TH EDITION
Student Solutions Manual to Accompany
PHYSICAL CHEMISTRY PETER ATKINS • CHARLES TRAPP CARMEN GIUNTA • MARSHALL CADY
STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY
PHYSICAL CHEMISTRY EIGHTH EDITION
STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY
PHYSIC~l
CHEMISTRY Eighth Edition
P. W. Atkins Professor of Chemistry, University of Oxford and Fellow of Uncoln College
C. A. Trapp Professor of Chemistry, University of Louisville, Louisville, Kentucky, USA
M. P. Cady Professor of Chemistry, Indiana University Southeast , New Albany Indiana, USA
C. Giunta Professor of Chemistry, Le Mayne College, Syracuse, NY, USA
II w. H. Freeman and Company New York
Student's solutions manual to accompany Physical Chemistry, Eighth Edition © Oxford University Press, 2006 All rights reserved ISBN13: 9780716762065 ISBNlO: 0716762064 Published in Great Britain by Oxford University Press This edition has been authorized by Oxford University Press for sale in the United States and Canada only and not for export therefrom. Library of Congress Cataloging in Publication Data Data available Second printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com
Preface This manual provides detailed solutions to all the endofchapter (b) Exercises, and to the oddnumbered Discussion Questions and Problems. Solutions to Exercises and Problems carried over from previous editions have been reworked, modified, or corrected when needed. The solutions to the Problems in this edition rely more heavily on the mathematical and molecular modelling software that is now generally accessible to physical chemistry students, and this is particularly true for many of the new Problems that request the use of such software for their solutions. But almost all of the Exercises and many of the Problems can still be solved with a modem handheld scientific calculator. When a quantum chemical calculation or molecular modelling process has been called for, we have usually provided the solution with PC Spartan pro™ because of its common availability. In general, we have adhered rigorously to the rules for significant figures in displaying the final answers. However, when intermediate answers are shown, they are often given with one more figure than would be justified by the data. These excess digits are indicated with an overline. We have carefully crosschecked the solutions for errors and expect that most have been eliminated. We would be grateful to any readers who bring any remaining errors to our attention. We warmly thank our publishers for their patience in guiding this complex, detailed project to completion. P. W. A. e.A.T. M. P.e. e. G.
Contents PART 1 Equilibrium
1
1
The properties of gases
3
Answers to discussion questions
3 4 13 13 18 20
Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
5
Simple mixtures
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
6
Phase diagrams
Answers to discussion questions
2
The First Law
22
Solutions to exercises Solutions to problems
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
22 23 33 33 41 47
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
7
Chemical equilibrium
Answers to discussion questions
3
The Second Law
50
Solutions to exercises Solutions to problems
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
4
50 51 58 58 68 74
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
PART 2 Structure 8
Physical transformations of pure substances
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
78 78 80 83 83 86 87
91 91 91 98 98 104 107
112 112 113 119 119 124 124
127 127 128 137 137 148 150
155
Quantum theory: introduction and prinCiples
Answers to discussion questions Solutions to exercises Solutions to problems
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
157 157 158 162 162 165 172
viii
9
Contents
Quantum theory: techniques and applications
176
Solutions to theoretical problems Solutions to applications
176 176 183 183 186 195
10 Atomic structure and atomic spectra
199
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems
14 Molecular spectroscopy 2: electronic transitions
Solutions to numerical problems Solutions to theoretical problems Solutions to applications
280 28 1 284 284 289 292
15 Molecular spectroscopy 3: magnetic resonance
297
Answers to discussion questions Solutions to exercises Solutions to problems
Answers to discussion questions Solutions to exercises
Solutions to applications
199 200 207 207 211 218
11 Molecular structure
221
16 Statistical thermodynamics 1 : the concepts
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
22 1 223 226 226 238 241
Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
12 Molecular symmetry Answers to discussion questions Solutions to exercises Solutions to problems Solutions to applications
Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
297 299 305 305 309 311
315
315 315 322 322 326 329
244
244 245 249 255
13 Molecular spectroscopy 1 : rotational 259 and vibrational spectra Answers to discussion questions Solutions to exercises
280
259 260 269 269 275 276
17 Statistical thermodynamics 2: applications
331
Solutions to applications
331 332 338 338 345 353
18 Molecular interactions
357
Answers to discussion questions
357 358 361
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems
Solutions to exercises Solutions to problems
Contents Solutions to numerical problems Solutions to theoretical problems Solutions to applications
361 366 368
22 The rates of chemical reactions
440
Answers to discussion questions
440 443 450 450 455 458
Solutions to exercises
19 Materials 1: macromolecules and aggregates
Solutions to problems
370
Solutions to numerical problems Solutions to theoretical problems
Answers to discussion questions Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
370 372 375 375 379 383
Solutions to applications
23 The kinetics of complex reactions
464
Answers to discussion questions
464 465 468 468 471 478
Solutions to exercises Solutions to problems
20 Materials 2: the solid state
389
Answers to discussion questions
389 390 398 398 405 408
Solutions to numerical problems Solutions to theoretical problems
Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applicati ons
ix
Solutions to applications
24 Molecular reaction dynamics
489
Answers to discussion questions
489 490 497 497 502 506
Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems
PART 3 Change
411
21 Molecules in motion
413
25 Processes at solid surfaces
509
Answers to discussion questions
413 414 424 424 430 433
Answers to discussion questions
509 511 521 521 531 534
Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
Solutions to applications
Solutions to exercises Solutions to problems Solutions to numerical problems Solutions to theoretical problems Solutions to applications
PART 1 Equilibrium
The properties of gases
Answers to discussion questions 01.1
An equation of state is an equation that relates the variables that define the state of a system to each other. Boyle, Charles, and Avogadro established these relations for gases at low pressures (perfect gases) by appropriate experiments. Boyle determined how volume varies with pressure (V ex lip), Charles how volume varies with temperature (V ex T), and Avogadro how volume varies with amount of gas (V ex n). Combining all of these proportionalities into one we find nT Vex  . p
Inserting the constant of proportionality, R, yields the perfect gas equation
V 01.3
RnT
= p
or
pV
= nRT.
Consider three temperature regions: (1) T < TB . At very low pressures, all gases show a compression factor, Z ~ I. At high pressures, all gases have Z > I , signifying that they have a molar volume greater than a perfect gas, which implies that repulsive forces are dominant. At intermediate pressures, most gases show Z < I, indicating
that attractive forces reducing the molar volume below the perfect value are dominant. (2) T ~ TB . Z ~ I at low pressures, slightly greater than I at intermediate pressures, and significantly greater than I only at high pressures. There is a balance between the attractive and repulsive forces
at low to intermediate pressures, but the repulsive forces predominate at high pressures where the molecules are very close to each other. (3) T > TB. Z > I at all pressures because the frequency of collisions between molecules increases with temperature. 01.5
The van der Waals equation 'corrects ' the perfect gas equation for both attractive and repulsive interactions between the molecules in a real gas. See Justification 1.1 for a fuller explanation. The Bertholet equation accounts for the volume of the molecules in a manner similar to the van der Waals equation but the term representing molecular attractions is modified to account for the effect of temperature. Experimentally one finds that the van der Waals a decreases with increasing temperature. Theory (see Chapter 18) also suggests that intermolecular attractions can decrease with temperature.
4
STUDENT'S SOLUTIONS MANUAL
This variation of the attractive interaction with temperature can be accounted for in the equation of state by replacing the van der Waals a with a/ T o
Solutions to exercises E1.1(b)
(a) The perfect gas law is
pV
= nRT
implying that the pressure would be nRT P=V
All quantities on the right are given to us except n, which can be computed from the given mass of Ar.
n
=
25 g 39.95 g mol 1
= 0.626 mol
(0.626 mol) x (8.31 x 10 2 dm 3 bar K 1molI ) x (30 + 273 K)
so P =
1.5dm
3
1

1
= . 10.5 bar .
not 2.0 bar. (b) The van der Waals equation is P
RT a V  b  V2
=
m
sop
=
m
10 2 dm 3 bar K 1molI ) x (30 + 273) K (1.53 dm 3 / 0.626 mol)  3.20 x 10 2 dm 3 mol  1
(8 .31
X
(1.337dm 6 atmmol 2 ) x (1.013baratm
E1.2(b)
3
1 )
_I 10.4 bar 1 .

( 1.5 dm / 0.626 mo1)2
(a) Boyle's law applies: PV
= constant so
p f Vf
= Pi Vi
and Pi
prVr
=  = Vi
(1.97 bar) x (2.14dm (2.14 + 1.80) dm 3
3
)
=
1
1.07 bar
1
(b) The original pressure in bar is Pi
E1.3(b)
= (1.07 bar)
x ( 1 atm) x (760 TOrr) 1.013 bar I atm
= I803 Torr I
The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV
= nRT
so P ex: T
and
Pi
Ti
Pr Tr
THE PROPERTIES OF GASES
5
The final pressure, then, ought to be
= pjTr = Pr E1.4(b)
Tj
= 1120 kPa I
(125 kPa) x ( II + 273) K (23 + 273) K
According to the perfect gas law, one can compute the amount of gas from pressure, temperature, and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV
= nRT
pV (1.00atm) x (1.013 x 105 Paatm I) x (4.00 x 10 3 m3 ) so n   .  RT (8 .3145J K Imol  I) x (20+273) K
and m E1.S(b)
= ( 1.66 x
lOS mol) x (16.04 g molI)
x 106 g
= 12.67
X
5 10 mol
10 3 kg 1
Identifying P ex in the equation P = Pex + pgh [1.3] as the pressure at the top of the straw and P as the atmospheric pressure on the liquid, the pressure difference is P  Pex
=
pgh
= ( 1.0 x 103 kg m 3 ) x (9.8 1 m s 2) x (0.15 m)
=11.5 x 10 Pa 1(=1.5 3
E1.6(b)
= 2.67
= 1.66 x
X
10 2 atm)
The pressure in the apparatus is given by
= Palm + pgh [1.3] P alm = 760 Torr = I atm = 1.013 x
P
pgh = 13.55 g cm
P = 1.013 E1.7(b)
X
3
x
105 Pa
(/o~gg) x (1O:~m3) x
105 Pa + 1.33 x 104 Pa
= 1.146 x
0.100 m x 9.806 m s2 = 1.33
105 Pa
X
104 Pa
= 1115 kPa I
All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R. m The molar mass is obtained from PV = nRT =  RT M . mRT RT which upon rearrangement gives M =   = p V P
P
The best value of M is obtained from an extrapolation of p / P versus P to P = 0; the intercept is M / RT. Draw up the following table
0.750000 0.500000 0.250000
0.082 00 14 0.082 0227 0.082 0414
From Figure l.l (a), (PVm) T p=O
1.428 59 1.428 22 1.427 90
= I0.082 0615 dm 3 atm K I molI
I
6
STUDENT'S SOLUTIONS MANUAL
[1.• ::: "'M
: ,.: .I · •.· •.
:,.!.:.• .
:I, . · ·•.
•. · •.
·· i.. ·!···!···i .. ·!· .. !··· .. · ··· ··· .. . . 'i" '!"' !"'! '" .. ,." ..... .
",,,.,,,.! ...
···'···I···I···,···j·· ·j··
~ ,,;8.202,
...
'· !.!j~!~.f !• !•• • . • . • • ••
· .r ·.• .• •. ri,.:· .•= · ·•. .•·: .!, •• .·:,i
•.• •.•
";"';";"';";"';"';" ... i ... i ... i ... ! .
··! ··!··! ·! ·· !···, ··· ,··!· ··.··! .. ·i .. ·!
... ............ . ! . .. ! ... ! . ..
. ~~ . ~ ...~ . .~ . .... ~ . .~ ...!... !...j ··i"!"'! "! "!". 'i"+" .. ·!· .. i.. ·!···!··· ..
"M
,,!"' ! '''! . ... . ,',,' ! , .. !. ,, !',, '
". :2 ";8.200' ·; .. ·;··;· ·;.. ·;.. ·:.. Y·i ...:... :... :.. :
::!:··!:.·!:::!: : !::i:::.:::·:::~.~~:· : : ·~.~~;.: ::'.:~.~~: ...;... '1.0
.:t::::ttt:::::J:::':::;:::':::':::: ::.f.!~ti:!:':::::::':::'::t:· :::i:::
From Figure 1.l(b),
(~) p
 1.42755 g dm 3 atm I
p=o
..
...:
i .4288 ..
"': ·':.. ·1.4286'
..:.
.,: ... ~ .. ':
,
.. ~ .. .., ..., .. ~ . ,! .. 'r"'~ ... ... "f " '~ .
"': .,,?
Figure 1.I(a)
..~
~
~
... ; .. ! .. ;....: .. i . . i ... .: .. j ••. ~
Figure 1.I(b) M = RT
(~) p
(0.0820615 dm3 atm molI K 1) x (273.15 K) x (1.42755 g dm 3 atm
l
)
p=o
= 131.9987 g molII The value obtained for R deviates from the accepted value by 0.005 percent. The error results from the fact that only three data points are available and that a linear extrapolation was employed. The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors. E1.8(b)
The mass density p is related to the molar volume Vrn by
M p
where M is the molar mass. Putting this relation into the perfect gas law yields pVrn
= RT
so
pM =RT p
THE PROPERTIES OF GASES
7
Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule
+
(8.314 Pa m3 molI) x [(100 273) K] x (0.6388kgm 3 ) ~~
RTp
MP 
1.60xl()4Pa
= 0. 124 kg mol I = 124 g mol I The number of atoms per molecule is 124g mol  I ',I = 4.00 31.0g molsuggesting a formula of ~ E1.9(b)
Use the perfect gas equation to compute the amount; then convert to mass. PV
pV
= nRT
so
n
= RT
We need the partial pressure of water, which is 53 percent of the equilibrium vapor pressure at the given temperature and standard pressure. p = (0.53) x (2.69 x 10 3 Pa) = 1.43 x 103 Pa
so n =
(1.43 x 103 Pa) x (250 m3 ) 2 = 1.45 x 10 mol (8.3145 J K I molI) x (23 + 273) K
or m = (l.45 x 102 mol) x (l8.0g molI) = 2.61 x 103 g = 12.61 kg 1 E1.10(b)
(a) The volume occupied by each gas is the same, since each completely fills the container. Thus solving for V we have (assuming a perfect gas) V
nJRT 0.225 g = nNe = =,I
20.18 g mol
PJ
= 1.115x 10 2 mol, V=
(!.l15 x
10 2
PNe= 8.87kPa,
T=300K
mol) x (8.314 kPa K I molI) x 300 K) 3 =3 .137dm 8.87 kPa dm 3
=13 .14dm3 1 (b) The total pressure is determined from the total amount of gas, n = nCH4 nCH4 =
0.320 g 2 I = 1.995 x 10 mol 16.04 g mol
n = (1.995
+ 0.438 + 1.115)
nAr =
+ nAr + nNe.
0.175 g = 4.38 x 1O 3 mol 39.95 g mol I
x 1O 2 mol = 3.548 x 1O 2 mol
p = nRT [1.8] = (3.548 x 10 mol) x (8.314 d_m 3 kPa K I molI) x (300 K) 2
3.137dm 3
V
= 128.2 kPa 1
8
E1.11(b)
STUDENT'S SOLUTIONS MANUAL
This is similar to Exercise l.ll(a) with the exception that the density is first calculated.
RT M = p  [Exercise 1.8(a)] p
33.5mg =0.1340gdm 3 , 250cm
p=
3
M= E1.12(b)
p= 152 Torr,
T=298K
(0.1340gdm 3 ) x (62.36dm3 TorrK 1 molI) x (298K) I II = 16.14gmol 152 Torr
This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature. Thus V = Vo
+ aVoO =
+ bO , b =
Vo
aVo
At absolute zero, V = 0, or 0 = 20.oodm3 + 0.0741 dm 3 °C I x O(abs. zero) 3
O(abs. zero) = 
E1.13(b)
20.00 dm 3 I = t 270 °C 0.0741 dm °C
t
which is close to the accepted value of 273°C. nRT (a) P= V
n = 1.0mol T = (i) 273.15K; (ii) 500K V = (i) 22.414dm 3 ; (ii) 150cm 3
(i)
(ii)
(1.0 mol) x (8.206 x 10 2 dm3 atmK I molI) x (273.15K) P= 22.414dm 3 = t 1.0 atm t (1.0mol) x (8.206 x 10 2 dm 3 atmK I molI) x (5OOK) P= 0.150dm 3 = t 270 atm t (2 significant figures)
(b) From Table (1.6) for H2S
a = 4.484 dm 6 atm mol I
b = 4.34 x 10 2 dm 3 mol I
nRT an 2 P=VnbV2'
(i)
(1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) x (273. 15 K) P= 22.414 dm 3  (1.0 mol) x (4.34 x 102 dm 3 mol I) (4.484 dm 6 atm molI) x (1.0 mol)2 (22.414 dm 3 )2
=
t
0.99 atm
t
THE PROPERTIES OF GASES
9
( 1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) x (500K)
(ii)
p
=
0.150dm 3
(1.0 mol) x (4.34 x 10 2 dm 3 mol I)

(4A84dm 6 atmmor
1 )
x (1.0mol)2
(0.150 dm 3 )2
= 185.6atm ~ 1190 atm 1(2 significant figures). E1.14(b)
The conversions needed are as follows:
Therefore,
E1.1S(b)
a
= 1.32 atm dm 6 mol 2 becomes, after substitution of the conversions
a
= 11.34 x
b
= 0.0436 dm 3 mol  I becomes
b
= 14.36 X
10 1 kg m5s 2 mol 2
~
and
10 5 m 3 mol 1 I
The compression factor is pVm Vm z==RT
V;:'
= V;:' + 0.12 V;:' = (I. 12)V;:', we have Z = [Iill IRepulsive Iforces dominate.
(a) Because Vm
(b) The molar volume is
V
= (1.12)V~ = (1.12)
x
(R;) 3
V=(1.12)x (
E1.16(b)
(a)
RT
o
V  
p
m 

0.08206dm atmK I molI ) x (350K») 127 d 3 II 1 12atm =.' m mo .
(8.314JK I mol I) x (298.15 K)
~~(200 bar) x (l05Pabar l )
I
= 1.24 x 104 m3 mol I = 0.124 dm 3 mol I
I
(b) The van der Waals equation is a cubic equation in Vm. The most direct way of obtaining the molar
volume would be to solve the cubic analytically. However, this approach is cumbersome, so we proceed as in Example 104. The van der Waals equation is rearranged to the cubic form Vm3

(
b+
RT) V 2 + (a)p V P m
m 
abp = 0 or x
3
 (b+
RT) P x + (a) p x  ab p=0 2
10
STUDENT'S SOLUTIONS MANUAL
The coefficients in the equation are evaluated as
b + RT P
=
(3.183 x 1O2 dm3 mol  I)
= (3.183
X
+
3 2 (8.206 x 1O dm mol I) x (298.15 K) (200 bar) x (1.013 atm barI)
10 2 + 0.1208) dm 3 mol I
1.360 dm 6 atm mol 2
a
:: = 6.71 (200 bar) x (1.013 atm barI)
P
= 0.1526 dm 3mol  1
3 3 I 2 x 10 (dm mol  )
(1.360 dm 6 atmmol 2) x (3.183 x 1O 2dm 3 mol I) (200 bar) x (1.013 atmbar I )
ab
= 2.137
~.:......__;_.:.
P
4 3 I 3 x 10 (dm mol )
Thus, the equation to be solved is x3  0.1526x 2 + (6.71 x 1O 3)x  (2.137 x 10 4 ) = O. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112
or
Vrn = 10.112 dm 3 molII
The difference is about 15 percent. E1.17(b)
The molar volume is obtained by solving Z
Vrn
= pVrn / RT [1.17] , for Vrn , which yields
ZRT
(0.86) x (0.08206dm 3 atmK l mol l ) x (300K)
p
20atm
=  =
(a) Then, V
= n Vrn = (8.2
x 10 3 mol) x (1.059 dm 3 mol I)
= 8.7
=
3 I 1.059dm mol
x 10 3 dm 3 = 18.7 cm 3 1
(b) An approximate value of B can be obtained from eqn 1.19 by truncation of the series expansion after the second term, B/Vrn , in the series. Then,
B
= Vrn (p;;
 I)
= Vrn
= (1.059 dm 3 mol I) E1.18(b)
X
(Z  I)
x (0.86  I)
= \ 0.15 dm 3mol 1 \
(a) Mole fractions are nN
XN== ntotal
Similarly, XH
(2.5
2.5 mol ~6 + 1.5) mol =~
= 10.371
(c) According to the perfect gas law Ptotal V
so
Ptotal
=
= ntotal RT ntotal RT V
I _ (4.0 mol) x (0.08206 dm 3 atm mol K I ) x (273.15 K) 140 1  . . atm. 22.4dm 3
THE PROPERTIES OF GASES
11
(b) The partial pressures are PN
=
XNPtot
= (0.63) x (4.0 atm) = 12.5 atm 1
and PH = (0.37) x (4.0 atm) = 11.5 atm 1 E1.19(b)
The critical volume of a van der Waals gas is Vc
= 3b
I
so b = 1 Vc = 1( 148cm 3 mol  I) = 49.3cm 3 mol I = 0.0493 dm 3 mol I
I
By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centers of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); that volume times the Avogadro constant is the molar excluded volume b
4n(2r)3 )
r=
so
b=NA (   3
2
3(49.3cm 3 mol
I
r=
~ (~)1 /3
2 ( 4n(6.022 x
10 23
l
4nNA IP
)
I mol ) )
=1.94 xlO S cm =11.94 xlO IOml
The critical pressure is a
Pc
=
27b2
so a = 27Pcb2
= 27(48.20 atm) x
(0.0493 dm 3 mol I) 2
= 13.16 dm 6 atm mol 2 1
But this problem is overdetermined. We have another piece of information
8a T.   c  27Rb According to the constants we have already determined, Tc should be
However, the reported Tc is 305.4 K, suggesting our computed alb is about 25 percent lower than it should be. E1.20(b)
(a) The Boyle temperature is the temperature at which Iim vm~oo dZ /(d(l I Vm to the van der Waals equation
a )
RT
Z
=
pVm RT
=
(
v:=h  ~ RT
Vm
Vm a    Vm b
VmRT
» vanishes. According
12
STUDENT'S SOLUTIONS MANUAL
so
dZ d(l/Vrn )
=
( dZ) ( dVrn ) dVrn x d(l/Vrn )
= V~ Cd:rn ) = V~ V~b
a
(Vrn  b)2
RT
CV:~rnb)2 + Vrnlb + vtRT)
In the limit of large molar volume, we have dZ
. I 1m
Vm ..... OO
and T
dO/Vrn )
a RT
=b
=0
so
a RT

=b
a
(4.484 dm 6 atm mol 2)
Rb
(O.08206dm 3 atmK I molI) x (O.0434dm 3 molI)
= =
= 11259 K I
(b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter ofthose spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b 47r(2r)3 ) b=NA (   3
so
r
= ~ (~)1 /3 2
47rNA
1/3 1 3(0.0434 dm molI) r =I = 1.286 x 102 (47r(6.022 x 1023 mol ) ) 3
E1.21 (b)
9
dm
= 1.29 x
10 10 m
= I0.129 nm I
States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25°C are
Pr
1.0 atm = Pc P = 33.54 = 0.030 atm
and
T __ (25 + 273) K __ 2.36 Tc 126.3 K
Tr __ _ I,
The corresponding states are (a) For H2S
= (0.030) x (88.3 atm) = 12.6 atm I T = TrTc = (2.36) x (373.2K) = 1881 K I
P = PrPc
(Critical constants of H2S obtained from Handbook of Chemistry and Physics.) (b) ForC02
= (0.030) x (12.85 atm) = 12.2 atm I T = TrTc = (2.36) x (304.2 K) = 1718 K I
P = PrPc
(c) For Ar
= (0.030) x (48.00atm) = 11.4 attn I T = TrTc = (2.36) x (l50.12K) = 1356 K I
P = PrPc
THE PROPERTIES OF GASES
E1.22(b)
13
The van der Waals equation is
which can be solved for b
RT
4
b = Vm  a = 4.00 x 10
3
_,
m mol

P+V2 m
= 11.3 x 104 m3 mol'
(8.3145JK' mol' ) x (288K) 2 ) 4.0xI06 Pa+ 0.76m Pamol(4.00 X 10 4 m3 mol')2
(
6
I
The compression factor is
z=
4
6
3
pVm = (4.0 x 10 Pa) x (4.00 x 10 m mol i) = 10.671
RT
(8.3145 J K' mol') x (288 K)
Solutions to problems Solutions to numerical problems P1.1
Since the Neptunians know about perfect gas behavior. we may assume that they will write p V = nRT at both temperatures. We may also assume that they will establish the size of their absolute unit to be the same as the oN. just as we write 1K = 1°C. Thus
pV(T,) = 28.0dm 3 atm = nRT, = nR x (T, + OON) . 0 pV(T2) = 40.0dm 3 atm = nRT2 = nR x (T, + 100 N). ° 40.0 dm 3 atm T, + 100 N=    nR
28.0dm 3 atm orT, =     nR
T, + 1000N
40.0dm 3 atm
° . 3 = 1.429 or T[ + 100 N = 1.429T,. T, = 233 absolute UOltS. T, 28.0 dm atm As in the relationshi between our Kelvin scale and Celsius scale T = ()  absolute zero(ON) so absolute zero (ON) = 233°N . Dividing.
COMMENT.
=
To facilitate communication with Earth students we have converted the Neptunians' units of
the pV product to units familiar to humans, which are dm3 atm. However, we see from the solution that only the ratio of pV products is required , and that will be the same in any civilization.
Question. If the Neptunians' unit of volume is the lagoon (L). their unit of pressure is the poseidon (P). their unit of amount is the nereid (n). and their unit of absolute temperature is the titan (T). what is the value of the Neptunians' gas constant (R) in units of L. p. n. and T? P1.3
The value of absolute zero can be expressed in terms of a by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature. Hence 0= Vo[l + a(}(abs. zero)].
14
STUDENT'S SOLUTIONS MANUAL
1
Then () (abs. zero) = . ex All gases become perfect in the limit of zero pressure, so the best value of ex and, hence, () (abs. zero) is obtained by extrapolating ex to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, ex = 3.6637 x 1O 3°C I, or ()(abs.zero)=
I 3 I =1272.95°cl, 3.6637 x 10 °C· .
which is close to the accepted value of 273.15°C. .... _ · ..... ........ . . .... . .. .. . .. ..... ..:... :. ...: .. .
3.672··
~
.. . .;. ...... ; ...i···;'··; . . . . . . . . . . . ', .. !' ... ..... .. : .. . .. : ...:.. ,! .. .. ,! . . .~
~
~
~
~
~
~
~
..; .. , . . , "!
i··~··
. · .. . : .. . ': .. : .. .:.. .. :. ... ..· ...., ................. . . . . ,. .. ~ .. ;.. .~ .. :.. ~... ; .. ~ . .~
~
~
:··" 3.670'
. ,',. ,! ..
. .;.. . ! .. :... ~ .. . . . ~ . . ! ...... ! .. : .. ! .. ~ .. ~ •. ~ ... ~ .! ..:
....,.
'.,
. ...~ .. :.. ';' .. :.. ':' .. :.. ~ .. .~ ... ... _."
.;.: .. ~ .. ;.. ;... ; . .
;f'3. ~$ ·
'U
i
···;···i···;.. ·j· ··;···i.·,:..
~
......
..
. :. ':' . ;. ':' . ;. ;... ~ .. ... ;..... .. :...: ..:.. :... :.. ...; ••
~ G:~::rT: .~
! ~.:
"
~
~o
3.666·
: 
.: .. : ... ~ ..
~
i.·{·····I···:···I··
~
.~ .. ~ .. ~ .. :.. ~... "
.: .. ;... ~ .. ~ .. :
· ··~ ··:··7·· :
'''3.664 ' . . . . . ' .: .. .. . ':. .. .••. .
.  ..•. • ~ •. .• ••• . ~
"!"': ":"~'
~
~
~
Figure 1.2
P1.S
p nR. p P3 .  = constant, If n and V are constant. Hence,  =  , where P IS the measured pressure at V T T3 T temperature, T, and P3 and T3 are the triple point pressure and temperature, respectively. Rearranging,
P=(~~)T. ~
~~
I
. . .
The ratio  is a constant = = 0.0245 kPa K . Thus the change III P, t.p, IS proportIOnal to T3 273.16K the change in temperature, t.T : t.p = (0.0245 kPaKI) x (t.T) .
I
(a) t.p = (0.0245 kPaKI) x (LOOK) = 0.0245 kPa
I·
(b) Rearranging,p= (!.)P3 = (373.16K) x (6.69kPa) =19.14kPal. T3 273.16K (c) Since
f
is a constant at constant n and V, it always has the value 0.0245 kPa K I ; hence T 1 t.p = P374.15K  P373.15K = (0.0245 kPa K ) x (1.00 K) = 0.0245 kPa
I
_ RT _ (8.206 x 10 2 dm atm K I molI) x (350 K) 1125 d 3 II I (a) Vrn  . . m mo . P 2.30atm RT a RT (b) Fromp =    2 [1.2Ib], we obtain Vrn = ( ) + b [rearrangeI.21b]. Vrn b Vrn + a p V2 3
P1.7
I·
rn
THE PROPERTIES OF GASES
15
Then, with a and b from Table 1.6, (8.206 x 1O 2 dm 3 atmK I mol 
Vm "'"
(2.30 atm)
"'"
+ (6.260dm
3
28 .72dm mol2.34
I
l
)
x (350K) 2
6
+ ( 5.42
2
3
+ (5.42 x
2
3
atm mol  )/ (12.5 dm mol I) )
I
I
x 10 dm mol I "'" 12.3dm 3 molI . 2
3
)
Substitution of 12.3 dm 3 mol I into the denominator of the first expression again results in Vrn = 12.3 dm 3 mol I, so the cycle of approximation may be terminated. P1.9
As indicated by eqns 1.18 and 1.19 the compression factor of a gas may be expressed as either a virial
:m ).The virial form of the van der Waals equation is derived in Exercise 1.20(a)
expansion in p or in ( and is p =
~: { 1 + (b  RaT)
x ( :
m) + ...}
Rearranging Z = pV = I + (b  ~) x (_1_) ' RT RT Vm m
+ ...
On the assumption that the perfect gas expression for Vm is adequate for the second term in this expansion, we can readily obtain Z as a function of p .
(a)
Tc = 126.3 K .
V (R;) m =
x Z =
R; + (b  :T) + ...
(0.08206dm 3 atm K I mol  I) x (126.3 K) 1O.Oatm
+ {(0.0387 dm3 mol  I) _
6
(
_ Z 
(l!...) RT
x
(V) _ m 
2
) 1.352 dm atm mol(0.08206 dm 3 atm K I mol I) x ( 126.3 K)
= (1.036 0.092) dm 3 mol  1 =10.944dm 3 mol  l
}
l.
3
(1O.0atm) x (0.944dm molI) =0.911. (0.08206dm 3 atm KI molI) x (126.3 K)
(b) The Boyle temperature corresponds to the temperature at which the second virial coefficient is zero, hence correct to the first power in p , Z = 1, and the gas is close to perfect. However, if we assume that N2 is a van der Waals gas, when the second virial coefficient is zero,
=0' (b~) RTB TB =
(0.0387dm 3
or
a TB =  . bR
1.352 dm 6 atm mol 2 = 426K. mol I) x (0.08206dm 3 atrnKI molI )
I
10 dm mol  )
16
STUDENT'S SOLUTIONS MANUAL
The experimental value (Table 1.5) is 327.2 K. The discrepancy may be explained by two considerations.
1. Terms beyond the first power in p should not be dropped in the expansion for Z. 2. Nitrogen is only approximately a van der Waals gas. WhenZ
=
I , Vrn
RT
= , p
and using TB
= 327.2K
(0.08206 dm 3 atm K I molI) x 327.2 K 10.Oatm
= 12.69dm3 molII and this is the ideal value of Vm . Using the experimental value of TB and inserting this value into the expansion for Vrn above, we have 0.08206 dm 3 atm K1mol 1 x 327.2 K
Vrn
= 10.Oatm
+ { 0.0387 dm 3 mol I = and Z (c) TI
6 2 1.352 dm atm mol) } 1 0.08206dm 3 atmKl mol x 327.2K
(2.685  0.012) dm 3 mol 1 = 12.67 dm 3 molII
Vrn
=
(
V;;'
=
2.67 dm 3 molI 2.69dm 3 molI
= 0.992 ~
I.
= 621 K [Table 2.9]. Vrn
0.08206dm 3 atm KImol I x 621 K
= ~~1O.Oatm + { 0.0387 dm 3 mol  I =
and Z
=
2 6 ) } 1.352 dm atm mol3 0.08206 dm atm K I molI x 621 K
(
(5.096 + 0.012) dm 3 molI
5.11 dm 3 mol5.lOdm3
= 15.11 dm 3 molII
1
molI
=
1.002
~
I.
Based on the values of TB and TJ given in Tables 1.4 and 2.9 and assuming that N2 is a van der Waals gas, the calculated value of Z is closest to I at but the difference from the value at TB is less than the accuracy of the method.
I!i],
P1.11
(a)Vm
18.02g molI 10 353 d 3 II 1 _ molar mass _ M _ mmo . .  ..1 p 1.332 x \0 2 gdm 3 densIty

_pVm _ (b) Z [1.l7b] 
RT
l 3 (327.6atm) x (0.1353dm mol ) 1069571 I . , 3 (0.08206 dm atm KI mol ) x (776.4 K)
THE PROPERTIES OF GASES
17
(e) Two expansions for Z based on the van der Waals equation are given in Problem 1.9. They are
= 1 + (0.0305dm mol I)3
{
(
6
2
5.464dm atmmol)} (0.08206dm3 atm KI molI) x (776.4 K)
J x 0.J353dm 3 mol 1 = 10.4084=0. 5916~0.59.
Z
= 1+ (RIT)
x (b  :T) x (P) + ...
I = I + ,;;3
(0.08206dm atmK1 molI) x (776.4K)
x
{
(0.0305 dm 3 molI) 
= 1 0.2842 ~
(
6
2
5.464 dm atm mol)} x 327.6 atm (0.08206dm 3 atm KI molI) x (776.4 K)
lo.nl. I
In this case the expansion in p gives a value close to the experimental value; the expansion in Vrn is not as good. However, when terms beyond the second are included the results from the two expansions for Z converge. P1.13
Vc
= 2b,
a Tc = 4bR [Table 1.7]
Hence, with Vc and Tc from Table 1.5, b =
~ Vc
=
~
x (118.8 cm 3 molI) = 159.4 cm 3 molI
a = 4bRTc = 2RTc Vc
= (2) x (8.206 x 10 2 dm 3 atmK 1 molI) x (289.75K) x (118.8 x 10 3 dm 3 molI) = 15 .649 dm 6 atm mol 2 1. Hence
(1.0 mol) x (8.206 x 10 2 dm 3 atm K I molI) x (298 K) (1.0dm 3 ) x exp

(1.0 mol) x (59.4 x 10 3 dm 3 molI)
) (l.Omol) x (5.649dm6 atmmol 2 ) ( (8.206 x 10 2 dm 3 atmK1 molI) x (298K) x (1.0dm 6 atmmol l )
= 26.0atm x e O.23T = 121 atm
I.
I.
18
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems PU5
This expansion has already been given in the solutions to Exercise 1.20(a) and Problem 1.14; the result is
· expansion .. Compare thIS wlthp
I
and hence find B = b Since C
=
B)
Vm
iT I
and
1200 cm6 mol 2 ,
= RT(b 
a
P1.17
=
= RT
b
= (8.206 x
( 1+  B +  C + ... ) [1.19] Vm Vm2
Ie = b
=
2
C I/ 2
1.
= 134.6 cm 3 mol  II
10 2 ) x (273 dm 3 atm molI) x (34.6 + 21.7) cm 3 molI
(22.40 dm 3 atm mol I) x (56.3 x 10 3 dm 3 mol  I)
= 11.26 dm 6 atm mol 2 1.
The critical point corresponds to a point of zero slope that is simultaneously a point of inflection in a plot of pressure versus molar volume. A critical point exists if there are values of p , V, and T that result in a point that satisfies these conditions.
I"
th, ,ri,kal poiOl.
. Th at IS,
RTc V; + 2BVc  3C 2 RTcVc  3BVc + 6C
which solve to Vc
~
=0 =0
}
=  C~2 , Tc =   . B
3RC
Now use the equation of state to find Pc
Pc
RTc
= V; 
B C V2 + vg
=
(RB2) (B) ( B 3RC x 3C  B 3C 3
It follows that Zc P1.19
=
PcVc RTc
=
(B 27C2 ) x (3C) Ii x
(I) R
)2 + C (3CB)3 = iB3l ~.
x (3RC) fi2
For a real gas we may use the virial expansion in terms of p [1 .18] P
nRT =( I + B'p + V
RT I ... ) = p(I + B P + . .. ) M
= III llJ·
THE PROPERTIES OF GASES
RT P which rearranges to  = p M
RTB'
+   p + .... M
Therefore, the limiting slope of a plot of B'RT M
19
B'RT . 3 th limi· · I . p against p IS. ~. From Fig. I. e tmg s ope IS
p
4 2 (5.84  5.44) x 10 m s2 _ 44 . (10.132  1.223) x IQ4Pa
~'...,...,..:,;:::
X
102 k  I 3 g m .
RT From Fig. 1.3, = 5.40 x 104 m 2 s2 ; hence M l 3 2 , _4.4xlO kg m 081 106p 1 B22 . x a, 5.40 x IQ4 m s
B' = (0.81 x 1O6 Pa
l)
x (1.0133 x 105 Paatm l ) =10.082atm
l
l.
B = RTB' [Problem 1.18]
= (8.206 x 10 2 dm 3 atm K I molI) x (298 K) x (0.082 atm I ) = 12.0dm3 mol  I

y
I.
= 5.3963 + 0.046074x R = 0.99549
5.9
~
5.8
I",
E
5.7
b
:§: 5.6 ,::, 5.5 5.4 0
2
4
6 p/( 104 Pa)
P1.21
8
10
12
Figure 1.3
The critical temperature is that temperature above which the gas cannot be liquefied by the application of pressure alone. Below the critical temperature two phases, liquid and gas, may coexist at equilibrium, and in the twophase region there is more than one molar volume corresponding to the same conditions of temperature and pressure. Therefore, any equation of state that can even approximately describe this situation must allow for more than one real root for the molar volume at some values of T and p, but as the temperature is increased above Te , allows only one real root. Thus, appropriate equations of state must be equations of odd degree in Vm. The equation of state for gas A may be rewritten V~  (RT / p) Vm  (RTb / p) = 0, which is a quadratic and never has just one real root. Thus, this equation can never model critical behavior. It could possibly model in a very crude manner a twophase situation, since there are some conditions under which a quadratic has two real positive roots, but not the process of liquefaction.
20
STUDENT'S SOLUTIONS MANUAL
The equation of state of gas B is a firstdegree equation in Vrn and therefore can never model critical behavior, the process of liquefaction, or the existence of a twophase region. A cubic equation is the equation of lowest degree that can show a crossover from more than one real root to just one real root as the temperature increases . The van der Waals equation is a cubic equation in V rn . P1.23
The two masses represent the same volume of gas under identical conditions, and therefore, the same number of molecules (Avogadro's principle) and moles, n. Thus, the masses can be expressed as nMN
= 2.2990 g
for 'chemical nitrogen' and
for 'atmospheric nitrogen ' . Dividing the latter expression by the former yields so
X Ar
(MAr _ MN
and XA
(2.3102/ 2.2990)  1 
(MAr/MN)  1
r 
I) =
2.3102  1 2.2990
(2.3102/ 2.2990)  I
=
(39.95gmol 1)/ (28.013gmol
I
~
 I)
= L.2:2.!..!J .
This value for the mole fraction of argon in air is close to the modem value.
COMMENT.
Solutions to applications P1.25
I t = 103 kg. Assume 300 t per day. n(S02)
V
=
300
103 kg
64 x 10 3 kg mol
= nRT = p
P1.27
X
I
= 4.7
x 106 mol.
3 1 6 (4.7 x 10 mol) x (0.082 dm atm K mol  I) x 1073 K 1.0atrn
= 14.1 x
108 dm 3
1.
The pressure at the base of a column of height H is p = pgH (Example 1.1). But the pressure at any altitude h within the atmospheric column of height H depends only on the air above it; therefore p = pg(H  h) and dp = pg dh .
Since p
pM
= ~
[Problem 1.2], dp
This relation integrates to p
pMgdh .
.
dp
Mgdh
= , LmplYIng that p = RT ~
= poe Mgh/RT
For air M ~ 29 g mol 1 and at 298 K
THE PROPERTIES OF GASES
(a)
(b)
P1.29
h
= 15 cm.
P
= Po
x
e ( O.15m )x( 1.l5 x lO 
4
m
1 )
21
= 0.99998 Po ;
h = II km = 1.1 X 104 m . P = Po x e ( l.l x 1O 4)x( 1.15 x IO4 m 
l)
= 0.28 Po;
Refer to Fig. 1.4. F,op
1
~1
T h
Air . l (envlronment)
F bottom
Ground
77777777777
Figure 1.4
The buoyant force on the cylinder is
Fbuoy
= Fbottom 
FlOP
= A(Pbottom
 Ptop)
according to the barometric formula. Ptop = Pbotlom e
 Mgh / RT
where M is the molar mass of the environment (air). Since h is small, the exponential can be expanded in a Taylor series around h
Ptop
= Pbottom ( I

