
0.0 1 7 g
(a) The quantities
t:JfO yap
and t.,.so
yap
can be calculated using the
relationship In
P
=
�o R
ya p
1 + t.,.so .T R
yap
'-
-
Because we have two temperatures with corresponding vapor pressures, we can set up two equations with two unknowns and solve for �o
yap
and
P must be expressed in atm, which is the standard reference state. Remember that the value used for P is really activity which, for pressure, is P divided by the reference state of 1 atm, so t.,.so
ya p .
If the equation is used as is,
that the quantity inside the In term is dimensionless.
SM-2 1 7
Mfo vap 3 5 Torr 8 . 3 1 4 J . K- I . mol- l x ln = + L1So yap 760 Torr 1 6 1 .2 K
8 . 3 1 4 J . K -I . mo 1- 1
x
Mf o vap 253 Torr =+ L1S ° yap 760 Torr 1 89 . 6 K
1n
which give, upon combining terms, -25 . 59 J . K- I . mor l = -0.006 203 K- I -9. 1 45 J · K- I · mol- 1 = -0.005 274 K- 1
X
Mfo va p + L1Sovap
x
Mfo vap + L1So va p
Subtracting one equation from the other will eliminate the L1So Yap term and allow us to solve for MfO ya p -1 6.45 J . K -I . mol- I = -0 . 000 929 x Mfo yap Mfo = + 1 7. 7 kJ · mol- I yap
(b) We can then use Mfo vap to calculate L1So vap using either of the two equations: -25 . 59 J · K- I · mol- I = -0.006 203 K- I L1So = 84 J . K- I . mor l
x
( + 1 7 700 J · mor l ) + L1Sovap
I -9 . 1 45 J . K- I . mor l = -0.005 274 KI l L1So = 84 J . K- . mor
x
l ( + 1 7 700 J · mor ) + L1So yap
yap
yap
( c) The L1Go yap is calculated using L1Go r = MfO r - TL1SO /::" G or L1Gor
r
= +1 7 . 7 kJ · mor l - (298 K) (84 J . K- I . mOr l )/(l OOO J . kr l )
=
-7.3 kJ · mor l
Notice that the standard L1Gor is negative, so that the vaporization of arsine is spontaneous, which is as expected; under those conditions arsine is a gas at room temperature . (d) The boiling point can be calculated from one of several methods. The easiest to use is that developed in the last chapter: T. -0 L1Go yap - MfO yap - B L1So Yap A U Llll O ya p
T.
B
- T. L1So ya p 0 r T. B
Mfo Yap
_ _
B -
L1S yap
1 7 . 7 kJ . mor l x 1 000 J . kr l = 21 1 K = 84 J . K - I . mor l SM-2 1 8
· h·lp I n � Alternatively, we could use th e reIatlons
-
= -
1'1H0 yap
�
R
[
1 -
-
T;
1 -
I;
]
.
Here, we would substitute, in one of the known vapor pressure points, the value of the enthalpy of vaporization and the condition that P
=
1 atm at
the normal boiling point.
8.7
(a) The quantities
1'1H0 yap
and !1Soyap can be calculated, using the
relationship In
P
= _
1'1H0 yap R
1 . _
T
+
!1SO yap R
Because we have two temperatures with corresponding vapor pressures (we k now that the vapor pressure
=
1 atm at the boiling point), we can set
up two equations with two unknowns and solve for the equation is used as is,
P
1'1H0 yap and
!1SoYpa If •
must be expressed in atm which is the
standard reference state. Remember that the value used for P is really activity which, for pressure, is P divided by the reference state of 1 atm, so that the quantity inside the In term is dimensionless. 8.314 J· K-1 • mol-I xIn 1 = 8. 314 J .K-1. morl xln
1'1H0 yap
315.58K
140 Torr
=_
760 Torr
+!1SO yap
l'1H°vap 273.2K
+!1SO yap
which give, upon combining terms, O J·K-I . mol-I -0.003 169K-I x 1'1H0 yap +!1SO yap =
-14.06J.K-1 • mol-I -0.003 660K-1 X l'1H° vap +!1Sovap =
Subtracting one equation from the other will eliminate the !1SO yap term and allow us to solve for -14.06 J . K-1 • mol-I
(b) We can then use
1'1H0 =
vap.
- 0.000 491K-1 x 1'1H0
1'1H0 yap to
yap
calculate !1Soyap using either of the two
equations: SM-219
0 = -0.003169K-Ix (+28 600J·morl) + �SO yap
�SO yap = 90.6J.K-I . morl -14.06J.K-I . morl = - 0.003 660K-I x (+ 28 600J. morl) + �SO yap �SO yap = 90.6J.K-I . morl (c) The vapor pressure at another temperature is calculated using In
� ;:
=
-
Miovap R
[�-�l I;
7;
We need to insert the calculated value of the enthalpy of vaporization and one of the known vapor pressure points: In
P.!25.0°C
_
[
28 600J·mol-I
1 atm
p.! 25.00C 8.9
_
1
8.314J·K-I ·morI 298.2K = 0.53 atm or 4.0 x 102 Torr
-
1 315.58K
1
Table 6.2 contains the enthalpy of vaporization and the boiling point of methanol (at which the vapor pressure = 1 atm). Using this data and the equation In
In
� ;:
=
� 50°C
[
MiO yap _1 � R
_
1
_
I;
7;
1
35 300J·mol-I 8.314J·K-I·mol-1
[
1 298.2K
-
1 337.2K
1
2 � 5.00C = 0.19 atm or 1.5 x 10 Torr 8.11
(a) vapor;
(b) liquid;
(c) vapor
8.13
(a) 2.4 K;
(b) about 10 atm;
8.15
(a) At the lower pressure triple point, liquid helium-I and -II are in
(c) 5.5 K;
(d) no
equilibrium with helium gas; at the higher pressure triple point, liquid helium -I and -II are in equilibrium with solid helium.
8.17
(b) helium-I
The pressure increase would bring CO2 into the solid region. SM-220
8.19
(a) KCI is an ionic solid, so water would be the best choice; is non-polar, so the best choice is benzene;
(b) CCl4
(c) C H3COOH is polar, so
water is the better choice.
8.21
(a) hydrophilic, because NH2 is polar, and has a lone pair and H atoms that can participate in hydrogen bonding to water molecules; (b) hydrophobic, because the C H3 group is not very polar; (c) hydrophobic, because the Br group is not very polar; (d) hydrophilic, because the carboxylic acid group has lone pairs on oxygen and an acidic proton that can participate in hydrogen bonding to water molecules.
8.23
(a) The solubility of 02(g) in water is 1. 3x10-3 mol·C' . atm-'. solubility at 50 k Pa
50 kPa
=
x1.3x10-3 mol . C' . atm-' , 101.325 k Pa·atm-
=6.4x10-4 mol·L-' (b) The solubility of CO2 (g) in water is 2. 3x10-2 mol·C' . atm-' . soI U b'l' 1 Ity at 500 Torr
500 Torr
=
=
(c) solubility
8.25
=
, 2 x 2.3x10- moI ·L-' . atm, 760 Torr·atm2 1.5x10- mol·C' 2 mol·L-' . atm-'
O.l 0 atmx 2. 3x10-
=
2.3x10-3 mol·L-'
.
(a) 4 mg·L-' x1000 mL ·L-'x1.00 g·mL-'x1kgIl 000 g =
4 mg·L-' or 4 ppm
(b) The solubility of 02 (g) in water is 1.3 x10-3 mol·C' . atm-I which can be converted to parts per million as follows: In 1. 00 L (corresponding to 1 kg) of solution there will be 1.3x10-3 mol 02 1.3x10-3mol·kg-l·atm-' 02 x32.00 g·mor' 02 x103 mg·g-I SM -221
=42 mg·kg-J atm-J or 42 ppm·atm-J •
4 mg -;- 41 mg· kg-J atm-l = 0.1 atm •
(c) p =
8.27
0. l atm 0.21
=0.5 atm
(a) By Henry's law, the concentration of CO2 in solution will double. (b) No change in the equilibrium will occur; the partial pressure of CO2 is unchanged and the concentration is unchanged.
8.29
The solubility of CO2 can be determined using Henry's law: S
ea2
nco
2
=
kH
2 x P = 2. 3 X 10- mol·L-1 ·atm-J x 3. 60 atm
=S x 0.420 L
=
2 3. 48 X 10- mol CO 2
This is equal to 1. 5 g of CO2•
8.31
(a) Because it is exothermic, the enthalpy change must be negative; (b) L i2 S04 (s) ----+ 2 Li+ (aq) + SO/-(aq) + heat;
(c) Given that
t:J{ + t:J{hydration = t:J{ of solution, the enthalpy of hydration should be L
larger. If the lattice energy were greater, the overall process would be endothermic.
8.33
To answer this question we must first determine the molar enthalpies of solution and multiply this by the number of moles of solid dissolved to get the actual amount of heat released. (a)
6H;ol 3.9 kJ . mol-I =
t:J{ = (b)
10. 0 g NaCl 2 x (+3. 9 kJ·mol-J )=0. 67 kJ or 6.7xlO J J 58.44 g·mor
t:J{ ;ol =-7.5 kJ· mol-I
t:J{ =
10. 0 g NaI 149. 89 g . mol-I
x(-7. 5 kJ . mol-I) =- 0. 50 kJ = -5. 0 x102 J
SM-222
(c)
M;ol =-329 kJ . mol-I
M= (d)
133.33 g·mol- I
x (-329 kJ·morl) = -24.7 kJ
M;ol =6.6 kJ. mol-I
M=
8.35
10.0 g AICl)
10.0 g NH 4 NO3 2 x(+6.6 kJ·mol-I)=+0.82 kJ= +8.2x10 J 1 . 80.05 g mol-
(a) mNaC1 =
(
(b) mNaOH =
25.0 g NaCl
58.44 g . mol-I
)
0.5000 kg
= 0.856 m
( 4:'��S N:7_ ) 1
g
0.345 kg
= 0.18m
mass NaOH = 2.5 g
(
0.978 g urea
)
60.06 g . mol-I (c) mNaCI = I =0.0571m 3 (285 mL )(1 g . mL-1 )(10- kg . g - )
8.37
(a) 1 kg of 5.00% K)P04 will contain 50.0 g K)P04 and 950.0 g H20.
(
50.0 g K)P04 212.27 g·mol-I
K)P04
1
--'--------"------"---'-
0.950 kg
= 0.248m
(b) The mass of 1.00 L of solution will be 1043 g, which will contain
(
1
1043 gx 0.0500 =52.2 g K)P04• 52.2 g K)P04 212.27 g·mol-I
K)P04
--'--------'----'--
1.00 L
8.39
=0.246 M
(a) If XMgC l is 0.0120, then there are 0. 0120 mol M gC12 for every 0.9880 l mol H20. The mass of water will be 18.02 g . mol-I x 0.9880 mol = 17.80 g or 0.01780 kg. SM -223
mcl- =
(
2 mol Cl1 mol MgCl2
(b) mNaOH =
(
J
(0.0 120 mOI MgCI 2)
0.0 1780 kg solvent) 6.75 gNaO H
40.00 g . mol-I
J
0.325 kg solvent)
(c) 1.000 Lof 15.00
M
= 1.35m
=0.5 19m
HCI (aq) will contain 15.00 mol with a mass of
15.00x36.46 g·mol- I =546.9 g. The density of the 1.000 Lof solution is 1.0745 g . cm-3 so the total mass in the solution is 1074.5 g. This leaves 1074.5 g - 546.9 g 15.00 mol HCI
0.5276 kg solvent
8.41
=
=
527.6 g as water.
28.43m
(a) Molar mass of CaCl2 ·6 H2 0
=
2 19.08 g . morl, which consists of
1 10.98 g CaCl2 and 108. 10 g of water. mC C , aI
=
x
mol CaCl--=2·6 H20 ---"---
------
0.500 kg
+ x
-
---
(6x0.0 18 02 kg H20)
Note: 18.02 g H 20 = 0.0 18 02 kg H20 = 1.000 mol H20 x =
number of moles of CaCl2·6 H20 needed to prepare a solution of
molality mCaC ,' in which each mole of CaCl 2 ·6 H20 produces 6 (0.0 18 I 02 kg) of water as solvent (assuming we begin with 0.250 kg H 20). For a 0. 125m solution of CaCl2 ·6 H2 0,
SM-224
0. 125
m
=
0.500 kg
x
x (6)(0.01802 kg H20)
_ ___________
x =0.0625 mol
+
+
x (0.0135 mol)
x -0.0 135x =0.0625 mol
0.986 x = 0.0625 mol
x =0.0634 mol CaCl2 · 6 H20
:. (0.0634 mol CaCl2 ·6 H20)x
(
2 l9.08 gcaCI2 . 6 H20 1 mol CaCl2·6 H20
)
= 13.9 g CaCI2 ·6 H20
(b) Molar mass of Ni S04 ·6 H20 = 262.86 g . morl, which consists of 154.77 gNi S 04 and 108.09 g H20. mNiSO 4
x molNi S04 ·6 H20 = ----------'---''---0.500 kg + x (6 x 0.0 1802 kg H20)
where x
=
number of moles of Ni S04 ·6 H20 needed to prepare a solution
of molality
mNiSO,'
in which each mole of Ni S04·6 H20 produces
6 (0.0 18 02 kg) of water as solvent. Assuming we begin with 0.500 kg H20, for a 0.22 0.22
m
m
solution ofNi S04·6 H20,
x = ---0.500 kg + x (6)(0.0 1802 kg H2))
x =0. l 1 mol
+
x (0.0238 mol)
x -0.0238 x = 0.1 1 mol 0.976 x = 0. 11 mol x =0.1 1 molNi S04 ·6 H20
(
262.86 gNi SO4 ·6 H20 . :. (0. 1 1 molNl S 04 · 6 H20) X . 1 molNl S04 ·6 H20 8.43
)
. = 29 gNl S04 ·6 H20
The free energy of the solvent in theNaCI solution will alway s be lower than the free energy of pure water; if given enough time, all of the water from the "pure water" bea ker will become part of the NaCI solution, leaving an empty bea ker. An equilibrium will be established between the solvent and the water vapor." SM-225
8.45
(a) P
xsolvent X F;, uresolvent
=
At 100DC, the normal boiling point of water, the vapor pressure of water is 1.00 atm. If the mole fraction of sucrose is 0.100, then the mole fraction of water is 0.900: P
=
0.900x1.000 atm=0.900 atm or 684 Torr
(b) First, the molality must be converted to mole fraction. If the molality is 0. 100 mol· kg- I, then there will be 0.100 mol sucrose per 1000 g of water. 1000 g 18.02 g . mol- I
1000 g + 0. 100 mol 18.02 g. morl P = 0.998x
8.51
1.000 atm
=
=
0.998
0.998 atm or 758 Torr
The mole fraction of benzene in the mixture can be calculated using Raoult's law : p.b
enzene50111
Xb enzene =
=
xbenzenex Ppureb enzene'. so xb enzene = � ellzene50ln F;, ureb enzene
75.0 Torr
94.6 Torr
=
0.793
Since the sum of the mole fractions of solute and solvent must equal 1 and since we know the moles of benzene that are present calculation of the moles of solute added is straightforward:
Xunknown
1
=
xunknown =
-
xb enzene=1 - 0.793 = 0.207
nunknown nunknown + nb enzene
---""'=-"---
Solving this gives
8.47
n�mknown
n-"u""n"" kn,-" ow -,,- n
_ _
_ _
nunknown 0.300 +
fraction of the solvent: =
0.207
= 7.80 x 10-2 mol .
(a) From the relationship P = xsolvent X
94.8 Torr
=
xsolventxl 00.0 Torr
xsolvent 0.948 =
SM -226
F;, uresolventwe can calculate the mole
The mole fraction of the un known compound will be 1.000 -0.948
=
0.052. (b) The molar mass can be calculated by using the definition of mole fraction for either the solvent or solute. In this case, the math is slightly easier if the definition of mole fraction of the solvent is used: xsolvent
0.948
nsolvent
=
=
nunknown
+
nsolvent
100 g 78 . 1 1 g . morl
--[(
100 g =-----,78 . 1 1 g . morl =
M unknown
8.05 g
+ ---=-
Munknown
100 g
78 . 1 1g· morl
8.51
( ]
8.05 g
Jr
1 0 .948 -
100 g
78.1 1g·morl
J
=1 15 g . mo 1-1
(a) Pure water has a vapor pressure of 760 .00 Torr at 100 cC. The mole fraction of the solution can be determined from P = xsolvent 751 Torr Xsolvent
=
=
Xsolvent X
X
�ure solvent.
760 .00 Torr
0.988
The mole fraction needs to be converted to molality:
Because the absolute amount of solution is not important, we can assume that the total number of moles = 1.00.
nH 20
=
0.988 mol;
. molalIty =
(
nsolute
= 0 .0 12 mol
0.0 12 mol
0.988 mol x18.02 g . morl 1000 g . kg -I
)
=
0.67 mol· kg- I
Knowing the mole fraction, one can calculate I1Tb :
SM-227
I1Tb kb m I1Tb 0.51 K·mol· kg- I x 0.67 mol· kg- I 0.34 K or 0.34°e =
=
=
Boiling point
100.00oe + 0.34°e 100.34°e
=
=
(b) The procedure is the same as in (a). Pure benzene has a vapor pressure of7 60. 00 Torr at 80.1°C. The mole fraction of the solution can be determined from P
=
xsolvent X �uresolvent
740 Torr
=
Xsolvent X7 60. 00 Torr
Xsolvent 0.974 =
The mole fraction needs to be converted to molality:
Xsolvent 0.974 =
=
---'==-
-
nb enzene
-
nb enzene + nsolute
Because the absolute amount of solution is not important, we can assume that the total number of moles 1.00: =
0.974
nb enzene
1.00 mol
=
nb enzene 0.974 mol; nsolute 0.026 mol =
mo Ia I·Ity
=
=
(
0. 026 mol
0.974 mol x7 8.11 g. morl 1000 g· kg-I
J
=
0 34 mo · I kg -I .
Knowing the mole fraction, one can calculate I1Tb :
I1Tb
=
kb m
I1Tb 2.53 K· kg· mol-I x 0.34 mol· kg-I =
Boiling point
8.53
=
=
80.1°e + 0. 86°e 81.0oe =
111'r 17 9.8°e - 17 6.9°e 2.9°e or 2.9 K =
=
SM-228
0.86 K or 0.86°e
!::..1'r = kfrn
(
)
2.9 K = (39.7 K· kg· morl)m
2.9 K = (3 9.7 K.kg . mol-') 0.100 kg x 2.9 K
39 .7 K· kg· mo l- ' Munknown
8.55
=
1.14 g Munknown
0.100 kg
1.14 g Munknown
1.14 g x 39.7 K· kg·morl
=
0.100 kg X 2.9 K
2 1.6 x10 g . morl
=
(a) A 1.00% aqueous solution ofNaCI will contain 1.00 g ofNaCI for 99.0 g of water. To use the freezing point depression equation, we need the molality of the solution:
molality =
(
1.00 g 58.44 g . mol-'
)
0.0990 kg
=0.173 mol· kg-'
!::.T . f = ikfm
!::.T . f = i (1.86 K· kg· mol-' ) (0.173 mol·kg -I) =0.593 K i =1.84 (b) molality of all solute species (undissociatedNaCI (aq) plusNa+ (aq) + CI - (aq)) =1.84 x 0.1 73 mol· kg-' = 0.318 mol· kg -' (c) If all the NaCI had dissociated, the total molality in solution would have been 0.346 mol· kg- I, giving an i value equal to 2. If no dissociation had taken place, the molality in solution would have equaled 0.173 mol· kg-'. NaCI (aq) 0.173 mol· kg -' 0.173 mol· kg-' 0.173 mol· kg- I X
-
x
x
x + x + x
+ X
x
= 0.318 mol· kg-'
= 0.318 mol· kg-'
= 0.145 mol· kg -'
0.145 mol· kg-' . . · 01 10 dIssoclatlOn 1 00 = 83. 80Yo = I X 0.173 mol· kgSM-229
8.57
The compound that freezes at the lower temperature will have the largest
I�Tfl (assuming that i
=
1 ).
Since the same mass of each compound is
present in the same amount of solvent for each solution, the compound with the lower molar mass will have the larger number of moles and therefore the larger
I6.Tfl (since the compound with the largest molar amount present will
have the largest amount of solute particles present). Since compound A caused the water to freeze at a lower T than compound B, compound Amust have the lower molar mass. compound B must therefore have the greater molar mass.
8.59
For an electroly te that dissociates into two ions, the van't Hoff i factor will be 1 plus the degree of dissociation, in this case 0.075. This can be readily seen for the general case MX. Let A initial concentration of MX (if none =
is dissociated) and let Y
=
the concentration of MX that subsequently
dissociates:
A -Y
Y
Y
The total concentration of solute species is (A -Y)
Y +Y
+
=
A +Y
The value of i will then be equal to A +Y or 1.075. The freezing point change is then easy to calculate:
6.Tr = ikf m = 1.075 x 1.86 K· kg·morl x 0.10 mol· kg-I =0.20 Freezing point of the solution will be O .OO°C -0.20°C
8.61
(a) [1 =
=
-0.20°e.
iRT x molarity
= 1 x 0.082 06 L·atm · K-I ·morl x 293 K x 0.010 mol·L-1 = 0.24 atm or 1.8 x l02 Torr (b) Because HCl is a strong acid, it should dissociate into two ions, H + and Cl -, so = 2.
i
SM-230
2 x 0.082 06 L · atm · K- ' · mor l x 293 K x 1 .0 mol · L- ' = 48 atm ( c) CaCl 2 should dissociate into 3 ions in solution, therefore i 3. II = 3 x 0.082 06 L · atm · K- ' . mol-I x 293 K x 0.0 1 0 mol · L- ' = 0.72 atm or 5.5 x 1 02 Torr The polypeptide is a nonelectrolye, so i = 1 . II iRT x molarity II
=
=
8.63
( 0.40 g J
=
II
=
3.74 Torr = 1 x 0.082 06 L · atm · K- ' . mol- I x 300 K x Munknowil 760 Torr · atm- ' 1 .0 L
Munknown
8.65
l . = 0.082 06 L · atm · K- ' mOr x 300 K x 0.40 g x 760 Torr · atm-I 3.74 Torr x 1 .0 L = 2.0 x l 03 g · mol- '
We assume the polymer to be a nonelectrolyte, so i = 1 . II = iRT x molarity II
=
6.3 Torr 1 x 0.08206 L . atm . K -I . mol-I x 293 K x M unknown 0. 1 00 L 760 Torr · atm -I
M unknown
8.67
( 0.20 g J
=
x 293 K x 0.20 g x 760 Torr · atm-I - 0.08206 L · atm · K . mol-I 6.3 Torr x 0. 1 00 L -I
�-------------
-------------------------------
= 5.8 x l 0 3 g · mol-I
(a) C I2HnOl l should be a nonelectrolyte, so i 1 . II iRT x molarity = 1 x 0.082 06 L · atm · K- ' · mor l x 293 K x 0.050 mol · L- ' =1 .2 atm (b) NaCI dissociates to give 2 ions in solution, so i = 2. II iRT x molarity = 2 x 0.082 06 L · atm · K- ' . mol-I x 293 K x 0.00 1 0 mol · L- ' =0.048 atm or 36 Torr =
=
=
SM-23 1
(c) AgCN dissociates in solution to give two ions (Ag+ and CN- ) , so i = 2. II iRT x molarity We must assume that the AgCN does not significantly affect either the volume or density of the solution, which is reasonable given the very small amount of it that dissolves. II = 2 x 0.082 06 L · atm · K - I . mol - I x 293 K x 2.3 X 1 0-5 g 1 33.89 g· mol-I =
(
=
8.69
)
8.3 x 1 0-5 atm
II = n
RT V
=
� RT M V
where m is the mass of unknown compound.
M= M=
8.7 1
mRT VII
(
(0. 1 66 g) (0.08 2 06 L . atm · K -I . mol - I) (293 K) = 2.5 x 1 05 g . mol- I 1 .2 Torr I (0.01 0 L) 760 Torr · atm J -
I
(a) To determine the vapor pressure of the solution, we need to know the mole fraction of each component. 1 .50 mol xbenzene = 0.75 1 .50 mol + 0.50 mol 0.25 Xtoluene = 1 - xbenzene x �otal = (0.75 x 94.6 Torr) + (0.25 29. 1 Torr) 78.2 Torr The vapor phase composition will be given by �enzene 0.7 5 x 94. 6 Torr = 0.91 xbenzene . vapor phase 78 . 2 Torr Ptotal Xtoluene in vapor phase = 1 - 0.91 0.09 The vapor is richer in the more volatile benzene, as expected. =
=
=
=
III
=
=
SM-232
(b) The procedure is the same as in (a) but the number of moles of each component must be calculated first: nbenzene =
15.0
g
=0.192
g . mol- I 6 5 .3 g = 0.709 ntoluene = . 92 14 g . mol-I 0.192 mol = 0.213 xbenzene = 0.192 mol + 0.709 mol 78.11
Xtoluene = 1 - xbenzene = 0.787 ;:otal = (0.213 x 94.6 Torr) + (0.787 x 29.1 Torr)= 43.0
Torr
The vapor phase composition will be given by �enzene 0.213 x 94.6 Torr = 0 469 = Xbenzene in vapor phase = p 43.0 Torr total .
Xtoluene in vapor phase =1- 0.469 = 0.531
8.73
To calculate this quantity, we must first find the mole fraction of each that will be present in the mixture. This value is obtained from the relationship .F;otal = [XI.i.dichloroethane X �ure I.I-dichloroethane ] +[XI.I-dichlorotetrafluoroethane
X
�ure I.I-dichlorotetrafluoroethane]
Torr= [XI.I-dichloroethane X 228 Torr] + [XI.I-dichlorotetrafluoroethane 79 Torr] 157 Torr= [XI.I-dichloroethane X 228 Torr] + [(1- XI.I-dichloroethanJ X 79 Torr] 157 Torr= 79 Torr + [(XI,I-dichloroethane X 228 Torr) - (XI.I-dichloroethane X 79 Torr)] 78 Torr= XI.I-dichloroethane X 149 Torr
15 7
X
XI,i-dichloroethane = 0.52
XI I-dichlorotetrafluoroethane
=1- 0.52 = 0.48
To calculate the number of grams of1, I-dichloroethane, we use the definition of mole fraction. Mathematically, it is simpler to use the XI,I-dichlorotetrafluoroethane defi ni ti on:
SM-233
1 00.0 g = 0.5851 mol 1 70.92 g . mor l 0_.5_8_5_1_m_o_I = 0.48 x1,I·dichlorotetranuoroethane = m 0,5851 mol + 98.95 g . mol- I
nl,l.dichlorotetranuoroethane
=
___
__ _
[
0.5851 mol = 0.48 x 0.5851 mol +
m
98.95 g · mor I
0.585 1 mol - (0.48 x 0.5851 mol)= 0.48 x 0,3042 mol= 0.004 851 mol · g-I x m m = 63 g 8.75
]
[ 98.95 g · mor ] m
I
Raoult ' s Law applies to the vapor pressure of the mixture, so positive deviation means that the vapor pressure is higher than expected for an ideal solution. Negative deviation means that the vapor pressure is lower than expected for an ideal solution. Negative deviation will occur when the interactions between the different molecules are somewhat stronger than the interactions between molecules of the same kind (a) For methanol and ethanol, we expect the types of intermolecular attractions in the mixture to be similar to those in the component liquids, so that an ideal solution is predicted. (b) For HF and H 2 0 , the possibility of intermolecular hydrogen bonding between water and HF would suggest that negative deviation would be observed, which is the case. HF and H 2 0 form an azeotrope that boils at 1 1 1 °C , a temperature higher than the boiling point of either HF (19.40c) or water. ( c) Because hexane is nonpolar and water is polar with hydrogen bonding, we would expect a mixture of these two to exhibit positive deviation (the interactions between the different molecules would be weaker than the intermolecular forces between like molecules). Hexane and water do form an azeotrope that boils at 61 .6 ° C, a temperature below the boiling point of either hexane or water. SM -234
8.77
A/oam is a colloid formed by suspending a gas in a liquid or a solid matrix, while a sol is a suspension of a solid in a liquid. Some examples of foams
are styrofoam and soapsuds; examples of sols are muddy water and mayonnaIse. 8.79
Colloids will reflect or scatter light while true solutions do not (this is known as the Tyndall effect).
8.81
Water is a polar molecule and as a result will orient itself differently around cations and anions, aligning its dipole in such a way as to present the most favorable interaction possible:
8 8�8+ ®� ~ 8+ 88� 8- 8-� + 8- ct o+�o8- 8- 88+ ~ ~ 8+ ~ 8+ ~ 8- ~ ~ 8+ 8 8H
0
H
H
o
0
®
o
8.83
8.85
8.87
H
H
0
H
H
(a) stronger; (b) low; (c) high; (d) weaker; (e) weak, low; (f) low; (g) strong, high (a, b) Viscosity and surface tension decrease with increasing temperature; at high temperatures the molecules readily move away from their neighbors because of increased kinetic energy. (c, d) Evaporation rate and vapor pressure increase with increasing temperature because the kinetic energy of the molecules increase with temperature, and the molecules are more likely to escape into the gas phase. (a) Butane (a nonpolar molecule) will dissolve in the nonpolar solvent (tetrachloromethane). SM-235
H
(b) Calcium chloride will dissolve in the polar solvent, water, since the solvation ofthe ions formed will be aided by water's dipole moment:
8.89
8.91
(a) 25.0 Torr x 1 00 78.5%; (b) At 25 °C the vapor pressure of water 3 1 .83 Torr is only 23.76 Torr, so some of the water vapor in the air would condense as dew or fog. =
t1T
=
kf m,
m
nsolute
'--= ----" = -
masssolvent(kg)
n solute =
mass s l t
o u e
M solute
or mass solute ��---M solute x mass solvent(kg) Solving for Msolute, kr x mass solute M solute = 'T' il1 r X mass s olvent(kg) (a) If maSSsolute appears greater, Msolute appears greater than actual molar mass, as maSSsolute occurs in the numerator above. Also, the t1T measured will be smaller because less solute will actually be dissolved. This has the same effect as increasing the apparent Msolute. il
A T kf =
-----
A
SM-236
(b) Because the true masssolvent = d x V, if dso vent is less than 1 .00 g . cm-3 , then the true maSSsolvent will be less than the assumed mass. Msolute is inversely proportional to maSSsolvenl> so an artificially high maSSsolvent will lead to an artificially low Msolute. ( c) If true freezing point is higher than the recorded freezing point, true f.. T < assumed f.. T , or assumed f.. T > true f.. T , and Msolute appears less than actual Msolute' as f.. T occurs in the denominator. (d) If not all solute dissolved, the true maSSsolute < assumed mass solute or assumed maSSsolute > true maSSsolute, and Msolute appears greater than the actual Msolute, as maSSsolute occurs in the numerator. I
8.93
The water Coleridge referred to was seawater. The boards shrank due to osmosis (a net movement of water from the cells of the wood to the saline water). You can ' t drink seawater: osmosis would cause a net flow of water from the cells of the body to the saline-enriched surround solution and the cells would die.
8.95
(a) The 5.22 cm or 52.2 mm rise for an aqueous solution must be converted to Torr or mmHg in order to be expressed into consistent units. 3 52.2 mm x 0.998 g · cm -3 = 3.83 mmHg or 3.83 Torr 1 3.6 g· cmThe molar mass can be calculated using the osmotic pressure equation: JJ = iRT x molarity Assume that the protein is a nonelectrolyte with i = 1 and that the amount of protein added does not significantly affect the volume of the solution. 0.01 0 g 17= 1 x O.082 06L ' atm . K-1 ' mol-1 x 293 K x Mprotein = 3.83 Torr 0.0 1 0L 760 Torr · atm-I
[ ]
1 x O.082 06L . atm . K- 1 . mol-1 x 293 K x O.01 0 g x 760 Torr · atm- 1 = Mprotelll 0.0 1 0L x 3.83 Torr .
SM-237
= 4.8 x 1 03 g . mol-I
Mprotein
(b) The freezing point can be calculated using the relationship !J.Tr = ikr m
(
0.01 0 g I 3 'J = 1 x 1 .86 K · kg · mol-I x 4.8 x 1 0 g · mol1 0 mL x 1 .00 g . mL-' 1 000 g · kg- ' = 3.9 X 1 0-4 K or 3.9 x 1 O- 4 oc The freezing point will be O.OO°C - 3.9 X 1 O-4 oC = -3.9 X 1 O-4 oC. (c) The freezing point change is so small that it cannot be measured accurately, so osmotic pressure would be the preferred method for measuring the molecular weight.
(
8.97
J
!J.T = 77. 1 9°C - 76.54°C = 0.65 ° 0.65 K
( J b kg solvent msalute
A T.b -l·kbm -l·k
Ll
Msalute
8.99
-
-
Msa'ute
= (1)(4.95 K · kg · mol- I )(0.30 g) (kg solvent)(!J.Tb) (0.0300 kg)(0.65 K) = 76 g·mor l
=
ikb msa'ute
(a) The data in Appendix 2A can be used to calculate the change in enthalpy and entropy for the vaporization of methanol: CH 3 0H(l) � CH 3 0H(g) M = MOr(CH3 0H, g) - MOr(CH 3 0H, 1) = ( - 200.66 kJ · mol- I ) - (-238.86 kJ · mol- ' ) = 38.20 kJ . mor l O
vap
!J.SO = Sam (CH 3 0H(g)) SO m (CH 3 0H(l)) = 239.81 J . K- ' . mor l - 126.8 J . K- ' . mol-I = 1 1 3 .0 J · K- ' · mol- ' yap
-
SM-238
To derive the general equation, we start with the expression that I1G o yap = RT In P , where P is the vapor pressure of the solvent. Because -
= MfO yap - TI1S0yap ' this is the relationship to use to detennine the temperature dependence of In P : MfO yap - TI1S o yap = RT In P This equation can be rearranged to give MfO yap 1 . + I1Soyap In P = I1G o yap
-
_
R
T
_
R
To create an equation specific to methanol, we can plug in the actual values of R, MfO vap' and I1So yap: 38 200 J · mol- I .-.!... + 1 13.0 J · K-I · mol-I 8.3 1 4 J . K - I . mol- I T 8.3 1 4 J . K - I . mor l = 4595 K + 1 3.59 T
In P =
_
_
(b) The relationship to plot is In P versus -.!.... This should result in a T
straight line whose slope is
MfO yap -
R
and whose intercept is
I1S0 yap
R
. The
pressure must be given in atm for this relationship, because atm is the standard state condition. (c) Because we have already detennined the equation, it is easiest to calculate the vapor by inserting the value of o.ooe or 273.2 K : InP = - 4595 K + 1 3.59 =_ 4595 K + 1 3.59 = -1 6.82 + 1 3 .59 = -3.23 T 273.2 K P 0.040 atm or 30. Torr (d) As in (c) we can use the equation to find the point where the vapor pressure of methanol = 1 atm. In P = In 1 = 0 = 4595 K + 1 3.59 =
-
T= 338.1
T
K
SM-239
8.101
(a) fJ.Tr. = 1 .72 K = i · kr. .m = ( 1 .86 K · kg · morl )
naa
0.95 kg
where naa are the moles of acetic acid in solution. Assuming i = 1 :
( 1 .72 K )( 0.95 kg ) = 0.878 mol = aa ( 1 .86 K . kg · mol- I ) Experimentally, the molar mass of acetic acid is, therefore: 50 g = 56 9 g . mo1- 1 0.878 mol This experimental molar mass of acetic acid is less than the known molecular mass of the acid (60.0 g . mol- I ) indicating that the acid is dissociating in solution giving an i > 1 .
n
--=--
.
(b) As in part (a) the experimental molar mass is first found: fJ.Tj
= 2.32 K = i · k j .m = (5. 1 2 K . kg . mol-I )
[ 0.95 kgJ naa
where naa are the moles of acetic acid in solution. . 1. = 1 : = (2.32 K)(0.95 kg) = 0.430 mol. Assummg naa (5. 1 2 K · kg · mol-I ) Expelimentally, the molar mass of acetic acid is, therefore: _5_0-=g'---- = 1 16 g . mol-I. 00430 mol This experimental molar mass of acetic acid is significantly higher than the known molar mass of the acid (60.0 g·morl) indicating that the acid is dissociating in solution giving an i < 1 . A van't Hoff factor less than 1 indicates that acetic acid is not completely dissolved in the benzene, or acetic acid molecules are aggregating together in solution. 8 . 1 03
(a) The plot of the data is shown below. On this plot, the slope = and the intercept =
fJ.SO a yp R
SM-240
_
WO ap y R
Temp. (K) 1 90 228 250 273 0
_ _ . .. r� ;��··'I. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . .� ._ . �_. _-= . . -""""+
-2
.. .. ,'- ..... . .. . .. .... ...............................................
-3 1- . --................,...............
c
.
-4 1-...-..-
.
..
...
-......---
... .
... ._.
"-<;:....................
............................ .
-
_
.. .. ..
.. . .. _
__+
_
..+
_
.._.....
. .
.
.......... ;-_
--_
,',
.. ......
,
..
_______
........
.
_.
_
.._
_
"'-
......,
_
.......
..
.. .. .
..
.
_
_...---.
..,.." .,' .... . ... . . ......... ..-
. ............. _
....__
.. _
��
-5
InP -5.47 -2.41 -1.15 -0. 1 23
Vapor pressure (Torr) V.P. (atm) 3.2 0.0042 68 0.089 240 0.3 1 6 672 0.884 _
-1
Q
T- ' (K- ' ) 0.005 26 0.004 38 0.004 00 0.003 66
.. .
-
......
--
. .
I---...............---;.-...----....-.-+-......-..-...-..--...........-. +.-�............- --. -
-6 ������ 0.0035 0.004 0.0045 0.005 0.0055 1/T
(K1)
From the curve fitting program: y -3358.7 1 4x + 12.247 =
(b) -
t.J{0 yap
R
t.J{0 yap
(c)
=
=
-3359
(3359)(8.3 1 4 J . K- ' . mol-' )= 28 kJ · mol-'
/j,so y...!. ap ...
__
R
=
1 2.25
(12.25)(8.3 1 4 J . K -I . mol-I ) 1 .0 x I 02 J . K -I . mar' (d) The normal boiling point will be the temperature at which the vapor pressure 1 atm or at which the In 1 O. This will occur when T- ' = 0.0036 or T = 2.7 x 1 02 K. (e) This is most easily done by using an equation derived from t.J{°yap /j,so yap
=
=
=
=
and /j,SO yap '. SM-241
fjJ{0 ya
InP R In 1 5 Torr 760 Torr T = 2. 1 X 1 02 K
.1
p
=
=
_
T
+
!1S0 yap
R
28 000 J · mol- I .� + 1 00 J . K- I · mor l 8.3 1 4 J · K - I · mor l T 8.3 1 4 J · K - I · mol- 1
8.105
The critical temperatures are Compound Tc (OC) CH4 -82. 1 C2H 6 32.2 C3HS 96.8 C4H 1 0 1 52 The critical temperatures increase with increasing mass, showing the influence of the stronger London forces.
8 . 1 07
(a) If sufficient chloroform and acetone are available, the pressures in the flasks will be the equilibrium vapor pressures at that temperature. We can calculate these amounts using the ideal gas equation: 1 95 Torr (1 .00 L) 760 Torr· atm- I
[ J ( 1 19.37 g · mol- chloroform J (0.08206 L · atm · K-I . mol-I )(298 K) =
�
mChloro onn
mchlorofonn
= 1 .25 g
[ 760225Torr·Torratm-I J (1 .00 L) J (0.08206 L · atm · K -I . mol- I )(298 K) . 58.08 g mol acetone ( 0.703 g =
o��
macel
macelolle
=
In both cases, sufficient compound is available to achieve the vapor pressure; flask A will have a pressure of 1 95 Torr and flask B will have a pressure of 222 Torr. SM-242
(b) When the stopcock is opened, some chlorofonn will move into flask B and acetone will move into flask A to restore the equilibrium vapor pressure. Additionally, however, some acetone vapor will dissolve in the liquid chlorofonn and vice versa. Ultimately the system will reach an equilibrium state in which the compositions of the liquid phases in both flasks are the same and the gas phase composition is unifonn. The gas phase and liquid phase compositions will be established by Raoult's law. It is conceptually most convenient for this calculation to start by putting all the material into one liquid phase. Such a solution would have the following composition: 35.0 g 58.08 - g . mol- 1 acetone ---Xacelone = - --- - - - - - ----==-- ----- - - - - 3 3 0 0 5 5 g . . g ---------=-------=---- + --:58.08 g . mol- 1 acetone 1 1 9.37 g . mol- 1 chlorofonn Xacelone = 0.67 35.0 g 1 1 9.37 g·mol- 1 chlorofonn or 1 Xacelollc Xchlorofonn 35.0 g g 35.0 -----"---- -- + 58.08 g . mol- 1 acetone 1 1 9.37 g . mol-1 chlorofonn Xchlorofonn = 0.33 -
--
--
----------'=------ ------:------
-
-
---------=- --:----
This gives the composition of the liquid phase. The composition of the gas phase will be detennined from the pressures of the gases: pacetone = Xacetone. liqUid ' po acetone = (0.67)(222 Torr) = 1 49 Torr pchlorofonn = Xchlorofonn. hquld. . po chlorofonn = (0.33)(1 95 Torr) = 64 Torr .
.
.
Xacetone, gas =
P.CClone
+ pchlorofonn 1 49 Torr = 0.70 Xacelone, gas = 1 49 Torr + 64 Torr Xchlorofonn. gas = 1 - Xacetone. gas = 0 . 30 The gas phase composition will, therefore, be slightly richer in acetone than in chlorofonn. The total pressure in the flask will be 2 1 3 Torr. (c) The solution shows negative deviation from Raoult's law. This means that the molecules of acetone and chlorofonn attract each other slightly more than molecules of the same kind. Under such circumstances, the SM-243 pacelone
vapor pressure is lower than expected from the ideal calculation. This will give rise to a high-boiling azeotrope. The gas phase composition will also be slightly different from that calculated from the ideal state, but whether acetone or chloroform would be richer in the gas phase depends on which side of the azeotrope the composition lies. Because we are not given the composition of the azeotrope, we cannot state which way the values will vary. 8.109
The vapor pressure is more sensitive if Mi is small. The fact that va
p
Miva
is small indicates that it takes little energy to volatilize the sample, which means that the intermolecular forces are weaker. Hence we expect the vapor pressure to be more dramatically affected by small changes in temperature. p
8.1 1 1
The initial information tells us that the detector is more responsive to A than to B. Under these conditions, the response is 5.44 cm 2 7 0.52 mgA = 1 1 cm 2 . mg- 1 A whereas the response for B is 8.72 cm 2 72.30 mg = 3.79 cm2 mg-1 B. The detector is thus 1 1 cm2 · mg- 1 73.79 cm2 · mg- 1 = 2.9 times more sensitive for A than B. Because the conditions are different for determining the unknown amount of A present, we cannot use the area ratios directly to determine the quantity of A. Instead, we use the standard B as a reference. (X = mg A) 3.52 cm2 7.58 cm 2 = 2.9 2.00 mg (B) X 2 X = 3.52 cm x 2.00 mg2 2.9 7.58 cm X = 0.32 mg A •
SM-244
8. 1 1 3
= iRT x molarity Assuming i= 1 7.7 atm = ( 0.0821 L . atm .K- 1 . mol-1 )( 3 1 0 K)(M) M = 0.303 mol · Ll If 0.500 L of solution is needed: ( 0.303 mol · Ll )( 0.500 L ) = 0. 1 5 1 mol of glucose Molecular mass of glucose is 8 1 . 1 58 g . mol- 1 ( 0. 1 5 1 mOI )( 8 1 . 1 58 g . mol-1 ) = 27.3 g
8. 1 1 5
First, calculate the weight % of acetic acid in a solution with Xacetic acid 0. 1 5: If you have 1 mol total of a solution with Xacelic acid = 0. 1 5 and XH20 = 0.85, you would have 0. 1 5 mol of acetic acid (60.053 g . mol- I ) in 0.85 mol of H 2 0 (1 8.0 1 5 g · mol- 1 ). The weight % of acetic acid in such a solution is: ( 0. 1 5 mol ) (60.053 g . mol-I ) = 9.01 g of acetic acid ( 0.85 mol ) (1 8.0 1 5 g . mol-I ) = 1 5.3 g of H 2 0 9.01 g ( 1 00% ) = 37.0% acetic acid by weight 9.01 g + 1 5.3 g
II
If a solution with Xacetic acid = 0.1 5 is 37.0% acetic acid by weight, 30 g of such a solution will contain: ( 30 g )( 0.370 ) = 1 1 . 1 g acetic acid 1 1.1 g = 0. 1 85 mol of acetic acid 60.053 g . mol- I given that acetic acid is a monoprotic acid which will react in a 1 :1 fashion with NaOH, 0. 1 85 mol ofNaOH will completely react with 0. 1 85 mol of acetic acid. The volume of a 0.0 1 0 M solution ofNaOH needed to completely consume the acetic acid 0. 1 85 mol = 1 8.5 L IS: 0.0 1 0 mol · Ll
SM-245
=
8.117
The non-polar chains of both the surfactant and pentanol will interact to form a hydrophobic region with the heads of the two molecules pointing away from this region towards the aqueous solution. To prevent the heads of the shorter pentanol molecules from winding up in the hydrophobic region, the layered structure might be comprised of a water region, a surfactant layer (heads pointing toward the water) a pentanol layer (with tails pointing toward the hydrophobic tails of the surfactant) and back to an aqueous regIOn.
8.1 1 9
(a) The hybridization of the carbon atoms in citric acid are: o
sp3 sp2 IIC HO / --CH2
;0 sp3 CH2-C1/sp2 \/ \ C OH p� /s OH HO-C 2 sp �
o
(b) yes; all four -OH groups present can participate in hydrogen bonding; the remaining three oxygens can accept H-bonds. (c) since citric acid can hydrogen bond with itself it should have large intermolecular forces and should be a solid. The ability to both donate and accept hydrogen bonds should also make it soluble in water. (d) 1 00.0 g ofa 0.9% NaCI solution will contain 0.9 g ofNaCI (MM 58.44 g/mol), which is 1 .54 x 1 0-2 mol ofNaCl. Assuming that NaCI completely dissociates, the moles of solute present is two times this amount or 3.08 x 1 0 -2 mol of solute in 1 00.0 mL. Therefore, the concentration of normal saline will be 0.3 M. (e) To make a 500.0 mL sports drink isotonic, you need it to have a total molarity of 0.3 M (from part d). This means that the amount of solute needed is: 0.3 M x 0.5000 L 0.1 5 mol total solute. Addition of 1 .0 g of NaCI to the 500 mL sports drink will account for 0.034 mol of solute (twice the moles ofNaCI added), leaving 0. 1 3 mol of glucose being =
=
SM-246
needed to achieve isotonicity. Since glucose has a molar mass of 1 80 g/mol, this means that in addition to the 1 .0 g ofNaCI, 2 x 1 0 1 g of glucose needs to be added to the 500.0 mL sports drink. (f) 300.0 mL of a 1 .00% boric acid 300.0 g of solution (assuming the density of the solution is l .00 g cm-\ meaning there are 3.00 g ofB(OH)3 or 4.85 x 1 0-2 mol ofB(OH)3 in the solution (since MM8(OH1 = 6 l .8 1 g · mol- I ). To be isotonic, 300.0 mL should have =
0.3 M x 0.3000 L = 0.09 mol total solute. Subtracting out the amount of boric acid present means that 0. 04 mol of solute needed to be added to achieve isotonicity; since 1 mol ofNaCI provides 2 mol of solute, 0.02 mol, or 1 g, ofNaCI must be added to the boric acid solution.
SM-247
CHAPTER 9 CHEMICAL EQUILIBRIA
9.1
(a) False. Equilibrium is dynamic. At equilibrium, the concentrations of reactants and products will not change, but the reaction will continue to proceed in both directions. (b) False. Equilibrium reactions are affected by the presence of both products and reactants. (c) False. The value of the equilibrium constant is not affected by the amounts of reactants or products added as long as the temperature is constant. (d) True.
9.3
0.35 OJ 0.25 0.2 0.15 0.1 0.05 0
[PCb]
=
[Cb]
[PCIs]
2 time
0
3
9.5 (a) Flask 3 represents the point of reaction equilibrium. (b) % decomposition = (6/1 1 ) x 1 00 54.5% 2X (c) Decomposition: X2 =
====
SM-248
4
K=
9.7
9.9
(PX) 2 = [(mole fraction of PX) J:;J2 (mole fraction of Px,) � Px,
(a) K
c
=
= 0. 1 6
[COCI][CI] . [CO][CI2 ] '
(a) Because the volume is the same, the number of moles of 0 2 is larger in the second experiment. (b) Because Kc is a constant and the denominator is larger in the second case, the numerator must also be larger; so the concentration of 02 is larger in the second case. (c) Although [0 2 f /[0 3 ] 2 is the same, [° 2 ]/[° 3] will be different, a result seen by solving for Kc in each case. (d) Because Kc is a constant, reciprocal must be the same.
9.1 1
(0.01 37) 2 = 48.8 (6.47 x 1 0-3 )(0.594x 1 0-3 ) 2 (0.0169)� C" · · 2, K --48.9 � lor condIhon (3.84xl 0- 3 )(1 .52 x l 0- 3 ) (0 . 01 00) 2 for condition 1 , K = = 48.9 (l .43x1 o-3)(l .43x1 0- 3 )
for condition 1 , K= =
9.13
--
--
(a) 1 3 R 0,
SM-249
--:--
-
9.15
To answer these questions, we will first calculate IJ.GO for each reaction and then use that value in the expression IJ.Go = -RT In K. (a) 2 H 2 (g) + 0 2 (g) � 2 H 2 0(g) IJ.Gor = 2(IJ.Gor (H 2 0, g) = 2( -228.57 kJ · mor l ) = -457. 14 kJ IJ.Gor = -RT In K or IJ.Go In K = - P RT -457 140 J In K = = +1 84.42 . (8.3 14 J K -I . mor l ) (298. 1 5 K) K = 1 x 1 08 0 --
(b) 2 CO(g) + 02 (g) � 2 CO 2 (g) IJ.Gor = 2 x IJ.Gor (C02 , g) - [ 2 x IJ.Gor (CO, g)] = 2( -394.36 kJ · mor l ) - [2( -1 37. 1 7 kJ . mol- I )] = -5 14.38 kJ -514 380 J = +207.5 In K = (8.3 14 J · K-I . mor l ) (298. 1 5 K) K = 1 X 1 090 (c) CaC0 3 (s) � CaO(s) + CO 2 (g) IJ.GOr = IJ.Gor (CaO, s) + IJ.Gor (C0 2 , g) - [IJ.Gor (CaC03 (s)] = ( - 604.03 kJ · mOr l ) + ( - 394.36 kJ · mOr l ) - [ - 1 1 28.8 kJ · mol-I] = +1 30.4 kJ +1 30 400 J = -52.6 In K = . (8.3 14 J K-I. mOr l ) (298 K) K = 1 X 1 0-2 3 _
_
9.17
(a) IJ.Gor = -RT In K = -(8.314 J · K-I . mor l ) ( 1 200 K) ln 6.8 = - 1 9 kJ · mol-I SM-250
(b) I1.Gor = -RT In K -(8.3 1 4 J . K-I . mor l ) (298 K) In 1 . 1 x 1 0-1 2 = +68 kJ . mor l =
9.19
First we must calculate K for the reaction, which can be done using data from Appendix 2A: I1.Go r 2 x I1.Gor (NO, g) = 2 x 86. 55 kJ ·mor l 1 73 . 1 kJ · mor l I1.Go = -RT In K I1.Go In K = -- RT +1 73 1 00 J · mol- 1 = -69.9 In K = (8. 3 1 4 J . K- I . mor l ) (298 K) K = 4 x 1 0- 3 1 Because Q > K, the reaction will tend to proceed to produce reactants. =
=
9.2 1
The free energy at a specific set of conditions is given by I1.Gr 11. GO r + RT In Q I1.Gr = -RT In K + RT In Q I1.G = -RT ln K + RT ln [ lf [12 ] -(8.3 14 J . K-I . mor l ) ( 1 200 K) In (6. 8) 2 + (8.3 14 J · K-I . mol-I) ( 1200 K) In (0.98) (0. 1 3) 8.3 x 1 0. 1 kJ . mol-I Because I1.Gr is positive, the reaction will be spontaneous to produce h. =
r
=
=
9.23
The free energy at a specific set of conditions is given by
SM-25 1
�Gr = �GOr +RT In Q �Gr -RT In K + RT In Q �G -RT In K + RT In [NH 3 f 3 [N 2 ][H 2 ] -(8.3 1 4 J · K- ' · moI- ' ) (400 K) In (41 ) + (8.3 14 J . K-' · mor ' ) (400 K) In =
r
=
=
(21 ) 2 (4.2) ( 1 .8) 3
Because �Gr is negative, the reaction will proceed to form products.
9.25
(a)
K=
( -
P.
2p �'l = 1.8 x l 0-2
NO PNOC1
)
n 6
T K1 2.027 K KC (3 - 2 ) 1 2.027 K 1 2.027 K K= 1 .8 x 1 0-2 = 4.3 x 10-4 KC = T 500 K (b) K = PeOl = 1 67
(
6/1
(
1 2.027 K KC = T 9.27
) (
)
) ( 1 ) 12.027 K ( ) 1 073 K 1 67 = 1 .87 6
/1
K=
For the reaction written as N 2 (g) + 3 H 2 (g) -)- 2 NH 3 (g) Eq. 1 ,
(a) For the reaction written as 2 NH 3 (g) -)- N 2 (g) + 3 H 2 (g) Eq. 2, P. P. 3 Nl Hl = K PNH 2 •
3
1 = -1 = 0.024. ThIS. IS. -41 KEq, ,
SM-252
Note that Eq. 3 = 1 Eq. 1 and thus KEq3 = KEq.l l l2 . (c) For the reaction written as 2 N 2 (g) + 6 H 2 ( g) � 4 NH 3 ( g) Eq. 4,
Note that Eq. 4 = 2 Eq. 1 and thus KEq3 = KEq} . 9.29
K = 377 2 BrCl (g) = Br2(g) + Ch (g) (1) H2(g) + Ch (g) = 2 HCI (g) K = 4.0 x 1 031 (2) ( 1 ) + (2): 2 BrCl(g) + H 2(g) Br2(g) + 2 HCl( g) K for this reaction = KI K2 = 377 x 4.0 x 1 03 1 1 .5 x 1 034 . =
x
3 9. 1
H 2 ( g) + I 2 (g) � 2 HI (g)
=
Kc
= 1 60
= [HI f [H 2 ] [I 2 ] 32 1 60 = ( 2.21 x 1 0- ) 3 [H 2 ] (1 .46 X 1 0- ) -3 2 [ H 2 ] - ( 2.21 x l 0 ) -3 (1 60) ( 1 .46 x 1 0 ) [H 2 ] = 2.1 x 1 0-5 mol · LI Kc
_
33 9.
K=
P.
PCI,
P.
CI2
PrC! ;
5.43) PrC!3 (-25 = -,1.18 P. = (25) (1 . 1 8) = 5 . 4 PCI) 5.43 PrC! = 5.4 bar 1
SM-253
9 .35
(a) K = 2 =160 (0.10)2 = 0.50 Q (0.20) (0.10) (b) Q K, . . the system is not at equilibrium. (c) Because Q K, more products will be formed. PHI
=
7=
9.37
'
<
]2 =1.7x106 (a) Kc = [S0[SO)2 f [OJ [1.0x10-4]2 0.5002 Qc [1.20.05X00lO-)] [5.00.X500I0-4] 6.9 (b) Because Qc Kc, more products will tend to form, which will result in the formation of more SO) . 1.90 HI = 0.0148 mol HI 127.91 marl 0.0172mol-2x x x 0.0148 mol = 0.0172 mol-2x x = 0.0012 mol [0.0012][0.0012] [0.0012]2 2.00 = 2.00 0.0066 or 6.6xl0-) K c = 2.00 [0.2.001480 ]2 [0.2.00148] 02 K Kc Kc (since �n =0) =
=
<
9.39
g
g.
=
=
(RT)/::,.11
=
SM-254
0.0174 g C�2 = 3.95 x 1 0-4 mol CO 2 44.01 g . mol- CO2 2 mol NH3 are formed per mol of CO2 , so mol NH3 = 2 x 3.95 X 1 0-4 = 7.90 x 1 0-4. 2 X 1 0-4 7.90 3.95 x l 0-4 = 1 .58 x 1 0-8 2 KC = [NH3 ] [CO 2 ] 0.250 0.250 =
9.43 9.45
(
)(
)
(c) K = x2/(l .0 - 2x)2 (a) The balanced equation is Cl 2 (g) � 2 CI(g). The initial concentration of Cl2 (g) is 0.0020 mol Cl 2 = 0.00 1 0 mol · Cl. 2.0 L Concentration (mol · L-1) Cl 2 (g) � 2 CI(g) 0.00 1 0 0 initial change -x +2 x equilibrium O.OO I O - x +2 x [Cl (2 X) 2 = 1 .2 x 1 0-7 K = f= C [CI 2 ] (0.00 1 0 - x) 4x2 = (1 .2 x 1 0-7 ) (0.00 1 0 - x) 4x2 + (1 .2 x 1 0-7 )x - (1 .2 x 1 0- 1 ° ) = 0 7 ) ± �(1 .2 x 1 0-7 ) 2 - 4(4) (-1 .2 x 1 0-1°) -(1 .2 x 1 0----------�---------------------x= 2x4 x 1 0-7 ) ± 4.4 x 1 0 -5 -(1-.2 -----' ------x = ----'8 -x = -5.5 x 1 0-6 0r + 5.5 x l 0-6 The negative answer is not meaningful, so we choose x = + 5.5 x 1 0-6 mol · L- 1 • The concentration of Cl 2 is essentially unchanged because 0.0010 - 5.5 x 1 0-6 � 0.00 1 0. The concentration of Cl atoms is 2 x (5.5 x I 0-6 ) = I . l x I 0-5 mol · L-I• (b) The balanced equation is: F2 (g) � 2 F(g).