= 0 (e x =
1 x
+ ~x2 + .. ). Keeping the firstorder term only yields
Mgh) RT .
The buoyant force becomes
Fbuoy
(pbotto mM) g RT = Ah RT ( I  I + Mgh) M Pbottom VM) RT g =n g (
= APbottom =
n is the number of moles of the environment (air) displaced by the balloon, and nM = m, the mass of the displaced environment. Thus Fbuoy = mg . The net force is the difference between the buoyant force and the weight of the balloon. Thus
F net
= mg 
mballoon
g
= (m 
This is Archimedes ' principle.
mballoon)g
The First Law
Answers to discussion questions 02.1
Work is a precisely defined mechanical concept. It is produced from the application of a force through a distance. The technical definition is based on the realization that both force and displacement are vector quantities and it is the component of the force acting in the direction of the displacement that is used in the calculation of the amount of work, that is, work is the scalar product of the two vectors. In vector notation w =  f . d =  fd cos e, where e is the angle between the force and the displacement. The negative sign is inserted to conform to the standard thermodynamic convention. Heat is associated with a nonadiabatic process and is defined as the difference between the adiabatic work and the nonadiabatic work associated with the same change in state of the system. This is the formal (and best) definition of heat and is based on the definition of work. A less precise definition of heat is the statement that heat is the form of energy that is transferred between bodies in thermal contact with each other by virtue of a difference in temperature. At the molecular level, work is a transfer of energy that results in orderly motion of the atoms and molecules in a system; heat is a transfer of energy that results in disorderly motion. See Molecular interpretation 2.1 for a more detailed discussion.
02.3
The difference results from the definition H = U + PV; hence I'1.H = I'1.U + 1'1. (PV). As I'1.(PV) is not usually zero, except for isothermal processes in a perfect gas, the difference between I'1.H and I'1.U is a nonzero quantity. As shown in Sections 2.4 and 2.5 of the text, I'1.H can be interpreted as the heat associated with a process at constant pressure, and I'1.U as the heat at constant volume.
02.5
In the louIe experiment, the change in internal energy of a gas at low pressures (a perfect gas) is zero. Hence in the calculation of energy changes for processes in a perfect gas one can ignore any effect due to a change in volume. This greatly simplifies the calculations involved because one can drop the first term of eqn 2.40 and need work only with dU = Cv dT. In a more sensitive apparatus, louie would have observed a small temperature change upon expansion of the ' real' gas. louie's result holds exactly only in the limit of zero pressure where all gases can be considered perfect. The solution to Problem 2.33 shows that the louieThomson coefficient can be expressed in terms of the parameters representing the attractive and repulsive interactions in a real gas. If the attractive forces predominate, then expanding the gas will reduce its energy and hence its temperature. This reduction in temperature could continue until the temperature of the gas falls below its condensation point. This is the principle underlying the liquefaction of gases with the Linde refrigerator, which utilizes the louIeThomson effect. See Section 2.12 for a more complete discussion.
THE FIRST LAW
02.7
23
The vertical axis of a thermogram represents Cp , and the baselines represent the heat capacity associated with simple heating in the absence of structural transformations or similar transitions. In the example shown in Fig. 2.16, the sample undergoes a structural change between TI and T2, so there is no reason to expect Cp after the transition to return to its value before the transition. Just as diamond and graphite have different heat capacities because of their different structures, the structural changes that occur during the measurement of a thermogram can also give rise to a change in heat capacity.
Solutions to exercises E2.1(b)
The physical definition of work is dw
= F dz [2.4]
In a gravitational field the force is the weight of the object, which is F
= mg
If g is constant over the distance the mass moves, dw may be intergrated to give the total work
W
=
Zf 1
Fdz
=
1Zf
Zi
w
E2.2(b)
mg dz
= mg(Zf 
Zi)
= mgh
where
h
= (Zf 
Zi)
Zi
= (0.120 kg)
x (9.81 m S
2)
x (50m)
= 59 J = 159 J needed 1
This is an expansion against a constant external pressure; hence w = Pex l:!. V
[2 .8]
The change in volume is the crosssectional area times the linear displacement:
l:!.V
so w E2.3(b)
=
(50.0cm
= (121
2)
x (15cm) x
(~)3 = 7.5 x 1O4m3 , 100 cm
x 103 Pa) x (7 .5 x 10 4 m 3 )
= 191 J 1as 1 Pa m3 = 1 J.
For all cases l:!. U = 0, since the internal energy of a perfect gas depends only on temperature. (See Molecular interpretation 2.2 and Section 2. 11 (b) for a more complete discussion.) From the definition of enthalpy, H = U + pV, so l:!.H = l:!.U + l:!.(pV) = l:!.U + l:!. (nRT) (perfect gas). Hence, l:!.H = 0 as well, at constant temperature for all processes in a perfect gas. (a)
1
l:!.U
w
=
l:!.H
=01
= nRT In (~) [2.11] = (2.00 mol) x (8.3145 J K 1 molI) x (22 + 273) K x In 31.7 dm: = 11.62 x 10 3 J 1 22.8dm
q (b)
= w = 11.62 x
1 l:!.U
w
=
=
l:!.H
103 J 1
=0 1
Pexl:!. V [2.8]
where P ex in this case can be computed from the perfect gas law pV
= nRT
24
STUDENT'S SOLUTIONS MANUAL
(2.00mol) x (8.3145J K Imol I) x (22 + 273) K 3 31.7 dm
sop=
 (1.55
X
and w = q
I 3 (lOdmm  ) = 1.55
X
105 Pa
105 Pa) X (31.7  22.8) dm 3 _ / _ 3 / (lOdmm  I )3 _ . 1.38 x 10 J.
= w = [ 1.38
/ ~U
(C)
X
X
10 3 J [
= ~H = 0 1
/ w = 0 1[free expansion] q =
~U 
w = 0  0 =@]
COMMENT. An isothermal free expansion of a perfect gas is also adiabatic.
E2.4(b)
The perfect gas law leads to PIV = nRTI P2 V nRT2
or
P2 = PI T 2 = (111kPa) X (356K) =1 143kPai TI 277 K
There is no change in volume, so 1w = 0 I. The heat flow is q=
f
Cy dT
~ Cy~T =
(2.5)
X
(8.3145J K I molI)
X
(2.00mol)
X
(356  277) K
= [ 3.28 x 103 J [
~U = q + w =
[ 3.28
X
10 3 J [
(a)
_  (7.7 x 103 Pa) x (2.5 dm 3 ) wpex~V(V) (lOdmm  I)3
(b)
w=nRTln
_
E2.S(b)
~
6.56 g
_
~
~
[2.11]
w= ( I 39.95 g mol
)
( ) (2.5 + 18.5) dm x 8.3145JK l mol  1 x (305K)x ln 3 18.5dm
3
= /52.8 J /
E2.6(b)
~H = ~condH =  ~vapH =  (2.00 mol) x (35.3 kJ mol  I) = /70.6 kJ / Since the condensation is done isothermally and reversibly, the external pressure is constant at 1.00 atm. Hence, q = qp =
~H = /70.6 kJ /
w = Pex ~ V [2.8]
where
~V
= Vliq 
Vvap ~  Vvap
because
Vliq «
On the assumption that methanol vapor is a perfect gas, Vvap = nRT / p and P condensation is done reversibly. Hence, w
~ nRT =
Vvap
=
(2.00 mol) x (8.3145 J K I molI) x (64 + 273) K = [ 5.60 x 103 J [
and ~U = q + w = (70.6 + 5.60) kJ = / 65.0kJ /
Pex,
since the
THE FIRST LAW
E2.7(b)
25
The reaction is
so it liberates 1 mol of H2(g) for every 1 mol Zn used. Work at constant pressure is w =
Pex~V
=( E2.8(b)
= pVgas = nRT
5.0g _I) 65.4 g mol
X
~H .
(a) At constant pressure, q = q=
f
CpdT=
l
(8.3145JK l mol l ) x (23+273) K=I188JI
l°O+273 K
[20.17+ (0.400I)T/K]dTJK 1
0+273 K
I = [ (20.17) T + (0.4001) x (T2)JI373K J K I 2 K 273 K
= [(20.17) x (373  273) + w =
~(0.4001) x (373 2 
273 2)] J = 114.9 x 103 J 1=
~H
p~V = nR~T =  (1.00 mol) x (8.3145JK 1 molI) x (lOOK) = 1831 J I
kJl
~U=q+w=(14.90. 831)kJ=114.1
(b) The energy and enthalpy of a perfect gas depend on temperature alone. Thus, ~H = 114.9
~U = 114.1 kJ Ias above. At constant volume, w =@] and ~U = q, so q = 1+14.1 kJ I. E2.9(b)
For reversible adiabatic expansion
V)I /C
Tr
= Tj ( V~
[2.28a]
where c
CVm
Cp,m R
=  ' = ',=R
R
so the final temperature is Tr = (298.15 K) x
E2.10(b)
(37 .11  8.3145) J KImol I          :     ;  1 
8.3145J KImol
3 3)
500 x 10 dm
(
2.00dm
= 3.463,
1/ 3.463
= 1200 K I
3
Reversible adiabatic work is w =
Cv~T
[2.27] = n(Cp,m  R) x (Tr  Tj)
where the temperatures are related by [solution to Exercise 2.15(b)] Tr = Tj
(~~) lie [2.28a]
where
c =
C~m
=
Cp,~
R = 2.503
kJ Iand
26
STUDENT'S SOLUTIONS MANUAL
and w = ( E2.11 (b)
3
(
= 156 K
)
2.00dm
3.12 g I) x (29.125  8.3145) J K I mol I x (156  296) K = 1325 J 1 28.0 gmol
For reversible adiabatic expansion
Pr = Pi
so
E2.12(b)
1/ 2.503
400 x 1O 3 dm 3
So Tr = [(23.0 + 273.15) K] x
qp
= nCp,m!:!,.T
C
p,m
(Vi) 
Y
Vr
= (8.73 Torr) x
500 x 10 3 dm 3)1.3 (
3.0dm
3
= 18.5 Torr I
[2 .24]
=~= n!:!"T
178J =153JK I mol I 1 1.9 mol x 1.78K
CV ,m = Cp,m  R = (53  8.3) J K I mol  I = 145 JK I mol II E2.13(b)
!:!"H = qp = Cp!:!"T [2.23b, 2.24] = nCp,m!:!,.T !:!"H = qp = (2.0 mol) x (37.1 I J K I molI) x (277  250) K = 12.0 x 103 J molI I !:!"H = !:!,.U
+ !:!,.(pV)
= !:!,.u
+ nR!:!"T
so
!:!,.U = !:!"H  nR!:!"T
!:!,.U = 2.0 x 10 J mol  I  (2.0 mol) x (8.3145 J K I mol  I) x (277  250) K 3
= 11.6 x 10 3 J mol  I E2.14(b)
In an adiabatic process, q =
w
= Pex !:!" V =
I
@]. Work against a constant external pressure is
(78.5x1Q 3 Pa)x(4xI515)dm 3 _1_ 3 1 I 3  . 3.5 x 10 J. (IOdm m )
!:!,.U = q + w = 13.5 x 10 3 J I
One can also relate adiabatic work to !:!"T (eqn 2.27): w = Cv!:!"T = n(Cp,m  R)!:!"T
!:!,.T= !:!"H
=
!:!"T =
w n(Cp,m  R)
,
3.5 x 103 J
=124KI. (5.0mol) x (37.11  8.3145)JKI mol  I !:!,.U
+ !:!"(PV) =
= 3.5 x 103 J E2.1S(b)
so
!:!,.U
+ nR!:!"T,
+ (5.0 mol)
x (8.3145JK I molI) x (24K) =14.5 x 103 J I
In an adiabatic process, the initial and final pressures are related by (eqn 2.29)
!
pr V = Pi Vr
where
Cp,m Cp,m  CV,m  Cp,m R
y
I
I
6 20.8JK mol;;I = I. 7 I (20.8  8.31) JK mol
THE FIRST LAW
27
Find Vi from the perfect gas law: l I . _ nRTi _ (1.5mol)(8.3IJK mol )(315K) =00171 3 V,. m 230 x 103 Pa
Pi
so
Vr =
Vi
(Pi) I/ y = pr
(Om 71m3) (230 kPa) 1/ 1.67 = I 0.0205 m31. 170kPa
Find the final temperature from the perfect gas law:
Adiabatic work is (eqn 2.27) w = Cv/'o,.T = (20.8  8.31) JK I molI x 1.5 mol x (279  315) K = 16.7 x 102 J E2.16(b)
I
At constant pressure q = /'o,.H = n/'o,.vapW = (0.75 mol) x (32.0 kJ molI) = 124.0 kJ
I
and w = p/'o,.V ~ pVvapor = nRT = (0.75 mol) x (8 .3145JK I molI) x (260K) w = 1.6 x 103J = 11.6kJ /'0,. U
I
= w + q = 24.0  1.6 kJ = 122.4 kJ
I
COMMENT. Because the vapor is here treated as a perfect gas, the specific value of the external pressure
provided in the statement of the exercise does not affect the numerical value of the answer.
E2.17(b)
The reaction is C6HsOH(l)
+ 702(g)
~
/'o,.cW = 6/'o,.rW(C02) = [6(393.15) E2.18(b)
6C02(g)
+
3H20(l)
+ 3/'o,.rW(H20) 
+ 3( 285 .83) 
/'o,.rW(C6HsOH) 7/'o,.rW (02)
( 165.0)  7(0)] kJ mol I = 13053.6 kJ molI
We need /'o,.rW for the reaction (4)
2B(s)
+ 3H2(g) ~
reaction(4) = reaction(2) Thus,
B2H6(g)
+3 x
/'o,.rW = /'o,.rW{reaction(2)}
reaction(3)  reaction(l)
+3 x
/'o,.rW{reaction(3)}  /'o,.rW{reaction(l)}
= [2368 + 3 x (241.8)  (1941)] kJ molI = 11152 kJ molI
I
I
28
E2.19(b)
STUDENT'S SOLUTIONS MANUAL
For anthracene the reaction is
"""e cr = """ell"  """ngRT
[2.21],
"""ecr = 7061kJmol

=
7055kJmol 
1
"""ng
=  ~ mol
(~ x 8.3 x 1O 3 kJK l mol 1 x 298K)
1
3
Iql = Iqvl =
2.25 x 10 g ) In"""ecrl = ( 172.23gmol _I
c=M =
0.0922 kJ
"""T
1.35 K
( ) x 7055kJmol 1
= 0.0922kJ
= 0.0683 kJ K I = 168.3 J K I I
When phenol is used the reaction is
"""ell"
"""eU
=
3054kJmol 1 [Table2.5]
= """eH  """ngRT,
= (3054kJmol =
"""ng l
)
= ~
+ (~) x (8.314
X
1O 3 kJK 1 molI) x (298K)
3050 kJ molI 135 x 10 ~) x (3050kJmOI I) 94.12gmol3
Iql
=(
"""T
=C =
Iql
4.375kJ 0.0683 kJ KI
= 4.375kJ
= 1+ 64 .1 K I
COMMENT. In this case f'>. ci f and f'>. cH" differed by about 0.1 percent. Thus, to within 3 significant figures,
it would not have mattered if we had used f'>.cH" instead of f'>. cif, but for very precise work it would.
E2.20(b)
The reaction is AgBr(s) + Ag+(aq) + Br(aq) """solJr
=
"""rJr(Ag+,aq) + """rJr(Br,aq)  """rJr(AgBr, s)
= [105.58 + (121.55)  (100.37)] kJ molI = 1+84.40 kJ mol l
E2.21 (b)
I
The combustion products of graphite and diamond are the same, so the transition C(gr) + C(d) is equivalent to the combustion of graphite plus the reverse of the combustion of diamond, and
"""= 11" =
[393.51  (395.41)] kJ molI = 1+ 1.90 kJ molI
I
THE FIRST LAW
E2.22(b)
(a)
reaction(3)
= (2)
x reaction(l) + reaction(2)
and
L'l.ng
29
= 1
The enthalpies of reactions are combined in the same manner as the equations (Hess's law). L'l.r~(3)
= (2) x L'l.r~(l) + L'l.r~(2) = [(2) x (52.96) + (483 .64)] kJ moll = 1589.56kJmOI 1 1
L'l.rlr
= L'l.r~  L'l.ngRT = 589.56kJmol 1 =
(3) x (8.314JK 1mol 1) x (298K)
589.56kJ mol 1 + 7.43 kJmol 1 = 1582.13 kJ molII
(b) L'l.fH>7 refers to the formation of one mole of the compound, so
L'l.f~(HI) = L'l.f~(H20) E2.23(b)
! (52.96 kJ molI)
= 126.48 kJ molII
= ! (483.64kJmOI 1) = 1241.82kJmol 1 1
L'l.r~ = L'l.rlr + RT L'l.ng [2.21]
=
772.7kJmol 1 +(5) x (8 .3145 x 1O3kJKlmoll) x (298K)
= 1760.3 kJ molII E2.24(b)
Combine the reactions in such a way that the combination is the desired formation reaction. The enthalpies of the reactions are then combined in the same way as the equations to yield the enthalpy of formation.
!N2(g) + !02(g) ~ NO(g) NO(g) + !CI2(g) ~ NOCl(g)
+ 90.25 !(75.5) +52.5
Hence, L'l. f H B(NOCI , g) E2.25(b)
= 1+52.5 kJ moll
1
According to Kirchhoff's law [2.36]
L'l.r~(lOO°C) = L'l.r~(25 °C) +
100°C
{
L'l.rC;dT
125°C
where L'l.r as usual signifies a sum over product and reactant species weighted by stoichiometric coefficients. Because Cp,m can frequently be parametrized as Cp,m
= a + bT + c/T2
30
STUDENT'S SOLUTIONS MANUAL
the indefinite integral of Cp,m has the form
Combining this expression with our original integral, we have
Now for the pieces /).rH"(25 °C) = 2( 285.83 kJ mol  I)  2(0)  0 = 571.66 kJ mol 
1
/).ra
= [2(75 .29)  2(27.28)  (29.96)]JK 1 mol  1 = 0.06606kJK  1 mol  1
/).rb
= [2(0)  2(3.29)  (4.18)] x 10 3 J K 2 mol  1 = 10.76 x 10 6 kJ K 2 mol 1
/).rC
= [2(0)  2(0.50)  (1.67)] x 105 J K mol 1 = 67 kJ K mol 1
/).rH"(1oo 0c) = [571.66 + (373  298) x (0.06606) + x (10.76 x 10 6 )

~ (373 2 
298 2)
(67) x (_1_  _1_)] kJ mol 1 373 298
= 1566.93kJmol 1 1 E2.26(b)
The hydrogenation reaction is
The reactions and accompanying data which are to be combined in order to yield reaction (1) and /).rW(T) are (2)
H2(g) + !02(g)
(3)
C2!it(g) + 302 (g)
+
2H20(l) + 2C02(g)
(4)
C2H2(g) + ~02(g)
+
H20(l) +2C02(g)
+
H20(l)
/).cH"(2) = 285.83 kJ mol 
1
/).cH" (3) = 1411 kJ mol 1 /).cH"(4) = 13ookJmol
1
reaction (1) = reaction (2)  reaction (3) + reaction (4) (a) Hence, at 298 K: /).rH" = /).cH"(2)  /).cH"(3) + /).cH"(4) = [(285.83)  (1411) + (1300)] kJ mol 1 = 1175 kJ molII /).r if' = /).rH"  /).ngRT
[2.21];
/).n g = 1
= 175kJmol 1  (1) x (2.48kJmol 1) =1173kJmOI 1 1
THE FIRST LAW
(b) At 348 K:
to rF'(348 K) = to rF'(298 K) torCp
=
+ to r C;(348 K 
L vJC;m(J) [2.37] = C; m(C2IL!,g) 
298 K)
[Example 2.6]
Cp~m(C2H2,g)  Cp~m (H2, g)
J
= (43 .56 to rF'(348K)
43.93  28.82) x 10 3 kJ K I mol I
= (175kJmol l ) 
=
29.19 x 10 3 kJ K I mol I
(29.19 x 1O 3 kJK I molI) x (50K)
= 1176kJmol 1 1 E2.27(b)
NaCI, AgN03 , and NaN03 are strong electrolytes; therefore the net ionic equation is Ag+(aq)
+ Cl(aq) +
= tofF'(AgCI) 
torF'
= [(127.D7) E2.28(b)
AgCI(s) tofF'(Ag+)  tofF'(CI)
(105.58)  (167.16)]kJmol 1
= 165.49kJmol 1 1
The cycle is shown in Figure 2. 1.
~,
Ionization
Ca(g) + 2Br(g) ~
Dissociation
Ca(g) + Br2(g)
Electron gain Br
~
Vaporization Br
Ca(g) + Br2(1)
Ca 2+(g) + 2Br (g) t
~
Sublimation Ca
Ca(s) + Br2(1) ~~
Formation
Hydration BrCa 2+(g) + 2Br (aq)
,~
CaBr2(s) Hydration ci+
~I'
Solution
,~
Figure 2.1
to hyd F'(Ca 2+)
= tosolnF'(CaBr2) 
=
+ tosubF'(Ca)
+ tovapF' (Br2) + todissF' (Br2) + toionF' (Ca) + toionF'(Ca+) + 2to eg F'(Br) + 2to hyd F'(Br) [(  103. 1)  (682.8) + 178.2 + 30.91 + 192.9 + 589.7 + 1145 + 2(33\.0) + 2(337)] kJ molI
= 11587 kJ molII so tohyd W (Ca2+)
tofF'(CaBr2, s)
= 11587kJmol 1 1
31
32
E2.29(b)
STUDENT'S SOLUTIONS MANUAL
The JouleThomson coefficient Jt is the ratio of temperature change to pressure change under conditions of isenthalpic expansion. So Jt  (aT) "'" 6.T = _ __I_O_K_ _ = I 0.48 Katm I I ap H 6.p (1.00  22) atm
E2.30(b)
The internal energy is a function of temperature and volume, Urn = Um(T, Vm), so
dUm
aUm) =(aT
Vm
dT+ (aUm) dVm aVm T
For an isothermal expansion dT = 0; hence
a 22.1 dm 3 molI
a 1.00dm 3 molI
:;;+;;;
2l.la 3 = 0.95475adm mol 22.1 dm mol
;;3;1
From Table 1.6, a = 1.337 dm 6 atm molI 6.Um = (0.95475 mol dm 3 ) x (1.337 atm dm 6 mol 2 )
= 129Pam 3 mol
w =fpdVm
where
1
=1129JmOI 1 1 RT
a
Vm  b
Vm
p =   2
for a van der Waals gas.
Hence,
w
= /
(~) dVm + / Vm  b
;'dVm Vm
=
q+ 6. Urn
Thus
q
=
22.1 dm 3 mol / l.00dm 3 mol  1
1
(
RT
Vm  b
) dVm
= RTln(Vm 
122. 1 dm
b)
3 mol  1 3
I
l.00dm mol 
2
I
22.1  3.20 x 10 ) = (8.314 JK 1 molI) x (298K) x In ( 2 = +7 .7465kJmol  ~ 1.00  3.20 x 10and w = q
+ 6.Um =
(7747 J molI)
+ (l29Jmol l )
I
= 176181 molI I = 17.62kJmol 1 1
THE FIRST LAW
E2.31(b)
33
The expansion coefficient is V' (3 .7
X
1O 4 K 1 +2 x l.52 x 1O 6 TK 2 ) V
V' [3.7 x 104 + 2 x 1.52 x 10 6 (T / K)] K I = V' [O.77 + 3.7 x 1O 4 (T / K) + 1.52 x 1O 6 (T / K) 2]
[3.7 x 104 + 2 x 1.52 x 10 6(310)] K I 0.77 + 3.7 x 10 4 (310) + 1.52 x 10 6(310)2
= E2.32(b)
=
I l.27 x
3
10 K
I
I
Isothermal compressibility is so
Lip
LiV
= VKT
A density increase of 0.08 percent means Li V /V applied is Lip 0.0008  2.21 x 10 6 atm I E2.33(b)
= 13.6 x .
= 0.0008. So the additional pressure that must be
102 at~ 1 .
The isothermal JouleThomson coefficient is ( 8H) 8p T
=
J1,Cp
=
(1.11 K atm I) x (37.11 J K I molI)
= 141.2J atm 1 mol  I 1
If this coefficient is constant in an isothermal JouleThomson experiment, then the heat which must be supplied to maintain constant temperature is LiH in the following relationship LiH / n = 41.2Jatm1 molI Lip LiH
=
so
LiH = (41.2Jatm 1 moll)nLip
(4l.2J atm I molI ) x (l2.0mol) x (55 atm)
= 127.2 x
103 J I
Solutions to problems Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298 K.
Solutions to numerical problems P2.1
The temperatures are readily obtained from the perfect gas equation, T
TI
=
(1.00atm) x (22.4dm 3 ) (1.00 mol) x (0.0821 dm 3 atm mol I Kl)
Similarly, T2 = 1546 K I.
~
= ~ = T3
= PV , nR
[isotherm).
34
STUDENT'S SOLUTIONS MANUAL
In the solutions that follow all steps in the cycle are considered to be reversible. Step I + 2 W
= Pexf!. V = pf!. V = nRf!.T
W
= (1.00 mol) x (8.314J K I molI) x (546  273) K = 12.27 x 103 J I.
f!.U = nCv.mf!.T = (l.OOmol) x q = f!.U 
W
[f!.(pV) = f!.(nRT)],
23 x
(8.314J K I molI) x (273 K) = +3.40 x 103 J.
= +3.40 x 103 J  (2.27 x 10 3 J) = 1+5.67 x 103 J I. 3
f!.H = qp = 1+5.67 x 10 J
I·
If this step is not reversible, then w, q, and f!.H would be indeterminate. Step 2 + 3
I W = 0 I [constant volume]. 3 qv = f!.U = nCv,mf!.T = (1.00 mol) x (2) x (8.314J K I molI) x (273K)
= 13.40 x 10 3 J I. FromH
==
U +pV
f!.H = f!.U + f!.(pV) = f!.U + f!.(nRT) = f!.U + nRf!.T
= (3.40 x 103 J) + (1.00 mol) x (8.314JK 1 molI) x (273K) =15 .67 x 10 3
Jj.
Step 3 + 1
I
I
f!.U and f!.H are zero for an isothermal process in a perfect gas; hence for the reversible compression 3
q =
W
= nRT In VI = (1.00 mol) x (8.314JK I mol I) x (273K) x In (22.4dm 3 )
= 1+ 1.57
~
X
~8~
103 J I, q = 11.57
X
103 J I·
If this step is not reversible, then q and w would have different values which would be determined by the details of the process. Total cycle
State
p/atm
Vldm 3
TIK
2 3
1.00 1.00 0.50
22.44 44.8 44.8
273 546 273
THE FIRST LAW
35
Thermodynamic quantities calculatedfor reversible steps
Step
Process
q/kJ
w/kJ
t:.U j kJ
t:.H j kJ
1+2 2+3 3+1 Cycle
p constant = Pex V constant
+5.67 3.40 1.57 +0.70
2.27 0 +1 .57 0.70
+3.40 3.40 0 0
+5 .67 5.67 0 0
Isothermal, reversible
COMMENT. All values can be determined unambiguously for the reversible cycle. The net result of the overall
process is that 700 J of heat has been converted to work.
P2.3
Since the volume is fixed , 1w Since t:.u t:.H
= 0 I.
= q at constant volume, 1t:.u = +2.35 kJ I.
=
t:.u
+ t:.(pV ) = t:.u + Vt:.p
[t:.V
= 0].
From the van der Waals equation [Table 1.6] so
Therefore, t:.H
= t:. U +
Rt:.T t:.p=   Vm b
[t:. Vm
= 0 at constant volume] .
RVt:.T
. Vm b
From the data, Vm
=
15.0dm 3 2.0 mol
t:.T
= (341
 300) K
RVt:.T
(8.3141 K I molI) x (15 .0dm 3 ) x (41 K)
Vm b
7.46 dm molI
3
Therefore, t:.H P2.S
= 7.5 dm 3 molI,
=
(2.35 kJ) + (0.68 kJ)
= 41
K.
= 0.68 kJ.
= 1+3.03 kJ I.
This cycle is represented in Figure 2.2. Assume that the initial temperature is 298 K . (a) First, note that 1 w = 0 1 (constant volume). Then calculate t:.U since t:.T is known (t:.T and then calculate q from the First Law. t:.U
= nCv.mt:.T [2.16b];
t:.U
=
q
(1.00 mol) x
= qV =
t:. U  w
G)
CV,m
= Cp,m 
R
=
7
'2R  R
=
0
= 1+6.19 kJ I.
298 K)
5
'2R ,
x (8.3141 K I molI) x (298 K)
= 6.19 kJ 
=
= 6.19
x 103 1 = 1+6.19 kJ I.
36
STUDENT'S SOLUTIONS MANUAL
2 2.0
S '"
(a)
~
1.0
3 100
50
150
Figure 2.2
I':1.H
=
I':1.U + I':1.(PV)
= (6.19kJ) + (b) 1q
=
I':1.U + I':1.(nRT)
=
I':1.U + nRI':1.T
(1.00 mol) x (8.31 x 10 3 kJ mol  I) x (298 K)
= 1+8.67 kJ I.
= 0 1(adiabatic).
Because the energy and enthalpy of a perfect gas depend on temperature alone, I':1.U(b)
=
I':1.U(a)
= 16.19 kJ I, since I':1.T(b) =
Likewise I':1.H(b) = I':1.H(a) = 18.67 kJ
=
w (c) I':1.U
1':1. U
I':1.T (a).
I.
= 16.19 kJ 1[First Law with q = 0].
= I':1.H = 0 [isothermal process in perfect gas]. q = w [First Law with I':1.U = 0];
=
V2
VI
W
=
VI
nRT l ln  [2.11] . V3
nRTI
(1.00 mol) x (0.08206dm 3 atm K I molI) x (298K)
PI
1.00atm
=  =
where
so V3
=
_ 3 ((2) x (298 K) (24.45 dm ) x 298 K
q
10 3 J
= 14.29 kJ I.
= 1+4.29 kJ I·
R
)5/2= 138.3_dm 3.
w = (1.00 mol) x (8.314J K I molI) x (298K) x In
= 4.29 x
CVm
5 2
c = ' = 
22.45 dm3) 3 ( 138.3dm

= 24.45 dm
3 .
THE FIRST LAW
P2.7
37
The fonnation reaction is ~fW(298 K) = 84.68 kJ molI.
In order to determine ~fW (350 K) we employ Kirchhoff's law [2.36] with T2
where ~rCp
=L
vJCp,m(J)
= Cp,m(C6H6)
= 350 K, TI = 298 K,
 2Cp,m(C)  3Cp,m(H2).
J
From Table 2.2
l
T2
T,
C
~ dT ~I p _I J K mol
= 100.83
X
(T2  TI) +
2  ( 1.56 x 106 K)
=
(
(I) (0.1079K I) 2
(Ti 
T~)
 I  I ) T2
100.83 x (52 K) +
TI
(~) (0.1079) (3502 
298 2 ) K
6 (  I   I ) K  (1.56 x 10) 350 298
= 2.65
X
103 K.
Multiplying by the units J KImol  I, we obtain
r ~rCpdT =  (2.65 iT, T2
= Hence ~fW(350 K)
Cr(C6H6h(s) ~rW
+
=
=
2.65 x 103 J molI
~fW(298 K)  2.65 kJ molI
84.68 kJ molI  2.65 kJ molI
Cr(s) + 2C6H6(g) ,
= ~r~ + =
103 K) x (J KImol I)
2.65 kJ molI .
=P2.9
X
~ng
= 187.33 kJ molI I.
= +2 mol.
2RT, from [2 .21]
(8.0 kJ mol  I) + (2) x (8.314 J K I mol I) x (583 K)
= 1+17.7 kJ mol I I.
38
STUDENT'S SOLUTIONS MANUAL
In terms of enthalpies of formation /';.rlF or
=
(2) x /';.flF(benzene, 583 K)  /';.flF(metallocene,583 K)
/';.rH~(metallocene, 583 K) = 2/';.fH~(benzene, 583 K)  17.7 kJ molI .
The enthalpy of formation of benzene gas at 583 K is related to its value at 298 K by /';.flF(benzene, 583 K)
=
/';.flF(benzene, 298 K)
+ (Tb 
298 K)Cp.m(l) + /';. vaplF + (583 K  Tb)Cp.m(g)
 6 x (583 K  298 K)Cp,m(gr)  3 x (583 K  298 K)Cp,m(H2 , g) where Tb is the boiling temperature of benzene (353 K). We shall assume that the heat capacities of graphite and hydrogen are approximately constant in the range of interest and use their values from Table 2.7. /';.flF(benzene,583 K)
= (49.0 kJ mol  I) +
(353  298) K x (136.1 J K I molI)
+ (30.8 kJ molI) + (583  353) K x (81.67 J K I molI)  (6) x (583  298) K x (8.53 J K I molI)  (3) x (583  298) K x (28.82 J K I molI)
= {(49.0) + (7.49) + = +66.8 kJ mol  I. Therefore P2.11
/';.fH~ (metallocene, 583 K) = (2
(18.78) + (30.8)  (14.59)  (24.64)} kJ molI
x 66.8  17.7) kJ mol I = 1+ 116.0 kJ mol 1 I.
(a) and (b). The table displays computed enthalpies of formation (semiempirical, PM3 level, PC Spartan ProTM), enthalpies of combustion based on them (and on experimental enthalpies of formation of H20(l) and C02 (g),  285.83 and 393.51 kJ molI respectively), experimental enthalpies of combustion (Table 2.5), and the relative error in enthalpy of combustion.
ClL;(g) C2 H6(g) C3 Hg(g) C4 H 10 (g) C5 H I2(g)
54.45 75.88 98.84 121.60 142. 11
910.72 1568.63 2225.01  2881.59 3540.42
890 1560 2220  2878 3537
The combustion reactions can be expressed as:
The enthalpy of combustion, in terms of enthalpies of reaction, is
2.33 0.55 0.23 0.12 0.10
THE FIRST LAW
where we have left out l'lr1F (02) %error
=
39
= O. The % error is defined as:
l'lcHG(caIc.)  l'lcHG(expt.) G x 100%. l'lcH (expt.)
The agreement is quite good . (c) If the enthalpy of combustion is related to the molar mass by
then one can take the natural log of both sides to obtain:
Thus, if one plots In ll'lcHGI vs. In [M/(g molI )), one ought to obtain a straight line with slope nand yintercept In Ikl. Draw up the following table. Compound
M/(g molI)
l'lcH j kJ molI
In M/(g mol  I)
In ll'lcHGj kJ mol  II
CH4 (g) C2 H6(g) C3 H8(g) C4H IO(g) CSHI2(g)
16.04 30.07 44.10 58.12 72.15
910.72 1568.63 2225.01 2881 .59 3540.42
2.775 3.404 3.786 4.063 4.279
6.814 7.358 7.708 7.966 8.172
The plot is shown in Fig 2.3. 9 I
"0
E 8 :2
:t:
7
6
V
~
2
~
~
3

~f"'
~
4
5
Figure 2.3 The linear leastsquares fit equation is:
These compounds support the proposed relationships, with n
= 10.9031
and
k
=
_e 4 .30 kJ mol  I
= 173.7 kJ molI I.
The agreement of these theoretical values of k and n with the experimental values obtained in P2.1O is rather good.
40
P2.13
STUDENT'S SOLUTIONS MANUAL
The reaction is
Because the reaction does not change the number of moles of gas, lleH = lleV [2.21]. Therefore
Now relate the enthalpy of combustion to enthalpies of formation and solve for that of C60.
IlfW(C60) = 60IlfW(C02)  60IlfW(02)  lleW = [60(393.51)  60(0)  (25968)] kJ mol 1 = 12357 kJ mol 1 I. P2.1S
(a)
IlrH" = IlfW(SiH2)
+ IlfW(H2) 
L\fW (SiH4)
= (274 + 0  34.3) kJ mol 1 = 1240 kJ mol 1 I. (b)
IlrW = L\fW(SiH2)
+ L\fW(Si~) 
L\fW(Si2H6)
= (274 + 34.3  80.3) kJ mol 1 = 1228 kJ mol 1 I. P2.17
The temperatures and volumes in reversible adiabatic expansion are related by eqn 2.28a:
Vf)l /C ( Vi
Tf = Ti 
CVm where c =  ' . R
From eqn 2.29, we can relate the pressures and volumes: Pf = Pi
Vf)Y ( Vi where y =
Cp,m . CV,m
We are looking for Cp,m, which can be related to c and y.
,m) cy _ (Cv R
X
(Cpo m) _ Cp,m . CV,m R
Solving both relationships for the ratio of volumes, we have so
Pf Pi
Therefore
C =R p,m
( In (202.94kPa)) In ( Pf) :81 840 kPa P, =(8314JK 1 mol 1)x . =141.40JK  1 mol 1 1. (Tf) ' .. (298.15K) ~ ~ Ti 248.44K
THE FIRST LAW
P2.19
Hm = Hrn(T,p). aHm) dHm= (  dT+ (aHm)  dp. aT p ap T Since dT = 0, aHm) dHm= (  dp where ( aHm) =/LCp ,m[2.53]= (2a   b) . ap T RT ap T tlHm=
I
Pf
~
dH rn =
l
Pf
~
(2a RT
b) dp =  (2aRT  b)
(2) x (1.352 dm 6 atm mol  2 )
3
':c,;' 
(0.08206
dm 3
atm K I molI) x (300 K)
x (1.00 atm  500 atm)
= (35.5 atm)
x
(~) 3x 10 dm
(Pf  Pj)
5
(1.013 x 10 pa) I atm
I
(0.0387 dm mol )
= 3.60 x
3= [
10 J
+3.60 kJ [.
COMMENT. Note that it is not necessary to know the value of Cp ,m.
Solutions to theoretical problems P2.21
(a)
dz = (a z) d.x + (a z) dy [definition of total differential]. ax y ay x az ) = (2x  2y + 2), ( az ) = (4y ( ax ay x )' dz =
1
(2x 
2y + 2) d.x+(4y2x4)dy I·
(b)
a (a z) = a (2x  2y + 2) =
(c)
az) =(y +~) , (a z) =(xI) , ( ax )' x ay x
ay ax
dz
2x  4) ,
ay
2,
a axa (aayz) =(4y2x4)=2. ax
= (y + ~) d.x + (x  I) dy .
A differential is exact if it satisfies the condition
a (a z) a (a z) ax ay = ay ax ' a (a z) a ~ ay (aaxz) = ~ ay (y + x~) = I , ax ay =ax(xI)=1. COMMENT. The total differential of a function is necessarily exact.
41
42
P2.23
STUDENT'S SOLUTIONS MANUAL
(a)
U= UCT. V)
dU= ( au ) dT + ( au ) dV = CvdT + ( au ) dV. aT v av T av T
For
U= consta nt, dU= 0 , and
so
CvdT = _( au ) dV av T
Cv = _(au) (dV) = _( au ) (av) . av T dT u av T aT u
or
This relationship is essentiall y Eu ler's chain re lati on [Further information 2.2 ].
(b)
H= H(T,p)
dH = ( aH) dT + ( aH) dp = CpdT + (aH) dp. aT p ap T ap T
so
According to Euler's chain re lati on
so, using the reciprocal identity [Further information 2.2 ],
P2.2S
Ca) H
=
U + pV
_l +p( av ) ( aH) au p au p
so
(
(a Hjav)" (au jav)"
(b)
so
1+
caCU + PV) ) av "
P
caU jav)"
(au jav)" + p cau jav)p
l+P( ~V ) . au "
aH) P ( au p I + caU jav)p
P2.27
V Inserting  = Vm into the virial equation fo r P we obtain
n
p
1
= nRT ( V +
2
nB V2
Therefore, W = nRT
n C
l
v2
VI
(
I
V
v? + n2 RTB w =  nRT In =. VI For n
)
+ V3 + ". . 2
C + 2nB + n 3 + .") dV, V
(
= 1 mol: nRT = ( 1.0 mol) x
V
 1   1) V2
VI
I 3 RTC ( + n 2
21  2I ) V2
V2
(8.3 14 J K I mol  I) x (273 K)
+ ""
= 2.27 kJ .
THE FIRST LAW
43
From Table 1.4, B = 21.7 cm 3 mol 1 and C = 1200 cm 6 mol2, so n 2 BRT
=
(1.0 mol) x (21.7 cm 3 molI) x (2.2'7 kJ)
~n3CRT = ~(\.O mol)2 2
2
=
49.3 kJ cm 3 ,
x (l200cm 6 mol  2) x (2.27 kJ)
=
+1362 kJ cm 6 .
Therefore, (a)
W

= 2.27 kJ In 2 =

(49.3 kJ) x
(I
I)
1000  500
(1.57) + (0.049)  (4.1 x 10 3 ) kJ
=

(I
+ (1362 kJ) x
I)
10002  5002
= 11.5 kJ I.
\.52 kJ
(b) A perfect gas corresponds to the first term of the expansion of p , so
W
P2.29
= 1.57 kJ = 11.6 kJ
I.
J1, = (aT) = _~(aH) [2.51 ap H Cp ap T
and 2.53],
J1, = ~ {T (a V)  V} [See problem 2.34 for this result] . Cp aT p But V
= nRT + nb
or
p
nR
(av) aT p
p
Therefore,
J1,=~{nRT _v}=~{nRT
Cp
p
Since b > 0 and
Cp
Cp >
P
_nRT nb}= nb . P Cp
0, we conclude that for this gas
J1, <
0 or
(aT) ap H
< O. This says that when the
pressure drops during a JouleThomson expansion the temperature must increase""l. rI
P2.31
p
nRT
n2a
= V _ nb  V2
Hence 1 T =
[Table 1.7].
(ffR) x (V  nb) + (~) x (V  nb) I, nb nR
V 
Vm b
R
. relatIOn, . we need to show that For Euler ,s cham
(aT) (a p ) (av) = I. ap v av T aT p
44
STUDENT'S SOLUTIONS MANUAL
p . addItIon . . to ( aT) we need (a Hence, In  ) ap v aV
ap ) ( av
T
=
nRT (V  nb)2
a n(av) daT p
T
2n 2a
+ \.f3
which can be found from
( aaT) V p aT) ( av p
= (l!...) + (~) _ (2na) nR
=
Rv2
RV3
x (V  nb)
'
( _ T ) _ (2na) x (V _ nb). V  nb RV3
Therefore,
nRT x ( (V _ nb)2
(v~nb) (V
~Tnb)
(V ~ nb)
2
2n a)
+ \.f3
 ( 2na) x (V nb) RV3
2na) + ( RV3 
x (V  nb)
2na) ( RV3 x (V nb)
= I. P2.33
j1Cp
= T ( av) aT p
V

=
(a TT) av
aT) ( a V p
= T V  nb
 V [reciprocal identity, Further information 2.2] p
2na RV3
 ( V  nb) [Problem 2.31]
Introduction of this expression followed by rearrangement leads to
= P
/LC
(2na) x (V  nb)2  nbRTV 2 xV. RTV3  2na(V  nb)2
Then, introducing
j1Cp
=
~
=
RTV 3 ~ to simplify the appearance of the expression 2na(V  nb)
I_nb~ ( V
~l
)
V =
(1 ~)
~ V. ~l
THE FIRST LAW
4S
For xenon, Vm = 24.6 dm 3 mol  I, T = 298 K , a = 4.137dm 6 atmmol 2 , b = 5.16 x 10 2 dm 3 mol  I, nb
b
V
Vm
~ =
(8.206 (2)
X
X (4. 137 dm 6 atm
1  (73.0)
Therefore, !tCp =
(298 K) X (24.6 dm 3 mol  I) 3 I 3 (24.6 dm mol  5.16 X 10 2 dm mol
10 2 dm 3 atm K  I mol  I)
X
2
mol  ) (2 .09
X
X
103)
72.0
X 3
= 73.0.
1)2
x (24.6 dm 3 molI ) = 0.290dm 3 mol I.
Cp = 20.791 K I mol  I [Table 2.7], so
0.290dm 3 mol  I !t = 20.791KI mol  I
0.290 x 10 3 m3 mol I 20. 791 KI mol 1
The value of!t changes at T = TI and when the sign of the numerator I positive). Hence b~

Vm
= I at
T
=
TI
or
RTbV 3 2na(V  nb)2Vm
;: =
b ) 2a ) ( that is, TI = ( Rb x 1  Vm
2
27 ( ="4 Tc I 
1 implying that TI =
nb~ V
changes sign
(~
 I is
2a(Vm  b)2 2
RbVm
'
b )2 Vm
X 2)2=11946KI.
516 10andsoTI=(1954K)x ( 1  ' 24.6
Question. An approximate relationship for !t of a van der Waals gas was obtained in Problem 2.30. Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to the result above. P2.35
2
Cp .m  CV.11l
a TV =nKT
[2.49]
(a p )
aTV = n
nR  [Problem 2.31]. (aTap )v = V  nb aV
(av) aT
I'
aT v
[2.57].
46
STUDENT'S SOLUTIONS MANUAL
Substituting,
aT)
Also ( av
p
= T 
2na

V  nb
RV
3 (V  nb) [Problem 2.31].
Substituting,
RT C
C
p,m 
with A =
V,m =
_=_'(_V:;;:_n_b...:..)_ _ _ T 2na ::::~  X (V  nb) (V  nb) RV3
I
I or I  ~ x (V _ nb)2 A (RTV3) 2
=I
= AR
2a(Vm  b)3 3
RTVm
. . 8a Now mtroduce the reduced vanables and use Tc = 27Rb ' Vc = 3b. After rearrangement,
For xenon, Vc = 118. I cm 3 molI, Tc = 289.8 K. The perfect gas value for Vm may be used as any error I introduced by this approximation occurs only in the correction term for . A Hence, Vm ~ 2.45 dm 3 , Vc =118 .8 cm 3 molI , Tc = 289.8 K, and Vr = 20.6 and Tr =1.03 ; therefore I
 = A
I
(6 1.8  1)2 (4) x ( 1.03) x (20.6)3
= 0.90
,
giving A.
~
I.l
and
Cp.m  CV,m ~ 1.1 R
P2.37
(a)
JL
= 19.2 J K 1 molI I.
I ( aH) =  I {(aVm) = T 
Vm
Cp
ap
= RT +aT2 p
T
so
Cp
aT
aVm) ( aT p
=
p
 Vm } [2.53 and Problem 2.34].
R p + 2aT.
THE FIRST LAW
(b)
Cv
p
a ) v aTVOl ( aT
= cp 
= Cp 
47
p
T (aVOl) aT p (aaT ) v '
RT
But,p =
VOl aT
ap ) ( aT v
2'
R
=
RT(2aT)
VOl  aT2  (VOl  aT2)2
Therefore
RT( 2aPT) ( 2aPT) =Cpp l+~ x l+~ x
(p) T '
2aPT)2 Cv=CpR ( 1 +~
Solutions to applications P2.39
Taking the specific enthalpy of digestible carbohydrates to be 17 kJ gI (Impact 12.2), the serving of pasta yields q = (40 g) x ( I 7 kJ g  I) = 680 kJ .
Converting to Calories (kcal) gives: I Cal q = (680kJ) x 4.184kJ = 162CaJ.
As a percentage of a 2200Calorie diet, this serving is I 62 Cal r::;:;;;;:l 2200 Cal x 100% = ~.
P2.41
(a)
q=n6.cH~ =
1.5g 342.3grnol 
1
x (5645kJrnol I)=125kJI.
(b) Effective work available is ~ 25 kJ x 0.25 = 6.2 kJ. Because w = mgh , and m ~ 65 kg h ~
6.2x 103J ~ 2 =~' 65 kg x 9.81 rn s
48
STUDENT' S SOLUTI ONS MANUA L
(c) The energy released as heal is
q
=
6. r H
= n6. c H " = 
(
2.5g ) x (2808 kJ mol  1 ) 180 g mol  1
~
= ~.
(d) If onequarter of thi s energy were available as work a 65 kg person could climb to a he ight h given by 1
4q =
W
= Illgh
so
h
=~= 4mg
3
39 x 10 J 4 (65 kJ) x (9.8 m s2)
= ~.
P2.43
First, with the pure sample, record a thermogram over a temperature range within which P undergoes a structural change, as can be inferred fro m a peak in the thermogram. The area under the thermogram is the enthalpy change associated with the structural change fo r the g iven quantity of P. Then, with an identical mass of the suspected sample, record a thermogram over the same te mperature range. Ass umin g that the impurities in P undergo no structural change over the temperature rangea reasonable assumption if the impurities are monomers or oligomers and if the te mperature range is suffi cientl y narrowthen the peak in the test sample thermogram is attributable onl y to P. The rati o of areas under th e curve in the test sample to the pure sample is a measure of the purity of the test sample.
P2.45
The coeffi cient of thermal ex pansion is
so
6.V""' a V 6.T.
This change in volume is equ al to th e change in height (sea level ri se, 6.h ) times the area of the ocean (ass uming that area rem ains constant). We will use a of pure water, althou gh the oceans are co mplex solutions. For a 2 a C rise in te mperature
so 6.h
=
6.V A
= 1.6 X
10 3 km
= ~.
Since the rise in sea level is direc tl y proportional to the ri se in temperature, 6. T
6.h
= I ac would lead to
= 10.80 m 1and 6.T = 3.5 a C would lead to 6.h = 12.8 m I.
COMMENT. More detailed models of climate change predict somewhat smaller rises, but the same order
of magnitude.
P2.47
We compute M from
and we estimate
(88pH)
fro m the enthalpy and pressure data. We are given both enth alpy and heat T
capacity data on a mass basis rath er than a molar bas is; however, the masses will cancel, so we need not convert to a molar bas is.
THE FIRST LAW
(a) At 300 K. The regression analysis gives the slope as 18.0 J gI MPa 1 ~ so J1
18.0kJkg 1 MPa 1 0.7649 kJ kg I K I
=
I
I
(aH) ap T ,
I
= . 23.5 K MPa.
(b) At 350 K. The regression analysis gives the slope as 14.5 J gI MPa 1 ~ so J1
(a)
=
1
14.5kJkg MPa1.0392kJkg 1 K I
1
(aapH) ,
=1 14.0KMPa 1 I·
426.6 426.4 426.2
H /(kJkgl) 426.0 425 .8 425.6 L...'....:....''_''....:''_:..~ 0.07 0.08 0.09 0.10 0.11 0.12 0. 13
plMPa (b)
462
... .x.=..473.....2..~.!.~ . 4.59:x.
... ...
R2",
461
1.000
·
,
.. ,.
.
...... ; .. .
460
H l(kJ kgI) 459 458 457 456 0.8
0.9
1.0
l.l
plMPa
1.2
1.3
Figure 2.4
T
49
The Second Law
Answers to discussion questions 03.1
We must remember that the Second Law of Thermodynamics states only that the total entropy of both the system (here, the molecules organizing themselves into cells) and the surroundings (here, the medium) must increase in a naturally occurring process. It does not state that entropy must increase in a portion of the universe that interacts with its surroundings. In this case, the cells grow by using chemical energy from their surroundings (the medium) and in the process the increase in the entropy of the medium outweighs the decrease in entropy of the system. Hence, the Second Law is not violated.
03.3
All of these expressions are obtained from a combination of the First Law of Thermodynamics with the Clausius inequality in the form TdS :::: dq, as was done at the start of Justification 3.2. It may be written as dU  PexdV + dWadd
+ TdS
:::: 0
where we have divided the work into pressurevolume work and additional work. Under conditions of constant energy and volume and no additional work, that is, an isolated system, this relation reduces to dS:::: 0 which is equivalent to
~Stot
= ~ Suniverse :::: O. (The universe is an isolated system.)
Under conditions of constant entropy and volume and no additional work, the fundamental relation reduces to dU:::: O. Under conditions of constant temperature and volume, with no additional work, the relation reduces to
ciA:::: 0, where A is defined as V  TS. Under conditions of constant temperature and pressure, with no additional work, the relation reduces to dG:::: 0,
where G is defined as V
+ pV 
TS
=H
 TS.
THE SECOND LAW
51
In all of the these relations, choosing the inequality provides the criteria for spontaneous change. Choosing the equal sign gives us the criteria for equilibrium under the conditions specified. 03.5
The Maxwell relations are relations between partial derivatives all of which are expressed in terms of functions of state (properties of the system). Partial derivatives can be thought of as a kind of shorthand for an experiment. Therefore, the partial derivative (aS j aV)T tells us how the entropy of the system changes when we change its volume under constanttemperature conditions. But, as entropy is not a property that can be measured directl y (there are no entropy meters), it is important that the derivative (and hence the experiment) be transformed into a form that involves directly measurable properties. That is what the following Maxwell relation does for us.
Pressure, temperature, and volume are easily measured properties. 03.7
The relation (aG j ap)r = V shows that the Gibbs function of a system increases with p at constant T in proportion to the magnitude of its volume. This makes good sense when one considers the definition of G, which is G = U + pV  TS. Hence, G is expected to increase with p in proportion to V when T is constant.
Solutions to exercises Assume that all gases are perfect and that data refer to 298. 15K unless otherwise stated. E3.1(b)
(a)
(b) E3.2(b)
At 250 K, the entropy is equal to its entropy at 298 K plus I'1.S where
f
I'1.S 
so
E3.3(b)
dqrev  T
f
CV,m dT _ Cv mIn Tf T ' Tj
S = 154.841 K ' mol  '
+ [(20.786 
S = 1152.651K' mol  '
I
250K 8.3145) 1 K 'mol'] x I n  298K
However the change occurred I'1.S has the same value as if the change happened by reversible heating at constant pressure (step I) followed by reversible isothermal compression (step 2)
52
STUDENT'S SOLUTIONS MANUAL
For the first step
f
AS 
'' I 
=
f>"SI
f
dqrev T

Cp,m dT  C L Tf T p,m n Ti
(2.00 mol) x ( 7) x (8.3l45JK 1 mol  I) x In (l35 + 273) K 2 (25+273)K
=
18.31K 1
and for the second f>"Sz =
where qrev
f =
dqrev = qrev
T
T
=
 w
JpdV = nRT In Vf
= nR In Pi =
so f>"Sz
Pf
=
f>"S
Vi
= nRT In
Pi

Pf
I I L50atm (2.00 mol) x (8.3 145 J K mol  ) x In 7.00atm
(18.3  25.6)JK 1
=
 25 .61 K
I
= !7.31K 1 !
The heat lost in step 2 was more than the heat gained in step 1, resulting in a net loss of entropy. Or the ordering represented by confining the sample to a smaller volume in step 2 overcame the disordering represented by the temperature rise in step I. A negative entropy change is allowed for a system as long as an increase in entropy elsewhere results in f>"Stotal > O. E3.4(b)
q
= qrev =0 [adiabatic reversible process] =
f>"S
if i
f>"U
dqrev
=
T
W
= nCv ,mf>"T = (2.00 mol )
x (27.5 J K I molI) x (300  250) K
= 2750J = 1+2.75 kJ 1
w
=
f>"H
f>"U  q
= 2.75 kJ 
E3.S(b)
= 12.75 kJ 1
= nCp.mf>"T
Cp,m = CV ,m + R
So f>"H
0
=
=
(27.51 K I molI + 8.3141 K I molI)
(2.00 mol) x (35.8141 K I mol  I) x (+50 K)
=
35.814J K I mol  I
= 358l.41 = 13.58 kJ 1
Since the masses are equal and the heat capacity is ass umed constant, the final temperature will be the average of the two initial temperatures,
The heat capacity of each block is C
= mCs
where Cs is the specific heat capacity
so f>"H (individual)
= mCsf>"T =
1.00 x 103 g x 0.4491 K I gI x (±87.5K)
= ±39kJ
THE SECOND LAW
I
These two enthalpy changes add up to zero: 6. H tot
= mCs In
6.S
6.S1
(i);
200 °C
= 473.2 K; 2S °C =
6.S2 = (1.00 x 10 3 g) x (0.449 J K I g I)
E3.6(b)
=0I
(a)
q = 0 [adiabatic]
(b)
w
X
298.2 K; 112.5 °C
= 38S.7 K
G!~:~)
=
IIS .5J K I
38S.7) In ( 473.2
=
9 1.802J K I
( 1.00 x 103 g) x (0.449J K I g I) x In
=
5
=
Pex6.V
=
 ( I.Satm) x
=
227.2 J
(.
1.0 I x 10 pa) x (lOO.Ocm 2 ) x (lScm) x ~
6.U
= q+w=0
(d)
6.U
= nCv.m6.T
6. T
227.2 J = 6.U =     :
nCV.m
(
Im
3
63
)
10~
= 1230 J I
(c)
230 J
53
= 1230 J I
( I.Smol) x (28.8 JK I mol I)
=
IS.3K I
(e) Entropy is a state function , so we can compute it by any convenient path. Although the spe
cified transformation is adiabatic, a more convenient path is constantvolume cooling followed by isothermal expansion . The entropy change is the sum of the entropy changes of these two steps : 6.S
= 6.S1 + 6.S2 = nCV.m In
Tr = 288. IS K  S.26 K
~
nRT
nR In
(~)
[3.19 and 3.13]
= 282.9K
( I.S mol) x (8.206 x 10 2 dm 3 atm K I molI) x (288.2 K)
=  = ~~~~~~~ Pi
9.0atm
= 3.942 dm
Vr
(i) +
3
= 3.942dm 3 + ( IOO cm2 )
3
x (lScm) x
1 dm 3 ) ( lOOOcm
54
STUDENT'S SOLUTIONS MANUAL
/'<,.S = ( 1.5 mol ) x
{
(28.8 JK  1 mol  I) x In
?282 .9) ? ( _8L
+(8.314J K I mol  I) x In ( 5.44)} 3.942
E3.7(b)
(a)
/'<,. vap H
/'<,. vapS =
y:;; =
35.27 x 103 J mol  I _I (64. 1 + 273.15) K = + I 04.58 J K =
I
_I 104.6 J K
I
(b) If vaporization occurs reversibly, as is generally assumed
/'<,.Ssys + /'<,.Ssur = 0 E3.8(b)
(a)
so /'<,.Ssur = 1 104.6 J K I
I
/'<,.rS & = S,~(Zn 2+ , aq) + S,~(CU , s)  S,~ (Zn , s)  S,~(CU 2+, aq) = [112.1 + 33.15  41.63 + 99.6] J K I mol  I = 121.0J K Imol I
(b)
I
/'<,.rS & = 12S:' (C0 2, g) + IIS,~ (H 2 0, 1) S,~(CI 2 H22 011 'S)  12S:'(0 2, g) = [( 12 x 213 .74) + ( II x 69.91 )  360.2  ( 12 x 205.14)] J K I mol I = 1+5 12.0JK l mol  1 1
E3.9(b)
(a)
/'<,.rW = /'<,.rW(Zn2+, aq)  /'<,.rW(Cu 2+ , aq) =  153.89  64.77 kJ mol  I = 2 18.66kJmol  1 /'<,.rG & = 218.66kJmol 1  (298 .15K) x (2 1.0JK  1 mol  I) = 1212.40kJmol  1
(b)
/'<,.rW = /'<,. c H t> = 5645 kJ mo l I /'<,.rG& = 5645 kJ mol  I  (298.15 K) x (5 12.0J K  I mol  I) = 15798 kJ mol  I
E3.10(b)
(a)
I
/'<,.rG & = /'<,.rG &(Zn 2+ , aq)  /'<,.rG &(C u 2+ , aq) = 147.06  65.49kJmol  1 = 12 12.55kJmOI  1
(b)
I
I
/'<,.rGt> = I 2/'<,.r Gt> (C0 2, g) + 11 /'<,.rG &(H20 , I)  /'<,.rG &(C 12H220 11 , s)  12/'<,.rG&(02, g) = [12 x (394.36) + II x (237. 13)  (1543)  12 x 0] kJ mol  I = 15798kJmol 1
I
COMMENT. In each case these values of L'. rG& agree closely with the calculated values in Exercise 3.9(b).
THE SECOND LAW
E3.11(b)
=
6. r W
L
v6.r W
L

Products
v6. r W [2.32]
Reactants
= 484.5 kJ molI  (238.66 kJ molI)  (110.53 kJ molI) = 135.31kJmol 1
Products
Reactants
= 159.8JK I mol I 126.8JK I molI 197.67JK I molI = 164.67 J K I mol  I 6. r c B
= 6. r W
 T 6. r SB
= 135.31kJmol 1  (298K) x (I64.67JK I molI) = 135.31kJ mol  I + 49.072 kJ molI = 186.2 kJ molII E3.12(b)
The formation reaction of urea is
The combustion reaction is
+ ~02 (g) + C02Cg) + 2H20(l) + N2(g) = 6.rW(C02, g) + 26.rW(H20, 1)  6.rW(CO(NH2h. s)
CO(NH2 )z (S) 6. c H
+ 26.rW(H20, I)  6.cH(CO(NH2)z , s) 393 .51 kJ molI + (2) x (285.83 kJ mol  I)  C632 kJ mol  I)
6.rW(COCNH2h. s) = 6.rWCC02 , g)
=
= 333.17kJmol 1 6. rSB
= S!CCO(NH2)z, s) 
S!CC, gr)  !S!(02 , g)  S!(N2 , g)  2S!CH2, g)
= 104.60J K I molI  5.740J K I mol  I  !(205.138J K I molI)  191.61 J K I mol  I  2(130.684J K I molI) = 456.687JK I molI 6.rcB = 6.rW  T 6. rS B
= 333.17kJmol 
1
= 333.17kJ molI = 1197kJmol 1 1

(298K) x C456.687JK  I molI )
+ 136.093 kJ mol  I
55
56
E3.13(b)
(a)
STUDENT'S SOLUTIONS MANUAL
~S(gas) =
nR In (Vf) [3 .13] = ( Vi
21 g I) x (8.314J K 1 molI) In 2 39.95gmol
= 3.029JK 1 = 13.0JK I
~S(surroundings) ~S(total) = (b)
~S(gas)
=
= 13.0JK I 1[reversible]
@]
= 1+3.0JK I 1[S is a state function]
~S(surroundings) =
(c)
~S(gas)
I
@] [no change in surroundings]
~S(total)
= 1+3 .0JK I 1
qrev = 0
so ~S(gas) =
~S(surroundings) =
~S(total) =
@]
@] [No heat is transfered to the surroundings]
@]
C3Hg(g) + 502(g) + 3C02(g) + 4H20(l)
E3.14(b)
~rG~ = 3~fG~(C02 , g) + 4~fG~(H20, I)  ~fG~(C3Hg, g)  0
= 3(394.36kJmol l ) + 4(237.13 kJmol
l
1(23.49kJmol l )
) 
= 2108.11 kJ molI The maximum nonexpansion work is 12108.11 kJ molII since IWaddl = E3.15(b)
I
(a)
I I
I~GI.
f = 1 _ Tc [3.10] = 1 _ 500 K = 0.500 Th 1000 K ,_, (b) Maximum work = f lqhl = (0.500) x (1.0 kJ) = 0.50 kJ (c) f max = frey and Iwmaxl = I%IIqc,minl
I
Iqc,minl = 1%1  Iwmaxl = 1.0 kJ  0.50 kJ = I 0.5kJ I E3.16(b)
~G = nRTln e~) [3 .56] = nRTln (~;)
[Boyle's law]
~G = (2.5 x 103 mol) x (8.314JK I molI) x (298 K) x In U~) = 12.0J I E3.17(b)
C~)p =
S [3.50] ;
hence
Ca~:)p =
Sf, and
(~i)p =
aT p+(aGi) aT p=_(aCGfGJ) aT p =  (~)p =  aa (73 .1 J +42.8J x ~) T
~S=Sf_Si=_(aGf)
= 142.8 J K I 1
Si
THE SECOND LAW
E3.18(b)
dG = 5 dT f'..G =
l
+ V dp [3.49);
57
at constant T , dG = V dp ; therefore
Pf
Vdp
Pi
The change in volume of a condensed phase under isothermal compression is given by the isothermal compressibility (eqn 2.44). KT = I
V
(B  V) Bp
= 1. 26 x 10 9 Pa
I
T
This small isothermal compressibility (typical of condensed phases) tells us that we can expect a small change in volume from even a large increase in pressure. So we can make the following approximations to obtain a simple expression for the volume as a function of the pressure
where Vi is the volume at I atm, namely the sample mass over the density, m/ p . f'..G=
=
=
=
lOO MPa m  ( IKTP) dp f 100 kPa P m (fIOOMPa fIOOMPa pdp ) dp  KT P 100 kPa 100kPa m ( 1IOOMPa I 21 100 MPa) p  KTP P 100 kPa 2 100kPa 250 '" 3 0.791 gcm
(I
9.99 x 107 Pa   (1.26 x 1O 9 Pa l ) x ( 1.00 x 1016 Pa 2) 2
(~) 3 x 9.36 100cm
= 31.6cm 3 x
X
)
107 Pa
= 2.96 x 1031=i3.0k1 i E3.19(b)
f'..Gm = Gm.r  Gm,i = RT In
(~~)
[3 .56)
= (8.3141 K I mol  I) x (323 K) x In ( 252.0) = 12.71 kJ mol  I 1 92.0 E3.20(b)
c: + RT In ( ; ) [3.56 with Gm = G2) But for a real gas, G m = c: + RT In (;e) [3.58)
For an ideal gas, G2 =
So G m

G2 = RT In
£p [3.58 mjnus 3.56) ; £p = ¢
= RTln¢ =
(8.3141 K I molI ) x (290K) x (In 0.68) = 10.93kJmol 1 1
58
E3.21(b)
STUDENT'S SOLUTIONS MANUAL
"'G = nVm"'p [3.55) = V "'P
"'G = ( 1.0dm
E3.22(b)
"'G
m
3)
x
(~ 3) 10 3 dm
3
x (200 x 10 Pa) = 200Pa m
= RT In (Pf) = (8.314 J K I mol  I) ~
3=
1200 J 1
x (500K) x In (loo.OkPa) mO~a
= 1+2.88 kJ mol i 1
Solutions to problems Solutions to numerical problems P3.1
(a) Because entropy is a state function "'lfSS(I + s, 5°C) may be determined indirectly from the
following cycle
= ",SI +
Thus "'trsS(I + s, 5°C) Tf
where "'SI = Cp.m (I) In T and "'Ss
[3.19;
"'lfSS(I + s, O°C) + "'Ss,
ef
= O°C,
e=
5 °C]
T
= Cp,m (s) In Tf '
Thus, "'lfSS(I + s, T)
"'rus H
=  Tf
T
"'Cp In . Tf
° 6.01 x 103 J mol i 268 I "'trsS(I + s, 5 C) = 273K  (37.3 J K mol  I) x In 273
= 121.3 J K I mol  i I. "'SSUf
=
"'fusH(T) T .
= "'HI + "'fusH (Tr)  "'Hs· "'HI + "'Hs = Cp,m(l)(Tf  T) + Cp,m(s)(T "'fusH(T) = "'fusH(Tf)  "'Cp(Tf  T) . "'fusH (T)
Tr)
=
"'Cp(Tf  T).
THE SECOND LAW
Thus, !'::,.Ssur
!'::,.Ssur
=
!'::,.fusH (T) T
=
6.01 kJ mol 268 K
=
!'::,.fusH (Tf ) (T  Tr ) T + !'::,.Cp T ' I + (37.3 J K
= 1+ 2 1.7 J K I mol  I
!'::,.Stotal = !'::,.Ssur + !'::,.S
59
mol
I
) x
(268  273) 268
I· 21.3) J K Imol  I
= (21.7 
Since !'::,.Stotal > 0, the transition I +
I
S
= 1+0.4 J K I mol  I I.
is spontaneous at 5°C.
(b) A similar cycle and analysis can be set up for the transition liquid + vapor at 95°C. However, since the transformation here is to the high temperature state (vapor) from the low temperature state (liquid), which is the opposite of part (a), we can expect that the analogous equations will occur with a change of sign.
!'::,.trsS( 1 + g, T)
= !'::,.trsS(1
+
T
g, Tb) + !'::,.Cp In Tb
!'::,. vap H T I I =   + !'::,.Cpln  , !'::,.Cp = 41.9JK  mol  . Tb Tb I 40.7kJ mol !'::,.trsS(I + g, T) =  (4 1.9J K I molI ) x In (368) 373 K 373
= 1+109.7 J K I mol I I·
!'::,.Ssur =
 !'::,. vap H (T)
=(
T
= 
!'::,. vapH(Tb)
=
!'::,.Cp(T  Tb )  'T
I 40.7kJmOI  ) I I (368373) 368 K  (41.9JK mol ) x 368
= 1 111.2 J K I mol  I
!'::,.Stotal
T
I.
( 109.7  111.2) J K I mol I
= 1 1.5 J K
I mol  I I.
Since !'::,.Stotal < 0, the reverse transition, g + I, is spontaneous at 95°C. P3.3
(a)
= q(H20) + q(Cu) = 0; hence  q(H20) = q(Cu) . q(H20) = n(  !'::,.vap H ) + nCp.m (H20 , I) x (f)  100°C) q(total)
where f} is the final temperature of the water and copper. q(Cu)
= mCs(f} 
Setting q (H20 )
0)
= mCsf},
Cs
= 0.385 J K IgI .
= q (Cu) allows us to solve for f} .
60
STUDENT'S SOLUTIONS MANUAL
Solving for
e=
e yields: + Cp,m(H20 , I) x 100°C} mCs + nCp,m(H20, I)
n( t:. vapH
( 1.00 mol) x (40.656 x 10 3 J mol  I + 75.3JoC l mol 2.00 x 103 g x 0.385JOC l g 
1
+ 1.00 mol
1
x 100°C)
x 75 .3 J oC l mo l
1
= 57.0°C = 330.2 K. q (Cu ) = (2.00 x 10 3 g) x (0.385 J K I g I) x (57.0 K) = 4.39 q (H20) = 143.9 kJ
t:.S (total) = t:.S(H20 ) t:.S(H20 ) =
X
104 J = 143.9 kJ
I·
+ t:.S(Cu ).
nt:.vap H Tb [3.16]
+ nCp .m In (Tr) Tj
[3.19]
( 1.00 mol) x (40.656 x 10 3 Jmol  I ) 373.2 K
+ (1.00 mol)
330.2 x (75.3J K I molI) x In ( K) 373.2 K
= 108.9J K I  9.22J K I = 1118.1 J K I I.
Tr
t:.S(Cu) = mCs In 
Tj
K) = (2.00 x 10'3 g) x (0.385J K I g I) x In (330.2  273.2 K
= 1145.9 J K I I. t:.S(total) = 118.1 J K I + 145.9J K I = 128 J K I I. This process is spontaneous since t:.S(surroundings) is zero and, hence, t:.S(universe) = t:.S(total) > O. (b) The volume of the container may be calculated from the perfect gas law. nRT ( 1.00 mol) x (0.08206 dm 3 atm K I mol  I) x (373.2 K) 3 V= = = 30.6 dm 1.00atm P
At 57°C the vapor pressure of water is 130 Torr (Handbook of Chem.istry and Physics). The amount of water vapor present at equilibrium is then ( 130 Torr) x ( 1 atm ) x (30.6 dm 3 ) n=pV = 760 Torr =0.193mol. RT (0.08206 dm 3 atm K I mol  I) x (330.2 K) This is a substantial fraction of the original amount of water and cannot be ignored. Consequently the calcu lation needs to be redone taking into account the fact that only a part, Ill , of the vapor condenses
I.
THE SECOND LAW
61
into a liquid while the remainder (1.00 mol  nl ) remains gaseous . The heat flow involving water, then, becomes q(H 2 0 ) =  nl t:. vapH + nl Cp,m(H20, 1)t:.T(H20)
+ (1.00 mol  nl)Cp,m(H20, g)t:.T(H20). Because nl depends on the equilibrium temperature through n I = 1.00 mol  p V, where p is the
RT
vapor pressure of water, we will have two unknowns (p and T) in the equation q(H20 ) = q(Cu ) . There are two ways out of tills dilemma: (1) p may be expressed as a function of T by use of the Clapeyron equation (Chapter 4), or (2) by use of successive approximations. Redoing the calculation yields:
e=
nl t:.vapH + nlCp,m(H20, I) x 100°C + (1.00  nl)Cp,m(H20, g) x lOO°C . mCs + nCp,m(H20, I) + ( \.00  nl)Cp,m(H20 , g)
With nl = (1.00 mol)  (0.193 mol ) = 0.807 mol
(noting that Cp.m(H20, g) = 33.6Jmol 1 K 1 [Table 2.7]), vapor pressure of water is 80AI Torr, corresponding to
e=
47.2°C. At this temperature, the
nl = (1.00 mol)  (0.123 mol) = 0.877 mol.
This leads to e = SO.8°C. The successive approximations eventually converge to yield a value of 1 49.9°C = 323.1 K 1 for the final temperature. (At this temperature, the vapor pressure is 0.123 bar.) Using this value of the final temperature, the heat transferred and the various entropies are calculated as in part (a).
e=
q(Cu ) = (2 .00 X 103 g) x (0.38S J K 1 g I) x (49.9 K) = 138A kJ 1 = q(H20) .
t:.S(Cu) = mC, In
~
= 1129.2 J K
1
I.
t:.S (total) = 119.8JK 1 + 129.2JK 1 =19JK 1 1. P3.5
q
w t:.U t:.H t:.S
t:.Stol t:.G
Step 1
Step 2
Step 3
+l1 .S kJ l1 .S kJ 0 0 +19.1 J K0 11.S kJ
0 3.74 kJ 3.74 kJ 6.23 kJ 0 0 ?
S .74 kJ 0 +S .74 kJ +3.74 kJ 0 +3.74 kJ 0 +6.23 kJ 19.1 J K 1 0 0 0 ? +11.S kJ
1
Step 4
Cycle
S.8 kJ S .8 kJ 0 0 0 0 0
62
STUDENT'S SOLUTIONS MANUAL
Step J
!':1U
w
= !':1H = @] [isothermal].
= nRT In (~~) = nRT In (~~)
[2. I I, and Boyle's law]
= (1.00 mol) x (8.314J K I molI) x (600 K) x In (1.00 atm) = 1 11.5 kJ I. 10.0 a t m · .
q = w = 111.5 kJ I. !':1S
= nR In ( ~~ )
[3.15]
= nR In e~
)
[Boyle 's law]
x (8.314J K I mol  I) x In ( I.ooatm) = 1+19.1J K I 1. 10.Oatm· . I !':1S (system) [reversible process] = 19.1 JK .
= (1.00 mol) !':1S(sur) !':1Stol
=
= !':1S(system) +
!':1G =!':1H  T!':1S
!':1S(sur)
=0
= @].
(600K) x (19.1 J K I)
= ll1.5kJ I.
Step 2 q
= @] [adiabatic] .
!':1U
= nCv,m!':1T[2.16b] = (1.00 mol)
x
G)
x (8.314J K I molI) x (300 K  600 K)
= 13.74 kJ I·
w = !':1U = 13.74kJ I.
!':1H =!':1U
+ !':1(pV) = !':1U + nR!':1T
= (3.74kJ) +
(1.00 mol) x (8.314J K I molI) x (300K)
= 16.23 kJ I. !':1S = !':1S(sur) = @][reversible adiabatic process]. !':1S101
= @].
!':1G = !':1(H  TS) = !':1H  S!':1T [no change in entropy). Although the change in entropy is known to be zero, the entropy itself is not known, so !':1G is
I indeterminate I·
THE SECOND LAW
63
Step 3
These quantities may be calculated in the same manner as for Step 1 or more easily as follows. f!.U = f!.H = @] [isothermal). Srev
Tc 300K qc = I   [3 .10] = I    = O.Soo = 1+  [3 .9] . Th 600K qh
qc = O.SOOqh = (O.Soo ) x (1I.SkJ) = S.74kJ. qc = ls.74kJ I,
f!.S =
qrev T
w = qc = IS.74kJ I· 3
[isothermal] =
I
'I
S .74 x 10 J 300K = 19.1 JK.
f!.S (sur) = f!.S(system) = +19. 1 J K'.
f!.G = f!.H  Tf!.S = 0  (3OO K) x (19.1 JK  ' ) = I +II.SkJ I. Step 4 f!.U and f!.H are the negative of their values in Step 2. (Initial and final temperatures reversed.) f!.U
= 1+3.74 kJ I,
f!.H
= I +6.23 kJ ~
q = @][adiabatic].
w = f!.U = I +3.74kJ I·
f!.S = f!.S(sur) = @][reversible adiabatic process] . f!.Stot = @]. Again f!.G = f!.(H  TS ) = f!.H  Sf!.T [no change in entropy] but S is not known, so f!.G is I indeterminate I. Cycle f!.U = f!.H = f!.S = f!.G = @]
[f!.(state function) =
ofor any cycle].
f!.S(sur) = 0 [all reversible processes]. f!.Stot =@]. q(cycle) = ( II .S  S.74) kJ = IS .8 kJ ~ w(cycle) = q(cycle) = IS.8 kJ I. P3.7
~(T)
= ~(298 K) + f!.S .
f!.S =
l
T2
TJ
dT = T
Cp ,m 
lT2(a + TJ
T
I (I I)
c ) T2 b+ 3 dT = a In  + b(T2  T,)  c T T, 2
2'  2' T2
T,
.
64
STUDENT'S SOLUTIONS MANUAL
(a)
S!(373 K)
= (l92.45J K 1mol l ) +
(29.75J K I molI) x In (373) 298
+(25.IOx 1O 3 JK 2 mol 
l
)
x (75.0K)
+(21) x (1.55 x 105 JK  l mol = 1200.7 J K I molI
l
)
I _(298.15)2 I)
x (
(373.15)2
I.
(b)
+ (25.10 x 10 3 J K 2 molI) x (475K) +
(~) 2
x (1.55 x 105 J K Imol  I) x
(~2  ~2) 773 298
= 1232.0J K I mol I I. P3.9
""S
=
nCp,m In
~ + nCp,m In
In the present case, Tf
t
1
= 2(500 K +
[3 .19] [Tf is the final temperature, Tf
250 K)
g)
Sm(T)
= Sm (O) +
~ (Th + Tc) ] .
= 375 K.
_ Tl _ (Th + Tc)2 ( 500 ",,S  nCp.m In    nCp,m In = 3 Th Tc 4ThTc 63.54 g cm
P3.11
=
II x (24.4 J K mol )
loT Cp,mdT  [3 .18]. o
T
From the data, draw up the following table. TIK
10
15
20
25
30
50
Cpom I(J K 2 molI ) T
0.28
0.47
0.540
0.564
0.550
0.428
TI K
70
100
150
200
250
298
Cp,m 1(1 K 2 molI) T
0.333
0.245
0.169
0.129
0.105
0.089
Plot CpomIT against T (Fig. 3.1). This has been done on two scales. The region 0 to 10 K has been l constructed using Cpom = aT 3 , fitted to the point at T = 10K, at which Cp,m = 2.8JK  l mol , 4 3 so a = 2.8 x 10 J K mol I. The area can be determined (primitively) by counting squares. Area A = 38.28J KImo]I .
THE SECO ND LAW
65
0.6 0.5
, "0
0.4
E 'I :,( 0.3 . ~
f.. 0.2 .. E
<..;"
0.1 0 0
20
40
60
80
100
200
300
T /K
Area B up to O°C
P3.13
Figure 3.1
= 25.60J K I mol I; area B up to 25°C = 27.801 K I mol I. Hence
(a) Sm (273 K)
= Sm (O) + 163 .88 J K I mol I I,
(b ) Sm(298 K)
= Sm (O) + 166.08 J K I molI I.
Sm (T) = Sm (O)
+
T
loo
C
dT
~ [ 3 . 1 81.
T
Perform a graphical integration by plotting Cp,m/ T against T and determining the area under the curve. Draw up the foll owing table. (The last two columns come from determining areas under the curves described below.)
TIK 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 80.00 90.00 100.00 110.00 150.00 160.00 170.00 180.00 190.00 200.00
s.;,  s.;,
Cp . m 1 K I mol I
Cp,m/ T 1 K 2 mol I
(0) 1 K I mol I
H~  H~ (O)
0.00 2.09 14.43 36.44 62.55 87.03 111.00 131.40 149.40 165.30 179.60 192 .80 23 7.60 247. 30 256.50 265 .10 273 .00 280.30
0.00 0.2 1 0.72 1. 21 1.56 1.74 1.85 1.88 1.87 1.84 1.80 1.75 1.58 1.55 1.5 1 1.47 1.44 1.40
0.00 0.80 5.6 1 15.60 29.83 46.56 64.62 83.29 102 .07 120.60 138.72 156.42 222 .9 1 238.54 253.79 268 .68 283 .2 1 297. 38
0.00 0.0 1 0.09 0.34 0.85 1.6 1 2.62 3.84 5.26 6.84 8.57 10.44 19.09 2 1.52 24.05 26.66 29.35 32.13
kJ mol I
66
STUDENT'S SOLUTIONS MANUAL
Plot Cp,m/ T against T (Fig. 3.2(a»). Extrapolate to T = 0 using C,l.m = aT 3 fitted to the point at T = 10K, which gives a = 2.09 mJ K 4 mol  I. Determine the area under the graph up to each T and plot Sm against T (Fig. 3.2(b)).
(a)
2.0 ~.........
1.8
,.
1.4
E "I
1.2
2
1.0
1
~
E
S
ci.
I
0.8
1
0.6 0.4 0.2 0.0
~
j _L
(3
:.::
~
j
1.6
I
j
V o
50
150
100
200
Figure 3.2(a)
TIK
(b)
300 250
I
'0
,.:.::
E
200
L
~
>:: 2:
150
'1 100 ~
~ ~
50
o
o
/
/
r
L
¢I E
L
V
~
100
50
T/K
150
200
Figure 3.2(b)
The molar enthalpy is determined in a similar manner from a plot of Cp,m against T by determining the area under the curve (Fig. 3.3)
~ (200 K)  ~ (0)
{ 200 K
=
10
Cp,mdT = 132.1 kJ molI
I·
THE SECOND LAW
67
300 250 I
"0
200
,.:.::E
ISO
(.)"'
100
...., E
50
o
./
/
~
/
V
~
I
/ o
50
100
150
200
TIK
P3.1S
Figure 3.3
The entropy at 200 K is calculated from ~ ( 200K) = ~ (I00K)
+
2°O K
1
C dT ~.
lOOK
T
The integrand may be evaluated at each of the data points ; the transformed data appear below. The numerical integration can be carried out by a standard procedure such as the trapezoid rule (taking the integral within any interval as the mean value of the integrand times the length of the interval). Programs for performing this integration are readily available for personal computers. Many graphing calculators will also perform this numerical integration. TIK
100
120
140
150
160
180
200
Cp,m / (J K I mol  I)
23.00
23.74
24.25
24.44
24.61
24.89
25.11
Cp ,m / (JK 2 mol  l )
0.230
0.1978 0.1732 0.1629 0.1538 0.1383 0.1256
T
Integration by the trapezoid rule yields
~ (200 K) =
(29 .79 + 16.81 ) J K Imol I
= 146.60 J K 1 mol 1 I.
Taking Cp,m constant yields ~ (2ooK)
= ~ (I00K) + Cp,mln (2ooK/ looK) =
[29.79 + 24.441n (2ooK/ loo K)] J K 1mol 1 = 146.60 J K 1 mol 1 I.
The difference is slight. P3.17
The GibbsHelmholtz equation [3 .52] may be recast into an analogous equation involving f':.G and since
= ( af':.G) aT p
(aGf) _(aGi) aT aT p
p
f':.H,
68
STUDENT'S SOLUTIONS MANUAL
and I"1H Thus
,
= Hr 
Hi.
c? )
I"1 r a(aT T
p
T  Te '
r 
For the reaction
500 (a) At 500 K, r = = 1.678, 298 so
I"1 r G'(500K)
= {( 1.678) x 2 x
(16.45 )
+ (I 
1.678) x 2 x (46. 11 )} kJmol 
1
1=7kJmOI 1 1· (b) At 1000 K, r so
1000 298
=
I"1 r G'( 1000 K)

= 3.356,
= {(3.356)
x 2 x (16.45)
+ (1 
3.356) x 2 x (46. 11 )}kJ mol 
1
I = +107 kJ mol  1 I. Solutions to theoretical problems P3.19
The isotherms correspond to T = constant, and the reversibly traversed adiabats correspond to S = constant. Thus we can represent the cycle as in Fig. 3.4. In this figure, paths I, 2, 3, and 4 correspond to the four stages of the Carnot cycle listed in the text following eqn 3.6. The area within the rectangle is
(isothermal expansion from VA to VB , stage 1).
THE SECOND LAW
.. f .
... ,A
Entropy
But, w(cycle)
69
Figure 3.4
Th  Tc) VB = £qh = (   nRTh In VB [text Fig. 3.6] = nR(Th  Tc) In. ~ ~ ~
Therefore, the area is equal to the net work done in the cycle. P3.21
The thermodynamic temperature scale defines a temperature P (where the superscript a is used to distinguish this absolute thermodynamic temperature from the perfect gas temperature) in terms of the reversible heat flows of a heat engine operating between it and an arbitrary fixed temperature T~ (eqn3 .11)
where the efficiency of a heat engine is given in terms of work and heat flows: f
=~
[3.8]
= I + qc [3.9]. qh
%
The problem asks us to show that the thermodynamic and perfect gas temperatures differ by at most a constant numerical factor. That amounts to showing that
where the superscipt g indicates the perfect gas temperature defined by the perfect gas law. The subscripts c and h represent two reservoirs between which one might run a heat engine and whose temperatures one might characterize using either temperature scale. Justification 3.1 relates the ratio of two perfectgas temperatures to reversible isothermal heat flows in a Carnot cycle run between the two temperatures. T,g
qh =
qc
_% T~
[3.7, Justification 3.1]
so
The corresponding ratio of thermodynamic temperatures is:
T ; T~ =
I
f
rev
= I
(IWI) 1  q;
re v
=
(qc ) qh
rev·
70
STUDENT'S SOLUTIONS MANUAL
As Section 3.2(b) shows, the efficiency of any reversible heat engine (including one that uses a perfect gas as a working fluid ) is the same, and therefore the ratio of heat flows to the two reservoirs is the same. That is, the ratio qc is the same in the expression for the perfectgas temperature ratio and the %
thermodynamic ratio; since the two ratios are equal to the same heat ratio, they are equal to each other. The constant numerical factor becomes I if Th and T~ are both assigned the same value, say 273.16 at the triple point of water.
P3.23
(avas ) = (a aT p
)
v
T
[Table 3.5].
(a) For a van der Waals gas
Hence,
(:~)T =G~)v =1
VmR_b
I·
(b) For a Dieterici gas RTe a/ RTVm p=
Vm b
R
(I + _a_) ea/RVmT RV",T
For an isothermal expansion, 6.S
so we can simply compare (
= { Vr dS = ( Vr
as)
av
lv;
l v;
(as) av
dV , T
expressions for the three gases. For a perfect gas, T
as ) is certainly greater for a van der Waals gas than for a perfect gas, for the denominator is smaller ( av T
for the van der Waals gas. To compare the van der Waals gas to the Dieterici gas, we assume that both have the same parameter b. (That is reasonable, for b is an excluded volume in both equations of state.) In that case,
R
(I + _a_) ea/RVmT =(as) RVmT Vm  b
Now notice that the additional factor in ( is always less than I. Clearly (I
+ x)e
av
as )
av X
(1+_a_)ea /RVmT . T ,vdW
has the form (1
RVmT
+ x)e
X
,
where x > O. This factor
T,Die
< I for large x, for then the exponential dominates. But
THE SECOND LAW
71
( I +x)e X < 1 even for small x, as can be seen by using the power series expansion for the exponential:
( l +x)(lx+x? / 2+ . . . ) =Ix 2 /2+ .. · So
(as) av
<
T,Die
(as)
.
. To summanze, for

av
T ,vdW
isothermal expansions:
I Ll.Svdw
> Ll.SDie I and
1Ll.Svdw > Ll.Sperfecl I·
The comparison between a perfect gas and a Dieterici gas depends on particular values of the constants a and b and on the physical conditions. P3.2S
H
== U + PV.
dH = dU + pdV + V dp = TdS  pdV [3.43] + pdV + V dp = TdS + V dp.
Since H is a state function , dH is exact, and it follows that
(aasH)
T
and
p
==
Similarly, A
U  TS.
dA = dU  TdS  SdT = TdS  pdV [3.43] TdS  SdT = pdV  SdT.
Since dA is exact,
P3.27
(avas) = (a aT p
T
dS =
S=
(~) av
)
v
7rT
f f
=T
V
dV [constant temperature] = nR dV = nRd In V .
V
T
dS =
nR d In V.
S = nR In V + constant P3.29
nR
[Maxwell relation] ;
or
(aaTp ) v  p [3.48],
Sex R In V.
72
STUDENT'S SOLUTIONS MANUAL
P=
RT
BRT
V + \12 m
Hence, lfT
[first two terms of the vi rial expansion, 1.19].
m
= R~2
(aB) "" R~2 !'!.B . aT v VOl!'!.T
Vm
Since lfT represents a (usually) small devi ation from perfect gas behaviour, we may approximate VOl'
RT Vm " " 
p
From the data !'!.B
{( IS .6)  (28.0) } cm 3 mol 
=
1
=
+12.4 cm 3 mol
1
Hence, (a)
(b)
ex. p2;
lfT
so at p
=
10.0 atm,
lfT
= 10.30 atm I.
In (a) Jrr is 0.3 per cent of p ; in (b) it is 3 per cent. Hence at these pressures the approximation
COMMENT.
for Vm is justified . At 100 atm it would not be.
Question. How would you obtain a reliable estimate of lfT for argon at 100 atm? P3.31
lfT
p
=T
(aaT p
= nRT V  nb
ap ) T(aT v
Hence, llf T lfT c>
0 as p
)
v
 p
[3.48].
x ean/RTV[Table l.7].
 all/RTV nap = nRT   x e an/RTV + na x nRT  x e =p+ .
V  nb
RTV
V  nb
RTV
= ~ I. c>
0, V
c> 00 ,
a
c>
0, and T
c> 00 .
The fact that lfT > 0 ( because a > 0) is consistent
with a representing attractive contributions, since it implies that
(au) av
.

> 0 and the Internal energy T
rises as the gas expands (so decreasing the average attractive interactions).
THE SECOND LAW
P3.33
If S
73
= S(T,p)
then dS
=
( as ) dT+ ( as) dp , aT p ap T
=T (
Td S
as ) dT aT p
=_ T
dp . T
x C
p
[
(~~)
p
= T,
Problem 3.25] ,
( av ) [Maxwell relation]. aT P
= CpdT 
Hence, TdS
as ) ap
(~ ) ( aH)  ~ aH p aT p  T
Use ( as) _ aT p 
as ) ( ap
+T (
G~) p dp =
T
1
CpdT  aTVdp I·
For reversible, isothermal compression, TdS = dqrev and dT = 0; hence dqrev qrev
= aTVdp .
=
l
Pf
aTV dp
= laTV t:..p I[a and Vassumed constant].
Pi
For mercury qrev = (1. 82 P3.35
In!/>
=
10 4 K I) x (273 K) x (1.00
X
(Z;
foP
104 m 3) x (1.0 x 108 Pa)
= 10.50 k.J I·
I) dp [3 .60],
8
C
Vm
Vm
Z=I++2 8 with 8 ' =  , RT
X
C' =
,, 2
=1+8p+Cp
+ ...
C8 2 R2T 2 [Problem U8].
Z I , ,   =8 +Cp+ ···. p
{P
{P
1
8p
Therefore, In!/>
= 10 8' dp + 10 C'pdp + .. . = B'p + "2 C'p2 + . . . = lIT +
8p For argon, RT
=
(C  8 2)p2
2R2T2
+ .. .
(21.1 3 x 10 3 dm 3 mol  l ) x (l.00atm) _ 4 2 3 I I  9.43 x 10 , (8.206 x 10 dm atm K mol ) x (273 K)
(C  8 2)p2 _ {( 1.054 x 10 3 dm 6 mol  2)  (2 1.13 x 10 3 dm 3 moll)2) x (l.OOatm)2 2R2T2 (2) x {(8.206 x 10 2 dm 3 atm K I molI ) x (273 K) }2
=
6.05
X
10 7
74
STUDENT'S SOLUTIONS MANUAL
Therefore, In¢ = (9.43 x 10 4 )
+ (6.05
x 10 7 ) = 9.42 x 104 ; ¢ = 0.9991.
Hence,f = (1.00atm) x (0.9991) = 10.9991 atm
I.
Solutions to applications P3.37
6. r G [3.38] .
wadd, max =
6. r G"(37°C) = r 6. rG"(Tc)
= (
+ (1 r)6. rfF(Tc)
310K ) x (6333kJmol l ) 298.15K
[problem 3.17, r =
+ (1
~J
310K ) x (5797kJmol l ) 298.l5K
= 6354kJmol l . The difference is 6. r G"'(37°C)  6. r G"'(Tc) = {6354  (6333») kJmol 1 = 121 kJ molI I. Therefore an
additional 21 kJ mol I of nonexpansion work may be done at the higher temperature. As shown by Problem 3.16, increasing the temperature does not necessarily increase the
COMMENT.
maximum nonexpansion work. The relative magnitude of t:. rGB and t:. rW is the determining factor.
P3.39
The relative increase in water vapor in the atmosphere at constant relative humidity is the same as the relative increase in the equilibrium vapor pressure of water. Examination of the molar Gibbs function will help us estimate this increase. At equilibrium, the vapor and liquid have the same molar Gibbs function. So at the current temperature Gm,liq(To) = Gm,vap(TO)
so
~, liq(To) = ~,vap(TO)
+ RTo In PO,
where the subscript 0 refers to the current equilibrium and p is the pressure divided by the standard pressure. The Gibbs function changes with temperature as follows
and similarly for the vapor. Thus, at the higher temperature
Solving both of these expressions for C:,liq (To)  ~,vap (To) and equating them leads to (6.T)(s;rq  S:p)
+ R(To + 6.T) Inp =
RTo Inpo.
Isolating p leads to Inp
(6.T)(S!p 

p=exp (
s;rq) +ToInpo 
R(To+6.T)
To+6.T'
s;rq»)
(6.T)(S!p R(To+I'!T)
(To /( To + t.T) )
Po
.
THE SECOND LAW
7S
1 I _ (2.0K) x (188 .8369.91 ) Jmol K ) 0089 (290K / (290+2.0)K) SO p  exp I x ( . 1 ) , (8.3145Jmol KI) x (290+2.0)K
= 0.0214 which represents a 113 per cent 1increase.
p
P3.41
The change in the Helmholtz energy equals the maximum work associated with stretching the polymer. Then
d WmaJ<
= cIA =
Jdt.
For stretching at constant T
Assuming that (aU / al)T =
J=T
o(valid for rubbers)
(asat ) =T (a) at T
T
{3kBt2 + C} 2Na 2
=T{_3kBt} =_(3kBT)t. Na 2 Na 2
This tensile force has the Hooke's law formJ P3.43
= kHt with kH = 3kBT /Na 2.
The Otto cycle is represented in Fig. 3.5. Assume one mole of air.
P
C (2) B
D
(4)
v Figure 3.5 c = Iwlcycle [3 .8].
Iq21 Wcycle q2 C
=
=
=
WI + W3
~U2
=
=
~UI
+
~U3
[ql
= q3 = 0] = CV(TB
Cv(Te  TB)·
ITB  TA + To  Te l ITeTBI
=
1 _ (To  TA ).
Te TB
 TA) + Cv(To  Tc) [2.27] .
76
STUDENT'S SOLUTIONS MANUAL
We know that
T
~ =
and
(~)I /C
.£ Vo
Te
[2.28a].
. TA To TATe SInce VB = Ve and VA = Vo ,  =  , orTo =   . TB Te TB
7 5 2 Given that Cp,m = "2R, we have CV ,m = "2R [2.26] and c = S. V For ~ VB
=
10,
£
=
I 
( I
10
)2/5=10.471.
~Sl = ~S3 = ~Ssur. 1 = ~Ssur,3 =
@]
[adiabatic reversible steps].
~S2 = CV ,m In (;:). At constant volume (;:) =
~S2 =
G)
~Ssur,2
=
X
e:)
= 5.0.
(8.3 14JK 1 molI) x (In 5.0) =1+33JK I I·
~S2
= 133 J K I I.
llS4=llS2 [Te = TB]=1_33JK I. ' To TA llSsur,4
P3.45
= llS4 = 1+33 J K I I·
1.00
11.00 1
In case (a), the electric heater converts kJ of electrical energy into heat, providing kJ of energy as heat to the room. (The Second Law places no restriction on the complete conversion of work to heatonly on the reverse process.) In case (b), we want to find the heat deposited in the room Iqhl :
Iqhl = Iqcl + Iwl
so
where
Iqcl = Iwl

Tc Th  Tc
c = [ImpactI3.I]
_ ~ _ 1.00 kJ x 260 K = 4 kJ (295 _ 260)K 7. .
Iqcl  Th _ Tc 
THE SECOND LAW
77
The heat transferred to the room is Iqlrl = Iqcl + Iwl = 7.4 kJ + 1.00 kJ = 18.4 kJ I. Most of the thermal energy the heat pump deposits into the room comes from outdoors. Difficult as it is to believe on a cold winter day, the intensity of thermal energy (that is, the absolute temperature) outdoors is a substantial fraction ofthat indoors. The work put into the heat pump is not simply converted to heat, but is ' leveraged' to transfer additional heat from outdoors.
Physical transformations of pure substances
Answers to discussion questions 04.1
Consider two phases of a system, labeled ct and 13. The phase with the lower chemical potential under the given set of conditions is the more stable phase. First consider the variation of J.t with temperature at a fixed pressure. We have
= ( aJ.ta) aT p
S
a
and
= Sf3. ( aJ.tfJ) aT p
Therefore, if Sf3 is larger in magnitude than Sa , the 13 phase will be favored over the ct phase as temperature increases because its chemical potential decreases more rapidly with temperature than for the ct phase. We also have
aJ.ta) m,a ( ap T v
and
(
aJ.tf3) ap T = Vm .f3.
Therefore, if Vm ,a is larger than V rn ,f3 the 13 phase will be favored over the ct phase as pressure increases because the chemical potential of the 13 phase will not increase as rapidly with pressure as that of the ct phase. See Example 4.1. 04.3
Refer to Fig. 4.1 below and Fig. 4.5 in the text. Starting at point A and continuing clockwise on path peT) toward point B, we see a gaseous phase only within the container with water at pressures and temperatures peT). Upon reaching point B on the vapor pressure curve, liquid appears on the bottom of
the container and a phase boundary or meniscus is evident between the liquid and less dense gas above it. The liquid and gaseous phases are at equilibrium at this point. Proceeding clockwise away from the vapor pressure curve the meniscus disappears and the system becomes wholly liquid. Continuing along peT) to point C at the critical temperature no abrupt changes are observed in the isotropic fluid. Before point C is reached, it is possible to return to the vapor pressure curve and a liquidgas equilibrium by reducing the pressure isothermally. Continuing clockwise from point C along path peT) back to point A, no phase boundary is observed even though we now consider the water to have returned to the gaseous state. Additionally, if the pressure is isothermally reduced at any point after point C, it is impossible to return to a Liquidgas equilibrium. When the path peT) is chosen to be very close to the critical point, the water appears opaque. At near critical conditions, densities and refractive indices of both the liquid and gas phases are nearly identical. Furthermore, molecular fluctuations cause spatial variations of densities and refractive indices on a scale large enough to strongly scatter visible light. This is called critical opalescence.
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
79
Vapor pressure curve of water Explorati on path peT)
22 1.2 bar p
647.4 K
T
04.5
Figure 4.1
Vapor pressure curve of water
The supercritical fluid extractor consists of a pump to pressurize the solvent (e.g. C02), an oven with extraction vessel, and a trapping vessel. Extractions are performed dynamically or statically. Supercritical fluid flows continuously through the sample within the extraction vessel when operating in dynamic mode. Analytes extracted into the fluid are released through a pressuremaintaining restrictor into a trapping vessel. In static mode the supercritical fluid circulates repetitively through the extraction vessel until being released into the trapping vessel after a period of time. Supercritical carbon dioxide volatilizes when decompression occurs upon release into the trapping vessel.
Advantages
Disadvantages
Current uses
Dissolving power of SCF can be adjusted with selection of T and p
Elevated pressures are required and the necessary apparatus expensive
Extraction of caffeine, fatty acids, spices, aromas, flavors, and biological materials from natural sources
Select SCFs are inexpensive and nontoxic. They reduce pollution
Cost may prohibit large scale applications
Extraction of toxic salts (with a suitable chelation agent) and organics from contaminated water
Thermally unstable analytes may be extracted at low temperature
Modifiers like methanol (110%) may be required to increase solvent polarity
Extraction of herbicides from soil
The volatility of SCC02 makes it easy to isolate analyte
SCC02 is toxic to whole cells in biological applications (C02 is not toxic to the environment)
SCH20 oxidation of toxic, intractable organic waste during water treatment
SCFs have high diffusion rates, low viscosity, and low surface tension
Synthetic chemistry, polymer synthesis and crystallization, textile processing
02 and H2 are completely miscible with SCC02. This reduces mUltiphase reaction problems
Heterogeneous catalysis for green chemistry processes
80
04.7
STUDENT'S SOLUTIONS MANUAL
Firstorder phase transitions show discontinuities in the first derivative ofthe Gibbs energy with respect to temperature. They are recognized by finite discontinuities in plots of H, V, S, and V against temperature and by an infinite discontinuity in Cpo Secondorder phase transitions show discontinuities in the second derivatives of the Gibbs energy with respect to temperature, but the first derivatives are continuous. The secondorder transitions are recognized by kinks in plots of H, V, S, and V against temperature, but most easily by a finite discontinuity in a plot of Cp against temperature. A Atransition shows characteristics of both first and secondorder transitions and, hence, is difficult to classify by the Ehrenfest scheme. It resembles a firstorder transition in a plot of Cp against T , but appears to be a higherorder transition with respect to other properties. At the molecular level firstorder transitions are associated with discontinuous changes in the interaction energies between the atoms or molecules constituting the system and in the volume they occupy. One kind of secondorder transition may involve only a continuous change in the arrangement of the atoms from one crystal structure (symmetry) to another while preserving their orderly arrangement. In one kind of 8transition, called an orderdisorder transition, randomness is introduced into the atomic arrangement. See Figs. 4.18 and 4.19 of the text.
Solutions to exercises E4.1(b)
Assume vapor is a perfect gas and p* /:;.vapH In=+p R
/:;. vapH
is independent of temperature
( ~ ~) T
T*
1 1 R p* =+InT
T*
p
/:;. vapH
1 8.314J K I mol I (S80) =+ xln  ' 293.2K 32.7 x 10 3 Jmol 1 66.0 = 3.378 x 10 3 K I T =
3.378 x 10 3 KI
= 296 K = 123 °C 1
E4.2(b)
assuming
/:;.fusS
_
and
/:;.fus V
independent of temperature.
3
/:;.fus S (lS2.6cm mol
= (l0.6crn 3 rnol 
I _
5 II x (1.2 x 106 Pa)  (1.01 x 10 Pa) 142.0crn rno) 429.26K427. ISK
l
(~)
)
x
3
x (S.21 x 105PaKI)
= S.S2Parn 3 K I molI =IS.SJKIrnol II
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
I1rusH
=
Trl1S
=
(427.15 K) x (5 .521 K I molI)
= 12.4 kJ molII E4.3(b)
Use
f
d Inp
=
f 11~;2H
= constant 
In p
dT
6. vapH RT
Terms with I / T dependence must be equal, so _ I1 vapH
3036.8 K T/K
I1vapH
RT
= (3036.8 K)R = (8 .3141 K I molI)
x (3036.8 K)
= 125.25 kJ molII E4.4(b)
(a)
logp
= constant 
I1vapH /(RT(2.303))
Thus
I1 vapH
x (8.3141 K I molI) x (2.303)
= (1625 K)
= 131.11 kJ mol  I I
(b) Normal boiling point corresponds to p log(760) 1625 T/K
T/ K Tb
E4.5(b)
6. T
= 8.750 
= 8.750 =
1625 T/ K
log(760)
1625 8.750  log(760)
11 rusS
x I1p
=
Tr 6. f.US V x I1p 6.rusH
[Tr
= 3.65 + 273 .15 = 269.50 K]
6.T
=
I1rusH
x 11 ( I ) P
106 K Pari mol ) x (M) x (+0.01899cm 3 / g) x
(+ 5.889 x 10 2 K Pa m 3 rig I mol)M
= (46.07 g mol  I)
= 269.50K +
=
2.71K
= 1272 K 1
(~) 6 3 10 cm
(+ 5.889 x 10 2 KgI mol)M
x (+5 .889 x 10 2 KgI mol)
= +2.71K Tr
Tr I1pM =
(269.50 K) x (99.9 MPa)M x ( I _ 1 ) 8.68kJmol  1 0.789gcm 3 0.801 gcm 3
= (3 . 1017 x
I1T