SM -255
The problem is worked in an identical fashion to (a) but the equilibrium constant is now 1 .2 x 1 0-4 • . 0f F ( g) IS. 0.0020 mol F2 = 0 . 00 1 0 mo I · L- 1 . . . . I concentratIOn The ImtIa 2 2.0 L Concentration (mol · L-1 ) F2 (g) � 2 F(g) o 0.0010 initial +2 x change -x +2 x 0.001 0 - x equilibrium [Ff = (2 X) 2 Kc = 1 .2 X 1 0-4 [F2 J (0.00 1 0 - x) 4 x2 = (1 .2 x 1 0-4 ) (0.0010 - x) 4 x2 + (1 .2 x 1O-4 )x - (1 .2 x 1 0-7 ) = 0 (l .2 x 1 0-4 ) 2 - 4(4) (- 1 .2 x 1 0-7 ) -(1 .2 x 1 0-4 ) ± �----------------------x = ---------2.4 x 1 0-4 ) ±--1 .4----x 1 0-3 -(1-.2 -----'--x = ---'8 4 x = -1 .9 x 1 0- 0r +1 .6 x 1 0-4 The negative answer is not meaningful, so we choose x +1 .6 x 1 0-4 mol · L- 1 • The concentration of F2 is 0.001 0 - 1 .6 x 1 0-4 = 8 x 1 0-4 mol · L- 1 • The concentration of F atoms is =
�
=
(c) Cl 2 is more stable. This can be seen even without the aid of the calculation from the larger equilibrium constant for the dissociation for F2 compared to CI? 9.47
Concentration (bar) 2 HBr(g) initial 1 .2 X 1 0-3 -2 x change 1 .2 X 1 0-3 - 2x final SM-256
o
+x +x
o
+x +x
7.7 x 1 0- 1 1 =
(x) (x) X2 _ = (l .2 X 10-3 - 2 X) 2 (l .2 X 1 0-3 - 2 X) 2
�7.7 x 10- 1 1
=
___
_ _ _
(l .2 X 1 0-3 - 2 X) 2 = 8.8 X 1 0-6
x (l .2 X 10-3 - 2 x) X = (8.8 X 1O-6 ) (l .2 X 1 0-3 - 2 x) x + 2(8.8 X 1O-6 )x = (8.8 X 1 O-6 ) (l .2 X 10-3 ) x � l .1 X 10-8 = 1.1 X 10-8 bar; the pressure of HBr is essentially unaffected PH2 = PBr2 by the formation of Br2 and H 2 . 9.49
(a) Concentration of PCl5 initially =
( 208.221.0gg. mol-I PCls PCls J
= 0.019 mol · L- I .
0.250 L Concentration (mol · L-1 ) PCls (g) 0.019 initial -x change 0.019 - x final
�
x2 = l .1 10-2 ---(0.019 - x) X
x2 = (1.1 X 10-2 ) (0.0 1 9 - x) x2 + (l .1 X 1O-2 )x - 2.1 X 1 0-4 = 0
SM -257
PCl 3 (g) o
+x +x
+
Cl 2 o
+x +x
1 0-2 ) ± �(1 . 1 X 1 0-2 ) 2 - (4)(1)(-2. 1 X 1 0-4 ) -(1 . 1 x X= 2. 1 2 -(1 . 1 X 1 0- ) ± 0.03 1 = +0.01 0 or -0.02 1 x= 2. 1 The negative root is not meaningful, so we choose x= 0.010 mol· L-' . ----------�----------------------
(b) The percentage decomposition is given by 0.0 1 0 X 1 00= 53%. 0.0 1 9 9.51
Starting concentration of NH ) = OAOO mol = 0.200 mol · L-' 2.00 L Concentration (mol· L- ' ) NH 4 HS(s) � initial 0.200 o change +x +x final 0.200 + x +x Kc = [NH J [H 2 S] = (0.200 + x) (x) 1 .6 X 1 0-4 = (0.200 + x) (x) x2 + 0.200x - 1 .6 x 1 0-4 = 0 2 -( +0.200) ± �(+0.200) - (4) (1) ( -1 .6 x 1 0-4 ) --------� --------------------x= 2.1 -0.200 ± 0.201 6= +0.0008 or - 0.2008 x= 2. 1 The negative root is not meaningful, so we choose x= 8 x 1 0-4 mol · L- ' (note that in order to get this number we have had to ignore our normal significant figure conventions). [NH)]= +0.200 mol · L- ' + 8 x 1 0-4 mol · Cl = 0.200 mol · L-' [H 2 S] 8 x l 0-4 mol · L- ' Alternatively, we could have assumed that x « 0.2, the 0.200x= 1 .6 x l 0-4 , X = 8.0 x 1 0-4 . =
SM-258
9.53
The initial concentrations of N 2 and 02 are equal at 0.114 mol · Cl because the vessel has a volume of 1.00 L. Concentrations (mol · C ' ) N 2 (g) + 02 (g) � 2 NO(g) 0.114 initial 0.114 o -x change -x +2 x 0.114 - x 0.114 - x +2x final (2 X) 2 (2 X) 2 [NO]2 Kc [N2 ][02 ] (0. 1 14 - x) (0.114 - x) (0. 1 1 4 - x) 2 2 (2 X) 5 1 .00 x 10(0.114 - xt? 2 (2 X) -5 x 10 1.00 � (0.114 - x) 2 (2 x) 3.16 x 10-3 (0.114 - x) 2 x (3. 1 6 x 10-3 ) (0. 1 14 - x) 2.003 16 x = 3.60 X 10-4 x = 1 .8 X 10-4 [NO] 2 x 2 x 1.8 X 1 0-4 = 3.6 X 10-4 mol · L-' ; the concentrations of N 2 and 0 2 remain essentially unchanged at 0.1 1 4 mol · L- ' . =
=
=
=
=
=
=
=
9.55
=
The initial concentrations of H 2 and 12 are [H 2 ] 0.400 mol = 0.133 mol . L- ' · [I 2 ] = 1 .60 mol 0.533 mol · Cl ' 3 .00 L 3.00 L Concentrations (mol · L- ' ) H 2 (g) + 12 (g) � 2 HI(g) o 0.533 0. 1 3 3 initial +2 x -x -x change +2 x 0.133 - x 0.533 - x final At equilibrium, 60.0% of the Hz had reacted, so 40.0% of the H 2 remams: =
=
SM-259
(0.400) (0. 1 33 mol · C') = 0. 1 33 mol · C' - x x = 0. 1 33 mol · L- ' - (0.400) (0.133 mol · L- ' ) x = 0.080 mol · L- '
At equilibrium: [ H 2 ] = 0.1 33 mol · C' - 0.080 mol · L- ' = 0.053 mol · L- ' [1 2 ] = 0.533 mol · L-' - 0.080 mol · L-' = 0.453 mol · L- ' [HI] = 2 x 0.080 mol · C' = 0. 1 6 mol · L-' 0. 1 62 [HI]2 = K = =1.1 C [H 2 ] [IJ (0.053) (0.453) 9 .57
( 28.010.28g · gmorCO'CO J 3 = 5.0 x 1 0- mol · L- ' [CO] = 2.0 L ( 32.00.032 g °2 0 g · mol - ' ° J
Initial concentrations of CO and 02 are given by
[0 2 ] =
2 = 5.0 x 1 0-4 mol . L- ' .
2.0 L Concentration (mol · L- ' ) 2 CO(g) + 0 2 (g) � 2 CO2 (g) initial 5.0 x 1 0-3 5.0 X 1 0-4 o change -2 x -x +2 x final 5.0 X 1 0-3 - 2 x 5.0 x l O-4 - x +2 x [C02 ] 2 K = C [COf[0 2 ] (2 X)2 = ------�--�--------(5.0 x 1 0-3 -2 x) 2 (5.0 x 1 0-4 - x) 4x2 = (4 x2 - 0.020 x + 2.5 x 1 0-5 ) (5.0 x 1 0-4 - x) -�------------------------
SM-260
4 x2 0.66 = - 4 x3 + 0.022 x2 - 3.5 x l 0-5 x + 1 .25 x l 0-8 4x2 = (0.66) (-4x3 + 0.022 x2 - 3.5 x l 0-5 x + 1 .25 x l 0 -8 ) 6.06 x2 = -4 x3 + 0.022 x2 - 3.5 X 1 0-5 X + 1 .25 X 1 0-8 0 = -4x3 - 6.04 x2 -3.5 x l 0-5 x + 1 .25 x l 0-8 X = 4.3 X 1 0-5 [C0zJ = 8.6 X 1 0-5 mol · L- ' ; [CO] = 4.9 X 1 0-3 mol · L- ' [°2 ] = 4.6 x l 0-4 mol · L- ' 2 = [BrCl] [CI 2 ][BrJ 2 0.03 1 = (0.1 45) (0.495) [Br2 ] 2 [Br2 ] = (0. 145) = 1 .4 mol . C ' (0.495) (0.03 1)
9.59
Kc
9.61
We can calculate changes according to the reaction stoichiometry: Amount (mol) 3.00 2.00 o o initial -3 x +x -X +x change 0.478 2.00 - X 3.00 - 3 X +x final According to the stoichiometry, 0.478 mol = X; therefore, at equilibrium, there are 2.00 mol - 0.478 mol = 1 .52 mol CO, 3.00 - 3(0.478 mol) = 1 .57 mol H 2 , and 0.478 mol H 2 0. To employ the equilibrium expression, we need either concentrations or pressures; because Kc is given, we will choose to express these as concentrations. This calculation is easy because V= 1 0.0 L: [CO] = 0. 1 52 mol · C ' ; [H 2 ] = 0. 1 57 mol · C ' ; [CH 4 ] = 0.0478 mol · L- ' ; [H 2 0] = 0.0478 mol · L- ' [CH 4 ][H 2 0] = (0.0478) (0.0478) K = c [CO][H 2 ] 3 (0. 1 52) (0. 1 57)3 = 3.88 SM-261
9 .63
First, we calculate the initial concentrations of each species: [SO 2 ]= [NO]= 0.1 00 mol= 0.0200 mol · L-' '· 5.00 L [NO 2 ]= 0.200 mol= 0.0400 mol . L- ' · [SO 3 ]= 0. 1 50 mol = 0.0300 mol · L- ' 5.00 L 5.00 L We can use these values to calculate Q in order to see which direction the reactions will go: Q= (0.0200) (0.0300) = 0.75 (0.0200) (0.0400) Because Q < Kc' the reaction will proceed to produce more products. Concentration (mol · L- ' ) S0 2 (g) + N02 (g) ---+ NO(g) + S03 (g) initial 0.0200 0.0400 0.0200 0.0300 -x change -x +x +x final 0.0200 - x 0.0400 - x 0.0200 + x 0.0300 + x [NO][S03 ] K = C [S0 2 ][N0 2 ] 85.0= (0.0200 + x) (0.0300 + x) (0.0200 - x) (0.0400 - x) x2 + 0.0500 x + 0.000 600 x2 - 0.0600 x + 0.000 800 85.0(x2 - 0.0600x + 0.000 800) = x2 + 0.0500x + 0.000 600 85.0 X2 - 5.1 0 x + 0.0680 = x2 + 0.0500x + 0.000 600 84.0 x2 - 5.1 5 x + 0.0674 = 0 -( -5.1 5) ± �(-5.1 5) 2 (4) (84.0) (0.0674) = +5.1 5 ± 1 .97 X= (2) (84.0) 1 68 x = +0.0424 or +0.0 1 89 The root 0.0424 is not meaningful because it is larger than the concentration of N0 2 . The root of choice is therefore 0.01 89. At equilibrium: '
=
-------
-
----
SM-262
[S0 2 ]= 0.0200 mol · L-' - 0.0 1 89 mol · L-' = 0.001 1 mol · L-' [N0 2 ] = 0.0400 mol · L- ' - 0.01 89 mol · L- ' = 0.02 1 1 mol · L- ' [NO] = 0.0200 mol · L-' + 0.01 89 mol · L- ' = 0.0389 mol · L-' [SOJ = 0.0300 mol · L-' + 0.01 89 mol · L- ' = 0.0489 mol · L- ' To check, we can put these numbers back into the equilibrium constant expreSSIon: [NO][SO} ] K= C [S0 2 ][N0 2 ] (0.0389) (0.0489) = 82.0 (0.001 1) (0.02 1 1) Compared to Kc = 85.0, this is reasonably good agreement given the nature of the calculation. We can check to see, by trial and error, if a better answer could be obtained. Because the Kc value is low for the concentrations we calculated, we can choose to alter x slightly so that this ratio becomes larger. Ifwe let x= 0.01 90, the concentrations of NO and S03 are increased to 0.0390 and 0.0490, and the concentrations of S0 2 and N0 2 are decreased to 0.00 1 0 and 0.0200 (the stoichiometry of the reaction is maintained by calculating the concentrations in this fashion). Then the quotient becomes 91 .0, which is further from the value for Kc than the original answer. So, although the agreement is not the best with the numbers we obtained, it is the best possible, given the limitation on the number of significant figures we are allowed to use in the calculation. 9.65
(a) The initial concentrations are [PCl 5 ]= 1 .50 mol = 3 . 00 mol . L-' · [PCl3 ] = 3.00 mol= 6.00 mol · L-' '· 0.500 L ' 0.500 L 0.500 mol = 1.00 mol· L-1 [el2 ]= 0.500 L First calculate Q: •
SM-263
Because Q -:f:. K, the reaction is not at equilibrium. (b) Because Q < Kc, the reaction will proceed to form products. (c) Concentrations (mol · L-1 ) PCI3 (g) + CI 2 (g) � PCls (g) 3.00 1 .00 6.00 initial -x +x -x change 3 .00 + x 1 .00 - x 6.00 - x final
3.00 + x = 2 3.00 + x (6.00 - x) (1 .00 - x) x - 7 x + 6.00 (0.56) (x2 - 7 x + 6.00) = 3.00 + X 0.56 x2 - 3.92 x + 3.36 = 3.00 + X 0.56x2 - 4.92 x + 0.36 = 0 2 -(-4.92) ± )(-4.92) - (4) (0.56) (0.36) +4.92 ± 4.48 = ---x= (2) (0.56) 1 .12 x = 9.2 or 0.07 Because the root 9.2 is larger than the amount of PCl 3 or Cl 2 available, it is physically meaningless and can be discarded. Thus, x 0.071 mol· L- 1 , gIvmg [PCls ] = 3.00 mol · L- 1 + 0.07 mol · L- 1 = 3.07 mol · L- 1 [PCI3 ] = 6.00 mol · L- 1 - 0.07 mol · L- 1 = 5.93 mol · L- 1 [CI 2 ] = 1 .00 mol · L- 1 - 0.07 mol · L- 1 = 0.93 mol · L-1 The number can be checked by substituting them back into the equilibrium constant expression: K = [PCls] C [PCI3 ] [CIJ (3.07) ,; 0.56 (5.93) (0.93) 0.56::10.56 0.56 =
----:-----
=
SM-264
9.67
Pressures (bar) 2 HCl(g) H 2 (g) + CI 2 (g) initial 0.22 0 0 change -2 x +x +x final 0.22 - 2 x +x +x p.p. H 2 CI2 K= PHCI2 3.2 X 1 0-34 = (x) (x) 2 (0.22 - 2 X) Because the equilibrium constant is small, assume that x « 0.22: X2 3 4 = X 1 03.2 (0.22) 2 2 3 x2 = (3.2 x 1 0- 4 ) (0.22) X = �(3.2 x 1 0-34) (0.22) 2 x = ± 3.9 x l O- ' 8 The negative root is not physically meaningful and can be discarded. x is small compared to 0.22, so the initial assumption was valid. The pressures at equilibrium are PHCI = 0.22 bar;PH 2 = Fel2 =3.9 X 1 0-1 8 bar The values can be checked by substituting them into the equilibrium expressIOn: (3.9 x l O- ' 8 ) (3.9 x l O- ' 8 ) :b3.2 x l O-34 (0.22) 2 3 . 1 X 1 0-34 :!::. 3.2 X 1 0-34 PHCI = 0.22 bar;PH 2 = Fe 1 2 = 3.9 x l O- ' 8 bar The numbers agree very well for a calculation of this type.
9.69
(a) To determine on which side of the equilibrium position the conditions lie, we will calculate Q:
SM-265
mol = 0. 1 14 mol . L- ' · [H ] = 0.2 1 5 mol = 0.07 1 7 mol . L- ' ; [CO] = 0.342 ' 2 3.00 L 3.00 L [CH 3 OH] = 0. 125 mol = 0.0417 mol · L-' 3.00 L 0.0417 = 71.1 x 1 03 Q = [ CH 3 0H]2 = 2 [CO ][H 2 ] (0. 1 14) (0.07 17 ) Because Q > Kc ' the reaction will proceed to produce more of the reactants, which means that the concentration of methanol will decrease. (b) Concentrations (mol · Cl ) CO(g) + 2 H 2 (g) � CH 3 0H(g) initial 0.1 14 0.07 1 7 0.04 1 7 change +2 x -x +x final 0.01 14 + x 0.07 1 7 + 2 x 0.04 1 7 - x 0.041 7 - x Kc = = 1 . 1 x 1 0-2 (0.01 14 + x) (0.071 7 + 2 X) 2 0.04 1 7 - x = (1 . 1 x 1 0-2 ) (0. 1 14 + x) (0.07 1 7 + 2 X) 2 = (1 .2 X 1 0-3 + 1 . 1 X 10-2 x) (4x2 + 0.287 x + 5. 14 x 1 0-3 ) = 4.4 x l 0-2 x3 + 8.0 x l 0-3 x2 + 4.0 x l 0-4 X + 6.2 x l 0-6 0 = 4.4 x 1 0-2 x3 + 8.0 x 1 0-3 x2 + 1 .00x - 0.04 1 7 This equation can be solved approximately, simply by inspection: It is clear that the x term will be very much larger than the x3 and the x2 terms because their coefficients are very small compared to 1 .00. This leads to a prediction that x = 0.0417 mol · L- ' to within the accuracy of the data. Essentially all of the CH 3 0H will react, so that [CO] = 0. 1 14 mol · L- ' + 0.0417 mol · L- ' = 0.1 56 mol · L- ' ; [H 2 J 0 . 07 1 7 mol · L-' + 2 ( 0 .04 1 7 mol · Cl ) 0 . 1 5 5 mo l · L- ' . The mathematical situation is odd in that clearly a [CH 3 0H] = 0 will not satisfy the equilibrium constant. Knowing that the methanol concentration is very small compared to the CO and H 2 concentrations, we can now =
=
SM -266
back-calculate to get a concentration value that will satisfy the equilibrium expreSSIOn: y K = = 1 . 1 x 1 0-2 C (0. 1 56) (0. 1 55) 2 y = (1 . 1 X 1 0- 2 ) (0.1 56) (0.1 55) 2 = 2.6 X 1 0-4 Alternatively, the cubic equation can be solved with the aid of a graphing calculator like the one supplied on the CD accompanying this book. 9.7 1
Since reactants are strongly favored, it is easier to push the reaction as far to the left as possible then start from new initial conditions. Pressures (bar) 2 HCI(g) ---+ H 2 (g) + CI2 (g) 3 .0 1 .0 2.0 original 2.0 0 4.0 new initial +x -2 x +x change 2.0 +x 4.0 - 2 x +x final K=
P.H 2 P.C 2
I PHCI 2 � = 3.2 x 1 0-34 = (x) (2.0 + x)2 (x) (2.0) 2 (4.0 - 2 X) (4.0) 1 6 x = 2.6 x 1 0-33 bar 2 = nR T (1 mole)(8.3 1 4 x l 0- L bar K -I mol- ' )(298K) V = 2.6 x l 0-33 bar p = 9.7 x 1 0 33 L �
9.73
(a) According to Le Chatelier' s principle, an increase in the partial pressure of CO 2 will result in creation of reactants, which will decrease the H 2 partial pressure. (b) According to Le Chatelier' s principle, if the CO pressure is reduced, the reaction will shift to form more CO, which will decrease the pressure of CO2 • SM-267
(c) According to Le Chatelier' s principle, if the concentration of CO is increased, the reaction will proceed to fonn more products, which will result in a higher pressure of H 2 • (d) The equilibrium constant for the reaction is unchanged because it is unaffected by any change in concentration. 9.75
9.77
(a) According to Le Chatelier's principle, increasing the concentration of NO will cause the reaction to fonn reactants in order to reduce the concentration of NO; the amount of water will decrease. (b) For the same reason as in (a), the amount of O2 will increase. (c) According to Le Chatelier's principle, removing water will cause the reaction to shift toward products, resulting in the fonnation of more NO. (d) According to Le Chatelier's principle, removing a reactant will cause the reaction to shift in the direction to replace the removed substance; the amount of NH3 should increase. (e) According to Le Chatelier's principle, adding ammonia will shift the reaction to the right, but the equilibrium constant, which is a constant, will not be affected. (f) According to Le Chatelier's principle, removing NO will cause the fonnation of more products; the amount of NH3 will decrease. (g) According to Le Chatelier' s principle, adding reactants will promote the fonnation of products; the amount of oxygen will decrease. As per Le Chatelier' s principle, whether increasing the pressure on a reaction will affect the distribution of species within an equilibrium mixture of gases depends largely upon the difference in the number of moles of gases between the reactant and product sides of the equation. If there is a net increase in the amount of gas, then applying pressure will shift the reaction toward reactants in order to remove the stress applied by increasing the pressure. Similarly, if there is a net decrease in the amount SM-268
of gas, applying pressure will cause the formation of products. If the number of moles of gas is the same on the product and reactant side, then changing the pressure will have little or no effect on the equilibrium distribution of species present. Using this information, we can apply it to the specific reactions given. The answers are: (a) reactants; (b) reactants; (c) reactants; (d) no change (there is the same number of moles of gas on both sides of the equation); ( e) reactants. 9.79
(a) If the pressure of NO (a product) is increased, the reaction will shift to form more reactants; the pressure of NH 3 should increase. (b) If the pressure of NH 3 (a reactant) is decreased, then the reaction will shift to form more reactants; the pressure of 02 should increase.
9.81
If a reaction is exothermic, raising the temperature will tend to shift the reaction toward reactants, whereas if the reaction is endothermic, a shift toward products will be observed. For the specific examples given, (a) and (b) are endothermic (the values for (b) can be calculated, but we know that it requires energy to break an x-x bond, so those processes will all be endothermic) and raising the temperature should favor the formation of products; (c) and (d) are exothermic and raising the temperature should favor the formation of reactants.
9.83
Even though numbers are given, we do not need to do a calculation to answer this qualitative question. Because the equilibrium constant for the formation of ammonia is smaller at the higher temperature, raising the temperature will favor the formation of reactants. Less ammonia will be present at higher temperature, assuming no other changes occur to the system (i.e., the volume does not change, no reactants or products are added or removed from the container, etc.).
9.85
To answer this question we must calculate Q: SM-269
Because Q -:F K, the system is not at equilibrium, and because Q < K, the reaction will proceed to produce more products. 9.87
Because we want the equilibrium constant at two temperatures, we will need to calculate !'1J{ °r and J".,.S or for each reaction: (a) NH 4 Cl � NH 3 (g) + HC 1(g) !'1J{ °r = !'1J{ °r(NH 3 , g) + !'1J{ °r(HC1, g) - !'1J{ °r(NH 4 Cl, s) !'1J{ ° r = (-46. 1 1 kJ · mar ' ) + (-92.3 1 kJ · mol-i ) - (-3 1 4.43 kJ · mar ' )
J".,.S or = 1 92.45 J . K- ' . mol -' + 1 86.91 J . K- ' . mol- ' - 94.6 J K -' . mol- ' J".,.S or = 284.8 J . K- ' · mar ' .
At 298 K: J".,. Gor(298 K) = 1 76.01 kJ - (298 K) (284.8 J . K- ' )/(1 000 J . kr' ) = 9 1 . 1 4 kJ · mar ' J".,. Gor(298 K) = - RT In K J".,.Go In K= - - r(298 K) RT 91 140 J = = -36.8 (8.3 14 J . K- ' ) (298 K) K= 1 X 1 0- ' 6 At 423 K: J".,. Gor(423 K) = 1 76.01 kJ - (423 K) (284.8 J K - ' )/(1 000 J . kr ' ) = 55.54 kJ · mar ' J".,.Go r(423 K) = - RT In K --'---'-
.
SM-270
In K =
_
I1GO ,(423 K) RT
55 540 J = - 1 5.8 (8.3 1 4 J · K- I ) (423 K) K = 1 X 1 0- 7 (b) H 2 (g) + D 2 0(l) � D 2 (g) + H 2 0(l) Mfo, = Mfor (H 2 0, 1) - [Mfor (D 2 0, 1)] Mfo, = (-285.83 kJ · mol-I ) - [-294.60 kJ · mor l ] Mfo, = 8.77 kJ · moI- 1 I1So, = SO(D 2 , g) + SO(H 2 0, 1) - [SO(H 2 ' g) + SO(D 2 0, 1)] I1So, = 1 44.96 J . K - I . moI -1 + 69.91 J . K-I . mOr l - [1 30.68 J · K-I . mor l + 75.94 J · K - I . mOr l ] I1SO , = 8 . 25 J . K mOr l At 298 K: I1Go,(298 K) = 8.77 kJ · mOr l - (298 K) (8.25 J · K-I ' moI- I )/(l OOO J . kJ-I ) = 6.3 1 kJ · mor l I1GO ,(298 K) = - RT In K =
-I .
In K = -
I1Go r(298 K)
----'-----'-
RT
-
63 1 0 J = -2.55 (8.3 14 J . K- I ) (298 K) K = 7.8 x l 0-2 At 423 K: I1G°r(423 K) = 8.77 kJ · moI- 1 - (423 K) (8.25 J · K- I ' mor l )/(1000 J · kr l ) = 5.28 kJ · mOr l =
In K =
_
5280 J = -1 .50 (8.3 14 J · K- I ) (423 K)
K = 0.22
SM-271
[In ((RTR� )/;Iltil KKcc2 = - R [�-T _T12 J lJ K 2 =-f':,}{ro [ 1 1 � + In l:.n ln J [ [ KCcI J R � J � - � J - l:.n ln [ � J [ In ( Kc 2 =Kc l J R T2 f':,}{ro
I
I
--
T;
T;
f':,}{ro
9.91
9.93
- - -
T;
T;
The change increased the concentration ofX(g) and decreased the concentration ofX2 (g)in Flask 2. Increasing the temperature (a) would increase the formation of X(g) because the reaction is endothermic. Adding X(g) atoms or decreasing the volume (increasing pressure) would increase the formation ofX2 (g), which did not occur. Adding a catalyst would not favor the formation of either X(g) or X2 (g). Also, the value of the equilibrium constant is larger in Flask 2 than in Flask 1 , which is consistent with an increase in temperature of an endothermic reaction. (a) According to Le Chatelier' s principle, adding a product should cause a shift in the equilibrium toward the reactants side of the equation. (b) Because there are equal numbers of moles of gas on both sides of the equation, there will be little or no effect on compressing the system. (c) If the amount of CO 2 is increased, this will cause the reaction to shift toward the formation of products. (d) Because the reaction is endothermic, raising the temperature will favor the formation of products. (e) If the amount of C6 H I 2 06 is removed, this will cause the reaction to shift toward the formation of products. (f) Because water is a liquid, it is by definition present at unit concentration, so changing the amount of water will not affect the SM -272
reaction. As long as the glucose solution is dilute, its concentration can be considered unchanged. (g) Decreasing the concentration of a reactant will favor the production of more reactants. 9.95
(a) 2A(g)
�
B(g) + 2C(g) (5 / 1 00) (1 0/100) 2 = 1 .54 x 1 0-2 (1 8/1 00)2 Note: The partial pressure units are changed to bar
(c) K = (RT/nKc; Kc = (RTILJnK = (RTliK =KIRT= 2 1 .54 1 0= 6.21 x 1 0-4 (8.3 1 45 1 0-2 L · bar · K- 1 · mol- I )(298. 1 5 K) x
x
9.97
(a) In order to solve this problem, we will manipulate the equations with the known K ' s so that we can combine them to give the desired overall reaction. First, reverse equation ( 1 ) and multiply it by t : (4) Multiply equation (2) by t also: CO2 (g) � CO(g) + t 02 (g) Ks = (1 . 3 x 1 0- 1 °)" 2 Adding equations (4) and (5) gives the desired reaction. The resultant equilibrium constant will be the product of the K' s for (4) and (5):
(5)
(3) (b) To obtain the K value for a net equation from two (or more) others, the K' s are multiplied, but I'1Go:s are added: (1)
SM-273
= -RT In K = -(8 . 3 1 4 J . K- I . mol-I ) (1 565 K) In 1 .6 x 1 0-1 1 = +3.2 X 1 02 kJ · mol-I 2 CO2 (g) � 2 CO(g) + 0 2 (g) !1 G o r -RT In K = -(8.3 1 4 J . K - I . mol -I ) (1 565 K) In 1 .3 x 1 0-1 0 = +3.0 x l 02 kJ · mol- 1 The corresponding values for (4) and (5) are 2 2 !1Gor(4 ) = - 1 (3.2 x 10 kJ) = -1 .6 x 1 0 kJ 2 2 !1Go r( ) = t (3.0 x 1 0 kJ) = 1 .5 x 1 0 kJ. 5 Summing these two values will give 2 2 !1 G o r(3 ) = -1 .6 x 1 0 kJ + 1 .5 x 1 0 kJ = -1 0 kJ · mol -1 !1GO r
(2)
=
In K = - !1G or = RT
-1 0 000 J · mol-1 (8.3 14 J . K -I . morl ) (1 565 K) = +0.8
K=2. This value is in reasonable agreement with the one obtained in (a), given the problems in significant figures and due to rounding errors. 9.99
Pressure initial change final
�
PCIs (g)
PCI 3 (g)
-na
+na
+na
n(1 - a)
+na
+na
n2a2
na2
=
(1 - a) P = n(1 - a) + na + na = n(1 + a) n(1 - a )
n(1 - a)
P
1+a
--
K=
na '
(1 - a )
Cl 2 (g)
n
K = (na) (na) = n=
+
=
(1 ) (1 - ) = 1 P +a
a'
a
Pa ' a2
SM-274
(1 - a 2 )K = Pa 2 K - K a 2 = Pa 2 K = Pa 2 + K a 2 a2 = K P+K
��
a- p K (a) For the specific conditions K = 4.96 and P = 0.50 bar, a = / 4.96 = 0.953 . 0.50 + 4.96 (b) For the specific conditions K = 4.96 and P = 1 .00 bar, 4.96 = 0.91 2 . a= 1 .00 + 4.96 ----
----
9.101
(a) If K = 1 .00, then � G O must be equal to 0 ( �Go - RT In K). (b) This can be calculated by determining the values for MO and �SO at 25 ° C. MO = - 393.5 1 kJ ' mol- ' - [( -1 1 0.53 kJ · mor ' ) + ( - 241 .82 kJ ' mol- ' )] = - 4 1 . 1 6 kJ . mar ' �SO = 1 3 0.68 J . K- ' . mol-' + 2 1 3.74 J . K- ' . mar ' - [1 97.67 J · K- ' · mol- ' + 1 88.83 J · K -' · mar ' ] = - 42.08 J . K- ' · mar ' �G = ( -4 1 . 1 6 kJ ' mol- ' ) ( 1 000 J . kr ' ) - T( -42.08 J . K- ' · mol- I ) = 0 T = 978 K (or 705° C) H 2 (g) (c) CO(g) + H 2 O(g) � CO 2 (g) + 5.00 bar 1 0.00 bar 5.00 bar 1 0.00 bar -x change -x +x +x net 1 0.00 - x 5.00 + x 5.00 + x 1 0.00 - x x = 2.50 bar All pressures are equal to 7.50 bar. =
SM-275
(d) First, check Q to determine the direction of the reaction: Q (1 0.00) (5.00) = 2.08 (6.00) (4.00) Because Q is greater than 1 , the reaction will shift to produce reactants. CO 2 (g) + 1 0.00 bar 6.00 bar 4.00 bar -x -x +x change +x net 4.00 + x 1 0.00 - x 5.00 - x 6.00 + x (1 0.00 - x) (5.00 - x) = 1 (6.00 + x) (4.00 + x) (1 0.00 - x) (5.00 - x) (6.00 + x) (4.00 + x) 2 x - 1 5.00 x + 50.0 x2 + 1 0.00 x + 24.0 25.00 x 26.0 x= 1 .04 bar PCO(g) 7.04 bar ; PH 0(g) = 5.04 bar; Pco (g) = 8.96 bar; Pr' (g) = 3.96 bar 2 2 2 =
=
=
=
=
9 . 1 03
(a) These values are easily calculated from the relationship 110°= - R T In K. For the atomic species, the free energy of the reaction will be t of this value because the equilibrium reactions are for the formation of two moles of halogen atoms. The results: Bond Dissociation Energy 110° Halogen (kJ . mol- I ) (kJ · mor l ) fluorine 1 46 1 9. 1 chlorine 230 47.9 bromine 181 42.8 iodine 139 5.6
SM-276
(b) 60 •
•
• •
o +-----�-----,--�--� o
50
1 00
1 50
200
250
300
bond dissociation energy (kJ·mor l )
There is a correlation between the bond dissociation energy and the free energy of formation of the atomic species, but the relationship is clearly not linear.
=
.9 .....
.c::i 0 '" -'"""' '" c:=
:a � '0 �
;.., "'":1 I:)J) � '-" s..
:=
.;:
1 20
1 00
•
80
•
60
40
•
20 0
• 0
10
20
30
40
atomic number
50
60
For the heavier three halogens, there is a trend to decreasing free energy of formation of the atoms as the element becomes heavier, but fluorine is anomalous. The F-F bond energy is lower than expected, owing to
SM-277
repulsions of the lone pairs of electrons on the adjacent F atoms because the F-F bond distance is so short. 9 . 1 05
(a) Using the thermodynamic data in Appendix 2A: Br2 (g) � 2 Br(g) flGO = 2(82.40 kJ . mol-I ) - 3 . 1 1 kJ . mol- I = 1 61 .69 kJ . mol - I K e- I'1Co I RT = 4.5 x 1 0-29 For equilibrium constant calculations, this is reasonably good agreement with the value obtained from part (a), especially if one considers that M-[0 will not be perfectly constant over so large a temperature range. (b) We will use data from Appendix 2A to calculate the vapor pressure of bromine: Br2 (1) � Br2 (g) flGo = 3 . 1 1 kJ · mol-I K = e- I'1Co /RT = 0.285 The vapor pressure of bromine will, therefore, be 0.285 bar or 0.289 atm. Remember that because the standard state for the thermodynamic quantities is 1 bar, the values in K will be derived in bar as well. =
PBr(g) 2
= 3.6 x l 0- 15 bar or 3.6 x l 0- 15 atm (d) Use the ideal gas law: PV = nRT (0.289 atm) V = (0.01 00 mol) (0.082 06 L · atm · K- 1 mol- I ) (298 K) V 0.846 L or 846 mL •
=
9 . 1 07
First, we calculate the equilibrium constant for the conditions given. 2 . wntten . as K (23.72) 3 = 41 . 0, WhIC' h correspond s to th e reactlOn (3. 1 1) (1 .64) N 2 (g) + 3 H 2 (g) � 2 NH ) (g) =
SM-278
We then set up the table of anticipated changes upon introduction of the nitrogen: N 2 (g) + 3 H 2 (g) ---+ 2 NH 3 (g) initial 4.68 bar 1 .64 bar 23.72 bar -3 x change -x +2 x total 4.68 - x 1 .64 - 3x 23.72 + 2 x 2 (23.72 + 2 x) 4 1 .0 = (4.68 - x) (1 .64 - 3 X)3 The equation can be solved using a graphing calculator, other computer software, or by trial and error. The solution is x = 0.0656 and the pressures of gases are PN2 = 4.61 bar, PH2 = 1 .44 bar, PNHJ = 23.85 bar. 9.109
The equilibrium reaction across the membrane is Na+ (outside of the cell) � � Na+(inside the cell) . cons tant Kc [Na + Lsidelhecell = 1 0 mM = 7 . 1 x 1 0-2 The eqUl·l1· b flum [Na + ]oulsidelhecell 140 mM i1GrO = -RT lnK = - (8.3 14 J · K- 1 · mol-I)(3 10.1 5 K) ln(7.1 x 1 0-2) 6.8 x 1 03 J = 6.8 kJ A positive i1Gro means that the reaction above is not favorable to the right side (the Na+ inside the cell). Note: K = Kc in this problem so Kc is directly used in i1Gro calculation. =
=
9.1 1 1
250
300 T IK
350
(b) If N0 2 is added, the equilibrium will shift to produce more N 2 04 . The amount of N0 2 will be greater than initially present, but less than the 3 . 1 3 mol · L- ' present immediately upon making the addition. Kc will not be affected. (c) Concentrations (mol · L- ' ) N 2 04 --'> 2 N0 2 0.405 3.13 initial -2 x change +x 3. 1 3 - 2 x 0.405 + x final 2 1 1 .2 = (3. 1 3 - 2 X) 0.405 + x (1 1 .2) (0.405 + x) = (3. 1 3 - 2 X) 2 1 1 .2 x + 4.536 = 4 x2 - 12.52 x + 9.797 4x2 - 23.7 x + 5.26 = 0 2 (-23.7) ± J(-23.7) - (4) (4) (5.26) 23.7 ± 2 1 .9 = X= (2) ( 4) 8 x = 5.70 or 0.23 At equilibrium, [N 2 0 4 ] = 0.405 mol · L- ' + 0.23 mol · L-' = 0.64 mol · C' [N0 2 ] = 3 . 1 3 mol · L- ' - 2(0.23 mol · L- ' ) = 2.67 mol · L- ' . These concentrations are consistent with the predictions in (b). ----
9.1 15
To find the vapor pressure, we first calculate 11 0 ° for the conversion of the liquid to the gas at 298 K, using the free energies of formation found in the appendix:
SM -280
/)'G O H20(I) -> H20(g)
/)'G O 020(1) ->
= /),G O f(H20(g)) - /),GO f(H20(I)) = ( - 228.57 kJ · mol-I ) - [ - 237. 1 3 kJ · mol-I ] = 8.56 kJ . mol-I /)'G O f(020(g)) - /)'G O f(020(l )) 020(g) ( - 234.54 kJ · mol- I ) [ - 243.44 kJ . mor l ] = 8.90 kJ . mor l =
=
-
The equilibrium constant for these processes is the vapor pressure of the liquid: Using /)'G o = - RT In K, we can calculate the desired values. In K =
_
/)' G o
RT
=
8560 J . mor l = - 3 .45 (8.3 1 4 J . K - I . mol -I ) (298 K)
K = 0.032 bar 0.032 bar x
In K =
_
1 atm x 760 Torr = 24 Torr 1 .0 1 3 25 bar 1 atm
/)' G o
RT
=
8900 J . mol - I = - 3.59 (8.3 1 4 J . K - I . mor l ) (298 K)
K 0.028 bar =
1 atm x 760 Torr = 2 1 Torr 1 .0 1 3 25 bar 1 atm The answer is that D has a lower zero point energy than H. This makes the D 2 0 - D 2 0 "hydrogen bond" stronger than the H 2 0 - H 2 0 hydrogen bond. Because the hydrogen bond is stronger, the intermolecular forces are stronger, the vapor pressure is lower, and the boiling point is higher. Potential energy curves for the O-H and O-D bonds as a function of distance: M = energy required to break O-H or O-D bond 0.028 bar x
SM-28 1
9. 1 1 7
First, we derive a general relationship that relates I1GO f values to nonstandard state conditions. We do this by returning to the fundamental definition of I1G and I1GO : I1GO f = "L. G o m(prodUCIS) - "L. G o m(reaClanlS) Similarly, for conditions other than standard state, we can write I1Gf = "L. Gm(ProduCIS) - "L. Gm(reaClanIS) · Because
Gm
= GO m + R T In Q,
But because all reactants or products in the same system refer to the same standard state, I1Gf = "L. G o m(producls) - "L. G o m(reaClanIS) + I1n ( RT In Q(reaClanIS» ) The value I1 Gf , which is the nonstandard value for the conditions of I atm, 1 mol · L-I , etc. becomes the new standard value. (a) t H 2 (g) + t 1 2 (g) � HI(g), I1n = 1 mol - (t mol + t mol) = 0 1 atm = 1 .01 3 25 bar I1Gf = 1 .70 kJ · mol- I + (0) (8.3 14 J . K- I . mol-I ) (298 K) (In 1 .0 1 3 25)/(1 000 J . kr l ) = 1 .70 kJ · mol- I (b) C(s) + t 0 2 (g) � CO(g), I1n = 1 mol - t mol = t mol = -1 37. 1 7 kJ . mor l + (t) (8.3 1 4 J . K-I . mol-I ) (298 K) (In 1 .0 1 3 25)/(1 000 J . kr l ) = -1 37. 1 5 kJ ·mol- I (c) C(s) + t N 2 (g) + t H 2 (g) � HCN(g), I1n = 1 mol - (t mol + t mol) =0 I Torr x l atm x 1 .0 1 3 25 bar = 1 . 333 x 1 0-3 bar 760 Torr 1 atm I1Gf
SM-282
1 24. 7 kJ . mol- I + (0) (8.3 1 4 J . K- ' . mor ' ) ( 298 K) (In 1 .333 x l 0-3 )/(1 000 J . kr' ) = 1 24.7 kJ · mol - I (d) C(s) + 2 H 2 (g) � CH 4 (g), I1n= 1 mol - 2 mol = - 1 mol 1 Pa = 1 0-5 bar I1 Gf =
I1Gf =
-SO.72 kJ · mol-I +( - 1 ) ( 8.3 1 4 J . K- ' · mor ' ) (298 K) (In 1 0-5 )/(1 000 J . kr ' ) = -22.2 kJ · mor l
9.1 19
(a) (i) Increasing the amount of a reactant will push the equilibrium toward the products. More N02 will form. (ii) Removing a product will pull the equilibrium toward products. More N02 will form. (iii) Increasing total pressure by adding an inert gas not change the relative partial pressures. There will be no change in the amount ofN02 . (b) Since there are two moles of gas on both sides of the reaction, the volume cancels out of the equilibrium constant expression and it is possible to use moles directly for each component. At equilibrium, S03 = 0.24S moles - 0.240 moles = O.OOS moles, S02 0.240 moles and N02 0.240 moles since the reaction is 1 : 1 : 1 : 1 . The original number of moles of NO is set to x. K = 6.0 x 1 03 = (0.240)(0.240)/(0.00S)(x - 0.240) = ( 1 1 .S2)1(x - 0.240) x = ( 1 1 .S2/6.0 x 1 03 ) + 0.240 x = 0.242 moles of NO =
=
SM-283
CHAPTER
10
ACIDS AND BASES
10.1
1 0 .3
(a) CH J NH J + (b) NH2NH J + (c) H2CO) (d) COJ 2 (e) C6 HsO- (f) CH 3 C0 2For all parts (a) to (e), H2 0 and H3 0+ fonn a conjugate acid-base pair in which H20 is the base and H 3 0 is the acid. (a) H 2 S0 4 (aq) + H 2 0(l) � H 3 0+ (aq) + HS0 4 - (aq) H 2 S0 4 and HSO 4 - (aq) fonn a conjugate acid-base pair in which H 2 S04 is the acid and HSO 4 - (aq) is the base. (b) C6 H S NH 3 + (aq) + H 2 0(l) � H ) O+ (aq) + C 6 H S NH 2 (aq) C6 H S NH/ and C6 H S NH 2 (aq) fonn a conjugate acid-base pair in which C 6 HsNH/ is the acid and C6 H S NH 2 (aq) is the base. (c) H 2 P04 - (aq) + H 2 0(l) � H 3 0+ (aq) + HPO/- (aq) H 2 P0 4 - (aq) and HPO /- (aq) fonn a conjugate acid-base pair in which H 2 P0 4 - (aq) is the acid and HPO/ - (aq) is the base. (d) HCOOH(aq) + H 2 0(l) � H 3 0+ (aq) + HC0 2 - (aq) HCOOH(aq) and HCO? - (aq) fonn a conjugate acid-base pair in which HCOOH(aq) is the acid and HC0 2 - (aq) is the base. (e) NH 2 NH/ (aq) + H 2 0(l) � H 3 0+ (aq) + NH 2 NH 2 (aq) NH 2 NH 3 + (aq) and NH 2 NH 2 (aq) fonn a conjugate acid-base pair in which NH 2 NH/ (aq) is the acid and NH 2 NH 2 (aq) is the base.
SM-284
1 0 .5
(a) Bf0nsted acid: HN03 Bf0nsted base: HP0 4 2 (b) conjugate base to HN03 : N0 3 -
1 0.7
(a) HCI03 (chloric acid) ; conjugate base: CI03 -
(b) HN02 (nitrous acid); conjugate base: N02 -
1 0.9
10.11
10.13
(a) proton transferred from NH/ to H20, NH/ (acid), H 2 0 (base); (b) proton transferred from NH/ to 1, NH/ (acid), 1 (base); ( c) no proton transferred (d) proton transferred from NH/ to NH2-, NH/ (acid), NH2- (base) (a) HC03- as an acid: HC03 - (aq) + H 2 0(l) � H 3 0+ (aq) + cot (aq), HC03- (acid) and CO/- (base), H20 (base) and H 3 0+ (acid); HC03- as a base: H 2 0(l) + HC0 3- (aq) � H 2 C0 3 (aq) + OH - (aq), HC03- (base) and H 2C03 (acid), H2 0 (acid) and OH- (base) (b) HP042- as an acid: HP0 4 2 - (aq) + H 2 0(l) � H 3 0+ (aq) + P04 3- (aq), HPO/- (acid) and pol- (base), H2 0 (base) and H 3 0+ (acid); HPO/- as a base: HP0 4 2- (aq) + H 2 0(l) � H 2 P0 4 - (aq) + OH- (aq), HP042- (base) and H2P04- (acid), H20 (acid) and OH- (base) (a) Two protons can be accepted (one on each N);
SM-285
(b) o
o
or (c) Each of the two nitro- groups will show amphiprotic behavior in aqueous solution (can either accept a proton and donate a proton). o
10.15
The Lewis structures of (a) to (e) are as follows: F H -N - H I . .i B I :F: Ag+ F/ """--- F H (a) Lewis base
1 0. 1 7
(a) f. "'" /f.: P .
:F : i :F. . "", I / F: ..
:
: :p/ .F. .. 1: F ", . : Lewis acid
(c) Lewis acid
(b) Lewis acid
..i -
:F:
Lewis base
:F. . · ·
p-1
/ "'
: F. : .
. ·
Lewis acid
.. i-
:Cl:
Lewis base
· ·
product
(b) Q=S· \ :0:
F: . .
i
-
:CI: Q =S : \ :0: /
product
SM-286
(d) Lewis base
. .i -
H
(e) Lewis base
10.19
(a) basic; (b) acidic; (c) amphoteric; (d) basic
1 0.2 1
In each case, use Kw = [H 3 0+ ] [OH- ] = l .0 x 1 0-1 4 , then
4 (a) [OH- ] = l .0 x l 0-1 = 5.0 x l 0-1 3 mol · Cl 0.02 -' 4 l .O x 1 0 = = l .O x 1 0-9 mol · L-' (b) [OH- ] l .0 x 1 0-5 14 (c) [OH- ] = 1 .0 X 1 0- 3 = 3.2 X 1 0-1 2 mol · L- ' 3 . 1 x 1 0-
= �2. 1 x 1 0- 1 4 = 1 .4 x 1 0-7 mol · L- 1 pH = -log[H3 0+ ] = 6.80 (b) [OH - ] = [H 3 0+ ] = l .4 x 1 0-7 mol · L-1 x
1 0.25
Because Ba(OH) 2 is a strong base, Then nominal concentration of Ba(OH) 2 . 0.25 g 3 = 1 .5 x 1 0 - mol moles 0 f Ba(OH) 2 = I 1 7 1 .36 g · mor 3 [Ba(OH) 2 ] = 1 .5 x 1 0- mol = 1 .5 x 1 0-2 mo! . L- 1 = [Ba 2 + ] 0. 1 00 L [OH- ] = 2 x [Ba(OH) 2 ]0 = 2 x 1 .5 x 10-2 mol · L- ' = 2.9 x 1 0-2 mol · L-' 4 1 K 1 .0 x 1 0w _ = 3.4 x 1 0- 1 3 mol . L- l [H O+ ] 2 [OH - ] 2.9 X 1 00
3
-
SM-287
1 0.27
Because pH = -log[H 3 0+ ], log[H 3 0+ ] = -pH. Taking the antilogs of both (a) [H 3 0+ ] = 1 0-3 3 = 5 X 1 0-4 mol · L- ' (b) [H 3 0+ ] = 1 0-6 7 mol · L-' = 2 x l 0-7 mol · L- ' (c) [H 3 0+ ] = 10-44 mol · L- ' = 4 x l 0-5 mol · L- ' (d) [H 3 0+ ] = 1 0-5 3 mol · L-' = 5 x l 0-6 mol · L- '
1 0. 2 9
(a) pH = 1 3 .25, [Ht = 1 0 - 1 3 .2 5 = 5.6 x 1 0 - 1 4 mol · L - I, [OH - ]= 0. 1 8 mol · L - 1 (b) [OH - ] = 0. 1 8 x 50 0. 0mL = 1 8 mol · L - 1 5.00mL (c) The reaction is Na2 0(s) + H2 0(1) 7 2 NaOH(aq) The mass ofNa2 0 I S 1 8 mol · L - 1 x (0.200 L) x 1 mol Na 2 0 2 mol NaOH
(
J
(
•
[
J
x 6 l .98 g Na 2 0 - I I 0 grams Na2 0 . I mol Na 2 0 1 0.3 1
(a) [HN0 3 ] = [H 3 0+ ] = 0.0146 mol · L- ' pH = -log(0.0146) = l .84, pOH 14.00 - (-l .84) = 12. 1 6 (b) [H e l] = [H 3 0+ ] = 0. 1 1 mol · L- ' pH = -log(0. 1 1) = 0.96, pOH = 14.00 - 0.96 = 1 3.04 (c) [OH- ] = 2 x [Ba(OH)J = 2 x 0.0092 M = 0.01 8 mol · C l pOH = -log(0.0 1 8) = 1 .74, pH = 14.00 - l .74 = 12.26 =
(
J
[OH- ] = 2.00 mL x (0. l 75 mol · L- ' ) = 7 .0 X 1 0-4 mol · L- ' 500 mL
SM-288
J
pOH = -log(7.0 x 1 0-4 ) = 3. 1 5, pH = 14.00 - 3 . 1 5 = 1 0.85 (e) [NaOH]o = [OH - ] number of moles of NaOH = 0.0 1 36 g = 3.40 X 1 0-4 mol 40.00 g · mor 4 [NaOH]o = 3.40 x 1 0- mol = 9.71 x l 0-4 mol · L- 1 = [OH - ] 0.350 L pOH = -log(9.7 1 x 1 0-4 ) = 3.01, pH = 14.00 - 3.01 = 1 0.99 (f) [HBr] o = [H 3 0+ ] I
[
]
[H 3 0+ ] = 75.0 mL X (3.5 X I 0-4 mol · L-1 ) = 5.2 x l 0-5 mol · C I 500 mL pH = -log(5.3 x 1 0-5 ) = 4.28, pOH = 14.00 - 4.28 = 9.72 1 0 .33
= -log Ka l ; therefore, after taking antilogs, Ka l = 1 0- pK"I. Ka pKa Acid l 2. 1 2 7.6 x 1 0-3 (a) H 3 P0 4 2.00 1 .0 x 1 0-2 (b) H 3 P0 3 3.5 x 1 0-3 (c) H 2 Se0 3 2.46 1 .2 x 1 0-2 1 .92 (d) HSe0 4 (e) The larger Ka l , the stronger the acid; therefore H 2 Se03 < H 3 P04 < H 3 P0 3 < HSe0 4 - · pKa l
I
1 0 .35
(a) HCI0 2 (aq) + H 2 0(I) � H 3 0+ (aq) + CI0 2 - (aq) [H 3 0+ ][CI0 2 - ] Ka = [HCI0 2 ] CI0 2 - (aq) + H 2 0(l) � HCI0 2 (aq) + OH- (aq)
SM-289
(b) HCN(aq) + H 2 0(l) � H 3 0+ (aq) + CN- (aq) K
a
=
-] 0_+'--'. ] [_ C.. N__ -[H =-----=:3_ [HCN]
CN- (aq) + H 2 0(l) � HCN(aq) + OH - (aq) Kb [HCN][OH- ] [CN- ] (c) C6 HsOH(aq) + H 2 0(l) � H 3 0+ (aq) + C 6 HsO- (aq) =
Ka [H 3 0+ ][C 6 HsO- ] [C6 HsOH] C6 HsO- (aq) + H 2 0(l) � C6 HsOH(aq) + OH- (aq) Kb = [C6 HsOH][OH-] [C6 HsO- ] ----' --=: '--' '--'----' � -''= -=---
1 0.37
Decreasing pKa will correspond to increasing acid strength because pKa = -log Ka . The pKa values (given in parentheses) determine the following ordering: (CH 3 ) 2 NH/ (14.00 - 3.27 1 0.73) < +NH 3 0H (14.00 - 7 97 = 6.03) < HN0 2 (3.37) < HCI0 2 (2.00). Remember that the pKa for the conjugate acid of a weak base will be given by pKa + pKb = 14. =
1 0.39
.
Decreasing pKb will correspond to increasing base strength because pKb = -log Kb . The pKb values (given in parentheses) determine the following ordering: F- (14.00 - 3.45 = 1 0.55) < CH 3 COO- (14.00 - 4.75 9.25) < CsHsN (8.75) « NH 3 (4.75). Remember that the pKb for the conjugate base of a weak acid will be given by pKa + pKb = 14. =
SM-290
1 0. 4 1
Any acid whose conjugate base lies above water in Table 1 0.3 will be a strong acid; that is, the conjugate base of the acid will be a weaker base than water, and so water will accept the H+ preferentially. Based on this information, we obtain the following analysis: (a) HCI03 , strong; (b) H 2 S, weak; (c) HS04 - , weak (Note: Even though H 2 S0 4 is a strong acid, HS0 4 - is a weak acid. Its conjugate base is SO4 2 - . ); (d) CH3 NH 3 +, weak acid; (e) HC0 3 -, weak; (f) HN0 3 , strong; (g) CH 4 , weak.
1 0.43
For oxoacids, the greater the number of highly electronegative ° atoms attached to the central atom, the stronger the acid. This effect is related to the increased oxidation number of the central atom as the number of ° atoms increases. Therefore, HI03 is the stronger acid, with the lower pKa •
1 0.45
(a) HCI is the stronger acid, because its bond strength is much weaker than the bond in HF, and bond strength is the dominant factor in determining the strength of binary acids. (b) HCI0 2 is stronger; there is one more ° atom attached to the CI atom in HCI0 2 than in HCIO. The additional ° in HCI0 2 helps to pull the electron of the H atom out of the H-O bond. The oxidation state of CI is higher in HCI0 2 than in HCIO. (c) HCI0 2 is stronger; CI has a greater electronegativity than Br, making the H-O bond HCI02 more polar than in HBr0 2 • (d) HCI04 is stronger; CI has a greater electronegativity than P. (e) HN03 is stronger. The explanation is the same as that for part (b). HNO) has one more ° atom. SM-291
(f) H 2 C03 is stronger; C has greater electronegativity than Ge. See part (c). 1 0 .47
(a) The -CCI3 group that is bonded to the carboxyl group, -COOH, in trichloroacetic acid, is more electron withdrawing than the - CH 3 group in acetic acid. Thus, trichloroacetic acid is the stronger acid. (b) The - CH 3 group in acetic acid has electron-donating properties, which means that it is less electron withdrawing than the -H attached to the carboxyl group in formic acid, HCOOH. Thus, formic acid is a slightly stronger acid than acetic acid. However, it is not nearly as strong as trichloroacetic acid. The order is CCl3 COOH » HCOOH > CH 3 COOH.
1 0 .49
The larger the Ka , the stronger the corresponding acid. 2,4,6Trichlorophenol is the stronger acid because the chlorine atoms have a greater electron-withdrawing power than the hydrogen atoms present in the unsubstituted phenol.
1 0 .5 1
The larger the pKa of an acid, the stronger the corresponding conjugate base; hence, the order is aniline < ammonia < methylamine < ethylamine. Although we should not draw conclusions from such a small data set, we might suggest the possibility that ( 1 ) arylamines < ammonia < alkylamines (2) methyl < ethyl < etc. (Arylamines are amines in which the nitrogen of the amine is attached to a benzene ring.)
1 0.53
(a) Concentration (mol · L- 1 ) CH}COOH initial 0.29
+
H2 0 o SM -292
0
change -x equilibrium 0.29 -x
+x x
+x x
x = [H ) O + ] = 2.3 X 1 0-) mol · L- 1 pH = -log(1 .6 x 1 0-) ) = 2.64, pOH = 14.00 - 2.64 = 1 1 .36 (b) The equilibrium table for (b) is similar to that for (a). [ H)O+ ][CCI ) C0 2 -] = x2 K = 3.0 X 1 0- 1 = [CCI) COOH] 0.29 - x 2 or x + 3.0 X 1 0-1 X - 0.087 = 0 1 ± )(3.0 x l 0- 1 ) 2 - (4) (-0.087) -3.0 x l 0= 0. 1 8, -0.48 x= 2 The negative root is not possible and can be eliminated. x = [H ) O+ ] = 0. 1 8 mol · L- 1 pH = -log(0. 1 8) = 0.74, pOH = 14.00 - 0.74 = 1 3 .26 (c) Concentration (mol · Cl ) HCOOH + H 2 0 � H 3 0+ + HC0 2 o 0.29 o initial -x change +x +x x x 0.29 - x equilibrium a
x = [H 3 0+ ] = )0.29 x l .8 x l 0-4 = 7.2 x l 0-3 mol · L- 1 pH = -log(7.2 x 1 0-3 ) = 2. 14, pOH = 14.00 - 2. 14 = 1 l .86 (d) Acidity increases when the hydrogen atoms in the methyl group of acetic acid are replaced by atoms that have a higher electronegativity, such as chloline.
1 0.55
(a) Concentration (mol · L-1 ) H 2 0 + NH) SM -293
NH + 4
+
OH-
o o 0.057 initial +x +x -x change x x 0.057 - x equilibrium [NH / ][OH- ] = x · x � = 1 .8 X 1 0-5 Kb = 0.057 - x 0.057 [NH 3 ] x = [OH - ] = �0.057 x 1 .8 x l 0-5 = 1 .0 x l 0-3 mol · L- 1 pOH = -log(1 .0 x 1 0-3 ) = 3.00, pH = 14.00 - 3.00 = 1 1 .00 . = 1 .0 x 1 0-3 x 1 00% = 1 .8% percentage protonatlOn 0.057 (b) Concentration (mol · Cl ) NH 2 0H + H 2 0 � +NH 3 0H + OHinitial 0. 162 0 0 change -x +x +x equilibrium 0. 1 62 - x x X 2 X x2 Kb = 1 . 1 X 1 0- 8 = 0.1 62 - x 0. 1 62 x = [OH- ] = 4.2 x 1 0-5 mol · L- 1 pOH = -log(4.2 x 1 0-5 ) = 4.38, pH = 1 4.00 - 4.38 = 9.62 . = 4.2 x 1 0-5 x 1 00% = 0.026% percentage protonatlOn 0. 1 62 (c) Concentration (mol · C l ) (CH 3 ) 3 N + H 2 O � (CH 3 ) 3 NH+ + initial 0.35 0 change -x +x equilibrium 0.35 - x +x �
� --
...--
SM-294
OH0 +x +x
x2 0.35 - x Assume x « 0.35 Then x = 4.8 x 1 0-3 mol · L- 1 [OH- ] = 4.8 X 1 0-3 mol · L- 1 pOH = -loge 4.8 x 1 0-3 ) = 2.32, pH = 14.00 - 2.32 = 1 1 .68 . = 4.8 x 1 0-3 x 1 00% = 1 .4% percentage protonatlOn 0.35 (d) pKb = 14.00 - pKa = 14.00 - 8.21 = 5.79, Kb = 1 .6 X 1 0-6 6.5 X 1 0-5 =
---
2 x2 ::::; 0.073 - x 0.073 x = [OH- ] = 1 . 1 x l O-4 mo1 · L- 1 pOH = -log(1 . 1 x 1 0-4 ) = 3.96, pH = 14.00 - 3.96 = 1 0.04 . = 1 . 1 x 1 0-4 x 1 00% = 2.5/0 01 percentage protonatlOn 0.0073 Kb = 1 .6 X 1 0-6 =
10.57
X
--
(a) HCI0 2 + H 2 0 � H 3 0+ + CI0 2 [H 3 0+ ] = [CI0 2 -] = 1 0-pH = 1 0-12 = 0.06 mol · L-1 2 K = [H 3 0+ ] [C I02 -] = (0. 06) = 0.0 9 (1 sf) 0. 1 0 - 0.06 [HCI0 2 ] pKa = -log(0.09) = 1 .0 (b) C3 H 7 NH 2 + H 2 0 � C 3 H 7 NH 3 + + OH pOH = 1 4.00 - 1 1 .86 = 2.14 [C3 H 7 NH 3 +] = [OH- ] = 1 0-2 1 4 = 7.2 X 1 0-3 mol · L- 1 32 +][OHNH C H (7.2 ] [ 3 X 1 0- ) 3 7 = = 5.6 X 1 0-4 Kb = [C3 H 7 NH 2 ] 0. 1 0 - 7.2 x 1 0-3 pKb = -log(5.6 x 1 0-4 ) = 3.25 a
Let x = nominal concentration of HCIO, then SM -295
Concentration (mol · C ' ) HCIO + nominal x o equilibrium x - 2.5 x 1 0-5 2.5 x 1 0-5 (2.5 X 1 0-5 ) 2 K = 3.0 X 1 0-8 = x - 2.5 x 1 0-5 8 52 S o Ive fior x,. x - (2.5 X 1 0- ) + (2.5 X 1 0-8 5 ) (3.0 X 1 0- ) 3.0 x 1 02 = 2. 1 x 1 0- mol · L- ' = 0.02 1 mol · L- ' (b) pOH = 14.00 - pH = 14.00 - 1 0.20 = 3.80 [OH- ] = l O- poH = 1 0-3 .80 = l .6 X 1 0-4 Let x = nominal concentration of NH 2 NH 2 , then Concentration (mol · L- ' ) nominal x o equilibrium x - 1 .6 x 1 0-4 l .6 X 1 0-4 6 (l .6 x 1 0-4 ) 2 K = 1 .7 x 1 0- = x - 1 .6 x 1 0-4 Solve for x; x = 1 .5 x 1 0-2 mol · L- '
+
2.5 x 1 0-5
a
_
o l .6 X 1 0-4
b
1 0.61
Concentration (mol · L- ' ) C 6 H 5 COOH + H 2 O H 3 0+ initial 0. 1 1 0 0 -x change +x equilibrium 0. 1 1 0 - x x H 2 0 + octylamine � octylamineH+ + OH ---'. ...-
SM-296
+
C 6 H 5 C0 2 0 +x X
x = 0.024 x 0. 1 1 0 mol · L- 1 = [H 3 0+ ] = [C6 HsC0 2 - ] a [H 3 0+ ] [C 6 HsC OO- ] = (0.02 4 x 0. 1 1 0) 2 = 6.5 1 0-s K = x (1 - 0.024) x 0. 1 10 [C 6 HsCOOH] pH = -log(2.6 x l 0-3 ) = 2.58 1 0 .63
The change in the concentration of octylamine is x = 0.067 x 0.1 0 = 0.0067 mol · L-1 • Thus the equilibrium table is (mol · L- 1 ) H2O + octylamine � octylamineH+ + OHinitial 0. 1 00 a a change -x +x +x equilibrium 0. 1 00-x x x x = 0.067 x 0. 1 00 mol · L- 1 = [OH-] = [octylamine H+] b [OH -][octylamine H+] = (0.06 7 x 0. 1 00) 2 = 4.8 X l O-s K = (1 - 0.067) x 0. 1 00 [octylamine] pH = 1 4 - [ -log(6.7 x 1 O-3 ) J = 1 l .83
1 0.65
K
[N0 2 - ][NH / ] Ka(HN02 ) Kb(NH3 ) (4.3 x 1 O-4)(l .8 x 1 0-5) [HNOJ[NH 3 ] 7.7 x 1 0-9 This means that the reaction is in favor of the left. =
=
x
=
=
1 0 .67
(a) less than 7, NH/ (aq) + H 2 0(l) � H 3 0+ (aq) + NH 3 (aq) (b) greater than 7, H 2 0(l) + C03 2 - (aq) � HC03 - (aq) + OH- (aq) (c) greater than 7, H 2 0(I) + F- (aq) � HF(aq) + OH- (aq) (d) neutral (e) less than 7, AI(H 2 0) 63+ (aq) + H 2 0(l) � H 3 0+ (aq) + AI(H 2 0)s OH 2+ (aq) (f) less than 7, Cu(H 2 0)/+ (aq) + H 2 0(l) � H } O+ (aq) + Cu(H 2 0)s OH+ (aq) SM -297
1 0.69
(a)
K
b
=
Kw Ka
Concentration
-1 4 = 1 .00 X 1 0 5 5.6 X 1 0-1 0 1 .8 X 1 0=
a
initial 0.63 -x change equilibrium 0.63 - x
+x x
x
a
+x x
= 1 .9 X 1 0-5 = [OH- ], pOH = -log(1 .9 x 1 0-5 ) 4.72 pH = 14.00 pOH = 14.00 - 4.72 = 9.28 4 (b) K = Kw = 1 .00 X 1 0-1 = 5.6 X 1 0-1 0 a Kb 1 .8 X 1 0-15 =
-
Concentration (mol · C l ) NH/ (aq) + H 2 0(l) � H 3 0+ (aq) + NH 3 (aq) initial 0.1 9 0 o change +x +x -x equilibrium 0. 1 9 - x x x 2 2_ [H 0+ ][NH x ] 3 3 0 _X _ a 1 K = 5.6 X 1 0- = [NH 4 CI] 0.1 9 - x 0.1 9 x = 1 .0 X 1 0-5 mol · C l [H 3 0+ ] pH = -log(1 .0 x 1 0-5 ) = 5.00 (c) Concentration (mol · L- 1 ) AI(H 2 0)t (aq) + H 2 0(l) � H 3 0+ (aq) + AI(H 2 0) 5 OH 2 + (aq) initial 0.055 o a change -x +x +x equilibrium 0.055 - x x x 2+ ] 2 x2 OH 0+ ][AI(H?O) [H 3 Ka = = I A x l 0 -5 = x - 5 0.055 - x 0.055 [AI(H 2 0)t ] x = 8.8 X 10 4 mol · L- 1 = [H 3 0+ ] =
�
=
� --
-
SM-298
pH = -log(8.8 x 1 0-4 ) 3.06 (d) Concentration (mol · L- 1 ) H 2 0(l) + CN- (aq) � HCN(aq) + OH- (aq) 0 0.065 0 initial +x -x +x change equilibrium 0.065 - x x x 2 2 K 1 .00 x l 0- 1 4 2.0 x l 0-5 = [HCN] [OH- ] x x w = K = Ka 0.065 - x 0.065 b 4.9 x 1 0-1 0 [CN- ] x [OH- ] 1 . 1 X 1 0-3 mol · L- 1 pOH - log(I . 1 x 1 O-} ) 2.96, pH 1 1 .04 =
=
=
�
:r=
=
1 0. 7 1
=
=
Concentration (mol . L- 1 ) CH } NH } + (aq) 0.5 1 0 initial -x change equilibrium 0.5 10 - x
=
+
H 2 °(l)
---'>. ...---
H } O+ (aq) 0
+
CH 3 NH 2 (aq) 0
+x
+x
x
x
0.5 1 0
pH = -log(3.8 x 1 0-6 ) = 5.42 1 0.73
pH of the NaA is 1 0.35, [H+] =1 0- 1 0 .35 , [OH-] = 1 0-3 .65 2.2 x 1 0-4 Hydrolysis reaction: A- (aq) + H2 0(l) � HA(aq) + OH- (aq) [HA ][OH - ] = (2.2 x 1 0-4 ) 2 = 4.84 x 1 0-6 Kb = (0.01 0) [A- ] 9 Ka = KwlKb = 2.1 x 1 0- . The fOTInula of the acid is HBrO. =
SM-299
1 0.75
(a) 0.020 mol · Cl NaCH) C02 x 0.1 50 L = 0.00600 mol . L- 1 0.500 L Concentration o o 0.00600 initial +x +x -x change x x 0.00600 - x equilibrium H- ] 14 Kb = Kw = 1 .00 X 1 0--5 = 5.6 X 1 0- 1 0 = [CH)COOH][O Ka 1 .8 X 1 0 [CH) C02 - ] 2 2 x X 0 1 --= 5.6 x l O0.00600 - x 0.00600 x = 1 .8 x l 0-6 mol · L- 1 = [CH ) COOH] ;::;
J( J (
(
J
1 mol NH 4 Br (b) 2.1 6 g NH 4 Br 1 mL ) 400 mL 1 0- L 97.95 g NH 4 Br = 0.055 1 (mol NH Br) · L-1 4 Concentration (mol · L- 1 ) o 0.0551 initial -x +x change x 0.055 1 - x equilibrium 14 Ka = Kw = 1 .00 X 1 0-5 = 5.6 X 1 0- 1 0 = [NH ) ][H ) O+ ] [NH 4 +] Kb 1 .8 x 1 02 5.6 X 1 0- 1 0 = x 0.0551 - x 0.0551 ;::;
1 0.77
pH = + (pKa + pKa I
2
) = -21 (2.340 + 9.886) = 6. 1 1 3
SM-300
o
+x x
1 0.79
(a) H 2 S04 ( aq) + H 2 0(l) � H 3 0+(aq) + HS04- (aq ) HSO 4- ( aq) + H 2 0(l) � H 3 0+ (aq) + SO /- (aq) (b) H 3 As04 (aq) + H 2 0(l) � H 3 0+ (aq) + H 2 As04- ( aq ) H 2 As04- ( aq) + H 2 0 ( l) � H 3 0+ (aq) + HAsO/- (aq) HAsO/- (aq) + H 2 0(l) � H 3 0+ (aq) + AsO/- ( aq) ( c) C6 H4 (COOH) 2 ( aq) + H 2 0( I) � H 3 0+ (aq) + C6 H4 (COOH)C0 2 - (aq) C6 H4 (COOH)C0 2 - (aq) + H 2 0 (I) � H 3 0+ ( aq) + C 6 H4 ( C0 2 ) 2 2- ( aq)
10.81
The initial concentrations of HS04- and H 3 0+ are both 0. 1 5 mol · L-' as a result of the complete ionization of H 2 S04 in the first step. The second ionization is incomplete. Concentration HS04 - + H 2 O ---'. ( mol · L-' ) ..,-- H 3 0+ + S04 2initial 0. 1 5 0. 1 5 0 change -x +x +x equilibrium 0. 1 5 - x 0. 1 5 + x x 2 -] (0. 1 5 + x) ( x) 4 0+ [H ][S0 2 3 = Ka2 = 1 .2 X 1 0- = 0.1 5 - x [HS04 -] x2 + 0.1 62x - 1 .8 x 1 0-3 = 0 -0. 1 62 + )(0. 1 62) 2 + (4) (1 .8 x 1 0-3 ) x= = 0.01 04 mol · L 2 [ H 3 0+ ] = 0. 1 5 + x = (0. 1 5 + 0.01 04) mol . L-' = 0. 1 6 mol · L-' pH = - log(0. 1 6) = 0.80 _I
1 0.83
( a) Because Ka2 « Ka l ' the second ionization can be ignored. Concentration H 2 C0 3 + H 2 O ---'. (mol · L-' ) ..,-- H 3 0+ + HCO3 initial 0.0 1 0 0 0 change -x +x +x SM-301
x x equilibrium 0.010 - x 2 ] � = 4.3 X 1 0-7 Ka l = [H 3 0+ ][HC03- = X [H 2 C03 ] 0.01 0 - x 0.0 1 0 x = [H 3 0+ ] = 6.6 X 1 0-5 mol · L- 1 pH = - log(6.6 x 1 0-5 ) = 4.1 8 (b) Because Ka2 « Kai ' the second ionization can be ignored. Concentration H 3 0+ + (COOH)C0 2 (mol · L- 1 ) (COOH) 2 + H 2 O 0.1 0 initial 0 0 -x change +x +x x equilibrium 0. 1 0 - x x 2 Ka l = S . 9 X 1 0-2 = [H 3 0+ ][(COOH)C0 2 -] = x 0. 1 0 - x [(COOH) 2 ] 3 x2 + S.9 x 1 0-2 x - S.9 x 1 0- = 0 -S.9 x 1 0-2 + �(S.9 x 10-2 ) 2 + (4)(S.9 x 1 0-3 ) x= = 0.OS3 mol · L 2 pH = - log(0.OS3) = 1 .28 ( c) Because Ka2 « Kai ' the second ionization can be ignored. �
--'..-----
_I
+
equilibrium
x 0.20 - x 2 _X_2_ Ka l = 1 · 3 X 1 0-7 = [H 3 0+ ] [HS- ] = x 0.20 - x 0.20 [H 2 S] x = [H 3 0+ ] = 1 .6 X 1 0-4 mol · L- 1 pH = - log(1 .6 x 1 0-4 ) = 3.80
x
�
1 0 .85
( a) The pH is given by pH = + (pKa l + pKa2 ). From Table 1 0.9, we find Ka l = 1 .S x 1 0-2 pKa l = 1 .82 Ka2 = 1 .2 x 1 0-7 pKa2 = 6.92 pH = + (1 .82 + 6.92) = 4.37
SM-302
(b) The pH of a salt solution of a polyprotic acid is independent of the concentration of the salt; therefore, pH = 4.37. 1 0.87
(a) The pH is given by pH 1(pKa ' + pKa2 ). For the monosodium salt, the pertinent values are pKa ' and pKa2 : pH = 1(3 . 1 4 + 5.95) = 4.55 (b) For the disodium salt, the pertinent values are pKa2 and pKa) : pH = 1(5.95 + 6.39) = 6. 1 7
1 0.89
The equilibrium reactions of interest are H 2 CO)(aq) + H 2 0(l) � H ) O+(aq) + HCO ) -(aq) Ka ' = 4.3 X 1 0-7 HCO)- (aq) + H 2 0(l) � H ) O+ (aq) + CO) 2 -(aq) Ka2 = 5.6 X 1 0-11• Because the second ionization constant is much smaller than the first, we can assume that the first step dominates:
=
Concentration initial 0.0456 o change -x +x final 0.0456 - x +x Ka l = [H ) O+] [HCO) - ] [H 2 CO) ] X2__ 4.3 x 1 0-7 = (x) (x) = ___ 0.0456 - x 0.0456 - x Assume that x « 0.0456. Then x2 = (4.3 x 1 0-7 ) (0.0456). x = 1 .4 X 1 0-4 Because x < 1% of 0.0456, the assumption was valid. x = [H}O+ ] = [HCO)- ] = 1 .4 x 1 0-4 mol · L- '
SM-303
o
+x +x
This means that the concentration of H2C03 is 0.0455 mol · L-I - 0.00014 mol · L- I = 0.0455 mol · C l . We can then use the other equilibria to determine the remaining concentrations: [H 3 O+ ][CO32 - ] Ka 2 = [HC03 -] 4 32- ] 5.6 x l 0- 1 1 = (l .4 x l0- ) [C0 (1 .4 x l 0-4 ) [C03 2- ] = 5.6 x 1 0-1 1 mol · L- I Because 5.6 x 1 0-1 1 « 1 .4 x 1 0-4 , the initial assumption that the first ionization would dominate is valid. To calculate [OH - ], we use the Kw relationship: = [H 3 0+ ] [OH- ] 1 .00 x 10- 1 4 = 7. 1 x l 0- 1 1 mol . L- 1 [ OH - ] - Kw [H 3 0+ ] 1 .4 x l O-4 In summary, [H 2 C03 ] = 0.0455 mol · L-I , [H 3 0+ ] = [HC0 3 - ] = 1 .4 x 1 0-4 mol · L-I , [C0 3 2 - ] = 5.6 x 1 0- 1 1 mol · L-I , [OH- ] = 7.1 X 1 0-1 1 mol · L-I • The equilibrium reactions of interest are now the base forms of the carbonic acid equilibria, so Kb values should be calculated for the following changes: C03 2-(aq) + H 2 0(l) � HC03- (aq) + OH - (aq) K 1 .00 X 1 0- 1 4 = 1 .8 X 1 0-4 w = K = b l Ka2 5.6 x 1 0- 1 1 K
w
-
1 0. 9 1
HC03- (aq) + H 2 0(l) � H 2 C03 (aq) + OH-(aq) K 1 .00 x l 0- 1 4 = 2.3 x 1 0-8 w = K 2 = b Ka l 4.3 x 1 0-7 Because the second hydrolysis constant is much smaller than the first, we can assume that the first step dominates:
SM-304
Concentration (mol · L-I ) cot (aq) + H 2 °(l) HCO) - (aq) + OH- (aq ) initial 0 0 0.0456 -x change +x +x final 0.0456 - x +x +x -] Kb l = [HCO) ][OH 2 [CO) -] X_2 1 .8 X 1 0-4 = (x) (x) 0.0456 - x [0.0456 - x] Assume that x « 0.0456. Then x2 (1 .8 x 1 0-4 ) (0.0456). x 2.9 X 1 0-) Because x > 5% of 0.0456, the assumption was not valid and the full expression should be solved using the quadratic equation: x2 + 1 .8 X 1 0-4 x - (1 .8 X 1 0-4 ) (0.0456) 0 Solving using the quadratic equation gives x = 0.0028 mol · L-I . x = [HCO) -] = [OH- ] 0.0028 mol · L-I Therefore, [CO) 2-] = 0.0456 mol · L- I - 0.0028 mol · L-I = 0.0428 mol · L- I . We can then use the other equilibria to determine the remaining concentrations: Kb 2 [H 2 CO) ] [OH- ] [HCO) -] 2.3 x 10-8 = [H2C03 ] (0.0028) (0.0028) ----'. �
=
_ _
_ _
=
=
=
=
=
----' =--=---
Because 2.3 x 1 0-8 « 0.0028 , the initial assumption that the first hydrolysis would dominate is valid. To calculate [H,O+ ] , we use the Kw relationship: SM-305
Kw = 1 .00 x 1 0-1 4 = 3.6 x l 0-1 2 mol . L- 1 [OH- ] 0.0028 In summary, [H 2 C03 ] 2.3 X 1 0-8 mol · C l , [OH - ] = [HC03 -] = 0.0028 mol · L- 1 , [C0 3 2-] = 0.0428 mol · L- 1 , [H 3 0+ ] 3.6 x l 0-1 2 mol · C I . [ H 3 0+ ] =
=
=
1 0.93
(a) Phosphorous acid: The two pKa values are 2.00 and 6.59. Because pH = 6.30 lies between pKa l and pKa2 ' the dominant form will be the singly deprotonated HA - ion. (b) Oxalic acid: The two pKa values are 1 .23 and 4. 1 9. Because pH = 6.30 lies above pKa 2 ' the species present in largest concentration will be the doubly deprotonated A 2 - ion. (c) Hydrosulfuric acid: The two pKa values are 6.89 and 1 4. 1 5. Because pH = 6.30 lies below both pKa values, the species present in highest concentrations will be the fully protonated H 2 A form.
1 0.95
The equilibria present in solution are H 2 S03 (aq) + H 2 0(lh =� H 3 0+ (aq) + HS03 -(aq) Ka l = 1 .5 x 1 0-2 HSO] -(aq) + H 2 0(l) � H 3 0+ (aq) + S03 2-(aq) Ka2 = 1 .2 x l 0-7 • The calculation of the desired concentrations follows exactly after the method derived in Equation 25, substituting H 2 S03 for H 2 C0 3 , HS03- for HC03- , and S03 2- for CO] 2- . First, calculate the quantity f (at pH 5.50 [H 3 0+ ] = 1 0-5 .5 = 3.2 X 1 0-6 mol · L- 1 ) : f = [H 3 0+ ] 2 + [H 3 0+ ] Ka l + Ka l Ka 2 = (3.2 X 10-6 ) 2 + (3.2 x 1 0-6 ) (1 .5 x 1 0-2 ) + (1 .5 x 1 0-2 ) (1 .2 x 1 0-7 ) = 5.0 x 1 0-8 The fractions of the species present are then given by =
SM-306
a (H 2 SO } ) = [ H}O + ] = (3.2 x 1 O-6 f 2.1 x 1 0-4 5.0 x l 0-8 f 2) ]K + O0.5 1 1 (3.2 (l x x a ) [H}O 6 l = a (HSO} -) 0.96 5.0 x l O-8 f 2 a ( SO} 2 - ) Ka l Ka2 = ( 1 .5 x 1 O- ) ( l .28 x 1 0-7 ) = 0.036 5.0 x l Of =
=
=
=
Thus, in a 0. 1 50 mol · L- 1 solution at pH 5.50, the dominant species will be HSO} - with a concentration of (0. 1 50 mol · L- 1 ) (0.96) = 0. 1 4 mol · L- 1 • The concentration of H 2 SO} will be (2. 1 x 1 0-4 ) (0. 1 50 mol · L- 1 ) 3.2 x l 0-5 mol · L- 1 and the concentration of SO} 2- will be (0.036) (0. 1 50 mol · L-1 ) 0.0054 mol · C l • =
=
1 0 .97
(a) Concentration (mol . L-1 ) B(OH)}
+
2 H2O
----'�
H } 0+
+
B (OH) 4 -
0 0 initial l .0 x 1 0-4 -x change +x +x X X equilibrium l .0 x l 0-4 - X 2 )4 + ][B(OH ] x [H}O 0 = = Ka 7.2 X 1 0- 1 1 .0 x 1 0-4 - x l .0 X 1 0-4 [B ( OH)} ] X [ H}O+ ] = 2.7 X 1 0-7 mol · L-1 pH = -log(2.7 x 1 0-7 ) = 6.57 Note: This value of [ H}O+ ] is not much different from the value for pure water, 1 .0 x 1 0-7 mol · L- 1 ; therefore, it is at the lower limit of safely ignoring the contribution to [H} 0+ ] from the autoprotolysis of water. The exercise should be solved by simultaneously considering both equilibria. =
=
SM-307
� ---
Concentration B(OH)3 + 2 H 2 O (mol · L- 1 ) equilibrium 1 .0 x 1 0-4 - X
---"---
---"---
H 3 0+ X
+
B(OH) 4 Y
Concentration (mol · L-1 ) 2 H 2 O H 3 0+ + OHx equilibrium z Because there are now two contributions to [H 3 0+ ], [H 3 0+ ] is no longer equal to [B(OH) 4 - ], nor is it equal to [OH- ], as in pure water. To avoid a cubic equation, x will again be ignored relative to 1 .0 x l 0-4 mol · L- 1 • This approximation is justified by the approximate calculation above, and because Ka is very small relative to 1 .0 x 1 0-4 . Let a = initial concentration of B(OH)3 ' then Ka = 7 2 X 1 0-1 0 = � .xy or y = aKa x a-x a 4 Kw = 1 .0 x l 0- 1 = xz. Electroneutrality requires x = y + z or z = x - y; hence, Kw = xz = x(x - y). Substituting for y from above: aKa = Kw xx x-� ;::::
•
(
)
x- - aKa = Kw x2 = Kw + aKa x = �Kw + aKa = �1 . 0 x 1 0- 1 4 + 1.0 X 1 0-4 x 7 . 2 X 1 0- 1 0 x = 2.9 X 1 0-7 mol · L- 1 = [H 3 0+ ] pH = -log(2.9 x 1 0-7 ) = 6.54 This value is slightly, but measurably, different from the value 6.57 obtained by ignoring the contribution to [H 3 0+ ] from water. (b) In this case, the second ionization can safely be ignored; Ka 2 « Ka l · ?
SM-308
Concentration o o 0.0 1 5 initial +x +x change -x x x equilibrium 0.0 1 5 - x 2 Ka l = 7.6 x 1 0-3 = X 0.0 1 5 - x 3 2 x + 7.6 X 1 0- x - 1 . 14 X 1 0-4 = 0 -7.6 X 1 0-3 + �(7.6 X 1 0-3 ) 2 + 4.56 X 1 0-4 x = [H 3 0+ ] = 2 3 = 7.5 X 1 0- mol · L- 1 pH = -log(7.5 x 1 0-3 ) = 2. 1 2 (c) In this case, the second ionization can safely be ignored; Ka2 « Ka 1 . Concentration H 3 0+ + HSO3 (mol . L- 1 ) H 2 S03 + H 2 O 0 0 initial 0. 1 -x change +x +x x x equilibrium 0. 1 - x 2 Ka l = 1 .5 x 1 0-2 = X O. l O - x x2 + 1 .5 X 1 0-2 x - 1 .5 X 1 0-3 = 0 3 22 2 -1 .5 X 1 0- + �(1 .5 X 1 0- ) + 6.0 X 1 0= 0.032 mol · L- 1 X = [H 3 0 + ] = 2 pH = -log(0.032) = 1 .49 ----" ---
1 0.99
The three equilibria involved are
SM-309
= 6.2 x 1 0-8 = [H 3 0+ ] [HPO�- ] [H 2 PO�] 3 [H 3 0 + ][PO!- ] K = 2. 1 X 1 0- 1 = � [HPO� ] We also know that the combined concentration of all the phosphate specIes IS [ H3 PO4 ] + [H 2 PO� ] + [HPO�- ] + [PO!-] = 1 .5 x 1 0-2 and the hydronium ion concentration is [H] = 1 0-pH = 1 0-2 25 = 5.62 x l 0-3 mol · L-I • At this point it is a matter of solving this set of simultaneous equations to obtain the concentrations of the phosphate containing species. We start by dividing both sides of the equilibrium constant expressions above by the given hydronium ion concentration to obtain three ratios: Ka,
Through rearrangement and substitution of these three ratios, we can obtain the following expressions: [H 2 PO� ] = 1 .35 · [H 3 PO4 ] ' [HPO�-] = 1 . 1 0 x 1 0-5 . [H 2 PO� ] = 1 . 1 0 x 1 0-5 · 1 .35 · [H 3 PO4 ] = 1 .48 x 1 O-5 . [H 3 P04 ], and [PO!- ] = 3.74 x 1 O- 1 1 · [HPO�-] = 3.74 x 1 0- 1 1 . 1 .48 x 1 0-5 · [H 3 P04 ] = 5.54 x 1 0-1 6 ·[H 3 P04 ] Substituting these expressions back into the sum:
SM-3 1 0
[ H 3 PO4 ] + [H 2 PO� ] + [HPO�- ] + [PO !-] = = [ H 3 PO 4 ] + (1 .35 · [H 3 P04 J) + ( 1 .48X 1 0-5 . [H 3 PO4 ] ) + ( 5.54 x 1 0- 1 6 . [H 3 PO 4 ] ) 2 = 1 .5 x 1 0we find: [ H 3 P0 4 ] = 6.4X I O-3 mol · L-I , [H 2 PO� ] = 1 .35 · [H 3 P04 ] = 8.6x 1 0-} mol · L-I , [HPO�-] = 1 .48 x l O-5 . [H)P04 ] = 9.5x l 0-s mol · L- I , and [PO!- ] = 5.54 x 1 0- 1 6 . [H 3 P04 ] = 3.5x 1 0- 1 8 mol · L-I. 10.101
We can use the relationship derived in the text, [H ) o+ ] 2 - [HAL,itial [H 3 0+ ] - Kw = 0 , in which HA is any strong acid. Solving using the quadratic equation gives [H 3 0+ ] = 6.70 x l 0-7 , pH = 6. 1 74. This value is slightly lower than the value calculated, based on the acid concentration alone (pH = - log(6.55 x 1 0-7 ) = 6.1 84).
1 0 . 1 03
We can use the relationship derived in the text, [H 3 0+ f + [Blnitial [H 3 0+ ] - Kw = 0 , in which B is any strong base. Solving using the quadratic equation gives [H3 0+ ] = 6.24 x l 0-8 , pH = 7.205. This value is higher than the value calculated, based on the base concentration alone (pOH = - log(9.78 x 1 0-8 ) = 7.009).
1 0. 1 05
(a) In the absence of a significant effect due to the autoprotolysis of water, the pH values of the 1 .00 x l 0-4 M and 1 .00 x 1 0-6 M HBrO solutions can be calculated as described earlier. For 1 .00 x 1 0-4 mol · L- I : SM-3 1 1
Concentration (mol · C l ) HBrO(aq) initial 1 .00 x 1 0-4 -x change final 1 .00 x 1 0-4 - X K = [H 3 0+ ][BrO- ] [HBrO] 2.0 x 1 0-9 = (x)(x)
H 2 O(l)
+
� --
H 3 0+ (aq) + BrO- (aq) a
+x +x
a
+x +x
a
LOa x
Assume x «
1 0-4
-x
1 .00 x 1 0-4•
=
2 _X_
_ _
_ _
[1 .00 X 1 0-4
- x]
x = 4.5 X 1 0-7 Because x < 1 % of 1 .00 x l 0-4 , the assumption was valid. Given this value, the pH is then calculated to be - loge4.5 x 1 0-7 ) = 6.35. For 1 .00 x 1 0-6 mol · L- ' : Concentration (mol · L- ' ) a initial 1 .00 X 1 0-6 -x change +x 1 .00 X 1 0-6 - X +x final K = [H 3 0+ ][BrO- ] [HBrO] 2 2.0 x 1 0-9 = (x)(x) = ___X_----:-_ 1 .00 X 1 0-6 - x [1 .00 X 1 0-6 - X] Assume x « 1 .00 x 1 0-6. x2 = (2.0 X 1 0-9 ) (1 .00 X 1 0-6 ) X 4.5 X 1 0-8 a
=
SM-3 1 2
a
+x +x
x is 4.5% of x the assumption is less acceptable. The pH is calculated to be -10g(4.5 x =7.35. Because this predicts a basic solution, it is not reasonable. (b) To calculate the value taking into account the autoprotolysis of water, we can use Equation x3 + Ka x2 -(Kw + Ka · [HALitial )x -Kw ·Ka wherex=[H30+] To solve the expression, you substitute the values of Kw = xl , the initial concentration of acid, and Ka into this equation and then solve the expression either by trial and error preferably, using a graphing calculator such as the one found on the CD accompanying this text. Alternatively, you can use a computer program designed to solve simultaneous equations. Because the unknowns include [H30+], [OH-], [HBrO], and [BrO-], you will need four equations. As seen in the text, pertinent equations are [BrO-] a = [H3[0+]HBrO] Kw =[H30+][OH-] [H30+] [OH-] [BrO-]. [HBrO);"ilial =[HBrO] + [BrO-] Both methods should produce the same result. The values obtained are [H30+]=4.6 x mol· L-1, pH=6.34 (compare to 6.35 obtained in (a» [BrO-] 4.4 mol· L-l1 [HBrO] xl mol· C ] mol·L-I • Similarly, for [HBrO);"ilial xl 1.00
10-6,
SO
10-8)
22:
=0,
1.00
= 2.0 x 10-9
Of,
K
+
=
10-7
xl 0-7
=
::=
1.0
0-5
[OH- =2.2 x l 0-8
=1.00
0-6,
SM-313
0- 1 4
[H30+] =1.1 x10-7 mol· L-', pH =6.96 (compare to 7.35 obtained in (a)) [BrO-]=1.8 x10- 8 mol· [HBrO] 9.8 x10-7 mol· L-' [OH-] =9.1 x10-8 mol· L-' Note that for the more concentrated solution, the effect of the autoprotolysis of water is very small. Notice also that the less concentrated solution is more acidic, due to the autoprotolysis of water, than would be predicted if this effect were not operating. (a) In the absence of a significant effect due to the autoprotolysis of water, the pH values of the 8.50 x10-5 and 7.37 x10- 6 HCN solutions can be calculated as described earlier. For 8.50 x 10-5 mol· L-' : Concentration (mol· L- ' ) HCN(aq) CN-(aq) initial 8.50 10-5 change final 8.50 10-5 = [H}O+][CN-] [HCN] 4.9 xl 0-1 0 = 8.5 10-5 = [8.5 10-5 - ] Assume 8.5 x10-5. =(4.9 xl0-1 0)(8.5 xl0-5) =2.0 10-7 Because of 8.50 x10-5, the assumption was valid. Given this value, the pH is then calculated to be -10g(2.0 xI 0-7 ) =6.69. For 7.37 x10-6 mol· L-': Concentration L-'
==
10.107
M
M
+
+
o
X
-x
X
X
Ka
(x) (x)
X
X
+x
+x
+x
2
_ __ _ __
-x
X
x«
2 X
x
+x
X
x < 1%
SM-314
X
L-)
(mol· HCN(aq) CN-(aq) initial 7.37 10-6 change final 7.37 10-6 [H)O+] [CN-] [HCN] 4.9 xl 0-10 7.37 10-6 [7.37 10-6 - ] Assume 7.37 x10-6. (4.9 xl0-10)(7.37xlO-6) 6.0 10-8 is 1% of 7.37 x10-6, so the assumption is still reasonable. The pH is then calculated to be -log(6.0 xl 0-8) 7.22. This answer is not reasonable because we know HCN is an acid. (b) To calculate the value, taking into account the autoprotolysis of water, we can use Equation 21: 0, where [H)O+] To solve the expression, you substitute the values of 1.00 x10-14 , the initial concentration of acid, and 4.9 x10-10 into this equation and then solve the expression either by trial and error or, preferably, using a graphing calculator such as the one found on the CD accompanying this text. Alternatively, you can use a computer program designed to solve simultaneous equations. Because the unknowns include [H)O+], [OH-], [HBrO], and [BrO-], you will need four equations. As seen in the text, the pertinent equations are +
+
o
X
-x
X
X
+x
+x
+x
+x
Ka = -=----=
(x) (x) X
X2 = __ _ _ -,--_ x -x X
x«
x2 =
X
x= X
<
=
x) + Ka x2 - ( Kw + Ka . [HAlni'iaJx - Kw . Ka =
x=
Kw
Ka =
SM-315
=
+][CN-] Ka = [H)[OHCN ] Kw =[H)O+][OH-] [H)O+] =[OH-] [CN-] [HCNlnitia' =[HCN] [CN-]. Both methods should produce the same result. For [HCN]=8.5 x 10-5 mol· L-', the values obtained are [H3 0+] = 2.3 10-7 mol· L-' , pH=6.64 (compare to 6.69 obtained in (a)) [CN-]=1.8 xl 0-7 mol· L-' [HCN]=:8.5 xlO-5 mol·L-' [OH-]=4.4 xlO-8 mol·L-' . Similarly, for [HCNlnitia' =7.37 x10-6, [H)O+]=1.2 x10-7 mol· L-' , pH=6.92 (compare to 7.22 obtained in (a)) [CN-]=3.1xlO-8 mol·L-' [HCN]=: 7.3 x10-6 mol· L-' [OH-]=8.6 xl 0-8 mol· L-' . Note that for the more concentrated solution, the effect of the autoprotolysis of water is smaller. Notice also that the less concentrated solution is more acidic, due to the autoprotolysis of water, than would be predicted if this effect were not operating. (a) Acid (1) is a strong acid because it totally dissociates. (b) Acid (3) has the strongest base because it is the weakest acid. (c) Acid (3) has the largest pKa. Since pKa -logKa. The smallest Ka value will result the largest pKa. The apparent motion of hydronium and hydroxide ions is not as dependent on the diffusion of individual ions as other ions; they are formed and reformed as protons are transferred from and to water molecules in solution. Autoprotolysis allows rapid proton transfer between water molecules. SM-316 +
+
X
10.109
=
10.111
10.113
(a) We begin by finding the empirical fonnula of the compound: C : 44.0.091142 gg.COmol-2 I 0.0214 mol CO2 :. 0.2 14 mol C (0.214 mol C)(12.0 11 g·moll) 0.257 g C 964 g H20 l 0.00535 mol H20 :. 0.0107 mol H H: 18.0.00158 g·mor (0.0107 mol H)(1.008 g·mol-I) 0.0108 g H g. H2 0 0.0107 mol Na Na: 22.0.246 99 g mol-I (0. 0107 mol Na)(22.99 g . mol-I) 0.0246 g Na 0: mass ofO 1.200 g-0.257 g-0.0108 g-0.0246 g 0.686 g ofO 0.686.gO 0.0429 mol ° 16.00 g mol-I Dividing through by 0.0107 moles, we find the empirical fonnula to be: C2HNa04. A molar mass of 112.02 g·mol- indicates that this is also the molecular fonnula. =
=
=
=
=
=
=
=
=
I
SM-317
(b)
..
.. :0: \
•
C-C
II
· 0:
i-
:O-H
I
\\ .
-.0 ·
(c) The dissolved substance is sodium oxalate, it is capable of gaining or losing a proton and, therefore, amphiprotic. pH 1 (p a , p a ) = 2.7. The Lewis structure of boric acid is H"'-." H " I H-O: :O-C-O: B-O: I .. H H-Q: H ""'-.H There is no resonance in boric acid (there is no double bond). B(OH)3 is a very weak acid because it does not have a conjugate system to delocalize the electrons on the oxygen to weaken the O-H bond. (b) In that reaction, boric acid acts as a Lewis acid because it accepts electron pairs (OH -) from water. But, its acidity is not due to its dissociation. (a) Nitrous acid will act like a strong acid because its conjugate base, the nitrite ion, is a weaker base than the acetate ion. The presence of acetate ions will shift the proton transfer equilibrium of nitrous acid toward the products and H30+) by consuming H30+, which will increase the apparent of nitrous acid. Carbonic acid is a weaker acid than acetic acid, so no shift in equilibrium occurs. (b) Ammonia will act like a strong base because its conjugate acid, the ammonium ion, is a weaker acid than acetic acid. The presence of acetic acid will shift the proton transfer =
10.115
0:
\
..
\
I
boric acid
10.117
I
:0
conjugate base
(N02-
Ka
SM-318
I
K + K2
10.119
equilibrium of ammonia with water toward the products (NH4+ and OH-) by consuming OH-, which will increase the apparent Kb of ammonia. We wish to calculate Ka for the reaction This equation is equivalent to This latter writing of the expression is simpler for the purpose of the thermodynamic calculations. The �Go value for this reaction is easily calculated from the free energies given in the appendix: (-278.79 kJ 'mol-I) - ( -296.82 kJ · mOrl) 18.03 kJ ·mol-I -RTln K K 6.9 K (a) D 2 0 + D 2 0 D30+ + OD(b) K020 [D30+][OD-] 1.35 10-15, pK020 -log K02 14.870 (c) [D30+] [OD-] �1.35 10-15 3067 10-8 mol· (d) pD -log(3.67 10-8) 7.435 pOD (e) pD + pOD pK020 14.870 =
=
= =
10.121
e-t.COIRT
' ' e-(I8030Jomol- )/[80314JoK-
0
' mol- )(298 K)j
=
xl 0-4
�
x
=
=
=
=
x
x
=
=
=
=
=
0
L-
X
=
I
=
=
10.123
CH3CH(OH)COOH(aq) + H2 0(I) Concentration (mol· 1.00 initial -x change equilibrium 1.00-x Cl)
SM-319
o
o
+x x
+x x
)CH(OH)COO-] =� =8.4 X 10-4 = [H)O[CH)+][CHCH(OH)COOH] 1.00-x =0.029 Percent deprotonation=0.1.00290 (100%) = 2.9% deprotonated. To calculate TJfor this solution we need the concentration in moles of solute per kilogram of solvent. Assuming a density of 1 g . cm - 1.00 of solution will weigh 1000 g. One liter of solution will contain 1 mole (90.08 g) oflactic acid. Therefore, one liter of solution will contain 1000 g -90.08 g = 910 g or 0.9 10 kg of solvent. solute =l.l3 · . 0.1.092910 mol The moI·anty f thi e so utlOn kg solvent f:.,.Tr =(1.86 K . kg . morl)(1.13 mol . Kg-I)=2.1 K The temperature at which the solution will freeze is 271 K. T 30°C 303 K, T= 20°C 293 K In(�:)=��n� - ;) =�r�-/) This is the van't Hoff equation. 4 X I.768 I0-293K] ( In 1.765x10-4 ) -_ 8.3 14J·f:.,.KH-OI· morl (303K 293Kx303K In (1.1.776865 xx10-10-44 ) (8.31 4J . K-I . mol- I) 2 f:.,.HO= =1.3x10 J ·mol-I 303 K -293 K ( 293 K x303 K ) (a) The equilibrium constant for the autoprotolysis of pure deuterium oxide is given by = = -84800 J . mol-I/(I,8.3145J.K-I.mol-I)(298 K) =1.37 x10-15 The concentration ofD30+(aq) at equilibrium is found in the familiar way: Ka X
1,
0
10.125
10.127
=
K
=
e-Mj".,
IS
=
e
SM-320
m .
L
Concentration (mol· L-1 ) initial change equilibrium K [D30+][OD-] 1.37 10-1 5 [D30+] 3.7 10-8 pD -log(3.7 10-8) 7.4 (b) Given the expression for K in part (b) above, it is apparent that, as increases, K will increase resulting in an increase in [D30\aq)] and a decrease in pD. (a) and (b) Buffer regions are marked A, B, and C. =
=
=
=
10.129
o
o
+x
+x
+x
+x
X
X
=
x
14
pH
(c) Region A: H3P04 and H2 P04 Region B: H2 P04- and HPO/Region C: HP04 2- and P04 3(d) 1.0
c
0.5
o
o
pH
14
SM-321
T
(e) The major species present are similar for both H3P04 and H3As04 : H2 E04 - and HE04 2 -, where E P or As. For As, there is more HAs04 2- than H2 As04-, with a ratio of approximately 0.63 to 0.37, or 1.7 : 1. For P, the situation is reversed, with more H2 P04 - than HP04 2in a ratio of about 0.61 to 0.39, or 2.2 : 1. (a) The H+ in lactic acid can interact with Hb02-to produce HHb and release more oxygen molecules, which means that the concentration of Hb02-will be lower in the tissues. (b) Since more Hb - will return to the lung, more oxygen will be bound to Hb - to produce Hb02-, which will increase the concentration of Hb02-. H2A(aq) H20(1) HA2-(aq) H30+(aq) (1) HA-(aq) H20(l) A -(aq) H30\aq) (2) The [H+] concentration from the first dissociation is 0.020 M (total dissociation). Since the pH of the H2A solution is 1.66, [H+]solution 2 2.2 2 10- M. The [H+]3 from the2 second dissociation will be (2.2 10- ) - 3 3 0.020 2.0 10- M; [A -] 2.0 10- M; [HA - ] 0.020-2.0 100.0 18 M [H3O+][A2-] 2.0x10-3 2 2.2 10- 4 [HA-] 0.0 18 (a) Given that CO2 will react with water to form carbonic acid, H2C03, it only remains to determine the concentration ofH30+ due to the deprotonation ofH2C03. l -4 atm) =7.0xlO-6 mol· L-1 [C02 ]=( 2.3xlO-2 mol. L-1. atm-1)(3.04xO initial 7.0 x 10- 6 change =
10.131
10.133
-7
+
+
-7
+
+
=
=
x
=
x
x
x
=
=
Ka 2
10.135
=
=
(
)
=
x
+
+
+x
-x
SM-322
+x
x
equilibrium 7.0 x1 0.6
-x
x
K a = 1 0 -PKa = 1 0-6. 3 7 = 4.27X I O -7 =
( Xl J
X = [H30+ ] = 1.73x1 0-6 pH = - log(1.73x10-6) = 5.75 (b) The equilibria involved are: HlC03(aq) HlO(l) HC03-(aq) +
�
x
7.0 x l O-6
+
H 3 0+
Ka l
=
Kal
=
.3
4 X 1 0- 7
5.6X 1 0-"
Given these two equilibria, two equations can be written: [HCO)-][ H)O+] K = [ H2 CO) ] [ CO/- ][ H)O+] K a2 = [ HCO)- ] Also, enough information is provided to construct two other equations: ) [ H2CO) ] + [ HCO)-]+[ cot] = 4.5 xl O[ H)O+] = 1 0-4 8 =1.6 xlO-5 Solving this set of simultaneous equations for the three unknown concentrations, one finds: [HlC03 ] 4.4 x 1 0-3 mol· L- 1 [HC0 3-] 1.2 x 10-4 mol· L- 1 l [C03 -] 4.2 x 1 0- 10 mol· L- 1 (c) Assuming all sulfur is converted to S02, the amount ofS02 produced ( 1 .00 x 1 0) kg )( 0.025 )( 1 000 g . kg-I ) -- 32.07 g·mor -'- 780 mol of S Therefore, (780 mOI)( 64.07 g . mol-I ) 50, 000 g, or 50 kg, of S0 2 is produced (d) To determine the pH, we first must calculate the volume of water in which this 50 kg of S02 is dissolved: al
=
=
=
IS
--'--
----'-----'-1 ---
=
=
SM-323
2( 100cm ) 2( 1 3 ) =5.2XI07 1000m ) ( cm )( 2. 6 Km ) = ( 2 V 1 m 1000 cm 1 The concentration of S02(aq) is then ( 780 mol)/( 5.2 107 ) 1.5X10-5 M. We can assume that upon solution the S02(aq) is converted to sulfurous acid and the pH is calculated as described earlier: Concentration + (mol . H2S03 (aq) + initial 1.50 10 5 +x +x -x change +x +x final 1.50X10-5 H 0+][HS03-] K [ 3 [ H 2S03 ] )(x) 1.55 10-2 1.5X(x10-5 -x Employing the quadratic formula we find x=[H30+] 1.5 10-5, giving a pH of pH log(1.5 10-5) 4.82. (e) When dissolved in water, S03 will form sulfuric acid: The fist deprotonation of sulfuric acid is complete, the concentration of hydronium ion due to the second deprotonation of sulfuric acid may be found as described earlier: Concentration (mol . HSO- (aq) + H20(l) H30+(aq) + S0 2- (aq) 1.50X10-5 0 initial 1 50 10-5 +x +x -x change 1.50X10-5 +x final 1.50 10-5 - x L
L
Km
L
x
=
L- 1 )
x
-
X
a
=
=
x
x
=
=
=
x
L-1)
�
4
.
4
x
x
-
SM-324
X
[H30+ ][ SO/-] [ HS04 - ] 1.2 x10-2 (1.51.5 10-510-7-x)(x) Employing the quadratic formula we find x [H30+] 1.495 10-5, and the total hydronium ion concentration to be 1.495 x10-5 1.5 x10-5 3.0 xl0-5. Therefore, pH -log(3.0 x10-5) 4.52. (f) Because the process is only 90% efficient, to remove 50.0 kg of S02 one must supply enough CaC03 to remove 55.6 kg of S02 (50.0 kg is 90% of 55.6 kg). The moles of S02 to be removed is given by 55600 g 868 mol S02. 64.06 g mol-1 1 mole of S02 is consumed by 1 mole ofCaC03. Therefore, (868 mol CaC03 )(100.09 g mol-I ) 86.9 kg. K
a=
X
=
X
X
=
=
=
=
=
SM-325
=
X
+
=
CHAPTER 11 AQUEOUS EQUILIBRIA
11.1
(a) When solid sodium acetate is added to an acetic acid solution, the concentration of H3 0+decreases because the equilibrium shifts to the left to relieve the stress imposed by the increase of [C 2H3 02-] (Le Chatelier's principle). (b) When HCI is added to a benzoic acid solution, the percentage of benzoic acid that is deprotonated decreases because the equilibrium shifts to the left to relieve the stress imposed by the increased [H3 0+] (Le Chatelier's principle). (c) When solid NH4 CI is added to an ammonia solution, the concentration of OH- decreases because the equilibrium NH3 (aq) H2 0(1) NH4+(aq) OH-(aq) shifts to the left to relieve the stress imposed by the increased [NH4 +] (Le Chatelier's principle). Because [OH- ] decreases, [H)O+] increases and pH decreases. +
�
+
pKa pH. pH pKa 3.08, Ka 8.3 10-4 (b) Let [lactate ion] [Cl] and [H3 0+] =
=
=
x =
=
=
x
y=
SM-326
Concentration H)O+(aq) + L-(aq) (mol· L-') HL(aq) + H2O(I) 2x x initial +y +y -y change equilibrium 2x- y y y +x 0+][L-] (y)(y + x) (y)(x) 8.3 10-4 Ka [H3[HL] 2x - y) (2x) y 2(8.3 10-4 ) 1.7 10- ) mol· L-' [H)O+] pH 2.77 In each case, the equilibrium involved is HS04 -(aq) + H2 0(I) H)O+(aq) + SO/-(aq). HSO4-(aq) and SO42-(aq) are conjugate acid and base; therefore, the pH calculation is most easily performed with the Henderson-Hasselbalch equation: 2 S0 e ba [ ] [ 4 0 1 0 1 � J pH pKa + g( [acId] pKa g ( [HS04-}] J mol· L-' 1.62, pOH 14.00 -1.62 12.38 (a) pH 1.92 + log ( 0.0.255mol· LJ L-' 1.22, pOH 12.78 (b) pH 1.92 + log ( 0.0.5100 mol· mol· L- J (c) pH pKa 1.92, pOH 12.08 See solution to Exercise 11.3. 1 mol NaF J 0. 17 mol· L-' ( 0.30.56050g NaF J( L 41.99 g NaF --" -.;--
=
=
�
X
=
�
(
=
X
�
X
�
11.5
�
+
=
=
=
=
=
11.7
=
_I
1
=
=
=
=
SM-327
=
=
=
Concentration (mol· HF(aq) + H2 °(l) H30+(aq) + F-(aq) initial OAO 0 0.17 change -x +x +x equilibrium OAO-x 0.l7 + x x 0+][F-] (x)(O.17 + x) (x)(O.l7) 3.S x10-4 Ka = [H3[HF] (OAO -X) (OAO) x 8.2 10-4 mol· [H30+] pH -log[H30+] -log(8.2 x10-4 ) 3.09 change in pH 3.09-1.93 1.16 (a) HCN(aq) H2 0(l) H30+(aq) CN-(aq) total volume = 100 mL 0.100 moles ofHCN 0.0300 LxO.O SO mol· 1.S x10-3 mol HCN moles ofNaCN 0.0700 Lx 0.030 mol· 2.1 x10-3 mol NaCN . . . I [HCN]0 1.S0.x11000-3 mol l .S x10-2 moI . lmha 3 mol 21. x10-2 moI . .Inlha .. I [CN-]0 2.10.x10100 Concentration (mol· HCN(aq) H2O(I) H30+(aq) CN- (aq) initial I.S x10-2 0 2.1 10-2 change -x +x +x equilibrium I.S x10-2 -X x 2.1 10-2 + X 2 2 CN-] [ H 0+ (x)(2. 101 + (x)(2.1 10x) ][ = 3 K [HCN] (1.SxlO-2 -x) (1.Sx10-2 ) ) 4.9 10-10 x 3. S x 10-1 0 mol· [H,O+] pH -log[H,O+] -log(3.S x10-1 0) 9A6 L-I )
� �
�
=
==
L-I
X
=
==
=
=
=
11.9
=
=
�
+
+
L
=
L-I
=
=
L-I
=
=
=
L
Cl)
L-I
=
L
=
L-I
=
� �
+
+
X
X
X
=
a
L-I
�
=
�
�
=
=
SM-328
X
=
X
(b) The solution here is the same as for part (a), except for the initial concentrations: [HCN]o 0.0400 0.1000.030L mol· L-1 1.2 10-2 mol . [CN-]o 0.0600 L0.10.00050L mol· L-1 3.0 10-2 mol . L-1 Ka 4.9 0- 10 (x(1.)(3.2 0 10-0-2 )2) x [H30+] 2.0 0-10 mol· L-1 pH -log(2.0 10-10) 9.71 (c) [HCN]o [NaCN]o after mixing; therefore, Ka 4.9 10-10 (x)[[HNCN]aCN]O o x [H30+] pH pKa -log(4.9 10-1 0) 9.31 In a solution containing HCIO(aq) and CIO-(aq), the following equilibrium occurs: The ratio [ClO-]/[HCIO] is related to pH, as given by the Henderson. pH pKa log( [CIO-] J , or Hasselbalch equatlOn: [HCIO] O-] J pH -pKa 6.50 -7.53 -1.03 10g ( [[HCICIO] [CIO-] 9.3 10-2 [HCIO] The rule ofthumb we use is that the effective range of a buffer is roughly within plus or minus one pH unit of the pKa of the acid. Therefore: (a) pKa 3.08; pH range 2-4 SM-329 Lx
=
x
=
=
xl
=
x
=
=
x
xl
xl
=
x
=
x
=
=
=
=
X
=
=
11.11
=
=
x
=
=
=
11.13
=
x
=
=
=
+
=
Cl
(b) pKa 4.19; pH range 3-5 (c) pKa3 = 12.68; pH range 11.5-13.5 (d) pKa2 7.21; pH range 6-8 (e) pKb 7.97, pKa 6.03; pH range 5-7 Choose a buffer system in which the conjugate acid has a pKa close to the desired pH. Therefore: (a) HCl02 and NaCl02 , pKa =2.00 (b) NaH2 P04 and Na 2 HP04 , pKa2 = 7.21 (c) CH2 ClCOOH and NaCH2 ClC02 , pKa = 2.85 (d) Na 2 HP04 and Na3P04 , pKa =12.68 =
=
=
11.15
=
pH=pKa?- log ( [[HCC0O3 23-]-]J log( [[HCC0O3 32-]-] J =pH - pKa2 11.0 -10.25 = 0.75 O3_2- -] =5.6 _[[HCC0---'3 -] (b) [C03 2-] =5.6 x[HC03 -] =5.6 x0.100 mol· =0.56 mol· L-' moles of cot =moles of K2 C03 =0.56 mol· L-' xI L =0.56 mol mass of K2 C03 = 0.56 mol x(138.1 mol21 gKK2 C02 CO3 3 J=77 g K2 C03 2 [C0 3 _ CO (c) [H ]= 5.6 -] = 0.1005.mo6 l·L-' =1.8 xlO-2 mol . L-' +
=
Cl
3
SM-330
moles of HCO)- = moles of KHCO) 1.8 x10-22 mol· L-1 x1 l.8 10- mol mass KHCO) 1.8 x10-2 mol x100.12 g. mol-I 1.8 g KHCO) L
=
=
X
=
=
moles of HC03- moles KHC03 0.100 mol·L-1 x0.100 L 1.00 x 10-2 mol Because the final total volume is the same for both KHCO) and K2 CO), the number of moles of K2 CO) required is 5.6 x 1.00 x10-2 mol 5.6 xl0-2 mol. Thus, 0-2 mol 0.28 L 2.8 x102 mL. volume of K2 C03 solutIon. 0.5.2600xlmol·C =
=
=
=
1
=
11.19
=
=
)CO2 -] ) (see ExercIse. 1l.23) (a) pH pK log ( [[CH CH)COOH] 0.100) 4.75 (.ImtIa. . I pH) pH pK Iog ( -0.100 final pH: (0.0 100 L) (0.950 mol· L- ) 9.50 xl 0-3 mol NaOH (strong base) produces 9.50 x10-) mol CH3C02 - from CH)COOH 0.100 mol· L-1 x 0.100 L 1.00 x10-2 mol CH)COOH initially O.l 00 mol· L-1 x 0.100 L l.00 10-2 mol CH)C02- initially After adding NaOH: 3 ) mol 5 x 10-) mol· L-1 2 (l. 0 0 x 10-9. 5 x 10 [CH3COOH] 0.110 L [CH)C02 -] (l.00 x 10-20.1 109.5Lx10-)) mol l.77 x10- mol . L-1 77 10-13 mol· L- ) 4.75 l.5 6.3 pH 4.75 log(l.5x10 - mol·L =
=
a
+
a
=
+
1
=
=
=
X
=
=
+
=
=
+
X
1
=
_I
1
SM-331
=
+
=
(b) (0.0200 L)(0.100 mol· L-1 )=2.00 xl 0-3 mol HN03 (strong acid) produces 2.00 10-3 mol CH3COOH from CH3C02-. After adding HN03 [see part (a) of this exercise]: [CH3COOH]=(1.00 10-20.12.200L0 x10-3) mol = 1.00 10-1 mol . L-1 (1.00 x10-2 -2.00 10-3 ) mol [CH3 CO 2-]= =6.7 10 2 mol .L-1 0.120L pH=4.75+log ( 1.6.070 x10-10-21 mol·L-1 mol· L-1 J = 4.75-0.17=4.58 i'1pH=-0.17 (a) The titration curve is as follows: x
+
x
x
x
x
-
X
11.21
14
12
I
10 J: 0.
=
/
(c) pH at half-way titration
8 6
(b) pH at Stoichiometric point 7.0
=
11.4
4 2
O �--�----�--� 5
o
11.23
10
15
20
Vol. Hel (mL)
25
30
(b) and (c) are labeled within the titration curve. HCI(aq) NaOH(aq) H?O(l) Na+ (aq) cr (aq) (a) (t)(25 0 mL{\o:�J( 0.110 �� NaOH J (11molmolNaOH HCI J( 1 L HCI J 0.150 mol HCI 3 = 9.17 10- L HCI SM-332 +
VHc.
x
�
�
+
+
35
(b) 2x9.17x10-3 L=0.0183L (c) volume (0.0250 0.0183) L =0.0433 L 1mol Na+ ( 1 [Na+]=(0.0250L) ( 0.1 10mo1 LINaOH J( 1 mol NaOH J 0.0433 L J =0.0635mol· L-J (d) number ofmoles of H,o· (from acid) =(0.0200 L{ 0.1�OLmol J =3.00 x10-3mol H30+ Na+ J number ofmoles of OH- (from base) (0.0250 L) ( 0.1 10mol 1 L = 2.75 10-3 molOHexcess H30+ = (3.00 - 2.75) x10-3 mol =2.5 x10-4 mol H30+ mol = 5.6 xl 0-3mol .L-J [H30+] = 2.50.x010-4 450 L3 pH = -log(5.6 x10- ) =2.25 Themoles ofOH- added are equiv alent to the number ofmoles ofHA present: (0.350mol · L -1)(0.052 L) = 0.0182mol OH Therefore, 0.0182mol HA were present in solution (a) Iarmass f HA 0.04.18225 gmol = 234 g . 1(b) pKa pH at half-way titration 3.82 mass of pure NaOH=(0.0342 L HCl) ( 0.06951 LmoHCll HCl ) NaOH )( 40.00 g NaOH )( 300mL ) ( 1mol 1mol HCl 1mol NaOH 25.0mL = 1.14g percent pun.ty 1.1.41364gg x 100% = 79.4 SM-333 =
+
=
X
11.25
rna
0
=
11.27
rna
:
=
=
1
11.29
(a) pOH=-log(0.110)=0.959, pH=14.00 -0.959 13.04 (b) initial moles of OW (from base)=(0.0250 L{ 0.1:�m a l ) =2.75 10-3 molOHmoles of H,o' added (0.0050 L)(0.1�OLm a l ) = 7.5 x10-' mol H,o' excess OH-=(2.753 -0.75) x10-3 mol=2.00 x10-3 molOH[OH-]= 2.00.0 xl03000- Lmol =0.0667 mol.L-1 pOH=-log(0.0667) 1.176, pH 14.00 -1.176=12.82 (c) moles of H30+ added =2 x 7.5 10-4 mol=1.50 x10-3 mol H30+ excess OH-=(2.753 -1.50) xl0-3 mol=1.25 x10-3 molOH10- mol =0.0357 mol .L-1 [OH-]=1.250.0350 L pOH=-log(0.0357)=1.447, pH=14.00 -1.447=12.55 (d) pH 7.00 Hel )( 1 L Hel ) VHC1 =(2.75 10-3 mol NaOH)(11molmolNaOH 0.150 mol Hel =0.0183L 1 ( ) ) . (0.150mol (e) [H30+]=(00050L) 1 L (0.0250 0.0183 0.0050) L =0.016 mol· L-1 pH=-log(0.016)=1.80 ( 0.010L ) (0.150mol) =0.028mol .L-1 (f) [H3 0+ ]= 0.0533 L 1 L pH=-log(0.028)=1.55 =
X
=
=
=
X
X
=
X
+
SM-334
+
11.31
(a) Concentration (mol· L-1 ) CH)COOH(aq) + H20(l) H)O+(aq) + CH)C02-(aq) initial O.l 0 change -x +x +x equilibrium 0.10 -x x x �
o
o
x2=1.8 10-6 x=1.3 x10-) mol·L- 1 = [H)O+] initial pH -log(1.3 x10-))=2.89 (b) moles of CH)COOH=(0.0250 L)(O.l 0 =2.50 10-) mol CH)COOH moles of NaOH=(0.0100 L)(0.10 = 1.0 10-) molOHAfter neutralization, 1.50 10-) mol CH)COOH =4.29 x10-2 mol . L- 1 CH)COOH 0.0350 L ) 1.0 x1 0- moICH)C02- =2.86x10-2 mol . L-1 CH)CO-2 0.0350 L [H O+][CH)CO 2] .. . Then consider eqUlhbnum, [CH)COOH] Concentration 2.86 10-2 initial 4.29 10-2 +x change -x +x equilibrium 4.29 x10-2 X x 2.86 10-2 + X (X (X + 2.86 xl 0-2 assume +x and -x negII'gl'ble. 1 . 8 x10-5= (4.29 0-- -x) [H)O+ ]=x=2.7 x10-5 mol· L-1 and pH -log(2.7 x10-5)=4.56 (c) Because acid and base concentrations are equal, their volumes are equal at the stoichiometric point. Therefore, 25.0 mL NaOH is required to SM-335 X
=
M)
X
M)
X
X
Ka =
)
-
)
xl
x
o
X
?