= 276.87
= 1276.9 K 1
I1rus =V
=
= 1.000 atm = 760 Torr
81
82 E4.6(b)
STUDENT'S SOLUTI ONS MANUAL
dm dn  = dl dt
X
MH, O
dn
dq / dt
dt
!:!"vapH

q where n =  !:!" vapH
(0.87 x 103 Wm 2) x ( 104 m2) 44.0 x 103 J molI = 197.7 J s I rl mol = 200 mol sI
dm I I  = (197.7mol s ) x (l8.02gmol  ) dt
=13 .6kgs  1 1 E4.7(b)
The vapor pressure of ice at 5 °C is 0.40 kPa. Therefore, the frost will sublime. A partial pressure of 0.40 kPa or more will ensure that the frost remains.
E4.8(b)
(3) According to Trouton 's rule (Section 3.3(b), eqn 3.16)
(b) Use the ClausiusClapeyron equation [Exercise 4.8(a)]
At T2 = 342.2 K, P2 = 1.000 atm; thus at 25 °C In I = P
29.T x 103JmOII) ( 8. 314J K I mol I
X
(I 298.2 K

I) = 1.509 342.2 K

PI = 10.22 atm 1= 168 Torr
At60 °C, In
PI
= 
29.T x 103JmOI  I) x (1 I) = 0.276   ( 8.3 14JKImol 1 333.2K 342.2K
PI = 10.76 atm 1= 576Torr E4.9(b)
!:!,.T=Tfus(lOMPa) Tfus(O. IMPa) = Tfu s!:!"pM !:!,. !:!,. fusH
(I) 
p
[See Exercise4.5 (b)]
!:!"fusH = 6.01kJmol  1
!:!,.T=
(273.15K) x (9.9 x 106 Pa) x (18 x 1O3 kgmOI  I) } { 6.01 x 103 J mol I x {9.98
X
1~2 kg m3 
9.15 x
1~2 kg m3 }
= 0.74K Tfus(lO MPa) = 273.15 K  0.74 K = 1272.41 K 1
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
E4.10(b)
fl vapH = fl vapV + fl vap(PV) fl vapH = 43.5 kJ mol  I flvap(P V) = pfl vap V = p(Vgas  Vliq) = pVgas = RT [per mole, perfect gas] fl vap(PV) = (8.3 141 K I molI) x (352 K) = 29271 mol  I . flvap(P V) 2.927kJmol 1 FractIOn = = I fl vap H 43.5 kJ mol= 16.73 x 10 2 1= 6.73 percent
Solutions to problems Solutions to numerical problems P4.1
At the triple point, T3, the vapor pressures of liquid and solid are equal; hence 1425.7 K =8.3186; T3 = 1196.oK I. T3 T3 1871.2 K log(P3/ Torr) = 196.0 K + 10.5916 = 1.0447; P3 = 111.1 Torr I·
10.5916
P4.3
(a)
(b)
1871.2 K
dp fl vapS fl vapH  =  = [4.6, Clapeyron equation] dT fl vapV hfl vapV I 3 14.4 x 10 1 molI II = = +556 kPa K(180 K) x (14.5 X 10 3  1.15 x 10 4 ) m3 mol  I .
$]
fl ~H ~ = 2 x p [ 4.II,withdlnp=dT RT P =
(14.4 x 103 1 mol  I) x (1.013 x 105 Pa) (8.3141 K I mol I) x (180 K) 2
The percentage error is 12.5 per cent P4.5
(a)
_(BILBp(S» ) ( BIL(1») Bp T
I = +5 .42 kPa K .
I.
= Vm (I)  Vm (s) [4.13] = Mfl = (18.02 g molI) x ( = 11.63 cm 3 mol  I
(b)
( BIL(g» ) Bp
_ T
(BIL(I») Bp
(~) P
T
= Vm (g) 
I _ _ _1_.".) 1.000 g cm 3 0.917 g cm  3
I.
Vm(l)
T
= (18.02 g molI) x ( = 1+30.1 dm 3 mol  I
I.
I _ _ _ _1:_0) 0.598 g dm 3 0.958 x 103 g dm 3
83
84
STUDENT'S SOLUTIONS MANUAL
At 1.0 atm and 100o e, JL(I) (as in Problem 4.4)
= JL(g); therefore, at 1.2 atm and 100o e , JL(g) 
JL(l) "" b. Vvapb.p =
(30.1 x 10 3 m 3 molI) x (0.2) x (1.013 x 105 Pa) "" 1+ 0.6 kJ mol I I. Since JL (g) > JL (I), the gas tends to condense into a liquid. P4.7
The amount (moles) of water evaporated is ng = The heat leaving the water is q
AT __
u
_
2_
RT
= nb. vapH.
The temperature change of the water is b. T
Th eref ore,
PH oV
= ~, nCp,m
n
= amount of liquid water.
P H20 V b.vapH RTnCp,m  (3.17 kPa) x (50.0 dm 3 ) x (44.0 x 103 J mol I) (8.314kPadm 3 K I molI) x (298.15 K) x (75.5 J K  I molI) x
=
C8.0~5g0~OI
I)
2.7K.
The final temperature will be about 122°e I. P4.9
(a) Follow the procedure in Problem 4.8, but note that Tb
=
1
227.5°e
1
is obvious from the data.
(b) Draw up the following table.
Or e TI K 1000 KI T Inpl Torr
57.4
100.4
133.0
157.3
203 .5
227.5
330.6 3.02 0.00
373.6 2.68 2.30
406.2 2.46 3.69
430.5 2.32 4.61
476.7 2.10 5.99
500.7 2.00 6.63
The points are plotted in Fig. 4.2. The slope is 3
6.4 x 10 K, so implying that b.vapH P4.11
b.vapH R
= 6.4
3
x 10 K,
= 1+53 kJ mol 1 I.
(a) The phase diagram is shown in Fig. 4.3 . (b) The standard melting point is the temperature at which solid and liquid are in equilibrium at 1 bar. That temperature can be found by solving the equation of the solid liquid coexistence curve for the temperature 1 = P3/bar + 1000(5.60 + 11 .727x)x.
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
85
i:::::·;··:::::I::::::f::::::r:::::I::::::r:::::r::::r:::::I::::::f:::::j
6
i·· ·· · ·~·· · ··· ; · ·
..;. ...... ! .......; ...... ~ .......: ...... .;. ... .. .;...... .; ..... ~
! 41·····+·····1······j······: · ····[·····+···+····+·····1··· ·+···1
;'IFlll l iiF!l , o :......L. ... L .... i. ....L .....~ ...... L.. ....t. ..... L.....L .... : .... J 2.0
2.2
2.4
2.6
2.8
3.0
3
Figure 4.2
(10 /T)K
2
. . · ·:i· · · · ~· ···+·····
o ::::::!::::::! ::::::;::: ·r~~;~··r:··::r····;
j
i
2 4
:::~~:;J:::::I::: : f : : t:· ·:f···::F::t:::l::::r~: l: : : : ::::::C:::L:::::::: :r·.c::L::L::L::t:::::L:::L:::
6 ••
S
• Liquid...Vapor • SolidLiquid
iF.:•.i • •tJ.·• FIFEL ~
1
o
100
;
;
l
200
j.
~
300 1}K
.~
400
l
.;
l
500
600
Figure 4.3
So 11727x2 + 5600x + (4.362 x 10 7

I)
= o.
The quadratic formula yields 5600 ± 1(5600)2  4 (11727) x (I)} 1/ 2 X =
1 ±
2 (11 727)
/I + (4(l1727)/56002)}1 / 2 2 ((11727)/5600)
The square root is rewritten to make it clear that the square root is of the form {I a
«
I; thus the numerator is approximately I
+ (1 + ~a )
~a,
+ a} 1/2,
with
and the whole expression
reduces to x ;;:,; 1/5600
=
1.79 x 104 .
Thus, the melting point is T
=
(I +x) T3 = ( 1.000179) x (178.15 K)
= 1178.18 K I.
(c) The standard boiling point is the temperature at which the liquid and vapor are in equilibrium at
I bar. That temperature can be found by solving the equation of the liquidvapor coexistence curve
86
STUDENT' S SOLUTIONS MANUAL
(d) The slope of the liquidvapor coexistence curve is given by !:;. vapHG T!:;. vapVG
dp dT so
!:;. vapW
=
dp dT
T !:;.vap ~ 
The slope can be obtained by differentiating the equation for the coexistence curve. dp dT
d In p dy
d In p
= P;rr = Pd.Y dT '
2dp = (10.413 dT y
x
15.996 + 2(14.015)y  3(5.0120)i

 (1.70) x (4.7224) x (1 y )070)
(:J.
At the boiling point, y = 0.6458, so dp = 2.851 x 10 2 bar K 1 = 2.851 kPa KdT
1
3
and
!:;. vapW
= (383.6 K) x
30.3  0.12) dm mOlI) ( _ I) 3 x 2.851 kPa K ( 1000 dm m 3
= 133.0 kJ mo]  I I.
Solutions to theoretical problems P4.13
a!:;'G) =(aGfJ ) ( ap T ap
T
_( aCet ) ap
=VfJ Vet .
T
Therefore, if VfJ = Vet , !:;.C is independent of pressure. In general, VfJ f= Vet , so that !:;'C is nonzero, though small, since VfJ  Vet is small. P4.1S
Amount of gas bubbled through liquid = pV
RT
(p
=initial pressure of gas and emerging gaseous mixture). .
Amount of vapor earned away
m = .
M
. f· . mj M M oIe f ractIOn 0 vapor III gaseous mixture =       (m j M)
.
Partial pressure of vapor = p =
mj M (m j M)
+ (PV j RT)
+ (PV j RT)
xp =
p (mRT / PVM)
mPA
= , (mRT j PVM) + 1 rnA + 1
RT A = . PVM
For geraniol, M = 154.2 g mol  I, T = 383 K, V = 5.00 dm 3 ,p = 1.00 atm, and m = 0.32 g, so A=
(8.206
X
10 2 dm 3 atmK
(1.00 atm) x (5 .00
dm3 )
1
mol I) x (383 K)
x (154.2 x
10 3
kg molI)
I =40.76kg .
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
87
Therefore p
P4.17
=
(0.32 x 10 3 kg) x (760 Torr) x (40.76 kgI) (0.32 x 103 kg) x (40.76kg')
+1
=
I98 Torr I .
.
In each phase the slopes are given by
= Sm [4. 1] . ( al1) aT p The curvatures of the graphs of 11. against T are given by 2
a 11) ( aT2
p
td
=  (aSm) aT = T p
x C
p,m [Problem 3.26].
Since Cp,m is necessarily positive, the curvatures in ail states of matter are necessarily negative. Cp,m is often largest for the liquid state, though not always, but it is the ratio Cp,m/T that determines the magnitude of the curvature, so no precise answer can be given for the state with the greatest curvature. It depends upon the substance. P4.19
Sm
= Sm(T,p).
dSm
= ( aSm ) aT
p
dT
+ (aSm) ap
dp. T
asm ) Cp ( aT p = T [Problem 3.26] , dqrev Cs
= T dSm = Cp dT 
= ( aq ) = Cp aT s
m) (aS ap
=  (aVm) T
aT
p
[Maxwell relation].
aVm) T(dp. aT p
p TVma ( a ) aT s
= Cp aVm x
trs
!:1 H [4.7]. !:1 trs V
Solutions to applications P4.21
(a) The Dieterici equation of state is purported to have good accuracy near the critical point. It does fail badly at high densities where V m begins to approach the value of the Dieterici coefficient b. We will use it to derive a practical equation for the computations.
Substitution of the Dieterici equation of state derivative (apr/aTr )vr reduced form of eqn 3.48 gives
= (2 + Tr Vr)pr/T? Vr into the
88
STUDENT'S SOLUTIONS MANUAL
Integration along the isotherm T, from an infinite volume to V, yields the practical computational equation.
I'1.U,(T" V,)
=_
2 (T,
1
00
p,
Trconstant
~)
" , dV,. T, V,
The integration is performed with mathematical software. (b) See Fig. 4.4(a). (c) 8(T" V,) = JPcI'1.Ur/Vr where Pc = 72.9 atm. Carbon dioxide should have solvent properties similar to liquid carbon tetrachloride (8 when the reduced pressure is in the approximate range 10.85 to 0.90 when T,
(a)
4 l;U r
\ Tr =1
5
Tr= 1.2
T,= 1.5
6
7 1.5
0.5
(b)
2
2.5 p,
3
3.5
4
4.5
Solubility parameter of carbon dioxide
20
15
.L atm!!>
10
5 0.75
0.8
0.9
0.85 Pr
8
~
9)
= 1 I. See Fig. 4.4(b).
2
3
~
0.95
Figure 4.4
PHYSICAL TRANSFORMATIONS OF PURE SUBSTANCES
P4.23
C (graphite) ;= C (diamond), D.r~
89
= 2.8678 kJ molI at T . T = 25 °C.
We want the pressure at which D.rG = 0; above that pressure the reaction will be spontaneous. Equation 3.50 determines the rate of change of D.rG with p at constant T.
(1)
(aD.rG)
ap
T
= D.rV = (Vo  VG)M
where M is the molar mass of carbon; Vo and VG are the specific volumes of diamond and graphite, respectively. D.rG( T , p) may be expanded in a Taylor series around the pressure p<7 = 100 kPa at T . (2)
D.rG( T ,p)
= D.rLrrof7 ( T ,p(7) + (aD.r +I ( a2D.r~(T,p<7» ) ap2
2
T
G (T,p<7 )) ( ap T p  P(7)
(pp(7)2 +8pp ( (7)3 .
We will neglect the third and higherorder terms ; the derivative of the firstorder term can be calculated with eqn I. An expression for the derivative of the secondorder term can be derived with eqn I . (3)
2
a D.rG) 2= {(avo)  (avG)} M ( ap T ap T ap T
= {VGKT (G)  VOKT (D)} M [2.52 with 2.22].
Calculating the derivatives of eqns I and 2 at T and p<7,
(4) (
(5)
(
g )) T = (0.284  0.444) x (cm3) aD.G(T,p<7 r ap g x (12.01 ~ ) = 1.92cm3 molI ,
(7)) a2 D. r G(T ap2 ,p T = {0.444(3.04 x 3
x ( cm
=
10 8 )  0.284(0.187 x 1O 8 ) }
~aI) x C~~~ g)
1.56 X 10 7 cm 3 (kPa)1 molI.
It is convenient to convert the value of D.r ~ to the units cm 3 kPamol l ,
rof7 = 2.8678 kJ mol _I
D.rLr
(6) D.r~
= 2.8678 x
2 3
l
{8.3 15 x 1O dm barK molI I 8.315J K mol 
1 x
3 3)
(10 Cm   3dm
x
S (IO pa)} 
bar
106 cm 3 kPa mol  I.
Setting X = P  p<7 , eqns 2 and 36 give 2.8678 x 106 cm 3 kPa mol I  (1.92 cm 3 mol l )x
+ (7.80
X
10 8 cm 3 kPa 1 molI )X 2
=0
90
STUDENT'S SOLUTIONS MANUAL
when
~rG(
X p
l' . p) = O. One real root of this equation is
= l.60 X =
106 kPa
=p 
/ ' or
6
2
6
= 1l.60 x
l.60 x 10 kPa  10 kPa
= l.60 x
10 kPa
104 bar!.
Above this pressure the reaction is spontaneous. The other real root is much higher: 2.3 x 107 kPa.
Question. What interpretation might you give to the other real root?
5
Simple mixtures
Answers to discussion questions OS.1
At equilibrium, the chemical potentials of any component in both the liquid and vapor phases must be equal. This is justified by the requirement that, for systems at equilibrium under constant temperature and pressure conditions, with no additional work, 6.G = 0 [see Section 3.5(e) and the answer to Discussion question 3.3] . Here 6.G = 11;(v)  11; ( 1) , for all components, i, of the solution; hence their chemical potentials must be equal in the liquid and vapor phases.
OS.3
All of the colligative properties are a function of the concentration of the solute, which implies that the concentration can be determined by a measurement of these properties. See eqns 5.33, 5.34, 5.36, 5.37, and 5.40. Knowing the mass of the solute in solution then allows for a calculation of its molar mass. For example, the mole fraction of the solute is related to its mass as follows:
The only unknown in this expression is MB which is easily solved for. See Example 5.4 for the details of how molar mass is determined from osmotic pressure. OS.S
A regular solution has excess entropy of zero, but an excess enthalpy that is nonzero and dependent on composition, perhaps in the manner of eqn 5.30. We can think of a regular solution as one in which the different molecules of the solution are distributed randomly, as in an ideal solution, but have different energies of interaction with each other.
Solutions to exercises ES.1(b)
Total volume V Total mass m
n
=
= nA VA + nB VB = n(XA VA + XB VB)
= nAMA + nBMB
1.000kg(103 g/kg) (0.3713) x (241.1 g/mol) + (l  0.3713) x (I98.2g/mol)
= 4.6701 mol
92
STUDENT'S SOLUTIONS MANUAL
v = n(XA VA + XB VB) (4.6701 mol) x [(0.3713) x (188.2 cm 3 molI)
=
+ (1
 0.3713) x (176.14 cm 3 molI)]
= 1843.5 cm 3 1 E5.2(b)
Let A denote water and B ethanol. The total volume of the solution is V
= nA VA + nB VB
We know VB; we need to determine nA and nB in order to solve for VA. Assume we have 100 cm 3 of solution; then the mass is m
= pV = (0.9687 g cm 3 )
=
nB
=
77.496 g 19.374 g
V  nB VB
 =
VA
Check that PB/XB
(PB /XB) / kPa
KB E5.4(b)
= 77.496 g is water.

= 0.4205 mol ethanol
I
100 cm 3

= 18.15 cm
3
(0.4205 mol) x (52.2 cm 3 molI)
= =4.30 mol
nA
E5.3(b)
x (96.87 g)
= 4.30 mol H20
18.02 g mol 1 46.07 g mol
= 96.87 g
= 19.374 g is ethanol and (0.80)
of which (0.20) x (96.87 g)
nA
x (100 cm 3 )
= a constant (KB)
0.010 8.2 x 103
0.015 8.1 x 10 3
= p ix, average value is 18.2 x
0.020 8.3 x 103
103 kPa
I
In Exercise 5.3(b), the Henry's law constant was determined for concentrations expressed in mole fractions. Thus the concentration in molality must be converted to mole fraction .
meA)
= 1000 g,
corresponding to n(A)
=
1000 g 74.1 g mol
Therefore, XB
=
using KB
0.25 mol 0.25 mol
= 8.2

+ 13.50 mol
= 0.0182
x 103 kPa [Exercise 5.3(b)] 3
P = 0.0182 x 8.2 x 10 kPa
= 11.5 x
102 kPa
I
I

= 13.50 mol
nCB)
= 0.25 mol
SIMPLE MIXTURES
E5.5(b)
We assume that the solvent, 2propanol, is ideal and obeys Raoult's law. 49.62 50.00
= p jp* =   = 0.9924
xA(solvent)
MA(C3H80) = 60.096 g mol I nA
=
XA
=
250g 60.096 g molnA nA
nA
nA +nB
+ nB
= nA C~
nB

= 4.1600 mol
I
=XA
1)
= 4.1600 mol (0.9~24  I) = 3.186
=
MB E5.6(b)
8.69 g 2 3.186 x 10 mol
X
2
10 mol
= 273 g molI = 1270 g molI .
1 .
Kr = 6.94 for naphthalene mass ofB
MB=nB
nB = mass of naphthalene · bB
E5.7(b)
t3.T Kr
bB
=
MB
=
t3.T
= KrbB
P
=
(mass of B) x Kr (mass of naphthalene) x t3.T
.,::c:::::=
(5.00 g) x (6.94 K kg molI) (0.250 kg) x (0.780 K )
nv nB = RT
Tr
MB
and
=
1
178 g mol
nB =mass of water
bB
_I 1
nB
Vp
103 kg m 3 (density of solution ~ density of water)
=
t3.T
so
=
t3.T
= Kr nRTp
Kr
= 1.86 K mol I kg
(1.86 K kg molI) x (99 x 103 Pa) (8. 314 JK I molI) x (288K) x (l03kgm 3)
= 10.077 °C I
E5.8(b)
t3.mi X G
= pV(! In!
+! In!)
=
pVln2
3
3
= (100 x 10 Pa) x (250cm )
C;6
: :3) In 2
= 17.3 Pa m3 = 117.3 J 1 t3. . S mix
=
t3. mi x G
T
=
17.3 J 273 K
= 16.34 X .
10 2 J K I 1 .
= 7.7
2
x 10 K
93
94
ES.9(b)
STUDENT'S SOLUTIONS MANUAL
t.mi xG
= nRT LXJ InxJ
[5 . 18]
J
t.mi xS
" InxJ = nR '~xJ
[5.19]
G
t.mix =
T
J
n = 1.00 mol + 1.00 mol = 2.00 mol x (Hex ) = x (Hep) = 0.500
Therefore, t.mix G
= (2.00 mol)
x (8.314 J K 1 molI) x (298 K) x (0.500 In 0.500 + 0.500 In 0.500)
= 13.43 kJ 1 t.mi xS
kJ 1 = +3.43 298 K = +11 .5 J K
I
1
t.mixH for an ideal solution is zero as it is for a solution of perfect gases [7.20]. It can be demonstrated from
ES.10(b)
Benzene and ethyl benzene form nearly ideal solutions, so
To find maximum t.mixS, differentiate with respect to XA and find value of XA at which the derivative is zero. Note that XB
=
1  XA so
dln x use   =: dx x d (t.mixS) dx
= nR(lnxA + I = 0
!n(l  XA)  I)
=  n R l nXA I XA
I
whenxA = 2
Thus the maximum entropy of mixing is attained by mixing equal molar amounts of two components. nB
mB / MB
nE
mE / ME
 = I = mB mE ES.11 (b)
mE x mB
ME
=
MB
106.169
=  = 78 .115
1.3591
= 10.73581
With concentrations expressed in molalities, Henry's law [5.26] becomes PB Solving for b, the molality, we have bB
= PB / K = xPlolal/K
and Plolal
= bBK.
= Palm
SIMPLE MIXTURES
For N2 , K
b
For 0 2, K
b
E5.12(b)
=
bB
1.56 x 105 kPa kg molI [Table 5.1]
=
0.78 x IOl.3kPa 1.56 x 105 kPa kg molI
=
= 10 .51
mmo Ik g  I I
7.92 x 104 kPa kg mol  I [Table 5.1]
=
0.2 1 x 101.3 kPa 7.92 x 104 kPa kg molI
=
95
PB
=
K
= I 0.27 mmol kg I I
2.0 x 101.3 kPa 3.0 1 x 103 kPa kg mol  I
= 0.067 mol kg I
The molality will be about 0.067 mol kg  I and, since molalities and molar concentrations for dilute aqueous solutions are approximately equal, the molar concentration is about 10.067 mol dm 3 1
E5.13(b)
The procedure here is identical to Exercise 5. 13(a) .
In XB
/: ;.rusH = X (IR
T*
I)
 T
[5 .39; B, the solute, is lead]
1)
3
XB
=
=
(I 5.2X 10 J mol  I ) ( 8.314 J K I molI x 600 K  553 K
=
0.0886, implying that XB
= 0.92
n(Pb) , implying that n(Pb) n(Pb) + n(Bi )
. . For I kg of bismuth, n(BI ) =
XBn(Bi)
= I
XB
1000 g I = 4.785 mol 208.98 g mol 
Hence, the amount of lead that dissolves in I kg of bismuth is
n(Pb)
COMMENT.
=
(0.92) x (4. 785 mol) I _ 0.92
= 55 mol ,
~
or
~
It is highly unlikely that a solution of 11 kg of lead and 1 kg of bismuth could in any sense be
considered ideal. The assumptions upon which eqn 5.39 is based are not likely to apply. The answer above must then be considered an order of magnitude result only.
E5.14(b)
Proceed as in Exercise 5. 14(a) . The data are plotted in Figure 5.1, and the slope of the line is 1.78 cm/(mg cm 3 ) = 1.78 cm/(g dm 3 ) = 1.78 x 10 2 m4 kgI .
96
STUDENT'S SOLUTIONS MANUAL
Figure 5.1
Therefore,
M =
ES.1S(b)
ES.16(b)
(1.000
x 10 3
(8.314 J K I molI) x (293.15 K) 1 I 1 = 14.0 kg mol kgm 3 ) x (9.81 m s2) x (1.78 x 10 2 m 4 kgI)
Let A = water and B = solute.
B
= 0.02239 atm = I0.9701 I
aA
= PA 
nA
=
0.920 kg 0.01802 kg molI
XA
=
5105 . 51.05 + 0.506
P~
[5.43]
0.02308 atm
= 51.05 mol
= 0.990
and
and
YA
nB
=
0.122 kg 0.241 kg molI
= 0.506 mol
0.9701 = =~ 0.980 0.990
= Benzene /LB(I) = /LB(I) + RT InxB [5.25, ideal solution] RTlnxB
= (8.314 J K I molI)
x (353 .3 K) x (In 0.30)
= 13536 J molII
Thus, its chemical potential is lowered by this amount. PB
= aBPB
[5.43]
= YBXBPB = (0.93)
x (0.30) x (760 Torr)
= 1212 Torr 1
Question. What is the lowering of the chemical potential in the nonideal solution with Y = 0.93? ES.17(b)
YA
=
PA PA
+ PB
PA
101.3 kPa
PA
= (101.3 kPa)
x (0.314)
PB
= 101.3 kPa 
31.8 kPa
= 0.314 = 31.8 kPa
= 69.5 kPa
SIMPLE MIXTURES
97
31.8kPa ~ =~ 73.0 kPa
PA
aA== p~
PB 69.5 kPa ~ aB== =~
Ps
92.1 kPa
YA = aA = 0.436 = XA 0.220
0.755 0.780
aB
[ill ~
YB===~ XB
ES.18(b)
I
= ~ "2:,;(bdb") z1 [5 .71] = pb/ b£>, b_ / b" =
and for an MpXq salt, b+ / b" I = ~(Pz~
qb/b" , so
+ qz~) b/b"
1= I (K3[Fe (CN)6D
+ I (KCI ) + I (NaBr) = ~ (3 + 32) b (K3 [F:~CN)6D + b(~:I) + b(::Br)
= (6) x (0.040) + (0.030) + (0.050) = 10.320 1 Question. Can you establish that the statement in the comment following the solution to Exercise 5. 18(a) holds for the solution of this exercise? b 1= I (KN03) = b" ( KN0 3) = 0.110
ES.19(b)
Therefore, the ionic strengths of the added salts must be 0.890. (a)
(b)
I (KN0 3)
= 0.890 mol kg I
so b(KN03)
and (0.890 mol kg I) x (0.500 kg)
= 0.445 mol KN03
So (0.445 mol) x (10 1.11 g mol I)
= 145.0 g KN03 1must be added .
I 2 b b I (Ba( N0 3h) = 2 (2 + 2 x 12) b" = 3 b" = 0.890
b
ES.20(b)
b
= b'"
0.890 " = 3b = 0.2967 mol kg I
and (0.2967 mol kg I) x (0.500 kg)
= o.1484 mol Ba(N03h
So (0. 1484 mol) x (261.32 g molI)
= 138.8 g Ba(N03h 1
Since the solutions are dilute, use the DebyeHiickel limiting law log Y± I 1= 2
=
lz+z_IAlI /2
L z1(bdb
£»
I
= 2 {I x
(0.020) + 1 x (0.020) + 4 x (0.035) + 2 x (0.035)}
I
= 0.125 logy±
= I
x 1 x 0.509 x (0. 125)1 /2 = 0. 17996
(For NaCI) y± = 100. 17996 = 10.6611
98
ES.21(b)
STUDENT'S SOLUTIONS MANUAL
The extended DebyeHtickellaw is log y ±
z_IIl/2 =  Alz ''+~1 + BII/2
Solving for B I B =  ( [Ifi
+
AIZ+Z_I) log y±
(1 = 
0.509)
(b/bG )I /2
+ log y±
Draw up the following table
y±
5.0 x 10 3 0.927
B
1.32
b/(mol kgI)
10.0 x 10 3 0.902 1.36
50.0 x 10 3 0.816 1.29
Solutions to problems Solutions to numerical problems PS.1
PA
= YAP and PB = YBP (Dalton's law). Hence, draw up the following table. 0
1.399
XA
0
0.0898 0.2476 0.3577 0.5194 0.6036 0.7188 0.8019 0.9105
YA
0
0.0410 0.1154 0.1762 0.2772 0.3393 0.4450 0.5435 0.7284
PBikPa
0
4.209
XB
0 0
0.0895 0.1981 0.2812 0.3964 0.4806 0.6423 0.7524 0.9102 0.2716 0.4565 0.5550 0.6607 0.7228 0.8238 0.8846 0.9590
YB
3.566
8.487
5.044
6.996
7.940
9.211
10.105 11.287 12.295
PAikPa
11.487 15.462 18.243 23.582 27.334 32.722 36.066
The data are plotted in Fig. 5.2. We can assume, at the lowest concentrations of both A and B, that Henry's law will hold. The Henry's law constants are then given by KA
= PA = 115.58 kPa Ifrom the point at XA = 0.0898. XA
KB
= PB = 147.03 kPa Ifrom the point at XB = 0.0895. XB
PS.3
Vsalt
=
(av) ab
mol I [Problem 5.2] H20
= 69.38(b 
0.070)cm 3 mol I
with b
== b/(mol kgI).
SIMPLE MIXTURES
Figure 5.2
Therefore, at b = 0.050 mol kgI , Vsalt = 11.4 cm 3 molI
I.
The total volume at this molality is
v = (1001.21) + (34.69)
x (0.02)2 cm 3
=
lool.22cm 3 .
Hence, as in Problem 5.2, V(H20)
=
(1001.22 cm 3 )

(0.050 mol) x (1.4cm 3 mol l ) 55.49 mol
=
1
2 _I 1 18.04cm mol .
Question. What meaning can be ascribed to a negative partial molar volume? PS.S
Let E denote ethanol and W denote water; then
v = nE VE + nw Vw
[5.3].
For a 50 per cent mixture by mass, mE
which solves to nE
Furthermore, XE
=
= mw , implying that
V ""77;;
MEVW' VE+Mw
nE = nE +nw
M
I+~
·
Mw
Since ME XE
= 46.07 g molI
= 0.2811 ,
XW
and Mw
= 1
XE
=
18.02 gmol I,
= 0.7189.
ME = 2.557. Therefore Mw
99
100
STUDENT'S SOLUTIONS MANUAL
At this composition 3
Vw = l7 .5cm mol I [Fig.5. l ofthetextl.
Therefore, /lE =
lOOcm 3 1
1
(56.0cm 3 mol  ) + (2.557) x (l7 .5cm 3 mol  )
= 0.993 mol ,
/lw = (2.557) x (0.993 mol ) = 2.54 mol. The fact that these amounts correspond to a mixture containing 50 per cent by mass of both components is easily checked as follows : mE = /l EM E = (0.993 mol ) x (46.07 g molI ) = 45.7 g ethanol, mw = /lWMW = (2.54 mol ) x (I8 .02gmol 
l
)
= 45 .7 g water.
At 20 0 e the densities of ethanol and water are, PE = 0.789gcm  3,
Pw = 0.997gcm 3. Hence,
45.7g =!57.9 cm3! ofethanol, VE = mE = PE 0.789 gcm  3 mw 45 .7g 1 31 = 3 = 45.8cmofwater. Pw 0.997 gcm 
Vw = 
The change in volume upon adding a small amount of ethanol can be approximated by
where we have assumed that both VE and Vw are constant over this small range of /lE. Hence 3
L'l V "" (56.0 cm molI) x
PS.7
L'lT
bB
= 
Kf
=
3 3 (l.oocm ) x (0.789 1g cm » )=! +0.96 cm 3 !. (46.07 g mol  )
0.0703 K 1
1.86 K / (mol kg )
= 0.0378 mol kg
_I
.
Since the solution molality is nominally 0.0096 mol kgI in Th (N03)4, each formula unit supplies 0.0378 ~ . . . . . 0.0096 ",, ~. (More careful data, as descnbed m the ongmal reference gIves \J "" 5 to 6.) PS.9
The data are plotted in Figure 5.3. The regions where the vapor pressure curves show approximate straight lines are denoted R for Raoul! and H for Henry. A and B denote acetic acid and benzene respectively. As in Problem 5.8, we need to form YA =
PA * and YB = PB * for the Raoult's law activity coefficients XAPA XBPB
and YB = PB for the activity coefficient of benzene on a Henry's law basis, with K determined by xBK extrapolation. We use p*A = 7.3 lcPa, Ps = 35.2 lcPa, and K*B = 80.0 kPa to draw up the following table.
SIMPLE MIXTURES
0.2
0.4
0.6
0. 8
1.0
Figure 5.3
XA
0
XA
0 35.2 0 1.00
PA/kPa PB / kPa aA(R) aB(R) YA(R) YB(R) aB(H) YB(H)
1.00 0.44 0.44
101
0.2
0.4
0 .6
0.8
1.0
2.7 30.4 0 .36 0.86 1.82 1.08 0.38 0.48
4.0 25.3 0.55 0.72 1.36 1.20 0.32 0 .53
5.1 20.0 0.69 0.57 1.15 1.42 0.25 0.63
6.7 12.4 0.91 0.35 1.14 1.76 0.16 0.78
7.3 0 l.oo[PA/p~]
O[PB/PS] l.oo[PA /XAP~]
[PB/XBPS] O[PB/KB] l.oo[PB /xs KBl
GEjs defined as [Section 5.4] GE = 6mixG(actual)  6mixG(ideal) and, with a
=
 nRT(xA InxA
+ XB InxB)
yx ,
GE = nRT(xA In YA For n
= nRT(xA In aA + XB In aB)
+ XA In YB).
= I, we can draw up the following table from the information above and RT = 2.69 kJ mol  I.
XA In YA XB In YB GE /(kJ molI )
0 0 0
0.2
0.4
0.6
0.8
1.0
0. 12 0 .06 0.48
0.12 0.11 0.62
0.08 0.14 0.59
0.10 0.11 0.56
o o o
102
PS.11
STUDENT'S SOLUTIONS MANUAL
(a) The volume of an ideal mixture is
so the volume of a real mixture is
v=
Videal
+ VE .
We have an expression for excess molar volume in terms of mole fractions. To compute partial molar volumes, we need an expression for the actual excess volume as a function of moles.
= nlVrn.1 +n2Vrn ,2 + nln2   ( ao + al(nln2») .
so V
nl +n2
nl +n2
The partial molar volume of propionic acid is _ ( av )
VI 
1
ani
=
VI
aon~
_

P,T ,1I2
Vrn,1 + aoxi
 Vrn I + '
(nl
+ n2)
2
+
al (3nl  n2)n~ (nl
+ n2)3
,
+ al (3x l  X2)X~ I·
That of oxane is
(b) We need the molar volumes of the pure liquids, Vrn I and Vrn2 ,
MI
74.08gmol 1
= PI = 0.97174gcm  3 = 76.23cm =
86.13 gmol I 3 0.86398 g cm
31 mol
1 = 99.69cm3mol
In an equimolar mixture, the partial molar volume of propionic acid is VI
= 76.23 + (2.4697)
x (0.500)2
+ (0.0608)
x [3(0.5)  0.5] x (0.5)2cm3mol1
x (0.500)2
+ (0.0608)
x [0.5  3(0.5)] x (0.5)2 cm 3 molI
= 1 75 .63cm3 molII and that of oxane is V2
= 99.69 + (2.4697) = 1 99.06 cm3 molI
PS.13
I.
Henry's law constant is the slope of a plot of PB versus XB in the limit of zero XB (Fig. 5.4). The partial pressures of C02 are almost but not quite equal to the total pressures reported.
SIMPLE MIXTURES
103
80
60
....
OJ
.D
o
;::; 40 U 'i:(
20
o
0.1
0.0
0.2
0.3
Figure 5.4
Linear regression of the lowpressure points gives KH
= 1371 bar I.
The activity of a solute is PB aB = =XBYB
KH
so the activity coefficient is
where the last equality applies Dalton's law of partial pressures to the vapor phase. A spreadsheet applied this equation to the above data to yield
PS.1S
pfbar
Ycyc
Xc yc
Ye02
10.0 20.0 30.0 40.0 60.0 80.0
0.0267 0.0149 0.0112 0.00947 0.00835 0.00921
0.9741 0.9464 0.9204 0.892 0.836 0.773
1.01 0.99 1.00 0.99 0.98 0 .94
CE
= RTx(l 
x){0.4857  0.1077 (2x  I)
+ 0.0191(2x 
1)2}
E
with x = 0.25 gives C = 0.1021RT. Therefore, si nce
~mix C(actual)
~mi x C(ideal)
+ nCE, ~mi xC = nRT(XA In xA + XB InXB) + nCE = nRT (0.25 In 0.25 + 0.75 In 0.75) + nCE = 0.562nRT + 0.1021nRT = 0.460nRT. =
Since n = 4 mol and RT = (8 .314 J K  I mol  I) x (303 .15 K) = 2.52 kJ molI ,
~mix C = (0.460) x (4 mol) x (2.52 kJ mol  I) = 14.6 kJ I.
104
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems PS.17
= gRTxA (\ = gRT(1 Therefore, 1 /LA PS.19
nAdVA
 XA)
 XA)2
+ (1
 xA)gRT(l  2xA)
= gRTx~.
= /L~ + RTlnxA + gRTx~ I·
+ nBdVB = 0 [Example 5.1].
nA Hence  dVA nB
= dVB .
Therefore, by integration,
Therefore, VB(XA , XB) = VB (0, 1) 
j
VA(XA ) XA dVA
VA (0)
. 1 XA
We should now plot XA / (\  XA) against VA and estimate the integral. For the present purpose we integrate up to VA (0.5 , 0.5) = 74.06 cm 3 molI [Fig. 5.5], and use the data to construct the following table. 74.11
73.96
73 .50
72.74
0.60 1.50
0.40 0.67
0.20 0.25
o o
The points are plotted in Fig. 5.5, and the area required is 0.30. Hence, V (CHCI3; 0.5 , 0.5)
= 80.66 cm3 molI
 0.30 cm 3 molI
= 180.36cm3 molI I. PS.21
InaA
(a)
¢=. r
Therefore, d¢
d InaA
I
1
=  r d In aA + 2" In aA dr, r 1
=r
InaAdr  rd¢.
(b)
SIMPLE MIXTURES
105
=
J1.G +
1.5
1.0
'< 'i'
f
~ ><
I
0.5
~
1;( 0
o
~
~ 0: ~
73
72
75
Figure 5.5 From the GibbsDuhem equation, XA dJ1.A = Xs dJ1.s RT In a, dJ1.A = RTd In aA , dJ1.s = RTd In as )
=
dlnas
XA  dlnaA Xs
=
0, which implies that (since J1.
d InaA
= r
I
1
= "2 In aA dr = dl/> [from(b)] = I/>dr = dr = dl/> [from(a)] r r = I/>dlnr +dl/> . Subtract din r from both sides, to obtain as din r
= (I/> 
I)d Inr+dl/>
(I/>  I)
= dr+dl/>. r
. . and notmg . that In (as) Then, by mtegratlOn r r=O
A(s)
PS.23
;='
r
r=O
= In (YS)r=O = In I = 0,
A(I) .
J1.~(s)
=
J1.~(I) + RTlnaA
and ~fus G
=
J1.~ (I)  J1.~ (s)
Hence, In aA
= In (YSXS) 
=
 RT In aA·
~fu s G
= ~.
din aA I d (~fus G) ~fusH  = T =  [GibbsHelmholtzeqn]. dT R dT RT2 For
~T
= T( 
T , d~T
=  dT and
106
STUDENT'S SOLUTIONS MANUAL
Therefore, dlnaA
MA
MAdt:.T
dt:.T
Kf
Kf
  =   and dlnaA =    According to the GibbsDuhem equation
which implies that
nB
and hence that d In aA =   d In aBo nA
din aB nAMA I Hence    =   =  'dt:.T
nBKf
bBKf
We know from the GibbsDuhem equation that XA d In aA + XB d In aB = 0
and hence that J d In aA = Therefore In aA = 
f
f
XB

d In aBo
XA
xB d In aB. XA
The osmotic coefficient was defined in Problem 5.21 as I
XA
r
XB
¢ =   In aA =   In aA . Therefore, ¢ =XA xB
f
XB I dlnaB=
lob
1 bdlnaB=bob
xA
lob 0
lob
bdlnyb=1 bdlnb+1 bob
lob
=1+1 bdlny . b 0 From the DebyeHiickellimiting law,
In y
= A ' b I /2
[A'
= 2.303AJ.
1
Hence, dIn y = 2A' b1 / 2db and so
¢
= 1+ bI ( 2I A ') 10{b b 1/ 2 db = I  21 (AI) b
COMMENT.
x _2b3/ 2
3
11
For the depression of the freezing point in a 1, 1electrolyte
_IAII /21.
3
lob bdlny 0
SIMPLE MIXTURES
(1 1) (~ 2.) _
D.fusH and hence rl/> = R _ D.fusHxA
Therefore, I/> 
R
Xs
T
  T T*
_
T*
.
D.fusH xA

R
Xs
(T*  T) "" IT*
D. T R T*2
D.fuSH xA
Xs
D.fusHD. T
"" vRbsT*2MA where v
MRT*2
= 2. Therefore, since Kf =  , H D.fus
T
~
1/> 2bsKf .
Solutions to applications PS.2S
In thi s case it is convenient to rewrite the Henry 's law expression as
(1) At PN2
= 0.78
x 4.0 atm
= 3.1
= 3.1 atm
mass ofN2
atm,
x lOOgH20 x 0. 18 ~gN2 / (g H20atm)
= IS6~gN2 1.
(2) At PN 2 = 0.78 atm, mass of N2 = 114 ~g N21. (3) In fatty tissue the increase in N2 concentration from 1 atm to 4 atm is
PS.27
(a) i = I only, N,
= 4, K, = 1.0 X
107 dm 3 mol  ' ,
4 x IOdm 3 ~mol  '
v
[AJ
I + IOdm3 ~mol' x [A(
The plot is shown in Fig. S.6(a) . (b)
i = I; N, = 4, N2 = 2; K, = 1.0 X 105 dm 3 mol' = 0.10 dm 3 ~mol',
K2
= 2.0
v
[AJ
X
10 6 dm 3 mol  '
4 x 0.IOdm3~mol'
2 x 2.0dm3 ~mol'
1+ 0.IOdm 3 ~mol' x [AJ + 1+ 2.0dm3
The plot is shown in Fig. S.6(b).
PS.29
= 2.0 dm 3 ~mol  '.
By the van ' t Hoff equation [S.40], cRT fl=[BJRT=  . M
~mol'
x [A(
107
108
STUDENT'S SOLUTIONS MANUAL
40.0 r                ,
30.0
<' "f
20.0
10.0
0.0 L.....L......L....l..L.1.L....I..L.......I.L....I..L.....L.L.'.L.1''' 0.0 0.2 0.4 0.6 0.8 1.0 [Aj/(dm3 I1moll)
Figure 5.6(a)
5., 4
3
2
2
8
10
Figure 5.6(b) Division by the standard acceleration of free fall, g, gives
n
c(R/g)T
g
M
(a) This expression may be written in the form cR' T
n ' =M ' which has the same form as the van't Hoff equation, but the unit of osmotic pressure (n ') is now force/area length/ time
(mass length)/(area time 2 )
mass
length/time 2
area
2
This ratio can be specified in g em units of R/g, energyK 1 molI length/ time
2
2.
Likewise, the constant of proportionality (R') would have the
(mass length 2 / time 2) K 1 molI
~=';;
length/ time2
= mass length K1mol l .
SIMPLE MIXTURES
This result may be specified in I g cm K I molI ,
R
8.3 1447 J K I molI
g
9.8066Sms 2
R=
= 0.847840 kg m K I mol I
I R'
= 84784.0 g cm K I molI
109
I.
(10kg3 '"0) x (102mCm)
I.
In the following we will drop the primes giving cRT
[]=
M
and use the [] units of g cm 2 and the R units g cm K ImolI. (b) By extrapolating the low concentration plot of [] Ie versus e (Fig. S.7(a» to e intercept 230 g cm 2 I g cm  3 . In this limit the van' t Hoff equation is valid so RT

M
=
M
intercept or M
RT
= intercept
(84784.0gcmK 1 mol I) x (298.ISK) (23 0gcm  2 )/(gcm 3 ) ,
~~~
1M = 1.1 x
10
5
g mol I I.
Polyisobutylene in chlorobenzene at low concentrations
500
Intercept: 230 gcm  2 Igcm J
450
;::400 I
§ 00 .......
I'
E u
•
350
~
.......
~
§:300
250
200
~~~~~~~~~
0.000
0.010
0.020 c/( gcm  J )
0.030
0.040
Figure 5.7(a)
=
0 we find the
110
STUDENT'S SOLUTIONS MANUAL
(c) The plot of n/c versus c for the full concentration range (Fig. 5.7(b» is very nonlinear. We may conclude that the solvent is good. This may be due to the nonpolar nature of both solvent and solute. (d) n/c = (RT /M)(l + B'c + C'c 2 ). Since RT / M has been determined in part (b) by extrapolation to c second and third virial coefficients with the linear regression fit
(n /c)/ (RT/M)  1 c R
= 0, it is best to determine the
= B' + C'c,
= 0.9791.
B' = 21.4cm 3 g', C' = 21lcm 6 g 2,
standard deviation standard deviation
= 2.4cm3 g'. = 15 cm6 g2.
(e) Using 114 for g and neglecting terms beyond the second power, we may write 1
(I
+ "2 B'c) . CH 3
oCI
t
~t
CH ' 
CH3
Polyisobutylene in chlorobenzene
,,
7~
•
6~
5~
au
~ 4~
•
§: 3~
2~
I~
o 0.00
0.050
0.100
0.150 0.200 c/(gcm 3 )
0.250
0.300
Figure 5.7(b)
SIMPLE MIXTURES
111
We can solve for 8 '; then g(8')2 = C',
RT 1M has been determined above as 230 g cm 2 / g cm 3 . We may analytically solve for 8 ' from one of the data points, say, IT Ic = 430 gcm 2I gcm  3 at c = 0.033 g cm 3. 430 g cm 2I g cm 3 ) 1/ 2  I ( 230'g'cm=2'1g=cm=3
8
,
=
2 x (1.367  I) 0.033gcm 3