X
)
;
=
reach the stoichiometric point, and 1 2 .S mL NaOH is required to reach the half stoichiometric point. (d) At the half stoichiometric point, pH = pKa
=
4.7S.
(e) 2S.0 mL; see part (c) (f)
The pH is that of O.OSO M NaCH ) C0 2 •
Concentration �
(mol · L-1 ) H 2 0(l) + CH 3 C0 2 - (aq) initial O.OSO
CH ) COOH(aq) + OH- (aq) 0
0
change
-x
+x
+x
equilibrium
O.OSO - x
x
x
Kw 1 .00 X 1 0- 1 4 X2 x2 S. X 1 0 - 1 0 = :::::: ___ Ka 1 .8 x 1 0- 5 O.OSO - x O.OSO 2 x 2.8 X 1 0- 1 1 x S. X 1 0-6 mol · L-1 [OH- ] pOH = S.28, pH 1 4.00 - S.28 8.72
Kb
=
=
=
11.33
=
3
(a) Kb
=
=
6
=
=
[NH 4 + ] [OH- ] [NH 3 ]
Concentration (mol . L-1 ) H 2 °(l)
+
=
=
1 .8 X 1 0-5
----'.".---
NH 3 (aq)
ini tial
O. I S
change equilibrium
NH 4 + (aq)
0
-x
+x
+x
O. I S - x
x
x
=
:::::: --
=
=
1.6
=
=
(b) initial moles of NH 3
=
OH- (aq)
0
x2 x2 O. 1 S - x O. IS ) [OH- ] = x x 1 0- mol · Cl pOH 2.80, initial pH 1 4.00 - 2.80 1 1 .20 1 .8 X 1 0-5
+
(0.01S0 L) (O. lS mol · L- 1 ) = 2. x 1 0-3 mol NH 3
3
SM-336
moles of HC I = (0.0 1 50 L) (0. 1 0 mol · Cl) = 1 .5 X 1 0-3 mol HC I 3 ) mol NH 1 . 5 x 1 0---(2.3--x 1 0- 3 -------3 -"- = 2.7 x 1 0-2 mol · L- NH 3 0.0300 L 1 .5 x l 0-3 mol HCl = 5 .0 x l 0-2 mol . L-1 HCl� 5.0 x l 0-2 mol · CI NH 4 + 0.0300 L I
Then consider the equilibrium: Concentration (mol · L-1 ) H 2 0(l)
+
initial change
NH 3 (aq) 2.7 x 1 0-2
NH/(aq) 5.0 x l 0-2
-x 2.7 X 1 0-2 -x
equilibrium
+ +x
+x 5.0 X 1 0-2
+x
x
[NH4 +] [OH - ] = 1 .8 x l 0 -s [NH 3 ] (x) (5.0 x l 0- 2 + x) . . = ; assume that +x and -x are neghgIble 2 (2.7 x l O- -x)
Kb =
[OH- ] = x = 9.7 x l 0-6 mol · CI and pOH = 5.01 Therefore, pH = 1 4.00 - 5.01 = 8.99. (c) At the stoichiometric point, moles of NH 3 = moles of HC! . volume HC I added =
(0. 1 5 mol · L-1 NH 3 ) (0.0 1 50 L) = 0.022 L HC I 0. 1 0 mol · L-1 HC I
Therefore, halfway to the stoichiometric point, volume HC I added = 22/ 2 = 1 1 mL. (d) At half stoichiometric point, pOH = pKb and pOH = 4.75. Therefore, pH = 1 4.00 - 4.75 = 9.25 . (e) 22 mL; see pari (c) (f)
NH/ (aq) + H 2 0(l) � H 3 0+ (aq) + NH) (aq)
The initial moles of NH) have now been converted to moles of NH 4 + in a
SM-337
( 1 5 + 22.5 3 7.5) mL volume: 2.25 x 10-3 mol 0.060 mol. [: 1 [NH +] 0.0375 L =
=
4
Ka
=
Kw
=
=
Kb
1.00 X 10-14
1 .8 X 10-5
=
Concentration (mol · L- 1 ) NH/ (aq) initial
0.060
change
-x
5. 6 X 10-10
+
+
H20 (l)
o +x x
equilibrium 0.060 - x 5 6 X 10 1 0
=
X
2
=
x
[H30+] = 5.8 X 10-6 mol · L- 1 = - log(5.8 10-6) = 5.24
=
pH
-
0.060
+x x
X2
Ka
'
NH3 (aq)
-
x
� --
0.060
x
(g) Methyl red will be appropriate for this titration. 11.35
(a) The acid is a weak acid because the initial pH of the acid =5.0 and the stoichiometric point occurs at pH =10 (the anion is a base). (b) pH =5.0, [H30+] 1 x 10-5 mol ' L- 1 =
(c) At halfway titration, pH = pKa. It takes 5.0 mL base to reach halfway,
3
7.5 = pKa of the acid. Ka = 10-7.5 = (d) HA(aq) + H2 0(l) -7 H3 0+ (aq) + A- (aq)
the pH
=
x
10-8.
When HA self-dissociation reaches to an equilibrium: [H30+] = [A-] = 1 10-5 mol, L- 1 ; [HA]eq [HA]illitial - (1 x 10-5) x
Ka = 3
x
10-8 =
=
[H30+][A -]
[HA]eq
=
(1 x 10-5)2 [HA lnitial
-
(1 x 10-5)
Since Ka is small, the dissociation level of HA will be very small.
x
Therefore, [HA]illitial - (1 10-5) [HA]illitial. �
Solve for the equation: [HA]illitial = 3
x 10-3
mol, L - 1
(e) It takes 10.0 mL base to reach to stoichiometric point. [HA]illitial = 3 . 1 6 x 10-3 mol · L- 1 • If the volume of acid = 25. 0 mL, then
SM -338
[B] (f)
11.37
=
[HA] VHA Vbase
=
(3 x l 0· 3 )(25 mL) ( 1 0.0 mL)
=
8 x 1 0-3 M .
Phenolphthalein will be an appropriate indicatoL
At the stoichiometric point, the volume of solution will have doubled; therefore, the concentration of CH 3 C0 - will be 0. 1 0 M. The equilibrium 2 IS
Concentration (mol . L-1 ) initial
0. 1 0
o
o
change
-x
+x
+x
x
x
equilibrium K w K = b K a
0. 1 0 - x 1 .00 X 1 0 - 1 4
= 5.6 X 1 0-1 0 1 .8 X 1 0-5 [HC H 3 C0 2 ] [OH- ] x2 K � 5.6 X 1 0-1 0 = b = [CH 3 C0 - ] 0. 1 0 - x 0. 1 0 2 x = 7.5 x l 0-6 mol · L-1 = [OH- ] pOH log(7.5 X 1 0-6 ) = 5 . 1 2, pH = 1 4.00 - 5 . 1 2 = =
�
=
=
8.88
-
From Table 1 1 .3, we see that this pH value lies within the range for phenolphthalein and thymol blue; the others would not. 11.39
Exercise 1 1 .3 3: thymol blue or phenolphthalein; Exercise 1 1 .3 5 : methyl red or bromocresol green.
11.41
(a) To reach the first stoichiometric point, we must add enough solution to neutralize one H+ on the H 3 As0 4 . To do this, we will require 0.0750
Lx 0.137 mol ·
Cl
=
0.0 1 03 mol OH- . The volume of base
required will be given by the number of moles of base required divided by
SM -339
the concentration of base solution: 0.0750 L x 0. 1 37 mol . L-' = 0.0374 L or 37.4 mL , 0.275 mol · L(b) and ( c) To reach the second stoichiometric point will require double the amount calculated in (a), or 74.8 mL, and the third stoichiometric point will be reached with three times the amount added in (a), or 1 1 2 mL.
11.43
HP032- is the fully deprotonated form of phosphorous acid H3 P03 (the remaining H attached to P is not acidic). It will require an equal number of moles of HN03 to react with HP03 2- in order to reach the first stoichiometric point (formation of H2 P0 3 ). The value will be (a) The base
-
0.0355 L x 0. 1 58 mol · L- ' . = 0 02 20 L or 22 . 0 m L gIVen b y 0.255 mol · L.
I
(b) To reach the second stoichiometric point would require double the amount of solution calculated in (a), or 44.0 mL. 11.45
(a) Thiosulfuric acid titrated by KOH
- ---- -- -- - -- -- -- -------------
12 ----------- - 10 8
- --
-
-�---------------- ---- J -
---
-
o +---�--�----__. 2 4 o 14 12 10 6 8 16 Vol KOH
SM-340
Point (a): initial point, point (b): first halfway point, point (c): first stoichiometric point, point (d): second halfway point, point (e) second stoichiometric point Note: (i) The initial pH is higher than points b and c because we assume [H+]
that the initial
from the first dissociation only (actually second
dissociation contributes also); (ii) due to the close pKa values of thiosulfuric acid, only one obvious titration curve can be observed. In an actual aqueous titration, two protons will be titrated at the same time for this acid. (b) First stoichiometric point: (E) = 1 0. mL
(C)
=
5.0 mL; second stoichiometric point:
(c) First stoichiometric point: pH = 1 (pKaJ + pKa2 ) = 1 (0.6+ 1 . 74) = 1 .2 ; second stoichiometric point: S20/-
+ H20
Kb J
K w Ka 2
=
=
� HS20 -
3
5.56 x 1 0 -1 3 ,
2
X
0.003 3 - x
=
5.56
x
+ OH[S20/- ] initial
1 0-1 3 ,
X =
=
( 5.0
[OH- ]
=
x 0.0 1 )11 5.0 = 0.0033 M 4.30x l 0-8
<
[OH- Later
Therefore, pH = 7.0. 11.47
(a) This value is calculated as described in Example 1 0. 1 2. First, we calculate the molarity of the starting phosphorous acid solution: 0. 1 22 g /0.0500 L = 0.0298 mol · L-J• We then use the first acid 8 1 .99 g . mol-J /. dissociation of phosphorous acid as the dominant equilibrium. The KaJ is 1 .0 X 1 0-2 . Let H 2P represent the fully-protonated phosphorus acid. Concentration (mol · L- 1 ) initial change
H 2 P(aq) + H 2 0(l) 0.0298 -x S M - 341
o
o
+x
+x
+x
0.0298 - x
final
+x
[H 3 Q + ] [HP- ] = 1 .0 x 1 0-2 [H 2 P] __X2 X·X = _ __ 1 .0 X 1 0-2 = 0.0298 - X 0.0298 - X
K = a
If we assume x« 0.0298, then the equation becomes x2
= ( 1 .0 x l 0-2) (0.0298) = 2.98 x 1 0-4 X = 1 .73 x 1 0-2 .
Because this value is more than 1 0% of 0.0400, the full quadratic solution should be undertaken. The equation is x2 = (1 .0 x l 0-2 ) (0.0298 - x) or x 2 + (1 .0 x l 0-2 x) - (2.98 x l 0-4 ) = 0. Using the quadratic formula, we obtain x = 0.0 1 3 . pH = 1 .89 (b) First, carry out the reaction between phosphorous acid and the strong base to completion: H 2 P(aq) + OH- (aq) � HP- (aq) + H 2 0(l) moles of H 2 P = (0.0298 mol · L-J) (0.0500 L) = 1 .49 x 1 0-3 mol moles of OH- = (0.00500 L) (0. 1 75 mol · L-J) = 8.76 X 1 0-4 mol 8.75 X 1 0-3 mol OH- will react completely with 1 .49 x 1 0-3 mol H 2 P to give 8.75 x 1 0-3 mol HP- , with 6. 1 x 1 0 -4 moles of H 2 P remaining. [H 2 P] =
6. 1 X 1 0-4 mol = 0.01 09 mol . l:J 0.0565 L
[HP- ] =
8.75 x 1 0-3 mol = 0.01 55 mol . L-J 0.0565 L
Concentration initial
0.0 1 09
change
-x
0.01 55 +x
SM-342
o
+x
final
0.0 1 09 - x
0.0 1 55 + x
+x
The calculation is performed as in part (a): 1 .0 x 1 0-2 =
(0.0 1 55 + x)x 0.0 1 09 - x
x = 1 .08 X 1 0-2 pH = 1 .96 (c) moles of H P = 1 .49 x l 0- 3 2 moles of OH - = 8.75 x 1 0-4 + 8.75 x 1 0-4 =1 .75 x 1 0-3 Following the reaction between H P and OH- , 0 mol o f H P remain and 2 2 2.62 x 1 0 -4 mol OH- remain. [OH- ] = 4.37 x 1 0-3 M, :. [H 3 0+ ] =2.29 x 1 0-1 2 M. . 1 .49 x 1 0-3 mol 1 .49 x 1 0-3 mol of HP- remam, [HP- ] = = 2.48 x 1 0- M. 0.060 L 2
Concentration (mol . L-1 )
HP-
initial
2.48 x 1 0-2
2.29 x l 0- 1 2
change
-x
-x
+
final 2.48 x 1 0-2 - x
H 3 0+
� -.----
2.29 x 1 0-1 2 -x
H P 2 0
+
H O 2
+x X
The calculation is performed as in part (a): x 1 00=�------��--------� ( 2.48 x 1 0-2 - x )( 2.29 x 1 0-1 2 - x ) Using the quadratic equation, we find x = 5.6 x l 0- 16 , [H 3 0] = 2.29 x 1 0-12, and pH = - log(2.29 x l 0-1 2) = 1 1 .6. 11.49
(a) The reaction of the base Na HP0 4 with the strong acid will be taken 2 to completion first: HPO / - (aq) + H 3 0+ (aq) � H P0 4 - + H 2 0(l) 2 Initially, moles of HPO 4 2- = moles of H 3 0+ = 0.0500 L x 0.275 mol · L-1 = 0.0 1 3 8 mol
SM-343
Because this reaction proceeds with no excess base or acid, we are dealing with a solution that can be viewed as being composed of H 2 P0 4 - . The problem then becomes one of estimating the pH of this solution, which can be done from the relationship pH = t (pKa + pKa 2 ) ' pH t(2. 1 2 + 7.2 1 ) = 4.66. =
(b) This reaction proceeds as in (a), but there is more strong acid available, so the excess acid will react with H 2 PO 4 - to produce H 3 PO 4 . Addition of the first 50.0 mL of acid solution will convert all the HPO 4 2 into H 2 PO 4 - . The additional 25.0 mL of the strong acid will react with
0.0 1 3 8 mol H 2 P0 4 - will react with 0.006 8 8 mol H 3 0+ to give 0.0069 mol H 3 P0 4 , with 0.069 mol H 2 P0 4 - in excess. The concentrations will be 0.0069 mol [H 3 P0 4 ] [H?PO 0.055 . The appropriate relationship - 4 - ] = 0. 1 25 L =
=
to use is then Concentration initial
0.055
0.055
0
change
-x
+x
+x
final
0.055 - x
0.055 + x
+x
Ka
'
=
[H 2 P0 4 - ] [H 3 0+ ] [H3P0 4 ]
Because the equilibrium constant is not small compared to 0.055, the full quadratic solution must be calculated: x 2 + 0.055x 7.6 0-3 (0.055 - x)
xl x 2 + 0.063x - 4.2x 1 0-4 =
=
0 SM -344
x = 1 .6 X 1 0-3 pH = -log( 1 .6 x 1 0-3 ) = 2.80 (c) The reaction of Na 2 HP0 4 with strong acid goes only halfway to completion. 0.275 mol of HPO/- will react with (0.025 L x 0.275 mol · L-1) = 6.9 x 1 0-3 mol HCI to produce 6.9 x l 0-3 mol H 2 PO 4 - and leave 6.9 x l 0-3 HPO 4 2- unreacted. 6.9 x l 0-3 mol -7- 0.075 L 0.092 mol · L-1 =
Concentration (mol . Cl) intial
0.092
0.092
o
change
-x
+x
+x
final
0.092 - x
0.092 + x
+x
Ka 2
= 6.2 x l 0-8 =
[HPO/- ] [H 3 0+ ] [H 2 P0 4 - ]
[0.092 + X] [H 3 0+ ] = 6.2 X 1 0-8 [0.092 - x] assuming x « than 0.092 x = [H 3 0+ ]
=
6.2 X 1 0-8
pH = -log(6.2 x l 0-8 ) = 7.2 1
11.51
(a) The solubility equilibrium is AgBr(s) � Ag + (aq) + Br- (aq). [Ag+ ] = [Br- ] = 8.8 x 1 0- 7 mol · L-1 = S = solubility Ksp
=
[Ag+ ] [Br- ] = (8. 8 x 1 0-7 ) (8.8 x 1 0-7 )
=
7.7 x 1 0- 1 3
(b) The solubility equilibrium is PbCr0 4 (s) � Pb 2 + (aq) + Cr0 4 2 -(aq). [Pb 2 + ] = 1 .3 X 1 0-7 mol · L-1 = S, [Cr0 4 2 - ] = 1 .3 x 1 0-7 mol · L-1 = S Ksp = [Pb 2 + ] [Cr0 4 2 - ] = (1 .3 x 1 0-7 ) (1 .3 x 1 0-7 ) = 1 . 7 x 1 0-1 4 (c) The solubility equilibrium is Ba(OH) 2 (s) � Ba 2 + (aq) + 2 OH- (aq). SM -345
[Ba2+ ] = 0. 1 1 mol · L-1 = s,
[OH- ] = 0.22 mol · L-1 = 2S
= [Ba 2 + ] [OH- ] 2 = (0. 1 1) (0.22)2 = 5.3 X 1 0-3
Ks p
(d) The solubility equilibrium is MgF2 (S) [Mg 2+ ] = 1 .2 x 1 0-3 mol · CI = S, Ksp
11.53
�
Mg 2 + (aq) + 2 F- (aq).
[F- ] = 2.4 x l O-3 mol · L-1 = 2S
= [Mg2+ ] [F- f = (1 .2 x 1 0-3 ) (2.4 x l 0-3 ) 2 = 6.9 x l 0-9
(a) Equilibrium equation: BiI3 (s) 0 Bi 3 +(aq) 9 [Bi][C ] 3 (S)(3Si Ksp 7.71 x 1 0- 1 S 1 .3 0 x 1 0-5 mol · L- 1
3 C (aq)
=
=
=
+
=
(b) Equilibrium equation: CuCI(s) Ksp 1 .0 x 1 0- 6 [Cu+] [CI] S S 1 .0 x 1 0-3 mol · L- 1 =
=
0
Cu+ (aq)
+
CC (aq)
=
=
(c) Equilibrium equation: CaC0 3 (s) � Ca 2 + (aq) + cot (aq) Ks p = [Ca2+ ] [C0 2 -] = S x S = S 2 = 8.7 X 1 0-9 3 S = 9.3 x 1 0-5 mol · L-1
11.55
TI Cr0 4 (s) � 2 TI + (aq) + CrO/- (aq) 2 [Cr0 4 2 -] = S = 6.3 X 1 0-5 mol · L-1 [TI+ ] = 2S 2(6.3 x 1 0-5 ) mol · C l Ksp = [Ttf [Cr0 4 2 - ] = (2S) 2 x eS) =
K sp
11.57
= [2(6.3 x 1 O-5 )f x (6.3 x 1 0-5 ) = l .0 x 1 0-12
(a) Concentration (mol · L-1 ) AgCI(s)
�
Ag + (aq)
+
Cl- (aq)
initial
0
0.20
change
+S
+S
equilibrium
S
S + 0.20
Ks p
= [Ag+ ] [C I - ] = (S) x (S + 0.20) = 1 .6 x l 0-1 0
Assume S in (S + 0.20) is negligible, so 0.20 S = 1 .6 x 1 0-1 0 SM-346
S = 8.0 x I 0-10 mol · L- I
=
[Ag+ ]
=
molar solubility of AgCI in
0. 1 5 M NaCI Hg/+ (aq) + 2 cr (aq)
�
(b) Concentration (mol · L-I ) Hg Cl (s) 2 2 initial change
o
0. 1 50
+S
+2S
S
0. 1 50+ 2S
equilibrium
Ksp [Hg 2 +] [CI- ] 2 (S) x (2S + 0. 1 50) 2 2.6 x 1 0-18 2 Assume 2S in (2S + 0. 1 50) is negligible, so (0. 1 50iS = 2.6 x 1 0-18 • =
=
=
S = 1 .2 x 1 0- 1 6 mol · L-I = [Hg 2 2+ ] = molar solubility of Hg 2 Cl in 2 0. 1 50 M NaCl. (c) Concentration ( mol · L-I ) PbCI (s) 2 initial
�
Pb 2 + (aq)
change equilibrium Ksp
=
[Pb 2+ ] [CI- ] 2
=
S x (2S + 0.05) 2
=
+
2 Cr (aq)
o
2 x 0.025 = 0.05
+S
+S
S
S+ 0.05
1 .6 X 1 0-5
S may not be negligible relative to 0.05, so the full cubic form may be required. We do it both ways: For S3 + 0.20 S 2 + (0.0025 x 1 0- 2 S) - (1 .6 x 1 0-5 ) = 0, the solution by standard methods is S 4.6 x 1 0-3 mol · L- I =
•
If S had been neglected, the answer would be =6.4 x 1 0-3 . (d) Concentration (mol · L-I ) Fe(OH) 2 (s) initial
�
change equilibrium Ksp
=
S
Fe 2 + (aq) + 2 OH- (aq) 2.5 X 1 0-3
o
+S
+2S
2.5 x l 0-3 + S
2S
[Fe 2 + ][OH-f = (S + 2.5 x l 0-3 ) x (2S) 2 = 1 .6 x l 0-1 4
Assume S in (S+ 2.5 x 1 0-3 ) is negligible, so 4S 2 x (2.5 X 1 0-3 ) = 1 .6 X 1 0-1 . SM -347
-
S 2 1 .6 X 1 0-1 2 S = 1 .3 x 1 0-6 mol · Cl molar solubility of Fe(OH) 2 in 2.5 x l 0 3 M FeC1 2 . =
11.59
=
(a) Ag + (aq) + cr (aq) � AgCI(s) Ag+
Cl-
initial
0
1 .0 x 1 0-5
change
+x
0
equilibrium
x
1 .0 x 1 0-5
Concentration (mol · L-')
Ksp
[Ag+ ] [Cr ]
=
=
1 .6 x 1 0-10
=
x [Ag + ] 1 .6 X 1 0-5 mol· L-' =
=
(b) mass AgN0 3 = =
11.61
(x) ( 1 .0 x 1 0-5 )
(
(
J
1 .6 X 1 0-5 mol AgN0 3 1 69.88 g AgN0 3 (0. 1 00 L) 1 mol AgN0 3 1L 2.7 x 1 0 2 Jlg AgN0 3
J
(�J
(a) Ni 2+ (aq) + 2 OH- (aq) � Ni(OH) 2 (s) Concentration (mol · L- ' ) 0.060
initial
o
change Ksp
equilibrium [Ni 2 + ] [OH- ] 2
=
0.060 =
0 +x x
6.5 X 1 0- 18 (0.060) (X) 2 =
[OH- ] = x = 1 .0 x 1 0-8 mol · L- ' pOH -log( 1 .0 x 1 0-8) = 8.00, pH = 1 4.00 - 8.00 6.00 =
(b) A similar setup for [Ni 2 + ] 0.030 M. =
gives x 1 .5 X 1 0-8 pOH -log(1 .5 x 1 0-8) = 7.82 pH = 1 4.00 - 7.82 6. 1 8 =
=
=
SM-348
=
1 0-6 g
11.63
(
)
1 mL x 1 drop 20 drops
and (S x 1 0- 5
=
O.OS mL = O.OS X 1 0-3
L = S X 1 0-5 L
L) (0.0 1 0 mol · L- I ) = 5 X 1 0-7 mol NaCI = 5 x 1 0-7 mol CI-
(a) Ag+ (aq) + cr (aq) � AgCI(s), [Ag+ ] [CI- ] = Ksp Qs p =
[
(0.0 1 0 LHO.0040 mol . L-I) 0.0 1 0 L
][
] L
S x 1 0-7 m O I =2X l O-7 0.0 1 0
(b) Pb 2 + (aq) + 2 Cl- (aq) � PbCI 2 (s), [Pb 2 + ] [CI- ] 2 = Ks p
[
(0. 0 1 00 LHO.0040 mol . L-I) Qsp = 0.0 1 0
L
][
] L
5 x l O-7 mol 0.0 1 0
2 = l x 1 0-11
Will not precipitate because Qs p (1 x 1 0-1 1 ) < Ksp (1 .6 X 1 0-5 ).
This is the order for the solubility products of these hydroxides. Thus, the order of precipitation is (first to last) Ni(OH) 2 ' Mg(OH) 2 ' Ca(OH) 2 . (b) Ksp [Ni(OH) 2 ] 6.S x 1 0-1 8 = [Ni 2 + ] [OH- f =
6.5 x 1 0-1 8 = 6.5 x 1 0-15 0.00 1 0 [OH- ] = 8. 1 x 1 0-8 [OH- ] 2 =
pOH -log[OH- ] 7.09 pH 7 Ks p [Mg(OH) 2 ] 1 . 1 X 1 0-1 1 [Mg 2 + ] [OH- f 1 . 1 X 1 0-1 1 [OH - ] = = 1 .0 X 1 0-4 0.00 1 0 pOH = -log( 1 .0 x 1 0-4 ) = 4.00 pH = 1 4.00 - 4.00 = 1 0.00, pH Ks p [Ca(OH) 2 ] 5.5 x 1 0-6 = [Ca 2 + ] [OH- f =
=
=
=
�
�
10
=
5 .S X 1 0-6 = 7.4 X 1 0- 2 0.00 1 0 pOH = -log(7.4 x 1 0- 2 ) = 1 . 1 3 pH = 1 4.00 - l . 1 3 = 1 2.87, pH
[OH- ] =
SM-349
�
13
1 1.67
The Ksp values are: MgF2
6.4 x 1 0-9
BaF2
1 . 7 x 1 0-6 1 .0 x 1 0- 5
MgC03
8 . 1 x 1 0-9
BaC03
The difference in these numbers suggests that there is a greater solubility difference between the carbonates, and thus this anion should give a better separation. Because different numbers of ions are involved, it is instructive to convert the Ksp values into molar solubility. For the fluorides the reaction is: MF2 (s) � M 2 + (aq) + 2 F- (aq) Change Ks p
=
+X
+2x
X(2X) 2
Solving this for MgF2 gives 0.00 1 2 M and for BaF2 gives 0.0075 M . For the carbonates: MC0 3 (s) � M 2 + (aq) + cot (aq) +x +X
Solving this for MgC03 gives 0.0032 M and for BaC03 gives 9.0 x l 0-5 M. Clearly, the solubility difference is greatest between the two carbonates, and C032 - is the better choice of anion. 1 1 .69
Cu(I03) 2 (Ksp Pb(I0 3 ) 2 (Ksp
=
=
1 .4 x 1 0-7 ) is more soluble than 2.6 x 1 0-1 3), so Cu 2 + will remain in solution until
essentially all the Pb(I03) 2 has precipitated. Thus, we expect very little Pb 2 + to be the left in solution by the time we reach the point at which Cu(I0 J ) 2 begins to precipitate.
SM -350
The concentration of 10 3 - at which Cu 2 + begins to precipitate will be Ksp = [Cu 2 + ] [10 -f = I A x 1 0-7 = [0.00 1 0] [10 - ] 2 . 3 3 gIVen by [10 3 - ] = 0.0 1 2 mol · L-I . The concentration of Pb in solution when the [10 3 - ] = 0.0 1 2 mol · L- I is given by = [Pb 2 + ] [10 3 - f = 2.6 x l 0-1 3 = [Pb 2+ ] [0.0 1 2] 2 [Pb 2 + ] =1 .8 x 1 0-9 mol · L- I • Ksp
11.71
(a) pH = 7.0; [OH- ] = 1 .0 x 1 0-7 mol · L- I A1 3+ (aq) + 3 OH- (aq) � AI(OH) 3 (s) [AI 3 + ] [OH - ] 3 = Ksp = 1 .0 X 1 0-33 S X (1 0-7 )3 = 1 .0 X 1 0-33 1 .0 X 1 0- 33 = 1 .0 X 1 0-1 2 mo l . L-I = [AI 3+ ] S= 1 X 1 0-2 1 = molar solubility of AI(OH) 3 at pH = 7.0
(b) pH = 4.5; pOH = 9.5; [OH- ] = 3.2 x 1 0- 10 mol · C l [Ae + ] [OH- ] 3 = Ksp = 1 .0 X 1 0-33 S x (3 .2 X 1 0-10 ) 3 = 1 .0 X 1 0-33 LO x 1 0-33 = 3 . 1 x 1 0-5 mol . L-1 = [AI 3 + ] S= 3 .3 X 1 0-29 = molar solubility of AI(OH) 3 at pH = 4.5 (c) pH =7.0; [OH- ] = 1 .0 x 1 0-7 mol · L-1 Zn 2+ (aq) + 2 OH- (aq) � Zn(OH) 2 (s) [Zn 2 + ] [OH- f Ks 2.0 X 1 0-17 =
p
=
S x (1 .0 X 1 0-7 ) 2 = 2.0 X 1 0-17 2.0 x 1 0-1 7 = 2.0 x I 0- 3 mo l · CI = [Zn 2 + ] S= 1 .0 x 1 0-1 4 = molar solubility of Zn(OH) 2 at pH = 7.0 (d) pH = 6.0; pOH = 8.0; [OH- ] = 1 .0 x 1 0-8 mol · L-I
SM-3 5 1
[Zn 2 + ] [OH- ] 2 = 2.0 X 1 0-17 = Ksp S X (1 .0 X 1 0-8 ) 2 = 2.0 X 1 0- 1 7 2.0 x 1 0-17 = 2. 0 x 1 0-1 = 0.2 0 mo l · L-1 = [Zn 2 + ] X 1 0-16 1 .0 = molar solubility of Zn(OH) at pH = 6.0 2
s=
Ksp
= 4 . 0 X 1 0- 1 1
F- (aq) + H 2 0(l) � HF(aq) + OH- (aq) (a) Multiply the second equilibrium equation by 2 and add to the first equilibrium: CaF (s) + 2 H 0(l) � Ca 2 + (aq) + 2 HF(aq) + 2 OH- (aq) 2 2 2 K = K w . Kb = (4.0 X 1 0-1 1 ) (2.9 X 1 0-1 1 ) 2 = 3 .4 X 1 0- 3 2 (b) The calculation of Ksp is complicated by the fact that the anion of the salt is part of a weak base-acid pair. If we wish to solve the equation algebraically, then we need to consider which equilibrium is the dominant one at pH 7.0, for which [H 3 0+ ] = 1 x 1 0-7 . =
To determine whether F- or HF is the dominant species at this pH (if either), consider the base hydrolysis reaction: F- (aq) + H 2 0(l) � HF(aq) + OH- (aq) Kb
=
K b = 2.9 X 1 0-1 1
[HF] [OH- ] [F- ]
[HF] [ l x 1 0-7 ] [F- ] [HF] 2.9 x 1 0- 1 1 = 3 X 10-4 = [F- ] 1 x 1 0-7
2.9 x 1 0- 1 1 =
Given that the ratio ofHF to F- is on the order of 1 0-4 to 1 , the F- species is still dominant. The appropriate equation to use is thus the original one for the Ksp of CaF2 (s) :
SM-352
CaF2 (s) � Ca 2+ (aq) + F- (aq) Ksp [Ca 2 + ] [F-f 4.0 x l 0-11 X) 2 4X3 X = 2.0 X 1 0-4
2
=
=
x(2
molar solubility
Ksp
=
4.0 x l 0 - 1 1
=
=
2.0 x 1 0-4 mol · L-I
(c) At pH = 3 .0, [H 3 0+ ] = 1 x 1 0-3 mol · L- I and [OH- ] = 1 x 1 0- 1 1 mol · L-I • Under these conditions, [HF] [OH- ] [F- ] [HF] [ l x 1 0-1 1 ] x 1 0-1 1 [F- ] [HF] 1 0-1 1 = 3 [F- ] 1 x 1 0-1 1 [HF] 3 [F- ]. Kb
=
2.9
=
=
2.9
x
=
As can be seen, at pH 3 .0 the amounts of F- and HF are comparable, so the protonation of F- to form HF cannot be ignored. The relation
2 [Ca2+ ]
=
[F- ] + [HF] is required by the mass balance as imposed by the
stoichiometry of the dissolution equilibrium : 2 [Ca2 + ] = [F- ] + 3 [F- ] [Ca2+ ] 4 [F- ] rCa 2 + 1 = 2 [F- ]
2
=
Using this with KiP relationship:
(2 [F- ]) [F- f 4.0 1 0-1 1 2 [F- ]3 4.0 x4 1 0-1 1 [F- ] 2 x 10rCa 2+ (2)(2 x 1 0-4 ) 4 x 1 O-4 mol · L-I =
X
=
=
1=
=
The solubility is about double that at pH
SM-353
=
7.0.
11.75
AgBr(s) � Ag+ (aq)+ Br- (aq)
Ksp
Ag+ (aq)+ 2 CN- (aq) � Ag(CN) 2 -(aq) AgBr(s) + 2 CN- (aq) � Ag(CN) 2 -(aq) + Br- (aq)
Kf
Hence, K =
=
7.7
X
1 0- 1 3
= 5.6 X 1 08 K = 4.3 X 1 0-4
[Ag(CN) 2 -] [Br- ] = 4.3 x 1 0-4 [CN- f
Concentration initial change equilibrium
0. 1 0 - 2S 0. 1 0 - 2S
o
o
+S
+S
S
S
[Ag(CN) 2 - ] [Br- ] S2 = 4.3 x 1 0-4 = 2 (0. 1 0 - 2S) [CN- f S = �4.3 x l O-4 = 2. 1 x l O-2 0. l 0 - 2S S = (2. 1 x 1 0-3 ) - (4.2 x 1 0-2 S) 1 .042S = 2. 1 x 1 0-3 S = 2.0 x 1 0-3 mol · L-1 = molar solubility of AgBr 11.77
The two salts can be distinguished by their solubility in NH 3 . The equilibria that are pertinent are: AgCI(s)+ 2 NH 3 (aq) � Ag(NH 3 ) 2 + (aq)+ CI- (aq) K = Ksp . Kf = 2.6 X 1 0-3
K
= Ksp · Kf = 2.4 x 1 0-9
For example, let's consider the solubility of these two salts in 1 .00 M NH 3 solution:
SM-354
Concentration -'�
(mol · L-I ) AgCl(s) + 2 HN 3 (aq) initial
1 .00
change
-2 x
final
1 .00
-
Ag(NH 3 ) + (aq) + cr (aq) 2 0 0
2x
+x
+x
+x
+x
[A g(NH 3 ) 2 + ] [Cl- ] = 2.6 x 1 0-3 [NH 3 f [x] [x] x2 2.6 x 1 0-3 = = [1 .00 - 2xf [1 .00 - 2 X] 2 x 0.05 1 = 1 .00 - 2x x 0.046
K=
---
=
0.046 mol AgCI will dissolve in 1 .00 L of aqueous solution. The molar mass of AgCl is 1 43 .32 g . mol-I ; this corresponds to 0.046 mol · L-I x 1 42.32 g . mor l
=
6.5 g . L- I .
For AgI, the same calculation gives x 4.9 x 1 0-5 mol · L-I . The molar =
mass of AgI is 234.77 g . mor l , giving a solubility of 4.9 x 1 0-5 mol · L- I x 234.77 g . mol-I
=
0.023 g . L-I .
Thus, we could treat a 0. 1 0 g sample of the compound with 20.0 mL of 1 .00 M NH 3 . The AgCI would all dissolve, whereas practically none of the AgI would. Note: AgI is also slightly yellow in color, whereas AgCI is white, so an initial distinction could be made based on the color of the sample. 11.79
In order to use qualitative analyses, the sample must first be dissolved. This can be accomplished by digesting the sample with concentrated HN0 3 and then diluting the resulting solution. HCI or H 2 S0 4 could not be used because some of the metal compounds formed would be insoluble, whereas all of the nitrates would dissolve. Once the sample is dissolved and diluted, an aqueous solution containing chloride ions can be SM-3 55
introduced. This should precipitate the Ag+ as AgCI but would leave the bismuth and nickel in solution, as long as the solution was acidic. The remaining solution can then be treated with H 2 S . In acidic solution, Bi 2 S ) will precipitate but NiS will not. Once the Bi 2 S) has been precipitated, the pH of the solution can be raised by addition of base. Once this is done, NiS should precipitate.
11.81
The relation to use is pH = pKa + log acetic acid is 1 .8 x 1 0-5 and pKa
=
[Base form] . The K value for [Acid form] a
4.74. Because we are adding acid, the
pH will fall upon the addition, and we want the final pH to be no more than 0.20 pH units different from the initial pH, or 4.54. [Base form] [Acid form] [Base form] 0.20 = log [Acid form] [Base form] = 0.63 [Acid form]
4.54 = 4.74 + log _
We want the concentration of the base form to be 0.63 times that of the acid form. We do not know the initial number of moles of base or acid forms, but we know that the two amounts were equal. Let C initial =
number of moles of acetic acid and the initial number of moles of sodium acetate. The number of moles of H ) O+ to be added (in the form of HCI(aq)) is 0.00 1 00 L x 6.00 mol · L-1 volume will be 0. 1 0 1 0 L.
SM-356
=
0.006 00 mol. The total final
C - 0.006 00 O. l O I O L C + 0.006 00
=
0.63
O. I O I O L
C - 0.006 00 C + 0.006 00
=
0.63
C - 0.006 00 = 0.63 ( C + 0.006 00)
C - 0.006
00 = 0.63 C + 0.003 78 0.37 C 0.009 78 C 0.026 =
=
The initial buffer solution must contain at least 0.026 mol acetic acid and 0.026 mol sodium acetate. The concentration of the initial solution will then be 0.026 mol 7 0. 1 00 L
=
0.260 M in both acetic acid and sodium
acetate. 11.83
(a) Ma
=
0.0567, Mb 0.0296, =
Va
=
Vb
1 5 .0,
=
0.0 to 50.0
1 2.0
=§..
6 . 62
1 . 24
L==t====:JLL----_ -1 25 . 0
o
----_ -1
_
50.0
(b) 28.6 mL (c) 1 .24 (d) Because this is a titration of a strong acid with a strong base, the pH at the equivalence point will be 7.00. 11.85
The strong acid, HCI, will protonate the
HC0 2 - ion:
moles of HCI 0.0040 L x 0.070 mol · L- 1 2.8 x 1 0-4 mol HCI (H + ) moles of HC0 2 - 0.0600 L x 0. 1 0 mol · L-1 6.0 x 1 0-3 mol HC0 2 =
=
=
=
SM-357
After protonation, there are (6.0 - 0.28) x 1 0-3 mol HC0 2 -
=
5.7 X 1 0-3 mol HC0 2 -
and 2.8 x l 0-4 mol HCOOH: 5.7 x l 0 -3 mol 2 = 8.9 x l 0- mol · L-1 [HCO 2 - ] = 0.0640 L 2.8 x 1 0-4 mol = 4.4 x l 0- 3 mol . L-1 [HCOOH] 0.0640 L =
Concentration
initial
4.4 x 1 0-3
change
-x
equilibrium
4.4 x 1 0-3 - X
Ka
=
1 .8 X 1 0-4
=
8.9 X 1 0-2
o
+x
+x 8.9 X 1 0-2 + X
x
[H 3 0+ ] [HC0 2 - ] [HCOOH]
=
1 0-2 + x) (4.4 x l 0-3 - X)
(x) (8.9 X
· · bl e; so 1 .8 x l 0-4 Assume th at +x an d -x are neg1 Igl
= =
(8.9 x 1 0-2 ) x (x) 4.4 x 1 0-3 20.2x.
[H 3 0+ ] = X = 8.9 x 1 0-6 mol · L-1 [HCOOH] (4.4 x 1 0-3 ) - (8.9 x 1 0-6 ) 4.4 x 1 0-3 mol · L-1 pH - log(8.9 x 1 0-6 ) = 5.05 =
�
=
11.87
Let novocaine
=
N; N(aq)+ H 2 0(l) � HN+ (aq) + OH- (aq).
[HN+ ] [OH- ] [N] pKa = pKw - pKb 1 4.00 - 5.05 8.95 Kb
=
( : =
pH 10g
=
pKa + log
( J [N] [HN+ ]
=
=
[N]
[HN+ ]
pH - pKa
=
7.4 - 8.95 - 1 .55 =
SM-358
Therefore, the ratio of the concentrations of novocaine and its conjugate acid is [N]/[HN+ ] = 1 0-155 11.89
=
2.8 1 0-2. X
In addition to the reaction corresponding to the dissolution of PbF2 (s): K = 3 .7x l O- 8 The buffer will provide a source of H 3 0\aq) ions, which will allow the reaction
These two coupled reactions give two equilibrium expressions that must be simultaneously satisfied: and Given that all fluoride ions come from PbF2 (s) and wind up as either F-(aq) or HF(aq), and that for every 1 mole of Pb2+(aq) generated 2 moles of F-(aq) are also produced, we can write a third equation that relates the concentration of the fluoride-containing species to the concentration of dissolved barium: [Pb2+] = '/2 ([F- ] + [HF ) )
In the end, the concentration of Pb2+ (aq) will be equal to the solubility of PbF2 (s). To determine the equilibrium concentration of Pb2+ (aq), we first determine [H 3 0+] , which is fixed by the buffer system, and then use the three simultaneous equations above to solve for [Pb2+]eq. The buffer determines the equilibrium concentration of H 3 0+ (aq). The initial concentrations of H30+(aq) and NaCH 3 C0 2 (aq) are . (0.055 L) (0. 1 5 mol . C1 ) = 0.0825 M and [H ,O+ ]. = '
.
[
_
[CH ,C02 l .'
[
=
0. 1 0 L (0.045 L)(0.65 mol . C1 ) 0.2 93 M . 0. 1 0 L =
SM-359
To detennine their equilibrium concentrations we solve using the familiar method: Concentration (mol · L- ' ) CH3COOH(aq) initial 0 change
H 2 0(l)
�
0.0825
x
= 1 .8 x l 0-5 =
H ) O+ (aq) 0.0825
+
-x
+x
equilibrium Ka
+
CH 3 C0 2 - (aq) 0.293 -x
-
x
0.292
-
x
[H 3 0+ ] [CH ) C0 2 - ] (0.0 825 - x) (0.292 - x) = [CH ) COOH] (x)
Rearranging this expression we obtain 0.024072 - 0.3745 x + x2 = O. Using the quadratic fonnula, we find x = 0.082493, and [H ) O+ ] = 7 . 1 x 1 0-6 M. With this equilibrium concentration ofH 3 0\aq), we revisit the three simultaneous equations from above, namely [F- f [Pb ] = 3 .7 x 1 0-8 ; [Pb 2 + ] = H [F- ] + [HF] ) ; and ) [HF] = 2.86 x l 0- . [H ) O+ ] [F- ]
Due to the presence of the buffer, [H ) O+ ] = 7 . 1 x l 0-6 , and this last equation simplifies to [HF] = 2.03 x 1 0-2 • [F- ] Rearranging these three simultaneous equations we find that [Pb 2 + ] =
;;_��-8 ,
3.
[HF] = [F- ] x ( 2.03 x l 0-2 ) ,
and
[Pb 2 + ] = + ( [F- ] + [HF] ) 3.
�;_;?-8
=
+[[F- ] ( [F- ] x ( 2.03 x 1 0-2 ))J. +
3 .7 x l 0-8 = 4. 1 4 x l 0-). [F- ] = ) 0.5203
SM - 360
Solving this expression for [F- ] :
The equilibrium concentration of Pb 2+ (aq) is then: 3 .7 x 1 0-8 3 .7 x 1 0-8 = 2 . 1 6 x 1 0-3 M = [Pb 2+ ] = 3 2 (4. 1 4 x 1 0 ) [F- f Therefore, the solubility of PbF2 (s) is 2.2 x 1 0-3 M 11.91
The Ksp values from Tables 1 1 .5 and 1 1 . 7 are Cu 2+ , 1 .3 X 1 0-36 ; C0 2+ , 5 X 1 0-22 ; Cd 2+ , 4 X 1 0-29. All the salts have the same expression for Ks p ' Ksp = [M 2 + ] [S 2 - ], so the compound with the smallest Ksp will precipitate first, in this case CuS.
11.93
( 1 ) First stoichiometric point: OH- (aq) + H 2 S0 4 (aq) � H 2 0(l) + HS0 4 - (aq) Then, because the volume of the solution has doubled, [HS0 4 - ] = 0. 1 0 M , Concentration (mol . L- J ) initial
0. 1 0
o
o
change
-x
+x
+x
0. 1 0 - x
x
x
equilibrium
x2 0. 1 0 - x 0.00 1 2 - 0.0 1 2x = x 2 x2 + 0.0 1 2x - 0.00 1 2 = 0 -0.0 12 + -'--� (0.0 1 2) 2(4) (0.00 1 2) + --x= 2 x = [H 3 0+ ] = 0.029 mol · L- J pH = - log(0.029) = 1 .54
Ka 2
= 0.01 2 =
---
(2) Second stoichiometric point: HS0 4 - (aq) + OH- (aq) � SO/- (aq) + H 2 0( l) . Because the volume of the solution has increased by an equal amount:
SM-3 6 1
Concentration 1 10-7 initial 0.067 change 1 x10-7 equilibrium 0.067 = 8.3 10-1 3 = 0.06710-7 + (5.2 6 x10-14 ) -(8.3 x10-1 3 =4 1 10-7 X + X2 X (1 10-7 -(5.6 x10-1 )=0 4 -lx10-7 +)(1x10-7) ---------2 2 +(4)(5.6 x10-1 ) [HS04 -] = 1.9 x1O-7mol· L-I [OH-] (1.0 10-7) (1.9 10-7) = 2.9 10-7mol· L-I pOH=-log(2.9 x10-7)=6.54, pH 14.00 -6.54 = 7.46 (a) S02(g) + H20(l) H2S03(aq) (b) (0.050 x 1.0 x 10-4) -(0.0302 x 1.5 x 10-4) 5 x 10-7 mol NaOH (c) H2S03 ) ( 1 mol S02 5 x 10-7 mol NaOH x( l2mol mol NaOH 1 mol H2S03 J 2.5 x 10-7 mol S02 Mass ofS02 = (2.5 x 10-7 mol) x(64.06 glmol) 1.6 x 10-5 g S02 Moles of air: PV=nRT, Torr )(3.0L . h-Ix2.5 h)=n(0.08206L . atm . mol-I . K-I)(295K) ( 760753Torr/at m n 0.31 mole air Mass of air (assuming 78% N2, 22% O2 to simplify the calculation): 0.31 x(0.78) x(28.0 g N2/mol N2) + 0.31 x(0.22) x (32.0 g 02/mole) 9.0 g Concentration of S02 in the air (in parts per million (ppm» : o
-x
+x
( x) ( l X
X
x)
-
x)
+
X
x
X
x)
�--
x=
-------------
x=
=
X
+
X
X
=
11.95
�
=
=
=
=
=
SM-362
+x
+x
x
-x
Kb
X
( 11.97
1 .6
X
1 0-5 9.0
)
X
(1 .0
X
1 0-6 )
=
1 .8 ppm
The Ksp value for PbF2 obtained from Table 1 1 .5 is 3 . 7 X 1 0 -8 . Using this value, the I1GO of the dissolution reaction can be obtained from I1GO =
R
In K. I1Go = -(8.3 1 4 J . K- l . mol- l ) (298.2 K) In (3 .7 x 1 0-8 ) I1Go = +42.43 kJ . morl From the Appendices we find that I1G o r (F- , aq) = -278.79 kJ . mol- l and I1GO r (Pb 2 + , aq) = -24.43 kJ . mol- I . I1GO = +42.43 kJ . mor l = I1Go r (Pb 2+ , aq) + I1Go r (F- , aq) - I1Go r (PbF2 , s) +42.43 kJ . mor l = ( -24.43 kJ . mol-I ) + (-278.79 kJ . mor l ) -
T
- I1Go r (PbF2 , s) I1Gor (PbF2 , s) = -345.65 kJ · mor l 11.99
(a) The amount of C02 present at equilibrium may be found using the equilibrium expression for the reaction of interest: H 3 0+ (aq) + HCO�
�
2 H 2 0(l) + C0 2 (aq) [C0 2 ] K = 7.9 X 1 0-7 = [H 3 0+ ] [HCO � ] Solving for [COJ : [C0 2 ] = (7.9 x 1 O-7 )[H 3 0+ ] [HCO � ] Given: [H 3 0+ ] = 1 0-6. 1 = 8 x l 0-7 M, and [HCO � ] = 5.5 jlmol · L- l = 5 .5 x l 0 -6 M [C0 2 ] = (7.9 x 1 0 -7 )(8 x 1 0-7 )(5.5 x 1 0-6 ) = 3 x 1 O- l 8 mol · L- l In 1 .0 L of solution there will be 3 x l 0- l 8 mol of CO 2 (aq). (b) Adding 0.65 x 1 0-6 mol of H30+ (aq) to the equilibrium system in (a) will give an initial [H30+ ] of 1 .44 x 1 0-6 M. To determine the equilibrium concentration of H30\aq) we set up the familiar problem: Concentration (mol . L- l ) initial
1 .44 x 1 0-6
5 . 5 x 1 0- 6
SM-363
3 .45 x 1 0- 1 8
-x
change
+x
-x
equilibrium 1 .44 x 1 0-6 - x 5.5 x 1 0- 6 - x K = 7.9 x l O-7 =
3.45 x l O- 1 8 +X
3.45 x l O-1 8 + x ( 1 .44 x l O-6 - x) ( 5.5 x l O-6 + x )
Rearranging to obtain a polynomial in x : 2.83 x 1 0-1 8 - x + 7.9 x 1 0-7 x2 0 Using the quadratic forumla one finds: x = 2.825 x 1 0-1 8 Giving: [H 3 0 + ] = 1 .4 x 1 0-6 and pH = 5.8, the same pH as the initial solution. =
11.101 (a)
pH = pKa + log ([base]/[acid]) = 4.75 + log(0.300/0.200) = 4.93
(b) 6.0 = 4.75 + log{(0.300 L x 0.300 M + x)/(0.3 00 L x 0.200 M - x)} = 4.75 + log{(0.0900 mol + x)/(0.0600 mol - x) } log {(0.0900 mol + x)/(0.0600 mol - x)} = 1 .25; x
= mol of OH- = 0.0520 mol = mole ofNaOH
mass ofNaOH = (0.0520 mol) x (40.0 1 g/mol) = 2.08 g NaOH 11.103
(a) Equilibrium equation: Fe(OH) 3 (S) B Fe3 + (aq) + 3 Ksp = 2.0 x 1 0-39 = [Fe3 + ] [OH-] 3 = (S) (3S) = 27(S)4
3
OH- (aq)
1 0- 1 1 mol ' L- 1 (the OH- from water is ignored) (b) [OH-] iron hydroxide = 3S = 2.8 x 1 0- 10 M; pOH = 9.55; pH =4.45
S=
9.3
X
(c) Results in (b) is not reasonable because the OH- from water is ignored. [OH-] water = 1 .0 1 0-7 M; Actually, [OH -] = 1 .0 x 1 0- 7 M in a saturated x
solution of Fe(OH) 3 because the [OH-] from Fe(OHh is much smaller than the [OH-] from water; pH = 7.0. ............ ( 1 ) (d) Fe(OH) 3 (S) B Fe3 +(aq) + 3 OH- (aq) ................ (2) 2 H2 0(I) B H 3 0+ (aq) + OH- (aq) 3 3 x (2) - ( 1 ): Fe + (aq) + 6 H 2 0(I) B 3 H 3 0+ (aq) + Fe(OH) 3 (S) K = (Kw ) 3 /Ksp = ( 1 .0 x 1 0- 1 4 ) 3 /(2.0 1 0-39 ) = 5 x 1 0-4 x
(e) 1 0.0 g of Fe2 (S04) 3 x ( 1 .0 mol/399.9g) = 0.025 mol Fe2 (S04) 3 Mol of Fe3 + = 0.025 x 2 = 0.050 mol; SM - 3 64
when pH is maintained at 8.0, [H 3 0+] 1 .0 1 0- 8 , [OH-] 1 .0 1 0-6 ; 3 (0.050)( 1 .0 1 0-6 ) 3 5.0 1 0-20 » Ksp, all of the 3 Q = [Fe +] [OH-] Fe3 + x
=
x
=
=
x
x
=
will be precipitated as long as pH is maintained. The moles of Fe(OH) 3 will be the same as that of Fe3 + (0.050 mol). Mass of Fe(OHh (f) Moles
=
0.050 mol ( 1 06.9 glmol) x
=
5.3 grams.
of [cq in 25.0 mL NaCl sample:
(24.72 g NaCl)
x
( 1 .0 mol NaCl/58.44 g) (25.00 mLl 1 000 mL)
( 1 mol crn mol NaCl)
x
=
0.01 057 mol cr
x
The moles of Cl- left in solution: (0.604 g AgCl) ( 1 .0 mole AgCl/ 1 43 .32 g) ( 1 mol Cn1 mol AgCl) 4.2 1 1 0-3 mol x
=
x
Percentage of Cl- removed: [(0.0 1 057 - 4.2 1 =
60. 1 %
SM -365
x
1 0-3 )/0.01 057]
x
1 00%
CHAPTER 12 ELECTROCHEMISTRY 12.1
(a) Cr reduced from 6+ to 3+; C oxidized from 2- to 1 -; (b) C 2 Hs OH(aq) � C 2 H 4 0(aq) + 2 H+ (aq) + 2 e- ; (c) Cr2 0/- (aq) + 1 4 H+ (aq) + 6 e(d)
12.3
�
2 Cr 3+ (aq) + 7 H 2 0(l) ;
8 H+ (aq) + Cr2 0t (aq) + 3 C 2 Hs OH(aq) � 2 Cr 3+ (aq) + 3 C 2 H 4 0(aq) + 7 H 2 0(l)
In each case, first obtain the balanced half-reactions. Multiply the oxidation and reduction half-reactions by appropriate factors that will result in the same number of electrons being present in both half-reactions. Then add the half-reactions, canceling electrons in the process, to obtain the balanced equation for the whole reaction. Check to see that the final equation is balanced. (a) 4[C1 2 (g) + 2 e- � 2 c r (aq)] I [S 2 0 3 2 - (aq) + 5 H 2 0(1) � 2 S0 4 2 - (aq) + 1 0 H+ (aq) + 8 e- ] 4 C1 2 (g) + S 2 03 2 -(aq) + 5 H 2 0(1) + 8 e - � 8 Cl- (aq) + 2 S0 4 2 -(aq) + 1 0 H+ (aq) + 8 e4 CI 2 (g) + S 2 0 3 2 -(aq) + 5 H 2 0(1) � 8 Cl- (aq) + 2 S0 4 2 - (aq) + 1 0 H+ (aq) C1 2 is the oxidizing agent and S 2 0/- is the reducing agent. (b) 2[Mn0 4 - (aq) + 8 H+ (aq) + 5 e- � Mn 2 + (aq) + 4 H 2 0(l)] 5 [H 2 S0 3 (aq) + H 2 0(1) � HS0 4 -(aq) + 3 H+ (aq) + 2 e- ] 2 Mn0 4 -(aq) + 1 6 H+ (aq) + 5 H 2 S0 3 (aq) + 5 H 2 0(1) + 1 0 e- � 2 Mn 2 + (aq) + 8 H 2 0(1) + 5 HS0 4 -(aq) + 1 5 H+ (aq) + 1 0 e2 Mn0 4 - (aq) + H+ (aq) + 5 H 2 S0 3 (aq) � 2 Mn 2 + (aq) + 3 H 2 0(1) + 5 HS0 4 - (aq) MnO 4 - is the oxidizing agent and H 2 S0 3 is the reducing agent.