3
1 ,
= 2. 8
= 22.2 cm
x (0.033 g cm
).
31 g
Better values of 8 ' and C' can be obtained by plotting (
~) 1/2 / (~) 1/2 against c. This plot
is shown in Fig. S.7(c). The slope is 14.03 cm 3 g I. 8' = 2 x slope = 128.0 cm 3 g I I. C' is then 1196 cm 6 g21. The intercept of this plot should theoretically be 1.00, but it is in fact 0.916 with a standard deviation of 0.066. The overall consistency of the values of the parameters confirms that g is roughly 114 as assumed. 6.0
5.0
N ~
r...
4.0
~ I~
'""'
...........
3.0
N
r...
t::1
<.>
'""' 2.0
1.0
0.0 0.00
0.05
0.10
0.15
c/(gcm J )
0.20
0.25
0.30
Figure S.7(c)
Phase diagrams
Answers to discussion questions 06.1
Phase: a state of matter that is uniform throughout, not only in chemical composition but also in physical state. Constituent: any chemical species present in the system. Component: a chemically independent constituent of the system. It is best understood in relation to the phrase 'number of components' which is the minimum number of independent species necessary to define the composition of all the phases present in the system. Degree of freedom (or variance): the number of intensive variables that can be changed without disturbing the number of phases in equilibrium.
06.3
See Figs. 6.I(a) and (b).
Liquid Solid p
Critical point
Triple point T
Figure 6.1(a)
PHASE DIAGRAMS p
=constant
Liquid A and B
T*B
,/
t T
Eutectic 2 Eutectic 1 Solid B and solid AB
Solid AB and solid A 0.50
B
A ~
xA
06.5
See Fig. 6.2.
Liquid A and B
Liquid A and B SolidAB
SolidAB
Solid B SolidAB
Solid A SolidAB
o
0.5
A
Solutions to exercises E6.1(b)
1.0
B
Figure 6.2
Figure 6.1(b)
113
114
STUDENT'S SOLUTIONS MANUAL
19kPa  18kPa
~
~ A is 1, 2dimethylbenzene
XA
= 20kPa _ 18kPa =
YA
= Ps + (p~ 
YB
= 1  0.526 = 0.474 "'" 0.5
XAP~
Ps) XA
=
(0.5) x (20 kPa ) 18 kPa
+ (20 kPa 

18 kPa)0.5
~
= 0.526 "'" L.22J
PA = YAP = 0.612p = XAP~ = xA(68.8 kPa)
E6.2(b)
PB = YBP = (I  YA)P = 0.388p = XBPS = ( I  XA)
YAP
XAP~
YBP
XBPS
(0.388)
X
and
X
82.1 kPa
0.612
0.388
(68 .8)XA = (0.612) x (82. 1)  (0.612)(82. I)xA
26.694xA = 50.245  50.245xA
50.245
XA =
XAP~
P= E6.3{b)
26.694 + 50.245
+ XBPs
~ =~ XB
~
= 10.65 3 =~
= (0.653) x (68.8 kPa) + (0.347) x (82.1 kPa) = 173.4 kPa 1
(a) If Raoult's law holds, the solution is ideal. PA = XAP~ = (0.4217) x (110.1 kPa) = 46.43 kPa PB = XBPs = (I  0.4217) x (94.93 kPa) = 54.90 kPa
P = PA + PB = (46.43
+ 54.90) kPa =
101.33 kPa = 1.000 atm
Therefore, Raoult's law correctly predicts the pressure of the boiling liquid andl the solution is ideal (b)
VA
.
I
PA 46.43 kPa =  = = 0.4582 P \0 1.33 kPa
I
YB = I  YA = 1.000  0.4582 = 10.5418 1
E6.4{b)
Let B = benzene and T = toluene. Since the solution is equimolar l B = ZT = 0.500 (a) Initially XB = ZB and XT = P = XBPs
+ XTPT
ZT ;
thus
[6.3) = (0.500) x (9.9 kPa ) + (0.500) x (2.9 kPa)
= 4.95 kPa + 1.45 kPa = 16.4 kPa 1 (b)
PB 4.95kPa ~ ~ YB = [6.4) = = L222.J YT = I  0.77 = ~
P
6.4kPa
(c) Near the end of the distillation YB = ZB = 0.500 and YT = ZT = 0.500
I.
PHASE DIAGRAMS
115
Equation 6.5 may be sol ved for XA [A = benzene = B here] _ _ _C_O._500_)_x_C_2_.9_k_Pa_)_ _ _ (9.9 kPa) + (2.9  9.9) kPa x (0.500)
YBPT XB
= PB* + (PT*  PB*) YB
= 0.23
XT = 1  0.23 = 0.77
This result for the special case of ZB
= ZT = 0.500 could have been obtained directly by realizing that
Y B Cinitial) = XT (final) ; YT Cinitial) = XB (final)
p (final)
= XBPS + XTPT = (0.23)
x (9.9 kPa) + (0.77) x (2.9 kPa)
= 14.5
kPa 1
Thus in the course of the distillation the vapor pressure fell from 6.4 kPa to 4 .5 kPa E6.5(b)
See the phase diagram in Figure 6.3. (a)
YA
(b)
XA
=@1D = 10.671
YA
= 10.9251
155 150 145 140
or e 135 a
130
b
125 120 0
0.2
0.4
0.6 XA
E6.6(b)
0.8
1.0
Figure 6.3
AI3+, H+ , AICl), AI (OHh, OH , Cl , H20 giving seven species. There are also three equilibria AICI3 + 3H20 ;==> Al (OHh + 3HCI AICl) ;==> AI3+ + 3CI H20 ;==> H+ +OHand one condition of electrical neutrality
Hence, the number of independent components is C
=7
(3 + I )
= [IJ
116
STUDENT'S SOLUTIONS MANUAL
NH 4CI(s)
E6.7(b)
;='
+ HCI(g)
NH3(g)
= I 1[Example 6.1] and 1P = 21 (s and g). added before heating, 1C = 21 (because NH4CI, NH3
(3) For this system 1C (b) If ammonia is
are now independent) and
1P = 21 (s and g). E6.8(b)
(3) Still 1C
= 21 (Na2 S04 , H20) , but now there is no solid phase present, so 1P = 21 (liquid solution,
vapor).
0.
(b) The variance is F = 2  2 + 2 = We are free to change any two of the three variables, amount of dissolved salt, pressure, or temperature, but not the third. If we change the amount of dissolved salt and the pressure, the temperature is fixed by the equilibrium condition between the two phases. E6.9(b)
See Figure 6.4.
+ 10  10
 30
50
 70
 90
Figure 6.4 E6.10(b)
See Figure 6.5. The phase diagram should be labeled as in figure 6.5. (a) Solid Ag with dissolved Sn begins to precipitate at QI , and the sample solidifies completely at Q2. (b) Solid Ag with dissolved Sn begins to precipitate at bl, and the liquid becomes richer in Sn. The peritectic reaction occurs atb2, and
(a)
b
(b)
a
Liquid
800
L + Ag solid contaminated with Sn
U
~ Bi
460 °C b2
L+ Sn solid
L + Ag)Sn solid b)
Ag)Sr + Ag cdntaminated I wit Sn
Be Sn + Ag)Sn solids
200
Sn
Ag)Sn
Ag
Time
Figure 6.S
PHASE DIAGRAMS
117
as cooling continues Ag 3 Sn is precipitated and the liquid becomes richer in Sn. At b3 the system has its eutectic composition (e) and freezes without further change. E6.11 (b)
See Figure 6.6. The feature denoting incongruent melting is circled. Arrows on the tie line indicate the decomposition products. There are two eutectics: one at XB =10.531, T =1T21; another at XB =10.821,
T=@].
o
0.67
0.33
Xs
A
E6.12(b)
B
Figure 6.6
The cooling curves corresponding to the phase diagram in Figure 6.7(a) are shown in Figure 6.7(b). Note the breaks (abrupt change in slope) at temperatures corresponding to points G"b" and b2. Also note the eutectic halts at G2 and b3. (b)
(a)
   +"",.
o A
0.33
0.67
Xs
I
B
Figure 6.7
118 E6.13(b)
STUDENT'S SOLUTIONS MANUAL
Rough estimates based on Figure 6.41 of the text are (a) XB ~ 10.75 1 (b) XAB2 ~ @]]
E6.14(b)
(c) XAB2 ~ 10.61
The phase diagram is shown in Figure 6.8. The given data points are circled. The lines are schematic at best.
Liquid
1000
900
800
Solid
700
o
0.2
0.4
0.6
0.8
Figure 6.8
A solid solution with x(ZrF4) = 0.24 appears at 855 °e . The solid solution continues to form, and its ZrF4 content increases until it reaches x(ZrF4) = 0.40 and 820 °C. At that temperature, the entire sample is solid. E6.15(b)
The phase diagram for this system (Figure 6.9) is very similar to that for the system methyl ethyl ether and diborane of Exercise 6.9(a). The regions of the diagram contain analogous substances. The solid compound begins to crystallize at 120 K. The liquid becomes progressively richer in diborane until the liquid composition reaches 0.90 at 104 K. At that point the liquid disappears as heat is removed. Below 104 K the system is a mixture of solid compound and solid diborane.
Figure 6.9
PHASE DIAGRAMS
E6.16(b)
119
Refer to the phase diagram in the solution to Exercise 6.14(a). The cooling curves are sketched in Figure 6.10. (a)
(d)
(c)
(b)
(e)
95 93 91
:.:
h'
89 87 85 83
ES.17(b)
1
Figure 6.10
+
(a) When X A falls to 0.47, a second liquid phase appears. The amount of new phase increases as XA falls and the amount of original phase decreases until, at XA = 0.314, only one liquid remains . (b) The mixture has a single liquid phase at all compositions. The phase diagram is sketched in Figure 6.11 . 54
52 50 ;;.; 48
.... "" 46 44
42
40 38 0. 1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
XA
1.0
Figure 6.11
Solutions to problems Solutions to numerical problems PS.1
(a) The data, including that for pure chlorobenzene, are plotted in Fig. 6.12. (b) The smooth curve through the x, T data crosses x mixture.
= 0.300 at 1391.0 K I,
the boiling point of the
(c) We need not interpolate data, for 393.94 K is a temperature for which we have experimental data. The mole fraction of Ibutanol in the liquid phase is 0.1700 and in the vapor phase 0.3691. According
120
STUDENT'S SOLUTIONS MANUAL
400
ex TIK 395
O y
x (0.3, 393.94 K) 390
385
0.0
0.4 xor y
0.2
0.6
0.8
Figure 6.12
to the lever rule, the proportions of the two phases are in an inverse ratio of the distances their mole fractions are from the composition point in question. That is,
~ = ~ = 0.3691  0.300 = 10.5321. l
I1va p
0.300  0. 1700
PA = aAP~ = YAXAP~ [5 .45].
P6.3
PA YAP YA== · XAP~
XAP~
Sample calculation at 80 K: O. II(lookPa)( 760TOrr) 0.34(225 Torr) 10 1.325 kPa '
Y0 2 (80 K)
=
Y0 2 (80 K)
= 1.079.
Summary: T/ K
77.3
Y0 2
78 0.877
80 1.079
82 1.039
84 0.995
86 0.993
88 0.990
90.2 0.987
To within the experimental uncertainties the solution appears to be ideal (y = I) . The low value at 78 K may be caused by nonideality ; however, the larger relative uncertainty in y(02 ) is probably the origin of the low value. A temperaturecomposition diagram is shown in Fig. 6.13(a). The near ideality of this solution is, however, best shown in the pressurecomposition diagram of Fig. 6.13(b). The liquid line is essentially a straight line as predicted for an ideal solution .
P6.S
A compound with 1probable formula A3B exists I. It melts incongruently at 700°C, undergoing the peritectic reaction A3B(S)
~
A(s) + (A
+ B,
I).
PHASE DIAGRAMS
121
92
X8
:.:
;::;
84
SO
76
0
20
40
60
80
100
x(02) or y(02)
Figure 6.13(a)
700
S()()
300
100
o
20
40
60
80
100
Figure 6.13(b)
The proportions of A and B in the product are dependent upon the overall composition and the temperature. A eutectic exists at 400°C and XB ~ 0.83. See Fig. 6.14. P6.7
The information has been used to construct the phase diagram in Fig. 6.15(a). In MgCu2 the mass ~ 48.6 r::;:;l 24.3 percentage of Mg is (100) x =~, and in Mg 2Cu it is (100) x =~. The 24.3 + 127 48.6 + 63 .5 initial point is Q I , corresponding to a liquid singlephase system. At Q2 (at 720°C) MgCu2 begins to come out of solution and the liquid becomes richer in Mg, moving toward e2. At Q3 there is solid MgCu2 + liquid of composition e2 (33 per cent by mass of Mg). This solution freezes without further change. The cooling curve will resemble that shown in Fig. 6.15(b).
P6.9
(a)
I Eutectic: 40.2 at% Si at 1268°C I, 1Eutectic: 69.4 at% Si at 1030°C I. Congruent melting compounds: Ca2 Si mp = 13 14°C CaSi mp = 1324°C
122
STUDENT'S SOLUTIONS MANUAL
1300 LiquidA& B
1100
900
£!
Solid A
700
500
Solid A
LiquidA& B
SolidA3B
SolidA3B Liquid A3B & Solid B
300 0.20
0
0.60
0.40
1.00
0.80
XB
Figure 6.14
(a)
(b)
1200
800
400
Mg
1
Incongruent melting compound: CaSi2
Time
mp
=
1040°C
1
Figure 6.15
melts into CaSi(s) and liquid (68
at% Si). (b) At 1000°C the 1phases at equilibrium will be Ca(s) and liquid (13 at% Si) I. The lever rule gives the relative amounts: nCa
iJiq
nbq 
ICa 

0.2  0 ~ ~ 0.2  0.13 . .
(c) When an 80 at% Si melt it cooled in a manner that maintains equilibrium, Si(s) begins to appear at about 1250°C. Further cooling causes more Si(s) to freeze out of the melt so that the melt becomes more concentrated in Ca. There is a 69.4 at% Si eutectic at 1030°C. Just before the eutectic is reached, the lever rule says that the relative amounts of the Si(s) and liquid (69.4% Si) phases are: nSi
lliq
nliq
lSi
0.80  0.694
:::::::c
1.0  0.80
=
0.53
. .. = relative amounts at T slightly higher than 1030°C .
Just before 1030°C, the Si(s) is 34.6 mol% of the total heterogeneous mixture, the eutectic liquid is 65.4 mol %.
PHASE DIAGRAMS
123
At the eutectic temperature a third phase appearsCaSi2 (s). As the melt cools at this temperature, both SiCs) and CaSi2(s) freeze out of the melt while the concentration of the melt remains constant. At a temperature slightly below 1030°C, all the melt will have frozen to SiCs) and CaSh (s) with the relative amounts: 0.80  0.667 1.0  0.80
= 10.665 = relative amounts of T
slightly higher than 1030°C I.
Just under 1030°C, the SiCs) is 39.9 mol% of the total heterogeneous mixture; the CaSi2(s) is 60.1 mol%. A graph of mol% SiCs) and mol% CaSi2 (s) vs. mol% eutectic liquid is a convenient way to show relative amounts of the three phases as the eutectic liquid freezes. See Fig. 6.16. Equations for the graph are derived with the law of conservation of mass. For the silicon mass,
where n = total number of moles.
= Si fraction in eutectic liquid =0.694 = Si fraction in SiCs) = 1.000 YS i = Si fraction in CaSi2(s) = 0.667 ZSi = Si fraction in melt = 0.800 WSi
XSi
Freezing of eutectic melt at 1030°C
0.7 r.,,.,.. 0.6 0.5
0.4 mol fraction CaSi 1 ........ mol fraction Si 0.3 0.2
0.1
0.1
0.2
0.3
0.4
mol fraction liq Freezing proceeds toward left
Figure 6.16
0.5
0.6
0.7
124
STUDENT'S SOLUTIONS MANUAL
This equation may be rewritten in mole fractions of each phase by dividing by n: ZSi
= (mol fraction liq)WSi + (mol fraction Si)XSi + (mol fraction CaSi2)Ysi.
Since, (mol fraction Iiq) + (mol fraction Si) + (mol fraction CaSi2) = I or (mol fraction CaSi2) = I  (mol fraction liq + mol fraction Si), we may write: ZSi
= (mol fraction Iiq)WSi + (mol fraction Si)xsi + [I  (mol fraction liq + mol fraction Si)]YSi.
Solving for mol fraction Si: . S· (ZSi  YSi)  (WSi  YSi)(mol fraction Iiq) moI f ractIon 1 := , XSi  YSi mol fraction CaSi2 := I  (mol fraction Iiq
+ mol fraction Si).
These two eqns are used to prepare plots of the mol fraction of Si and mol fraction of CaSi2 against the mol fraction of the melt in the range 00.65.
Solutions to theoretical problems P6.11
The general condition of equilibrium in an isolated system is dS isolated system, which are in thermal contact with each other dS
= O. Hence, if ct and f3 constitute an
= dSa + dS,B = O.
(a)
Entropy is an additive property and may be expressed in terms of U and V .
S
= S(U, V).
The implication of this problem is that energy in the form of heat may be transferred from one phase to another, but that the phases are mechanically rigid, and hence their volumes are constant. Thus, dV = 0, and dS
aSa = ( a  )
But, dUa
Ua
v
dUa
+ (as,B  )
= dUf3 ; therefore
aUf3
1 Ta
=
dU,B v
I
=
I dU a Ta
1 + dU,B
[3.45] .
Tf3
r~
Tf3 orl Ta
= Tf3I·
Solutions to applications P6.13
(i) Below a denaturant concentration of 0. 1 only the native and unfolded forms are stable.
(ii) At denaturant concentration of 0.15 only the native form is stable below a temperature of about 0.70. At temperature 0.70 the native and moltenglobule forms are at equilibrium. Heating above 0.70 causes all native forms to become moltenglobules. At temperature 0.90, equilibrium between
PHASE DIAGRAMS
125
moltenglobule and unfolded protein is observed and above this temperature only the unfolded form is stable. C
P6.1S
=
I;
hence,
F
=C
P
+2 = 3
P.
Since the tube is sealed there will always be some gaseous compound in equilibrium with the condensed phases. Thus when liquid begins to form upon melting, P = 3 (s, I, and g) and F = 0, corresponding to a definite melting temperature. At the transition to a normal liquid, P = 3 (t, l', and g) as well, so again F = O. P6.17
To examine the process of zone levelling with the phase diagram below, Fig. 6.17, consider a solid on the isopleth through a I and heat the sample without coming to overall equilibrium. If the temperature rises to a2 , a liquid of composition b2 forms and the remaining solid is at a; . Heating that solid down an isopleth passing through a~ forms a liquid of composition b3 and leaves the solid at a~. This sequence of heater passes shows that in a pass the impurities at the end of a sample are reduced while being transferred to the liquid phase which moves with the heater down the length of the sample. With enough passes the dopant, which is initially at the end of the sample, is distributed evenly throughout. A
B liquid
solid
a: o Composition, xB
P6.19
Figure 6.17
The data are plotted in Fig. 6.18.
(b) From the tie line at 2200°C, the liquid composition is y(MgO) 10.351· The proportions of the two phases are given by the lever rule,
~ = n(liq) = 0.05 = I 0.41. 12
n(sol)
0.12
(c) Solidification begins at point c, corresponding to I 2640°C
I.
= ~ and the solid x(MgO) =
126
STUDENT'S SOLUTIONS MANUAL
2640
o
1.0
0.3 x(MgO)
Figure 6.18
10
8 v
';' 6 c..
..............
~
:::::: 4
e x ,
O y
.... ... .. ,.... .. ..• ...
~
x (0.500, 6.02 MPa)
2 ......... ·····T
o 0.2
0.4
0.6
0.8
1.0
x or y
P6.21
Figure 6.19
(a) The data are plotted in Fig. 6.19. (b) We need not interpolate data, for 6.02 MPa is a pressure for which we have experimental data. The mole fraction of C02 in the liquid phase is 0.4541 and in the vapor phase 0.9980. The proportions of the two phases are in an inverse ratio of the distance their mole fractions are from the composition point in question, according to the lever rule
~ n vap
= ~ = 0.9980 I
0.5000 0.5000  0.4541
= 110.851.
Chemical equilibrium
Answers to discussion questions 07.1
The position of equilibrium is always determined by the condition that the reaction quotient, Q must equal the equilibrium constant, K. If the mixing in of an additional amount of reactant or product destroys that equality, then the reacting system will shift in such a way as to restore the equality. That implies that some of the added reactant or product must be removed by the reacting system and the amounts of other components will also be affected. These adjustments restore the concentrations to their (new) equilibrium values.
07.3
(1) Response to change in pressure. The equilibrium constant is independent of pressiIre, but the indi
vidual partial pressures can change as the total pressure changes. This will happen when there is a difference, f:l.n g , between the sums of the number of moles of gases on the product and reactant sides of the chemical equation. The requirement of an unchanged equilibrium constant implies that the side with the smaller number of moles of gas be favored as pressure increases. (2) Response to change in temperature. Equation 7.23a shows that K decreases with increasing temperature when the reaction is exothermic; thus the reaction shifts to the left, the opposite occurs in endothermic reactions. See Section 7.4 (a) for a more detailed discussion. 07.5
(a) Consider the metals M and Z, which, for the sake of simplifying discussion, form I: I oxides having the formulas MO and ZOo Z will spontaneously reduce MO provided that the ZO line upon the Ellingham diagram lies above the MO line (this statement assumes that the vertical f:l.rG axis decreases upward). In this case the standard Gibbs energy for the reaction MO(s) + Z(s) + M(s) + ZO(s) will be negative. Figure 7.10 of the text indicates that Fe will reduce PbO, CuO, and Ag 20 . (b) Using f:l.rG"(ZnO) = 318 kJmol i at 25 °C (Table 2.7) and a slope that is common for all the oxides, we may add the approximate line for ZnO in the Ellingham diagram as shown in Fig.7 .!. The ZnO curve passes under the reaction (iii) curve at about 1300°C so that is the estimate of the lowest temperature at which zinc oxide can be reduced to the metal by carbon. See Fig. 7.1.
07.7
Electrode combinations that produce identical cell compartments with differing concentrations only (electrolyte concentration cells) have a cell potential dependence upon the liquid junction potential and the concentration difference. If the cell has identical compartments with either gaseous or amalgam electrodes (electrode concentration cell), the cell potential will depend upon the gas pressure differences or the amalgam concentration differences but will not have a liquid junction potential. Other electrode combinations produce cells for which the cell potential depends upon the halfreaction reduction potentials.
07.9
The pH of an aqueous solution can in principle be measured with any electrode having an emf that is sensitive to H+ (aq) concentration (activity). In principle, the hydrogen gas electrode is the simplest
128
STUDENT' S SOLUTIONS MANUAL
500
400
 300
,.
"0
E :;:: 200
.""
""  100
"l
0
+100
o
500
1500
2000
Temperature, OI OC
2500
Figure 7.1
and most fundamental. A cell is constructed with the hydrogen electrode being the righthand electrode and any reference electrode with known potential as the lefthand electrode. A common choice is the saturated calomel electrode. The pH can then be obtained by measuring the emf (zerocurrent potential difference), E , of the cell . The hydrogen gas electrode is not convenient to use, so in practice glass electrodes are used because of ease of handling.
Solutions to exercises E7.1(b)
N204(g)
~
2N02(g)
Amount at equilibrium
a)n
(I 
I a I +a
 
Mole fraction
a)P I+a
(I 
Partial pressure
Assuming that the gases are perfect, aJ K
=
(PNOzlp f»2 (PN20 4 I pf»
=
4a 2
Forp=pf> K=   
,
1 a 2
2
4a p (I  ( 2)pf>
2an 2a

I+a
laP
l+a
= p~ p
CHEMICAL EQUILIBRIUM
(a)
(b) (c)
E7.2(b)
(a)
1
~r G =0 1 at equilibrium
K = 4(0.201)22 = 10.16841 1 10.201 ~rG~ = RTInK = (8.314JK I molI) x (298K) x In(0.16841) 0'=0.201
Br2(g) .= 2Br(g)
a = 0.24
Amount at equilibrium Mole fraction Partial pressure
(I  a)n
2an
Ia 1+0' ( I  a)P
20' 1+0' 2aP 1+0'
1+0'
Assuming both gases are perfect aJ = PJ
P~
4a 2p
K = (PBrlp~)2 =
P Br 2 /p~
(I  0' 2)p~
4(0.24)2 _ ~ = 1 _ (0.24)2 = 0.2445 = ~ (b)
~rG~ = RTInK = (8.314JK I molI) x (1600K) x In(0.2445)
1= 19kJmol 1 1 (c)
lnK(2273K) = In K(l600 K) _
~rH~ R
_ (ll2
= In (0.2445) 
(_1_ _ _1_) 2273 K 1600 K 3
x 10 mOl  I)
8.314JK1 mol 
I
x (1.851 x 10 4)
= 1.084 K(2273 K) =
el.
v(CHCI3) = 1,
E7.3(b)
(a)
084 = 1 2.96 1 v(HCI) = 3,
V(CH4) = I ,
V(CI2) = 3
~rG ~ = ~fG~(CHCb , I) + 3~fG~ (HC1, g)  ~fG~ (CH4, g)
= (73 .66kJmol 
l
)
+ (3) x (95.30kJmol l )

(50.72 kJ mol  I)
= 1308.84kJmol  1 1 ~rG~ ( 308.84 x 103 Jmol  I) InK =    [ 7.8] = = 124.584 RT (8.3145JK I molI) x (298.15K)
K
= 1 I.3
x 10
54
1
129
130
STUDENT'S SOLUTIONS MANUAL
= 6.fH " (CHCI), l) + 36.fH"(HCI,
6.,H"
(b)
= ( 134.47kJmol l ) + (3)
g)  6.rH"(CH4 , g)
x (92.31 kJmol l )  (74.81 kJmol l )
= 336.59 kJ mol  I = InK(25 0C) _
InK(50 0C)
6.,H" (_1_ _ _ 1_) [7.25] R 323.2 K 298.2 K
1~3 J mOilI) x (2.594 x 104 K I) = 114.083 8.3145JK mol 
= 124.584 _ (336.59 x
K(50 °C)
= 13.5 x
6.,G" (50 °C)
=
1049 1
RT In K(50 °C) [7.17]
=
(8.3145JK 1 mol  I) x (323.15K) x (114.083)
= 1306.52kJmol 1 1
E7.4(b)
Draw up the following table. A Initial amounts/mol Stated change/mol Implied change/mol Equilibrium amounts/mol Mole fractions
B
+
~
2.00
1.00
7.09 1.21 0.1782
7.09 0.21 0.0302
C
+
0 +0.79 +7.09 0.79 0.1162
(a) Mole fractions are given in the table. (b)
Kx
= nx~J , J
(0.1163) x (0.6745)2 Kx =
(c) PI
(0.1782) x (0.0309)
~ =~
= XIP. Assuming the gases are perfect, QJ = PI/p" , so K
= (Pe /le, ) x
(Po / p,,)2 (PA / P"' ) x (PB / p")
= Kx
(.£) =
Kx
when P
= 1.00 bar
p"
K=Kx=~ (d)
6.,G"'
= RTlnK = (8.314JK 1 molI) = 15.6kJmol 1 1
x (298K) x In(9.609)
2D
Total
3.00
6.00
+1.58 4.58 0.6742
6.79 0.9999
CHEMICAL EQUILIBRIUM
E7.5(b)
At 1120 K,
/).rG f7
131
= +22 x 103 J mol I
/).rG f7 (22 x 103 Jmol I ) InK(1l20K) =   = = 2.363 1 RT (8.314JK molI) x (l120K)
K = e 2.363 = 9.41 x 10 2 InK2 = InKI _
H f7 /).r
R
(~
_
~) TI
T2
Solve for T2 at In K2 = 0 (K2 = I) RlnKI I (8.314JK 1 molI) x (2.363) I 4 =  +  = +   = 7.36 x 10l 3 T2 /).rH f7 TI (125 x 10 Jmol ) 1120K

I
3
T2 =11.4 xlO KI
E7.6(b)
d(lnK) /).rH f7 Use=d(l I T) R
(~ )
We have InK = 2.04  1176K
+ 2.1 x 107 K3
(~ Y
T = 450K so H f7 _
/).r
R
= 1l76K + (2.1 x 107 K3) x 3 (_1_)2 = 865K 450K
/).rH f7
= +(865K) x (8.314Jmol 1 K I ) =1 7. 19lkJ molI 1
/).rG f7
= RT In K
1176K 2.1 x 107 K3} 1 = (8.314JK  molI) x (450K) x { 2.04  450K + (450K)3 = 16.55 kJ molI
/).r S
f7
/).rH f7 
/).r G f7
= T
7.191 kJmol 1  16.55kJmol 1            = 20.79JK 1 molI 450K
=121JK l mol 1 1 E7.7(b)
U(s) + ~H2(g) ~ UH3(S),
/).rG f7 = RTIn K
At this low pressure, hydrogen is nearly a perfect gas, a(H2) =
(Plpf7).
The activities of the solids are 1.
132
STUDENT'S SOLUTIONS MANUAL
P Hence, InK = In ( pe b.fG
)3/2 In!!..... 3 =
2
pe
3 P RTln
e
=2
pB"
= ( 3) 2
X
(8 .314JK  1 mol  I) x (500 K) x In (
139Pa5 ) 1.00 x 10 Pa
= 141.0kJmol  1 I E7.8(b)
Kx =
nx/J
[analogous to 7.16]
J
The relation of Kx to K is established in Illustration 7 .5
Kx
=
n( PJ) VJ[ pe
J
Therefore, Kx v
= 1+ I 
.
7.16wlthaJ
PJ] = pe
= K (pi pe ) v, Kx ex p v
1 I
[K and pe are constants]
I
= 0,
thus KA2 bar) = KAl bar)
I
E7.9(b)
Initial moles N2
=
Initial moles 0 2 =
5.0g 28 .01 gmol 
I
= 0.2380 mol N2
2.0g 2 I = 6.250 x 10 mol 0 2 32.00gmol
Initial amount/mol Change/mol
N2
02
NO
Total
0.2380  z
0.0625  z
0 +2z
0.300
o
Equilibrium amount/mol
0.2380  z
0.0625  z
2z
0.300
Mole fractions
0.2380  z 0.300
0.0625  z
2z 0.300
( I)
0.300
[v = ~ VJ = 0] , then
K
= Kx ( ; )
K
= Kx = ,,';:::,,.,
v
(2zI0.300)2
0.2380 ( 0.300
z) x (0.0625  z) 0.300 2
4z (0.2380  z)(0.0625  z)
= 1.69 x
10 3
CHEMICAL EQUILIBRIUM
4z2 = 1.69 x 103{0.01488  0.300S z + Z2 ) X
10 5
X
10 3 = 4.00
= 2.514 4.00  1.69
4z2 + (S.078
S.078

(S.078
10 4 ± {(S.078
X
~(S.078
z> 0
X
10 4 ± 2.006
X
104 )2  4 8
In K ' K
(4)
X
=
6.J H f>
(~
R
T = 31OK,
_
T
~)
X
f>
so
T'
6.JH
=2
(b)
K
=
10
 21
Rln
(~)
= (~ _ ~)
let = K
6.JH f> = (SS.84 kJ mol I)
I
T'
K
_ l InK = SS.84kJmol lnK
X
X
(In 2) = 139 kJ mol II
6. r H f> = (SS .84kJmol l ) X (In!) =139kJmol 1 1
2
E7.11(b) p = p(NH3 )
(a)
K
10 5 »)1 /2
K'
T' = 32SK ;
(8.3 14JK 1 mol  I) Now 6.JH f> = « 1/ 3IOK) _ (1/ 32SK)) K
X
10 3
X
T
(a)
(2.SI4
10 2)
3 2z 2(2.444 X 10 ) 1 =  = = 1.6 0.300 0.300
XNO
X
[Z < 0 is physically impossible] so
z = 2.444
E7.10(b)
X
10 5 = 0
X
z= =
1O 3 )Z2
X
so
1O 4 )z  2.S14
X
1O 4 )z + (1.69
X
=
+ p (HCI)
na;i
= 2p(NH3)
PJ
a(gases) = G;
[7 .16];
p
J
K = (P(
NH3» ) pf>
X
[P(NH3) = p(HCI)]
(P(HCI») = p(NH 3) 2 = pf>
pf>2
~
X
4
At 427 °C (700 K) ,
K =
~ (~: ~:) 2 = 19.241
At4S9 °C (732K) ,
K =
4'I
(!!...) 2
X
X
(1IISkPa) 2 ~ lookPa =~
pf>
133
134
STUDENT'S SOLUTIONS MANUAL
(b) I'::..,G" = RTlnK [7.S] = (S.3141 K1mol  l ) x (7ooK) x (In 9.24) =112.9kJmo1  1 1 (at 427° C) (C)I'::..H "::::; R1n (K 'j K ) [7 .25 ] , (l j T  1j T')
::::;
E7.12(b)
(S.3141 K 1mol l ) x In (31.0Sj9.24) I =. +161 kJmol (11700 K)  ( 1/732 K)
I
I .
The reaction is
For the purposes of this exercise we may assume that the required temperature is that temperature at which K = 1, which corresponds to a pressure of 1 bar for the gaseous products. For K = 1, In K = 0, and I'::..,G" = O.
Therefore, the decomposition temperature (when K = I) is I'::.. H " T='I'::..,S & CUS04 . 5H20 (s) ;=' CUS04 (s) + 5H20 (g) I'::..,H& = [(771.36) + (5) x (241.S2)  (2279.7)] kJ mo1 1 = +299.2 kJmo1 1 I'::..,S" = [( 109) + (5) x (1SS.S3)  (300.4)] JK 1 mol I = 752 .8JK 1 mol I
Therefore, T =
299.2 x 1Q3Jmol  1 ~ 1 1 = ~ 752.SJK mol
Question. What would the decomposition temperature be for decomposition defined as the state at which K
E7.13(b)
=
I j2?
PbI2 (S);=' PbI2(aq)
Ks = 1.4
X
10 8
I'::..,G" =RTlnKs=(S.314JK  lmol  l) x (29S.15K) x In (1.4 x 108 ) = 44.S3 kJ mol 
I
I'::..fG" (PbI2 , aq) = I'::..,G& I'::.. + I'::..fG& (PbI2, s)
= 44.S3kJmol 1  173.64kJmol 1 =112S .S kJmol 1 1
CHEMICAL EQUILIBRIUM
E7.14(b)
135
The cell notation specifies the right and left electrodes. Note that for proper cancellation we must equalize the number of electrons in halfreactions being combined. For the calculation of the standard ernfs of the cells we have used E~ potentials from Table 7.2. (a)
(b)
R:
AgzCr04(S) + 2e
L:
Clz(g) + 2e
+
2Ag(s) + CrO~(aq)
L:
2Fe3+(aq) + 2e
+1.36 V
+
R:
MnOz(s) + 4H+(aq) + 2e
L:
Cu2+(aq) + 2e
+
+
2Ag(s) + CrO~(aq) + (CIZg)
2Fe2+(aq)
+
+
Sn2+(aq) + 2Fe3+(aq)
Mn2+(aq) + 2Fe3+(aq)
0.62 V +1.23 V
Cu(s)
+0.34 V
Cu(s) + MnOz(s) + 4H+(aq)
+ Cu 2+(aq)
+2HzO(l) COMMENT.
0.91 V +0.15 V +0.77 V
Sn4+(aq) + 2Fe2+(aq)
Overall (R  L) :
+0.45 V
2Cl(aq)
Overall (R  L): AgzCr04(S) + 2Cl(aq) R : Sn 4+(aq) + 2e + Sn2+(aq) Overall (R  L) :
(c)
+
= t;  Et, with standard electrode
+ Mn2+(aq) +0.89V
Those cells for which E ~ > 0 may operate as spontaneous galvanic cells under standard
conditions. Those for which E ~ < 0 may operate as nonspontaneous electrolytic cells. Recall that E~ informs us of the spontaneity of a cell under standard conditions only. For other conditions we require E.
E7.1S(b)
The conditions (concentrations, etc.) under which these reactions occur are not given. For the purposes of this exercise we assume standard conditions. The specification of the right and left electrodes is determined by the direction of the reaction as written. As always, in combining halfreactions to form an overall cell reaction we must write halfreactions with equal number of electrons to ensure proper cancellation. We first identify the halfreactions, and then set up the corresponding cell. (a) R:
2HzO(I) + 2e
L:
2Na+ (aq) + 2e
20H(aq) + Hz(g)
+ +
2Na(s)
 0.83 V 2.71 V
and the cell is
I +1.88V I or more simply
I Na(s)INaOH(aq)IHz(g)IPt I (b) R: L:
Iz(s) + 2e
+
2H+ (aq) + 2e
21(aq) +
Hz(g)
+0.54 V
o
and the cell is
I+0.54 V I or more simply
I PtIHz(g)IHI(aq)llz(s)1 Pt I
136
STUDENT'S SOLUTIONS MANUAL
(c) R:
L :
2H+(aq) + 2e ~ H2(g)
O.OOV
2H20(I ) +2e~H2(g)+20H(aq)
0.083 V
and the cell is 10.083 V
I
or more simply
IPtIH2(g) IH20(l)IH2(g) IPt I All of these cells have Ee. > 0, corresponding to a spontaneous cell reaction under standard
COMMENT.
conditions. If Ee. had turned out to be negative, the spontaneous reaction would have been the reverse of the one given, with the right and left electrodes of the cell also reversed.
E7.16(b)
(a)
E=E
Q
=
e.
RT vF
lnQ
n
VJ
aJ
v=2
2 2 = aH+a Cl 
[all other activities
=
1]
J
= a~a:' = (y+b+)2
x (y_b_)2
[b
==
: e. here and below]
RT ( ) 2RT Hence, E = E e.  2F In y1 b4 = E e.  Fin (y±b)
=
(b)
6. r G
(c)
logy±
vFE [7.27]
=
= (2)
x (9.6485 x 104 C molI) x (0.4658 V)
I Z+Z_WI / 2[5.69]
=
(0.509) x (0.010)1 /2 [I
y±
= 0.889
E e.
MT (y±b) = (0.4658 V) + = E + Fin
The value compares favorably to that given in Table 7.2. vFE e.
In each case In K (a)
="RT""
[7.30]
Sn(s) + CuS04(aq) ;=: Cu(s) + SnS04(aq)
L:
Cu2+ (aq) + 2e ~ Cu(s) + 0.34 V } Sn2+ (aq) + 2e ~ Sn(s) _ 0. 14 V + 0.48 V
In K
=
R:
(2) x (0.48 V) 25.693 mV
= +37.4,
K
= b for HCI(aq)] =
0.0509
(2) x ( 25 .693 x ) 10 3 V x In (0.889 x 0.010)
= I +0.223 V I
E7.17(b)
= 189.89 kJ molII
= 11.7 x
10
16
1
CHEMICAL EQUILIBRIUM
R:
Cu2+ (aq) + e + Cu(aq) + 0.16 V } _ 0.36 V Cu+ (aq) + e + Cu(s) + 0.52 V
L: InK E7.18(b)
0.36V
2Bi3+(aq) + 6e
L:
Bi2S3(S) + 6e
Overall (R  L):
K (a)

= 25.693 mV = 14.0,
R:
In K
137
+
+
2Bi(s)
2Bi(s) + 3S 2(aq)
2Bi3+(aq) + 3S 2(aq)
vFE<7
6(0.96 V)
RT
(25.693 x 10 3 V)
=  =
= e 224 K =
+
Bi2S3(S)
v
=6

= 224 5
_ e 224 M [Bi3+ ]2 [S2 ]3 
aBi2S3(S) _ 3 a 2Bi3+ (aq) aS2(aq)
In the above equation the activity of the solid equals I and, since the solution is extremely dilute, the activity coefficients of dissolved ions also equals 1. Substituting [S2] = 1.5[Bi3+] and solving for [Bi3+] gives [Bi3+] = 2.7 x 10 20 M. BhS3 has a solubility equal I to 1.4 x 10 20 M.I (b) The solubility equilibrium is written as the reverse of the cell reaction. Therefore, Ks
= K  I = l/e224 = 15.2 x
10 98 1.
Solutions to problems Solutions to numerical problems P7.1
(a)
!:ire?
= RTlnK = (8.314JK I mol I)
x (298K) x (lnO.I64)
= I +4.48kJmol 1 I· (b) Draw up the following equilibrium table.
h Amounts Mole fractions Partial pressure
Br2
IBr
(la)n
2an
(1  a)
2a
(1 + a) (la)p
(1 + a) 2ap
(1 + a)
(l + a)
= 4.48 x
103 Jmol I
138
STUDENT'S SOLUTIONS MANUAL
Withp
= O.l64atm, I 5'
2
a =
= 2a1+a
PIEr
X
P
=
a = 0.447.
(2) x (0.447) I + 0.447
X
(0.164atm)
= I0.101 atm.I
(c) The equilibrium table needs to be modified as follows. P
= Plz + PBr2 + PIEr, = XIErP ,
PIEr
.
wIth XBrz and XlEr
=
(1  a)n (1
[n
+ a)n + nl 2
Plz
= XI 2P
= amount of Br2 introduced into container]
=  2an 
(1 +a)n+nI2
K is constructed as above [7.16], but with these modified partial pressures. In order to complete the calculation additional data are required, namely, the amount of Br2 introduced, n, and the equilibrium vapor pressure of 12(S). nl 2 can be calculated from a knowledge of the volume of the container at equilibrium which is most easily determined by successive approximations since PI2 is small.
Question . What is the partial pressure of IBr(g) if 0.0100 mol of Br2 (g) is introduced into the container? The partial pressure of 12(S) at 25°C is 0.305 Torr. P7.3
U(s)
3 + "2H2(g)
6.fH
~ UH3(S), K
. = (P l pB) _ 3/2 [ExerCIse 7.7(b)] .
B= RT d InK [7.23a] = RT 2
dT
2
3 d (lnpl p dT 2
= ~RT2 d lnp 2
= ~RT2
dT
(1.464
2
= ~R(l.464
4
10 K _ 5.65) T2 T X
4
x 10 K  5.65T)
= 1(2.196 x 104 K S.4ST)R
or 6. r
P7.5
I.
a6. f H B) C;: = (  = IS.4S R I. aT p
K=~. pB
B)
CHEMICAL EQUILIBRIUM
139
Since t':;.rc& and In K are related as above, the dependence of t':;.rc& on temperature can be determined from the dependence of In K on temperature t':;.rc&
=
RTlnK
=
RTln.E..
p"
= (8.314J K Imol I) x (400 K) x In (17.1kPa) [p" = Ibar] 100.OkPa
= +13.5kJmol 1 at400K. t':;.rc&(T) t':;.rc&(T' ) T T'
Therefore, taking T' t':;.rc&(T)
=
=
t':;.r
I
T
I ) T' [7.25].
= 400 K,
(_T_) x (13 .5 kJ molI) + (78 kJ molI) x (I _ _ T_) ~K
~K
I
= (78 kJ mol ) +
That is, t':;.rC&(T) / (kJ molI) = P7.7
W (
(13 .578)kJmOI400
I
)
x
(T) K .
178  0.161 x (T /K) I.
The equilibrium we need to consider is A2(g) <=' 2A(g). A = acetic acid. It is convenient to express the equilibrium constant in terms of a, the degree of dissociation of the dimer, which is the predominant species at low temperatures.
At equilibrium
A
A2
2an
(1 + a)n
2a
Mole fraction

I+a 2ap
Partial pressure
l+a
Total (1
+ a)n
Ia

I+a
C~:)p
p
The equilibrium constant for the dissociation is
We also know that pV
= ntotalRT =
(1 + a)nRT, implying that
a
pV nRT
=  
I and n
m M
=
In the first experiment,
a
pVM
=  mRT
I
=
(101.9 kPa) x (21.45 x 1O 3 dm3 ) x (120.1 g mol I) (0.0519 g) x (8.314 kPa dm 3 K I mol I) x (437 K)

1 = 0.392.
140
STUDENT'S SOLUTIONS MANUAL
H K (4) ence, =
X
(0.392)2 X (764.3/750.1) ~ 1 _ (0.392)2 = ~.
In the second experiment, pVM
Ct
=   I = mRT
(101.9 kPa) x (21.45 x 1O 3dm 3 ) x (120.1 g mol I) (0.038 g) x (8.314 kPa dm3 K I molI) x (471 K)
C
64 3 . ) 750.1 I  (0.764)2
(4) x (0.764)2 x Hence, K
=
 1 = 0.764.
= §].
The enthalpy of dissociation is
,r&
I1rfl
=
K' RlnK
(~
_
T
· ~) [7.25, ExerCISe 7.1O(a)]
=
T'
5.71 Rln ( ) 0.740 (_1_ _ _ 1_) 437K
= +103 kJ mol
I .
471K
The enthalpy of dimerization is the negative of this value, or 1103 kJ mol  II (i.e. per mole of dimer). P7.9
The equilibrium I2(g) .= 2I(g) is described by the equilibrium constant
K If pO=
=
x(I)2
P
2 x ~
x(l2)
p
4
=
2 Ct
(p) pB
I 
nRT
V ' then p = (1 +
2 Ct
[Problem 7.7].
a )pO, implying that
We therefore draw up the following table. 937 K
1073 K
1173 K
p/atm 104 nl
0.06244 2.4709
0.07500 2.4555
0.09181 2.4366
p O/atm
0.05757
0.06309
0.06844
Ct
0.08459
0.1888
0.3415
11.109 x 10 2 1
3
K
11.800 x 10 1
I1Er
= RT2 x
14.848 x 10 2 1
(dinK) I 2 (3.027  (6.320») ~ = (8.314 J KImol ) x (1073 K ) x 200 K
= 1+158 kJ molI
I.
CHEMICAL EQUILIBRIUM
P7.11
141
The reaction is
The equilibrium constant is
Let h be the uncertainty in value is
KlowH
KlowH Khi ghH
h)
) (!::"rSS) (!::"r~gh) ( = exp M{~ RT ow exp ~ = exp RT exp RT
(
= exp
So
== exp
(
P7.13
(a)
(b)
(!::"rSS) ~
h ) RT .
I (289  243) kJ mol) = 11.2 x 1081. I KhighH (8.3145 x 10 3 kJ K molI) x (298 K) I Kl owH ( (289  243) kJ mol31 ) 1 At 700 K,   = ex = 2.7 x 10 . I KhighH P (8.3145 x 10 3 kJ K molI) x (700 K) KlowH
b
exp
(:T )KhighH.
(a) At 298 K,  
( )
!::"rFr so that the high value is h + the low value. The K based on the low
= exp (
I=~{(;t zi +(;t zq[5.711=4(:t». For CUS04 , I
= (4) x (1.0 x
For ZnS04 , I
= (4)
logy±
10 3 )
=14.0 x 10 3 1.
x (3.0 x 10 3 ) = 11.2 x 10 2 1.
= l z+z_IAII /2.
logy±(CUS04) = (4) x (0.509) x (4.0 x 10 3 )1 / 2 = 0.1288,
= 1 0. 74 1. logY±(ZnS04) = (4)
y±(CUS04)
y±(ZnS04)
x (0.509) x (1.2 x 10 2 )1 / 2
= 0.2230,
= 10.60 I·
(c) The reaction in the Daniell cell is Cu 2+ (aq)
+ SO~ (aq) + Zn(s) + Cu(s) + Zn2+ (aq) + SO~ (aq).
142
STUDENT'S SOLUTIONS MANUAL
a(Zn2+)a(SO~ , R) Hence, Q = ::=a(Cu2+)a(SO~, L)
where the designations Rand L refer to the right and left sides of the equation for the cell reaction and all b are assumed to be unitless, that is, b/bf7 . b+(Zn2+) = b_(SO~, R) = b(ZnS04). b+(Cu 2+) = b_(SO~ , L) = b(CUS04)'
Therefore,
(d)
(e)
rG
c.
I
I
6. r G" (212.7 x 103 J molI) = [7.28] = = +1.102 V . vF (2) x (9.6485 x 104 C molI)
E=r=
25 .693
X
v
10 3
VlnQ=(1.102V)
(25 .693 x 10 3 ) 2 V In(5.92)
= (1.102V)  (0.023V) = 1+1.079 V I· P7.15
The electrode halfreactions and their potentials are
r R : Q(aq) + 2H+(aq) + 2e + QH 2(aq) L: Hg2Cl2 (s) + 2e + 2Hg(1) + 2CI (aq) Overall (R  L): Q(aq) + 2H+(aq) + QH2(aq) + Hg2Cl2(s), ..
Q(reactlOn quotient) =
0.6994V 0.2676V 0.4318 V
a(QH2) 2 2 a(Q)a (H+)a (Cl
Since quinhydrone is an equimolecular complex of Q and QH 2, m(Q) = m(QH 2) and, since their activity coefficients are assumed to be I or to be equal, we have a(QH 2) ~ a(Q) . Thus E
In Q =
\!(~  E)
25 .7 mV
=
=r 
25 .7 mV
v
InQ [Illustration 7.10].
(2) x (0.4318  0.190) V 25.7 x 10 3 V
= 18.
82
'
Q
=

8
1.49 x 10 .
CHEMICAL EQUILIBRIUM
For HCI(aq), b+ hence
=
b_
=
b and, if the activity coefficients are assumed equal, a 2 (H+)
=
143
a 2 (CI);
1
Q = a 2 (H+)a 2 (CI )  a 4 (H+)' Thus a(H+) , pH
E
=
=r
a(H+)
(
1.49
log a(H+)
=
=r 
X
10 8
X
10 3
'
= [IQ).
RT ina(H+)a(CI) [Section 7.8]. F
y+b+
a(H+)a(Cn E
1)1/4= ( 1 )1/4= 9
= Q
=
=
y+b;
a(CI)
y+y_b 2
=
=
y_b_
=
y_b [b
=;
here and belOW] .
ylb 2 .
2RT 2RT  I n b   I n y±.
F
(a)
F
Converting from natural logarithms to common logarithms (base 10) in order to introduce the DebyeHUckel expression, we obtain ,..&
E = c.

(2.303) x 2RT (2.303) x 2RT F log b F log y±
=r
 (0.1183 V) 10gb = r  (0.1183 V) 10gb = r  (0.1183 V) 10gb +
(0.11 83 V) log y± (0.1183 V)
[l z+z_ IAII /2]
(0.1183 V) x A
X
b l / 2 [I
= b].
Rearranging, E + (0.11 83 V) 10gb
=r
+ constant x b l / 2 .
Therefore, plot E + (0.1183 V) log b against b 1/ 2, and the intercept at b following table.
= 0 is £'7 IV . Draw up the
bl(mmol kg I)
1.6077
3.0769
5.0403
7.6938
10.9474
(; y/2
0.04010
0.05547
0.07100
0.08771
0.1046
EIV + (0.11 83) log b
0.27029
0.27109
0.27186
0.27260
0.27337
The points are plotted in Fig. 7.2. The intercept is at 0.26840, so £'7 = +0.26840 V. A leastsquares best fit gives £'7 =I +0.26843 V Iand a coefficient of determination equal to 0.99895.
144
STUDENT'S SOLUTIONS MANUAL 0.274
0.272
¢ c
........
c 0;;
..2 r')
~
0
+
0.270
>
~
0.268
o
0.02
0.04 0.06 (b/ b'7 )1/2
0.08
0.10
Figure 7.2
For the activity coefficients we obtain from equation (a)
rE
b
In y± = 2RT I F In be =
0.26843E / V b 0.05139 In be
and we draw up the following table.
3.0769
1.6077 0.3465 0.9659
In y± y± P7.19
5.0403
0.05038 0.6542 0.9509 0.9367
7.6938
10.9474
0.07993 0.09500 0.9232 0.9094
The cells described in the problem are backtoback pairs of cells each of the type Ag (s) IAgX (s) IMX (bl) IMxHg (s) . R:
M+ (bl)
+ e ~ M xHg (s)
L:
AgX (s)
+ e
RL:
Ag (s)
+ Ag (s)
F
+ X (bJ)
+ M+ (bl ) + X
RT E=r lnQ.
(Reduction ofM+ and formation of amalgam)
Hg
(b)) + MxHg (s)
+ AgX (s) ,
v
=
1.
CHEMICAL EQUILIBR IUM
145
For a pair of such cells back to back, Ag (5) IAgX (5) IMX (bl) IMxHg (s) IMX (b2) IAgX (5) lAg (5) ,
ER
=r

Rt FIn QR,
RT
E = 
F
QL QR
 In 
EL =
RT
= 
F
In
r 
RT
Fin QL,
(a (M+) a (X))L (a (M+) a (X ))R
77i+;~
(Note that the unknown quantity a (MxHg) drops out of the expression for E.)
With L = (1) and R = (2) we have
2RT bl 2RT Y±(1) E =  In  +  I n   . F b2 F y±(2)
Take b2 = 0.09141 mol kgI (the reference value), and write b =
:~ .
2RT y± ) . E= ( Ib n   +ln  F 0.09141 y± (ref) For b = 0.09141, the extended DebyeHlickel law gives
logy±(ref) =
(1.461)
(0.09141) 1/ 2 1/2 + (0.20) x (0.09141) = 0.2735, (I) + (1.70) x (0.09141) X
Y±(ref) = 0.5328. y± ) b Then E = (0.05139 V) x ( In 0.09141 + In 0.5328 '
E b In y± = 0.05139 V  In (0.09141) x (0.05328)
We then draw up the following table.
bl
(mol/kg I)
EIV y
0.09141
0.1652
0.2171
1.040
1.350
0.0220
0.0000
0.0263
0.0379
0.1156
0.1336
0.572
10.5331
0.492
0.469
0.444
0.486
0.0555
A more precise procedure is described in the original references for the temperature dependence of (Ag, AgCl, CI) ; see Problem 7.20.
r
146
P7.21
STUDENT'S SOLUTIONS MANUAL
(a) From
(aG) = V [3.50] ,
ap
we obtain
T
aflrG) ( ap T =
flr V.
Substituting flrG = vFE [7.27] yields
(b) The plot (Fig. 7.3) of E against p appears to fit a straight line very closely. A linear regression analysis yields
xI03mVatm'1~ Slope ' I=2.480
standard deviation = 3 x 10 6 mV attnI .
Intercept= 8.5583 mY,
standard deviation = 2.8 x 10 3 mY.
R = 0.99999701 (an extremely good fit).
From
flrV ( 2.666 x 10 6 m 3 molI) 1 x 9.6485
Since J = VC = Pam 3 ,
X
Pam 3 C=
a?
T ,n 
(2.666 x 10 ) V 9.6485 X 104 Pa
Pa 5
6

V
or
V
Therefore
( aE)
104 C molI .
X
1.01325 x 10 Pa = 2.80 x 10 6 Vatm I atm = 12.80 x 10 3 mV atm I I.
This compares closely to the result from the potential measurements.
13
12
II
>
8 ~ 10
9
8
o
500
1000
p/atm
1500
Figure 7.3
CHEMICAL EQUILIBRIUM
147
(c) A fit to a secondorder polynomial of the form E
= a + bp + cp2
yields a
= 8.5592 mY,
standard deviation
b = 2.835 x 10 3 mVatm
c = 3.02 x 10 9 m V atm 2, R
= 0.999 997
= 0.0039 mV
standard deviation = 0.012 x 10 3 mVatm
l,
1
standard deviation = 7.89 x 10 9 m V atm I
II.
This regression coefficient is only marginally better than that for the linear fit, but the uncertainty in the quadratic term is > 200 per cent. = b + 2cp. (aE) ap T
The slope changes from to
(aE) ap
= max
(aE) ap
= min
b= 2.835 x 10 3 mVatm 1
b+ 2c(l 500 atm) = 2.836 x 10 3 mVatm
We conclude that the linear fit and constancy of
l.
(~:) are very good.
(d) We can obtain an order of magnitude value for the isothermal compressibility from the value of c. 2
a E ap2
= _~ vF
(aLlrV) ap T
= 2c. 2vcF V
2(1 ) x (3 .02 x 10 12 Vatm 2) x (9.6485 x 104 Cmol(KT )cell
I)
x (
82.058 cm3 atm) 8.3145J
= (:,:::)''(lcm3 j O.996g) x
= 13.2
18.016g I mol
x 10 7 atm I 1 standard deviation "'" 200 per cent
where we have assumed the density of the cell to be approximately that of water at 30°C. COMMENT. It is evident from these calculations that the effect of pressure on the potentials of cells involving
only liquids and solids is not important ; for this reaction the change is only::>:: 3 x 10 6 V atm  1 . The effective isothermal compressibility of the cell is of the order of magnitude typical of solids rather than liquids; other than that, little significance can be attached to the calculated numerical value.
P7.23
We need to obtain Ll rH f7 for the reaction ~H2 (g)
+ Uup+ (aq)
+ Uup (s)
+ H+ (aq) .
We draw up the thermodynamic cycle shown in Fig . 7.4.
148
STUDENT'S SOLUTIONS MANUAL
Ej(H)
13.6eV H(g) + Uup+(g)
11.3eV t. hydH B (H+) H+(aq ) + Uup+(g)
1 x 4.5eV
A
1H2 + Uup+(g)
5.52eV
3.22eV
H+(aq )+ Uup(g)
1H2 + Uup+(aq)
1.5eV
x
W(aq) + Uup(s)
B
Figure 7.4
Data are obtained from Tables 10.3, lOA, IIA, 2.7, and 2.7b. The conversion factor between eV and kJmol 1 is 1 eV
= 96A85 kJ molI
The distance from A to 8 in the cycle is given by 6. r W
=x =
(3.22 eV) +
G)
x (4.5 eV) + (13.6 eV)  (11.3 eV)  (5.52eV)  (1.5 eV)
= 0.75eV. 6. r s""
= s"" (Uup, s) + s"" (H+ , aq) = (0.69) + (0) 
6. r
cr = 6. W r
G)
T6. r s""
=

!s"" (H2, g)

s"" (Uup+ , aq)
x (1.354)  (1.34) meV K 1 = 1.33 meV K 1 •
(0.75eV) + (298.15K) x (1.33 meVK 1)
which corresponds to I + III kJ molI The electrode potential is therefore
= +1.l5 eV
I.
~~cr , with v = 1, or 11.15 V I.
Solutions to theoretical problems P7.25
We draw up the following table using the stoichiometry A + 38
Mole fraction
~
I 
~
I  ~ 2(2 ~)
=
vJ~.
C
Total
3
0
4
3~
+2~
Initial amount Imol Equilibrium amount Imol
2C and 6. nJ
8
A
Change,6.nJ / mol
~
3(1
~)
3(1 2(2 
2~
~)
~
~)
2~
2(2 
~)
CHEMICAL EQUILIBRIUM
149
Since K is independent of the pressure 27
a 2 =  K , a constant. 16
Therefore (2 
~)~ = a ( ; )
which solves to
~
=I
x (1
)1/2
I (
~)2,
1+ apjpB
We choose the root with the negative sign because shown in Fig. 7.5.
0.1
100
\0
~
lies between 0 and 1. The variation of ~ with p is
\000
ap/p&
Figure 7.S
P7.27
log y± = _AII /2 = _AC I / 2 y± =e 
2.303AC I / 2
Ks = S'(S'
+ C)
We solve S'
2
x
2 y±
= e 4606AC .
e4.606ACI / 2
+ S' C 
Ks
2
y±
In y± = 2.303AC I / 2
=0
I /2
150
STUDENT'S SOLUTIONS MANUAL
to get S
,
1
=
(
2
2 4Ks C + 
. Therefore, since Y:f?
Yf
1/ 2 )
I Ks   C "'"  2 CYf
I/ 2 = e  4606AC .
K e  4.606AC I/2 S,""' _ 5_ _ __
C
Solutions to applications P7.29
6.C
= 6.c? + RTlnQ
[7 .11] .
In equation 7.11 molar solution concentrations are used with 1 M standard states. The standard state (G) pH equals zero in contrast to the biological standard state (EEl) of pH 7. For the ATP hydrolysis
we can calculate the standard state free energy given the biological standard free energy of  31 kJ mol  I (impact 17.2).
= 6. c? +
6.C EB 6.G"
=
RTln(l07 M/ IM)
6.C EB  RT In(10 7M/ I M)
=
31 k1 mol  I  (8.3141 molI K I)(310 K) In(10 7)
= 31 kJmol 1 + 41.5 kJ mol  I = +11 kJ mol I.
This calculation shows that under standard conditions the hydrolysis of ATP is not spontaneous! It is endergonic . The calculation of the ATP hydrolysis free energy with the cell conditions pH [Pi] = 1.0 x 10 6 M, is interesting.
= 7, [ATP] = [ADP] =
6.C = 6. G" + RT In Q = 6. G" + RT In{[ADP][P; ][H+]/ [ATP](1 M)2 }
=
+11 k1 mol  I + (8.314 J mol  IK I)(310 K) In(l06 x 10 7)
=
+11 k1 mol  I 77 k1 molI
= 66 k1 mol  I. The concentration conditions in biological cells make the hydrolysis of ATP spontaneous and very exergonic. A maximum of 66 kJ of work is available to drive coupled chemical reactions when a mole of ATP is hydrolyzed. P7.31
Yes, a bacterium can evolve to utilize the ethanol/nitrate pair to exergonically release the free energy needed for ATP synthesis. The ethanol reductant may yield any of the following products.
ethanol
ethanal
elhano ic acid
The nitrate oxidant may receive electrons to yield any of the following products. NO) + NO; + nitrate
nitrite
N2 dinitrogen
+
NH3 . ammonia
CHEMICAL EQUILIBRIUM
151
Oxidation of two ethanol molecules to carbon dioxide and water can transfer 8 electrons to nitrate during the formation of ammonia. The halfreactions and net reaction are: 2[CH) CH20H(I) + 2C02(g) + H20(l) NO)"(aq) + 9H+ (aq) + 8e + NH) (aq)
+ 4H+(aq) + 4e l + 3H20(1 )
6.r ~ = 2331 .29 kJ for the reaction as written (a Table 2.5 and 2.7 calculation). Of course, enzymes must evolve that couple thi s exergonic redox reaction to the production of ATP, which would then be available for carbohydrate, protein, lipid, and nucleic acid synthesis.
P7.33
The halfreactions involved are: R:
cytox
+e
+
cytred
L:
Dox
+e
+
Dred
t:;yt ~
The overall cell reaction is :
R  L = cytox
+ Dred ;=: cytred + Dox E" = t:;yt 
~
(a) The Nernst equation for the cell reaction is
At equilibrium, E
= 0; therefore
leq ) against In ([cyt] [Dox [ ] ox ) is linear with a slope of one and an intercept Therefore a plot of In ( cyt red [Dredleq of
:T (~t  ~) .
(b) Draw up the following table. I ([Dox leq ) n [Dred leq ( [cytox leq ) In [cytredleq
 5.882
4.776
3 .661
3 .002
 2.593
1.436
0.6274
4.547
 3.772
2.415
 1.625
 1.094
0.2120
0.3293
152
STUDENT'S SOLUTIONS MANUAL
[Doxleq ) . ( [cytox]eq ) . Th e plot ofln ( [D] agamst In [ IS shown in Fig. 7.6. The intercept is 1 .2124. Hence red eq cyt,ed]eq
~t =
RT
F
X
(1.2 124) + 0.237 V
=0.0257V
X
(1.2124) +0.237 V
= 1 + 0.206 V I·
0 \
g
2
1

B 3 ~ B 4 0
:5
5 6 5
4
3
2
 I
o Figure 7.6
P7.35
A reaction proceeds spontaneously if its reaction Gibbs function is negative. 6. r G
= 6. r (7 + RT In Q
Note that under the given conditions, RT = 1.58 kJ molI. (i)
6. r G/(kJ mol I) = 6. r (7( I)  RTlnp H 2 0 = 23.6  1.58 In 1.3 x 10 7 = +1.5.
(ii)
6. r G(kJ mol I) = 6. r (7(2)  RTln(pH2oP HN0 3)
= 57 .2  1.58ln [(1 .3 x 10 7) x (4. 1 x lO IO (iii)
6. r G/(kJ mol  I) = 6. r (7(3)  RT In(p~20 PHN03)
= 85 .6  1.581n[(1.3 x 10 7 )2 (iv)
)J = +2.0.
X
(4.1
10
10
X
(4. 1 X 10
10
X
)]
= 1.3 .
)]
= 3.5 .
6. r G/(kJ molI ) = 6. r (7(4)  RT In(p~2 0PHN03)
= 85.6  1.58In[(1.3 x 10 7 )3
CHEMICAL EQUILIBRIUM
153
So both the dihydrate and trihydrate form spontaneously from the vapor. Does one convert spontaneously into the other? Consider the reaction
which may be considered as reaction (iv)  reaction (iii). Therefore t;.rG for this reaction is
We conclude that the dihydrate converts spontaneously to the 1trihydrate I, the most stable solid (at least of the four we considered).
PART 2 Structure
8
Quantum theory: introduction and principles
Answers to discussion questions 08.1
At the end of the nineteenth century and the beginning of the twentieth, there were many experimental results on the properties of matter and radiation that could not be explained on the basis of established physical principles and theories. Here we list only some of the most significant. (1) (2) (3) (4) (5)
The energy density distribution of blackbody radiation as a function of wavelength. The heat capacities of monatomic solids such as copper metal. The absorption and emission spectra of atoms and molecules, especially the line spectra of atoms. The frequency dependence of the kinetic energy of emitted electrons in the photoelectric effect. The diffraction of electrons by crystals in a manner similar to that observed for Xrays.
08.3
The heat capacities of monatomic solids are primarily a result of the energy acquired by vibrations ofthe atoms about their equilibrium positions. If this energy can be acquired continuously, we expect that the eq uipartition of energy principle shou ld apply. This principle states that, for each direction of motion and for each kind of energy (potential and kinetic), the associated energy should be ~ kT . Hence, for three directions and both kinds of motion , a total of 3 kT , which gives a heat capacity of 3 k per atom, or 3 R per mole, independent of temperature. But the experiments show a temperature dependence. The heat capacity falls steeply below 3 R at low temperatures. Einstein showed that, by allowing the energy of the atomic oscillators to be quantized accordi ng to Planck 's formula , rather than continuous, thi s temperature dependence cou ld be explained. The physical reason is that at low temperatures only a few atomic oscillators have enough energy to populate the higher quantized levels; at higher temperatures more of them can acquire the energy to become active.
08.5
If the wavefunction describing the linear momentum of a particle is precisely known, the particle has a definite state of linear momentum, but then, according to the uncertainty principle, the position of the particle is completely unknown as demonstrated in the derivation leading to eqn 8.21. Conversely, if the position of a particle is precisely known, its linear momentum cannot be described by a single wavefunction, but rather by a superposition of many wavefunctions, each corresponding to a different value for the linear momentum. Thus all knowledge of the linear momentum of the particle is lost. In the limit of an infinite number of superposed wavefunctions, the wavepacket illustrated in Fig. 8.31 turns into the sharply spiked packet shown in Fig. 8.30. But the requirement of the superposition of an infinite number of momentum wavefunctions in order to locate the particle means a complete lack of knowledge of the momentum.
158
STUDENT'S SOLUTIONS MANUAL
Solutions to exercises E8.1(b)
The de Broglie relation is h h A=  = p
mv
v = 11.3 E8.2(b)
X
so
v= 
h
=
mA
6.626 X 10 34 J s (1.675 x 10 27 kg) x (3.0 X 10 2 m)
::::=::,;;;,,_
10 5 m sI 1 extremely slow!
The moment of a photon is 34 6.626 x 1O J s 1 = 1.89 9 350 x 10 m
h A
p =  =
X
10 27 kgms I
I
The momentum of a particle is p
= mv
so
p 1.89 X 10 27 kg m sI v    :=: m  2(1.0078 x 10 3 kg molI 16.022 x 1023 molI)
v = I 0.565 m SI E8.3(b)
I
The uncertainty principle is
so the minimum uncertainty in position is
n
n
l.0546 x 10 34 J s
Llx =   =   = ::c::::::,::;;::,::::::,,....,::,::::=::;;cc:2Llp 2mLlv 2(9.11 x 10 31 kg) x (0.000010) x (995 x 10 3 ms I)
=15.8 xlO 6m l E8.4(b)
he E=hv =_.
A'
he = (6.62608 x 10 34 J s) x (2.99792 x 108 m sI) = 1.986
X
10 25 J m
NAhe = (6.02214 x 1023 molI) x (1.986 x 10 25 J m) = 0.1196J m mol  I
Thus, E =
1.986 x 1O 25 1m A
E(per mole) =
0.11961mmo[1 A
We can therefore draw up the following table A
(a) 200 nm (b) 150 pm (c) 1.00 cm
Ell
E/(kJ molI)
0.93 x 10 19 1.32 x 10 15 1.99 x 10 23
598 7.98 x 105 0.012
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
E8.5(b)
159
Assuming that the 4He atom is free and stationary, if a photon is absorbed, the atom acquires its momentum p achievi ng a speed v such that p = my. m = 4.00 x 1.6605 x 10 27 kg = 6.642 x 10 27 kg
f
v =
m h
P="i (a)
6.626 x 10 34 J s 27 1 200 x 10 9 = 3.313 x 1O kgmsm
p=
27
v=!!...
1
= 3.313 x 1O kgms =10.499m s 1 1 m 6.642 x 10 27 kg
(b)
6.626 x 10 34 J s 12 =4.417 x lO 24 kgm s150 x 10 m
p=
27
v= f
= 4.417 x 1O kg msm 6.642 x 10 27 kg
(c)
=1665 ms  II
6.626 x 10 34 J s = 6.626 x 10 32 kgm sI 1.00 x 10 2 m
p =
6.626 x 10 32 kg m sI 1 6 _I I = 9.98 x 10 m s 6.642 x 10 27 kg
p
v=  = m E8.6(b)
1
1
Each emitted photon increases the momentum of the rocket by h/ A. The final momentum of the rocket will be Nh/ A, where N is the number of photons emitted, so the final speed will be Nh/Amrocket . The rate of photon emission is the power (rate of energy emission) divided by the energy per photon (he/A), so
N=
IPA he
and
h tP v = ( IPA) x (  ) =  he Amrocket emrocket
( 1O.0yr) x (365dayyr l ) x (24 hdayl) x (3600sh l ) x (1.50 x 1O 3 W) (2.998 x IO s m s l ) x ( 10.0 kg)
v
= 1158 m s  I I E8.7(b)
E8.8(b)
Rate of photon emission is rate of energy emission (power) divided by energy per photon (he/A) (0. IOW) x(700xlO 9 m) (6.626 x 10 34 J s) x (2.998 x lOS m sI )
= I 352 .
(a)
rate =
PA = he
(b)
rate =
( 1.0 W) x (700 x 10 9 J s) I = 3.52 x lOIS s II (6.626 x 10 34 J s) x (2.998 x lOS m S I) . .
Conservation of energy requires Epholon
and
EK
=
= ~meV2
+ EK so
= hl! = he/A
so
2E ) 1/ 2
v
= ( m~
EK
= he/A 
x 10
17
s
II
160
STUDENT'S SOLUTIONS MANUAL
(a)
EK =
(6.626 x 1O 34 Js) x (2.998 x 108 ms 1)  (2.0geY) x (1.60 x 10 19 J ey 1) 650 x 10 9 m
But this expression is negative, which is unphysical. There is I no kinetic energy or velocity Ibecause the photon does not have enough energy to dislodge the electron. (6.626 x 10 34 J s) x (2.998 x 10 8 m sI) (b) EK =  (2.0geY) x (1.60 x 10 19 J eV 1) 195 x 10 9 m =16.84 x 10 19 J I 2(6.84 x 1O19J») 1/ 2
and v = ( 9.11 x 10 31 kg ES.9(b)
ES.10(b)
E
= 11.23 x 106 m s  I I
= hI! = h/ T, so
(a)
E = 6.626 x 10 34 J s/ 2.50 x 10 15 s = 12.65 x 10 19 J = 160 kJ mol 1 I
(b)
E = 6.626 x 10 34 J s/2.21 x 10 15 s = 13.00 x 10 19 J = 181 kJ mol 1
(c)
E = 6.626 x 1O 34 J s/ 1.0 x 10 3 s = 16.62
X
I
10 31 J = 4.0 x 10 10 kJ mol 1 I
The de Broglie wavelength is h
A= 
P
The momentum is related to the kinetic energy by p2 EK = 2m
so p = (2mEK)1 /2
The kinetic energy of an electron accelerated through 1 Y is leY = 1.60 x 10 19 J, so h A = """(2=m"""E=K"",,)71/M2
(a)
A=
(b)
A
=
6.626 x 10 34 J s If? »
(2(9.11 x 10 31 kg) x (100eY) x (1.60xlO 19 Jey I
6.626 x 10 34 J S
:~
(2(9.11 x 10 31 kg) x (1.0x10 3 eY) x (1.60 x 1O 19 Jey 1»1 /2
= 13.9 x 10 11 m I (c)
6.626 x 10 34 J s
A=,'"
(2(9.11 x 10 31 kg) x (100 x 103 eY) x (1.60 x 10 19 J eyl»1 /2
= 13.88 x 10 12 m I
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
ES.11 (b)
161
The upper sign in the following equations represents the math using the A+ ill operator. The lower sign is for the A  iB operator. r is a generalized coordinate.
f
± iB I1/1jdr
1/1/IA
f 1/1tIAI1/1
=
j dr
±
= {f 1/1n A11/1;dr
if 1/1tIBI1/1
j
r
±i
dr
{f 1/I/IBI1/I;dr
r
{f 1/I/IA 1/I;dr =f if 1/I/IBI1/I;dr = {f 1/I *IA =f iBl1/I;dr =
1
r
j
This shows that the
A±
r
A andB are hermitian [8.30]
iB operators are not hermitian. Lf they were hermitian, the result would be
{J 1/Ij*I A ± iBl1/I;dr j*. ES.12(b)
The minimum uncertainty in position is 1100 pm I. Therefore, since f1xf1p 34
Ii
f1p>   2f1x f1 v
ES.13(b)
=
=
f1p m
1.0546 x 10 1 s 2(100 x 10 12 m)
= 5.3 x
5.3 x 10 25 kg m Sl 9.11 x 10 31 kg
=
10 25 k
gms
I5.8 x 10 .
5
~ ~Ii
I
m s I
I
Conservation of energy requires Ephoton
= Ebinding + ~mev2 = hv = hC/ A (6.626
x 10 34 Js) x
(2.998 121 x 10 12 m
and Ebinding =
so
Ebinding
= hC/ A 
~mev2
x 108 m s  I)
 ~(9.11 x 10 31 kg) x (5.69 x 10 7 ms I ) 2
=1l.67
xlO
16
11
COMMENT. This calculation uses the nonrelativistic kinetic energy, which is only about 3 percent less than
the accurate (relativistic) value of 1.52 x 10
15
J . In this exercise, however. Ebinding is a small difference of
two larger numbers. so a small error in the kinetic energy results in a larger error in value is E binding = 1 .26 x 10  16 J. ES.14(b)
Ebinding:
the accurate
The quality QIQ2  Q2Q I [Illustration 8.3] is referred to as the commutator of the operators
QI
and
Q2. In obtaining the commutator it is necessary to realize that the operators operate on functions ; thus, we form
QIQ2!(X)  Q2 Q U(X) Px
Ii d dx
=i
Therefore a
=
(x + Ii!)
and a t
=
(x  Ii ! )
162
STUDENT'S SOLUTIONS MANUAL
Thenaatf(x)=~(x+n!) andataf(x)
= ~ (xn!)
x (xn!)f(x)
x (x+ n! )f(x)
The terms in x 2 and (d/dx)2 obviously drop out when the difference is taken and are ignored in what follows ; thus
"
I ( xndxd+ nx d) f(x) dx d) f(x) at af(x) = I (d xnx  nx dx aa 'f(x) = 2
2
d
These expressions are the negative of each other, therefore •
t
(aa '  a a)f(x)
d . = nxf(x)
dx
. d
 ru:f(x) dx
( d.. d)
= njof (x)
= n dx x  x dx f(x) Therefore, (aa t  a t a)
= [KJ
Solutions to problems Solutions to numerical problems PS.1
A cavity approximates an ideal black body ; hence the Planck distribution applies,
8'Jlhe ( P =):5
I
ehc/ AkT _
I
)
[8.5].
Since the wavelength range is small (5 nm ) we may write as a good approximation I'1E = pI'1A,
A ~ 652.5 nm .
(6.626 X 10 34 J s) x (2.998 x 10 8 m sI) (6. 525 X 10 7 m) x (1.38 1 x 10 23 J K )
he Ak
'=''=:.1 =
2.
20
0 K 5 x I 4 .
8'Jlhe _ (8'Jl) x (6.626 X 10 34 J s) x (2.998 x 108 m sI) _ 22 07 J 4  _  4. I x 1 m . AS (652.5 x 10 9 m)S I'1E
=
(4.221 x 10 7 J m 4 ) x (
~ K) / T 
e(2.205x 1 3
I
1
) x (5
X
10 9 m) .
I
0.211 JmQ4 = 1.6 x 10 33 J m 3 . e(2.205x I )/298  I 3 0.211 J m(b) T = 3273 K, ll.E = Q4 = 2.5 X 10 4 J m 3 . e(2.205 x 1 )/3273  I (a) T = 298 K , ll.E =
I
I
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
163
COM M E NT. The energy density in the cavity does not depend on the volume of the cavity, but the total energy
in any given wavelength range does, as well as the total energy over all wavelength ranges.
Question. What is the total energy in this cavity within the range 650655 nm at the stated temperatures? PS.3
hI!
Js
X S I
er: = T ' [Ih]= JK  I =K. In terms of er: the Einstein equation [8.7] for the heat capacity of solids is Cv = 3R ( ; ) 2 x
(e~7:2~ I) 2,
classical value = 3R.
hI!
IT « 1 as demonstrated in the text (Section 8.1). The criterion for classical behavior is therefore that 1T » er: I.
It reverts to the classical value when T
»
8E or when
hI! (6.626x I0 34 JHC I) xv _II er:=T= 1.38I xlO 23 JK1 =4.798xlO (v/Hz)K.
(a) For v = 4.65 x 10 13 Hz,
er: =
(4.798 x 10 11 ) x (4.65 x 10 13 K) =12231 KI.
(b) For v = 7.15 x 10 12 Hz,
er: =
(4.798 x 10 11 ) x (7.15 x 1012K) =1343 KI.
Hence Cv (223IK)2 (e223T/(2X298l )2 (a) 3R = 298K x e223 I/298 _ I = 10.031
I·
Cv _ (343 K)2 (e 223T / (2X298 ) 2 _ (b) 3R x 1 0 .89 7 1. 298K e 343/298  1 COMMENT.
For many metals the classical value is approached at room temperature; consequently, the fail
ure of classical theory became apparent only after methods for achieving temperatures well below 25°C were developed in the latter part of the nineteenth century.
PS.5
The hydrogen atom wavefunctions are obtained from the solution of the Schrodinger equation in Chapter 10. Here we need only the wavefunction that is provided. It is the square of the wavefunction that is related to the probability (Section 8.4).
4
3
8r = 3rrTo'
TO
= 1.0 pm.
If we assume that the volume 8r is so small that l/f does not vary within it, the probability is given by
164
STUDENT'S SOLUTION S MANUAL
l/I 2or =:34
(a) r = 0 :
l/I 2 or
(b) r = C/o:
=
(1.0)3 53 = 19.0 x 10
6
1.
~3 (~)3 e 2= 11.2 x 1061. 53
Question. If there is a nonzero probability that the electron can be found at r = 0 how does it avoid destruction at the nucleus? (Hint . See Chapter 10 for part of the solution to this difficult question. ) PB.7
According to the uncertainty principle,
where !:"q and !:"p are rootmeansquare deviations:
To verify whether the relationship holds for the particle in a state whose wavefunction is
we need the quantummechanical averages (x) , (x 2 ) , (p ), and
(x )
2a)
= (;
1/ 2
10000
xe
2m .2
dx
= 0;
/4e ax2dx _ 2) 1 00 (2a )1 /4e _ax2 x 2()1 2a (X = 00 IT IT
so!:,.q
(p )
I = ~ /2'
2a
=
10000
l/I '
(~dl/l) dx 1
dx
and
We need to evaluate the derivatives:
dl/l dx
= (2a) ; 1/4(2ax)e ax 2
(P2).
(2a)1 /21 00 xe2 
IT
oo
_ 2ax
2
dx ,
QUANTUM THEORY : INTRODUCTION AND PRINCIPLES
165
and
OO
So
(P)
J
=
 00
(2a)I /4 ;
e
ax2
(It) T
(2a)I /4
;
(2ax)e
aX2
dx
2It (2a) 1/ 2 J OO xe 2{u 2dx _ O., _  ;I rr 00
(p2)
=
2a) 1/2 (
(2alt2) ( rr
2a
rr
l
/
2(2a)
2
rr
l
2
)
~ /2 = alt2; (2a)
3/ 2 
/
and
Finally,
=
6.q 6.p
I 1/2 ""I /2 x a
2a
It =
I Ii.,
2
which is the minimum product consistent with the uncertainty principle.
Solutions to theoretical problems PS.9
p
Srrhe (
= ;:s
1 e hc / AkT
I
_
)
[S.5]
As ).. increases, he j AkT decreases, and at very long wavelength he / AkT exponential in a power series. Let x = hej AkT, then e
I 2 I 3 = I +x + x + x + ...
x
2!
3!
Srrhe [
P=;:S .
I
I
+x + 
2!
Srrhe [
A~mOO P = ;:s
I
x2
+x 
]
I
+  1x 3 + ... 3
I
I
'
]
I
Srrhc (
= ;:s
SrrkT )..4
Thi s is the RayleighJeans law [S.3].
'
I
I ) hej AkT .
«
1. Hence we can expand the
166
STUDENT'S SOLUTIONS MANUAL
[ 00
PS.11
£
Let x
£
= 10
[ 00
P (A) dA
= Srehe 10
~
~
~IT
AkT
A kT
he
=
SrekT
[ 00
10
A2dx A5 (eX I)
he
PS.13
(J
=
We require
10
[ 2n
0
SrekT
3dxx (eX  1)
[ 00 ::_dx __
10
A3 (eX l)
3 (re = 8rekT (kT) 
4 )
15'
he
J 1{!*1{! dr = 1, and so write 1{! = Nf and find N for the givenf·
00 r2
(a4
!
3
x (2) X (2re) = 1 if N =
)
2
= 10 [ 2n
by symmetry with
=
[ 2n
cos I/> dl/>
10
(rea)
1/ 2
fansin3 BdB fa2n cos2 1/>dl/>
x r 2 e r / a dr
We have used J sin 3 B dB
10
1
°O
=
I (2 re 5k4 / ISh3 e2 ) Iis the StefanBoltzmann constant.
= N2 (d) N 2
1)' [S.S].
=  . Then,dx = 2dA ordA =    d x.
= 8rekT ( kT) 3
where
dA A5 (ehc/ AkT _
[x
= r cos I/> sin B]
~ (cos B)(sin 2 B + 2) , as found in tables of integrals and
2 sin I/> dl/>
2 (cos I/>
+ sin 2 1/»
[ 2n
dl/> =
10
dl/> = 2re.
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
PB.1S
In each case form Qf. If the result is wf where w is a constant, then f is an eigenfunction of the operator Q and w is the eigenvalue [8. 2Sb]. dk dx e' ;f
(b)
d.x cos kx
d
!
k
Iyes; eigenvalue =ik.I
= ik e''kx ;
(a)
(c)
PB.17
.
= k SIO kx;
no.
= 0; I yes; eigenvalue = 0 I·
(d)
d d.x kx
1
(e)
d ax2 d.x e = 
= k = :; kx;
no [) /x is not a constant].
2ax e 
a 2 .1
;
no [  2ax is not a constant].
Follow the procedure of Problem 8.15 . (a)
; 2e ikx = _k2e ikx ; yes; eigenvalue = lk2
(b)
d.x 2 cos kx = _k 2 cos kx; yes; eigenvalue = ~.
(c)
d.x 2 k
1·
~
d2
d2
= 0;
yes; eigenvalue
= [2].
2
(d)
d d.x 2 kx
(e)
~ 2 d.x2 e ax
= 0;
yes; eigenvalue
= [2]. 2
= (2a + 4a 2x2 )e ax
;
no.
d2 d2 d Hence, (a, b, c, d) are eigenfunctions of  2 ; (b, d) are eigenfunctions of  2' but not of  . d.x d.x d.x PB.19
167
The kinetic energy operator, T, is obtained from the operator analog of the classical equation
that is, , (jJ) 2 T=
2m'
,
Px
Ii d d.x [8.26] ;
=i
hence
and
Then (T)
= N2
f
~
1/1* (p; ) 1/1 dr: 2m ~
=
J l{I* (p2/2m) l{I dr: J 1/1*1/1 dr:
.
[N 2=
)]
J 1/1*1/1 dr:
.
  J 1/I*(e'kx cosx + e'kx sin X)dr: 2m d.x 2 J 1/1*1/1 dr:
_1i2 2k;kx ~2::.cm=_J_1/I_*_(_k_)_x;;(e_'_·f_c_o_s_x_+_e_  '_s_in_x_ ) _dr: = 1i2 k2 J 1/1*1/1 dr: = 11i2 k 2 1 Jl{I*1/Idr:
2mJ1/I*1/Idr:
2m '
168
STUDENT'S SOLUTIONS MANUAL
PS.21 1/ 2
N
(a)
1
5
00
(r2)
= ( _13 )
[Problem 8.14].
32rca o
6
+ r 2 )
er/ll{] dr = 1( 4 x 4!  4 x 5! 8a0 0 ao a 8a 3 0 0
=  13
(
4r4

4r
= 142
+ 6!)a6
a~ I·
1/ 2
1J;
(b)
= Nr sin () cos t/> e r / 2ao ,
(r)
= 132rca6
PS.23
1
00
0
N
= ( _15 )
[Problem 8.14].
32rcao
r 5 e r / ao dr x 4rc 3
= I24a6
~ x 5!a~ =~.
The superpositions of cosine functions of the form cos(nx) can be chosen with n equal to any integer between 1 and m. For convenience, x can be examined in the range between rc/2 and rc/2. The normalization constant for each function is determined by integrating the function squared over the range of x [8.15]. Using MathCad to perform the integration, we find:
G.
G.
2 I ( 2 . cos rc . n) sin rc . n) + rc . n) cos(n·x) dx +  . ''''.''"n / 2 2 n n/ 2
1
When n is an even integer, sin(rcnI2) = 0 and, when n is an odd integer, cos(rcnI2) = O. Consequently, when n is an integer, the above integral equals rc/2 and we select (2/rc)I /2 as the normalization constant for the function cos(nx). The normalized function is t/>(n, x). The superposition, 1J;(m, x), is the sum of these cosine functions from n = I to n = m. Since the cosine functions are orthogonal, 1J;(m,x) has a normalization constant equal to (11m) 1/ 2. t/>(n,x) :=
A·
cos(n· x)
1J;(m, x) :=
(T m Y;;; . L t/>(n,x) 11=1
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
169
Constants and variables needed for MathCad plots and computations: IT
N:= 1000
Xmin :=
2
IT
Xmax :="2
i :=O .. N
I
Xi := xmi n
+ N . (Xmax
 Xmin)
Examination of Fig. 8.1(a) reveals that, when the superposition has few terms, the particle position is illdefined. There is great uncertainty in knowledge of position . However, when many terms are added to the superposition, the uncertainty narrows to a small region around x = O. A plot with m greater than 10 will further confirm this conclusion. Each function in the superposition has been assigned a weight equal to the normalization constant (l I m ) 1/ 2 [8.33] . This means that each cosine function in the superposition has an identical probability contribution to expectation value for momentum (see Justification 8.4). Each cosine function contributes with a probability equal to 11m. Furthermore, each cosine function represents a particle momentum that is proportional to the argument n because (d 2 ¢ (n , x)ldx 2 ) 1/ 2 (the differential component of the squared momentum operator) is proportional to n. The following plot [Fig. 8.1(b)] of momentum probability against momentum, as represented by n, is an interesting contrast to the plot of probability density against position. Variables needed for the MathCad plot: n :=
0..15
Prob(n, m) := if (n < m, ~ ,
0)
This plot shows that, when there are many terms in the superposition, the range of possible momentum is very broad even though the range of observed positions becomes narrow. Position and momentum are Probability density plots [Fig. 8.1(a)] for superpositions that have 1, 3, and 10 terms: 8r,~r_,
6
'i'(l,XI)2 'i'(3,xY
4
'i'(10,XI)2
2
......... ..... . .... ~
............ o.~
Figure 8.1(a) complementary variables. As location becomes more precise with the superposition of many function s, precise knowledge of momentum decreases. This illustrates the Heisenberg uncertainty principle [8.36a] .
170
STUDENT'S SOLUTIONS MANUAL
1
Prob(n , 1)
1
0.8

0.6

Prob(n ,S)

Prob(n , 10) 0.4 I
0.2


.... ...........
00
\
~
....j
IS
10
S
Figure 8.1(b)
n
The plot of probability density against position clearly indicates that the superposition is symmetrical around the point x = O. Consequently, the expectation position for all superpositions is x = O. The expectation value for position is independent of the number of terms in the superposition. The square root of the expectation value of x 2 is called the rootmeansquare value of x, x rms. A plot of against m [Fig . 8.1 (c)] indicates that this expectation value depends upon the number of terms in the superposition. However, it does appear to very slowly converge to a very small (zero?) value when the superposition contains many functions.
Xrms
Xrm s(m): =
,,/2 1 (
2
x .1/I(I1l, x)2 dx
)1 /2
m:= 1 . . . 50
,, /2
0.8
Xms(m)
0.2
10
20
30
m
40
so Figure 8.1(c)
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
PS.25
171
(a) In the momentum representation Px = Px x , consequently [x, Px] 1> = [x ,Px x] 1> = xPx x 1>  Px x x1> = ilup [8.39] Suppose that the position operator has the form x = a(d j dpx) where a is a complex number. Then,
d (Px x dpx dpx 1> dpx
iii 1»  Px x (d 1> ) = 1>, dpx
a
+ Px d1>
d1> iii  Px = 1> [Rule for differentiation ofJ(x)g(x)]. dpx dpx a
iii 1> = 1> a This is true when a = iii. We conclude that I x = ili(d j dpx) I'in the momentum representation. (b) The fact that integration is the inverse of differentiation suggests the guess that in the momentum representation
x  I1>=
(ilidpxd)
 I
If
( 1>=:Iii

1f
P "
dpx ) 1>=:00 Iii
PX
1>dpx,
 00
where the symbol J~'~ dpx is understood to be an integration operator which uses any function on its right side as an integrand. To validate the guess that xI = (1 j ili) J~"oo dpx we need to
I
confirm the operator relationship x I x = xxI = integrals:
=
(f
P '
 00
I
i. Using Leibnitz's rule for differentiation of
d1> ) +1>(Px) li~ d de 1>(oo)d dp x = ddpx Px c 00 Px Px
(f
PX
 00
d1> dpx) 1>(00) . d Px
Since 1> (  00) must equal zero, we find that
from which we conclude that xx  I =
x  I x1> =
(
If
ill
PX
00
dpx )
(
i.
d)
iii dpx 1> =
f
PX
00
d1> = 1>(Px) 1>(00) = 1>(Px) = 1>.
172
STUDENT'S SOLUTIONS MANUAL
Solutions to applications PS.27
(a)
Anonrelativistic
=
h (2m e V) 1/2 e 6.626 {2 (9.109
Anonrelativistic
= 5.48
1O 3I kg)
X
X
10 34 J s
X
(1.602
X
10 19 C)
X
X
10 3 V)} 112·
X
10 3 V)
(50.0
pm [8 .12 and Example 8.2]. h
Arelativistic
=
{(ev)}c eV +
/2 I=
2m e
1
2mec2
5.48 pm
Anonrelativistic
( eV) 1+2mec2
=
1/2
5.48 pm {l+0.0489}1 /2
I
+
(1.602
{
2(9.109
X
X
10
19
C) (50.0
10 31 kg) (3 .00
X
JI /2
108 ms I )2
= 1 535 m . p I·
(b) For an electron accelerated through 50 kV the nonrelativistic de Broglie wavelength is calculated to be high by 2.4%. This error may be insignificant for many applications. However, should an accuracy of 1% or better be required, use the relativistic equation at accelerations through a potential above 20.4 V as demonstrated in the following calculation. Anonrelativistic 
Arelativistic
=
Anonrelativistic _
Arelativistic
1= (1 + ~) 1/2 _ 1 2mec2
Arelativistic
1(
= , + 2:
ev)
2mec2
1 ·3
+ 2 .4 .6
(
2mec
2
eV )3
2mec2
eV )
= I (   2
eV )2
1 (  2 . 4 2mec2
 ...
71
because 2nd and 3rd order terms are very small.
The largest value of V for which the nonrelativistic equation yields a value that has less than 1% error: c2 V :::::: 2 ( 2m e ) e PS.29
(a)
C~(g) +
X
~
(Anonrelativistic ArelativistiC) Arelatlvlstlc
C(graphite)
= 2 ( 2me
c2
) (0.01)
= 20.4 kV.
e
+ 2H2(g).
b.rG'
= b.fG'(CH4) = (50.72kJmo\I) = 50.72 kJ mo\I
b.rW
= b.fW(CH4) = (74.81 kJ molI) = 74.81 kJ mol I at 'f .
at 'f
= 25°C.
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
173
We want to find the temperature at which !:!"rc?(T) = O. Below this temperature methane is stable with respect to decomposition into the elements. Above this temperature it is unstable. Assuming that the heat capacities are basically independent of temperature, !:!"rC;(T) "'" !:!"rC;cn
= [8.527 + 2(28.824) : 35.31]1 K I molI l
"'" 30.8651 K 1mol
!:!,.rW(T) = !:!"rW(T)
f:
+
.
!:!"rC;(T)dT [2.36]
= !:!"rW('f) + !:!"rC; X a (!:!"rc?)) ( aT Tp
= !:!"rW T2
(T  T).
[3.52].
At constant pressure (the standard pressure)
fT !:!"rH~ T f T d(!:!"rc? I T) =  T 2dT, T =
!:!"rc?(T)
!:!"rc?( T) _
T
T
fT !:!"rH~ ( T) + !:!"rC;
The value of T for which !:!"rc?(T) !:!"rc?(T)
T
X
(T  T) dT
T2
T
=
0 can be determined by examination of a plot (Fig. 8.2) of
.
agamst T.
!:!,. c?(T) r T
= 50.72 kJ mol  I1298.15 K = 0.1701 kJ K I molI .
(10 k1) 3
x (298K) x
1
= 65.16 kJ mol  I. !:!"rC;
= (30.865JK I molI)
( 10 kJ) = 0.030865kJK I molI. 3
x
J
With the estimate of constant !:!,.rC; , 1methane is unstable above 825 K I.
174
STUDENT'S SOLUTIONS MANUAL Methane decomposition
0.20
0.15 I
"0 E
.,
0.10
~
.
"'"
:::::::
h ........
0.05
h' ¢
0.00
"i 0.05
0.10 200
400
600
800
1000
1200 1400
1600
T/K
(b)
Amax =
Amax
=
Figure 8.2
1(l .44cm K) 5 T [See the solution to Problem 8.10],
~(I.44cm K) 1000 K
1Amax (1000 K)
(c) Excitance ratio
=
9
4
= 2.88 x
10
cm
(10 nm) 102 cm '
= 2880 nm I·
M (brown dwarf) M(Sun)
(1000K)4 (6000 K)4
a T,4
=
brow: dwarf
a TSun
=17.7
x 10 4
[See the solution to Problem
8.) I]
I.
. . p (brown dwarf) Energy denSity ratio = '',:,'p(Sun)