SM-366
H 2 S (aq) � S(S) + 2 H+ (aq) + 2 eCl 2 (g) + H 2 S (aq) + 2 e- � 2 C I - (aq) + S(s) + 2 H+ (aq) + 2 e Cl 2 (g) + H 2 S (aq) � 2 CI - (aq) + S(s) + 2 H+ (aq) Cl 2 is the oxidizing agent and H 2 S is the reducing agent.
2 H 2 0(l) + Cl 2 (g) � 2 HOCI(aq) + 2 H + (aq) + 2 e2 H 2 0(l) + 2 Cl 2 (g) + 2 e- � 2 HOCI(aq) + 2 H + (aq) + 2 Cl- (aq) + 2 eor H 2 0(l) + Cl 2 (g) � HOCI(aq) + H+ (aq) + CI- (aq) Cl 2 is both the oxidizing and the reducing agent. 12.5
(a) 0 3 (g) � 0 2 (g) 0 3 (g) � 0 2 (g) + H 2 0(I)
(balances O's) (balances H's) (cancels H 2 0 )
H 2 0(I) + 0 3 (g) + 2 e- � 0 2 (g) + 2 OH - (aq)
(balances charge);
Br - (aq) � Br0 3 - (aq) (balances O's) 6 OH - (aq) + 3 H 2 0(l) + Br- (aq) � Br0 3 - (aq) + 6 H 2 0(l) (balances H's) 6 OH- (aq) + 3 H 2 0(l) + Br- (aq) � Br0 3 - (aq) + 6 H 2 0(l) + 6 e (balances charge) Combining half-reactions yields 3[H 2 0(I) + 0 3 (g) + 2 e- � 0 2 (g) + 2 OH- (aq)] 6 OH- (aq) + 3 H 2 0(l) + Br- (aq) � Br0 3 - (aq) + 6 H 2 0(l) + 6 e6 H 2 0(l) + 3 0 3 (g) + 6 OH- (aq) + Br- (aq) + 6 e - � 3 0 2 (g) + 6 OH- (aq) + Br0 3 - (aq) + 6 H 2 0(l) + 6 e and 3 0 3 (g) + Br- (aq) � 3 ° 2 (g) + BrO ) - (aq)
SM-367
0 ) is the oxidizing agent and Br- is the reducing agent. (b) Br2 (1) + 2 e- � 2 Br- (aq)
(balanced reduction half-reaction)
Br2 (l) + 6 H 2 0(l) � 2 BrO ) - (aq)
(O's balanced); then
Br2 (1) + 6 H 2 0(1) + 1 2 OH- (aq) � 2 BrO ) -(aq) + 1 2 H 2 0(I) (H' s balanced); and Br2 (1) + 1 2 OH- (aq) � 2 BrO ) - (aq) + 6 H 2 0(1) + 1 0 e- (electrons balanced) Combining half-reactions yields 5[Br2 (1) + 2 e- � 2 Br- (aq)] I [Br2 (1) + 1 2 OH- (aq) � 2 BrO ) -(aq) + 6 H 2 0(1) + 1 0 e- ] 6 Br2 (l) + 1 2 OH- (aq) + 1 0 e- � 1 0 Br- (aq) + 2 BrO ) -(aq) + 6 H 2 0(1) + 1 0 e6 Br2 (l) + 1 2 OH- (aq) � 1 0 Br- (aq) + 2 BrO) -(aq) + 6 H 2 0(1) Dividing by 2 gives
Br2 is both the oxidizing agent and the reducing agent. (O's balanced); then
balanced); and Cr 3+ (aq) + 8 OH- (aq) � CrO/- (aq) + 4 H 2 0(l) + 3 e
(charge
balanced) Mn0 2 (s) � Mn 2 + (aq) + 2H 2 0(1); then Mn0 2 (s) + 4 H 2 0(1) � Mn 2+ (aq) + 2 H 2 0(I) + 4 OH- (aq)
(H' s
balanced); and Mn0 2 (s) + 2 H 2 0(l) + 2 e- � Mn 2+ (aq) + 4 OH- (aq) balanced) Combining half-reactions yields
SM-368
(charge
2[Cr 3 + (aq) + 8 OH- (aq) � CrO/- (aq) + 4 H 2 0(l) + 3 e- ] 3[Mn0 2 (s) + 2 H 2 0(l) + 2 e- � Mn 2 + (aq) + 4 OH- (aq)] 2 Cr3+ (aq) + 1 6 OH- (aq) + 3 Mn0 2 (s) + 6 H 2 0(l) + 6 e- � 2 Cr0 4 2 - (aq) + 8 H 2 0(l) + 3 Mn 2 + (aq) + 1 2 OH- (aq) + 6 e2 Cr3+ (aq) + 4 OH- (aq) + 3 Mn0 2 (s) � 2 CrO/- (aq) + 2 H 2 0(l) + 3 Mn 2 + (aq) Cr3+ is the reducing agent and Mn0 2 is the oxidizing agent. (d) 3[P4 (S) + 8 OH- (aq) � 4 H 2 P0 2 - (aq) + 4 e- ] P4 (s) + 1 2 H 2 0(l) + 1 2 e- � 4 PH 3 (g) + 1 2 OH- (aq) 4 P4 (s) + 1 2 H 2 0(l) + 24 OH- (aq) + 1 2 e - � 1 2 H 2 P0 2 - (aq) + 4 PH 3 (g) + 1 2 OH- (aq) + 1 2 e4 P4 (s) + 1 2 H 2 0(l) + 1 2 OH- (aq) � 1 2 H 2 P0 2 - (aq) + 4 PH 3 (g) or P4 (s) + 3 H 2 0(l) + 3 OH - (aq) � 3 H 2 P0 2 - (aq) + PH 3 (g) P4 (s) is both the oxidizing and the reducing agent.
For the oxidation of P4 S3 ' both the P and S atoms are oxidized. The assignment of oxidation states to the P and S atoms is complicated by the presence of P-P bonds in the molecule, which leads to non-integral values. As long as we are consistent in our assignments, the end result should be the same. We will assume that S in P4 S 3 is 2 - and, therefore, loses 8 electrons on going to S 6 + in the sulfate ion. Because P4 S 3 is a neutral molecule and, if S has an oxidation number of -2, then each phosphorus atom will have an oxidation number of +1 .5. Phosphorus in phosphoric acid has an oxidation number of+5 . so each P atom of P4 S 3 must lose 3.5 electrons. The total number of electrons lost is (4 x 3 .5) + (3 x 8)
=
38.
We balance the charge by adding H+ in an acidic solution: SM -369
The final balance is achieved by adding water to provide the oxygen and hydrogen atoms: P4 S ) (aq) + 28 H 2 0(l) � 4 H ) P0 4 (aq) + 3 S0 4 2 -(aq) + 44 H+ (aq) + 3 8 eThe other half-reaction is simpler. NO ) -(aq) � NO(g) N has an oxidation number of +5 in the nitrate ion and +2 in nitric oxide. Each nitrogen atom gains three electrons in the course of the reaction.
Charge balance is again achieved by adding H+ : NO ) - (aq) + 4 H+ (aq) + 3 e-
�
NO(g)
The number of hydrogen and oxygen atoms is completed by the addition of water: Combining the two half-reactions gives 3 8 [NO ) -(aq) + 4 H+ (aq) + 3 e- � NO(g) + 2 H 2 0(l)] +3 [P4 S ) (aq) + 28 H 2 0(l) � 4 H ) P0 4 (aq) + 3 S0 4 2-(aq) + 44 H+ (aq) + 3 8 e- ] 3 P4 S ) (aq) + 3 8 NO) -(aq) + 20 H+ (aq) + 8 H 2 0(l) � 1 2 H ) P04 (aq) + 9 S0 4 2- (aq) + 38 NO(g) 12.9
(a) Ag+ (aq) + e- � Ag(s) Ni 2+ (aq) + 2 e- � Ni(s)
EO(cathode)
=
+0.80 V
EO(anode) = -0.23 V
Reversing the anode half-reaction yields Ni(s) � Ni 2 + (aq) + 2 eand the cell reaction is, upon addition of the half-reactions, 2 Ag+ (aq) + Ni(s) � 2 Ag(s) + Ni 2+ (aq)
SM -370
EOce l l
+ 0.80 V = + 1 .03 V =
-
(-0.23) V
(b) 2 H+ (aq) + 2 e-
---+
£O(anode) = 0.00 V
H 2 (g)
CI 2 (g) + 2 e- ---+ 2 Cr (aq)
£O(cathode) = +1 .36 V
Therefore, at the anode, after reversal, and, the cell reaction is, upon addition of the half-reactions, £ocell
Cl 2 (g) + H 2 (g) ---+ 2 H+ (aq) + 2 CI- (aq) (c) Cu 2 + (aq) + 2 eCe4 + (aq) + e -
---+
---+
Cu(s)
Ce3 + (aq)
= +1 .36 V - 0.00 V = +1 .36 V
£O(anode) = +0.34 V £O( cathode) = +1 .61 V
Therefore, at the anode, after reversal, and, the cell reaction is, upon addition of the half-reactions, 2 Ce4 + (aq) + Cu(s) ---+ Cu 2 + (aq) + 2 Ce3+ (aq) £Oce = 1 .6 1 V - (0.34 V) ll
(d) 0 2 (g) + 2 H 2 0(l) + 4 e- ---+ 4 0H- (aq) 0 2 (g) + 4 H+ (aq) + 4 e- ---+ 2 H 2 0(l)
= +1 .27 V
£O(cathode) = OAO V
£O(anode) = 1 .23 V
Reversing the anode half-reaction yields and the cell reaction is, upon addition of the half-reactions, 4 H 2 0(l) ---+ 4 H+ (aq) + 4 OH- (aq)
£ocell
= OAO V - 1 .23 V = -0.83 V
or, H 2 0(l) ---+ H+ (aq) + OH- (aq) Note: This balanced equation corresponds to the cell notation given. The spontaneous process is the reverse of this reaction. (e) Sn 4+ (aq) + 2 e- ---+ Sn 2 + (aq) £O(anode) = +0. 1 5 V Hg 2 Cl 2 (s) + 2 e-
---+
2 Hg(l) + 2 cr (aq)
£O(cathode) = +0.27 V
Therefore, at the anode, after reversal, and the cell reaction is, upon addition of the half-reactions, SM-3 7 1
z Sn + (aq) + Hg z Cl 2 (s) � 2 Hg(l) + 2 Cl- (aq) + Sn 4 + (aq) EO ce))
12.11
=
0.27 V - 0. 1 5 V = 0. 1 2 V
z (a) Ni + (aq) + 2 e- � Ni(s) Zn 2 + (aq) + 2 e- � Zn(s)
EO(cathode)
=
-0.23 V
EO(anode) = -0.76 V
Reversing the anode reaction yields Zn(s) � Zn 2 + (aq) + 2 e- (at anode); then, upon addition, Ni 2 + (aq) + Zn(s) � Ni(s) + Zn 2 + (aq) EOce))
=
(overall cell)
-0.23 V - (-0.76 V) +0.53 V =
and Zn(s) I Zn 2 + (aq) I I Ni 2 + (aq) I Ni(s) (b) 2[Ce 4+ (aq) + e- � Ce3 + (aq)]
EO(cathode) = + 1 .6 1 V
I? (s) + 2 e- � 2 r (aq)
EO(anode)
=
+0.54 V
Reversing the anode reaction yields 2 1- (aq) � 2 e- + I z (s) (at anode); then, upon addition, 2 1- (aq) + 2 Ce 4+ (aq) � 2 Ce3+ (aq) + 1 2 (s) EOce))
=
(overall cell)
+ 1 .6 1 V - 0.54 V = + 1 .07 V
and Pt(s) I 1- (aq) I 1 2 (s) I I Ce 4+ (aq), Ce3+ (aq) I Pt(s) An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution. (c) Cl z (g) + 2 e-
�
2 Cr (aq) EO(cathode)
2 H+ (aq) + 2 e- � H z (g) EO(anode)
=
=
+ 1 .36 V
0.00 V
Reversing the anode reaction yields H 2 (g) � 2 H+ (aq) + 2 e-
(at anode); then, upon addition,
H z (g) + Cl z (g) � 2 HCI(aq)
(overall cell)
EO ce))
= =
+ 1 .36 V - 0.00 V +1 .36 V
and Pt(s) I H 2 (g) I H+ (aq) II Cl- (aq) I Cl 2 (g) I Pt(s) An inert electrode such as Pt is necessary for gaslion electrode reactions. SM-372
(d) 3[Au+ (aq) + e- � Au(s)] EO(cathode) +1 .69 V Au 3+ (aq) + 3 e- � Au(s) EO(anode) = +l AO V =
Reversing the anode reaction yields Au(s) � Au 3+ (aq) + 3 e- then, upon addition (anode), 3 Au+ (aq) � (overall cell) 2 Au(s) + Au 3 + (aq)
E Ocell = +l .69 =
V - l AO V +0.29 V
and Au(s) I Au 3 + (aq) II Au+ (aq) I Au(s) 12.13
(a) Ag+ (aq) + e AgBr(s) + e-
�
�
Ag(s) EO(cathode) +0.80 V =
Ag(s) + Br- (aq) EO(anode)
=
+0.07 V
Reversing the anode reaction yields Ag(s) + Br - (aq) � AgBr(s) + e-
then, upon addition,
Ag+ (aq) + Br - (aq) � AgBr(s) (overall cell) EOcell +0.73 V
= +0.80
V - 0.07 V
=
This is the direction of the spontaneous standard cell reaction that could be used to study the reverse of the given solubility equilibrium. A cell diagram for this favorable process is Ag(s) IAgBr(s) I Br- (aq) II Ag+ (aq) I Ag(s) (b) To conform to the notation of this chapter, the neutralization is rewritten as H+ (aq) + OH-
�
H 2 0(1)
0 2 (g) + 4 H+ (aq) + 4 e- � 2 H 2 0(1) EO(cathode) 0 2 (g) + 2 H 2 0(I) + 4 e-
�
4 OH- (aq) EO(anode)
= =
+ l .23 V +OAO V
Reversing the anode reaction yields
4 H+ (aq) + 4 OH- (aq) � 4 H 2 0(l) or H+ (aq) + OH- (aq) � H 2 0(l) (overall cell) EO = +1 .23 V - OAO V +0.83 V =
SM-373
and Pt(s) 10 2 (g) 10H- (aq) II H+ (aq) 102 (g) IP t(s) (c) Cd(OH) 2 (s) + 2 eNi(OH) 3 (s) + e-
�
�
Cd(s) + 2 OH- (aq) EO(anode) -0. 8 1 V =
Ni(OH) 2 (s) + OH- (aq) EO(cathode) +0.49 V =
Reversing the anode reaction and multiplying the cathode reaction by 2 yields Cd(s) + 2 OH- (aq) � Cd(OH) 2 (s) + 2 e2 Ni(OH) 3 + 2 e-
�
2 Ni(OH) 2 (s) + 2 OH- (aq) then,upon addition,
2 Ni(OH) 2 (s) + Cd(s) � Cd(OH) 2 (s) + 2 Ni(OH) 2 (s)
overall cell EO +1.30 V =
and Cd(s) I Cd(OH) 2 (s) I KOH(aq) II Ni(OH) 3 (s) I Ni(OH)2 (s) I Ni(s) 12.15
2 (a) Mn0 4 - (aq) + 8 H+ (aq) + 5 e- � Mn + (aq) + 4 H 2 0(l) (cathode halfreaction) 5[Fe 2 + (aq) � Fe3+ (aq) + e- ] (anode half-reaction) (b) Reversing the anode reaction and adding the two equations yields 2 2 Mn0 4 - (aq) + 5 Fe + (aq) + 8 H+ (aq) � Mn + (aq) + 5 Fe3+ (aq)+ 4 H 2 0(l) The cell diagram is 2 2 Pt(s) I Fe 3+ (aq), Fe + (aq) II H+ (aq), Mn0 4 -(aq) , Mn + (aq) I Pt(s)
12.17
A galvanic cell has a positive potential difference; therefore, identify as
cathode and anode the electrodes that make EO (cell) positive upon calculating £O(cell) £O(cathode) - £O(anode) =
There are only two possibilities: If your first guess gives a negative EO (cell), switch your identification. 2 (a) Cu + (aq) + 2 e- � Cu(s) £O(cathode) = +0.34 V 2 Cr3 + (aq) + e- � Cr + (aq) £O(anode) = -0.4 1 V £O( cell) +0.34 V - ( -0.41 V) +0.75 V =
=
SM-374
(b) AgCI(s) + e- � Ag(s) + CI- (aq) EO(cathode) = +0.22 V AgI(s) + e- � Ag(s) + 1- (aq) EO(anode) -0. 1 5 V EO(cell) = + 0.22 V - ( - 0. 1 5 V) +0.37 V =
=
(c)
Hg/ + (aq) + 2 e- � 2 Hg(l) EO(cathode) +0.79 V =
Hg 2 Cl 2 (s) + 2 e- � 2 Hg(l) + 2 cr (aq) EO(anode) +0.27 V EO(cell) +0.79 V - (+0.27 V) +0.52 V 2 (d) Pb 4 + (aq) + 2 e- � Pb + (aq) EO(cathode) = +1 .67 V 2 Sn 4 + (aq) + 2 e- � Sn + (aq) EO(anode) = +0. 1 5 V EO(cell) +1 .67 V - (+0. 1 5 V) +1 .52 V =
=
=
12.19
=
=
In each case, detennine the cathode and anode half-reactions corresponding to the reactions as written. Look up the standard reduction potentials for these half-reactions and then calculate EO cel l
=
EO(cathode) - EO(anode). If EOcell is positive, the
reaction is spontaneous under standard conditions. (a) EOcel'
=
EO( cathode) - EO(anode) = +0.96 V - (+0.79 V) +0. 1 7 V . =
Therefore, spontaneous galvanic cell: Hg(l) I Hg/+ (aq) II N0 3- (aq), H+ (aq) I NO(g) I P t(s) /).Oo r - nFEO -(6) (9.65 x 1 0 4 C · mol- I ) (+0. 1 7 J . C - I ) = -98 kJ . morl =
=
(b) EO cel'
=
EO(cathode) - EO(anode) + 0.92 V - (+1 .09 V) -0. 1 7 V =
=
Therefore, not spontaneous. (c) EO cell
=
EO(cathode) - EO(anode) = +1 .33 V - (+0.97 V) +0.36 V =
Therefore, spontaneous galvanic cell. Pt(s) I P u 3+ (aq), Pu 4+ (aq) II Cr2 0 �- (aq), Cr 3+ (aq), H + (aq) I Pt(s) /).0; = - nFE o -(6) ( 9.65 x 1 0 4 C · mol- 1 )(0.36 J . C- 1 ) = -208 kJ . mol- 1 =
SM-375
12.21
The cell, as written Cu(s) I Cu 2+ (aq) II M 2+ (aq) I M (s), makes the Cu/Cu 2 + electrode the anode, because this is where oxidation is occurring; the M 2 + 1M electrode is the cathode. The calculation is EO ( cathode) - E O ( anode) - 0.689 V EO ( cathode) - (+0.34 V) EO (cathode) -0.349 V EO
=
=
=
12.23
Refer to Appendix 2B. The more negative (less positive) the standard reduction potential, the stronger is the metal as a reducing agent. (a) (b) (c) (d)
12.25
Cu < Fe < Zn < Cr Mg < Na < K < Li V < Ti < Al < U Au < Ag < Sn < Ni
In each case, identify the couple with the more positive reduction potential. This will be the couple at which reduction occurs, and therefore which contains the oxidizing agent. The other couple contains the reducing agent. (a) C0 2+ ICo EO -0.28 V, C0 2 + is the oxidizing agent (cathode) =
Ti 3 + ITi 2 + EO -0.37 V, Ti 2 + is the reducing agent (anode) P t(s) I Ti 2 + (aq), Ti 3+ (aq) I I C0 2+ (aq) I Co(s) EOcell EO(cathode) - EO(anode) -0.28 V - (-0.37 V) +0.09 V (b) U 3 + I U EO -1 .79 V, U 3+ is the oxidizing agent (cathode) =
=
=
=
=
La 3 + ILa EO -2.52 V, La is the reducing agent (anode) La(s) I La 3 + (aq) I I U 3 + (aq) I U(s) EO cell -1 .79 V - ( -2.52 V) +0.73 V (c) Fe3+ IFe 2 + EO +0.77 V, Fe3 + is the oxdizing agent (cathode) =
=
=
=
H+ IH 2 EO = 0.00 V, H 2 is the reducing agent (anode) P t(s) I H 2 (g) I H+ (aq) II Fe2 + (aq), Fe3+ (aq) I Pt(s) EOcel l +0.7 7 V - 0.00 V +0.7 7 V =
=
SM-376
(d) 0 3 / 0 2 ,OH- EO +1 .24 V, 0 3 is the oxidizing agent (cathode) Ag+ lAg EO +0. 80 V, Ag is the reducing agent (anode) Ag(s) I Ag+ (aq ) II OH- (aq) I 0 3 (g), 0 2 (g) I P t(s) EO cell +1 .24 V - 0.80 V +0.44 V =
=
=
=
12.27
(a) EO(CI 2 , Cl- ) +1 .36 V (cathode) EO(Br2 , Br- ) +1 .09 V (anode) Because EO(CI 2 , Cr ) > EO(Br2 , Br- ) the reaction favors products. =
=
EOcel l
=
+1 .36 V - 1 .09 V
=
+0.27 V
C1 2 (g) is the oxidizing agent. (b) EO(Ce4 + ICe3 + ) + 1 .6 1 V
(anode)
EO(Mn04 - IMn 2 + ) = + 1 .5 1 V
(cathode)
=
products. (c) EO( P b 4 + I Pb 2 + ) = + 1 .67 V
(anode)
EO( P b 2+ IPb) = -0. 1 3 V
(cathode)
Because EO(Pb 4 + I Pb 2 + ) > EO( Pb 2+ I P b), the reaction does not favor products. EO(Zn 2 + IZn) -0.76 V
(anode)
=
products. EOcell +0.80 V - (-0.76 V) +1 .56 V =
=
N0 3 - is the oxidizing agent. 12.29
(a) 3 Au+ (aq) � 2 Au (s) + Au 3+ (aq ) (b) Au+ (aq) + e-
�
Au(s) EO + 1 .69 V =
SM-377
Au 3+ (aq) + 3 e-
�
Au(s)
EO + 1 .40 V =
Multiplying the first equation by three and subtracting the second equation gives the net equation desired. The potential is given simply be subtracting the second from the first: EO = 1 .69 V - 1 .40 V
=
+0.29 V
Because EO is positive, the process should be spontaneous for standard state conditions. 12.31
The appropriate half-reactions are: EO -0.6 1
(A)
EO = -1 .79
(B)
=
(A) and (B) add to give the desired half-reaction (C): (C) In order to calculate the potential of a half-reaction, we need to convert the EO values into I'1Go values: I'1GO(A) -nFEO(A) -IF( -0.61 V) I'1GO(B) = -nFEO(B) = -3F( -1 .79 V) I'1GO(C) = -nFEO(C) -4FEO(C) I'1GO(C) = I'1GO(A) + I'1GO(B) -4FEO(C) = -I F( -0.61 V) + [-3F( - 1 .79 V)] =
=
=
The constant F will cancel from both sides, leaving: -4EO(C) = - 1(-0.6 1 V) - 3( -1 .79 V) EO(C) = -[0.6 1 V + 5.37 V] /4 = -1 .50 V
12.33
(a) Ti 2 + (aq) + 2 eMn 2 + (aq) + 2 e-
�
�
Ti(s) EO(cathode) = -l .63 V EO(anode) -1 . 1 8 V
Mn(s)
=
Note: These equations represent the cathode and anode half-reactions for the overall reaction as written. The spontaneous direction of this reaction under standard conditions is the opposite of that given. EO cell
=
EO(cathode) - EO(anode) -1 .63 V - (-l . 1 8 V) =
SM-378
=
-0.45 V , and
In K
=
nFE o RT
: . In K
=
At 25 ° C In K
.
(2)(-0.45 V) 0.02569 V
=
-35 and K 6 x 1 0- 16 . =
(b) In 3+ (aq) + 2 e- � In 2+ (aq) U 4 + (aq)
E�ell
=
+ e-
�
EO (cathode)
U 3+ (aq)
=
-0.49 V
EO (anode)
EO ( cathode) - EO ( anode) = -0.49 V (-0.61 V) -
at 25°C I n K
12.35
nEO 0.02569 V
= ----
=
(2)(+0. 1 2 V) 0.02569 V
=
=
-0.6 1 V
=
+0. 1 2 V , and
+9.3 .
(a) Pb 4 + (aq) + 2 e- � Pb 2 + (aq) EO( cathode) +1 .67 V =
Sn 2 + (aq) � Sn4+ (aq) + 2e- EO(anode) +0. 1 5 V Pb4+ (aq) + Sn 2 + (aq) � Pb 2 + (aq) + Sn 4 + (aq) EO ce lJ 1 .67 V - (0. 1 5 V) = +1 .52 V =
=
(
)
0.025 693 V In Q; 1 .33 V 1 .52 V n 0. 1 9 V 1 .52 V - 1 .33 V 1 5 Q 1 06 In Q 0.0 1 29 V 0.0 1 29 V 2[Cr2 0/- (aq) + 1 4 H+ (aq) + 6 e- � (b) 2 Cr3+ (aq) + 7 H 2 0(l)] EO( cathode) = 1 .33 V Then, E EO =
=
=
=
=
_
(
)
0.025 693 V In Q 2
=
3[2 H 2 0(I) � 0 2 (g) + 4 H+ (aq) + 4 e- ] EO(anode) = +1 .23 V 2 Cr2 0t (aq) + 1 6 H + (aq) � 4 Cr3+ (aq) + 8 H 2 0(l) + 3 ° 2 (g) EO cell 0. 1 0 V =
Then, E
=
EO -
( � ) 0.02 7 V
In Q; 0. 1 0 V = +0. 1 0 V
In Q 0.00 Q 1 .0 =
12.37
=
(a) Cu 2+ (aq, 0.0 1 0 M) + 2 e- � Cu(s) (cathode) Cu 2 + (aq, 0.00 1 0 M) + 2 e- � Cu(s) (anode)
SM-379
_
( 0��7 ) 0.
V
In Q
Cu 2 + (aq,O.O l ° M)
�
( )
Cu 2 + (aq, 0.00 1 0 M) , n = 2
(
)
EO cell = EO( cathode) - EO( anode) = ° V RT In Q _ - 0.025 693 V In Q at 25 C Ecell - E cell nF 2 0.00 1 0 M 0.025 693 V = +0.030 V Ecell = In 2 0.0 1 0 M _
°
(
) [
J
°
(b) at pH = 3 .0, [H+ ] = 1 X 1 0 -3 M at pH = 4.0, [H+ ] = 1 X 1 0-4 M Cell reaction is H+ (aq, 1 x 1 0-3 M)
( )
�
H+ (aq, 1 x 1 0-4 M), n
(
1
) [ =
l X l O -4 0.02 5 693 V ln EO ce ll = O V Ecel l = E o ce l _ RT ln Q = _ nF 1 l x l O-3 = +6 X 1 0-2 V l
12.39
J
In each case, EO cel l = EO( cathode) - EO( anode). Recall that the values for EO at the electrodes refer to the electrode potential for the half-reaction written as a reduction reaction. In balancing the cell reaction, the half reaction at the anode is reversed. However, this does not reverse the sign of electrode potential used at the anode, because the value always refers to the reduction potential. (a) 2 H+ (aq, 1 .0 M) + 2 e-
�
H 2 (g, 1 atm) EO(cathode) = 0.00 V
H 2 (g, 1 atm) � 2 H+ (aq,0.075 M) + 2 e- EO(anode) = 0.00 V 2 H+ (aq, 1 .0 M) + H 2 (g, 1 atm) � 2 H+ (aq, 0.075 M) + H 2 (g, 1 atm) EO cel l = 0.00 V Then, E = EO -
(
(
) [ ) [
0.025 693 V n
[H+ , 0.075 M] 2 PH In [H+ , 1 .0 M] 2 PH 0.075 M�2 x l atm
0.025 693 V In 2 ( 1 .0 M) x l atm E = - 0.0 1 29 V In (0.075) 2 = +0.067 V E = 0.00 V _
(b) Ni 2 + (aq) + 2 e-
�
J 2
2
Ni(s) EO(cathode) = -0.23 V
SM -380
J
Zn(s) � Zn 2 + (aq) + 2 e- EO(anode) -0.76 V =
Nih (aq) + Zn(s) ---+ Ni(s) + Zn 2 + (aq) EO cell = +0.53 V Then, E
�
E = 0.53 V
-
(
(
) (����:: J ) ( )
0 02 5 69 3 V In " 0.025 693 V 0.37 In 0.059 2
E" -
(c) 2 H+ (aq) + 2 e 2 Cl- (aq) ---+ C1 2 (g)
---+ +
=
H 2 (g) EO(cathode)
0.53 V - 0.02 V =
=
_
(
0.5 1 V
0.00 V
2 e- EO(anode) +1 .36 V =
2 H+ (aq) + 2 Cl - (aq) ---+ H 2 (g) + C1 2 (g) EO cel
E -1 .36 V
=
'
=
-1 .36 V
) )( ( )
� 250 0.025 693 V I 760 760 n (1 .01 325) 2 2 2 2 (0.85) (1 .0)
E -1 .36 V + 0.03 V -1 .33 V =
=
(d) Sn 4 + (aq, 0.867 M) + 2 eEO(cathode)
=
---+
Sn 2 + (aq, 0.55 M)
+0. 1 5 V
Sn(s) ---+ Sn 2 + (aq, 0.277 M) + 2 e- EO(anode) - 0. 1 4 V =
Sn 4 + (aq, 0.867 M) + Sn(s) ---+ Sn 2 + (aq, 0.55) + Sn 2 + (aq, 0.277 M) EOcel '
=
0.29 V
(
) (
0.025 693 V (0.55) (0.277 ) In (0.867) 2 E 0.29 V + 0.02 V = 0.3 1 V E = EO =
12.41
_
J
In each case, obtain the balanced equation for the cell reaction from the half-cell reactions at the electrodes, by reversing the reduction equation for the half-reaction at the anode, multiplying the half-reaction equations
SM -38 1
by an appropriate factor to balance the number of electrons, and then adding the half-reactions. Calculate EO ceil
=
EO(cathode) - EO(anode).
Then write the Nemst equation for the cell reaction and solve for the unknown. (a) Hg 2 CI 2 (s) + 2 eH 2 (g)
---)
0.3 3 V
�
=
2 Hg(l) + 2 Cr (aq) EO(cathode)
=
+0.27 V
2 H+ (aq) + 2 e- EO(anode) = 0.00 V
H 2 (g) + Hg 2 C1 2 (s) E = EO -
---)
(
---)
2 H+ (aq) + 2 Hg(l) + 2 Cl- (aq) EO cel l
0.025 693 V n
(
) [ In
[H+ f [cr f [H 2 ]
=
Ji:(I)'
) [ ) J
[W 0.025 693 V In 2 0.27 V - (0.0 1 29 V) In [H+ ] 2 0.2 7 V -
0.06 V = - 0.0257 V In [H+ ] - 0.0257 V x (2.303 log [H+ J) 0.06 V = 1 .0 PH = (2.303) (0.025 693 V) =
(b)
2[Mn0 4 - (aq) + 8 H+ (aq) + 5 e- ---) Mn 2 + (aq) + 4 H ? O(l)] EO(cathode) = +1 .5 1 V
5 [2 Cl - (aq) ---) Cl 2 (g) + 2 e - ] EO(anode)
=
2 Mn0 4 -(aq) + 1 6 H+ (aq) + 1 0 Cr (aq) ---) 5 C1 2 (g) + 2 Mn 2 + (aq) + 8 H 2 0(l) EO cell E -- EO -
(
0.02 57 V n
) [ In
+1 .36 V =
+0. 1 5 V
[CI 2 ] 5 [Mn 2 + ] 2 lo [Mn0 4 f [H+ ]1 6 [Cl- ]
SM-382
J
+0.27 V
-0.30 V -0.45 V
=
=
=
(
) (
0.0257 V ( 1 ) 5 (0. 1 0) 2 In 10 (0. 0 1 0) 2 (1 X 1 O-4 )16 ( CI - ) 1 0 1 x 1 0-2 -(0.002 5693 V) log (1 x 1 0- ) (1 x 1 0-64 ) [ Cl- ] 1 0
+0. 1 5 V -
[
(
4
-0.002 5693 V In ( 1 x 1 066 ) + In
-0.390 V + (0.0025 693 V) In -0.0594 V 0.002 5693 V In[Cr ] 10 (0.025 693 V) In[CI - ] -0.06 V 2 In[CI- ] 0.025 693 V [Cr ] 1 0-1 mol · L-I =
( J] 1
[CI - ] [Cr ] 1 0
10
J
J
=
=
=
=
_
=
12.43
Since the reduction potential of tin(II) is negative relative to the S .H.E., we will assume the tin electrode to be the anode such that the standard cell potential would be positive. Then we can use the Nemst equation to solve for the hydrogen ion activity in order to calculate the pH. The cell reaction is Sn(s) + 2 H+ � Sn 2 + + H 2 (g) . [Sn 2 + ] PH2 0.025 693 V EO E In n [H+ f
(
=
0.0 6 1 V 0.079 V 0.01 284 V -1 0.352 In[H+ ] pH 12.45
= = = = = =
0. 1 4 V
_
(
) ( J ) ( J
0.025 693 V [0. 01 5] [1] In 2 [H + f
0. 1 4 V - (0.01 284 V) (In(0.0 1 5) - ln [H+ f ) In(0.0 1 5) - 2 In [H+ ] 2 In [H+ ] -5. 1 76, [H+ ] 5.650 x I 0-3 - log(5.650 x 1 0 -3 ) 2.25 =
=
To calculate this value, we need to determine the solubility reaction:
SM-3 83
EO
value for the
+2
Hg 2 CI 2 (s) ---+ Hg/+ (aq) Cr (aq) EO ? The relationship !1 Go -nRT In K = -nFEo can be used to calculate the value of Ksp . =
=
The equations that will add to give the net equation we want are
+ 2 2 + 2 Cr (aq) +2
Hg 2 CI 2 (s) e- ---+ Hg(l) Hg(l) ---+ Hg/+ (aq) e-
2
EO
=
EO =
+0.27 V 0.79 V
Notice that the second equation is reversed from the reduction reaction given in the Appendix, and consequently the EO value is changed in sign. Adding these two equations together gives the desired net reaction, and summing the EO values will give the EO value for that process:
=
(+0.29 V) + (-0.79 V) -0.50 V = In = (2)(9.65 1.0 C .· mol-I ) ( -0.50 V) = -38.95 (8.314 J mol-I )(298 = 1.2 1 0(b) This value is a factor of 1 ° greater than the value in Table 1 1 .6 (1.3 1 0- ) . EO
Ksp
RT
17
X
Ksp
4
K-I
K)
18
x
12.47
x
nFEO
=
This cell uses two silver electrodes, so EO = ° and E is determined by the
I
is less than and E >
!1G=w = -nFE .
0, so the cell can do work because
V ) (5.0 X 10-3 ( 0.025693 V ) In [ [Ag[Ag++ Lnode ) ( 0.025693 1 In 0.15 J LlilOde = 0.0874 V !1G=wmax -(1 mol)(96 485 J V -I mol - I )(0.0874 V) = 8. 4 kJ Therefore, the maximum work that the cell can perform is 8.4 kJ per mole max
E=
_
=
n
=
-nFE
_
=
of Ag.
12.49
+0.27 V. If this were set equal to 0, all other potentials would also be decreased by 0.27 V. (a) Therefore, SM-384
For the standard calomel electrode, EO
=
the standard hydrogen electrode's standard reduction potential would be 0.00 V - 0.27 V or -0.27 V. (b) The standard reduction potential for 2 Cu + ICu would be 0.34 V - 0.27 V or +0.07 V. 12.51
The strategy is to consider the possible competing cathode and anode reactions. At the cathode, choose the reduction reaction with the most positive (least negative) standard reduction potential (EO value). At the anode, choose the oxidation reaction with the least positive (most negative) standard reduction potential (EO value, as given in the table). =
Then calculate EOceJ l
EO(cathode) - EO(anode). The negative of this
value is the minimum potential that must be supplied. 2 (a) cathode: Ni + (aq) + 2 e- -+ Ni(s) EO = -0.23 V (rather than 2 H 20 (l) + 2 e-
-+
H 2 (g) + 2 OH- (aq) EO = -0.83 V)
(b) anode: 2 H 2 0(l) -+ 02 (g) + 4 H+ (aq) + 4 e- EO = +1 .23 V (the SO/- ion will not oxidize) (c) EOcel' = EO(cathode) - EO(anode) = -0.23 V - (+1 .23 V) = -1 46 V .
Therefore E (supplied) must be > +1 .46 V (1 .46 V is the minimum). 12.53
In each case, compare the reduction potential of the ion to the reduction potential of water (EO = -0.42 V) and choose the process with the least negative EO value. 2 (a) Mn + (aq) + 2 e(b) AlJ+ (aq) + 3 e-
-+
-+
Mn(s)
Al(s)
EO = -1 . l 8 V EO -1 .66 V =
The reactions in (a) and (b) evolve hydrogen rather than yield a metallic deposit because water is reduced, according to 2 H2 0 (l) + 2 e- -+ H 2 (g) + 2 OH- (aq) (EO = -0.42 V, at pH = 7) 2 (c) Ni + (aq) + 2 e- -+ Ni(s) EO = -0.23 V (d) AuJ+ (aq) + 3 e-
-+
EO = +1 .69 V
Au(s) SM-3 85
In (c) and (d) the metal ion will be reduced. 12.55
4500 C .;- 9.65 X 1 0 4 C · F-I 0.047 F = 0.047 mol e=
(a)
(0.047 .;- 3) mol Bi 3+ + 0.047 mol e- -) (0.047 .;- 3) mol Bi, or 0.0 1 6 mol Bi = 3 .3 g
(b) 0.047 mol H+ + 0.047 mol e- -)- (0.047 .;- 2) mol H 2 0.024 mol H 2 x 24.45 L · mOr l (at 298 K) = 0.59 L (0.047 .;- 3) mol Co3+ + 0.047 mol e- -) (c) (0.047 .;- 3) mol Co = 0.0 1 6 mol Co or 0.94 g 12.57
(a) Ag+ (aq) + e- -)- Ag(s) . tIme = ( 1 .5 0 g Ag)
(
9.6 5 X 1 04 C 1 mol e-
(
1 mol Ag 1 07.98 g Ag
)(
)( )( ) l A.S 1C
4
mass Cu = (9.9 x 1 0 s) (0.0 1 36 A)
12.59
0.50 mol CU 1 mol e-
)(
)
1 = 9.9 X 1 04 s or 27 h 0.0 1 36
(b) Cu 2+ (aq) + 2 e- -)- Cu(s)
(
1 mol e1 mol Ag
63 .5 g cu 1 mol Cu
)
=
( --)( 1C 1 A·s
1 mol e9.65 x 1 0 C 4
)
0.44 g Cu
(a) Cr(VI) + 6 e- -)- Cr(s) current =
charge time
-=-
(
)(
)(
)
1 mol Cr 6 mol e - 9.65 X 1 04 C 52.00 g Cr 1 mol Cr 1 mol e= ------�----��----�---------1 2 h x 3600 s · h- I = 0.64 C . S- I = 0.64 A 2.5 g cr
(b) Na+ + e- -)- Na(s)
SM-386
current
=
=
12.61
(
J(
J(
1 mol e1 mol Na 22.99 g Na 1 mol Na ......:....1 2 h x 3600 s · h- I 0.24 C ' s- I = 0.24 A 2.5 g Na
-
Run+ (aq) + n e- � Ru(s); solve for n moles of Ru (0.03 1 0 g Ru) =
(
total charge (500 s) (1 20 rnA moles of e-
=
(60 C)
J
� ----.:. .... .. ..:....--' � �
-
=
9.65 x 1 04 C 1 mol e-
(
J
1 mol 1 0 1 .07 g Ru )
1 mol e96 500 C
[ J( J 1 0-3 A 1 rnA =
3 .07 x 1 0-4 mol
=
l C ' S- I 1A
J
=
60 C
6.2 x 1 0-4 mol e-
6.2 X 1 0-4 mol e- 2 mol charge 3 .07 X 1 0-4 mol 1 mol 2 Therefore, oxidation number of Ru + is +2. n
12.63
- --= =---
-
-
Hfn+ + n e-
�
Hf(s); solve for n.
charge consumed 1 5.0 C · S- I X 2.00 h x 3600 s · h-I =
moles of charge consumed = (1 .08 x 1 05 C) moles of Hfplated Then,
n=
=
(50.0 g Hf)
(
(
= 1 .08 x
1 mol e9.65 x 1 0 4 C
1 mol Hf 1 78.49 g Hf
J
=
J
=
l 05 C
1 . 1 2 mol e-
0.280 mol Hf
1 . 1 2 mol e= 4.00 mol e- /mol Hf 0.280 mol Hf
Therefore, the oxidation number is 4, that is, Hf 4 + . 12.65
MCl 3
�
M 3 + + 3 cr
F irst, determine the number of moles of electrons consumed; the number of moles of M 3 + reduced is one-third of this number. SM-387
charge used ( 6. 63 h) =
( J( 36 00 S 1h
number of moles of e -
=
0.700 C
Is
( 1 . 6 7 x 1 0 4 C)
(
J
=
1 .67 X 1 04 C
1 mol e9.65 x l 0 4 C
number of moles of M 3 (and M) 0. 1 73 mol e +
= =
mo Iar mass M
12.67
=
x
J
=
0. 1 73
1 mol M 3 + ----
3 mol e-
0.0577
3 00 g 52.0 g . mol - (Cr) 0.0577 mol .
I
=
(a) Assuming alkaline conditions: 6 0 B" (aq) + 2AI(s)
oxidation/anode:
--+
2AI(OH) 3 (aq) + 6e-
reduction/cathode:
overall : (b)
EO
cell
=
EO = =
12.69
cathode
- Eo
anode
( +0.40 V) ( 1 66 V) -
-
.
+ 2.06 V
Assuming all the energy comes from reduction of oxygen focuses attention on this half reaction: Body conditions are far from standard state values, so the actual value of E would be reduced by about 0.5 V if we take pH,
POl
and T into
account. However, we are only estimating an average current to one
SM-388
significant digit, so E= 1 .23 ± 0.5 V
�
1 V is adequate. With these
approximations in mind, we can calculate the current. It = nF = I1G -E
-
or
(-l O x 1 0 6 J) I = I1G = . 1 V·C -Et -(1 V)(24 h)(3600 s · h - ' ) 1 J
12.71
=
1 15 A
�
1 00 A
(a) The electrolyte is KOH(aq)/HgO(s), which will have the consistency of a moist paste. (b) The oxidizing agent is HgO(s). (c) HgO(s) + Zn(s) � Hg(l) + ZnO(s)
12.73
See Table I 2. 1 . The anode reaction is Zn(s) � Zn 2 + (aq) + 2 e- ; this reaction supplies the electrons to the external circuit. The cathode reaction is Mn0 2 (s) + H 2 0(l) + e- � MnO(OH) 2 (s) + OH- (aq). The OH- (aq) produced reacts with NH 4 + (aq) from the NH 4 Cl(aq) present: complexes with the Zn 2 + (aq) produced in the anode reaction
The overall reaction is complicated. 12.75
See Table 1 2. 1
(a) KOH(aq) (b) In the charging process, the cell
reaction is the reverse of what occurs in discharge. Therefore, at the anode, 2 Ni(OH) 2 (s) + 2 OH- (aq) � 2 Ni(OH) 3 12.77
Fe3+ (aq) + 3 e- � Fe(s)
EO = -0.04 V
Cr 3+ (aq) + 3 e- � Cr(s)
EO = -0.74 V
SM -389
+
2 e- .
Fe2 + (aq) + 2 e -
-+
Fe(s)
EO = -0.44 V
Cr 2+ (aq) + 2 e-
-+
Cr(s)
EO -0.9 1 V =
Comparison of the reduction potentials shows that Cr is more easily oxidized than Fe, so the presence of Cr retards the rusting of Fe. At the position of the scratch, the gap is filled with oxidation products of Cr, thereby preventing contact of air and water with the iron.
12.79
(a)
n
e
_
=
n
Ag
+
=
It
F
1 .43 x l 0-2 mol Ag
(3 .5 A)(395.0 s) = 1 .43 x 1 0-2 mol Ag (96 485 C · mor ' )
=
(
]
1 07.87 g Ag = 1 .55 g Ag mol Ag
1 .55 g x 1 00 = 57.4% Ag 2.69 g
(b) 2.69 g- 1 .55 g= 1 . 1 4 g X Since the salt is 1 : 1 Ag:X, the molar mass of X is 1 . 14 g --..:=-- = 79 . 7 g · mo 1-' 1 .43 x l O-2 mol This molar mass is closest to bromine, so the formula is AgBr. 12.81
(a) Fe2 0 3 • H 2 0
(b) H 2 0 and 0 2 jointly oxidize iron.
(c) Water is
more highly conducting if it contains dissolved ions, so the rate of rusting is increased. 12.83
(a) aluminum or magnesium; both are below titanium in the electrochemical series. (b) cost, availability, and toxicity of products in the environment EO +0.34 V (c) Cu 2+ + 2 e- -+ Cu(s) =
Cu+ + e- -+ Cu(s) Fe3+ + 3 e- -+ Fe(s)
EO = +0.52 V EO = -0.04 V EO - 0.44 V =
SM-390
Fe could act as the anode of an electrochemical cell if Cu 2 + or Cu+ were present; therefore, it could be oxidized at the point of contact. Water with dissolved ions would act as the electrolyte. 12.85
12.87
A I, Zn, Fe, Co, Ni, Cu, Ag, Au 2[Zn 2 (aq) + 2 e- -+ Zn(s)] M(s) -+ M 4 + (aq) + 4 e-
EO(cathode)
+
=
EO( anode) = x
M(s) + 2 Zn 2 + (aq) -+ 2 Zn(s) + M 4 + (aq) EOcell EOcell
=
12.89
-
=
0. 1 6 V
EO( cathode) - EO( anode)
+ 0. 1 6 V x=
-0.76 V
- 0.76 V - (x)
=
0.92 V = EO(M 4 + 1M)
The strategy is to find the EO value for the solubility reaction and then find appropriate half-reactions that add to give that solubility reaction. One of these half-reactions is our unknown, the other is obtained from Appendix 2B:
Cu(s)
EO = -0.34 V EO
EO =
RT
=
(B)
R T I n Ksp
nF
(C)
In Ksp
nF
(8 .3 1 4 ] . K . mor ' ) (298.2 K) In 0 .4 x 1 0-7 ) 2(9.65 x 1 04 C · mor ' ) - 0.20 V
-, = � ------� --� -� --�
=
-0.20 V EO (A) = 12.91
=
EO(A) + (-0.34 V)
+ O. 1 4
V
A negatively charged electrolyte flows from the cathode to the anode. SM-3 9 1
12.93
(a) In acidic solution, the relevant reactions are EO + 1 .23 V =
EO
=
- 0.80 V
Overall reaction: 0 2 (g) + 4 H + (aq) + 4 Ag(s) -+ 4 Ag + (aq) + 2 H 2 0(l)
EO +0.43 V =
Because the potential is positive, the reaction should be spontaneous and would be expected to occur. We should also consider the conditions; because air is only 20.95% ° 2 , the potential may be different from that calculated for standard conditions. If air is the source of oxygen, then it will be present at 0.2095 x 1 .0 1 3 25 bar 0.2 1 23 bar. =
E
=
EO
_
0.05 92 [Ag+ t 10 g 4 PO , [H+ ] 4
+0.43 V -
=
= = =
[LOt 0.0592 1 og 4 (0.2 1 23) [1 .0t
0.0592 1 10g 4 O.2 1 23 +0.43 V - 0.0 1 0 V +0.42 V +0.43 V
_
The potential is still positive and the reaction is expected to be spontaneous. (b) In basic solution, the relevant reactions are EO
=
+0.40 V
EO
=
-0.80 V
Overall reaction: 0 2 (g) + 2 H 2 0(1) + 4 Ag(s) -+ 4 Ag+ (aq) + 4 OH - (aq) EO
=
- 0.40 V
This process as written is nonspontaneous and is not predicted to occur. However, AgOH forms an insoluble precipitate, changing the nature of the reaction . The Ksp value for AgOH is 1 .5 x 1 0-8 . We use the Nemst equation to calculate the potential under these conditions: SM-392
E
=
=
=
EO
_
0.0 592
4
10g
[Ag + t [OH - t PO I
-OAO V _
0.0592
-OAO V _
0.0592
4
POI
4 = -OAO V + OA5 V =
K 4
P 10g _S_ 10 g
(1 .5 x 1 0-8 )4 0.2 1 32
+0.05 V
Under these conditions, the potential is slightly positive and the oxidation should be spontaneous. 12.95
(a) Reduction takes place at the electrode with the higher concentration, which would be the chromium electrode in contact with the 1 .0 M CrCh. (b)
Yes
M
Yes
(c)
Ag
+
=
M r Vr V g+
(d)
No
(0.0 1 5 M)(l 6.7 mL) --'(25.0 mL)
=-
A
-
2 = 1 .0 x l 0- M
(b) We can find [Ag+ ] by using the Nemst equation appropriately, E=
EO -
( 0.025 693 V ) In ( n
J
.
1 . . Since the standard reductIon [Ag ]
+ -
potential of silver(l) is + 0.80 V, it will be the reduction half reaction versus the S.H.E., so [Ag + ] appears in the denominator of Q. In addition, n
=
( 0.025 693 V ) In (_I_ J
1 and EO 0.080 V .
0.3 25 V
=
=
0.8 0 V -
1 [Ag+ ] -OA75 V=( - 2.56 7 x l 0-2 V)(ln l - ln[Ag+ ]) -1 8.50 In[Ag+ ] [Ag + ] = 9.23 x 1 0-9 M =
SM -393
Recalling that
Ksp
=
[Ag + ] p- ] , and assuming [Ag + ]=[r ] at the
stoichiometric point of the titration,
12.99 F2 (g) + 2 e- � 2 F- (aq) EO(cathode)
=
+2.87 V
2 HF(aq) � F2 (g) + 2 H+ (aq) + 2 e - E O (anode)
=
+3 .03 V
2 HF(aq) � 2 H+ (aq) + 2 F- (aq) EOcel l EO(cathode)-EO(anode) +2.87 V - (+3 03 V) -0. 1 6 V =
=
For the above reaction, at 250C K
=
Ka
=
nEe
0.025 69 V
K
=
=
[H+ f [F- ] 2 and In [HF] 2
(2) (-0. 1 6 V) 0.025 69 V
=
.
=
K
nFEo
= --
RT
-1 2
1 0-5
=
JK
=
J1(i5
=
1 0-3
12.101(a) L1 Grl ° -nFE, ° = Wr, ° - T, L1S./ where r, represents the reaction at T, . =
L1 Gr2°
=
-nFE2° = L1Hr2° - T2L1Sr2° where r2 represents the reaction
On subtracting the first reaction from the second, we obtain -nFE2° + nFE,o
=
Wr2° - T2L1Sr2° - [W./ - T, L1S./] .
Since Wr, ° L1Hr2 ° we obtain, =
nFE,o - nFE2° =
-
T2L1S./ + T, L1S./
which can be rewritten as -nFE,o + nFE2° = + T2L1Sr2° - T, L1Sr,o. Since L1Srl ° L1Sr2° L1S.o we obtain, =
-nFE, o + nFE20
=
=
L1SrO( T2 - T, )
nFE2° = nFE, ° + L1S.0(T2 - T, ) E2°
=
E,o + L1SrO( T2 - T, )/nF
(b) The redox reaction is 2H2(g) + 02(g)
�
2H 20(l).
Using the standard potentials at 25 ° C to calculate E,o,
SM-394
at T2.
g = + 1 .23 V
2H2 0(l) 02 (g) + 4H\aq) + 4e 2H2(g) 4H+(aq) + 4e-
--+
g=OV
--+
!1S� = I S,:, producls - I s,:, reaclanlS
= 2(69.9 1 J.K- I .mOr l ) - [2( 1 30.68 l.K- I .mor l ) + (205. 1 4 J.K- I .mor l )] = 1 39.82 J.K- I .mOr l - 466.50 l.K- I .mor l = -326.68 l.K- I .mor l
= + 1 . 23 V +
-326.68 l.K -I . mol - 1 (353 . 1 5 K - 298. 1 5 K) 4(9.6485 x 1 0 4 J.V-I .mol-I )
-326.68 J.K- I .mor l (55.0 K) 3 . 8594 x 1 0 5 J.V-I .mol-I ) = + 1 .23 V - 4.65 X 1 0-2 V
= + 1 .23 V +
= +1 . 1 8 V 12.103
The wording of this exercise suggests that K+ ions participate in an electrolyte concentration cell reaction. Therefore, EO Ceil
=
0.00 V, because
the two half cells would be identical under standard conditions. Then, E
=
=
EO
_
(
0.0257 V n
+0.09 V
and E
=
0.00 V -
(
) ( :] In
[K ou ] [K in ]
=
0.00 V
) ( )
0.0257 V 1 In I 20
=
The range of potentials is 0.08 V to 0 . 09
SM -395
_
(
+0.08 V V.
) ( )
0.0257 V 1 In _ 30 1
12.105
Ag+ (aq ) + e- � Ag EO(cathode) +0.80 V Fe2 + (aq) � Fe3+ (aq) + e- EO(anode) +0.77 V Ag+ (aq) + Fe 2 + (aq) � Fe3+ (aq) + Ag(s) EOcell +0.03 V =
=
)
(
0.0257 V Ecell - Eocel l _ ln n
( (
=
0.03 V - (0.0257 V) In
=
-0.2 1 V
Comm ent:
=
[Fe3+ ] [Ag+ ] [Fe2 + ]
J
0.20 (0.020)(0.001 0)
)
=
-
0.03 V 0.24 V
The cell changes from spontaneous to nonspontaneous as a
function of concentration. 12.107
( 1 ) Cl0 4 - + 2 H+ + 2 e-
�
Cl0 3 - + H 2 0
EO
=
+l .23 V
(2) Cl0 4 - + H 2 0 + 2 e- � Cl0 3 - + 2 OH- EO +0.36 V (a) The Nemst equation can be used to derive the potential as a function =
of pH: For ( 1 ), E' (1)
=
1 .23 V
_
[ClO;- ] 0.059 1 6 V log 2 [ClO; ] [ H + f
We are only interested in varying [H+ ] , so the [Cl0 3 - ] and [Cl0 4 - ] will be left at the standard values of 1 M. E' (1)
l .23 V
=
=
=
1 _ 0.05921 6 V 10g _ _ [H + ] 2
1 .23 V
_
1 0.059 1 6 V x 2 10 g __ 2 [H+]
1 .23 V - (0.059 1 6 V) (- log[H+ ])
Similarly, for (2): _ 0.059 1 6 V 10g [CIO;- ] [OH - ]2 E' (2) 0.36 V =
=
l .23 V (0.05916 V) pH
2
[CIO; ]
SM-396
As above, we are only interested in varying [OH- ], so [CIO] - ] and [CI0 4 -] will be left at the standard value of 1 M . £' (2) 0.36 V =
_
0.05 9 1 6 V 2 log [OH - ] 2
0.059 1 6 V x 2 10g [OH- ] 2 0.36 V - (0.059 1 6 V) x 10g [OH - ] 0.36 V + (0.059 1 6 V) pOH
= 0.36 V =
=
Because pOH + pH = pKw
=
1 4.00, we can write:
pOH = 14.00 - pH £' (2) 0.36 V + (0.059 1 6 V) (1 4.00 - pH) 0.36 V + 0.83 V - (0.059 1 6 V) pH 1 . 1 9 V - (0.059 1 6 V) pH =
=
=
If we compare this to £' (1) , we find that the equations are essentially the same. They should be identical, the difference being due to the limitation of the number of significant figures available for the calculations. (b) From the discussion above, we can see that the potential in neutral solution should be the same, regardless of which half-reaction we use to calculate the value. Using £' (1) = + 1 .23 V - (0.059 1 6 V) pH, £' (1)
=
+ 1 .23 V - (0.059 1 6 V) (7.00)
=
+0.82 V.
Using £' (2) = +0.36 V + (0.059 1 6 V) pOH , £' (2) = +0.36 V + (0.059 1 6 V) (7.00) = +0.77 V. Although these numbers differ slightly, they should be identical; again the difference lies in the limitation of the number of significant figures. 12.109
Buffer system.
SM-397
Note: (H+ ) , as opposed to [H+ ] , indicates a nonequilibrium molarity. Because in a buffer system (A) (HA) , we can write �
0.060 V = EOcel' Because 10g (H+ ) 0.060 V
=
0.060 V
=
0.060 V 0.060 V EO
=
= =
( 0.02: 693 ) (2.303) (log(H+ ))
=
- [-log(H+ )] -pH, we have =
EOcel'
- 0.0592 x (-pH)
EOcel'
+
0.0592 x pH
EOcell +
0.0592 x 9.40
EOcel'
0.556 V
+
0.060 V - 0.556 V -0.496 V =
Similarly, 0.22 V -0.496 V + 0.0592 V x pH =
PH
12.111
=
0.22 V + 0.496 V 0.0592 V
=
12
Using iJ.Go = - nFg and iJ.Go = - RT In K , one obtains the relationship RT g= ln K . nF
Must have g = 0 when pH = 7. pH = 7 when concentration of H+ and OR are equal to 1 .0 When K = 1 , then g = O. Therefore at pH = 7: K = At P H = 1 : g =
RT
At pH = 1 4: g =
nF
[H + ] [OH- ]
=
1 .0 x 1 0-7 mo l.L ' 1 .0 x l 0-7 mol.L'
In K = 0.025693 V In
RT
nF
1
1 0.7 mol.L- I .
and
g=0
1 .0 mol.L ' = +0.828 V 1 .0 x 1 0-1 4 mol.L '
In K = 0.025693 V In
SM -398
=
x
1 .0 x 1 0-1 4 m�I.L ' = -0.828V 1 .0 mol.L
12.113
(a) The more dilute electrolyte in a concentration cell is always the anode. The anode generates the electrons (in this case Ag(s) and the concentration of Ag+(aq) increases.
�
Ag\aq) + 1 e-)
In all chemical reactions the species that increases in concentration is the product. Therefore in the Nemst equation, for a concentration cell, the more dilute electrolyte is always the product. _
E EO =
RT
nF
In
[Ag+ ]anode + L'hode
[ Ag
A plot of cell voltage, E, versus In [Ag+]a
ode would
n
be a linear increase
with positive slope because a larger difference in concentration results in a larger cell voltage. In a concentration cell the larger the concentration difference between the anode and cathode the higher the cell voltage. In other words, a decreasing value for the numerator,
+ [Ag ] anode,
term increasingly negative and therefore the term,
_
RT
nF
In
makes the In
[ Ag Janode , [Ag L'hode ++
increasingly positive. (b) E
=
EO
_
RT
nF
In Q
=
For this redox reaction n
EO =
_
0.025693 V In Q n
1.
The slope is 0.025693 V, and this value corresponds to the terms
RT
nF
.
What is being done here is a rearrangement of the Nemst equation to E y
=
0.025693 V
-
n =
mx
. . In Q + EO, resultmg In a straight line plot of the form
+ c where the slope, m, is
RT
nF
and the y-intercept, c, is EO.
(c) y-intercept is EO, and for all concentration cells EO (d) Provided we know the values for R , RT
nF
=
0.025693 V.
SM-399
T and
n,
=
o.
the answer is yes since,
1 2. 1 1 5
The strategy for working this problem is to create a set of equations that will add to the desired equilibrium reaction: From Appendix 2B, we find 2 HBrO + 2 H+
+
2 e-
---+
Br2
+
2H 0 2
EO
=
+1 .60 V
EO
=
+0.76
On examination of these equations, it is clear that we will also need a half reaction that, when combined with the two above, will eliminate Br2 and Br- . The obvious choice is We combine these by adding twice the reverse reaction to the other two: 2 HBrO + 2 H+ + 2 e-
---+
Br2 + 2 H 2 0 2(Br- + 2 OH- ---+ BrO- + H 2 0 + 2 e- ) Br2 + 2 e- ---+ 2 Br-
Caution:
We must be careful here in adding the EO values-we have
created essentially a new half-reaction by summing these reactions, which requires that we convert to !:1G values. Whenever one sums more than two half-reactions, it is necessary to convert to the !:1G values using !:1Go
=
nFE o ,
in order to work the problem:
2 HBrO + 2 H+ + 2 e-
---+
Br2 + 2 H 2 0 !:1Go -2(9.65 x 1 04 C · morl )( + 1 .60 V) -309 kJ . mol - I BrO- + H 2 0 + 2 e- ---+ Br- + 2 OH!:1Go -2(9.65 x 1 0 4 C · mol-' )(+0.76 V) - 1 47 kJ · mol-' Br2 - + 2 e- ---+ 2 Br!:1Go -2(9.65 x 1 04 C · mor l )( + 1 .09 V) -2 1 0 kJ . mol- I For 2 HBrO + 2 H+ + 4 0H- ---+ 2 BrO- + 4 H 2 0 !:1Go -309 kJ + 2(+ 1 47 kJ) - 2 1 0 kJ -225 kJ · mol-' We now see that we will need to eliminate OH- from the left side of the =
=
=
=
=
=
=
=
equation. This can be done in one of two ways: we can use the Kw value SM-400
for the autoprotolysis of water or, equivalently, we can use appropriate half-reactions that sum to the autoprotolysis of water. The appropriate half-reactions are EO
= -
1 23 V .
EO +0.40 V
0 2 + 2 H 2 0 + 4 e- � 4 OH
=
These sum to give EO -0.83 V =
This is a 4 e- reaction. Alternatively, one can write the 1 e- process that will have the same EO value. EO -0.83 V =
I1GO -(1)(9.65 1 0 4 C · mol-' )( -0.83 V) +80 kJ . mol- I =
=
x
I1GO -225 kJ . mol-I =
4(H 2 0 � H+ + OH- )
4(I1Go +80 kJ . mol-I )
2 HBrO � 2 H+ + 2 BrO-
I1Go
=
= =
-225 kJ mol- I + 4(+80 kJ . mol-I ) +9 5 kJ . mol-I .
The desired reaction is half of this, for which I1Go +48 kJ . mol-I . =
Using I1Go
=
-RT
In K, we obtain K
=
4 X 1 0-9 , which is in reasonable
agreement for this type of calculation with the value of 2 1 0-9 given in x
Table 1 0. 1 . 12.117
In order to detennine the current applied, we need to find the number of moles of electrons transferred. The electrolysis of water to produce gaseous oxygen and hydrogen,
2 H 2 0(I) � 0 2 (g) + 2 H 2 (g) , transfers 4 moles of electrons for each mole of oxygen gas produced: 4 OH- (aq) � 0 2 (g) + 2 H 2 0(l) + 4 e- . We can detennine the number of moles of oxygen from its volume, partial pressure, and temperature.
SM-40 1
1L 1 atm (722 Torr- 1 9.83 Torr)(25.0 mL) . (0.08206 L · atm · K- ' mor ' )(295 K) 1 000 mL 760 Torr = 9.542 x 1 0-4 mol 0 2 produced =
---
4 mol e 4 mol e= 9.542 x l O-4 mol 0 2 x mol O mol 02 2 ) = 3 _8 1 7 x 1 0- mol e-
n _ = no e
It
2
x
= nF
) C (3.8 1 7 x 1 0- mol e- ) . 96 485 = t (30.0 min)(60 s · min-' ) mol e= 0.205 A
(
I = nF
12.119
(a) (i) 2H+ (aq) + 2eEff! = EO
_
RT nF
-+
)
H 2 ( g)
In Q
Eff! = 0 - 0.4 1 V = -0.41 V
-
(ii) NO) (aq) + 4H+ (aq) + 3e Eff! = EO
_
RT nF
-+
NO(g) + 2H 2 0
EO
=
+0.96
V
In Q
-
Eff! = + 0.96 V 0.55 V = + 0.4 1 V (b) NAD+ (aq ) + H + (aq) + 2 eEff! = E "
_
RT nF
-+
NADH(aq)
ln Q
SM-402
EO = -0.099 V
[1] ln E(JJ = Eo _ 0.025693V 2 [1 ] [1O-7 J V In I 07 E(JJ -0.099 0.025693 2 E(JJ -0.099 V -0.207 V -0.3 1 V (c) V
=
_
=
=
(d)
E(JJ = +0.3 1 V NADH(aq) � NAD+ (aq) + H+ (aq) + 2epyruvate + 2H+(aq) + 2e- � lactate E(JJ =-0.190V NADH(aq) + pyruvate + H+(aq) � NAD+(aq) lactate E(JJ +0.12 V E(JJ =Eo _ RT ln Q E" =E(JJ + RT In Q [Iactate] [NAD+(aq)J E = E(JJ + 0.025693V In [NADH(aq) ][pyruvate] [ H+(aq) J [1][1 ] EO =+0.12V + 0.025693V In 2 [1][1] [1O-7 J E" =+0.I2V + 0.O I28465V InI07 E" =+0.12 V + 0.207 V E" +0.33 V (e) - Eo -2(9.64853 1 +0.33 V) -63.7 = -64 - FEo = -RT In and +
nF
nF
=-=-=---"---= �
°
-
n
=
6G,�
6G,�
=
6G,�
=
(f) 6 G,�
=
nF
x
klmo\- I K
04 c.mor l )(
kJ.mor l 6G,�
=
SM-403
n
=
In K =
Therefore, K
nFE o RT
liFE' =
e
RT
( +63700 J.moI,I ) ) (8.3 1 4 J.K,I .moI'1 )(298. 1 5 K) K = e2 5 .69 K- e(
K = + 1 .43
x
lOl l
SM-404
CHAPTER 13 CHEMICAL KINETICS 13.1
(a) rate (N ?- ) = rate(H?- ) x (b) rate ( NH 3 )
13.3
(
=
rate(H? ) x -
)(
)
1 mol N 2 1 = - x rate(H 2 ) 3 3 mol H 2
(
2 mol NH 3 3 mol H 2
)
) 8
=
(a) The rate of formation of dichromate ions 0. 1 4 mol cr2 0 � · L·s
2 x rate(H 2 ) 3
-
=
2 mol cro �· ----� = 0.2 mo I · L · s 1 mol Cr2 0 2 7
(b) 0. 1 4 mol · L-1 . S - I -;.- 1
13.5
(
=
(a) rate of fOimation of O?-
-
0. 1 4 mol · L-1 . S -I
= =
(
6.5 x 1 0-3
)(
mol N0 2 1 mol ° 2 x L.s 2 mol N0 2
3.3 X 1 0-3 (mol ° 2 ) L-1 •
(b) 6.5 x 1 0-3 mol · L- 1 . S - I -;.- 2 = 3.3 X 1 0- 3 mol · L-1 13.7
I
•
. S
S
-I
-I
(a) and (c) '"' 1 2 10
"7 .J
--0--
-
0
E
-S = 0
....
e.:
.:: = ..
c.> = 0
u
�
8
[h], mmo1·L- 1 [HI], mmol·L- 1
6 4 2 0 0
1 000
2000
3000 Time (s )
SM-405
4000
5000
6000
)
Note that the curves for the [1 2 ] and [H 2 ] are identical and only the [1 2 ] curve is shown. (b) The rates at individual points are given by the slopes of the lines tangent to the points in question. If these are determined graphically, there may be some variation from the numbers given below. time, s rate, mmol · L- ' . s- ' 20.0060
0
1 000 20.003 2000 20.000 98 3000 20.000 6 1 4000 20.000 40 5000 20.000 3 1 13.9
For A
�
products, rate
=
(mol A) . L-' . s- '
(a) [(mol A) · L- ' . s- ' ] ko [A t =
=
ko ' so units of ko are (mol A) . C ' . s- '
(same as the units for the rate, in this case) (mol A) · C ' . s- ' . (b) [(mol A) · L-' · s- ' ] k, [A], so umts of k, are (mol A) · L =
_
,
=
s- '
(mol A) . L- ' . s- ' . [(mol A) . L- ' . s-' ] k, [A] 2 , so umts of k, are [ (mol A) · C' J 2 (c) =
=
13.11
L · (mol Ar ' · s- '
From the units of the rate constant, k, it follows that the reaction is first order, thus rate k[N 2 0s J. =
rate 5.2 x 1 0-3 s - ' x 0.0426 mol · L-' =
SM-406
=
4 2.2 X 1 0- (mol N 2 0 s ) · C ' · s- '
13. 13
(a) From the units of the rate constant, it follows that the reaction is second order; therefore, rate = k[112 ][1 2 ]
(
)(
)(
)(
I1 = (0.063 L . mol- l . s- I ) 0.52 g 2 I mol112 0. 1 9 g1 2 Imol1 2 0.750 L 2.0 16 g112 0.750 L 253.8 g I? 2.2 X1 0-5 mol · L- 1 S- I (b) rate(new) = k x 2X[11 2 Litial [1 2 ] = 2 x rate(initial), so, by a factor of2 =
•
13.15
Because the rate increased in direct proportion to the concentrations of both reactants, the rate is first order in both reactants.
13. 1 7
When the concentration of ICI is doubled, the rate doubles (experiments 1 and 2). Therefore, the reaction is first order in ICI. When the concentration of112 is tripled, the rate triples (experiments 2 and 3); thus, the reaction is first order in 112 . (a) rate = k[ICI][11 2 ] (b)
Ie
(
= 22 x
�o-: mOI )( 3 0 I�-J mOI )( 45 I�-J mOl )
= 0 . 1 6 L · mol- I . S-I
( )(
x
x
)(
3 3 (c) rate = 0. 1 6 L 4.7XI0- mOI 2 . 7 x 1 0- mOl L L mol· s = 2.0 X1 0-6 mol · L- 1 • S- I 13. 19
)
(a) Doubling the concentration of A (experiments 1 and 2) doubles the rate; therefore, the reaction is first order in A. Increasing the concentration ofB by the ratio 3.02/1 .25 (experiments 2 and 3) increases the rate by (3.0211 .25) 2 ; hence, the reaction is second order in B. Tripling the
SM-407
)
concentration of C (experiments 3 and 4) increases the rate by 3 2 = 9; thus, the reaction is second order in C. Therefore, rate = k[A][B] 2 [Cf. (b) overall order = 5 (c) k = rate [A][Bf[Cf Using the data from experiment 4, we get 2 2 L L L k = 0.45 7 mo I 1 .25 X1 0-3 mol 3.02 x 1 0-3 mol 3.75 x 10-3 mol L ·s = 2.85 x 1 01 2 L4 . mor4 . S- I From experiment 3, we get 2 2 mOI L L k = 5.08 X1 0L.s 1 .25 X1 0-3 mol 3.02 x 1 0-3 mol
(
J(
[ C
J[ � moJ 12 4 . 4 .
x
J(
J
J(
J[
J
25 x l -'
= 2.85 x 1 0 L mor S- I (Checks!) 2 3 I2 4 -3 (d) rate = 2.85XI4 O. L 3.01Xl O- mOI 1 .00XIO mOl L L mol s l5 x mol x =
13.21
[ C
�-'
J[
J[ J
'
1 . 1 3 X10-2 mol · L- I . S- I
(a) k 0.693 0.693 6.93 x 1 0-4 s - I tl/2 1 000 s [A]0 kt and solve for k. = (b) We use In [AJ -I In 0.67 mol · L I 0.53 mol · Lk = In ([A]o /[Al) = 9.4 X1 0-3 s - I 25 s t =
=
=
( J __
(
J
SM-408
=
J
1 1 53 0 (c) [AJ, ( :0 A ) [(�:::�J( 0.034Lm OI B )] = 0. 0 85 (mol A) . L-I mol · L-II In ( 0.0.153 L- J = 5.1x 10- s- I k = 085115mol· s (a) t112 = 0.6k93 = (3.0.7 69310-s 5 ) ( 160misn ) ( 601 mih n ) = 5 .2 h [AJ = [ AJa e-k( t = 3.5 h x 3600 s · h-1 = 1 .3 x104 s _
�
3
1 3.23
X
(b)
(c) Solve for t from In ([A[AJJa) = kt, which gives 567 ) In ([A[AJJa( ] = In ([[ NN220055JaJ( ) 1n ( 0.0 00135 4 t= k = =3. 9 ' s 10 x 5 k 3. 7 x 10- S-I 1 min 1 = 6.5 x lO-mm = (3. 9 x 104 s) (-60 s (a) [ [AAJaJ = 2.4 = (2.2 )2 ; so the time elapsed is 2 half-lives. t = 2 x 355 s = 710 s Because 15% is not a multiple of t, we cannot work directly from the half-life. But k = 0.693/tll2 so k = 0.355693s = 1.95 x 10-3 S- I Then [see the solution to Exercise 13.23(c)J, ?
1 3 .25
(b)
SM-409
] A [ ( In [ A ]O( 1) In ( 0.151 ) = 9.7x102 s k 1. 95x10-3 I_-n ,,--[ aA_A_]]o-o =-- In 9 1.1 x 1 03 (c) k 1.95 10-3 (a) 1/2 0.6k93 2.81 0.693 10-3 min-I 247 min (b) See the solutions to Exercises 13.3 1(c) and 13.3 3(c). In ([[ SS0022CClI2z]Jo{ ) In 10 = = 2.81x 10-3 min-I 819min k (c) [Al [A]o e-k{ Because the vessel is sealed, masses and concentrations are proportional, and we write (mass left){ (mass)o e-k{ 14.0 gx 10.9 g Note: Knowledge of the volume of the vessel is not required. However, we could have converted mass to concentration, solved for the new concentration at 1.5 h. and finally converted back to the new (remaining) mass. But this is not necessary (a) We first calculate the concentration of A at 3.0 min. [AJ, [ AJo G::��)x [ B], 0.0 15 mol · L-I - ( 3l momolI AB)x 0.0 18 (mol B) · 0.009 mol · SM-410 t=
-l S
=
t=
1 3.27
t
=
=
=
X
-I S
=
S
=
X
t=
=
=
e - ( 2 . 8 I x 10-3 min-I x60 min
=
·il-I x 1.5
il)
=
1 3.29
�
-
L-I
=
=
L-I
The rate constant is then determined from the first-order integrated rate law. In ( [A[AJlo ) n ( 0.0 15 ) 3. 0D.OO9 min 0. 17 min-I (b) [Al 0.0 15 mol · - ( 31 molmol AB J 0.030 (mol B) · 0.005 mol · In ( [[ AAJJo( ) n ( 0.0 15 ) 0.17o.oos min-I 6.5 min additional time 6. 5 min -3.0 min 3.5 min (a) 1
k
=
=
t
=
=
L-1
=
1 L-
x
L-1
1
t=
=
k
=
=
13.3 1
=
35.00 30.00
-- 25.00
�
; 20.00
� '-'
-... -
15.00
.....
10.00
� --
l /[HIJ
(0.0078 . + 1.0 t)
=
5.00 0.00 2000
1000
0
3000
4000
5000
Time (s)
Equation 17.b in the text can be rearranged as _ 1 _ 1 [ A Jo 1 + [ A l [ A Jo [ A Jo Thus, if the reaction is second order, a plot of against time should give a straight line of slope As can be seen from the graph, the data fit the equation for a second-order reaction quite well. The slope is determined by a least squares fit of the data by the graphing program. SM-411 =
+
kt
=
__
kt
l /[HI]
k.
(b) (i). The rate constant for the rate law for the loss of is simply the slope of the best fit line, 7. 8 x 10-3 L . mor'. s-'. (ii). Since two moles of are consumed per mole of reaction, the rate constant for the unique rate law is half the slope or 3.9 x 10-3 L . mor'. s-'. It is convenient to obtain an expression for the half-life of a second-order reaction. We work with Eq. 17.b. 1 ( 17.b) --=21 1 1 = 2, or = 1, or Therefore, 1 1 and = 1 It is also convenient to rewrite Eq. 17.b to solve for We take reciprocals: 1 1 gIVIng 1 1 = (a) 1 (50.5 s)(0.814 mol · L-') = 0.024 L · mor' . s-' 1 1 16 15 _ = 15 . . (0.024 L mol-' s-') (0.84 mol · L-') =7.4 xl0 s HI
HI
1 3.33
[AJ t
[AJo
+ [AJokt
[Al'12
----
--
+
[AJo
[AJo ktl/2
+ [AJo kt'l2
tl/2 =
--
k[AJo
k
[AJo kt'/2
---
tl/2 [AJo
t.
--=-- + kt [Al [AJo
-----
[Al
t
[A Jo
.=..:. ...---"-"-=-----=-
k=
k
t'I2[AJo
----
[AJo [AJ �� -=--=--k
t
=
----[AJo [AJo k
k[AJo
2
=
SM-412
(b) ==
t=
1 4 ---[A]o [A]o k
=
3 k[A]o
3 1.5 x l 0 s . . (0.024 L mol- I s- I ) (0.84 mol · L-I ) 2
==
5 1 ----[A]o [A]o
4 k k[A]o 4 = = 2.0 x 10 2 s (0.024 L · mol-I . S-I ) (0.84 mol · L- I )
(c) t =
13 .35
=
See the solution to Exercise 1 3 .33 for the derivation of the formulas needed here. 1 1 lL lL mol 1 .7 x 102 min (a) t = [A] [A]o = 0.080 mol 0. 10 k 0.01 5 L · mol-I . min - I
[(
)(
=
[A] 0.15 mol A _ 0. 1 9 mol B 1 mol A L L 2 mol B = 0 . 055(mol A) . L-I = 0.37[A] o 1 1 [Al-'-----'[A]o t=k 1 1 0.055 mol · L-I 0. 1 5 mol · L-I = 0.0035 L · mOrl . min - I 3.3 X 103 min
(b)
=
---
=
1 3.37
-
rate = � d[A] = k[A] a dt d [A] = -ak dt [A] integrate from [A]o at t 0 to [Al at t : r A J, d[A] = -ak r dt .b �A j(J [A] =
SM-41 3
J
]
[A] ( = -akt, and [Al = [A] exp(-akt) Ino [A]o For the half-life. At t l12 , [Al = li [A]o . Therefore: [A]o = In 2 = akt , and In -l/ 2 [A]( In 2 t l /2 = ak 1 3.39
Given: d[A] = -k[A]3 , we can derive an expression for the amount dt of time needed for the inital concentration of A, [A]o , to decrease by 1/2. Begin by obtaining the integrated rate law for a third-order reaction by separation of variables: f�� [A]"3 d[A] = r -kdt = - [[Ar2 _[A] �2 ] = -kt
�
To obtain an expression for the half-life, let [Al = /i[A]o and t =tl l2 : - [ (HA]o r2 _ [A]�2 J = -kt1 / 2
�
solving for the half-life: t1 / 2 13.4 1
13.43
3
= --::-
2k[A]�
2AC + B A2B + 2C Intermediate is AB. �
The overall reaction is CH2=CHCOOH + HCI � CICH2CH2COOH. The intermediates include chloride ion, CH2=CHC(OHh and +
13.45
(a) (b)
+
Overall reaction: 03 0 202 Rate Law Step 1 rate = k[03][�0] Step 2 rate k[�02][0] Reaction intermediate: �02 =
(c)
SM-414
�
Molecularity bimolecular bimolecular
(d) 1 3.47
Catalyst:
NO
The first elementary reaction is the rate-controlling step, because it is the slow step. The second elementary reaction is fast and does not affect the overall reaction order, which is second order as a result of the fact that the rate-controlling step is bimolecular. rate k[NO][Br2 ] =
1 3.49
Ifmechanism (a) were correct, the rate law would be rate k[N02] [CO]. But this expression does not agree with the experimental result and can be eliminated as a possibility. Mechanism (b) has rate k[N02] 2 from the slow step. Step 2 does not influence the overall rate, but it is necessary to achieve the correct overall reaction; thus this mechanism agrees with the experimental data. Mechanism (c) is not correct, which can be seen from the rate expression for the slow step, rate k[N03] [CO] . [CO] cannot be eliminated from this expression to yield the experimental result, which does not contain [CO]. =
=
=
1 3.5 1
1 3.53
(a) True; (b) False. At equilibrium, the rates of the forward and reverse reactions are equal, not the rate constants. (c) False. Increasing the concentration of a reactant causes the rate to increase by providing more reacting molecules. It does not affect the rate constant of the reaction. The overall rate of formation of A is rate = - k[A] + k'[B] . The first term accounts for the forward reaction and is negative as this reaction reduces [A]. The second term, which is positive, accounts for the back reaction which increases [A]. Given the I : I stoichiometry of the reaction, if no B was present at the beginning of the reaction, [A] and [B] at any time are related by the equation: [A] + [B] = [AL where [A]o is the initial concentration of A. Therefore, the rate law may be written: SM-4 1 5
d[A] = -k[A] + k' ([A] -[AD = -(k + k' )[A] + k'[AJo dt The solution ofthis first-order differential equation is: k kl k' [A] = + 'ke-( '+ ) [A] o k +k As t � 00 the exponential tenn in the numerator goes to zero and the concentrations reach their equilibrium values given by: and [B]eq = [A] o - [At, = k[A] k + k' taking the ratio of products over reactants we see that: [B]eq = !5...- = K where K is the equilibrium constant for the reaction. [A]eq k' 0
0
__
1 3 .55 0
..-.. '-' �
.s
-2 -4
-6 -8 -10
0.0011
0.00115
0.0012
0.00125
0.0013
0.00135
(a) Given the Arrhenius equation, In k = In A - Ea /RT, we see that the slope of the best fit line to the data ( - 3 .2 7 x 104 K) is Ea / R and the y intercept (3 5.0) is In A. Therefore, E a =(3 .27xl04 K)(8.3 1xlO-3 kJ.mol-I ·K-1)=2.72xl02 kJ·mol-l•
(b) At 600°C (or 873 K), the rate constant is: 1 + 3 5.0 = In(k) = (-3.27 X 104 K ) --2.46 873 K k = 0.088
SM-416
1 3 .57
(�k ) ( T T'1 ) (T'T'-TT ) In (: ) In (��� ::: ) K-I000 K) - ( 8.3 1xl0-3 kJ · K-' · morl )( 1030 1030 Kxl000 K (8.3 1 x 10-3 kJ · K . moI-1 )(1000 K)(1 030 K) In[ 0. 87 ) (1030 K -1000 K) 0. 7 6
W e us e In
Ea �
=
_
=
_
R
Ea R
=
Ea
E
a
=
13.59
39 kJ . moI-1
W e use In
(�k )
Ea
=
R
( �T T'1 ) _
=
Ea
(T'TT'- T)
-I
In
(:' )
=
=
R
=
+
=
� (� ;, ) � (T�;,T) _
=
K -318 K 323 )( ) (8.3 14x10310-3kJkJ· mol-' 0.60 · K-' · mol - ' 318 K x 323 K =
�k = 1.8 k' 1. 8 x 5.1 =
_
=
=
13.63
S
700°C, T' (700 273) K 973 K . 973 K - 1073 K ( In (�) -- ( 315 kJ mol-' k 8.3 14xl0-3 kJ · mor' J 973 Kxl073 K J -3. 63; -kk' 0.026 =
k ' = rate constant at
13.6 1
-I S
-I
=
x
1O -4 s-'
=
9.2
X
1 O -4 s -'
(a) The equilibrium constant will be given by the ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction: L · mol- I . min - ' � K k' L . mol-I. min- ' =
=
265 392
=
0.676
SM-417
(b) The reaction profile corresponds to a plot similar to that shown in Fig. 13 .31a. The reaction is endothermic-the reverse reaction has a lower activation barrier than the forward reaction. (c) Raising the temperature will increase the rate constant of the reaction with the higher activation barrier more than it will the rate constant of the reaction with the lower energy barrier. We expect the rate of the forward reaction to go up substantially more than for the reverse reaction in this case. k will increase more than k' and consequently the equilibrium constant K will increase. This is consistent with Le Chatelier' s principle. 1 3 .65
(a) cat = catalyzed, uncat = uncatalyzed Ea,cal = l iss Ea,uncal = 0,60 Ea,uncal .,.,,", / RT rate( cat) kcat Ae / RT e rate( uncat) kuncat Ae E / RT / RT / RT =e =e -(0.60)£, .
-£",",
=
=
=
-
",IlIlC;lI
(0.40)£",.,.,,",
(-0.60+1.00)£, .•",,",
1
1
1
= e[(0.40)(I2S kJ·mol- )/(8.314xI0-3 kJ·K- ·mol- x 298 K)j = 6 x 108 (b) The last step of the calculation in (a) is repeated with T= 350 K. The rate enhancement is lower at higher temperatures. 1 3.67
cat = catalyzed k rate(cat) � 1000 = A e rate(uncat) kuncat Ae In 1000 Ea,cal = Ea - RT In 1000 = 98 kJ . mor l - (8.31 x 10-3 kJ K -'---'--
__
-E",", 1 RT
=
=
-E,
1 RT
.
=
1 3.69
=
-I
-E, ,", 1 RT .
e e
_ -=---:-::-=-- E, 1 RT
. mor l ) (298 K) (In 1000)
81 kJ·mol-I
The overall reaction is RCN + H 2 0 � RC(RO)NH 2 ' RC(=�)OH and RC(=NH)OH are intermediates. The hydroxide ion serves as a catalyst for the reaction. SM-418
13.71
(a) False. A catalyst increases the rate of both the forward and reverse reactions by providing a completely different pathway. (b) True, although a catalyst may be poisoned and lose activity. (c) False. There is a completely different pathway provided for the reaction in the presence of a catalyst. (d) False. The position of the equilibrium is unaffected by the presence of a catalyst.
1 3.73
(a) To obtain the Michaelis-Menten rate equation, we will begin by employing the steady-state approximation, setting the rate of change in the concentration of the ES intermediate equal to zero: d[ES] = k [E][S] - k' [ES] k [ES] = o. - 2 \ \ dt Rearranging gives: [E][S] = k2 + k'\ [ES] = KM [ES]. k\ The total bound and unbound enzyme concentration, [E]o, is given by: [E]o [E] [ES], and, therefore, [E] [ES] - [E]o. Substituting this expression for [E] into the preceding equation, we obtain: ([ES] - [E]o)[S] = KM [ES]. [E] [S] Rearranging to obtain [ES] gives: [ES] = o . K + [S] M From the mechanism, the rate of appearance of the product is given by rate k2[ES]. Substituting the preceding equation for [ES], we obtain:
(
=
+
J
=
=
the Michaelis-Menten rate equation, which can be rearranged to obtain: 1 K M = + rate k2 [E] o [S] k2 [E] o
---
SM-419
.
1 , the slope wlll be 1 versus If one plots [S] rate
will be
KM
k2 [E]o
. t and the y-mtercep
k [E]o 2
--
(b) slope =
k k2[Elo m
__ _
y-intercept o
13.75
=
1
k2[Elo
_ _
1/[S]
(a) rate k[NO]2, bimolecular (b) rate k[Ch], unimolecular (c) rate k[N02]2, bimolecular (d) b and c =
=
=
1 3 .77
(a) Use experiments 1 and 4 to show that [C] is independent of the rate. Use experiments 2 and 4 to solve for the order with respect to A: 10
Therefore a 1. =
2
Use experiments 2 and 3 to solve for the order with respect to B: 200b 1 6 Therefore b 2. -- = -
100
=
4
Overall order 3 (b) rate k[A][Bf (c) Using experiment 1: 2.0 mmol L-1 S-l k(lO mmol L-1) (l00 mmol L-1/ =
=
=
SM-420
k = 2 . 0 x 1 0-5 L2 mmo 1 -2 s- 1 (d) rate = k[A][B] 2 =(2.0 x 10-5 L2 mmor2 s- I )(4.62 mmol L-1)(0. 177 mmol L- I )2 = 2.9 X 10-6 mmol L-1 S-I 1 3.79
k/
S Ince K =
k
.
Then kloss 1 3.81
and
Given, K = 326
=
reverse
kQIIQch K
kallach
= 7.4 x 1 07 L.mol.s- I
=
kQUQch
=
7.4 107 L . mol.s· 1 = 2 . 3 105 L.mo 1 .s 1 326
k,oss
X
x
(a) t1/2 = 5 s since the flask contains 1 2 atoms at t = 0 and 6 atoms at t = 5s. (b) Since tll2 = 5 s = 0.693 then k = 0. 139 s-1 k First order, In[A] = In[AJo - kt Then at 8 s, In[A] = In 12 - (0. 1 39 s- I )(8 s) = 1.37 A = 3.94:::::; 4 There should be four molecules at t = 8 s.
1 3.83
>-. co �
�
.
c
Effect of catalyst (new pathway lowers Ea)
First activated complex
Second activated complex
......
1
+
M-!
Progress of reaction SM-421
B
1 3 .85
x =
amount of original sample = 25.0 mg n = number of half - lives
(l)n 2"
x x =
..
amount remammg
1 0.9 = 0.886 half lives 1 2.3
(1)°·886 2"
x
25.0 mg = 1 3.5 mg
1 3.87
The anticipated rate for mechanism (i) is: rate = k[CI2H2201 1], while the expected rate for mechanism (ii) is: rate = k[C1 2H22011 ][H2 0]. The rate for mechanism (ii) will be pseudo-first-order in dilute solutions of sucrose because the concentration of water will not change. Therefore, in dilute solutions kinetic data can not be used to distinguish between the two mechanisms. However, in a highly concentrated solution of sucrose, the concentration of water will change during the course of the reaction. As a result, if mechanism (ii) is correct the kinetics will display a first-order dependence on the concentration of H20 while mechanism (i) predicts that the rate of the reaction is independent of [H20].
1 3 .89
(a) The objective is to reproduce the observed rate law. If step 2 is the slow step, if step 1 is a rapid equilibrium, and if step 3 is fast also, then our proposed rate law will be rate = k2 [N 2 O 2 ] [H 2 ] ' Consider the equilibrium [N 2 0 2 ] = � rNO f Substituting in our proposed rate law, we have k,g-
The assumptions made above reproduce the observed rate law; therefore, step 2 is the slow step. SM-422
(b)
>->
�
I=: r.r..:l
.......... ......
mI
�
5
o p..
Reactants
Products
Progress of reaction Note: The dips that represent the formation of the intermediate N 2 0 2 and N 2 0 will not be at the same energy, but we have no information to determine which should be lower. 13.91
k' t' 484 hh k' ) _Ea (-1 --1 ) and solve for Ea ' We use In ( k' In ( k ) (8.314 1 0-3 kJ · K- ' . mor ' ) In (�) 4 rate at 28°C rate at 5°C
k
=
,
�eT
T
T'
x
R
E
R
:f
T'
(I ll 278 K 301 K
SM -423
=
75 kJ . mor '
1 3.93
(a) To obtain the Michaelis-Menten rate equation assuming a pre-equilibrium between the bound and unbound states of the substrate we begin with the expression for the equilibrium constant of the fast equilibrium between the bound and unbound substrate: K=
[���] '
solving for [ES] we obtain: [ES] = K[E][S]
as before in problem 1 3.69, the total bound and unbound enzyme concentration, [E]o ' is given by: [E]o = [E]+ [ES], and, therefore, [E]=[E]o - [ES] Substituting this expression for [E] into the equation above we obtain: K([E]o -[ES])[S] = [ES]. Rearrainging to obtain [ES]: [ES]= K[E]o [S] [E]o [S] l +K[S] K-' + [S] From the mechanism, the rate of appearance of the product is given by: rate = k2 [ES]. Substituting the equation above for [ES] one obtains: . 0 [S] , the Mlchaehs-Menten rate = k2 [E] rate equatIOn. K + [S] =
I
1 3.95
1 3.97
.
.
(a) three steps (b) first step (c) third step (d) two (e) none (a) CIO is the reaction intennediate; CI is the catalyst. (b) CI, CIO, 0, 02 (c) Step 1 and step 2 are propagating. (d) CI + CI � Cl 2
SM-424
1 3 .99
For a third-order reaction, constant t oc 1 ort = --112
112
--
[Aof
[ Ao]2
(a) The time necessary for the concentration to fall to one-half of the initial concentration is one half-life: first half-hfe = tl = t1 /2 = constant A ]2 o
[ o
(b) This time, t 1 /4 , is two half-lives, but because of different starting concentrations, the half-lives are not the same: secon d haIf-10I£e = t = constant = 4(constant) = 4t 2
(HAo])2
I
[Aof
(c) This time, t 1 / 1 6, is four half-lives; again, the half-lives are not the same: ° .c t constant 1 6( constant) - 1 6t thIro d haIf - llie - I 2 _
_
(HAo])
3
fourth half-life = t
_
4
=
[Aof
constant = 64(constant) = 64t
I
[AO]2
(t[Ao])2
If tl is known, the times tl/4 and tl/)6 can be calculated easilyo 13.101
1 3 . 103
The following plots are linear: (b)
(c)
(f)
(d)
(g)
(a) The overall reaction is: ocr + r � or + CI(b) The rate law will be based upon the slow step of the reaction: rate = k 2 [HOCl][1- ] Even though HOCI is a stable species because it is an intermediate in the reaction as written, technically we should not leave the rate law in this form The concentration of HOC1 can be expressed in terms of the reactants and products using the fast equilibrium approach: .
•
°
SM-425
K �' [HOCl][OH ] kl [OCl- ] [HOCl] � [ ocr ] kl' [OH-] rate k2 kl [OCI ] [I-] kl' [OH-] (c) An examination of the rate law shows that the rate is dependent upon the concentration of OH- , which means that the rate will be dependent upon the pH of the solution. (d) If the reaction is carried out in an organic solvent, then H 2 0 is no longer the solvent and its concentration must be included in calculating the equilibrium concentration of HOCl: K �' [HOCI][OH- ] kl [OCr ][H 2 0] =
=
=
=
= = [HOCI] = �' [OCl- ][H 2 0] kl [OH- ]
k rate = ll [OCI- ][r][H 2 0] kl' [OH- ] The rate of reaction will then show a dependence upon the concentration of water, which will be obscured when the reaction is carried out with water as the solvent. -=-- -=-----=."--::..=...-=-
1 3. 1 05
To get an expression for tl/2 in terms of n, we need to evaluate an integral such as: [AJ d[A] f[AJ" [A] " = k 0fdt = kt I
1 1 (1[Arl - [A]O1 "-I J = k -
n
-
-
-
t
An expression for tl/2 is then:
SM-426
( 2"�"-I - ; ,, I = ktl/2 [A] O [A O - ) 1 ( 2 "-1 = kt 2 [A]O"-1 ) J/ An expression for t3/4 could be found by setting up [A] = 1 ( 411-1 3,,- 1 [A]O"- I [AJo"- 1 ) = kt _-1 1 ((�[A]or�,'_� lJ = kt3/4 1
n -I
1
n-I
I
n-I
J,I
[A]o
3/4
n
[
The ratio t l /2 /t3/4 is then "- 1 - 1 tl /2 /t3/4 = 2 " 1 (4/ 3 ) - - 1
J
1 3 . 107 S]2
3
sp
+
H
�I I
+
H-C-N H
(a)
S]2 H
3
H
polar
C
I
methyl isonitrile
sp
--�I - + H
(b)
C:
C
N:
polar
cyanomethane To convert from one isomer to the other, a single C-N bond is broken and a single C-C bond is formed, giving an approximate enthalpy of reaction of: 305 kJ . mor l - 348 kJ . morl = -43 kJ · mol- I. Cyanomethane is the more stable isomer. SM-427
(c) The rate constant at 300 K is found using: k' Ea -1 --1 . S ub shtutmg . th'IS equatIOn . we fim d : ' . mto Ink R T T' l ' k 1 1 1 6 1 000 J· mor In 4 I 1 6.6 x 1 0- S-I 8.3 1 4S J · K - · mol - SOO K 300 K Solving for k ' : k' 4 . 04 1 0- 1 5 S-I At this rate, the time required to reduce the initial concentration of the isomer by 7S% is found by: C C In [ H ) N ]t = In(0.7S) = -k ·t = -4.04 x 1 0- 1 5 S-I. t [CH ) NC]o t = 7 . 1 2 x 1 01) s i (d) Since tJ-fl for this isomerization reaction is exothermic (- 43 kJ.mor ), the reaction profile has the product lower than the reactant. ( e) We begin by finding the rate constant at which the concentration of the isomer will reduce by 7S% in one hour: CH C In [ ) N ]t In(0.7S) -k . t -k . 1 h 60 min 60 s h min [CH ) NC]o Solving for k : k 7.99 x 1 0-5 S-I We can now determine at what temperature this rate constant is reached using the equation from part (c) above: ' In . 5 1 6 1 000 J · mol- 1 1 1 In 7 . 99 x 1 0--4 S-I 6.6 x 1 0 S-I 8 . 3 14S J· K-I . mol-I SOO K T Solving for T : T 474 K (f) Argon serves as a collision partner. Collisions between the C H3NC (g) and argon atoms provide the energy needed to overcome the activation energy and allow the isomerization reaction to proceed toward products.
)
= ( -
(
_
)
_
= x
= =
=
(
)( )
(
-)
=
: = � (� -; ) ,
=
=
SM-428
The argon atoms also serve as an energy sink, accepting the energy released during the isomerization reaction. (g) At high argon concentrations, collisions resulting in reaction are between the CH3NC(g) and argon atoms: CH3NC(g) + Ar(g) -7 CH3CN(g) +Ar(g). Given this elementary reaction one would predict that the reaction was first order with respect to CH3NC(g) concentration and would appear first order overall if the concentration of Ar(g) was large and unchanging. If, however, the concentration of Ar(g) was greatly reduced, the isomerization reaction would also proceed through collisions between CH3NC(g) molecules, making the reaction second order with respect to CH3NC(g).
SM-429
CHAPTER
14
THE ELEMENTS: THE FIRST FOUR MAIN GROUPS
14.1
(a) nitrogen; (b) potassium; (c) gallium; (d) iodine
1 4 .3
(a) sulfur; (b) selenium; (c) sodium; (d) oxygen
1 4.5
Tellurium < selenium < oxygen
1 4 .7
(a) antimony; (b) Antimony has higher effective nuclear charge.
14.9
(a) bromide ion ; (b) Bromide has larger size.
1 4. 1 1
(a) KCI because the ionic radius of K+ is larger than that ofLt (b) K-O. The higher charge on Ca2+ makes its ionic radius much smaller than that ofK+ .
14.13
2 K(s) + H 2 (g) � 2 KH(s)
14.15
(a) saline; (b) molecular; (c) molecular; (d) metallic
14.17
(a) acidic; (b) amphoteric; (c) acidic; (d) basic
SM-430
Oxidation number of C in C2 H 2 = -1 ; of C in H 2 CRCH 2 = -2, carbon has been reduced. (b) CO(g) + H 2 0(g) � CO 2 (g) + H 2 (g) (c) BaH 2 (s) + 2 H 2 0(l) � Ba(OH) 2 + 2 H 2 (g) 14.23
(a) CH 4 (g) + H 2 0(g) � CO(g) + 3 H 2 (g) r r = t!J r( C O, g) - [ Mfor( CH 4 , g) + Mfor(H 2 0, g)] = (-1 1 0.53 kJ · mol- I ) - [(-74.8 1 kJ · mol- I ) + (-24l .82 kJ · mol- I )] = +206. 1 0 kJ . mar l (b) ;').Sor = SaCCO, g) + 3 S0(H 2' g) - [ SO(CH 4 , g) - SO (H 2 0, g)] l = 1 97.67 J . K- I . mar + 3(1 30.68 J . K- I . mol- I ) - [1 86.26 J . K - I . mol- I + 1 88.83 J . K - I . mol-I] I = +2 1 4.62 J . K- . mar l (c) ;').G or = ;').Gor (CO, g) - [;').Gor (CH 4 , g) + ;').Gor (H 2 0, g)] = (-1 37. 1 7 kJ · mor l ) - [(-50.72 kJ . mar l ) + (-228 .57 kJ . mol- I )] = +1 42. 1 2 kJ . mar l
;').lfO
= +206.1 0 =
14.25
kJ . mol- I
+142. 1 4 kJ . mol- I
- (298 K) ( +21 4.62 J . K- I . mol-I )/(1 000 J . krl)
(a) H 2 (g) + Cl 2 (g) l ight ) 2 H CI(g) (b) H 2 (g) + 2 Na(l) � 2 NaH(s) (c) P4 (s) + 6 H 2 (g) � 4 PH 3 (g) (d) 2 Cu(s) + H 2 (g) � 2 CuH(s)
1 4 .27
•
The reason for this trend is mainly due to the trend of the electronegativity
SM-43 1
of the central atom (N
1 4.31
EO = 0.00 V (a) H 2 (g) � 2 H+ (aq) + 2 e02 (g) + 4 H+ (aq) + 4 e- � 2 H 2 0(1) EO = + 1 .23 V 2 H 2 (g) + ° 2 (g) � 2 H 2 0(l) EO = +l. 23 V The maximum potential possible is 1 .23 V. (b) The difficulty is isolating the two half cells but still maintaining electrical contact. Ions need to flow through the system to maintain charge balance in the reaction. In this case, a material that allows hydrogen ions but not hydrogen gas or oxygen gas to pass through would be necessary.
Lithium is the only Group 1 element that reacts directly with nitrogen to form lithium nitride: Lithium reacts with oxygen to form mainly the oxide: The other members of the group form mainly the peroxide or superoxide. Lithium exhibits the diagonal relationship that is common to many first members of a group. Li is similar in many of its compounds to the compounds of Mg. This behavior is related to the small ionic radius of Li+ , 58 pm, which is closer to the ionic radius of Mg2 + , 72 pm, but substantially less than that of Na+ , 1 02 pm .
1 4.33
(a) (b) (c) (d)
4 Na(s) + 02(g) 2 Na20(S) 6 Li(s) + N 2 (g) � 2 Li) N(s) 2 Na(s) + 2 H 2 0(l) � 2 NaOH(aq) + H 2 (g) 4 K0 2 (s) + 2 H 2 0(g) � 4 KOH(s) + 3 ° 2 (g) --+
SM-432
14.35
14.37
14.39
1 mol Na 2 C0 3 · 1 0 H 2 0 yields 1 mol Na 2 C03 in water. mass of Na 2 C03 · 1 0 H 2 0 = 0.500 L x 0.1 35 mol · L- l x 286. 1 5 g Na 2 C0 3 · 1 0 H 2 0 . mol-l = 1 9.3 g Na 2 C03 · 1 0 H 2 0 Mg(s) + 2 H 2 0(l) � Mg(OH) 2 + H 2 (g) (a) CaO(s) + H 2 0(I) � Ca(OH) 2 (s) (b) !)"GOr = !),.Gor (C a(OH) 2 ' s) - [!)"Gor(CaO, s) + !)"Gor(H 2 0, 1)] -898.49 kJ . mor l - [( -604.03 kJ . mor l ) + ( -237. 1 3 kJ . mol - I )] = -57.33 kJ · mol- l =
14.41
14.43
Be is the weakest reducing agent; Mg is stronger, but weaker than the remaining members of the group, all of which have approximately the same reducing strength. This effect is related to the very small radius of the Be2 + ion, 27 pm; its strong polarizing power introduces much covalent character into its compounds. Thus, Be attracts electrons more strongly and does not release them as readily as other members of the group. Mg2 + is also a small ion, 58 pm, so the same reasoning applies to it as well, but to a lesser extent. The remaining ions of the group are considerably larger, release electrons more readily, and are better reducing agents . 2 AI(s) + 2 OH- (aq) + 6 H 2 0(l) � 2[Al(OH) 4 r (aq) + 3 H 2 (g) 2 Be(s) + 2 OH- (aq) + 2 H 2 0(l) � [Be(OH) 4 ] - (aq) + H 2 (g) Be and Al are diagonal neighbors in the periodic table and exhibit similar chemical behavior.
SM-433
1 4 .45
(a) Mg(OH) 2 (s) + 2 HCI(aq) ---+ MgCl 2 (aq) + 2 H 2 0(l) (b) Ca(s) 2 H 2 0(l) ---+ Ca(OH) 2 (aq) + H 2 (g)
+
(c) BaCO) (s) � BaO(s) + CO2 (g) 14.47
(a)
: CI-
Be - CI :
2+ [ �l: ]-
Mg
2:
•
•
MgCl 2 is ionic; BeCl 2 is a molecular compound. (b) 1 80° (c) sp 1 4.49
+
(a) 3 BaO(s) + 2 AI(s) Ah03(s) 3 Ba(s) tJ/ro =tJ/ro (Ah03, s) - 3 tJ/ro (BaO, s) (-1 675.7 KJ/mol) - 3(-553.5 kJ/mol) -1 5.2 kJ For 1 .00 mol production ofBa(s), tJ/ro -5.07 kJ f:..Sro {f:..so (Ah03, s) + 3 f:..so (Ba, s)} - {3 f:..so (BaO, s) + 2 f:..so (AI, s)} (50.92 J + 3 x 62.8 J) - (3 x 70.42 J 2 x 28.33 J) -28.6 J For 1 .00 mol production of Ba(s), f:..Sro -9.53 J f:..Gro f:..Gro (Ah03, s) - 3 f:..Gro (BaO, s) (-1 582 . 35 kJ/mol) - (3 x 525.1 kJ) = -7.05 kJ For 1 .00 mol production of Ba(s), f:..Gro = -2.35 kJ (b) f:..GO Mr - Tf:..So; since M? is "-" and f:..SO is "-", decreasing in temperature will result a more negative f:..G and will increase the yield of barium. However, decreasing in temperature may slow down the reaction rate and may take longer time to produce same amount of Ba. �
=
=
=
+
=
=
=
=
=
=
25.0 g CaC2 x 1 mole = 0.3900 mol CaC2 64. 10 g 25.0 mL H 2 0 x1 .00 g x 1 mole = 1 3.87 moI H -? O 1 mL 1 8.02 g
SM-434
=
Since the reaction is 1 :2, only 0.780 mol of water would be needed to completely react with all of the calcium carbide. Calcium carbide is the limiting reagent while water is present in excess. mass ethyne CH CH 0.3900 mol CaC2 1 mol 2 2 26.04 g 2 2 1 0.2 g C 2 H ?1 mol CaC2 1 mol C2 H 2 =
14.53
14.55
(
J(
J
=
The overall equation for the electrolytic reduction in the Hall process is 4 Ae+ (melt) + 6 02 - (melt) + 3 C(s, gr) � 4 AI(s) + 3 CO 2 (g) (a) B 2 03 (S) + 3 Mg(l) �2 B(s) + 3 MgO(s) (b) 2 AI(s) + 3 Cl 2 (g) � 2 AICl 3 (s) (c) 4 AI(s) 3 02 (g) � 2 Al 2 03 (s)
+
14.57
(a) The hydrate of AICI 3 , that is, AICl 3 · 6H 2 0, functions as a deodorant and antiperspirant. (b) a- Alumina is corundum. It is used as an abrasive in sandpaper. (c) B(OH) 3 is an antiseptic and insecticide.
14.59
Since there are only 22 valence electrons, or 1 1 electron pairs, it is not possible to draw a good conventional Lewis structure for tetraborane, B4HIO, that includes four B-H -B bridges. For the suggested structure given below, each bridging H and each four-coordinate B would have a formal charge of }- while each of the six terminal H atoms and each three-coordinate B would have a formal charge of O. The total formal charge adds up to 6- in this case even though the molecule is neutral. However, if the bridges are viewed as 3-center 2 -electron bonds, then every atom can be assigned a formal charge of O.
SM-435
This drawing shows the structure of tetraborane See: www.chem.1eeds.ac.uklboronweb/artic1eslincredible boronlincredible.htm 1 4.6 1
The Lewis structure of GaBrc is: :Bf.
I
: Br-- Ga- Br:
I
:Br:
1 4 .63
The shape of the GaBrc is tetrahedral.
The cathode reaction is A13+ (melt) + 3 e- � Al(l).
(
J
charge consumed = (1 2.0 h) 3600 s C3 .0x1 06 C · S- I ) 1.3 X 1 01 1 C 1h
SM-436
=
mass of Al produced l mol AI 26.98 g AI = ( 1 .3 x 1 01 1 C) 1 mol e4 9.65 x 1 0 C 3 mol e- 1 mol Al = 1.2 x 1 07 g Al
(
14.65
J(
J(
J
(a) We want EO for T13 + (aq) + 3 e- � TI(s), n = 3. This reaction is the reverse of the formation reaction: Therefore, for the Te+ /TI couple, f:"Gor = -2 1 5 kJ · mol- l . 5 : . mol -I -2. 1 1 0 5 x = +0. 743 V = EO -3 x 9.65 x l 0 C · mol- I (b) Using the potential from part (a) and the potential from Appendix 2B for the reduction of TI+ , we can decide whether or not Tt will disproportionate in solution. The equation of interest is 3 Tt (aq) � 2 TI(s) + T13 + (aq). The half-reactions to combine are: TI+ � T13+ + 2 e(1) (2) TI+ + e- � TI The potential for reaction ( 1 ) must be obtained using the f:"Go values of the two known half-reactions: =
f:"Gor -nF
f:"GO = -nFEO
-1 (9.65 x 1 04 J . V- I . mor l )( 0.34 V)/(1 000 J . kr l ) = +33 kJ . mol- I f:"GO for the combined half-reaction Tt � T1 3 + + 2 e- is the sum of these two numbers: +2 1 5 kJ · mor l + 33 kJ . mol- I = +248 kJ . mol- I =
-
SMA37
We can now combine this with the reduction process for Tl+ to get the desired equation: I1GO = +248 kJ · mol -' I1Go 2(+33 kJ . mol- ' ) +2(Tl+ + e- -+ Tl) I1Go +248 kJ . mol - ' + 2(+33 kJ . mor ' ) 3 Tl+ -+ 2 Tl + Te+ = +3 1 4 kJ . mol- ' Since I1Go is large and positive, K« 1 . This disproportionation reaction does not favor products. =
=
1 4.67
1 4.69
Silicon occurs widely in the Earth' s crust in the form of silicates in rocks and as silicon dioxide in sand. It is obtained from quartzite, a form of quartz (Si0 2 ), by the following processes: ( 1 ) reduction in an electric arc furnace Si0 2 (s) + 2 C(s) -+ SiCs, crude) + 2 CO(g) (2) purification of the crude product in two steps SiCs, crude) + 2 Cl 2 (g) -+ SiCl 4 (1) followed by reduction with hydrogen to the pure element SiCl 4 (1) + 2 H 2 (g) -+ SiCs, pure) + 4 HCl(g) In diamond, carbon is Sp3 -hybridized and forms a tetrahedral, threedimensional network structure, which is extremely rigid. Graphite carbon is Sp2 -hybridized and planar. Its application as a lubricant results from the fact that the two-dimensional sheets can "slide" across one another, thereby reducing friction. In graphite, the unhybridized p-electrons are free to move from one carbon atom to another, which results in its high electrical conductivity. In diamond, all electrons are localized in Sp3 hybridized C-C (J- bonds, so diamond is a poor conductor of electricity.
SM-438
14.7 1
(a) +4; (b) +4; (c) +4
14.73
(a) MgC2 (S) + 2 H20(l) C2H2(g) + Mg(OH)2(S) (acid-base) (b) 2 Pb(N03)2 (s) 2 PbO(s) + 4 N02(g) + 02(g) (Redox) --+
--+
14.75
:0: I - O: :O-Si .. I :0: formal charges: Si 0, 0 - 1 ; oxidation numbers: Si +4, 0 -2; This is an AX 4 VSEPR structure; therefore, the shape is tetrahedral. =
1 4.77
=
=
=
Si0 2 (s) + 2 C(s) � SiCs) + 2 CO(g) WOr = WO (products) - WO f (reactants) = [(2) (-1 1 0.53 kJ . mor l )] - [-91 0.94 kJ . mol-I ] = +689.88 kJ . mol- I f...sor = SO(products) - SO(reactants) = [ 1 8.83 J . K -' . mOr l + (2) ( 197.67 J . K- ' . mol- I )] - [41 .84 J . K-' . mOrl + (2) (5.740 J . K- ' . mol-I)] = +360.85 J . K- ' . mOr l f...Gor = WOr T f...S or = 689.88 kJ . mOrl - (298. 1 5 K) (360.85 J . K-' . mol- I )/1 000 J . kr' = +582.29 kJ . mol- I The temperature at which the equilibrium constant becomes greater than 1 is the temperature at which f...GOr = RT In K 0 because In 1 o. Above f
-
-
=
=
this temperature, the equilibrium constant is greater than 1 . !1Gor 0 =
3 ' T = WOr = +68 9.88 x l 0 J · mol- = 1 9 1 2 K. f...sor +360.85 J · K- ' · mol-' SM-439
1 4 .79
mass of HF = =
1 4.8 1
g HF ) (3.00 10-3 g) ( 60.1 mol09 gSi0Si02 2 )( 16molmolSi0HF2 )( 20.01 1 mol HF 5.99 10-3 g HF 5.99 mg HF x
x
=
(a) The Si 2 0 t ion is built from two SiO/- tetrahedral ions in which the silicate tetrahedra share one ° atom. This is the only case in which one ° is shared. (b) The pyroxenes, for example, jade, NaAI(SiO) ) 2 ' consist of chains of Si04 units in which two ° atoms are shared by neighboring
units. The repeating unit has the formula SiO/- . See Figure 1 4.39. 14.83
SiF62-
14.85
Ionic fluorides react with water to liberate HF, which then reacts with the glass. Glass bottles used to store metal fluorides become brittle and may disintegrate upon standing on the shelf.
14.87
The iron ions impart a deep red color to the clay, which is not desirable for the manufacture of fine china; a white base is aesthetically more pleasing.
14. 89
(a) False. Four elements are metals (only B is a metalloid element) (b) False. B203 is an acidic anhydride. (c) False . The outer s- and p-orbitals become farther in energy going down the group because of the inter-pair effect.
14.91
In the majority of its reactions, hydrogen acts as a reducing agent. Examples are 2 H 2 (g) + 0 2 (g) � 2 H 2 0(l) and various ore reduction processes, such as NiO(s) + H 2 (g) � Ni(s) + H 2 0(g). With highly electropositive elements, such as the alkali metals, H 2 (g) acts as an
-0
SM 44
oxidizing agent and fOTITIS metal hydrides, for example, K(s) + H 2 (g) � KH(s).
2
2
1 4.93
(a) -2.5 �-------. -2.6 -2.7 • --;, ;,-- -2.8 kl -2.9 • • • -3 • -3.1 +-------�--�--� Na
Li
Rb
K
Cs
Group
1 elements
•
•
•
•
Ca
Sr
Ba
Ra
(b) -1.6 -1.8 -2 --;, -2.2 kl -2.4 -2.6 -2.8 -3
•
0
'-'
•
Be
Mg
Group 2 elements
(b) For both groups, the trend in standard potentials with increasing atomic number is overall downward (they become more negative), but lithium is anomalous. This overall downward trend makes sense because we expect that it is easier to remove electrons that are farther away from the nuclei. However, because there are several factors that influence ease SM-44l
of removal, the trend is not smooth. The potentials are a net composite of the free energies of sublimation of solids, dissociation of gaseous molecules, ionization enthalpies, and enthalpies of hydration of gaseous ions. The origin of the anomalously strong reducing power of Li is the strongly exothermic energy of hydration of the very small Li+ ion, which favors the ionization of the element in aqueous solution. 14. 95
(a) Acid-base reaction . The oxide ion, 02 - , in CaO acts as a Lewis base and reacts with the Lewis acid, Si0 2 , in a Lewis acid-base reaction: CaO(s) + Si0 2 (s) � CaSiOJ (1) Si0 2 , which is an impurity in iron ore, is removed by this reaction. The calcium oxide in this reaction can be obtained from limestone: [CaCOJ (s) � CaO(s) + CO2 (g)]. This is the reason that limestone is important in the iron industry. (b) CaO(s) + CO 2 (g) � CaC03 (s) (not an efficient preparation of CaCOJ because of the weak Lewis acidity of CO 2 ) .
1 4.97
(a) The reaction between diborane and ammonia is as follows: The cation is [H2B(NH3)2t, the anion is [BH4] (b) The hybridization of B in both the cation and the anion is sp3. (c) The reaction is a Lewis acid-base reaction because each N atom from ammonia provides a pair of electrons for a coordinate covalent bond. The oxidation numbers do not change (+ 3 for B and -1 for H).
1 4.99
In the majority of its reactions, hydrogen acts as a reducing agent; that is, H 2 (g) � 2 H+ (aq) + 2 e- , EO 0 V. In these reactions, hydrogen resembles Group 1 elements, such as Na and K. However, as described in the text and in the answer to Exercise 1 4.91, it may also act as an =
SM-442
oxidizing agent; that is, H 2 (g) + 2 e- � 2 H- (aq), EO = -2.25 V. In these reactions, hydrogen resembles Group 1 7 elements, such as CI and Br. Consequently, H 2 will oxidize elements with standard reduction potentials more negative than -2.25 V, such as the alkali and alkaline earth metals (except Be). The compounds fonned are hydrides and contain the H- ion; the singly charged negative ion is reminiscent of the halide ions. Hydrogen also fOnTIS diatomic molecules and covalent bonds like the halogens. The atomic radius ofH is 78 pm, which compares rather well to that ofF (64 pm) but not as well to that of Li ( 1 57 pm). The ionization energy of H is 1 3 1 0 kJ . mor I , which is similar to that of F (1 680 kJ . mol- I ) but not similar to that of Li (5 1 9 kJ . mol-I ). The electron affinity ofH is +73 kJ 'mor l , that ofF is +328 kJ ' mol- I , and that of Li is 60 kJ . mOr l . So in its atomic radius and ionization energy, H more closely resembles the Period 2 halogen, fluorine, in Group 1 7, than the Period 2 alkali metal, lithium, in Group 1 ; whereas in electron affinity, it more closely resembles lithium, Group 1 . In electronegativity, H does not resemble elements in either Group 1 or Group 1 7, although its electronegativity is somewhat closer to those of Group 1 . Consequently, hydrogen could be placed in either Group 1 or Group 1 7. But it is best to think of hydrogen as a unique element that has properties in common with both metals and nonmetals; therefore, it should probably be centered in the periodic table, as it is shown in the table in the text. 14.101
14.103
Because Lt has a smaller size than K+ and it has stronger ion-dipole interaction with H 20 than K+, which will make it move slower than K+. H 2 (g) + Br2 (1) � 2 HBr(g)
[
1 mol H 2 number of moles of HBr = (0. 1 3 5 L H?) - 22.4 L H 2 = 0.0 1 2 1 mol SM-443
J [ 21 molmol HBrH 2 J
molar concentratIon of HBr = 0.0121 mol = 0.0538 mol · L- I 0.225 L ·
14.105 The reaction no = 2
v
= =
14.107
14.109
is 2 Cs02 (s) + cO 2 (g) � CS 2 C03 (S) + t 0 2 (g).