8nhe/A 5
1)) 1))
(1/(e(hc/ AkTbrowndWarf) 
8nhe/A5 (1/(e(hc/AkTs un ) e (hc/!..kTS un )

[85]
.
1
(e(hc/AkTbrowndwarf) 
I)'
The energy density ratio is a function of A so we will calculate the ratio at Amax of the brown dwarf.
he Abrown dwarfk
(6.62 x 10 34 Js) x (3.00 x 10 8 ms l ) (2880 x 10 9 m) x (1.381 x 10 23 J Kl)
= 4998K.
QUANTUM THEORY: INTRODUCTION AND PRINCIPLES
Energy density ratio =
175
e 4998 K j TSun  I 4""'9"'"98"""j= T    
e
brown dwarf 
1
e(4998j6000)  I
1.300
e (4998j IOOO)  I
147
= 18.8
X 10
3
1.
(d) The wavelength of visible radiation is between about 700 nm (red) and 420 nrn (v iolet). (See text Fig. 8.2.)
Fraction of visible energy density
=
 I4 11420nm p(.l..)d.l.. aT 700nm
I
[8.3 , 8.5, and solution to problem 8.11]
=  C4 11420nm p(.l..)dA I . 4aT
700nm
As an estimate, let us suppose that peA) doesn ' t vary too drastically in the visible at lOOO K. Then, 420 nm
p(.l..)dA
I ~ p(560 nm)
x (700 nm  420 nrn)
11 700nm ~
8n:hc ) x ( I ) x ( 280 x 10 9 m ) ( (560 x 10 9 m )5 e«4998K j IOOOK)x(2880nm j 560nm))  1
34
J s) x (3.00 x 10 8 m S I) ( I ) 1.97 X 10 25 m 4 e 25 .70  I
= 8n: (6.626 x 10
= 1.75 x
10
10 J m  3 .
Fraction of vis ible energy den sity ~
(3.00 x 10 8m s l ) x (1.75 x 1O2
IO Jm 3 )
4 (5 .67 x 10 8 Wm  K4) x (lOOOK)
~ 12.31
x 10 7
1.
Very little of the brown dwarf's radi ation is in the visible. It doesn ' t shine brightly.
4
9
Quantum theory: techniques and applications
Answers to discussion questions 09.1
In quantum mechanics, particles are said to have wave characteristics. The fact of the existence of the particle then requires that the wavelengths of the waves representing it be such that the wave does not experience destructive interference upon reflection by a barrier or in its motion around a closed loop. This requirement restricts the wavelength to values A = 21n x L , where L is the length of the path and n is a positive integer. Then using the relations A = hi p and E = p2/2m, the energy is quantized at E = /1 2 h 2 18mL 2 . This derivation applies specifically to the particle in a box, the derivation is similar for the particle o n a ring; the same principles apply (see Section 9.6).
09.3
The lowest energy level possible for a confined quantum mechanical system is the zeropoint energy, and zeropoint energy is not zero energy. The system must have at least that minimum amount of energy even at absolute zero. The physical reason is that, if the particle is confined, its position is not completely un certain, and therefore its momentum, and hence its kinetic energy, cannot be exactly zero. The particle in a box , the harmonic oscillator, the particle on a ring or on a sphere. the hydrogen atom, and many other systems we will encounter, all have zeropoint energy.
09.5
Fermions are particles with halfintegral spin, 1/2, 3/2, 5/2, ... , whereas bosons have integral spin, 0, I , 2, .... All fundamental particles that make up matter have spin 1/2 and are ferrnions , but composite particles can be either fermion s or bosons. Ferrnions: electrons, protons, neutrons, 3He, .... Bosons: photons, deuterons.
Solutions to exercises E9.1(b)
E
= /72
2 2 /1 /7  2 [9.4a] 8meL
(6.626 x 10 34 J s)2
_
.,,'..,......,;::',,,;= 2.678 X 31 9 8(9.109
X
10
kg) x (1.50 x 1O m)2
10 20 J
The conversion factors required are leV
=
1.602 x 10 19 J ;
I cm 
I
=
1.986 x 10 23 J ;
leV
= 96.485 kJ
mol 
I
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 2
h 20 (a) E3  EI = (9  1 )  2 = 8(2.678 x lO J) 8m eL
I
= 12.14 x 10 19 J 1= 11.34eV = 11.08 x 104 cm  1 1= h2 (b) E7  E6 = (49  36)2 = 13(2.678 8me L
X
I 129kJmol 1 I
10 20 J)
I
I
= 13.48 x 10 19 J = 12.17eV = 11.75 x 104 cm 1 1= 12lOkJmol 1 1
E9.2(b)
The probability is
J1fr*1fr
P=
dx =
~
J
sin
2
C~X) dx ~ 2~X sin 2 C~X)
where t.x = 0.02L and the function is evaluated at x = 0.66 L.
E9.3(b)
(a)
For n = I
P=
(b)
For n = 2
P=
~
2(0.02L)
sin 2 (0.66JT) = ~
2(0.02L)
sin 2[2(0.66JT)] = ~
L
L
~
The expectation value is
J1fr*p1fr
(P) =
dx
but first we need p1fr A
p1fr
d (2) 1/2. nJT x sm ( L ) =
In dx L •
=
2innJT r uJo
L
A
so 1jJ) =
sin
. (2) 1/2 nJT
In L
L
nJT x cos ( L )
(nJTX) (nJTX) Inl L cos L dx = L2.J
for all n. So for n = 2
E9.4(b)
The zeropoint energy is the groundstate energy, that is, with nx = ny = n z = I: E=
(n 2 + n 2 + n 2 )h 2 3h 2 x Y 2 z [9.12b with equal lengths] =  2 8mL 8mL
Set this equal to the rest energy me 2 and solve for L: 3h2 me 2 =  8mL2
soL =
(~)1 /2 ~ 8
me
=
(~)1 /2 AC 8
where AC is the Compton wavelength of a particle of mass m.
177
178 E9.5(b)
STUDENT'S SOLUTIONS MANUAL
1/15
= ( L2)1 /2 sin (5JTX) L
· ' " III P() dP(x) M aXlma andrrumma x correspon d to  = 0 dx d P (x) ex d1/l dx dx2 ex sin (5JT L X) cos (5JTX) L ex sin (IOJTX) ~
sin e
L
= 0, x = L 11'
E9.6(b)
= sin 2ex]
= 0 when e = 0, JT , 2JT , . . . , n'JT (n' = 0, 1, 2, . .. ) 10JTX , =nJT
x
[2 sin ex cos ex
for
11'
< 10 
so
n'L
x = 
10
are minima. Maxima and minima alternate, so maxima correspond to
= 1, 3, 5, 7, 9
x
=1
~ I, 1~~ I, [TI,1~~ I, 1~~ 1
The energy levels are
where £1 combines all constants besides quantum numbers. The minimum value for all the quantum numbers is 1, so the lowest energy is
The question asks about an energy 14/3 times this amount, namely 14£1 . This energy level can be obtained by any combination of allowed quantum numbers such that
The degeneracy, then, is []], corresponding to (nl , n2 , 113) = (1 , 2,3), (I , 3, 2), (2, 1, 3), (2, 3, I), (3 , 1, 2), or (3, 2, I).
E9.7(b)
£
= ~kT is the average translational energy of a gaseous molecule (see Chapter 17). 3
£ = kT = 2
£
=
G)
8mL 2
n2h2 8mL2
x (1.381 x 10 23 JK 1) x (300K)
n2 =  £ 2
h
(n 2 + n2 + n2)h 2 1 2 3 8mL2
= 6.214 x
10 21 J
QUANTUM THEORY : TECHNIQUES AND APPLICATIONS
(6.626
X
10
34
J S)2
= 1.180 x 10 42 J
0.02802 kg molI ) 2 x 100m ( 6.022 x 1023 molI
(8) x
21 2 6.214 x 10 J 21. n = 42 = 5.265 x 10 , 1.180 x 10 J
t:;.E = (2n
179
+ I)
x
h2
(
 2
8mL
)
n= !7.26 x 101O!
= [(2) x (7.26 x 10 10 )
+ 1]
x
(
h
2
2
8mL
)
=
14.52 x 1O 2
IO 2 h
8mL
= (14.52 x 10 10 ) x ( 1.180 x 1042 J) = ! 1.71 X 10 31 J! The de Broglie wavelength is obtained from h h A =  =  [8.12] p mv The velocity is obtained from EK
= !mv 2 = ~kT = 6.214 x 10 21 J 6.214
v2 = 1) ( A=
2"
X
10 21 J
= 2.671
X
105 m 2 s 2;
v = 517 m S I
(0.02802 kg molI ) x 6.022 x 1023 mol I
6.626 x 10 34 J s =2.75 x 10 11 m=127.5pml (4.65 x 10 26 kg) x (517 m S I)
The conclusion to be drawn from all of these calculations is that the translational motion of the nitrogen molecule can be described classically. The energy of the molecule is essentially continuous,
t:;.E
E «< E9.8(b)
1.
The zeropoint energy is 1
1 (k)1 /2
Eo = !'u.v = Ii 22m
1 = (\.0546 2
X
10 34 Js) x
(
285 Nm I )1 /2 26 5.16 x 10 kg
= ! 3.92 x 10 21 J ! E9.9(b)
The difference in adjacent energy levels is
(
k)
t:;.E = Ev+ 1  Ev = !'u.v [9.26] = Ii;;;
so k
m(t:;.E)2
= ::1i2
(2 .88
X
1/ 2
[9 .25]
10 25 kg) x (3 . 17 X 10 21 J) 2 ! _I ! ( 1.0546 X 1034 J s)2 = 260 N m
180 E9.10(b)
STUDENT'S SOLUTIONS MANUAL
The difference in adjacent energy levels, which is equal to the energy of the photon, is ~E
= Iiw = hv
,,(_k )1 /2
so "
he A
m
and A=
h; (~y /2 = 2Jre(T)
= 27T(2.998 x
1/2
8 I ((l5.9949U) x (1.66 x 1O 27 kg U I»)1 /2 10 ms ) x I 544Nm
A = 1.32 x 10 5 m = 113.2 I1m 'l E9.11 (b)
The difference in adjacent energy levels, which is equal to the energy of the photon, is ~E
and A =
= Iiw = hv
he (k ) n 111
1/ 2
so
k) 1/2
he
n(;;;
A
1/ 2 = 27Te (m) 
k
Doubling the mass, then, increases the wavelength by a factor of 21 / 2. So taking the result from Exercise 9 .1O(b), the new wavelength is A = 21 /2 (1 3.2 11m) = 118.7 11m ~E
E9.12(b)
I
= Iiw = hv
(a)
~E =
(b)
~E = Iiw = n(~) 1/2
hv = (6.626
X
10 34 1 HzI ) x (33 x 103 Hz) = 12.2 x 10 29 1
meff
I [ m eff
=  1 + 1 m,
m2
with ml
I
= m2 ]
For a twoparticle oscillator meff , replaces 111 in the expression for w. (See Chapter 13 for a more complete discussion of the vibration of a diatomic molecule.) 2k)I /2
~E = n( ;;;
= (1.055
X
10
34
( 1 s) x
(2) x ( ll77Nml) )1 /2 (16.00) x (1.6605 x 1027 kg)
=13 .14 x 10 20 11 E9.13(b)
The first excitedstate wavefunction has the form 1{1
= 2NIyexp (!l
)
where NI is a collection of constants and y == x(mw/n) 1/2. To see if it satisfies Schrodinger's equation, we see what happens when we apply the energy operator to this function 2
. n2 d 1{1 1 2 X2 1{1 H1{I=+111W 2 2m dx
2
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
181
We need derivatives of l/f
(nut»
dl/f = dl/f dy = dx dy dx
and
:~ = ~:~
n
(:r
, =  n2m
x
X
(l _ i ) x exp ( ~i)
2
(m;)
(mw) x (y
2
So Hl/f
=
1/ 2(2NI)
2
n
x (2NI) x (3y
 3)1/f
1 2 = "2 nw x (y  3) x l/f
+
+ y3)
X
exp (
~i) =
C:
W ) x
(i 
3)l/f
1 2 2 + mw x l/f
2
1 2 "2 nwy l/f
3
= "2nwl/f
Thus, l/f is a solution of the SchrOdinger equation with energy eigenvalue
E9.14(b)
The harmonic oscillator wavefunctions have the form
1/fv(x )
1 )
= NvHv(y) exp ( "2 i
with y
x
= ;;;
and a
=
(
n2 ) 1/4
mk
[9 .28]
The exponential function approaches zero only as x approaches ±oo, so the nodes of the wavefunction are the nodes of the Hermite polynomials. Hs (y )
= 32i
 160y3
= 0, which leads to x = O. The other factor can be made into a quadratic equation
So one solution is y by letting z =
i
4 z2  20z + 15 so
z=
+ 120y = 0 [Table 9.1] = 8y (4l 20i + l5)
b ± .Jb2
=0 
4ac
20 ± .J202  4 x 4 x 15
2a
2 x 4
Evaluating the result numerically yields x=10,±0.96a, or ±2.02a I.
z=
5±
.JTO 2
0.92 or 4 .08, so y = ±0.96 or ±2.02. Therefore
COMMENT. Numerical values could also be obtained graphically by plotting H5(y )'
E9.1S(b)
The zeropoint energy is Eo
= ~nw = ~n(~) 1/ 2 2
2
meff
For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so I Eo =  ( 1.0546 2 Eo
= 12.3421
X
X
10 34 J s) x
10 20 J 1
( 2 2 9 3 . 8 N m 1 ~(l4.0031 u) x (1.66054 X 10 27 kgu I )
)1 /2
182
E9.16(b)
STUDENT'S SOLUTIONS MANUAL
Orthogonality requires that
f if m
1/1,;,1/1" dr
= 0
f=. n.
Performing the integration
If m
f=.
/1 , then
f
1/1* 1/1 dr
III"
2
=
27T
N e i (II_lIIl1 i(n  m) 0
2
=
N (I  1) = 0 i(/1  m)
Therefore, they are orthogonal. E9.17(b)
The magnitude of angular momentum is
Possible projections onto an arbitrary axis are
where
m, = 0 or ± I or ±2. So possible projections include
I0, E9.18(b)
± 1.0546 x 10 34 J sand ±2.1109 x 10 34 J s I
The cones are constructed as described in Section 9.7(d) and Figure 9.40(b) of the text; their edges are of length {6(6 + J)} 1/ 2 = 6.48 and their projections are mj = +6, +5 , ... , 6. See Figure 9.1 (a). The vectors follow, in units of n. From the highestpointing to the lowestpointing vectors (Figure 9. J(b », the values of are 6, 5, 4, 3, 2, 1,0, I ,  2,  3,  4, 5 , and 6.
m,
~=====! In = + 6 ~>.."~:z.. + 5
"'">~z + 4 ~~~~ + 3
~~~~~.+2
S~===:::::::====:::20+ I
~~~  1 ~~~ 23 r~  4  5
Figure 9.1(a)
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
183
Figure 9.1(b)
Solutions to problems Solutions to numerical problems P9.1
E
=
n2h2 8mL2 '
=
We take m(02)
E2  E 1 =
We set E
E2  EI
(32.000) x (1.6605 X 10 27 kg), and find
1 39 1 (3) x (6.626 x 10 34 J s)2 = 1.24 x 10 J. (8) x (32.00) x (1.6605 x 10 27 kg) x (5.0 X 10 2 m)2
n2 h 2 8mL2
=
=
3h 2 8mL2'
=
1
2kT and solve for n.
From above, h2 18mL 2 = (E2  E I) 13 = 4.13 x 10 40 J; then n 2 x (4. 13 x 10 40 J)
We find n
=
G)
x (1.381 x 10 23 JK  1) x (300K)
= 2.07 x
10 21J.
9 2.07 X lO 2I J)1 /2 40 = 12.2 x 10 I· 10 J . .
= ( 4. 13 x
At this level,
h2 h2 h2 E"  E,,_I = (n 2  (n  1)2 ) x  2 = (2n  I) x  2 ~ (2n) x  2 8mL 8mL 8mL
= P9.3
E
m 2Ji2
m 2Ji2
21
2mr
=  '[9.38a] =  '2
Eo = 0
EI
(4.4 x 109 ) x (4. 13 x 1O4o J) [l
~ 11.8
x lO 30 J 1 [or 1.1I!Jmol I ].
= mr2 ] .
[m, = 0].
Ji2
== 2mr2
(1.055 x 10 34 J s)2 (2) x (1.008) x (1.6605 x 10 27 kg) x (160 X 10 12 m)2
The minimum angular momentum is ±Ji I. 1
=
1
1 130 xlO 22 J . .
184 P9.S
STUDENT'S SOLUTIONS MANUAL
(a) Treat the small step in the potential energy function as a perturbation in the energy operator:
{O
H(I ) =
£
0
for :'S x :'S (l / 2)(L  a) and (l / 2)(L + a) :'S x :'S L for (l / 2)(L  a) :'S x :'S (l / 2)(L + a) .
The firstorder correction to the groundstate energy, EI, is
E~I )
L
=
Ioo
(2)
= j(l /2)(L+a) (I /2)(L a)
j(i /2)(L+a) . 2 (JtX) Sin dx L (I /2)(L a) L
EI
= 2 £
E (I )
=
( I)
o/iO)*H(i ) o/~O) dx
1/ 2
L
£
(2) 
L
1/ 2
Jtx sin (  ) dx, L
(JtX) . (JtX)) li / 2(L+ a) Jtx  L cos Sin , LJt L L i/2(La)
= £
(
sa _ ~ cos (Jt(L + a») sin (Jt(L + a») L Jt 2L 2L
i
Jtx sin (  ) L
+ ~ cos (Jt(L Jt
2L
a») sin (Jt(L  a)). 2L
This expression can be simplified considerably with a few trigonometric identities. The product of sine and cosine is related to the sine of twice the angle:
. (Jt(L±a») = 1. (Jt(L±a») = 1. ( Jta) , Sill Sin Jt ± cos ( Jt(L±a») Sin 2L 2L 2 L 2 L and the sine of a sum can be written in a particularly simple form since one of the terms in the sum is Jt : sin ( Jt
±
7) =
Thus E (I) = sa I L (b) If a
. (Jta) + Jt£Sin L
(~a) ± cos Jt sin (~a) = 'f sin (~a) . .
= L/ IO, the firstorder correction to the groundstate energy is EI(I )
P9.7
sin Jt cos
£ . (Jt) = 10 + ;£ Sin 10 = I0.1984 £.I
The secondorder correction to the groundstate energy, EI, is
where H
(I) _
(0) _
 mgx'o/lI

. nJtx Sin   ,
L
2 _ n2h and En    2 ' 8mL
The denominator in the sum is
E(O) _ E(O) i
n
=
_h_2_ _ _n2_h_2 8mL2 8mL2
= _(1__n;:2);h_2 8mL2
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
185
The integral in the sum is
L 1/I (O)* H
loo
where a
/I
( I ) 1/1 (0) dx I
2mg = loL x sinaxsinbx dx L
0
'
= nn l L and b = n i L.
The integral formulas given with the problem allow this integral to be expressed as 2mg d l o L . d (cos (a  b)x cos(a + b)X) IL  cos axsmbxdx = mg 'L da 0 da 2(a  b) 2(a + b) 0 _ 
 2mg L
(x Sin (ab)X 2(a  b)

cos(ab)x 2(a  b)2
L
+ xsin(a + b)x + cos(a+b)x)I :02(a + b) 2(a + b)2 0 .
The arguments of the trigonometric functions at the upper limit are: (a  b)L
=
(n  I)n and (a
+ b)L = (n + I) n.
Therefore, the sine terms vanish. Similarly, the cosines are ± 1 depending on whether the argument is an even or odd mUltiple of n ; they simplify to ( _I)Il+ I . At the lower limit, the sines are still zero, and the cosines are all I. The integral, evaluated at its limits with Tt l L factors pulled out from the as and bs in its denominator, becomes mgL (1)11+ 1  I _ ( 1)11 + 1 n2 (n  1)2 (n + 1)2
I)
= mgL[ (I)n+1  I] (n + 1)2  (n _ 1)2) n2 (n _ 1)2(n + 1)2 '
4mgL[(I)n+1  l]n
n 2 (n 2

1)2
The secondorder correction, then, is
Note that the terms with odd n vanish. Therefore, the sum can be rewritten, changing n to 2k , as
The sum converges rapidly to 4.121 x 10 3 , as can easily be verified numerically; in fact, to three significant figures , terms after the first do not affect the sum. So the secondorder correction is (2) _ 
£1
186
STUDENT'S SOLUTIONS MANUAL
The firstorder correction to the groundstate wavefunction is also a sum: ./. (I )
'PO
= '""" Cno/n ./. (0) ' ~
II 4mgL[(I )II+ 1  l]n
where c
=
rL ./. (0) 0 HI ./.(0) dx _JO 'I'll
'1'1
£,\0) _ £~O)
II
2
=_
2
2
(n  I ) «n2 _ I )h2 j 8mL2) Jt
32m2gL 3[(l)n + I]n Jt 2h 2(n 2  1)3
Once again, the odd n terms vanish. How does the firstorder correction alter the wavefunction ? Recall that the perturbation raises the potential energy near the top of the box (near L) much more than near the bottom (near x = 0); therefore, we expect the probability of finding the particle near the bottom to be enhanced compared with that of finding it near the top. Because the zeroorder groundstate wavefunction is positive throughout the interior of the box, we thus expect the wavefunction itself to be raised near the bottom of the box and lowered near the top. In fact, the correction terms do just this. First, note that the basis wavefunctions with odd n are symmetric with respect to the center of the box; therefore, they would have the same effect near the top of the box as near the bottom. The coefficients of these terms are zero: they do not contribute to the correction. The evenn basis function s all start positive near x = 0 and end negative near x = L; therefore, such terms must be multiplied by positive coefficients (as the result provides) to enhance the wavefunction near the bottom and dimini sh it near the top.
Solutions to theoretical problems PS.9
The text defines the transmission probability and expresses it as the ratio of IA11 2j 1A1 2, where the coefficients A and AI are introduced in eqns 9.14 and 9.17. Eqns 9.18 and 9.19 list four equations for the six unknown coefficients of the full wavefunction. Once we realize that we can set BI to zero, these equations in five unknowns are: (a) A + B = C + D, (b) Cel
(c) ikA  ikB = KC  KD , (d) KCel([  KDe I
We need AI in terms of A alone, which means we must eliminate B, C, and D. Notice that B appears only in eqns (a) and (c). Solving these equations for B and setting the results equal to each other yields B=C+DA=A
KC
KD
ik + ik ·
Solve this equation for C:
C=
2A+D(~ K
lk
ik + I
I)
2Aik
+ D(K K
ik)
+ik
Now note that the desired AI appears only in (b) and (d). Solve these for AI and set them equal:
QUANTUM THEORY: TEC HNIQUES AND APPLICATIONS
187
Solve the resulting equation for C, and set it equal to the previously obtained expression for C: (K
+ ik )De  2KL K 
2Aik
ik
+ D (K K + ik
ik )
Solve this resulting equation for D in terms of A: (K + ik) 2e 2KL  (K  ik)2 2Aik '','D=(K  ik ) (K + ik)
so
D
K + ik '
2Aik(K  ik) (K  ik) 2 .
= (K + ik )2e2KL _
Substituting this expression back into an expression for C yields
Substituting for C and D in the expression for A ' yields
The transmission coefficient is
The denominator is worth expanding separately in several steps. It is (K + ik )2(K  ik)2e  2KL  (K  ik )4  (K = (K2
=
+ k2)2(e2KL + e 2KL ) _
(K 4 + 2K2k 2 + k4) (e 2KL
+ ik )4 + (K
 ik )2(K
+ ik )2e2KL
(K2 _ 2iKk _ e)2 _ (K 2 + 2iKk _ k 2)2
+ e 2KL )
_ (2 K4 _ l 2K 2k 2 + 2k 2).
If the 12K 2k 2 term were 4K 2k 2 instead, we could collect terms still further (completing the square), but of course we must also account for the difference between those quantities, making the denominator
So the coefficient is
We are almost there. To get to eqn 9.20a, we invert the expression
188
STUDENT'S SOLUTIONS MANUAL
Finally, we try to express (K2 + k 2 ) / k 2 K2 in terms of a ratio of energies, c = £ / V. Egns 9.14 and 9.16 define k and K. The factors involving, 2, n, and the mass cancel leaving K ex ( V  £)1 / 2 and k ex £1 / 2, so [£
+ (V 
V2
£)f £(V  £)
£(V  £)
c (l  c)'
which makes the transmission coefficient
P9.11
We assume that the barrier begins at x = 0 and that the barrier extends in the positive x direction.
(a)
(b)
Question. Is N a normalization constant?
p2
,n d
withT= andp=. 2m i dx
P9.13
n
T=~~=  2ma n 2 d =~nw~ 2m dx 2 dyZ 2 dyZ' 2
2
[x =a y, a 2 =m
w]
which implies that
We then use 1jr = NHe  i /2 , and obtain
From Table 9.1 H~  2yH~ = 2vH v
i Hv
=y
GHv+I + VHv  I) = ~ (~Hv+2 + (v + l)Hv) + v (~Hv + (v 
= ~Hv+2 + v(v 2
d 1jr 1 Hence, 2 _ N [ Hv+2 dy 4
I)Hv2
+ v(v 
+ (v +
D
HI'.
I)H v2  ( v + I ) HI' ] e _ y2/2 . 2
1)Hv 2)
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
Therefore, (T)
= N 2 (~IUv) [dx
i:
Hv
[~Hv+2 + v(v  I)Hv2  (v +
D
i
Hv] e dx
= ady)
= aN2(11Uv) [0 + 0 (v + 1)n l / 22 v!] V
[i:
oo
P9.1S
(a)
(x)
HvHv' ey2dy
= faL
=0
if v' "1 v, Comment 9.2]
(~)'/2 sinC~X)x(~)'/2 sinC~X)dx
= (~) faL xsin 2 axdx [a = n:] =
(~)
L X
L
L
=
(X2 _ xsin2ax _ cos2ax)I = 4 4a 8a 2 0
(~) L
x (L2) 4
[by symmetry also].
2
L
2
ox = [ "3
I) 6n2n 2 "4 2
L ] 1/ 2
(
I 
(
=
L
I
T2 
1)
~
1/ 2
(P) = 0 [by symmetry, also see Exercise 9.2(a)),
(p2) = n2 h2/4L2 [from E
o p
=
opox (b)
(x)
= p2/2m, also Exercise 9.2(a)].
(n 2h2)1 /2 = \ nh \' 4L2 2L
=
I
= a2
(x 2) =
I)
nh ( 2L x L 12  2n2n2
i:
1jr2ydy [x
1/ 2
nh (
= 2v'3 I 
I)
24n2n2
Ii
1/ 2
>
"2 '
= ay] = 0 [by symmetry, y is an odd function),
I 2) = k 2 (V ) k2 ( "2kx
189
190
STUDENT'S SOLUTIONS MANUAL
since 2 (T ) or (V)
=
= b (V)
(T)
(X2 ) =
ox = (P)
=
== EK] = 2 (V)
[9.35, (T )
[V
= axb = ~kx2, b = 2]
~ (v+~) h.w [Problem 9.13] .
(v+Dx (¥) = (v+Dx (w~) = (v+ Dx (~:)'/2 [9.32].
[(v+ D~
r
/2
= 0 [by symmetry, or by noting that the integrand is an odd function of x].
(p2) = 2m (T ) = (2m)
x
G)
x
(v+ Dx h.w [Problem 9.13].
COMMENT. Both results show a consistency with the uncertainty principle in the form !!..p!!..q :::: ~ as given 2 in Section 8.6, eqn 8.36a.
P9.17
Use the first two terms of the Taylor series expansion of cosine: V = Vo(l 
cos3¢)
~ Vo (I
_ I+
(3~)2 )
=
9~0 ¢2 .
The SchrOdinger equation becomes
This has the form of the harmonicoscillator wavefunction (eqn 9.24). The difference in adjacent energy levels is: EI  Eo = h.w [9.26] where w
= (9~0 ) 1/2 [adapting 9.25] .
If the displacements are sufficiently large, the potential energy does not rise as rapidly with the angle as would a harmonic potential. Each successive energy level would become lower than that of a harmonic oscillator, so the energy levels will become progressively closer together.
Question. The next term in the Taylor series for the potential energy is  (27gVo) ¢ 4. Treat this as a perturbation to the harmonic oscillator wavefunction and compute the firstorder correction to the energy.
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS
P9.19
e2
4Jt t:Q r
= b (V )
Since 2 (T) 2 (T)
[10.4 with Z
=
I]
= ax b
with b
=
I
[x + r]
[9.35 , (T ) == Ed
=
(V ) .
I
=
Therefore, (T) P9.21
I
=   . 
V
191
~ (V ) I·
The elliptical ring to which the particle is confined is defined by the set of all points that obey a certain equation. In Cartesian coordinates, that equation is
as you may remember from analytical geometry. An ellipse is similar to a circle, and an appropriate change of variable can transform the ellipse of this problem into a circle. That change of variable is most conveniently described in terms of new Cartesian coordinates (X, Y) where
=x
X
and Y
= ay/ b.
In this new coordinate system, the equation for the ellipse becomes
which we recognize as the equation of a circle of radius a centered at the origin of our (X, Y) system. The text found the eigenfunctions and eigenvalues for a particle on a circular ring by transforming from Cartesian coordinates to plane polar coordinates. Consider plane polar coordinates (R , cP) related in the usual way to (X , Y):
= R cos cP
X
and Y
= R sin cP.
In this coordinate system, we can simply quote the results obtained in the text. The energy levels are
=
E
m2 n
dl
[9.38a]
where the moment of inertia is the mass of the particle times the radius of the circular ring 1
= ma 2 .
The eigenfunctions are eim/
l/f
= (2Jt) '/ 2 [9.38b].
It is customary to express results in terms of the original coordinate system, so express cP in terms first of X and Y, and then substitute the original coordinates: Y
 = tan cP X
so
cP = tan'
~
X
y
= tan  ' a . bx
192
P9.23
STUDENT'S SOLUTIONS MANUAL
The SchrOdinger equation is 1i2  \1 2 1{1 = £1{1 [9.48 with V = 0], 2m 2 I a2(r1{l) I 2 \1 1{1 =   2 + ZA 1{1 [Table 8,1 ] , r ar r
Since r =constant, the first term is eliminated and the Schrodinger equation may be rewritten 2
1i 2 1i2 2 2 2 2/£1{I    A 1{1 = £1{1 or  A 1{1 = £1{1 [l = mr ] or A 1{1 =   ~~ II ~ 2 I a2 I a , a where A =   +  ,  sm8  , sin 2 8 a¢2 sm 8 a8 a8
Now use the specified 1{1 = YI,III/ from Table 9,3, and see if they satisfy this equation, (a) Because Yo,o is a constant, all derivatives with respect to angles are zero, so A2Yo,O = ~ implying that £ = ~ and angular momentum = ~ [from (l(l + I)} 1/21ij, (b)
2 1 a 2Y2_ 1 1 a , l1aY21 ' n i¢ A Y2 I =      '  +    s m u   '  where Y2 ,_ I=Ncos8smue , ,sin2 8 a¢2 sin 8 a8 a8
I a ay~  I 1 a, '¢ 2 ' 2    sin 8 '  =    sm 8Ne' (cos 8  sm 8) sin 8 a8 a8 sin 8 a8 N i¢ 2 ) = ~ (sin 8( 4cos8 sin 8) + cos 8(cos2 8  sin 8) sm8
, (6cos8sin 8+ COS8) ,
=Ne'¢
Ncos8sin8e i¢
a2Y2, _1
sin2 8 ~
2
sin 8
sm8
Ncos8e i¢
sin 8
so A2Y2,_ 1 = Nei¢(6cos8 sin8 ) = 6 Y2 ,1 = 2(2 + I) Y2 ,1 [i,e, I = 2] and hence 2/£
~
 6Y2, 1 = /i2Y2,I , implyingthatCTII
and the angular momentum is {2 (2 + I)} 1/21i = 16 1/21i \' (c)
2 1 a2Y3,3 1 a , naY3,3 3 3¢ A Y33 =  2    +    sm u   where Y3 ,3 = N sin 8e • , , sin 8 a¢2 sin 8 a8 a8
aY33' = 3Nsm' 2 8cos8e 3'''''Y', a8
QUANTUM THEORY: TECHNI QUES AND APPLI CATI ONS
1 a aY3 '3    sin e sin e ae ae
I a =  3N sin 3 e cos ee 3'''"+'
sin e ae
a2 Y3 ,3 sin 2 e
a
so A2Y3,3 = 12N sin 3 e e 3ir/> and hence
=
 12Y3.3
=
1 2Y3,3 = 3(3
~
21E
/i2Y3.3 , implying that ~I
and the angular momentum is (3(3
P9.2S
+ I)Y3,3 [i .e. l = 3]
+ I)}1/2 II. = 12.)311.1.
= mt!(l(l + 1)}1 / 2 and hence e =arccos
From the diagram in Fig. 9.2, cose
m, (l(l+I ) }1 /2
II
Figure 9.2
For an
ct
electron, m , =
+2' s = 2 and (with m, + 1
I
m, = I:
The minimum angle occurs for lim
I> 00
P9.27
emin
=
lim arccos (
I> 00
i=r XP=
m,., 1 + s)
(l (l
j y
k
x Px
py
pz
Z
lim arccos~ = arccos 1 = [2]. +l I ) } 1/2 ) = I> l 00
[see any book treating the vector product of vectors]
193
194
STUDENT'S SOLUTIONS MANUAL
Therefore,
Iy
z
I =
,
~I (z ~ ax  x~) az
= (zPx  xpz) =

.
(ipy  YPx) = ~ (x~ y~ )
ay
I
Ii
ax
a
We have used Px = i ax ' etc. The commutator of Ix and Iy is (lxly  1), /.,), We note that the operations always imply operation on a function . We form A
A
2(a . =  Ii y az
/,Ix! =
AA
.I
Ii
AA
AA
 za )(z a  x a)  f
Ixl,!
AA
and
A
ay
ax
az
2( za  xa)(y a  za) f ax az az ay
Since multiplication and differentiation are each commutative, the results of the operation Ixl)' and Iylx differ only in one term. For lyl.J , x(af lay) replaces y(af lax). Hence, the commutator of the operations,
(Ixl)' 1)'lx) COMMENT.
P9.29
is _ 1i2
(y~ ax  x~) ay orl ~Iz I. I
We also would find
~A
We are to show that [I , IJ
~
~
~A
~
A
~
A
The three commutators are: A2
A
A2
A
[lz , lzl
/2
A
/2
A
= lz Iz 
A
/2
= lz 
A
/2
/2
Izlz
~
A
= [I; + l" + Iz , Izl = [Lx ' Izl + [I)', Izl + [Iz , Izl = 0 A3
A
A3
lz A
= 0, A
A
A
A
A
A
/2
[lx, lzl = lxlz l:'x = '.I z l.tlzlx + lxlzlx  Iz'.
= Ix(lxlz  Izlx) + (l.Jz  Izlx)lx = Ix[lx, Izl + [lx, Izl/x = Ix(ilily) + (ilil)')lx = 
ilia.l)'
+ I)x) [9.S6aJ,
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 12 '
[iy ,lzl
12'
, 12
12'
, , ,
, , ,
195
, 12
= ly Iz  Izly = Iy Iz  lyizly + Iyll ly  Il ly
= ly(/ylZIlly) + (Iyll IZly)ly = Iy[ly, III + [I.", IzJly = Ly(inLx) + (inLx)ly = in(Lylx + lx/y) [9.56al. Therefore, [/2, IzI =
in(LxLy+ Lvlx) + in(lx/y+ LyLx) + 0 = O. ,
,
,
,
,,'
'2
= 0 and [1 2, Iyl = 0 because Ix, Iy, and Iz occur symmetrically in I .
We may also conclude that [1 2 , Ix I
Solutions to applications P9.31
(a) The energy levels are given by
and we are looking for the energy difference between n
= 6 and n = 7:
Since there are 12 atoms on the conjugated backbone, the length of the box is 11 times the bond length, L
=
11(140 x 10 12 m)
=
1.54 x 10 9 m,
34
= 1330 10 19 J 1 _ (6.626 x 10 J s)2(49  36) so f.,E  8(9.11 x 10 31 kg)(1.54 x 109 m ) 2 · x. (b) The relationship between energy and frequency is f.,E =hv
so
v
= f.,E = h
19 3.30 x 10 J 6.626 x 10 34 J s
= 14 .95
X
1 10 14 sI.
(c) Look at the terms in the energy expression that change with the number of conjugated atoms, N. The energy (and frequency) are inversely proportional to L 2 and directly proportional to (n + 1)2  n 2 = 2n + I , where n is the quantum number of the highest occupied state. Since n is proportional to N (equal to N / 2) and L is approximately proportional to N (strictly to N  1), the energy and frequency are approximately proportional to N 1 • So the absorption spectrum of a linear polyene shifts to 1 lower !trequency as the number of conjugated atoms 1 increases I. P9.33
In effect, we are looking for the vibrational frequency of an 0 atom bound, with a force constant equal
to that of free CO, to an infinitely massive and immobile protein complex. The angular frequency is
where m is the mass of the 0 atom,
m
=
(I6.0u)(1.66 x 10 27 kgu  I)
=
2.66 x 1O 26 kg,
196
STUDENT'S SOLUTIONS MANUAL
and k is the same force constant as in Problem 9.2, namely, 1902 N m I :
W
P9.35
1902Nm1 )1 /2 = ( = 12.68 2.66 x 10 26 kg
X
10 14 sI I.
The angular momentum states are defined by the quantum number m[ = 0, ± I, ±2, etc. By rearranging eqn 9.42, we see that the energy of state m[ is
and the angular momentum is
(a) If there are 22 electrons, two in each of the lowest II states, then the highest occupied state is m[ = ±5, so
lz = ±5h = ±5 x (1.055 x 10 341 s) = 15.275 x 10 34 1 s
I
25h2 andE±5 =   .
21
The moment of inertia of an electron on a ring of radius 440 pm is 1 = mr2 = 9.11 x 10 31 kg x (440 x 10 12 m)2 = 1.76 x 1049 kg m 2.
Hence E±5
34 = 25 x (1.055 x 10 1 s)2 = 17.89 x 10 19 1 I. 2 x (1.76 x 10 49 kg m2)
(b) The lowest unoccupied energy level is
m[
= ±6, which has energy
34 _ 36 x (1.055 x 10 1 s)2 _ 1 14 10 18 1 E±6 2 . x . 49 2 x (1.76 x 10 kg m ) Radiation that would induce a transition between these levels must have a frequency such that _ hv  /""E
so
_ /""E _ (11.4 7.89) x 10v  34
h
6.626 x 10
Js
19
J 152
. x
1014H
I
z.
This corresponds to a wavelength of about 570 nm, a wave of visible light. P9.37
The Coulombic force is
F~ qlq2 _ ~ 
dr 4rrfor  4rrf or2'
For two electrons 2.0 nm apart, the force is
I
I
(1.60 x 10 19 C)2 92=5.8xlO II N .  4rr x (8.854 x 10 I2C2l I m I) x (2.0 x 10 m)
F
QUANTUM THEORY : TECHNIQUES AND APPLICATIONS
P9.39
197
(a) In the sphere, the Schrodinger equation is
2
2 a + A I 2) 1/1" = E1/1"  li (a + 
2m
2
ar2
r ar
r2
[9.5 Ia]
where A 2 is an operator that contains derivatives with respect to () and 4> only. Let 1/1"(r,(}, 4»
= X(r)Y«(},4»
.
Substituting into the Schrodinger equation gives 2 2 li (a Y X+ 2yaX   + xA 2 Y) =EXY. 2m ar2 r ar r2
Divide both sides by XY:
The first two terms in parentheses depend only on r , but the last one depends on both r and angles; however, multiplying both sides of the equation by r2 will effect the desired separation:
Put all of the terms involving angles on the righthand side and the terms involving distance on the left:
Note that the right side depends only on () and 4>, while the left side depends on r. The only way that the two sides can be equal to each other for all r, (), and 4> is if they are both equal to a constant. Call that constant (li2 1(l + 1))/2m (with 1 as yet undefined) and we have, from the right side of the equation, 
li2
2mY
2 li2 1(l + I ) A Y =   ...,'
so
2m
A2y
= 1(1 + I )Y.
From the left side of the equation, we have
_~ (r2 a x + 2r ax ) _ Er2 = _ li 1(1 + I). 2
2m
2
X ar2
X ar
2m
After multiplying both sides by X / r2 and rearranging, we get the desired radial equation
2 (a X 2m ar2
_~
+ ~ ax ) + li r ar
2
1(l + I ) X 2mr 2
= EX.
Thus the assumption that the wavefunction can be written as a product of functions is a valid one, for we can find separate differential equations for the assumed factors. That is what it means for a partial differential equation to be separable.
198
STUDENT'S SOLUTIONS MANUAL
(b) The radial equation with l
a2x
2 ax
= 0 can be rearranged to read
2mEX
+   IL2· ar2 r ar Form the following derivatives of the proposed solution:
ax ar
= (2rcR)1 /2 [coS(nrcr/R) (nrc) _ Sin(nrcr/R)] r
R
r2
an d a2x _ (2 rcR) 1 / 2 [  sin(nrcr/R) (nrc)2  2cos(nrcr/R) (nrc) 
ar2
r
R
r2
R
2Sin(nrcr/R)] + ;:r3 .
Substituting into the left side of the rearranged radial equation yields
(2rcR)  1/2 [_ sin(nrcr/R) (nrc)2 _ 2cos(nrcr/R) (nrc) r R r2 R
+ (2rcR)1 /2 [2COS(nrcr/ R) r2
= (2rcR) _1 /2sin(nrcr/R) r
+ 2Sin(nrcr/R)] r3
(nrc) _ 2Sin(nrcr/R)] R r3 (nrc)2 R
=  (nrc)2 x. R
Acting on the proposed solution by taking the prescribed derivatives yields the function back mUltiplied by a constant, so the proposed solution is in fact a solution. (c)
Comparing this result to the right side of the rearranged radial equation gives an equation for the energy
E
2 2 (.!!...)2 = nh . 2mR2 2rc 8mR2 2 2
= (nrc) 2 If. = n rc R
2m
Atomic structure and atomic spectra
Answers to discussion questions 010.1
The Schrodinger equation for the hydrogen atom is a sixdimensional partial differential equation, three dimensions for each particle in the atom. One cannot directly solve a multidimensional differential equation; it must be broken down into onedimensional equations. This is the separation of variables procedure. The choice of coordinates is critical in this process. The separation of the Schrodinger equation can be accomplished in a set of coordinates that are natural to the system, but not in others. These natural coordinates are those directly related to the description of the motion of the atom. The atom as a whole (center of mass) can move from point to point in threedimensional space. The natural coordinates for this kind of motion are the Cartesian coordinates of a point in space. The internal motion of the electron with respect to the proton is most naturally described with spherical polar coordinates. So the sixdimensional SchrOdinger equation is first separated into two threedimensional equations, one for the motion of the center of mass, the other for the internal motion. The separation of the center of mass equation and its solution is fully discussed in Section 9.2. The equation for the internal motion is separable into three onedimensional equations, one in the angle ¢ , another in the ang le e, and a third in the distance r . The solutions of these three onedimensional equations can be obtained by standard techniques and were already we ll known long before the advent of quantum mechanics. Another choice of coordinates would not have resulted in the separation of the Schrodinger equation just descri bed. For the details of the separation procedure, see Sections 10.1 and 9.7.
010.3
The selection rules are ,6,n
= ± I , ±2, ... ,
,6,[
=
± I,
,6,m{
= 0, ± I .
In a spectroscopic transition the atom emits or absorbs a photon. Photons have a spin angular momentum of I. Therefore, as a result of the transition, the angular momentum of the electromagnetic field has changed by ± I n. The principle of the conservation of angu lar momentum then requires that the angu lar momentum of the atom has undergone an equal and opposite change in angular momentum. Hence, the selection rule on ,6,{ = ± I. The principal quantum number n can change by any amount since n does not directly relate to angular momentum. The selection rule on ,6,m/ is harder to account for on bas is of these simple considerations alone. One has to evaluate the transition dipole moment between the wavefunctions representing the initial and final states involved in the transition. See Justification 10.4 for an example of this procedure.
200
STUDENT'S SOLUTIONS MANUAL
010.5
See Section 10.4(d) of the text and any general chemistry book, for example, Sections 1.101.13 of P. Alkins and L. Jones, Chemical principles, 2nd edn, W. H. Freeman, and Co., New York (2002).
010.7
In the crudest form of the orbital approximation, the manyelectron wavefunctions for atoms are represented as a simple product of oneelectron wavefunctions. At a somewhat more sophisticated level, the manyelectron wavefunctions are written as linear combinations of such simple product functions that explicitly satisfy the Pauli exclusion principle. Relatively good oneelectron functions are generated by the HartreeFock selfconsistent field method described in Section 10.5. If we place no restrictions on the form of the oneelectron functions , we reach the HartreeFock limit which gives us the best value of the calculated energy within the orbital approximation. The orbital approximation is based on the disregard of significant portions of the electronelectron interaction terms in the manyelectron Hamiltonian, so we cannot expect that it will be quantitatively accurate. By abandoning the orbital approximation, we could in principle obtain essentially exact energies; however, there are significant conceptual advantages to retaining the orbital approximation . Increased accuracy can be obtained by reintroducing the neglected electronelectron interaction terms and including their effects on the energies of the atom by a form of perturbation theory similar to that described in Further information 9.2 and Section 10.9. For a more complete discussion consult the references listed under Further reading .
Solutions to exercises E1 0.1 (b)
The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving it any kinetic energy it now possesses Ephoion
= I + Ekinelic
I = ionization energy
The energy of a photon is related to its frequency and wavelength he
Epholon
= hv = T
and the kinetic energy of an electron is related to its mass and speed, s
he So A
I 2
he A
2
= I + mes => I =  = (6.626
X
10
2
8
J s) x (2.998 x 10 m S
58.4
X
(1.79 x 106 m s 
I
I X
34
I
mes 2
I) _
10 9 m
~ (9.11 2
r
X
10 31 k ) g
?
=11.94 x 10 18 11= 12.leV E10.2(b)
The radial wavefunction is [Table 10.1] R30 = A (6  2p
,
+ ~9 p2) e p / 6 where p ==
2Zr , and A is a collection of constants. ~
[Note: p defined here is 3 x p as defined in Table 10.1]
ATOMIC STRUCTURE AND ATOMIC SPECTRA
201
Differentiating with respect to p yields
d:;,O= 0= (6 A
2p
+ ~p2)
x (
D
e
p 6 /
+
(2 + ~p)
Ae
p 6 /
This is a quadratic equation
0= ap2 + bp +c
where a
I
= 54 '
b
5
= 9 '
and c
= 3.
The solution is p =
so r
b ± (b 2  4ac)I / 2 2a
= (~± 3 (71 /2) ) 2
2
= 15 ±3J7
ao .
Z
= 7.65 and 2.35, so r = 111.5ao/Z Iand 13.53ao/Z \. Substituting m, r = 1607 pm 1and 1187 pm I·
Numerically, this works out to p Z
= I and ao = 5.292 x
10 11
The other maximum in the wavefunction is at 1r = 0 I. It is a physical maximum, but not a calculus maximum: the first derivative of the wavefunction does not vanish there, so it cannot be found by differentiation. E10.3(b)
The complete radial wavefunction, R4, 1 is not given in Table 10.1; however in the statement of the exercise we are told that it is proportional to (20  lOp
+ p 2 )p
where p
= 2Zr ao
[Note: p defined here is n x p as defined in Table 10.1]
The radial nodes occur where the radial wavefunction vanishes, namely where (20  lOp
+ p2)p = O.
The zeros of this function occur at p =0,
and when (20  lOp then r
+ p2) = 0,
with roots p 764a
= 2.764,
pao = 2z = 2pao = 2. 2 o = 1 1.382ao 1
and
orr=17.31 x 10 11 ml and 11.917 x 1O lO ml
and p
= 7.236
7.236a o
1
  2  = 3.618 a o
1
202
E10.4(b)
STUDENT'S SOLUTIONS MANUAL
Nonnalization requires
Integrating over angles yields 1= 47TN 2
=47TN 2
1 1
00
e r / aO (2  r/ao)2 r 2 dr
00
e r / aO (4  4r/ ao
+ r2 /a6) r 2 dr = 47TN2(8a6)
In the last step, we used
I
So N = :== 4J27TG6 E10.S(b)
The average kinetic energy is
where 1/1
= N(2 
p)e p / 2 with N
=~ 4
d!"
(~) 1/2
and p
27TG6
= r 2 sinBdrdBd¢ =
==
Zr here ao
a 3 p2 sin B dp dB d¢ 0
Z3
In spherical polar coordinates, three of the derivatives in V 2 are derivatives with respect to angles, so those parts of V21/1 vanish. Thus
ATOMIC STRUCTURE AND ATOMIC SPECTRA
203
and (EK)
=
[ 00 [ "
[2"
N(2 _ p)e p /
1o 1o 1o
2(~) 2 x (_n2 ) ao
x Ne p / 2(4/p
2m
+ 5/2 
a 3 d sin e de p2 dp p/4)'0'_::_ __ Z3
The integrals over angles give a factor of 41T , so
The integral in thi s last expression works out to 2, using
10
00
e P pn dp = n! for n = 1, 2, and 3. So
The average potential energy is (V )
f = 1° 1°"1° =
ljr*Vljrdr where V
~~? =  = 
41T cor
41T eoaop
3 ) N(2  p)e p / 2 a'0'_,,_ p2 sin e dp de _d _ 41T eoao p Z3 The integrals over angles give a factor of 41T , so and (V)
2 "
00
22 Z e
N(2  p)e p / 2 ( 
2 2 (V ) = 41TN2 ( Z e ) x 41T eOQ'O
(a~) [ 00 (2 _ Z 1o
p)2pe P dp
The integral in this last expression works out to 2, using
e
2 2
(V) =41T
E10.6(b)
Z33 ) x (  Z ) x ( 321Ta 41Teoao o
10
00
e P pn dp
= n! for n =
1, 2, 3, and 4. So
3
(a% Z
)
x (2)
=
The radial distribution function is defined as P
= 41Tr2ljr2
where p
2Zr
== 
nao
so P 3s
2Zr
= 
3ao
= 41Tr 2(Yo.oR3,0)2,
here.
But we want to find the most likely radius, so it would help to simplify the function by expressing it in terms either of r or p, but not both. To find the most likely radius, we could set the derivative of P3s equal to zero; therefore, we can collect all multiplicative constants together (including the factors of ao/Z needed to turn the initial r 2 into p2) since they will eventually be divided into zero P 3s
= C 2 p2(6 
6p
+ p2)2 e P
204
STUDENT'S SOLUTIONS MANUAL
Note that not all the extrema of P are maxima; some are minima. But all the extrema of (P3s)I /2 correspond to maxima of P3s. So let us find the extrema of (P3s) 1/2 d(P )1 /2 3s
dp
= 0
d
= Cp(6 dp
= C[p(6 
6p
6p
+ p2)e p / 2
+ p2)
X
(!) + (6 
12p + 3p 2)]e p / 2
Numerical solution of this cubic equation yields p
= 0.49,
2.79, and 8.72
corresponding to r
= I0. 74ao/ Z ,
COMMENT.
4. 19ao/Z, and 13. 08ao/Z
I
If numerical methods are to be used to locate the roots of the equation which locates the
extrema, then graphical/ numerical methods might as well be used to locate the maxima directly. That is, the student may simply have a spreadsheet compute P 3s and examine or manipulate the spreadsheet to locate the maxima.
E10.7(b)
The most probable radius occurs when the radial wavefunction is a maximum. At this point the derivative of the function wrt either r or p equals zero. dR 31) ( dp max
=0=
(d
(4 
2
p
p) pedp
/
[Table 10.1]
))
= (4 _ 4p +
max
p2) e p / 2 2
The function is a maximum when the polynomial equals zero. The quadratic equation gives the roots p = 4 + 2.J2 = 6.89 and p = 4  2.J2 = 1.17. Since p = (2Z/nao)r and n = 3, these correspond to R R (PI) r = 10.3 x ao/Z and r = 1.76 x ao/Z. However, 31 = 31 (1.17) = 4.90. So, we conclude
I I I R31 (10.3) that the function is a maximum at p = 1.17 which corresponds to Ir = 1. 76ao /z·1 IR31 (pz)
E10.8(b)
I
)
Orbital angular momentum is
There are 1 angular nodes and n  1  1 radial nodes
(L 2) 1/2 = 6 1/ 2/i = 12.45
x 10 34 J s 1 ~ angular nodes
OJ radial node
(a)
n
= 4,1 = 2,
so
(b)
n
= 2, 1 = I,
so (U ) 1/ 2 = 21 / 2/i
= 11.49 x
10 34 J s 1
OJ angular nodes @]radial nodes
(c)
n
= 3,1 = I,
so (U ) I/2 = 21 /2/i
= 11.49 x
10 34 Js 1
OJ angular node OJ radial node
ATOMIC STRU CTURE AND ATOMIC SPECTRA
E10.9(b)
For I > 0, )
=
=I±
205
1/ 2, so
(a)
I
I, so)
= 1r112or312,1
(b)
1=5 , so)
= 19/2 or 11/21
E10.10(b) Use the ClebschGordan series in the form
+hI ,·· ·,U,hl
1=), +Jz,),
Then, with), 1 =
= 5 andh = 3
18, 7, 6, 5, 4, 3, 21
E10.11(b) The degeneracy g of a hydrogenic atom with principal quantum number n is g
= /1 2
The energy E of
hydrogenic atoms is
so the degeneracy is
(a)
g=
(b)
g=
(c)
g
he (2) 2 RH 4heRH he (4) 2 RH
~heRH
=[] =~
= _ he(5 )2 RH = ~ heRH
E10.12(b) The letter F indicates that the total orbital angular momentum quantum number L is 3; the superscript 3
is the multiplicity of the term, 25 + I, related to the spin quantum number 5 indicates the total angular momentum quantum number 1 . E10.13(b) The radial distribution function varies as
The maximum value of P occurs at r
= ao since
dP ex (2r _ 2r2 ) e 2rjao = 0 at r dr
= ao
ao
P fall s to a fraction! of its maximum given by
and Pmax
= ~e2 ao
=
I ; and the subscript 4
206
STUDENT'S SOLUTIONS MANUAL
and hence we must solve for r in ? fl /  _ r  rl ao e
e
ao
f = 0.50
(a)
0.260
r
= ao e rl ao solves to r = 2.08 ao = 1110 pm Iand to r = 0.380ao = 120.1 pm I
f = 0.75
(b)
0.319
=
:0
e  rl ao solves to r
= 1.63ao = 186 pm Iand to r = 0.555ao = 129.4 pm I
In each case the equation is solved numerically (or graphically) with readily available personal computer software. The solutions above are easily checked by substitution into the equation for f . The radial distribution function is readily plotted and is shown in Figure 10.1.
0.15
0. 10
.,
~ C!.
Q..,
0.05
0.00 0.5
0.0
1.0
1.5 rl ao
2.0
2.5
Figure 10.1
E10.14(b) (a) 5d + 25 is ~ an allowed transition, for t../ = 2 (t../ must equal
= I. 5p + 3f is 1 not 1 allowed, for t../ = +2 (t..[ must equal ± I). 6h: I = 5; maximum occupancy = ~
(b) 5p + 3s is 1allowed I, since t../
(c) (d)
The only unpaired electrons are those in the 3d subshell. There are three. S=
[I] and
~
I=
For S
= ~,
Ms = ±
for S
= ~,
Ms = ±
IT],
I ~ and ± ~ I
I ~I
± I).
ATOMIC STRUCTURE AND ATOMIC SPECTRA
E10.16(b) (a) Possible values of S for four electrons in different orbitals are 12, I , and 0
207
I; the multiplicity is 2S + I ,
so multiplicities are IS , 3, and I 1respectively. (b) Possible values of S for five electrons in different orbitals are IS/2, 3/2 and 1121; the multiplicity is 2S + I, so multiplicities are 16, 4, and 21 respectively.
= I) and a d electron (l = 2) gives rise to L = 3 (F), 2 (D), and I (P) terms. Possible values of S include 0 and I. Possible values of J (using RussellSaunders coupling) are 3, 2, and I (S = 0) and 4, 3, 2, 1, and 0 (S = I) . The term symbols are
E10.17(b) The coupling of a p electron (l
Hund's rules state that the lowest energy level has maximum mUltiplicity. Consideration of spinorbit coupling says the lowest energy level has the lowest value of J(J + I)  L(L + I)  S(S + I). So the lowest energy level is 13F2 1.
= I and L = 2, so J =13, 2, and I 1are present. J = 3 has I2J states, with MJ = 0, ± I , ±2, J = 2 has [IJ states, with MJ = 0, ± I, or ±2; J = I has [IJ states, with MJ = 0, or ±l.
E10.18(b) (a) 3D has S
or ±3 ;
(b) 4D has S
= 3/ 2 and L = 2, so J
states, with MJ
=
= 17/2, S/2, 312, and 1121 are present. J
± 7/ 2, ±S/ 2, ±3 / 2 or ± I / 2; J
±S/ 2, ±3/ 2 or ± I / 2; J = 3/ 2 has possible states, with MJ = ± I / 2.
=
S / 2 has
0
= 7/ 2 has 0
possible
possible states, with MJ
=
0 possible states, with MJ = ±3 / 2 or ± I / 2; J = 1/ 2 has [I]
= 9/ 2 and 7/ 2 are present. J = 9/ 2 has [IQJ possible states, with 9/ 2, ± 7/ 2, ±S/ 2, ±3/ 2, or ± I / 2; J = 7/ 2 has 0 possible states, with MJ = ± 7/ 2,
(e) 2G has S = 1/ 2 and L = 4, so J MJ = ± ±S/ 2, ±3 / 2, or ±1 / 2.
E10.19(b) Closed shells and subshells do not contribute to either Lor S and thus are ignored in what follows.
(a) Sc[Ar ]3d 14s 2: S
= !,L = 2; J = ~ , ~ , so the terms are 12D5/2 and 2D3/21.
(b) Br[Ar ]3d I0 4s 2 4p 5. We treatthe missing electron in the4p subshell as equivalent to a single "electron" with l
= I , s = !.Hence L = I , S = !,and J = ~ , !, so the terms are 12 P3/2 and 2P I / 2 1.
Solutions to problems Solutions to numerical problems P10.1
All lines in the hydrogen spectrum fit the Rydberg formula
~A = RH (~~) nT n~
[10.1 , with ii
=~] A
RH
= 109677cm l .
208
STUDENT'S SOLUTIONS MANUAL
Find nl from the value of Amax , which arises from the transition
n~
AmaxRH 
Since
=
nl
+ +
n2(nl 1
2nl
1
A
n~(nl
1, 2, 3, and 4 have already been accounted for, try
1)2
1 =
+ 1) 2 =
(nl
=
+I_
nl .
+I + 1)2'
2nl