(
(30 . 0 g) 1 mol Cs02 1 64.9 g Cs02
nRT
P
=
3.34 L
]( 2tmolmolCs002 2 ] = 0. 1 364 mol
l I. . (0. 1 364 mol) (0.08206 L atm · K - mor )(298 K) ( l .OO atm)
Species (a), (b), (c), and (d) can all function as greenhouse gases, while (e) cannot. Any molecule other than a homonuc1ear diatomic can exhibit a changing dipole moment as it vibrates with certain vibrational modes. Since argon is monoatomic it has no covalent bonds, no vibrational modes, and no dipole moment. CH 3 0H(l) +
Y;
02 (g) � CO 2 (g) + 2 H 2 0(l)
(
](
] ]( 1
number of kg of CO2 =1 . 00 L CH 3 0H 1 000mL 0.791 g CH 3 0H 1 mLCH 3 0H 1L I mol CH3 0H I mol C02 44.02 g C0 2 l kg 32.04 g CH 3 0H I mol CH 3 0H I mol C0 2 1 000g = 1 .09kg C02 This mass of carbon dioxide is about half the amount generated by combusting an equivalent volume of octane (2. 1 6 kg per liter). However, we also need to consider how much energy is produced per liter of fuel and how the mass of carbon dioxide produced compares for a given amount of energy produced. Standard enthalpies of combustion given in Appendix 2 are tlHco = -547 1 kJ . mol - I for octane and tlHco = -726 kJ . mol- I for methanol.
x(
l(
SM-444
](
(
)[
J
energy per L methanol = 1.00 L CH)OH 1000 mL 0.791 g CH ) OH 1L 1 mL CH)OH 726 kJ x 1 mole CH)OH 32.04 g CH)OH 1 mole CH)OH 4 = 1. 79 x 1 0 kJ energy per L octane = 1.00 L Cs H I S 1 000 mL 0.703 g CS H I S 1L 1 mL Cs H I S 5471 kJ x 1 mole Cs H I S 114.22 g CsH I S 1 mole CsH I S 3.37 x 104 kJ
[
(
[
=
)[
J[
J[
J
J
So the combustion of octane produces almost twice as much energy per liter as methanol (octane/methanol=1 .88). For an equivalent amount of combustion energy, methanol produces 1 .88 L x 1.09 kg CO 2 • C l = .05 kg CO 2 , which is still slightly less than octane. (However, it requires that the vehicle carry about 90% more fuel by volume, 1.9 vs. 1 L, and more than twice as much fuel by mass, 1 .5 kg vs. 0.7 kg.)
2
Diborane, B 2 H 6 , and Al 2 Cl6 (g) have the same basic structure in the way in which the atoms are arranged in space. (b) The bonding between the boron atoms and the bridging hydrogen atoms is electron deficient. There are three atoms and only two electrons to hold them together in a 3center-2-electron bond. The bonding in Al 2 Cl6 is conventional in that all the bonds involve two atoms and two electrons. Here, the lone pair of a Cl atom is donated to an adjacent AI. (c) The hybridization is sp) at the B and Al atoms. (d) The molecules are not planar. The Group 1 3 element and the terminal atoms to which it is bound lie in a plane that is perpendicular to the plane that contains the main group element and the bridging atoms.
14.111 (a)
SM-445
J
1 4. 1 1 3
(a) By viewing the unit cell from different directions, it is clear that it belongs to a hexagonal crystal system. (b) There are eight carbonate ions on edges, giving ± x 8 = 2 carbonate ions, plus there are four carbonate ions completely within the unit cell. The total number of carbonate ions in the unit cell is six. Calcium ions lie at the comers of the unit cell (t x 8 1) as well as on the edges (4 x ±) and within the cell (4). The total number of calcium ions in the cell is six, agreeing with the overall stoichiometry of calcite, CaC03 . =
1 4. 1 15
(a) C: si; B: si; N: not hybridized (b) A tube with a diameter of about 1 .3 nm will have a circumference of 271T or nd. Thus, the circumference of a 1 .3 x 1 0-9 m diameter nanotube will be (1 .3 x 1 0-9 m)(n) 4. 1 x 1 0-9 m . =
To calculate the number of C6 rings that will be strung together, we need to calculate the distance across the C6 ring as shown by the arrow . 1 4 2 pill ?
�
�-'
;y
'"
�
•
1 4 2 1'111
.0-:;.;.>
� �
'C.0-
�.,,-
......
I::;
1 4 2 1'111
J
The total distance from one carbon to the opposite carbon on the ring will be given by d = 1 42 pm + 2 (142 pm x sin 30°) = 284 pm.
SM-446
One repeat unit will be one benzene ring plus one C-C bond, 284 pm + 1 42 pm 426 pm or 4.26 x 1 0- 1 0 m. There are thus about 4. 1 x 1 0-9 -:- 4.26 x 1 0- 1 0 = 1 0 of units strung together around the smallest =
nanotube known. (c) In C60 the carbon atoms are si -hybridized and are nearly planar. However, the curvature of the molecule introduces some strain at the carbon atoms so that there is some tendency for some of the carbon atoms to undergo conversion to sp3 hybridization. However, to make every carbon sp3 -hybridized would introduce much more strain on the carbon cage and, after a certain point, further addition of hydrogen becomes unfavorable. (d) The spherical structures require the formation of five-member rings (see structure 10, C60). Boron nitride cannot form these rings because they would require high-energy boron-boron or nitrogen-nitrogen bonds. ( e) The unit cell described will contain a total of 4 B atoms and 4 N atoms. The volume of the cell is (361 .5 pm)3 = (3.6 1 5 x 1 0-8 cm)3 = or 4.724 x 1 0-23 cm-3 . The mass in the unit cell will be (4 x 1 0. 8 1 g . mol- I + 4 x 1 4.01 g . mor l ) -:- 6.022 x 1 023 mol- I 22 g. = 1 .649 x 1 022 g 1 .649 x 1 0-3 = 3 . 49 1 g · cm = d 2 3 4.724 x 1 0- cm} (f) Because the density of cubic boron nitride is greater than that of hexagonal BN, we would expect the cubic form to be favored at high pressures, exactly as found for the cubic (diamond) and hexagonal (graphite) forms of carbon.
SM-447
CHAPTER 1 5 THE ELEMENTS : THE LAST FOUR MAIN GROUPS
15.1
-3 -
2
-1 0
+1 +2 +3 +4 +5 15.3
NH 3 , Li}N, LiNH2 ' NH2 H2NNH 2 N2H2, NH20H N2 N2O, N2F2 NO NF3 , N02 - , NO+ N02 , N20 4 HN03 , N0 3 - , N02F
As the molar mass of the halogen increases so does it size. Nitrogen is a small atom and as the halogens get larger it becomes increasingly harder for nitrogen to arrange three of them around itself in a trigonal pyramid (due primarily to lone pair/lone pair repulsions on the halogens); therefore NX3 becomes less stable as X gets larger (X F, CI, Br, I). =
1 5.5
CO(NH 2 h (aq) + 2 H 2 0(l) � (NH 4 h C03 (aq) , 1 mol 1 0 3 g ,) ( mass of (NH 4 )2 C03 = (4 . 0 kg urea) (l -1 kg l 60.06 (NH 4 hC0 3 ) ( 1 mol (NH 4 ) 2 CO} ' ( 96.09 g (NH 4 ) 2CO } ' l 1 mol urea j l 1 mol (NH 4 )2 CO} j 3 = 6 . 4 x 1 0 g (or 6.4 kg) (NH4) 2 c0 3 SM-448
15.7(a) 1 mole of N 2 (g) occupies 22.4 L at STP. For the reaction Pb(N3) 2 � Pb + 3 N 2 , the volume of N 2 (g) produced is 1 . 5 g Pb(N 3 ) 2
(
I moI Pb(N 3 )? ) ( 3 mol N 2 ) ( 22.4 L N 2 ) l 29 1 .25 g Pb(N 3- )2 J l 1 mol Pb(N) 2 J l 1 mol N 2 J
(b) Hg(N 3 ) 2 would produce a larger volume, because its molar mass is less. Note that molar mass occurs in the denominator in this calculation. (c) Metal azides are good explosives because the azide ion is thermodynamically unstable with respect to the production of N 2 (g) . This is because the N:=N triple bond is so strong and also because the production of a gas is favored entropically. 15.9
+ +
N 2 0: H 2 N 2 0 2 ; N 2 0(g) H 2 0(l) � H 2 N 2 0 2 (aq) N 2 03 : HN0 2 ; N 2 03 (g) H 2 0(l) � 2 HN0 2 (aq) N 2 0 S : HN03 ; N 2 0 S (g) + H 2 0(l) � 2 HN03 (aq)
15.1 1 Ammonia (NH3 ) can undergo hydrogen bonding with itself while NF3 can not. 15.13 The Lewis structures of phosphorous (III) oxide (P406) and phosphorous (V) oxide (P4010) are: ·0 ·
" : 0 -- \ 0 : 1 : 0 :P1 -=:: :::::::--- 0 -:;?' \ 0- I - \ . P \\ •
•
•
°
..
phosphorous ( I I I) oxide SM-449
•
/ p --
.
.
p
.
.
•
•
.
•
.
: .0-- \\-0 ..: . . 0 ·· . oxide phosphorous (V)
The basic structure of the two molecules is the same. The phosphorus atoms lie in a tetrahedral arrangement in which there are bridging oxygen atoms to the other phosphorus atoms . In phosphorus(V) oxide, there is an additional terminal oxygen atom bonded to each phosphorus atom. In phosphorus(III) oxide, each oxygen atom has a formal charge of 0 as does each phosphorus atom. In phosphorus(V) oxide, this is also true. According to the Lewis structures, all P-O bonds in phosphorus(III) oxide have a bond order of 1 , while in phosphorus(V) oxide, the terminal oxygen atoms have a bond order of 2 between them and the phosphorus atoms to which they are attached. If one examines the molecular parameters, one sees that all of the P- Obridging distances for phosphorus(III) oxide are slightly longer than those of phosphorus(V) oxide ( 1 63.8 pm versus 1 60.4 pm). This is expected because the radius of phosphorus(V) should be smaller than that of phosphorus(III). The terminal P = ° distances for phosphorus(V) oxide are considerably shorter (142.9 pm); this agrees with the higher bond order between phosphorus and these atoms. 15.15
(a) The reaction is 3 N0 2 (g) + H 2 0(l) � 2 HN03 (aq) + NO(g) I l t:J[ ° r 2( - 207.36 kJ · mor ) + 90.25 kJ · mol-[3(33. 1 8 kJ ' mol- I ) + (285.83 kJ ' mol- I )] -138. 1 8 kJ · mor l =
=
(b) HN03 (l) � HN0 3 (aq) t:J[ 0 r -207.36 kJ . mol- I - ( - 1 74. 1 0 kJ . mol-I ) = -33.26 kJ.mor l =
SM-450
15.17
(a) 4 Li(s) + 02 (g)
/), )
2 Li 2 0(s)
(b) 2 Na(s) + 2 H 2 0(l) � 2 NaOH(aq) + H 2 (g) (c) 2 F2 (g) + 2 H 2 0(I) � 4 HF(aq) + 0 2 (g) (d) 2 H 2 0(l) � 0 2 (g) + 4 H+ (aq) + 4 e-
(b) CaO(s) + H 2 0(l) � Ca(OH) 2 (aq) (c) 2 H 2 S(g) + S0 2 (g)
3 00°C,A I203
)
3 S(s) + 2 H 20(l)
15.21
Both water and ammonia have four groups attached to their central atom and therefore both possess a tetrahedral electronic (or VSEPR) geometry. However, H20 has two unshared electron pairs while NH 3 only has one, producing a larger dipole moment for H20.
15.23
(a) H I
• •
: 0-0 : I
H
Each ° in H 2 0 2 is an AX 2 E 2 structure; therefore, the bond angle is predicted to be < 1 09.5°. In actuality, it is 97°. (b)-(e), The reduction potential of H 2 0 2 is + 1 .78 V in acidic solution. It should, therefore, be able to oxidize any ion that has a reduction potential that is less than + 1 .78 V. For the ions listed, Cu+ and Mn 2 + will be oxidized. It would require an input of 1 .98 V to oxidize Ag+ to Ag 2+ and 2.87 V to oxidize F-.
SM-45 1
15.25
O�- + H2 0 � HO; + OH- essentially complete
00 1 0- 1 4 = 5.6 1 0-3 Ka l = 1 .8 1 0- 1 2 Kb 1 . 1 .8 1 0- 2 Because this Kb is relatively small, we can assume that essentially all the x
x
=
x
x
1
(J
(
OH- is formed in the first ionization; therefore, 1 mol OH[OH- ] = 2.00 g Na 2 02 1 mol Na 2 0 2 77.98 g Na 2 0 2 1 mol Na 2 0 2 0.200 L 0. 1 28 mol · L- 1 pOH -log(0. 1 28) 0 . 893 pH = 1 4.00 - 0.893 1 3. 1 1 If we do not ignore the second ionization, then the additional contribution to [OH-] can be approximately calculated as follows: ° Kb = [H 2 z J [OH - ] x(0. 1 28 + x) 5.6 X 1 0-3 (0. 1 28 - x) [H02 - ] To a first approximation, x 5.6 x 1 0-3 mol · L- 1
(
J
=
=
J
=
=
=
=
=
. . x = Kb (0. 1 28 - 0.0056) 0.005 To a second approxImatIOn, (0. 1 28 + 0.0056) =
Then [OH- ] = 0. 1 28 + 0.005 = 0. 1 33; pOH -log (0 . 1 33) = 0.876; and pH 1 3 . 1 2. The difference between calculations is slight. =
=
15.27
The weaker the H-X bond, the stronger the acid. H 2 Te has the weakest bond; H 2 0, the strongest. Therefore, the acid strengths are
acid. (Note: ,6.H°f (H+) is defined as 0).
SM-452
13.}[o,
= -887.34 kJ . mol-I -( -813.99 kJ . morl ) = -73. 3 5 kJ . mol- I
(b) The number of moles of H 2 S04 g . mol- I mol
IS
10.00798 . 07 = 0.1020 The amount of heat generated should be 0. l 020 mol (-73. 3 5 kJ · mo )= -7.482 kJ. The heat capacity of water is 4.18 J . (OCr I . g- I . Adding 7.4 82 kJ of heat to the water should raise the temperature by kJ 0000 J . kr l ) = (4.187.4J82. (Oct = 3.56° · g- I )000. 0 g ) The final temperature should be 25 . 0°C + 3 . 5 6°C = 28. 6 °C. x
rl
f'..t
15.31
Fluorine comes from the minerals fluorspar, CaF2 ; cryolite, Na 3 AIF6 ; and the fluorapatites, Ca S F(P04 )3 . The free element is prepared from HF and the KF by electrolysis, but the HF and KF needed for the electrolysis are prepared in the laboratory. Chlorine primarily comes from the mineral rock salt, NaCl. The pure element is obtained by electrolysis ofliquid NaCl. Bromine is found in seawater and brine wells as the Br- ion; it is also found as a component of saline deposits; the pure element is obtained by oxidation of Br-(aq) by Cl2 (g). Iodine is found in seawater, seaweed, and brine wells as the pure element is obtained by oxidation of (aq) by Cl 2 (g).
15.33
1(a) HIO(aq) H = + 1, = -2; therefore, 1 = + 1 (b) CI0 2 0 = -2; therefore, Cl = +4 (c) Cl 2 0 7 0 = -2; therefore, CI = +14/ 2 = +7 (d) NaJa} Na = + 1, 0 = -2; therefore, 1 = +5 °
SM-453
1- ion; the
15.35
!::. (a) 4 KCI03 (1) -�) 3 KCI04 (s) + KCI(s)
(b) Br2 (l) + H 2 0(l) � HBrO(aq) + HBr(aq) (c) NaCI(s) + H 2 S04 (aq) � NaHS04 (aq) + HCI(g) (d) (a) and (b) are redox reactions. In (a), CI is both oxidized and reduced . In (b), Br is both oxidized and reduced. ( c) is a Bf0nsted acid-base reaction; H 2 S04 is the acid, and Cl- the base. 15.37
15.39
15.41
(a) HCIO < HCI0 2 < HCI03 < HCI04 (HCI0 4 is strongest; HCIO, weakest) (b) The oxidation number ofCI increases from HCIO to HCI04 • In HCIO 4 ' chlorine has its highest oxidation number of + 7, so HCIO 4 will be the strongest oxidizing agent. :
..
CI- 0- CI : , AX 2 E 2 , angular, about 1 09° . The actual bond angle is
50 0 -50
•
0 E - 1 00 -:., � 0_ - 1 50
•
-250 -300
2
H F(g)
3
H B r(g)
HI(
9J
H CI(g)
Period
SM-454
4
5
The thennodynamic stability of the hydrogen halides decreases down the group. The 6Go values of HCI, HBr, and HI fit nicely on a straight line; whereas HF is anomalous. In other properties, HF is also the anomalous member of the group, in particular, its acidity. Also see Exercise 1 5.42. f
15.43
(a) ICls has too many highly electronegative chlorine atoms present on the central iodine. (b) IF2 has an odd number of electrons and as a result is a radical and highly reactive. (c) ClBr3 is sterically too crowded (too many large atoms on a small central atom).
15.45
(a) We can write the following general chemical equation for the process described: Since the pressure is directly proportional to the moles of each gas present and since we only need to find the combining ratio of iodine to fluorine to detennine the chemical fonnula of IFx, we can use pressure in lieu of moles for the purposes of this problem. Since all of the XeF2 is gone at reactions end, and the reaction started with 3.6 atm ofXeF2 to start, there must be 3.6 atm ofXe present in the vessel at the end of the reaction. Therefore P (total at start) 7.2 atm = P (XeF2 start) + P (h start); solving this gives us P (h start) 3.6 atm. At the reactions end, P (total at end) = 6.0 atm = P (Xe end) + P (h end); solving this gives P (12 end) 2.4atm. This means that 1 .2 atm of h was used in the fonnation of lFx. From this we can solve for x as follows: ( x atm XeF2 '1 j 3.6 mol XeF2 ; x = 3 1 .2 atm 1 2 l atm 1 2 Therefore the chemical fonnula is IF3. =
=
=
1
=
SM-455
1 5.47
CI2 (g) + 2 e- � 2 CI- (aq) EO = +1 .36 V Mn0 4- (aq) + 8 H+ (aq) + 5 e- � Mn 2+ (aq) + 4 H 2 0(I) EO + 1 .5 1 V EO cell = ( 1 .36 - 1 .5 1 ) V -0. 1 5 V Because EO is negative, Cl2 (g) will not oxidize Mn 2 + to fonn the =
=
cell
pennanganate ion in an acidic solution. 15.49
F- (aq) + PbCl 2 (s) � CI- (aq) + PbCIF(s)
(
)(
](
1 mol F· 0 f F- I· Ons = 0.765 g PbCIF 1 mol PbCIF moIanty 0.0250 L 261 .64 g PbCIF 1 mol PbCIF = 0. 1 1 7 mol · Cl
)
The number of moles of NH 4 CI04 is
Q .00 kg Q OOO g . kg- I )} 1 1 7.49 g . mol- I
=
8.5 1 mol
The number of moles of Al is 1 .00 kg Q OOO g . kg- I }- 26.98 g . mor l = 37.06 mol
The limiting reagent is the NH 4 CI04 . The standard enthalpy for the reaction is given by MlOr = ( - 1 675.7 kJ · mOr l ) + ( - 704.2 kJ · mol-I ) + 6( - 241 .82 kJ · mol- I ) +3(90.25 kJ · morl ) - [3( - 295.3 1 kJ · mol- I )] -2674. 1 kJ . mOrl This value is the amount of heat released for 3 mol NH 4 CI04 . The amount released for 8.5 1 mol will be ( -2674. 1 kJ . mOr l 1 ) -7.59 1 03 kJ 8.5 1 mol NH 4Cl0 4 l 3 mol NH 4CI04 There will be 7.59 1 03 kJ of heat released. =
=
x
SM-456
x
15.53
Helium occurs as a component of natural gases found under rock formations in certain locations, especially some in Texas. Argon is obtained by distillation of liquid air.
15.55
(a) KrF2 :F = -1 ; therefore, Kr = +2 (b) XeF6 :F = -1 ; therefore, Xe = +6 (c) KrF4 :F = -1 ; therefore, Kr = +4 (d) XeO /- :O = -2, Nox (Xe) - 8 = -2; therefore, Nox (Xe) = +6
15.57
XeF4 (aq) + 4 H + (aq) + 4 e- -) Xe(g) + 4 HF(aq)
15.59
Because H 4 Xe06 has more highly electronegative ° atoms bonded to Xe, we predict that H 4 Xe06 is more acidic than H 2 Xe0 4 .
15.61
In fluorescence, light absorbed by molecules is immediately emitted, whereas, in phosphorescence, molecules remain in an excited state for a period of time before emitting the absorbed light. In both phenomena, the energy of the emitted photon is lower than that of the absorbed photon (emitted photons have a longer wavelength).
15.63
Fluorescent dyes allow much smaller concentrations ofbiomolecules to be detected than normally possible.
15.65
The "top down" approach to manufacturing nanomaterials refers to physically assembling the nanopartic1es into the form factor desired; on the other hand, the "bottom up" approach utilizes specific intermolecular interactions to cause the nanomaterials to self-assemble.
15.67
Ion-ion forces are among the strongest intermolecular interactions. Therefore, the interactions in (a) are the strongest shown. Hydrogen SM-457
bonding is stronger than a dipole-dipole interaction indicating that the interactions in (c) are stronger than those shown in (d). 15.69
(a) The P4 molecule has the following tetrahedron-like Lewis structure; all phosphorous atoms are s/ hybridized:
(b) s/ hybridization requires bond angles of 1 09.5°; however, the internal bond angles required by this structure are 60° meaning that the bonds will be strained. 15.71
(a) 1 4 electron species: CN-, N 2 and C;2 ; 24 electron species: NO�, and °3 ; (b) Strongest Lewis bases: NO; and C;2 ; (c) Strongest reducing agents: NO; and C;2
15.73
The larger the value of EO for the reduction X 2 + 2 e- � 2 X- , the greater the oxidizing strength of the halogen X 2 . From Appendix 2B, F2 + 2 e- � 2 FCl 2 + 2 e- � 2 CI Br2 + 2 e- � 2 Br1 2 + 2 e- � 2 r Thus, 1 2 < Br2 < Cl 2 < F2 ·
15.75
EO = +2.87 V EO = +1 .36 V EO = +1 .09 V EO = +0.54 V
(a) 2 NH/aq) + CIO s (aq) � N 2 H 4 (aq) + CI s (aq) + H 2 0(l) (b) Mg3 N 2 (s) + 6 H 2 0(l) � 3 Mg (OH � (s) + 2 NH 3 (g) SM-458
(c) p3 S (S) + H 2 0(l) � PH 3 (g) + OH \aq) (d) CI2 (g) + H 2 0(l) � HCIO(aq) + HCI(aq) (a) and (d) are redox reactions. In (a), CIO S is the oxidizing agent and
3
3
NH3 is the reducing agent. In (d), Cl2 is both oxidized and reduced. (b) is a Lewis acid-base reaction; H 2 0 is the acid, and Mg3 N2 the base. ( c) is a Bf0nsted acid-base reaction; p 3 S is the base and H 2 0 is the acid. 15.77
(a) State A: kJ · mor l , State B: kJ · mOr l . State B is the higher energy state because it requires more energy to pair electrons in the same orbital than it does to force the spins of two electrons in different orbitals to be antiparallel (b)
94.72
h·c AE _
_
157. 8 5
(6.62608 1 0 34 J s 'f.'!..9979 108 m · S-I X6. 02214 �4720 J . morl ) 1.263 10 6 m or 1 .263 .um x
x
=
15.79
-
x
.
x
1 023 mol - I )
-
The structure ofthiosulfuric acid and sulfuric acid are: :0 : :0 :
II
H-O-S
S-H
II . . -
II
H - O-S-O-H
II
..
:0 : :0 : Due to the replacement of one of the -OH groups in sulfuric acid with an -SH , it should be expected that an aqueous solution of thiosulfuric acid should be slightly less acidic; in addition, the boiling point should also be expected to be slightly lower (due to reduced hydrogen bonding). 15.81
2
2
The reaction of interest is: XeF4 (s) + SF4 (s) � SF6 (S) + Xe(g)
SM-459
To detennine how much product is fonned, we must first identify the limiting reagent: moles ofXeF4 (s) = 330.0. g = 1 .592 mol 207.29 g mor 250.0 g moles of SF4 (s) 2.3 1 3 mol 1 08.07 g . mol' ,
=
,
=
Given the 1 :2 stoichiometry of the reaction, SF4(s) is the limiting reagent and 2.3 1 3 mol of SF6(S) will be produced. (2.3 1 3 mol ) (146.07 g · mol· ' ) 337.9 g of SF6(s) produced. =
15.83
Orpiment is As 2 S) and realgar is AS4 S 4 • Orpiment is yellow and realgar is orange-red. They are both used as pigments.
15.85
(a) AX2 , linear 1 800 (b) F- , 1 33 pm; N )- , 1 48 pm; cr , 1 8 1 pm; therefore, between fluorine and chlorine. (c) HCl, HBr, and HI are all strong acids. For HF, Ka = 3.5 1 0-4 , so HF x
is slightly more acidic than HN) . The small size of the azide ion suggests that the H-N bond in HN3 is similar in strength to that of the H-F bond, so it is expected to be a weak acid. (d) ionic: NaN) , Pb(N ) h AgN ) , etc.
SM-460
15.89
15.91
The solubility of the ionic halides is determined by a variety of factors, especially the lattice enthalpy and enthalpy of hydration. There is a delicate balance between the two factors, with the lattice enthalpy usually being the determining one. Lattice enthalpies decrease from chloride to iodide, so water molecules can more readily separate the ions in the latter. Less ionic halides, such as the silver halides, generally have a much lower solubility, and the trend in solubility is the reverse of the more ionic halides. For the less ionic halides, the covalent character of the bond allows the ion pairs to persist in water. The ions are not easily hydrated, making them less soluble. The polarizability of the halide ions, and thus, the covalency of their bonding, increases down the group. (a) Ag+ + e- � Ag EO = +0.54 V For 2 Ag+ + 2 r � Ag + 1 2 , E O +0.80 V - 0.54 V +0.26 V The process should be spontaneous. (b) The formation of AgI precipitate means that the concentration of Ag+ ions is never high enough to achieve the conditions necessary for the redox reaction to take place. The solubility product Ksp limits the concentrations in solution, so that the actual redox potential is not the value calculated, which represents the values when [Ag+ ] 1 M and [r ] 1 M. Ifwe use the concentrations established by the solubility equilibrium and the Nemst equation, we can calculate the actual redox potential: =
=
=
SM-461
=
where, in this case, Ksp =
Q
=
� for the reaction as written Ksp
1 .5 1 0- 1 6 for AgJ x
l ( . I. I 1 E = +0.26 V (8.3 1 4 J K- .morI . )(298I K) In 2 1 (2)(96 485 J V- mol- ) l (1 .5 x 1 0- 6 ) J l )(298 K) ( . I. 1 I I + O .2 6 V - (8.3 1 4 J K- mor n (96 485 J V- I . mor l ) l (1 .5 x 1 0- 1 6 ) J +0.26 V - 0.94 V -0.68 V The fact that the concentrations of Ag+ and r are limited in solution means that the redox potential for a spontaneous reaction is never achieved. _
=
.
=
=
15.93
(a) The molecular orbital diagram for NO+ should have the oxygen orbitals slightly lower in energy than the nitrogen orbitals, because oxygen is more electronegative. This will cause the bonding to be more ionic than in either N 2 or 0 2 . There is an ambiguity, however, in that the MO diagram could be similar to either that of N 2 or that of 0 2 . Refer to Figures 3.3 1 and 3.32 where you will see that the CJ" 2p and 1C 2p have different relative energies. There are consequently two possibilities for the orbital energy diagram:
SM-462
. -...,. ...'...... , 1t 2p
�>-.- +-+-
+t:-,,'-'1t zp
,
N
"
(using
.or !I
- - -
(} Zs
NO+ N MO diagram)
N
o
" '� I' - - -(} z., NO+
(using 0 MO diagram)
(b) The two orbital diagrams predict the same bond order (3) and same magnetic properties (diamagnetic), and so these properties cannot be used to determine which diagram is the correct one. That must be determined by more complex spectroscopic measurements. 15.95
(a) Pyrite adopts a face-centered cubic unit cell. (b) The iron atoms lie at the comers and at the face centers of the unit cell. Eight sulfur atoms lie completely within the unit cell. (c) The coordination number of iron is six (octahedral). (d) Each sulfur atom is bonded to one other sulfur atom and three iron atoms. ( e) The locations of the sulfur atoms can be considered in one of two ways. An examination of the structure shows that these are best thought of as S 2 2- ions. The locations of the midpoints of the S-S bonds are at the centers of each edge of the unit cell (1 2 si- ±) plus 1 S /- ion in the center of the unit cell (only half the ions have sulfur atoms within a given cell). This gives a total of four S 2 2- ions in the unit cell. Alternatively, identify eight S atoms within the unit cell. x
15.97
(a) Cloud: [NO] 860 ppt and [N02] 250 ppt; Clear air: [NO] = 480 ppt and [N02] 260 ppt; =
=
=
SM-463
o
2p
(b) Clouds have the higher concentration of NO; (c) in addition to jet engines and automobile emissions, NO is also formed when N02 dissolves in water: As a result one would expect to find more NO and less N02 in clouds,which also contain water vapor. (d) the most likely product formed when NO (which is a radical) reacts with hydroxyl radical is nitrous acid: O=N . . - O. . - H
(e) Nitrous acid would be expected to be more stable than either NO or hydroxyl radical since both of these species have unpaired electrons while nitrous acid does not. (f) A possible mechanism that fits the observed rate law would be:
� N0 2 + 0 2 �low) Step 2: N0 2 + � NO + 02 (fast) (g) from part (f), rate = k[03 ][NO]; given that we know what the Step 1 : NO + 03
°
concentration of ozone is and the value of k, to determine the rate at which ozone molecules are being destroyed we need to determine the concentration of NO molecules present in the clear air sample. The total number of molecules in the sample per cm3 can be calculated as follows: 1 atm kPa 1 bar = 2. 1 7 1 0-3 atm 220 mbar 1 000 mbar 1 bar 1 0 1 .325 kPa n P = it 2. 1 7 1 0-3 atm = ) = 1 .09 1 0-4 mol of molecules · L- I Y . I I V RT \0.08206 L · atm · mol- K- ,,243 K
(
-
=
=
-
(1 .09
x
) (1 ) (
)
x
x
x
1
1 0-4 mol of molecules · L-I 6.022 0 23 molecules · mol- I l 1 0 -3 L · cm -3 )
X
3.56 1 01 6 molecules · cm-3 x
SM-464
x
X
x
If the amount of NO present is 480 ppt, then the amount of NO in the sample in molecules per cm3 is: 6.56 1 0 1 6 mOl ec�les 480 �� mOleCU les 3 . 1 5 1 0 7 NO mo1 ecules . cm-3 cm 1 1 0 molecules 2 If [0 3 ] 4 1 0-1 mo1ec · cm-3 and k = 6 1 0-15 cm 3 . molec- I . sec-I ; then the rate at which ozone molecules is being destroyed in clear air is rate (6 1 0-15 cm 3 · molec- I . sec- 1 X4 1 0-1 2 molec 0 3 . cm-3 X3 . 1 5 1 07 molec NO . cm-3 ) rate 7.56 1 0- 1 9 molecules · cm-3 . sec-I
(
)(
x
x
=
=
=
)
x
=
x
x
x
x
x
SM-465
x
CHAPTER 1 6 THE ELEMENTS : THE d BLOCK
16.1
Elements at the left of the d block tend to have strongly negative standard potentials.
16.3
(a) Ti (b) One might expect gold to be larger than silver because it is a third row transition metal and silver is in row two; however, because of the lanthanide contraction, they are about the same size (Au, 1 44 pm; Ag, 1 44 pm) (c) Ta (d) Ir
16.5
(a) Fe; (b) Cu; (c) Pt; (d) Pd; (e) Ta
16.7
Hg is much more dense than Cd, because the shrinkage in atomic radius that occurs between Z 58 and Z 7 1 (the lanthanide contraction) causes the atoms following the rare earths to be smaller than might have been expected for their atomic masses and atomic numbers. Zn and Cd have densities that are not too dissimilar, because the radius of Cd is subject only to a smaller d-block contraction. =
16.9
=
(a) Proceeding down a group in the d block (for example, from Cr to Mo to W), there is an increasing probability of finding the elements in higher oxidation states. That is, higher oxidation states become more stable on going down a group. (b) The trend for the p-block elements is reversed. Because of the inert pair effect, the higher oxidation states tend to be less stable as one descends a group. SM-466
16. 1 1
In M03 , M has an oxidation number of +6. Of these three elements, the + 6 oxidation state is most stable for Cr. See Fig. 1 6.6.
16. 1 3
(a) Ti(s), MgCl 2 (s) TiCl 4 (g) + 2 Mg(l) � Ti(s) + 2 MgClz (s) (b) C0 2 + (aq), HC03- (aq), N03- (aq) CoC03 (s) + HN0 3 (aq) --+ C0 2+ (aq) + HC03- (aq) + N0 3- (aq) (c) Yes), CaO(s) V2 0 S (s) + 5 Ca(l) � 2 Yes) + 5 CaO(s)
16. 15
(a) titanium(IV) oxide, Ti0 2 (b) iron(III) oxide, Fe2 03 (c) manganese(IV) oxide, MnO? (d) iron (II) chromite, FeCr2 04
16. 1 7
EO +1 .33 V (a) Cr2 0/ s + 14 H+ + 6 e- --+ 2 Cr2+ + 7 H 2 0 EO -1 .09 V 2 Br s --+ Br2 + 2 eEO( overall) = +0.99 V =
=
Therefore, Br- will be oxidized to Brz. (b) Cr2 0/ s + 1 4 H+ + 6 e- --+ 2 Cr2+ + 7 H 2 0
EO = +1 .33 V EO = S1 .98 V
Ag+ --+ Ag2+ + 1 e-
EO( overall) SO.65 V Therefore, no reaction will occur. =
16. 1 9
V2 0 S (s) + 2 H 3 0 + (aq) --+ 2 VO ; (aq) + 3 H 2 0(l) (b) V2 0 S (s) + 6 OH - (aq) --+ 2 VO �- (aq) + 3 H 2 0(l) (a)
SM-467
16.21
Even though all three Group 1 BI1 1 metal atoms have the valence shell electron configuration ( n l)d ons' Cu is more reactive than Ag or Au. Metals ordinarily lose one or more electrons to form cations when they react with some other species. As the value of n increases, d and! electrons become less effective at shielding the outermost, highest energy electron(s) from the attractive charge of the nucleus. This higher effective nuclear charge makes it more difficult to oxidize the metal atom or ion. So, for example, Cu 2+ exists in many common compounds (and can be formed by Cu + disproportionation in water) while Ag 2+ does not. Furthermore, the valence electron orbital energies for most common Lewis bases match the orbitals of Cu and its cations more closely and would interact with them more favorably to form products. -
16.23
,
,
(a) Cr3+ ions in water form the complex [Cr(H 2 0) 6 ] 3+ (aq), which behaves as a Bmnsted acid: (b) The gelatinous precipitate is the hydroxide Cr(OH) 3 . The precipitate dissolves as the Cr(OH) 4- complex ion is formed: Cr3+ (aq) + 3 OH- (aq) --+ Cr(OH)3 (s) Cr(OH) 3 (s) + OH- (aq) --+ Cr(OH) 4- (aq)
16.25
(a) hexacyanoferrate(II) ion Let x = the oxidation number to be determined x(Fe) + 6 (-1) = -4 x(Fe) = 4 (-6) = +2 (b) hexaamminecobalt(III) ion x(Co) + 6(0) = +3 x(Co) = +3 -
-
SM-468
(c) aquapentacyanocobaltate(III) ion x(Co) + 5 ( 1 ) + 1(0) = -2 x(Co) = -2 - (-5) = +3 (d) pentaamminesulfatocobalt(III) ion x(Co) + 1 (-2) + 5 (0) +1 x(Co) = +1 - (-2) = +3 -
=
16.27
(a) K 3 [Cr(CN)6 ] (b) [Co(NH 3 )s (S0 4 )] CI (c) [Co(NH 3 ) 4 (H 2 0) 2 ] Br3 (d) Na [Fe(H 2 0) 2 (C2 04 ) 2 ]
16.29
(a) The molecule HN(CH 2 CH 2 NH 2 ) 2 has three nitrogen atoms, each with a lone pair of electrons that may be used for bonding to a metal center. The molecule can thus function as a tridentate ligand. (b) The CO/- ion can bind to a metal ion through either one or two oxygen atoms. It may, therefore, serve as a mono- or bidentate ligand. (c) H 2 0 is always a monodenate ligand. (d) The oxalate ion can bind through two oxygen atoms and is usually a bidentate ligand.
16.31
As shown below, only the molecule (b) can function as a chelating ligand. The two amine groups in (a) and (c) are arranged so that they would not be able to coordinate simultaneously to the same metal center. It is possible for each of the amine groups in (a) and (c) to coordinate to two different metal centers, however. This is not classified as chelating. When a single ligand binds to two different metal centers, it is known as a bridging ligand.
SM-469
(b)
( a)
H 2N-
�
n \J _
- NH 2 , M
0� /
H 2N
"" /
(c)
_ J C NH ,
2
N',
H2
M
M
2
M
1 6.33
( a) 4 (b) 2 (c) 6 (en is bidentate) (d) 6 (EDTA is hexadentate)
1 6.35
(a) (b) ( c) (d)
1 6.37
(a) yes
structural isomers, linkage isomers structural isomers, ionization isomers structural isomers, linkage isomers structural isomers, ionization isomers
, +
, +
and
CI trans-tetraamminedichlorcobalt( I I I ) chloride monohydrate
CI cis-tetraamminedichlorcobalt( I I I ) c hloride monohydrate
(b) no (c) yes
cis-diamminedichloroplatinum(lI )
SM-470
trans-diamminedi c hloroplatinum ( II)
16.39
(a) mIrror Image
complex
A
A
I�
B
C
I
R
/ I � / Ii � I
B
A
D
D
e
B
No rotation will make the complex and its mirror image match; therefore, it is chiral. ( b) mIrror Image complex A
I
D
I
c
/ IJ / I� C
B
A
I
D
I
A
A
c
B
c
A double rotation shows that the complex and
its mirror image are
superimposable; therefore, it is not chiral. The two complexes are not enantiomers; they are not even isomers. 16.41
(a) 1 ; (b) 6 ; (c) 5; (d) 3; (e) 6 ; (f) 6
16.43 ( a)
2; (b) 5;
( c)
8;
(d)
1 0;
( e)
SM-47 1
0 (or 8); (f) 1 0
16.45
(a) octahedral; strong-field ligand, 6 e-
diamagnetic, no unpaired electrons (b) tetrahedral: weak-field ligand, 8 e2 5 to. T
2 unpaired electrons (c) octahedral: weak-field ligand, 5 e-
�5 to.0
5 unpaired electrons (d) octahedral: strong-field ligand, 5 e-
one unpaired electron SM-472
[Co(enh] 3 + has no unpaired electrons.
t.
t.
t.
t
t
[Mn(CN)6] 3- has two unpaired electrons. 1 6.49
Weak-field ligands do not interact strongly with the d-electrons in the metal ion, so they produce only a small crystal field splitting of the d electron energy states. The opposite is true of strong-field ligands. With weak-field ligands, unpaired electrons remain unpaired if there are unfilled orbitals; hence, a weak-field ligand is likely to lead to a high-spin complex. Strong-field ligands cause electrons to pair up with electrons in lower energy orbitals. A strong-field ligand is likely to lead to a low-spin complex. Ligands arranged in the spectrochemical series help to distinguish strong-field and weak-field ligands. Measurement of magnetic susceptibility (paramagnetism) can be used to determine the number of unpaired electrons, which, in tum, establishes whether the associated ligand is weak-field or strong-field in nature.
16.51
Because P- is a weak-field ligand and en a strong-field ligand, the splitting between levels is less in (a) than in (b). Therefore, (a) will absorb light oflonger wavelength than will (b) and consequently will display a shorter wavelength color. Blue light is shorter in wavelength than yellow light, so (a) [COP6 ] 3- is blue and (b) [Co(en) 3 ]3+ is yellow.
SM-473
16.53
16.55
Eh
p oton
= (l 209molkJ )\ (l 6.02 x 110mol23photons\) = 3.47 x 10-22 kJ . photon'I
This wavelength is in the yellow region of the visible spectrum. Since the complex absorbs yellow light, it transmits the complement, or purple (a.k.a. violet). In Zn2+, the 3d-orbitals are filled (d lo ). Therefore, there can be no electronic transitions between the and e levels; hence, no visible light is absorbed and the aqueous ion is colorless. The dl o configuration has no unpaired electrons, so Zn compounds would be diamagnetic, not paramagnetic. (6.63x l O-34 J ·s-I )(3.900x l08 m·s-I ) = 2. 69 xl 0-19 J (a) 740 x 10- m t
16.57
!J. o
he
= - =
A
J·s-I )(3.900x l08 m · s-l ) = 3. 46 x l 0-1 9 J (c) = - = (6.63x lO-34 575x l0- m These numbers can be multiplied by 6.02 x 1023 to obtain kJ . mol-I . (a) 2.69 x 10-1 9 J Q OS3 kJ . J SI X6. 02 x 1023 mol-I)= 162 kJ . mol-I (b) 4.32 x 10-1 9 J Q OS3 kJ . J SI X6.02 x 1023 morl )= 260 kJ . mol-I (c) 3.46 x 10-19 J Q OS3 kJ . J SI ). 02 x 1023 morl 208 kJ . morl (spectrochemical series) !J. o
he A
=
SM-474
16.59 16.61
The eg set, which comprises the dx'_ y' and dz' orbitals. (a) The CN- ion is a 7r -acid ligand accepting electrons into the empty 7r orbital created by the C-N multiple bond. (b) The CI- ion has extra lone pairs in addition to the one that is used to form the (J -bond to the metal, and so it can act as a 7r -base, donating electrons in a p-orbital to an empty d-orbital on the metal. (c) H 2 0, like CI- , also has an "extra" lone pair of electrons that can be donated to a metal center, making it a weak 7r -base; (d) en is neither a 7r -acid nor a 7r -base, because it does not have any empty 7r -type antibonding orbitals nor does it have any extra lone pairs of electrons to donate. (e) CI- < H 2 0 < en < CN- . Note that the spectrochemical series orders the ligands as 7r -bases < (J -bond only ligands < 7r -acceptors. *
16.63
Nonbonding or slightly antibonding. In a complex that forms only (J bonds, the t-? g set of orbitals is nonbonding. If the ligands can function as weak 7r -donors (those close to the middle of the spectrochemical series, such as H 2 0), the t2 g set becomes slightly antibonding by interacting with the filled p-orbitals on the ligands.
16.65
Antibonding. The eg set of orbitals on an octahedral metal ion are always antibonding because of interactions with ligand orbitals that form the (J bonds. This i s true regardless of whether the ligands are 7r -acceptors, 7r donors, or neither.
16.67
Water has two lone pairs of electrons. Once one of these is used to form the (J -bond to the metal ion, the second may be used to form a 7r -bond. This causes the t2 g set of orbitals to move up in energy, making � o smaller; therefore, water is a weak-field ligand. Ammonia does not have
SM-475
16.69
this extra lone pair of electrons and consequently cannot function as a 1[ donor ligand. (a) CO (b) In Zones D & C, 3 Fe2 0 3 (s) + CO(g) � 2 Fe3 04 (s) + CO 2 (g) Fe3 0 4 (s) + CO(g) � 3 FeO(s) + CO2 (g) These reactions combine to give Fe2 0 3 (s) + CO(g) � 2 FeO(s) + CO 2 (g) In Zone B, Fe2 03 (s) + 3 CO(g) � 2 Fe(s) + 3 CO2 (g) FeO(s) + CO(g) � Fe(s) + CO 2 (g) (c) carbon
16.71
The major impurity is carbon; it is removed by oxidation of the carbon to CO 2 , followed by capture of the CO2 by base to form a slag.
16.73
Copper and zinc
16.75
Alloys are usually ( 1 ) harder and more brittle, and (2) poorer conductors of electricity than the metals from which they are made .
16.77
The compound is ferromagnetic below Tc because the magnetization is higher. Above the Curie temperature, the compound is a simple paramagnet with randomly oriented spins, but below that temperature, the spins align and the magnetization increases.
16.79
The green color suggests chromium or copper but the blue solution formed when the mineral is dissolved in sulfuric acid points to copper as the metal ( CuS04 is blue). The colorless gas that forms either upon heating or on treatment with acid is carbon dioxide; this is proven when the gas is SM-476
bubbled into limewater (the cloudiness in the solution is due to calcium carbonate formation). CO2 is formed when carbonates are either heated or treated with acid. The mineral is most likely basic copper carbonate ( CU 2C0 3 (OH 1 ). 16.81
(a) +3 is the most likely oxidation state of vanadium
16.83
Cr(OH)3 (s) + 3 e- � Cr(s) + 3 OH -
EO = -1 .34 V EO = +0.74 V EO = -0.60 V
f.. G O = - nFEo = -RTln K o . . l In K nFE = (3)(96 485 J. y-I morl )(-0.60 V) = -70. 1 (8.3 1 4 J K- I . mo r )(298 K) RT K = e-70 1 = 3.6 x 1 0-3 1 =
sp
C is-
Diamminebromochloroplatinum(II)
trans- Diamminebromochloroplatinum(II)
(b) If the compound were tetrahedral, there would be only one compound, not two. 16.87
(a) The first, [Ni(S04 )(en) 2 ]Cl 2 , will give a precipitate of AgCl when AgN03 is added; the second will not. (b) The second, [NiCI 2 (en) 2 ]1 2 , will show free 1 2 when mildly oxidized with, for example, Br2 , but the first will not. SM-477
1 6.89
(a) [MnCI6t5 e-, cr is a weak-field ligand
t
t
t
t
t
[Mn(CN)6tt. t t. 5 e-, CN- is a strong-field ligand (b) [MnCI6t : five; [Mn(CN)6] 4-: one. (c) Complexes with weak-field ligands absorb longer wavelength light, therefore, [MnCI6] 4- absorbs longer wavelengths. 16.91
H igh spin Mn 2 + ions have a d 5 configuration with 5 unpaired electrons, as shown below.
t
t
t
_t.l.....-
t
In order for light to be absorbed in the visible region of the spectrum, an electron from the t2g set has to be moved into the eg set of orbitals. Because each orbital is already singly occupied by an electron and all five electrons have parallel spins, the electron making the transition must change spin in order to spin pair in the upper orbital. This sort of transition is called "spin-forbidden" because it has a very low quantum mechanical probability of occurring, and so the complexes usually are only faintly colored.
SM-478
16.93
If the prismatic structure were the true structural form of [ CoCl 2 (NH 3 )J one would expect to get four possible isomers, not the two seen (X = NH3 ): X CI/1- -1-1--11----- -- - - - ------ - -- -----------CI/·1.1.1. . · . ... . . · - . . · -- · - -- - .. -- --_-.----:: X /:- : " " 11111// /
. ..
.-
• ..
/
..
. .•
•.
•
,,
'
\ x \\
• ..
. /
•
•
•
•
•
,:,� \\ ..
. .. .. ..
. -.
16.95
.. •.
•.
• ••
•.
.. ... . .. ..
.
.• .• ..
··X_'II� ..· ··. .·Co
. ..
.
� • •.
.. ..
\\ \\ . x t\ - .\-
" " " .. " - " . . - . . . -....· ·..· ·.·--· ·.·-·.·.·--· ·.· X ·' · · · · · · · · · · · · · · ·CI ··
x
/ ' -' - ' - '- ' : .. . o � X , : · - . . - - - - - - - · - · -- - - . - - - . -::, CI " . " '. �. - - - - - - - - - - - - - - --
CI,(..11..11- · - · - .. · · .. -- · - · - ·- .. - ·- ------- - -:-: X: 1II11I1111 l X- 1"' ..." . .. co � �: : \" CI \\\. .\-\·\· :: . . - · · -- .. - . . - -- - .... ·- .. - - ·-- X: " . " " �. - . . . . .. . - - . •
� \ \"
/
..
-·
li'� ?> o
11111I1111
x. 1/· -1·1-. . - -- · · - - -- - ------ · - - -- - · - - · - - - - - :::CI
;
:
l
.
:� \ \ \" ,
. 1.1:'X111-11111" .
\ . , \)���\.\ .: ,
..
• .• •
• ••
• ••
.•
· · -· ·C
.
..
..
..
.
.
.
The correct structure for [Co(NH 3 )6 ]C13 consists of four ions,
Co(NH 3 ) t , and 3 cr in aqueous solution. The chloride ions can be easily precipitated as AgCl. This would not be possible if they were bonded to the other (NH 3 ) ligands. If the structure were Co(NH 3 -NH 3 -CI)3 ' VSEPR theory would predict that the C0 3 + IOn would have a trigonal planar ligand arrangement. The splitting of the d orbital energies would not be the same as the octahedral arrangement and would lead to different spectroscopic and magnetic properties inconsistent with the experimental evidence. In addition, neither optical nor geometrical isomers would be observed. 16.97
The spectrochemical series given in Figure 1 6.29 shows the relative ligand field strengths of the halide ions to lie in the order r < Br- < cr < F- . Since their electronegativities follow the same trend, there i s a positive SM-479
correlation between ligand field strength and electronegativity for the halide ions. The value of L}o correlates with the ability of the ligand's extra lone pairs of electrons to interact with the
t2 g
set of the octahedral
metal ion. This means that the less electronegative the ligand is, the easier it is for the ligand to donate electrons to the metal ion, the more the energy ofthe t2g in the complex is raised, and the smaller L} o becomes (see Figure 1 6.38(a» . 1 6.99
In order to determine these relationships, we need to consider the types of interactions that the ligands on the ends of the spectrochemical series will have with the metal ions. Those that are weak-field (form high spin complexes, 7r -bases) have extra lone pairs of electrons that can be donated to a metal ion in a 7r fashion. The strong-field ligands (form low spin complexes, 7r -acids) accept electrons from the metals. The complexes that will be more stable will be produced in general by the match between ligand and metal. Thus, the early transition metals in high oxidation states will have few or no electrons in the d-orbitals. These metal ions will become stabilized by ligands that can donate more electrons to the metal-the d-orbitals that are empty can readily accept electrons. The more stable complexes will be formed with the weak-field ligands. The opposite is true for metals with many electrons, which are the ones at the right side of the periodic table, in low oxidation states. These metals generally have most of the d-orbitals filled, so they would, in fact, be destabilized by 7r donation. They form instead more stable complexes with the 7r -acceptor ligands (strong-field, 7r -acids) that can remove some of the electron density from the metal ions.
SM-480
16.101
(a) PtCl2 (NH 3 ) 2 ' cis-diamminedichloroplatinum(II); (b) The only other isomer is the trans form. Neither the cis nor the trans form is optically active. CI
-
NH 3
I
Pt -Cl
I
NH 3
( c) square planar 16.103
=
---'-
2. 1 1 g number of moles of complex 2 1 1 .42 g . mor ' 3 = 9.98 x 1 0 mol complex number of moles ofCr 2.87 g AgCl 1 mole AgCl 1 43.32 g . mor' 2.00 x 1 0-2 mole cr
-
-
=
(
-
[J
1 mole cr 1 mole AgCl
=
Therefore the compound contains 2 moles of cr counterion for every mole of complex (assuming that the incorrect formula gives the correct molar mass). Then the correct formula of the compound would be [CrNH 3 Cl(H 2 0)2 JCI 2 . Since the Cr3+ is d 3 and coordinated by four ligands, the complex cation is likely to be tetrahedral. There are no linkage isomers possible and no enantiomers for this complex ion.
2 CI However, the compound as a whole has several isomers, including one that corresponds to the incorrect formula, [CrNH 3 ClJ · 2 H 2 0 . NH3 I , ,1\ Cr '-...� .. H20 cI' ! "CI CI •
SM-481
2
J
The other reasonable isomers include octahedral species that result if the chloride ions as well as both water molecules are all attached directly to the metal ion.
Ltz
OH2
OH2
C
H2
Cr
I
H3
Cl
16.105
C
N H3
zt?' ztz' I I Cr
H3
C
Cr
H2
Cl
Cl
The disproportionation of Hg(I) ion is as follows: Hg�+ (aq) � Hg(l) + Hg2+ (aq) ; using Appendix 2B, the EO for this process can be determined:
=
This gives EO -0.1 3 V. The equilibrium constant can then be calculated as follows: o -RT ln K �GO = -nFE o . -I -' In K = nFE = (2)(96 485 J. V' · mol )(-0. 1 3 V) = - 1 0. 1 (8.3 14 J K - mor ' )(298 K) RT K = e-' o" = 4.0 1 0-5 =
.
c
16.107
x
The electron configuration expected for Ni2 + is [ Ar ] 3d8 . For this complex to have no unpaired electrons it would have to be square planar in its electronic geometry, as both the octahedral and tetrahedral geometries require a d8 species to have two unpaired electrons. Square planar does not:
SM-482
t
t+
---t-t-
-t-t-
octahedral d8 complex
tetrahedral d8 complex
square planar d8 complex 1 6.109
t
(a)
/;
(b) iron (II) is a d6 species; if the deoxygenated heme is a high spin octahedral complex it should have four unpaired electrons and the oxygenated heme as a low spin octahedral complex should have zero unpaired electrons:
SM-483
t
t
+-
t2g
high spin cfi complex low spin cfi complex (large splitting) (large splitting) ( c) for a ligand to bond to the iron atom it needs to be a Lewis base. Of the species listed, only BF3 will not be able to do so, as it is a Lewis acid (the boron only has six electrons around it and is electron deficient) (d) For deoxyhemoglobin (HbH): Hb S ] [ pH = pKa l + log �---==, [ HbH ] Given the pKa\ of HbH is 6.62, at a physiological pH of 7.4: Hb s l Hb s ] [ [ log 7.4 - 6.62 + log [ ]' HbH [ HbH ] = 0.78 [ Hb S ] = 1 0° 78 and [ Hb s ] 6 [ HbH ] [ HbH ] _
=
This means 86% of the deoxyhemoglobin is in its' anionic form ( Hb s ) and 1 4% remains as HbH at a pH of7.4. For oxyhemoglobin (Hb02H): pH = pKa l Given the pKa\ of Hb02H is 8. 1 8, at a physiological pH of 7.4: [ HbO/ l 7.4 - 8. 1 8 + 10g [ Hb02 H J ' [ HbO2 s ] 1 0 50 78 and i==-_� [ Hb0 2 H ]
[ HbO 2 S ] + log [ Hb0 2 H ]
_
=
Thus, 83% of the oxyhemoglobin remains in its' acidic form (Hb02H), with only 1 7% forming the anion as Hb O2 - at a pH of 7.4. SM-484
(e) According to the equilibrium calculations from part (d), prior to oxygenation Hb s is the predominant form of deoxyhemoglobin found. Upon oxygenation it forms Hb0 2 ; this then picks up a hydrogen ion to become Hb02H as the equilibrium shifts over to the left. As a result, it is expected that the H30+ concentration will drop and as a result the pH of the blood will increase .
S
SM-485
CHAPTER 1 7 NUCLEAR CHEMISTRY
17.1
We assume that all the change in energy goes into the energy of the ray emitted. Then, in each case, v = till energy f 1 MeV = [ 1061 MeVeV J [ 1.6021 xeV10- 1 9 1 J = 1.602 x 10-13 1· MeV- I x 10-1 3 1 )1= 2. 1 307 x 10-1 3 1 (a) till = (1. 3 3 MeV) l( 1.6021 MeV X 10-13 = 3.22 X 1020 s-I = 3.22 X 1020 Hz v = -till = 6.26. 1307 -· 1· s 26 x 10 A = �v = 3.3.0202Xx108102m0 s·S-I-I = 9.32 X 10-1 3 m ( 1 (b) till = (1. 6 4 MeV)l 1. 602 x 10- 3 1 )1 = 2. 6 28 x 10- 1 3 1 1 MeV v = till = 6.2.626628x x10-·1�:113 1· s = 3.97 X 1020 s-I = 3.97 X 1020 Hz A = 3.3.0907Xx108102m0 s·-SI -I = 7.56 X 10-1 3 m 1 J = 1.76X lO-1 3 1 1 3 (c) till = (1.10MeV) [ 1.602X101 MeV v= till = 6.61.2676 xX 10-·10��3 1·1 s = 2.66 X 1020 s-I = 2.66 X 1020 Hz A = �v = 3.2.00x65 x10108 20mS·s-- I I = 1.13x lO-'2 m y
h
'
0
h
34
1
h
h
SM-486
17.3
Potassium-40 has 2 1 neutrons; f� Ar and i6 Ca are isotones of K-40.
17.5
(a) 3IT � ° e + 23 He (b) 8393 y � OI e + 3883 Sr (c) 8367 Kr � 0 e + 8737 Rb (d) 22915 Pa � 42 a + 2289l Ac -I
-I
A = 8 - 0 = 8, Z = 5 - 1 = 4, E = Be
17.7
so, 58 B � 0I e + 48 B e (b) �� Ni � _�e + � E A= 63 - 0 = 63, Z = 28 - ( -I ) = 29, E = Cu so, 6238 N1· � -0I e + 6293 C U (c) 1 �� Au � � a + � E A = 1 85 - 4 = 1 8 1 , Z= 79 - 2 = 77, E Ir 4 181 so, 18579 AU -r ----". 2 a + 77 Ir (d) � Be + _�e � � E A = 7 + 0 = 7, Z = 4 - 1 = 3, E = Li so, 47 B e + 0 e � 73 L 1· =
-I
17.9
17.11
(a) 2I4I Na � (b) 1 5028 Sn � (c) 1 45°7 La � (d) 22980Th �
;; Mg + _� e; a � particle is emitted. 1 28 Sb + _� e; a � particle is emitted. 51 1 45°6 Ba + � e; a positron W+ ) is emitted. 2 �:Ra + �a; an a particle is emitted.
(a) l 5i B + 42 a � 2 oI n + 1 73 N (b) 3175 CI + 2 D � oI n + 3186 Ar (c) 9462 M ° + 2I D � 0 n + 9743T c 2 (d) 4251 S c + oI n � 24 a + 419 K I
I
SM -487
17.13
(a) AIZ 68/29 = 2.34 > (AIZ)based ; hence, �� Cu =
is neutron rich, and � decay is most likely. �� Cu � _� e + �� Zn
(b) AIZ = 1 03/48 2. 1 5 < (AIZ)based ; therefore, =
I �� Cd is proton rich, and �+ decay is most likely. I �� Cd � � e + I �� Ag
(c) 2;� Bk has Z > 83 and is proton rich; therefore, a decay is most likely. 243 Bk � 4 a + 239 Am 2 95 97 (d) ���Db has Z > 83; therefore, a decay is most likely. 260 Db � 4 a + 256 Lr 2 105 103 17.15
17.17
a 23925 U � 42 a + 29301 Th 23 23 � 901 Th � 0I e + 9 I1 Pa 2 3 1 a 91 Pa � 42 a + 22897 Ac 22 22 � 897 Ac � 0I e + 970 Th a 22970Th � 42 a + 22883 Ra a 22883 R a � 42 a + 281 69 Rn a 281 69 Rn � 42 a + 281 45 p 0 25 2 15 At � 8l 4 p O � 0I e + 85 2 15 a 85 At � 24 a + 2831 1 S1 2 2 � 81 31 B 1' � °I e + 8l 4l p o 2 1 1 a 84 p O � 24 a + 20872 Pb To determine the charge and mass of the unknown particle, it helps to write : p and �n for the proton and neutron, respectively; and _�e and �e for the � particle and positron, respectively.
SM-488
(b) 2946B C + I n 24979 Bk _I e ( ) 29435 Am + o n 24946 C _I e (d) I�C + � n I: C + (a) 210° Ne + 42 a 4B Be + 1 B6 0 (b) 2100 Ne 21 00 Ne � 1 6B 0 241 2 Mg (c) 4420 Ca 42 a 42B rI (d) 2173 Al + 21 II P 2B Al In each case, identify the unknown particle by performing a mass and charge balance as you did in the solutions to Exercises 1 7.5 and 1 7.7. Then write the complete nuclear equation. (a) '� N + � a I� O : p (b) 2399p4 oI n --,----'- 24905 Am _0I e (a) unbihexium, Ubh (b) untrihexium, Uth (c) binilnilium, Bnn actl.vl.ty = (5.3 x 1 05 Bq) (l 3.7 x1 1Ci010 Bq 1) = 1 .4 x 1 0-5 CI. m
�
a
I
C
+
m
�
�
17.19
+
�Y+
H
+
�
�
U
17.25
17.27
+ y
�
+
17.23
a
y
+
17.21
+
a
13
+
+
+
1 Bq = 1 disintegration per second (dps) -6 Ci ] ( 3.7 x 101 0 dPS ] = 9 . 2 x 1 04 d s 1 0 ( (a) (2.5 Ci) 1 Ci 1 Ci P = 9 2 1 04 Bq (b) 1 42 Ci = (1 42) (3.7 x 101 0 dps) = 5.3 1 012 Bq (c) (7.2 mCi) ( 10-1 mCI3 �i ]( 3.7 x110C1I.0 dPS ] = 2.7 x l OB dps = 2.7 x l 0B Bq /l
/l
.
X
SM-489
X
17.29
--
dose= 1 5 J = 0.75 J . kg- I 2.0 kg ( dose in rads = 0.75 J · kg- I x l
1 rad )) = 75 rad 2 1 0- J · kg- I dose equivalent in rems = Q x dose in rads ( 1 rem) = l--) (75 rad) = 75 rem 1 rad 75 rem 7 1 00 rem/Sv = 0.75 Sv
17.31
17.33
( ]
1 .0 rad · day- I = ( 1 .0 rad · day- I ) 1 rem l rad 1 00 rem = 1 rem · day- I x time time = 1 00 day
=
1 rem · day- I
k = In 2 t"2 In 2 = 5.64 X 1 0-2 a- I (a) k = -1 2.3 a (b) k = � = 0.83 s- ' 0.84 s (c) k = In 2 = 0.0693 min- I 1 0.0 min
17.35
We know that initial activity = No = 2 1 50 dpm and final activity = N = 1 324 dpm at t = 6 . 0 h. Therefore, ( N) In 2 In l ) = Skt and k = t"2 N o ( N ) t in 2 So In l ) = S t" . N 2 o v
v
-
SM-490
( 1 342 1 (6.0 h)n 2 V =S In l SO.48 2 1 50 ) tl / 2 (6.0 h )n 2 . S - 8.8 h. tl / 2 SOA8 , ti l -? _
v
_
v
_
v
-
17.37
_
In 2 , N = N e- kr , N In each case, k = -o N = e- kr , and the percentage tl / 2 o remammg In 2 = 1 .21 x l O-4 a- 1 5.73 x l O3 a percentage remaining = 1 00% x e- ( 1 21 1 0-4 a - ' x 3000 a ) = 69.6%
(a) k =
x
(b) k = �a = 0.0563 a - I 1 2.3 percentage remaining = 1 00% x e- (0056k' 1 2 0 a ) = 50.9% x
17.39
(a) From Table 1 7.5, k = In 2 = fr
tl / 2
tl / 2
9
= 4.5 X 1 0 a
In 2 = 1 .54 X 1 0 - 10 a- I 4.5 x 1 0 a 9
· N = e - kr . . =action remammg No = e- ( 1 .54 x I O- I I' a-' 3.9 1 09 a) = e-0 60 = 0.55 x
x
After 3.9 Ga, 55% remains. · N = -3 ; . . (b) firactIon remammg No 5 =
-
SM-49 l
tl / 2
= 1 .2 6 x 1 09 a,
k
-3 = e- kl 5 � = e - (5. 5 0 x 1 0-10 a -I x x l 5 8 X = 9.29 x 1 0 a 1 7.41
=
In 2 5.50 x 1 0- 1 0 a- I = 9 1 .26 x 1 0 a
Let dis = disintegrations 1 500 dis/0 .250 g = 600 dIS' ' g - I . h- I 1 . actIvIty fjrom "a ld" samp e = . 1 0.0 h activity from current sample = 92 1 dis · g- I . h- I In 2 In 2 = 1 . 2 1 x l 0-4 a- 1 k = -- = tl / 2 5.73 x 1 0 a �ld6activity rx. N, current activity rx. No �ld6activity = � = e- kl No = e kl In ( No 1 = kt ' current activity No N l N) 3
'
Solve for t (= age) :
1 7 . 43
I
. m . Bq = k x N n each case, k = In . 2 , actIvIty tl / 2 ( m s)
. . m . C 1. actIvIty
.
_ -
activity in Bq 3 .7 x 1 0 1 0 Bq ' Ci -I
Note: Bq (= disintegrating nuclei per second) has the units of nuclei · S- I
)
[ )[
23 nUclei 1 mOI 6.02 x 1 0 3 = 2.7 x 1 0 1 8 nuclei N = (1 .0 x l 0- g) 1 mol 226 g SM-492
activity =1.37 x 10-11 S-I 2.7 xl 018 nuclei xl3( .7 x110Cilo )' = 9.9 10-4 Ci X
Bq
X
=
mol' l (6.022 x 1023 nuclei' 1.3 1019 nuclei (1-) 1 mol l 90 g ) Ci0 activity =(7.80 x 10-10 S-I)(1.3 x 1019 nUcleil / 3.7 1101 ) = 0.28 Ci (c) l2.6 = ( In 2'a)l(3.171xy107 s), =8.4 10-9 S-I l mOl' (6.022 x 1023 nUclei' =1.8 x 1018 nuclei / 147 N =(0.43 10-3 gl g) l 1 mol ) Ci0 ' = 0.40 Ci activity =(8.4 x 10-9 S-I)(1.8 x 1018 nUcleil ) (3.7 x1101 ) N =(2.0
X
10-3 g)
X
k
X
Bq
'
X
X
Bq
17.45
k =In 2 =109In min2 = 6.36x tll2
lOs3
min-I
N =No e an-=e d NN o Taking natural log of both sides gives In :, -kr
-kr
( ) � -kl
Let No =100% and N =10%; we can then write (10% In l100%) Solving for t gives ( 1 ) (10% l( 1 _ , In (0.10)= 400 min (1 sf) t=- - In l-), =100% 6.36xl0 mm I)
'=-kt
k
S3
SM-493
.
17.47
In 2 = In 2 =0.131 a tl/2 5.27 a N = -kl = -(0 131 .-1)(250 =0.266g e e
k=
-
a)
No
0.370 g 0.370 g 100 =26.4% 1.40 g
No
No
=
X
17.49
Since radioactive decay follows first-order kinetics, the rate of loss of X is d[X]
kI
dt - [X] (1) Y, which is an intermediate, is lost in the first reaction but formed in the second one, so its rate equation can be expressed as d[Y] dt =k [X] k2 [Y] (2) Z is the final product of the two consecutive reactions so its rate law is =
I
d[Z]
-
dt k2 [Y] (3) As discussed in Chapter 13, the integrated form of equation(1) is [X] [X]o e (4) Substituting this expression into the rate law for Y and rearranging gIves =
=
-kll
(5) This linear first-order differential equation has the solution when [YJo (6) Since [X]+[Y]+[Z]=[X]o at all times, [Z]= [X]o -([X]+[Y]), or =
0.
(7) The values of the rate constants can be found from the half-lives: SM-494
k
k In 2 In 2 = 0.0371 dol = 0.0253 dol 2 I 27.4 d 18.7 d Using these constants and assuming [X]o=2.00 g, equations (4), (6) and (7) are graphed below. -
2
1.5
(J)
E
J,.,
� C)
_
_
1
�
\
I-omo...
17.5 1
!
-
\
oj
___
/ . ...._.___
�
/ ._!� ..
_ooo___
!.ooo.o o·· ·
f--1
___
�j ii
I
�_�/_/�
0.5
a
T
T
/� a
!
/
I
,
;' . ,
,
,
---
25
50
t
75
(days)
100
I
-+-I ,
125
I-
.::::::-,
,
150
If isotopically enriched water, such as H2180, is used in the reaction, the label can be followed. Once the products are separated, a suitable technique, such as vibrational spectroscopy or mass spectrometry, can be used to determine whether the product has incorporated the 18 O. For example, if the methanol ends up with the 0 atom from the water molecules, then its molar mass would be 34 g . mol-I, rather than 32 g . mol-I found for methanol with elements present at their natural isotopic abundance.
17.53
The vibrational freq uency is proportional to the reduced mass of the two atoms that form the bond according to the equation:
SM-495
A B where fl = mm
mA +mB
Because we are not given v, it is easiest to make a relative comparison by taking the ratio of molecule:
v
for theC-D molecule versus v for theC-H
1 (I VC_D = 2n = 1_ _ k ve_H 27r fle H 12.105 13.019 = = 0.73422 24.190 14.025
V;;:
( (
�
) )
��C-" �C-D
mm e H = me +mH mm e D me +mD
=
(12.011)(1.0078) 12.011 + 1.0078 (12.011)(2.0140) 12.011 + 2.0140
We would thus expect the vibrational frequency for theC-D bond to be approximately 0.73 times the value for theC-H bond (lower in energy). 17.55
To determine the effective half-life we need to determine the effective rate constant, k E. This constant is equal to the sum of the biological rate constant QCB ) and the radioactive decay rate constant QcR), both of which can be obtained from the respective half-lives:
In2 In2 =1.56x10 -2 d-I + 90.0 d 87.4 d In2 = 44.4 d tl/ (effective) 2 1.56x 10-2 dk E= k B + k R =
=
17.57
I
Remember to convert g to kg. (a) E=mc2 =(l.Oxl O-3 kg)(3.00xl OB m·s-I)2 =9.0x lO'3 kg·m2·s-2 =9.0x lO'3 J (b) E=mc2 =(9.109xl O-31 kg)(3.00x lOB m·s-I)2 =8.20x lO-' 4 kg·m2·s-2 =8.20x10- 14 J SM-496
E =me
2
(c)
E = (1.0 X 10-15 E =me2
(d)
kg)(3.00xI08 m·S-I)2 =90. kg· m2 . S-2 =90. J
27 kg)(3 .00 X 108 m·S- I )2 =1.51 X 10-10 J
E = (1.673 X 10-
-?
17.59
2 -3.9xl 0 6 J . s 1 M =-4 .3xl0 9 kg . s-I 11m = 2 = e (3.00xl08 m · s-I)-
17.61
1 m =1.6605 X 10 - 27 kg u
In each case, calculate the difference in mass between the nucleus and the free particles from which it may be considered to have been formed. Then obtain the binding energy from the relation Ebilld = I1me2 . (a) �� Ni: 28 I H + 34 n � �� Ni
(
kg]
11m=61.928346 mu -(28X1.0078 mu + 34 X1.0087 mu) = -0.586 mu 7 1.6605XI0-2 = -9.73x l0-28 kg 11m= (-0.586 mJ 1 mu Ebind = -9.73xl0-28 kg x(2.99792xl0 8 m·s-I ) 2
!
!
=8.74xl0-11 kg· m 2· s -2=8.74xI0-11 J 8.74x10-11 J = lA lx10-12 J . nucleon-I Ebind I nucleon= 62 nucleons
SM-497
i1m = 239.0522 mu -(94x1.0078 mu + 145x1.0087 mu) = -1.94 mu i1m (-1.94 mu> 1.66051xmu10-27 kg = -3.23x10-27 kg Ebind 1-3.23 x10-27 kgl (2.99792 xl 08 m·S-I)2 = 2.90x10-10 kg . m2 ·S-2 2.90x10-10J 2.90x10-IOJ = 1.21x10-12J. nucleon_I Ebind / nucleon = 239 nucleons
(
=
=
)
X
=
(c) �H: IH + n �H i1m 2.0141 mu -(1.0078 mu + 1.0087 mu) = -0.0024 mu i1m =(-0.0024 mu/ll .66051xmu10-27 kg1) -4.0 x 10-30 kg Ebind = 1- 4.0 xl 0-30 kgl x(2.99792 x 108 m· S-I)2 = 3.6 10-13 kg ·m2 . S-2 3.6 x 10-13J 3.6 x 10-13J = 1.8 x 10-13J. nucleon-1 Ebind / nucleon = 2 nucleons �
=
=
X
=
(d) �H: I H 2 n � �H i1m = 3.01605 mu -(1.0078 mu + 2 x 1.0087 mu) = -0.0092 mu x 10-27 kg1) = -1.5 x 10-29 kg i1m =(-0.0092 mu/ 1.6605 l 1 mu +
= 1-1.5x10-29 kglx(2.99792 x108 m ·s-I) 2 = 1.3 x 10-12 kg ·m 2 . S-2 = 1.3 x 10-12J 1.3 x 10-12J 4.3 x 10-13J. nucleonEbind / nucleon = 3 nucleons Ebind
=
1
(e) 62 Ni is the most stable, because it has the largest binding energy per nucleon. 17.63
In each case, we first determine the change in mass, i1m =(mass of products) (mass of reactants). We then calculate the energy released from M =(i1m)c2. x
SM-498
(a) D +D � 3 He + n 2.0141 mu +2.0141 mu � 3.0160 mu +1.0087 mu 4.0282 mu � 4.0247 mu (1.661x10--? 7 kg ) = -5.8 X 10-3 0 kg �m= (-0.0035 mu ) 1 mu 2 2 M = �mc = (-5.8 x 10-30 kg)(3.00 X 108 m ·S-I ) = -5.2 X 10-1 3 J ( -5.2 X 10-13 J )( ) 1 mu = -7 .8x l0 10 J . g -1 4.0282 mu 1.661x 10-24 g
l
l
)l
)
)
3.0160 mu + 2.0141 mu � 4.0026 mu +1 .0078 mu 5.0301 mu � 5.0104 mu ( 1.661 X 10-27 kg ) �m= (-0.0197 mu ) = -3.27 x 1O-29kg 1 mu 2 2 2 2 M = �mc = -(3.27 x 10- 9 kg)(3.00 x 108m . S-I) = -2.94 X 10-1 J ( -2.94x 10-12 J ) ( ) 1 mu = -3 .52x 101 \ J . g- 1 2 4 5.0301 mu 1.661 x 10 g
)
l
l
)l
j
7.0160 mu +1 .0078 mu � 2(4.0026 mu ) 8.0238 mu � 8.0052 mu (1.661 X 10-27 kg ) = -3.09 X 10-29 kg �m =(-0.0186 mu 1 mu 2 2 M = �mc = (-3.09 X 10- 9 kg)(3.00 x 108 m · S-I )2 = -2.78 X 10- 12 J ( -2.78x l 0-12 J )( ) Imu =-2.09x l O 1\ J . g-1 2 8.0238 mu 1.661x 10- 4 g
)l
l
)l
)
)
SM-499
2.0141mu +3.0160 mu �4.0026 mu +1.0087mu 5.0301mu
5.0113mu
�
�m = (-0.0188 mu !:ill=
)(l1.661 1m10 -72 kg),=-3.12 u X
10-92 kg
X
�mc2 = (-3.12 x10-29 kg)(3.00 X108m ·S-I)2= -2.81 X10-12J
)
' 1 mu (-2.81xlO-12J'( =-3.36x10 J.g5.0301 mu 1.661x10-24g
l
.
17 65
)l
I I
I
0
24 Na � 24Mg + e (a) II 12 mass ( �� Na) =23.990 96 mu _I
mass (��Mg) =23.985 04 mu The mass of the electron does not need to be explicitly included in the calculation because it is already included in the mass of Mg. �m = mass (��Mg) - mass ( �� Na)= 23.985 04 mu -23.990 96 mu = -5.92 x10-3mu -I
�m (in kg)= -5.92 X10-3mux1.661x10-27 k gmu = -9.83 X10-30 kg
(b)
!:ill=
(c)
!:ill (per
�mc2 = -(9.83x10-30 kg)(3.00 X108 m · S- I)2= -8.85 X10-13J
nucleon) =
1J -8.85x10-3 1 J. nucleon -I = -3.69x10-4 24 nucleons
This simple calculation works because the number of nucleons is the same on both sides of the equation. 17.67
I 1 (a) 24945 Am � 134 5 31 + 04 72 Mo +3 n (b) 3295 U + I n � 4906 Zr + 1 358 T e +2 0In 2 2 (c) 2�� U + � n � 1�� M o + 1 �� Sn +3 �n 0
0
SM -500
17.69
(a) False; the dose equivalent is either equal to or larger than the actual dose, due to the Q factor. (b) False; 1.8x 10 8 Bq 0.003 Ci which is much smaller than lO Ci. =
(c) True. (d) True. 17.7 1
(a) 1 Ci =3.7 x1010 decays per second (dps) decays per minute (dpm) for
( J
60 s 4 pCi=4x lO-12 Cix3 .7 x1010 dPS X . 1 mm =9 dpm
( J (--J(
103 L (b) volume(L) = (2.0x 3.0x 2. S) m3x 1 m3 4 PCi 1L =7 x10 5 decays
number of decays =(1. Sx10 4 L)
17.73
1 .Sx10 4 L
=
J
9 decays . min - I . ( S.O mm) . 4 pCI
No = number of 222Rn atoms =2.0x10-5 molx 6.0x1023 atoms·morl =1.2x10 19 atoms In 2 In 2 =0.1 81 d- I k =-= tll 3.82 d 2 (a) rate of decaY= kxN=
1d ( 0.181 )( (1.2X 1019 atoms) d 8.64x10 4 s J
= 2 .S2xl 013atoms·S- I (dps or Bq) . . = (2. S2x1 13 Bq . . . I actIvIty ImtIa 0 ) x
(
1 Ci 3.7x10 1 0 Bq
J J( ( � io��
200 m' =3. 4 X 108 pCi· L-I
SM- SOI
J(
1 pCi 10-12 CI'
J
(c) In t�
(
J
activity = -kt initial acitivity
-G) In Cnit�:i:i:::rv ty J
�-
=1 X 102 days
17.75
k=
(a) (b)
=
-G) In ( :, J
initial mass of 238 U
N No N No
=
34 10'
-= ---mass of 206 Pb 1+ mass of 238 U
1 = --, therefore age 1 +1.00 2.00
=
tl/2 = 4.5 Ga
1 =_--= 0.556 1 1 +1.25
t(=age)=
17.77
In
01 8 d'
In 2 =1.5x l0-lOa -1 4.5x 10 9 a
t<� age) �
No
( \ J ( : J
In (0.556) =3.8 Ga I 4. a
( ;� ) { �I�:O J
3.7
Bq
=6.4x l 0" Bq =6.4xl 0" nuclei· S-I ( 1u . 1 ( 1 nucleus 1 =5.0 X 1016 nucleI 10-6 g) 1 ) I ) 2 U.661x 10- 4 g \. 24 u activity 6.4x 10" nuclei· S-I k= = =1.3xl 0-5 s =1. 1 d -I 5.0 x 10 16 nuclei N In 2 In 2 = =5 .3x 10 4 s=15 h=0.63 d t /2 = 1 k 1.3 X 10-5 S-I N = (2.0
X
-I
SM-502
17.79
(a) Radioactive substances which emit
r
radiation are most effective for
diagnosis because they are the least destructive of the types of radiation listed. Additionally,
r
counted, whereas
and f3 particles are stopped by the body tissues.
(b)
a
a
rays pass easily through body tissues and can be
particles tend to be best for this application b ecause they cause the
most destruction. (c) and (d) 131 I, 8d (used to image the thyroid); 67 Ga, 78 h (used most often as the citrate complex); 99m Tc, 6 h (used for
various body tissues by varying the ligands attached to the Tc atom). 17.8 1
17.83
(a) 429 8 Mo + °In � 4299 Mo� 99m Tc+ SI.O n.t-' 43 (b) Tc-99m ( NIZ = 1 .30 vs 1 .36 for Mo-99) Radon-222 decays to polonium-218 by alpha emission with a half-life of 3 .824 days. t l/2
=
3 .824 d
An alpha particle is the nucleus of 4 He . Assuming that (1) the alpha particles are captured inside the container, (2) they behave as an ideal gas and (3) the temperature is constant at 298 K, we can find the volume of the container by calculating the number of moles of � a formed in 15 days and then applying the ideal gas law. n He = n Po
k
=
In 2 =
tl/2
=
ninilial Rn - nfinal Rn
In 2 3.824 d
=
O. 181 d·1
n.final Rn cx N1 5 days
SM-503
N days =N e- kl = (2.5 g)e- (O1 8 1 d-')( 15 d)=0.165 g 15 0 1 mol 222Rn nfinal Rn = (0 165 g 222 Rn) 222.0175 g 222 Rn = 7.43xl 0-4 mol 222 Rn 1 mol 222 Rn ninilial Rn = (2.5 g 222 Rn) 222. 0175 g 222Rn =1.13xl O- 2 mol 222Rn 2 4 nHe = ninilial RII - n(inal Rn =1.13x10- mol - 7.43x10- mol
(
•
J
(
J
=1.06xl 0-2 mol
PV = nRT V= n
RT
=
P = 0.26 L
17.85
mlol
(1 .06xl 0-2 mol)(0.08206 L atm · K- I . morl )(298 K) 1.00 atm -
=m e- +m e+ =2m _ =2(9.109 39xl0-31 kg) =1.821 88xl 0- 30 kg e
E=mc2 =(1.821 88xl 0-30kg)(2.997 92xl 0 8m·S-I ) 2 =1.637 42xl0- 13] 17.87
The average time to reach 8000 disintegrations or counts for the sample is 22.25 min; therefore the average disintegrations or counts per minute for the sample is 359.6 dpm. Average time to reach 500 backround disintegrations or counts = 5.11min; therefore the average disintegrations or counts per minute for the backround is 97.8 dpm. (a) Average level of radioactivity in the sample = 359.6 dpm - 97.8 dpm = 261.8 dpm= 262 dpm (3 sf). (b) (262 dpm/
') 1 dps ') ( 1 Ci =1.18 l OslO Ci l 60 dpm ) l 3.7 x10 10 dps ) X
SM-504
1.18 x 10
17.89
s
10
( 1 jiCi 1 56 ) 1.18 x 10S4' jiCl. 1 10 Ci
xl
Ci
=
X
Before beginning this problem it is important to recognize that essentially all of the radioactivity of the sample will still be present under the conditions employed for the Schilling test. This can be easily demonstrated using the half life of Co-S8: k
=
In 2 tl/2
=
=
In 2 9.6 10-3 d - I 72 d
fr
X
. . . actIOn remammg =- =e
N
No
-kl
= e - ( 9. 10-Jd-1 I d) = 0 .99 So 99% of the Co-S8 remains after 24 h. 6
x
(a) For the standard: 3.0 mL
x
x
0 .4 nCijmL= 1.2 nCi; this produces 910
cpm; for the first sample obtained without intrinsic factor supplementation : 83 cpm 3.0 mL urine sample 3.3 X 104 cpm
x
x
1200 mL urine volume 3.3 x 104 cpm for entire urine sample =
6
10 ,LLCi
1.2 nCi 910 cpm standard
0.044,LLCi 58CO excreted 58 O.S ,LLC i CO originally dosed
x
x
100
10 9 nCi =
= 0.044,LLC1.
58CO excreted
90(10 excreted
for the second sample obtained with intrinsic factor supplementation : 120 cpm 3.0 mL urine sample 4 . 8 X 104 cpm
x
. 1200 mL unne volume 4.8 104 cpm =
6
l.2 nCi x
10 ,LLCi
910 cpm standard
0.063,LLCi 58CO excreted 0.5,LLCi 58CO originally dosed
x
x
9
10 nCi
-
=
. 0.063 ,LLCI 58Co excreted
100 - 130(10 excreted
SM-SOS
x
X
(b) From before: k = 9.6 1O-3d-1 So the fraction remaining = � e -kt No =
e-(9.6xIO-Jd-1x d) = 7
= 0.94
94% of the activity reamins after 7 days, which would be 0.47 ,lICi . (c) kE =kB tl/2
+k R
(effective)
=
In 2 In 2 2 = -- + --=1.35xl0 S d180 d 72 d
�
In 2 1.35x10 d-
I
=
I
51.4 d
(d) If the first dose was 0.5,lICi and 0.044 ,lICi was excreted (from part a), 0.456,l1Ci of 58CO remained in the patient's system. One week (7 d) after the first dose, we can determine the amount of 58CO remaining in the patients system using k = 0.693/51.4 d =1.35x l Os2 d-I: E
( p
N = 0.91 .456JiCi 0.415JiCi; this is the new No. =
(
)
We can now determine how much of this is eliminated No SN in one day by using the biological half life:
No SN 0.41 5JiCiSN e-kt/ e- ((0693/180) x7d ) = = 0.973 = = No 0.415JiCi
[(
p
N = 0.415JiCiS 0.973 .415JiCi
J
=
0.011JiCi
From this we can determine the fraction of counts in the second sample due to the 58CO left over from the first sample: 0.011 JiCi 0.063,LLCi
=
0.17
(e) Only 1% of the 58CO decays in the one day, meaning that only 0.005,l1Ci of 58CO was converted to 59Fe. Since each disintegration of a
SM-506
58CO atom yields a single 59Fe atom, the mass of iron produced from this process can be determined as follows: ( 1 Ci 1 3( .7 x 1010 dps1 (1 mol 59Fe produced 1 (5 x 1053j.1C) l106j.1Ci ) l 1 Ci ) l 6.022 x 1023 dps ) (59 g 59Fe produced 1 -1.8 x 10520 g of 59Fe. l1 mol 59Fe produced ) Essentially none of the 2.5 g of iron in the patient will be due to 59Fe. 1
x
x
x
_
SM-507
x
CHAPTER 18 ORGANIC CHEMISTRY I: THE HYDROCARBONS
IS.1
H
I I
(a)
H-C
H -
C=== -C
I I
C-H
-
H
H alkyne
H
I I
(b)
H-C
H
I I
C
-
H
H
I I
C
-
I
H
H
H
I I
I
H
I I
C-H
-
H H alkane
H-C===C-C
(c)
H
-
H
H
I I
C-H H
alkene
(d)
I I
H-C
H
H
H -
H
C===C
I
H
I
H
I I
C
-
-
C
C
-
H
H
alkene and alkyne (e)
H
H
\ I C===C /
H
H -
I I
C
I I
C-H
H -
H alkene SM-508
I
H
/ \
C===C
H
18.3
(a) (CH3)3CH or C4HIO, alkane; (b) C6H7CH3 or C7HlO alkene; (c) C6H12, alkane; (d) C6H12 alkane
18.5
(a) C12H26, alkane; (b) C13H22, alkane; (c) C7H14, alkane; (d) C14Hg aromatic hydrocarbon
18.7
(a) propane; (b) butane; (c) heptane; (d) decane
18.9
(a) methyl; (b) pentyl; (c) propyl; (d) hexyl
18. 1 1
(a) propane; (b) ethane; (c) pentane; (d) 2,3-dimethylbutane
18.13
(a) 4-methyl-2-pentene; (b) 2,3-dimethyl-2-phenylpentane
18.15
(a) CH2=CHCH(CH3)CH2CH3; (b) CH3CH2C(CH3)2CH(CH2CH3)(CH2hCH3; (c) HC=C(CH2hC(CH3)3; (d)
CH3CH(CH3)CH(CH2CH3)CH(CH3)2
18.17
H3
(a)
I I I , I I I I I
H--C--C--C--C--C--C--C--C--C--H
I
H
I
I
H
I
H
I
I
CH3 H
I
H
I
H
I
H
H
CH3
(b)
H--C
I I
I
,
H2
I I I I I
C--C--C--C--C--C--C--C--C--H
I
H
I I I I I CH2 CH3 I
CH2 CH2 H
CH3
SM-509
I
H
I
H
I
H
I
H
H
I H-C I
(c)
H
CH3 H
I C I
-
I C I
-
CH3 H
CH3 H
I C I
-
H
-
I I
C-H
H
(d)
18.19 (a)
(b)
D
(e)
0
18.21 (a) hexenes : H3C
/
CH2,
/CH2, �CH2 CH CH2
I-Hexene
cis-2- Hexene
trans-2- Hexene
cis-3-Hexene
trans-3- Hexene Pentenes :
4-Methyl-l-pentene
3-Methyl-l-pentene SM-SIO
2-Methyl- I-pentene 2-Methyl-2-pentene CH:�CH / '" H3C-HC CH3
\
CH3 cis-4-Methyl-2-pentene
cis-3-Methyl-2-pentene (+ trans
(+ trans isomer)
isomer)
butanes :
3,3-Dimethyl- I-butene
2,3-Dimethyl- I-butene
2,3-Dimethyl-2-butene (b) cyclic molecules:
o
Cyclohexane
ethy1cyclo butane
Methy1cyclopentane
1 , I-dimethy1cyclobutane
SM- SII
4
The following structures are drawn to emphasize the stereochemistry H CH3 CH3 H
) H
H
(
P
H
H
cis- l,2-Dimethy1cyclobutane
trans- l,2-Dimethy1cyclo butane (nonsuperimposable mirror images
H
trans-l ,3-Dimethylcyclobutane H
H H
cis-l ,3 -Dimethy1cyclo butane
H H
H H Propy1cyclopropane
H
H H H
H
Isoropylcyclopropane or 2-cyclopropylpropane
H
l-Ethyl-l-methy1cyclopropane SM-S12
H
H
H
H
H
H
H
trans- l-Ethyl-2-methylcyclopropane (nonsuperimposable mirror images)
H
H H
H
H
1,1,2-Trimethylcyclopropane (nonsuperimposable mirror images)
H H
H H
cis-l -Ethyl-2-methylcyclopropane (nonsuperimposable mirror images)
SM-513
H
H
H
H
H
18.23
1,2,3-Trimethylcyclopropane
1,2,3-Trimethylcyclopropane
(all cis isomer)
(cis-trans isomer)
(a) Butane is C4H ,o' cyclobutane is C4H S. Because they have different formulas, they are not isomers. (b) Same formula, but different structures; therefore, they are structural Isomers. (c) Same formula
(CsH,o) '
same structure (bonding arrangement is the
same), but different geometry; therefore, they are geometrical isomers. (d) Not isomers, because only their positions in space are different and these positions can be interchanged. Same molecule. 18.25
(a)
H
HC 3
I I CH C
-
CH 3
-
3
(b) If only two isomeric products are formed and they are both branched, then the only possibilities are CI HC 3
I I CH
CI
C-CH 3
-
I
HC 2
3
SM-S14
H
I 'I CH C
-
CH 3
-
3
18.27
*
An designates a chiral carbon . (a) optically active,
H
I* I Br
H C-C-CH2 -CH 3
(b) not optically active,
H
3
H
I I I HI Cl
C l-C-C-CH 3
(c) optically active,
Br
I
Cl
I*
H-C-C-CH
(d) optically active,
H
I
H
Cl
Cl
I I H
I
1* I H
3
H C-C-C-CH2 -CH2 -CH 3
18.29
3
The difference can be traced to the weaker London forces that exist in branched molecules. Atoms in neighboring branched molecules cannot lie as close together as they can in unbranched isomers.
18.3 1
The balanced equations are C3H8(g) +5 0 2(g) � 3C O2 (g ) +4 H20(l) C4HIQ (g ) + 13/ 2° 2 (g ) � 4CO2 (g ) + 5 H20(l) C5HI2 (g) +8 0 2 (g ) � 5CO2 (g ) + 6 H20(l)
The enthalpies of combustion that correspond to these reactions are listed in Appendix 2:
SM-5 I5
(a) Enthalpy of combustion Compound Propane
kJ ·mol-I
kJ·g-I
2220
50.3
- 2878
49.5
-
Butane
(b) Heat released per g
49.0 Pentane 3537 The molar enthalpy of combustion increases with molar mass as might be -
expected, because the number of moles of CO2 and H 2 0 formed will increase as the number of carbon and hydrogen atoms in the compounds increases. The heat released per gram of these hydrocarbons is essentially the same because the H to C ratio is similar in the three hydrocarbons . 18.33
There are nine possible products: H H
I
H
I
H
C
C
H
H
-
I
-
I
CI
Cl
-
I
Cl
H
I
I
C
C
H
Cl
-
I
-
I
Cl H
-
C
H
H
I
-
Cl
C
C
H
H
I
I
-
I
I
Cl CI
-
I
I
C
C
Cl
Cl
I
I
-
I
-
I
H
Cl
Cl
C
C-ClH
H
H
I
I
-
I
I
I
C-Cl
I
H
Cl
Cl
C
C
H
Cl
I
-
I
-
I I
CI
Cl
-
Cl
Cl
C
C
Cl
Cl
-
I I
-
I
I
-
Cl
one hexachloro compound
one pentachloro compound
None of these form optical isomers.
SM-516
Cl
-
two tetrachloro compounds
Cl
H -
H-C
Cl
-
-
two trichloro compounds
CI
I
Cl
two dichloro compounds
Cl -
I
C
-
one monochloro compound H
H
H
18.35
CI
Cl
\
/
/
\
H
C===C
/
/
\
C===C
CH3
H
CI
\
CI
CH3
cis- l,2-Dichloropropene trans-l,2-Dichloropropene cis - l,2-Dichloropropene is polar, although trans- l,2-Dichloropropene is slightly polar also. 18.37
(a)
H H
H
I I H
--
C
--
I I H C
Br
--
H
I I H
H
I I H
I I H
C--C--C--H
3-Bromopentane H H
I I H
--
C
Br
H
C
C
I I H
--
--
I I H
H
I I H
--
C
H
I I H
--
C
H
--
2-Bromopentane (b) addition reaction
H
H
/H
\ H /C",-- / \ C H-C H / H
I C",--
c
/ H
II /C""
\
H
SM-S17
H
+ NaBr
+
HOCH2CH3
(c) elimination reaction 18.4 1
C2H4 + X 2
�C2H4X2
We will break one x-x bond and form twoC-X bonds. Using bond enthalpies: Halogen x-x
bond breakage (kJ·mol- I)
C-X bond
formation (kJ . mol-I)
Total (kJ·mol-I)
Cl
Br
I
+242
+193
+151
22(338)
22(276)
22(238)
2434
2359
2325
The reaction is less exothermic as the halogen becomes heavier. In general, the reactivity, and also the danger associated with use of the halogens in reactions, decreases as one descends the periodic table. 18.43
(a) 1-ethyl-3-methylbenzene; (b) pentamethylbenzene (1,2,3,4,5pentamethylbenzene is also correct, but, because there is only one possible pentamethylbenzene, the use of the numbers is not necessary)
18.45
(a)
6
(b)
Cl
Cl
SM-518
18.47
Cl
Cl
Cl
1,3-Dichlo ro-2-methylbenzene
Cl 1,4-Dichloro-2-methylbenzene
Cl
CI CI 1,5-Dichloro-2-methylbenzene
Cl
1,3-Dichloro-5-methylbenzene
Cl
Cl
CI
1,2-Dichloro-3-methylbenzene
CI 1,2-Dichloro-4-methylbenzene
(b) All of these molecules are at least slightly polar.
18.49
+
Electrophiles tend to avoid the ortho and para positions that develop slight + charges in the resonance forms.
SM-519
18.5 1
Two compounds can be produced. Resonance makes positions 1, 4, 6, and 9 equivalent. It also makes positions 2, 3, 7, and 8 equivalent. Positions 5 and 10 are equivalent but have no H atom. E
18.53
9 8� 10
1 �2
7� 6
#3
5
E
4
These hydrocarbons are too volatile (they are all gases at room temperature) and would not remain in the liquid state.
18.55
Cracking is the process of breaking down hydrocarbons with many carbon atoms into smaller units, whereas alkylation is the process of combining smaller hydrocarbons into larger units. Both processes are carried out catalytically, and both are used to convert hydrocarbons into units having from 5 to 11 carbon atoms suitable for use in gasoline.
18.57
(a) 4 a-type single bonds (b) 2 a-type single bonds and 1 double bond with a a- and an-bond (c) 1 a-type single bond and 1 triple bond with a a-bond and 2n-bonds
18.59
(a) substitution, CH4
+
(b) addition CH2=CH2
Ch +
-7
Br2
CH3CI + HCI
-7
CH2Br-CH2Br
18.61
Water is not used as the nonpolar reactants will not readily dissolve in a highly polar solvent like water. Also, the ethoxide ion reacts with water.
18.63
The double bond in alkenes makes them more rigid than alkanes. Some of the atoms of alkene molecules are locked into a planar arrangement by the n-bond; hence, they cannot roll up into a ball as compactly as alkanes can. SM-520
Because they do not pack together as compactly as alkanes do, they have lower boiling and melting points. 18.65
(a) 2-methyl-1-propene, no geometrical isomers; (b) cis-3-methyl-2pentene, trans-3-methyl-2-pentene; (c) I-hexyne, no geometrical isomers; (d) 3-hexyne, no geometrical isomers; (e) 2-hexyne, no geometrical isomers.
18.67
(a) CIOH1s; (b) naphthalene,
001"'" �
#
,
CIOHS; (c) Yes. Cis and trans
forms (relative to the C-C bond common to the two six-membered rings) are possible.
trans-Decalin
cis- Decalin
H
H
18.69
H
-
I
C===C
I
H
H -
I I H C
H
-
I I H C
-
H
+
HC]
-t
H
-
I I H C
-
C]
H
C
C
I I H
--
I I H
H --
I I H C
-
H
Bonds Broken: H-Cl, C=C Bonds Formed: C-Cl, C-H, C-C The reaction enthalpy will be equivalent to the enthalpy gained by the system in breaking bonds and the enthalpy lost during bond formation : Mfr= MfB(H-Cl) + MfB(C=C) - [MfB(C-Cl) + MfB(C-H) + MfB(C-C)] From tables 6.7 and 6.8, Mfr= 431 kJ·morl + 612 kJ·morl - [338 kJ·morl l + 412 kJ·mor + 348 kJ·morl] = - 55 kJ·morl
SM-521
18.71
number of moles of H
[
[J
J
J
J
2 mol H 4.48 g H 20 18.02 g H 20/mol H 20 1 mol H 20 = 0.497 mol H
=
number 0f moIes 0f C = =
[
[
9.72 g C02 Imol C 44.01 g CO2 /mol CO2 1 mol CO2 0.221 mol C
0.497 mol H 9 mol H 0.221 mol C 4 mol C
----- = ---
The empirical formula is C4H9; the molecular formula might be C 8H1 8, which matches the formula for alkanes (CnH2n+2). It is not likely an alkene or alkyne, because there is no reasonable Lewis structure for a compound having the empirical formula C4H9 and multiple bonds. 18.73
(a) 4-methyl-3-propylheptane The longest chain has eight carbon atoms in it. The systematic name of the compound is 4-ethyl-5-methyloctane. 4-ethyl-5-methyloctane; (b) 4,6-dimethyloctane The compound name is almost correct, but the numbering scheme with the lowest numbers would be 3,5dimethyloctane.
3,5-dimethyloctane;
(c) 2,2-dimethyl-4-propylhexane The longest carbon chain in the molecule is seven carbon atoms long. The systematic name is 2,2-dimethyl-4-propylheptane. 2,2-dimethyl-4-ethylheptane; SM-522
(d) 2,2-dimethyl-3-ethylhexane. The name is essentially correct except that ethyl should be listed first The systematic name is 3-ethyl-2,2-dimethylhexane 18.75
3 -ethyl-2,2-dimethylhexane
Bromine is an electrophile which will undergo an addition reaction with alkenes in the dark. The lack of a reaction in the dark with Br2 indicates that the molecule is not an alkene. In the presence of light, bromine will undergo a substitution reaction with alkanes. Therefore, the molecule is most likely an alkane and the only alkane with the molecular formula C3H6 is
18.77
cyclopropane.
The N02 group is a meta-directing group and the Br atom is an ortho, para-directing group. Because the position para to Br is already substituted with the N02 group, further bromination will not occur there. The resonance forms show that the bromine atom will activate the position ortho to it as expected. The N02 group will deactivate the group ortho to itself, thus in essence enhancing the reactivity of the position meta to the N02 group. This position is ortho to the Br atom, so the effects of the Br and N02 groups reinforce each other . Bromination is thus expected to occur as shown: Br
o � �N o/ 18.79
(a) and (b)
Br
H H
,
/
H
1
H
1*
o �N o/
--
CH3 H
1*
1
CH3
1
C==C -- C -- C -- C -- C--CH3
1
1
CH3 H
1
H
1
H
(c) No, there are no cis/trans isomers for this molecule.
SM-523
Br
18.8 1
For a molecule such as 1,2-dichloro-4-diethylbenzene, C6H3CdCH2CH3), 175.04 u, it is relatively easy to lose heavy atoms such as chlorine and groups of atoms such as methyl and ethyl fragments. Molecules can also lose hydrogen atoms. In mass spectrometry, P is used to represent the parent ion, which is the ion formed from the molecule without fragmentation. Fragments are then represented as P x, where x is the particular fragment lost from the parent ion to give the observed mass. Because the mass spectrum will measure the masses of individual molecules, the mass of carbon used will be 12.00 u (by definition) because 2 the large majority of the molecules will have all 1 C. The mass of H is 1.0078 u. Some representative peaks that may be present are listed below. Fragment Relation to Mass, u formula parent ion 174.00 C6H3oDCh(CH2CH3) P 176.00 C6H3J)ClJlCI(CH2CH3) P 177.99 C6H3J/Ch(CH2CH3) P C6H/)CI(CH2CH3) P-CI 139.03 141.03 C6H3JI CI(CH2CH3) P-CI C6H3J)Ch(CH2) 158.98 P-CH3 C6H/)CIJ/CI(CH2) P-CH3 160.97 162.97 P-CH3 C6H3J/Ch(CH2) C6H/)Ch 144.96 P-CH2CH3 C6H3J)Cf'CI 146.96 P-CH2CH3 C6H3J/Ch 148.96 P-CH2CH3 C6H3J)CI P-CH2CH3-CI 109.99 C6H3J/CI P-CH2CH3-CI 111.99 Etc. -
18.83
The presence of one bromine atom will produce in the ions that contain Br companion peaks that are separated by 2 u. Any fragment that contains Br will show this "doublet" in which the peaks are nearly but not exactly equal in intensity. Thus, seeing a mass spectrum of a compound that is known to have Br or that was involved in a reaction in which Br could have been added or substituted with such doublets, is almost a sure sign that Br is present in the compound. It is also fairly easy to detect Br atoms in the mass spectrum at 79 and 81 u, confirming their presence. If more SM-524
than one Br atom is present, then a more complicated pattern is observed for the presence of the two isotopes. The possible combinations for a molecule of unknown formula with two Br atoms is 79 Br79Br, 79 Br81 Br, 81 Br79Br, and 81 Br8I Br. Thus, a set of three peaks (the two possibilities 79 Br81 Br and 81 Br79 Br have identical masses) will be generated that differ in mass by two units . The center peak, which is produced by the 79 Br81 Br and 81 Br79Br combinations, will have twice the intensity of the outer two peaks, because statistically there are twice as many combinations that produce this mass. All modem mass spectrometers have spectral simulation programs that can readily calculate and print our the relative isotopic distribution pattern expected for any compound formulation, so that it is possible to easily match the expected pattern for a particular ion with the experimental result. 18.85
CgH10 will have an absorption maximum at a longer wavelength. Molecular orbital theory predicts that in conjugated hydrocarbons (molecules which contain a chain of carbon atoms with alternating single and double bonds) electrons become delocalized and are free to move up and down the chain of carbon atoms. Such electrons may be described using the one dimensional "particle in a box" model introduced in Chapter 1 . According to this model, as the box to which electrons are confined lengthens, the quantized energy states available to the electrons get closer together. As a result, the energy needed to excite an electron from the ground state to the next higher state is lower for electrons confined to longer boxes . Therefore, lower energy photons, i.e. photons with longer wavelengths, will be absorbed by the CgH10 molecule because it provides a longer "box" than C6Hg .
SM-525
CHAPTER 19 ORGANIC CHEMISTRY II: FUNCTIONAL GROUPS 19.1
(a) RNH 2 , R 2NH, R3N
19.3
(a) ether; (b) ketone; (c) amme; (d) ester
19.5
(a) 2-iodo-2-butene; (b) 2,4-dichloro-4-methylhexane; (c) 1,1,1,
(b) ROH (c) RCOOH (d) RCHO
triiodoethane; (d) dichloromethane 19.7
(a)
(
�
C1C6H4Cl(OH) , phenol;
OH
(b) CH3CH(CH3)CH(OH)CH 2CH3, secondary alcohol; (c) CH3CH 2CH(CH3)CH 2CH(CH3)CH 20H, primary alcohol; (d) CH3C(CH3)(OH)CH 2CH3, tertiary alcohol
19.1 1
(a) butyl propyl ether; (b) methyl phenyl ether; (c) pentyl propyl ether
19.13
(a) aldehyde, ethanal; (b) ketone, propanone; (c) ketone, 3-pentanone
19.15
(a)
(b)
o� butanal o
3-hexanone
SM-526
2-heptanone
(c) 19.17
19.19
(a) ethanoic acid; (b) butanoic acid; (c) 2-aminoethanoic acid (b)
(a)
(c)
(d)
H2 C / �
C H2
�C
19.21
o
II C� H / /
H2 C / �
0
C H2
(a) methylamine; (b) diethylamine; (c) o-methylaniline, 2methylaniline, o-methylphenylamine, or l-amino-2-methylbenzene
19.23
(b)
(a)
o (c)
CH3 H3C -
iI
l+
-CH3
CH3 SM-527
19.25
Only (a) and (c) may function as nUcleophiles, because they have lone pairs of electrons that will be attracted to a positively charged carbon center. CO2 and SiH4 have no lone pairs and cannot function as nucleophiles.
19.27
(a) ethanol; (b) 2-octanol; (c) 5-methyl-l-octanol. These reactions can be accomplished with an oxidizing agent such as acidified sodium dichromate, Na 2 Cr2°
19.29
7
•
(a)
(b)
(c)
(d)
19.31
The following procedures can be used: (1) Dissolve the compounds in water and use an acid-base indicator to look for a color change. Tollens reagent
SM-528
(3) CH3COCH3
Tollens reagent
)
no reaction
Procedure (1) will distinguish ethanoic acid; (2) and (3) will distinguish propanal from 2-propanone. 19.33
CH3CH 2COOH
19.35
(a) -CH 2 - C(CH3)2 - CH 2 - C(CH3 )2 - CH 2 - C(CH3)2 (b) -CH-CH2-CH-CH2-CH-CH2-
\
CN (c)
\
\
CN
CN
H H H H H H \ / I \ / �I H C-C C H H C / "'" / 1\ 1\1 \/ H C =C H C C C=C / \ / \ / \ / "" CH3 HH C =C CH3 H /
CIS versIon
trans version
19.37
(a) CHCI=CH 2 ; (b) CFCI=CF2
SM-529
\
H
19.39
(a) -OCCONH(CH2 )4NHCOCONH(CH2 )4NH - ; (b) -OC -CH(CH 3 )- NH - OC - CH(CH 3 )- NH -
19.41
block copolymer
19.43
Larger average molar mass corresponds to longer average chain length. Longer chain length allows for greater intertwining of the chains, making them more difficult to pull apart. This twining results in (a) higher softening points, (b) greater viscosity, and (c) greater mechanical strength.
19.45
Highly linear, unbranched chains allow for maximum interaction between chains. The greater the intermolecular contact between chains, the stronger the forces between them, and the greater the strength of the material.
19.47
Polymers generally do not have definite molecular masses because there is no fixed point at which the chain-lengthening process will cease. The chain stops growing due to lack of nearby monomer units of the appropriate kind or a lack of properly oriented smaller polymeric aggregates. A polymer is, in a sense, not a pure compound, bur rather a mixture of similar compounds of different chain length. There is no fixed molar mass, only an average molar mass. Because there is no one unique compound, there is no one unique melting point, rather a range of melting points. Thus, there is no sharp transition between solid and liquid, and we say the solid softens rather than melts. o
19.49
(a)
II
-C
'"
NH -
(b) amide; (c) condensation SM-530
19.5 1
Side groups that contain hydroxyl, carbonyl, amino, and sulfide groups are all potentially capable of participating in hydrogen bonding that could contribute to the tertiary structure of the protein. Thus, serine, threonine, tyrosine, aspartic acid, glutamic acid, lysine, arginine, histidine, asparagine, and glutamine satisfy the criteria. Proline and tryptophan generally do not contribute through hydrogen bonding, because they are typically found in hydrophobic regions of proteins.
19.53
NH 2 0 I � H-C-C I "NH-CH2-COOH CH2
OH 19.55
(a) The functional groups are alcohols and aldehydes. (b) The chiral carbon atoms are marked with astelisks ( H H OH OH I I I I OHC -C -C -C -C - CH20H I I I I OH OH H H *
*
*
*
).
*
19.57
(a) GTACTCAAT; (b) ACTTAACGT
19.61
(a) alcohol, ether, aldehyde; (b) ketone, alkene; (c) amine, amide
* An asterisk ( * ) denotes a chiral carbon atom.
SM-531
19.63
(b)
(a)
*
o 19.65
(a)
(b)
o
H2 C--'CH 2 * / \ CH 0 "
CH
o
19.67
(a)
/'
/ *
H H H H I I_ I I H-C-C O-C-C-H I I I I H H H H Diethyl ether
H H H I H-C-C- - -o-H I H
I
� b � � � I-butanol
SM-532
(b) I-Butanol can hydrogen bond with itself but diethyl ether cannot, so I-butanol molecules are held together more strongly in the liquid and therefore I-butanol has the higher boiling point. Both compounds can form hydrogen bonds with water and therefore have similar solubilities. 19.69
o
(a) CH2
I I
HC
--
CH2 (b)
HO
19.7 1
19.73
-
-
0 0 0
II
-
-
-
C - (CH2)16CH3
�
C - (CH2)16CH3
�
C - (CH2)16CH3
--Q--
CHO
(a) addition; (b) condensation; (c) addition; (d) addition; (e) condensation (a) HOCH 2CH 20H + 2 CH} (CH 2 ) COOH --+ CH}(CH 2 ) COOCH 2CH 200C(CH 2 ) CH 3 + 2 H 20 (b) 2 CH}CH 20H + HOOCCOOH --+ CH}CH 200CCOOCH 2CH} + 2 H 20 (c) CH}CH 2CH 2CH20H + CH}CH 2COOH � CH}CH 2COOCH 2CH 2CH 2CH} + H 20 16
16
19.75
16
(a) The polymer concentration is given by: 0.325 atm c = II = = 0.0133 M iRT (1)(0.08206����1)(298 K) The moles of polymer in solution are: npolymer = (0.01 33 M)(0.500 L) = 0.00664 mol and the molar mass of the polymer is: 47.7 g 7180-L M.M. = mol 0.00664 mol =
SM-533
(b) If a monomer in the polymer is -CH2CH(CN)-, then the molar mass of a monomer is 53.06 glmol and the average number of monomers in a polymer is: (7180g/mol ) / (53.06 g/mol ) = 135 monomers . (c) The pressure ofH20(g) above the mixture is given by:
.III 100 mL, n and, npolymer Therefore, XH 20 PH20
H 20
=
(100 mL )(1.00 ;L)
= 5.55 mol 18.02 mol (0.100 L )(0.01328 m�l ) 0.00133 mol. =
g
_
=
5.55 mol 0.9998 and 5.55 mol + 0.00133 mol (0.9998 )(23.76 Torr ) 23.75 Torr
=
=
=
=
(d) Measuring the change in osmotic pressure proves to be a better method in this case. The osmotic pressure developed by the resulting polymer solution is readily measured while the change in partial pressure ofH20(g) changes by less than 0.1 % upon addition of the polymer. 19.77
(a) Primary structure is the sequence of amino acids along a protein chain. Secondary structure is the conformation of the protein, or the manner in which the chain is coiled or layered, as a result of interactions between amide and carboxyl groups. Tertiary structure is the shape into which sections of the proteins twist and intertwine, as a result of interactions between side groups of the amino acids in the protein. If the protein consists of several polypeptide units, then the manner in which the units stick together is the quaternary structure. (b) The primary structure is held together by covalent bonds. Intermolecular forces provide the major stabilizing force of the secondary structure. The tertiary structure is maintained by a combination of London forces, hydrogen bonding, and sometimes ion-ion interactions. The same forces are responsible for the quaternary structure.
SM-534
19.79
+
(a) H3NCH2COOH(aq) + H20(l)�+ H 3NCH2COO-(aq) + H30+ (aq) +H3NCH2COO-(aq) + H20(l) � H2NCH2COO-(aq) + H30+ (aq) (b) pKal 2.35 =
pKa2 9.78 =
pH 2, +H3NCH2COOH =
19.81
Condensation polymelization involves the loss of a small molecule, often water or HCI, when monomers are combined. Dacron is more linear than the polymer obtained from benzene-l ,2-dicarboxylic acid and ethylene glycol, so Dacron can be more readily spun into yam.
19.83
19.85
:0: :0: :0: II .. I I .. II .. H2N-CH - C ---HN - CH - C --tIN - CH - C-OH ·· I I I CH2 CH2 CH2 I .. I I C=9 CH - CH3 CH2 I .. I I :OH CH3 C=9 .. I :OH
(a)
(b)
3 Sp ;
(c)
2 Sp ;
(d) each N atom carries one lone pair of electrons; (e)
Yes, the N atoms help to carry the current because the unhybridized orbital on each N atom is part of the extended J[
J[
conjugation (delocalized
-bonds) that allows electrons to move freely along the polymer. SM-535
p
19.87
Two peaks are observed with relative overall intensities 3 : 1. The larger peak is due to the three methyl protons and is split into two lines with equal intensities. The smaller peak is due to the proton on the carbonyl carbon atom and is split into four lines with relative intensities 1 : 3 : 3 : 1.
19.89
The peaks in the spectrum can be assigned on the basis of the intensities and the coupling to other peaks. The hydrogen atoms of the
CH
3
unit of
the ethyl group will have an intensity of 3 and will be split into a triplet by the two protons on the
CH 2
unit. This peak is found at
0
::::::
1.2. The
CH 2
unit will have an intensity of 2 and will be split into a quartet by the three protons on the methyl group. This peak is found at
0
=
4.1.. The
CH 2
group that is part of the butyl function will have an intensity of two but will appear as a singlet because there are no protons on adjacent carbon atoms. This is the signal found at
0
=
2.1. The remaining
CH3
groups are
equivalent and also will not show coupling. They can be attributed to the signal at the
CH2
0
=
1.0. Notice that the peak that is most downfield is the one for
group attached directly to the electronegative oxygen atom, and
that the second most downfield peak is the one attached to the carbonyl group. 19.91
If one considers the reaction, the products should be those arising from substitution of hydrogen atoms on the propane by chlorine atoms. We would expect then to form a chloropropane or, perhaps a dichloropropane. Remember that in the halogenation of alkanes, substitution becomes more difficult as more halogen atoms are introduced. If we then consider the NMR spectrum, we see that there is one large peak that sees a single proton, because it is split into a doublet. There is also a weaker feature, corresponding most likely to one proton, which is split into a septet. This indicates that the proton sees six equivalent protons. The structure that is consistent with this spectrum is 2-chloropropane. SM-536
19.93
(a)
13
two
13 C
13
C ; (b) 1 .1 1 %; (c) No. The reason is that the probability of finding nuclei next to each other is very low. A 1 2 C nucleus next to a
C nucleus will not interact with the 1 3 C nucleus because the
12 C
nucleus has no spin. Because the natural abundance of 1 3 C is 1.11 %, the probability of finding two 1 3 C nuclei next to each other in an organic x
x
molecule will be 0.0111 0.011 1 or 1.23 10-4• Although such coupling is possible, it is generally not observed because the signal is so much weaker than the signal due to the molecules with a single 1 3 C nucleus. (d) Maybe. Because most of the carbon to which the protons are attached 2
is 1 C, the bulk of the signal will not be split. The protons that are attached to the 1 3 C atoms will be split, but this amounts to only 1 .11 % of the sample, so the peaks are very small. Peaks that result from coupling to a small percentage of a magnetically-active isotope are referred to as satellites and may be observed if one has a very good spectrum. (e) Yes. Although the splitting of protons by 1 3 C may not be observed because the amount of 1 3 C present is low, the opposite situation is not true. If a
13 C
is attached to H atoms, the large majority of those H atoms
will have a spin and so the 13 C will show fine structure due to splitting by the H atoms. 19.95
(a) 1; (b) 1; (c) 1; (d) 2; (e) Free rotation around the C-C and C-CI bonds averages out the environments so that both CH2 hydrogens are equivalent and all three CH3 hydrogens are equivalent, resulting in just two different hydrogen signals.
SM-537