nl
nl
=
5, 6, .. . . With
=
6 we get
136. Hence, the Humphreys series is 1n2 _ 61 and the transitions are given by
1) (l09677cm 1) x (1  ~
~
,
n2
= 7,8""
and occur at 12372 nm, 7503 om, 5908 om, 5129 nm, ... , 3908 nm (at n2 nm as n2  00, in agreement with the quoted experimental result.
P10.3
nl
A Lyman series corresponds to nl
=
15), converging to 3282
= I; hence
Therefore, if the formula is appropriate, we expect to find that
v( I 
1
n2
)1 .
is a constant (R Li 2+ ).
We therefore draw up the following table.
n
v/cm I 1
V ( 1  n2
)1 /cm
I
2
3
4
740747
877924
925933
987663
987665
987662
Hence, the formula does describe the transitions, and 1RLi2+
= 987663 cm 
I
I. The Balmer transitions
lie at
v=
R
.2+ LI
(~4 ~) n2
= (987 663cm
n = 3, 4, . ..
l)
x
G~2) =
1137175 cm
The ionization energy of the groundstate ion is given by
II,
1185187 cm
II, ····
ATOMIC STRUCTURE AND ATOMIC SPECTRA
209
and hence corresponds to i! = 987 663cm l , P10.5
or
1122.5 eV
I.
The 7p configuration has just one electron outside a closed subshel1. That electron has / = I, s = 1/2, and j = 1/2 or 3/2, so the atom has L = I, S = 1/2, and J = 1/2 or 3/2. The term symbols are 12P I/2 and 2P3/21, of which the former has the lower energy. The 6d configuration also has just one
= 2, s = 1/2, and j = 3/2 or 5/2, so the atom has L = 2, S = 1/2, and J = 3/2 or 5/2. The term symbols are 120 3/ 2 and 205/21, of which the former has the lower energy. According to the simple treatment of spinorbit coupling, the energy is given by electron outside a closed subshell; that electron has /
E/,sJ
=
!hcAUU
+ I )  /(/ + I) 
s(s
+ I)]
where A is the spinorbit coupling constant. So
and Ee03 /2) = !hcA[~(3/2
+ I) 
2(2 + 1)  !(1/2 + I)] = ~hcA.
This approach would predict the ground state to be 1203/21. COMMENT.
The computational study cited finds the 2P1 / 2 level to be lowest, but the authors cau
tion that the error of similar calculations on Y and Lu is comparable to the computed difference between levels. P10.7
RH
= kJ.LH,
Ro
= kJ.Lo ,
R
= kJ.L
[10.16]
where R corresponds to an infinitely heavy nucleus, with J.L
= me.
R
Likewise, Ro
=
R
(I
+ (melmd»
where
mp
is the mass of the proton and md the mass of the deuteron .
The two lines in question lie at
~ = Ro Ao and hence
(1  ~) = ~ 4
4
Ro
210
STUDENT'S SOLUTIONS MANUAL
Then, since 1 + (melmd) 1 + (mel mp) '
and we can calculate md from
9.10939 31
1 (
+
9.1039 x 10 kg) (82259.098cm  1)1 1.67262 x 10 27 kg x 82281.476cm  1
= 13.3429 x Since I
10 3 1 kg
X
10 27 kg
I.
= Rhe, 82281.476cm 1 82259.098cm 1
P10.9
= 11.0002721·
(a) The splitting of adjacent energy levels is related to the difference in wavenumber of the spectral lines as follows :
helll! Ilv
=
IlE
= /tBB,
so Ill!
/tBB
(9.274 x 10 24 JT I )(2 T) (6.626 x 10 J s)(2.998 x 10 cm sI)
=   = ::::...,."...:34 10 he
= 10.9 cm I I·
(b) Transitions induced by absorbing visible light have wavenumbers in the tens of thousands of recip
rocal centimeters, so normal Zeeman splitting is 1 small 1 compared to the difference in energy of the states involved in the transition. Take a wavenumber from the middle of the visible spectrum as typical :
Or take the Balmer series as an example, as suggested in the problem; the Balmer wavenumbers are (eqn 10.1):
The smallest Balmer wavenumber is
v = (l09677cm  l )
x (1/41/9)
= 15233cm 1
and the upper limit is I!
= (109677cm  l )
x ( 1/ 4  0)
= 27419cm l .
ATOMIC STRUCTURE AND ATOMIC SPECTRA
211
Solutions to theoretical problems P10.11
Consider t/l2p, = t/l2,I,Owhich extends along the zaxis. The most probable point along the zaxis is where the radial function has its maximum value (for t/l 2 is also a maximum at that point). From Table 10. 1 we know that
and so dR dp
=
Therefore, r* COMMENT.
(I  ~p) e4
=
p/4
=0
when
p= 4.
2;0, and the point of maximum probability lies at z
= ± 2;0 = I ±I06 pm I.
Since the radial portion of a 2p function is the same, the same result would have been obtained
for all of them. The direction of the most probable point would, however, be different.
P10.13
J
(a) We must show that 1t/l3 px 12 d r coordinates (Fig. 8.22).
{2n
= 10
I . The integrations are most easily performed in spherical
r 10r'" 1R31(P) {
10
= 2r/ao, r =
wherep
I
=
YII 
.j2 YII
pao/2, dr
}
=
12 rsm(e)drded¢ [Table 1O. I, eqn 10.24] ?
(ao/2) dp .
{2" 10{" 10roo (2a0 ) 31[( 27(6)11 /2 )
= 210
( I
1 = 466 S6rr
)3/2(4 "?/ I) pe
1
6
p /
fa 2" cos2(¢ ) d¢ fa" sin 3 (e) de faoo (4 ~pr p 4e /3d p p
'   v  ' ' '   .   ''
"
We must also show that
ao
3 1/2 ] \2p2 sin Ce) dp de d¢ 2 sinCe) cos(¢) [ (8rr)
X
=I
.
Thus,
f
4/ 3
t/l3px is normali zed to I.
t/l3px t/l3d.n dr
= O.
•
34992
'
212
STUDENT'S SOLUTIONS MANUAL
Using Tables 9.3 and 10.1 , we find that I ( I ) 3/2 ( I) 6 o/3Px =S4(2n)1 /2 aD 4 P pep/ sin(8)cos(cf»
3
=
32(2~) 1/2 (~J 3/2 p 2e
p 6 /
sin 2(8) sin(2cf»
where p = 2r I ao,r = pao/2, dr = (ao / 2) dp.
f
o/3px o/3dxy dr
fa oo p5 e p /3 dp
= constant x
/a
2IT cos(cf» sin(2cf» dcf>, faIT sin4(8) d8
o Since the integral equals zero, o/3P., and o/3d.,), are orthogonal. (b) Radial nodes are determined by finding the p values (p = 2r I aD) for which the radial wavefunction equals zero. These values are the roots of the polynomial portion of the wavefunction. For the 3s
orbital, 6  6p
+ p2 = 0, when IPnode = 3 + J3 and Pnode = 3 
J31.
The 3s orbital has these two spherically symmetrical modes. There is no node at p conclude that there is a finite probability of finding a 3s electron at the nucleus.
=
0 so we
For the 3px orbital, (4  p) (p) = 0, when 1Pnode = 0 and Pnode = 41· There is a zero probability of finding a 3px electron at the nucleus. For the 3dX), orbital 1 Pnode (c)
(rhs
=
f
=0
2 IRJO Yool rdr
1
is the only radial node.
=
f
2 IRJO Yool r3 sin(8) drd8 dcf>
= ~ {OO (6 _2p + p2 19 )2 p 3e p / 3 dp . 3888
/0
,
52488
(r h,
27ao
= 2 .
(d) The plot Fig. 10.2 (a) shows that the 3s orbital has larger values of the radial distribution function for r < aD. This penetration of inner core electrons of multi electron atoms means that a 3s electron experiences a larger effective nuclear charge and, consequently, has a lower energy than either a 3p or 3dxy electron. This reasoning also leads us to conclude that a 3px electron has less energy than a 3dX), electron.
ATOMIC STRUCTURE AND ATOMIC SPECTRA Radial distribution functions of atomic hydrogen
0. 12
0.1
0.08
r? 0.06
t
0.04
0.Q2
0 10
0
15
20
25
30
rlao
(e) Polar plots with
e=
Figure lO.2(a)
90°.
The p orbital 90
The ... orbital 120
90 120
60
60 150
30
180 180
0
300
240
210
270
'"
330
240
300 270
'"
The tI orbital
90 120
60
150
30
180
0
210
330
240
300 270
'"
Figure lO.2(b)
213
214
STUDENT'S SOLUTIONS MANUAL
Boundary surface plots. sorbital boundary surface
porbital boundary surface
dorbital boundary surface
f orbital boundary surface
Figure lO.2(c) P10.15
The general rule to use in deciding commutation properties is that operators having no variable in common will commute with each other. We first consider the commutation of Iz with the Hamiltonian. This is most easily solved in spherical polar coordinates. ,
Ii
a
lz
=i
H
=
a¢ [Problem 9.28 and Section 9.6 and egn 9.46].
li 2
2JL
V2
+V
[Further information 10.1]
Since V has no variable in common with Iz• this part of the Hamiltonian and Iz commute. V2
= terms in r only + terms in () only +
I r2
2

a22
sin () a¢
,
[Justification 9.7]
a2
The terms in r only and () only necessarily commute with lz(¢ only) . The final term in V2 contains  2 a¢
which commutes with
a a¢' since an operator necessarily commutes with itself. By symmetry we can
ATOMIC STRUCTURE AND ATOMIC SPECTRA
215
deduce that, if H commutes with Iz it must also commute with Ix and Iy since they are related to each other by a simple transformation of coordinates. This proves useful in establishing the commutation of Z2 and H . We form
If H commutes with each of Ix ,Iy, and Iz it must commute with I;, I~ , and I~ . Therefore it also commutes with
72 . Thus H commutes with both f2 and fz.
COMMENT. As described at the end of Section 8.6, the physical properties associated with noncommuting
operators cannot be simultaneously known with precision. However, since H,P., and
/z
commute we may
simultaneously have exact knowledge of the energy, the total orbital angular momentum, and the projection of the orbital angular momentum along an arbitrary axis.
P10.17
With r = (nao/2Z)p and m = 1 , the expectation value is (r
(a)
_I
),,1
=
(nao) 2 r oo 2 2Z p IR"tI dp .
10
(rt, = G~)2 {2 (~) 3/2}210 = I~ I
(rI)2S = (c)
.(rI)
= 2p
because
10
00
P e P dp [Table 10.1]
00
p e P dp
= 1.
[§J (a o )2 {_I_ Z 241 /2
= ~ (6) 24ao
(~) 3/2 } 2 10r oo p3 ePdp ao
because
10ro oo p3 e
The general formula for a hydrogenic orbital is
P
dp
[Table 10.1]
= 6.
(r I ) 1 = ~ . "
n2ao
216 P10.19
STUDENT'S SOLUTI ONS MANUAL
The trajectory is defined, which is not allowed according to quantum mechanics. The angular momentum of a threedimensional system is given by II (l + I ) }1/ 2 n, not by nn. In the Bohr mode l, the ground state possesses orbital angular momentum (nn, with n = I ), but the actual ground state has no angular momentum (l = 0). Moreover, the di stributi on of the electron is quite different in the two cases. The two model s can be distingui shed experimentally by (a) showing that there is zero orbital angular momentum in the ground state (by examining its magnetic properties) and (b) examining the electron distribution (such as by showing that the electron and the nucleus do come into contact, Chapter 15).
P10.21
Justification 10.4 noted that the transition dipole moment, J.tfi , had to be nonzero for a transi tion to
be allowed. The Justification examined conditions that allowed the z component of this quantity to be nonzero; now examine the x and y components.
As in the Justification, express the relevant Cartesian variables in terms of the spherical harmonics, Y/ ,III'
Start by expressing them in spherical polar coordinates: x =r sin ecosl/>
and
y =r sin esinl/> .
Note that YI , I and YI ,_ I have factors of sin e. They also contain complex exponentials that can be re lated to the sine and cosine of I/> through the identities
These relations motivate us to try linear combinations YI .I + YI ,_ I and YI , I  Yl.I (from Table 9.3 ; note c here corresponds to the normalization constant in the table): Yl,l + YI ._ I = csine(e i ¢ +e i¢) = 2csi n ecosl/> = 2cx/r,
so
x
= (YI ,I + YI. _ I)r / 2c ;
Yl,l  Yl.I = c sin e(e i¢  e i¢) = 2ic sin e sin I/> = 2icy/ r ,
so
y
= ( YI , I

YI ,_ I)r /2 ic.
Now we can ex press the integrals in terms of radial wavefunctions RII.I and spherical harmonics Y/ ,I11/
The angul ar integral can be broken into two, one of which contai ns YI , I and the other YI , I . According to the 'triple integral ' relation in Comment 9.6, the integral ll
loo
10211 0
Ytf
Yl,l Y/ j •III /. sin e de dl/>
111/
'
f
'
vanishes unless lr = Ii ± I and mf = mi ± I . The integral that contai ns YI ,_ I introduces no further constraints; it vanishes unless lr = Ii ± I and /1//f = m/ j ± I. Similarly, the y component introduces no
ATOM IC STRUCTURE AND ATOMIC SPECTRA
217
further constraints, for it involves the same spherical harmonics as the x component. The whole set of selection rules, then, is that transitions are allowed on ly if
I t..l = ± 1 and P10.23
t..m/
= 0 or ± 1 I·
(a) The Slater wavefunction [10.32] is 1/Ia(2)a(2) 1/Ia(2),8(2) 1/Ib(2)aC2)
1/I,,(3)a(3)
1/IaCN)a(N)
1/Ia( I),8(l) 1/I"C I )a(1)
1/Ia(3),8C3)
1/I,,(3)aC3)
1/I"CN),8CN) 1/I"CN)aCN)
1/Iz(l),8(1)
1/Iz(2),8C2)
1/Iz(3),8C3)
1/IzCN),8CN)
1/Ia(l)a(1)
1/I(I , 2,3, .. . ,N)
=
1
CN!)1 /2
Interchanging any two columns or rows leaves the function unchanged except for a change in sign. For example, interchanging the first and second columns of the above determinant gives:
I
1/ICI , 2, 3, . . . , N)
= (N!) 1/2
=
1/I,,(2)aC2) 1/Ia(2),8C2) 1/I,,(2)aC2)
1/1,,(1)0'(\) 1/1,,(1),8(1) 1/I,,(I)aCl)
1/Ia(3),8C3) 1/I,,(3)a(3)
1/I"CN)a(N) 1/I"CN),8CN) 1/I"CN)a(N)
1/Iz(2),8 (2)
1/Iz(l),8(1)
1/Iz(3),8C3)
1/IzCN),8CN)
1/Ia(3)aC3)
1/I(2, 1, 3, ... , N) .
This demonstrates th at a Slater determinant is antisymmetric under particle exchange. (b) The possibility that 2 electrons occupy the same orbital with the same spin can be explored by making any two rows of the Slater determinant identical, thereby, providing identical orbital and spin functions to two rows. Rows I and 2 are identical in the Slater wavefunction below. Interchanging these two rows causes the sign to change without in any way changing the determinant.
I 1/I( 1,2, 3, ... ,N) = N!I /2
= 1/1 (2,
1/I"Cl)a(l) 1/IaCl)aC l ) 1/I,,(1)aCI)
1/Ia(2)aC2) 1/Ia(2)aC2)
1/1,,(3)0'(3) 1/Ia(3)aC3)
1/I"CN)aCN) 1/I,,(N)a(N)
1/I,,(2)a(2)
1/I,,(3)aC3)
1/1" (N)a (N)
1/Iz(1),8(I)
1/Iz(2),8 (2)
1/Iz(3),8C3)
1/IzCN),8(N)
1, 3, . . . , N)
=
 1/I(1,2, 3, .. . , N) .
Only the null function satisfies a relationship in which it is the negative of itself so we conclude that, since the null function is inconsi stent with existence, the Slater determinant satisfies the Pauli exclusion principle [Section 10.4 b]. No two electrons can occupy the same orbital with the same spin.
218
STUDENT'S SOLUTIONS MANUAL
Solutions to applications P10.25
The wavenumber of a spectroscopic transition is related to the difference in the relevant energy levels. For a oneelectron atom or ion, the relationship is
Solving for V, using the definition Ii = h/2 rt and the fact that Z = 2 for He, yields
Note that the wavenumbers are proportional to the reduced mass, which is very close to the mass of the electron for both isotopes. In order to distinguish between them, we need to carry lots of significant figures in the calculation. _
J.lHe(1.60218 x 1O 19 C)4  2(8.85419 X 10 12 JI C2 m I )2 x (6.62607 x 10 34 J s)3 x (2.99792 x 1010 cm s I)
\! 
X
(~~) n~ n 2
v/cm I = 4.81870
X
1035(J.lHe/ kg)
(~  ~) . n n 2
l
The reduced masses for the 4He and 3He nuclei are m emnuc
J.l=
me
where
moue
+ moue
= 4.00260 u for 4He and 3.01603 u for 3He,
4He moue = (4.00260 u) x (1.66054
X
10 27 kg u
3He moue = (3.01603 u) x (1.66054
X
10 27 kg
I
or, in kg )
= 6.64648
U I ) =
X
10 27 kg,
5.00824 x 1O 27 kg.
The reduced masses are 4He
27 31 _ (9.10939 x 10 kg) x (6.64648 X 10 kg) 27 31 J.l(9.10939 X 10 + 6.64648 x 10 ) kg
= 9.10814 x
1031 kg,
3He
27 31 _ (9.10939 x 10 kg) x (5 .00824 X 10 kg) J.l(9. 10939 X 10 31 + 5.00824 x 10 27) kg
= 9.10773
1031 kg.
Finally, the wavenumbers for n
= 3 +
n
X
= 2 are
4Hev
= (4.81870 x
10 35 ) x (9.10814 x 10 31) x (1/4  1/9)cm 1 = 160957.4 cm I
I,
3Hev
= (4.81870 x
10 35 ) x (9.10773 x 10 31) x (l/41/9)cm 1 =160954.7cm
l·
l
ATOMIC STRUCTURE AND ATOMIC SPECTRA
The wavenumbers for n
P10.27
219
= 2 + n = I are ( I l l  1/ 4)cm 
1
= 1329170cm 1 I,
1035 ) x (9.10773 x 10 31) x ( I l l  1/ 4)cm
1
= 1329155cm 1 I·
4Hev
= (4.81870 x 1035 ) x (9. 10814 x 10 31) x
3Hev
= (4.81870 x
(a) Compute the ratios vSlariv for all three lines. We are given wavelength data, so we can use: VSlar
A
V
ASlar
The ratios are: 438.392 nm 438.882nm
= 0.998884,
440.510 nm 44l.000nm
= 0.998889,
and
441.510nm 442.020nm
= 0.998846.
The frequencies of the stellar lines are all less than those of the stationary lines, so we infer that the star is 1receding 1from earth. The Doppler effect follows:
Vreceding
=
vf
f 2( 1 + sic)
where f
= (I
1
SIC ) 1/2
= ( I +slc
, so 1 f 2
s=  2c.
 sic) ,
I
+f
Our average value off is 0.998873 . (Note: the uncertainty is actually greater than the significant figures here imply, and a more careful analysis would treat uncertainty explicitly.) So the speed of recession with respect to the earth is:
s
=
(I 
0.997747) c 1+0.997747
= 11.128 x .
10 3 c 1= 13.381 x 105 m s I
I.
(b) One could compute the star's radial velocity with respect to the sun if one knew the earth's speed with respect to the sun along the sunstar vector at the time of the spectral observation. This could be estimated from quantities available through astronomical observation: the earth's orbital velocity times the cosine of the angle between that velocity vector and the earthstar vector at the time of the spectral observation. (The earthstar direction, which is observable by earthbased astronomers, is practically identical to the sunstar direction, which is technically the direction needed.) Alternatively, repeat the experiment half a year later. At that time, the earth's motion with respect to the sun is approximately equal in magnitude and opposite in direction compared to the original experiment. Averagingf values over the two experiments would yieldf values in which the earth's motion is effectively averaged out. P10.29
See Figure 10.3. Trends: (i) II <
h < 13 because of decreased nuclear shielding as each successive electron is removed.
220
STUDENT'S SOLUTIONS MANUAL
First three ionization energies of group 13 40
>
'"
35 30
;>.,
~
::? 25
'"c: '"c:0
'Zl to
20
•
•
15
N
10
~
5
'2
•
.
0 0
20
40
60
Atomic number, Z
80
100
Figure 10.3
(ii) The ionization energies of boron are much larger than those of the remaining group elements because
the valence shell of boron is very small and compact with little nuclear shielding. The boron atom is much smaller than the aluminum atom. (iii) The ionization energies of AI , Ga, In, and Tl are comparable even though successive valence shells are further from the nucleus because the ionization energy decrease expected from large atomic radii is balanced by an increase in effective nuclear charge.
Molecular structure
Answers to discussion questions 011.1
Our comparison of the two theories will focus on the manner of construction of the trial wavefunctions for the hydrogen molecule in the simplest versions of both theories. In the valence bond method, the trial function is a linear combination of two simple product wavefunctions, in which one electron resides totally in an atomic orbital on atom A, and the other totally in an orbital on atom B. See eqns Il.l and 11 .2, as well as Fig. 11 .2. There is no contribution to the wavefunction from products in which both electrons reside on either atom A or B. So the valence bond approach undervalues, by totally neglecting, any ionic contribution to the trial function. It is a totally covalent function. The molecular orbital function for the hydrogen molecule is a product of two functions of the form of eqn 11.8, one for each electron, that is,
1/1
= [A(l) ± B(I)][A(2) ± B(2)) = A(l)A(2) + B(1)B(2) +A(1)B(2) + B(l)A(2).
This function gives as much weight to the ionic forms as to the covalent forms. So the molecular orbital approach greatly overvalues the ionic contributions. At these crude levels of approximation, the valence bond method gives dissociation energies closer to the experimental values. However, more sophisticated versions of the molecular orbital approach are the methods of choice for obtaining quantitative results on both diatomic and polyatomic molecules. See Sections 11.611.8. 011.3
Both the Pauling and Mulliken methods for measuring the attracting power of atoms for electrons seem to make good chemical sense. If we look at eqn 11 .23 (the Pauling scale), we see that if D(AB) were equal to 112[D(AA) + D(BB)) the calculated electronegativity difference would be zero, as expected for completely nonpolar bonds. Hence, any increased strength of the AB bond over the average of the AA and BB bonds, can reasonably be thought of as being due to the polarity of the AB bond, which in turn is due to the difference in electronegativity of the atoms involved. Therefore, this difference in bond strengths can be used as a measure of electronegativity difference. To obtain numerical values for individual atoms, a reference state (atom) for electronegativity must be established. The value for fluorine is arbitrarily set at 4.0. The Mulliken scale may be more intuitive than the Pauling scale because we are used to thinking of ionization energies and electron affinities as measures of the electron attracting powers of atoms. The choice of factor 112, however, is arbitrary, though reasonable, and no more arbitrary than the specific form of eqn 11.23 that defines the Pauling scale.
222
011.5
STUDENT'S SOLUTIONS MANUAL
The Hi.ickel method parameterizes, rather than calculates, the energy integrals, IX and (J , that arise in molecular orbital theory. They are considered to be adjustable parameters; their numerical values emerge only at the end of the calculation by comparison to experimental energies. The overlap integral is neglected, set equal to zero. Three other rather drastic approximations, listed in Section 11.6(a) of the text, eliminate many terms from the secular determinant and make it easier to solve: all diagonal terms of the determinant are set equal IX  E ; nearestneighbor terms (that is, between bonded atoms) all have the same value, {J; and all other terms are zero. (Ease of solution was important in the early days of quantum chemistry before the advent of computers, and without the use of these approximations, calculations on polyatomic molecules would have been difficult to accomplish.) The simple Hi.ickel method is usually applied only to the calculation of rr electron energies in conjugated organic systems. The simple method is based on the assumption of the separability of the a and rr electron systems in the molecule. This is a very crude approximation and works best when the energy level pattern is determined largely by the symmetry of the molecule. (See Chapter 12.)
011.7
The ground electronic configurations of the valence electrons are found in Figures 11.3111.33 and 11.37.
Oz
z 1agz 1auz lrr42a u g lazlaz2a21rr4)rrz g u g u g
NO
laz2a23a2lrr42rr I
Nz
b=3
2S + 1= 0
b=2
2S+ I = 3
b= 2!
2S+I=2
The following figures show HOMOs of each . Shaded vs. unshaded atomic orbital lobes represent opposite signs of the wavefunctions. A relatively large atomic orbital represents the major contribution to the molecular orbital.
Nz
2a molecular orbital
NO
2rr
Dinitrogen with a bond order of three and paired electrons in relatively low energy molecular orbitals is very unreactive. Special biological, or industrial, processes are needed to channel energy for promotion of 2a electrons into high energy, reactive states. The high energy Irrg LUMO is not expected to form stable complexes with electron donors. Molecular nitrogen is very stable in most biological organisms, and as a result the task of converting plentiful atmospheric N 2 to the fixed forms of nitrogen that can be incorporated into proteins is a difficult one. The fact that Nz possesses no unpaired electrons is itself an obstacle to facile reactivity, and the
MOLECULAR STRUCTURE
223
great strength (large dissociation energy) of the N2 bond is another obstacle. Molecular orbital theory explains both of these obstacles by assigning N2 a configuration that gives rise to a high bond order (triple bond) with all electrons paired. (See Fig. 11 .33 of the text. ) Dioxygen is kinetically stable because of a bond order equal to two and a high effective nuclear charge that causes the molecular orbitals to have relatively low energy. But two electrons are in the high energy IJTg HOMO level, which is doubly degenerate. These two electrons are unpaired and can contribute to bonding of dioxygen with other species such as the atomic radicals Fe(lI) of hemoglobin and Cu(Il) of the electron transport chain. When sufficient, though not excessively large, energy is available, biological processes can channel an electron into this HOMO to produce the reactive superoxide anion of bond order I ~. As a result, 0 2 is very reactive in biological systems in ways that promote function (such as respiration) and in ways that disrupt it (damaging cells). Although the bond order of nitric oxide is 2~, the nitrogen nucleus has a smaller effective nuclear change than an oxygen atom would have. Thus, the one electron of the 2JT HOMO is a high energy, reactive radical compared to the HOMO of dioxygen. Additionally, the HOMO, being antibonding and predominantly centered on the nitrogen atom, is expected to bond through the nitrogen. Oxidation can result from the loss of the radical electron to form the nitrosyl ion, NO+ , which has a bond order equal to 2. Even though it has a rather high bond order, NO is readily converted to the damagingly reactive peroxynitrite ion (ONOO) by reaction with 0; without breaking the NO linkage.
Solutions to exercises E11.1(b)
E11.2(b)
E11.3(b)
Use Figure I 1.23 forH;, I l.33for N2 , and 1l.31 for 0 2. 110 220* 1 1 b = 0.5
(a)
H; (3 electrons) :
(b)
N2 (10 electrons) :
110 220*21JT 430 2 1 b=3
(c)
02(l2electrons) :
110 2 20*230 2 1n 4 2n*21 b=2
CIF is isoelectronic with F2, CS with N2 . 110 220 *230 2 In 42n*41
(a)
CIF(14electrons):
(b)
CS(IOelectrons):
110 220*21 n 430 2 1 b = 3
(c)
0;(13 electrons) :
110 220*230 2 In 42n*3 1 b
b
=
I
= 1.5
Decide whether the electron added or removed increases or decreases the bond order. The simplest procedure is to decide whether the electron occupies or is removed from a bonding or anti bonding orbital. We can draw up the following table, which denotes the orbital involved
(a) ABChange in bond order (b) AB + Change in bond order
N2
NO
02
2JT *
2n*
 1/2
112
30
2n*
2n*
 1/2
+112
+1/2
C2
F2
CN
2n*
30
40*
1/2
+1/2 In  1/2
 1/2
2JT*
30+112 30
+1/2
1/2
224
STUDENT'S SOLUTIONS MANUAL
I
I
(a) Therefore, C2 and CN are stabilized (have lower energy) by anion formation.
I
(b) NO, 02 and F21 are stabilized by cation formation; in each of these cases the bond order increases. E11.4(b)
Figure 11 .1 is based on Figure 11.31 of the text but with Cl orbitals lower than Br orbitals. BrCl is likely to have a shorter bond length than BrCl; it has a bond order of 1, while BrCl has a bond order of 112.
3p
3s
Figure 11.1 E11.S(b)
0i (11 electrons) :
10" 220" *230" 21 JT 42JT *1
b = 5/ 2
02 (12 electrons) :
J0"220"*230"21 JT 42JT*2
b=2
0; (13 electrons) :
J0"220"*230"21JT 42JT*3
b = 3/2
O~ (14 electrons) :
Each electron added to
I0i, E11.6(b)
02, 0;,
10"220"*230"21JT 42JT*4
0i is added to an anti bonding orbital, thus increasing the length. So the sequence
o~I has progressively longer bonds.
1
1/f2 dr = N 2 2
1
= N (l
(1/fA
+ A1/fB)2 dr
+ A2 + 2AS) 1
Hence N E11.7(b)
b= 1
= 1= N
[I
2
1
1/fA1/fsdr
(1/fl
=
+ A21/f~ + 2A1/fA 1/fs) dr
= 1
s]
) 1/ 2
= ( 1 + 2AS + A2
We seek an orbital of the form aA + bB, where a and b are constants, which is orthogonal to the orbital N(0.145A + 0.844B) . Orthogonality implies
1+ 1 (aA
N
bB)N(0.145A
[0. 145aA2
+ 0.844B) dr = 0
+ (0.145b + 0.844a)AB + 0.844bB2] dr = 0
MOLECULAR STRUCTURE
The integrals of squares of orbitals are I and the integral
0= (0.145
+ 0.844S)a + (0.145S + 0.844)b
so
JAB d r a=
225
is the overlap integral S, so
0. 145S + 0.844  0.145 + 0.844S b
This would make the orbitals orthogonal, but not necessarily normalized. If S = 0, the expression simplifies to a
= _ 0.844 b 0.145
and the new orbital would be normalized if a = 0.844N and b = 0.145N. That is
IN(0.844A 
0.145B)
I
E11.8(b)
The trial function 1/1 = x 2 (L  2x) does not obey the boundary conditions of a particle in a box, so it is I not appropriate I· In particular, the function does not vanjsh at x = L.
E11.9(b)
The variational principle says that the minimum energy is obtajned by taking the derivatjve of the trial energy with respect to adjustable parameters, setting it equal to zero, and solving for the parameters: Etrial
=
3an _ e (~) 1/2 2J.L eo 2JT 3 2
2
so
Solving for a yields:
Substituting this back into the trial energy yields the minimum energy:
E11.10(b) Energy is conserved, so when the photon is absorbed, its energy is transferred to tbe electron. Part of
it overcomes the binding energy (ionization energy) and the remainder is manifest as the now freed electron 's kinetic energy.
so
Epholon
= I + Ekinelic
Ekinelic
=
Epholon  /
he (6.626 x 10 34 1 s) x (2.998 X 10 8 m sI) =  / =  4.6geV A (584 x 1O 12 m) x (1.602 x 1O 19 1eV I)
=1211gevl=13.39 x 10 16 11 E11.11 (b) The molecular orbitals of the fragments and the molecular orbitals that they form are shown in Figure 11.2. E11.12(b) We use the molecular orbital energy level diagram in Figure 11.41 . As usual, we fill the orbitals starting
with the lowest energy orbital, obeying the Pauli principle and Hund 's rule. We tben write
226
STUDENT'S SOLUTIONS MANUAL 7r *
Figure 11.2
(a) C6H6 (7 electrons) :
E
= 2(a +
2,8) + 4(a +,8) + (a ,8)
(b) C6 H t(5 electrons):
E
= 1 7a +
7,8 I
Ia~Ue~g I
= 2(a + 2,8) + 3(a + ,8) = 1 5a + 7,8 I
E11.13(b) The secular determinants from E 11.l3(a) can be diagonaJized with the assistance of generalpurpose mathematical software. Alternatively, programs specifically designed for Hiickel calculations (such as the one at Australia's Northern Territory University, http://www.smps.ntu.edu.auimodules/mod3/ interface.html) can be used. In both molecules, 14 nelectrons fill seven orbitals. (a) In anthracene, the energies ofthe filled orbitals are a + 2.414 21,8, a + 2.00000,8, a + 1.41421,8 (doubly degenerate), a + 1.00000,8 (doubly de enerate), and a + 0.41421,8, so the total energy is 140: + 19.313 68,8 and the n energy is 19.313 68,8 . (b) For phenanthrene, the energies of the filled orbitals are a + 2.43476,8, a + 1.95063,8, a + 1.51627,8, a+ 1.30580,8, a+ 1.14238,8, a +0.76905,8, a+0.60523,8,sothetotalenergyis 140: + 19.44824,8 and the n energy isI19.44824,8 I.
Solutions to problems Solutions to numerical problems P11.1
VrA = cos kx measured from A, 1/rB Then, with Vr
= cos k' (x 
R) measuring x from A.
= VrA + Vrs,
Vr = coskx + cosk'(x  R) = cos kx + cos k'R cos k'x + sin k'R sin k'x
= cos a cos b + sin a sin b]. Jt k = k' = ~. cos k'R = cos ~ = O' sink' R = sin 2" = I. 2R' 2' [cos(a  b)
(a)
Jtx
Vr = cos 2R + sin
Jtx
2R'
MOLECULAR STRUCTURE
I
"2R,
For the midpoint, x =
. I 1/1 ( I R) = cos I Jl + SIn  Jl =
so
2
4
interference (1/1 > 1/1A, 1/IB)· Jl I 3Jl I 3Jl (b) k=2R' k =2R; coskR=cosT=O, JlX 2R
4
21 / 2
and there
227
. . IS constructive
sink'R = 1.
3Jlx
1/1 =cos  si n  . 2R
I
interference P11.3
(1/1
I
"2R,
For the midpoint, x =
= cos  Jl 4
. 3 SIn 
4
Jl = 0 and there is destructive
1/IA, l/Is).
<
The s orbital begins to spread into the region of negative amplitude of the p orbital. When their centers coincide, the region of positive overlap cancels the negative region. Draw up the following table. R/ao
0
S
0
00429
2
3
4
5
6
0.588
0.523
0.379
0.241
0.141
7 0.D78
8 0.041
9
10
0.021
0.01
1.0
0.8
5 0.6
1
l'y
(
0.4
/ 0.2
o
5(\5, 2pz)
"\. 't
/
II o
"\ ~ ........
2
4
6
8
R/ao
'10
Figure 11.3
The points are plotted in Fig. II .3. The maximum overlap occurs at 1 R = 2. lao I. P11.S
We obtain the electron densities from
p+ = 1/Ii and p_ =
1/1~ with 1/1+ and 1/1 as given in Problem 1104.
228
STUDENT'S SOLUTI ONS MANU AL
We evaluate the factors preceding the exponentials in 1fr+ and 1fr.
N+
(

I
:n:al
Likewise, N _
)
I~
= 0.561 x
(~) 1/ 2 :n:ao
1/ 2
I
(:n:
1216pm 3/ 2 ·
x (52.9 pm)3 )
621 pm 3/ 2 ·
Then and The 'atomic ' density is
=
p
~(1frls(A)2 + 1frls(B)2) = ~ 2
x
2
e  2rA / aO
9.30 x
+ e 21ll /ao
e2lzl/ao
105 pm 3
The difference density is op±
(_1_) n~
(e  2rA/aO
+ e  21ll /aO)
+ e  2Iz  RI/ao
9.30 x 105 pm 3
= P± 
p.
Draw up the following table using the information in Problem 14.4. z/pm
p+ x 107 / pm 3 p_ x 107 / pm 3 p x 107 / pm 3 op+ x 107 / pm  3 op_ x 107 / pm 3
100
80
60
40
20
0
20
40
0.20 0.44 0.25 0.05 0.19
0.42 0.94 0.53 0.11 0.41
0.90 2.01 1.13 0.23 0.87
1.92 4.27 2.41 0.49 1.86
4.09 9.11 5.15 1 .05 3.96
8.72 19.40 10.93 2.20 8.47
5.27 6.17 5.47 0.20 0.70
3.88 0.85 3.26 0.62 2.40
100
120
140
160
180
200
7.42 14.41 8.88 1.46 5.54
5.10 11 .34 6.40 1.29 4.95
2.39 5.32 3.00 0.61 2.33
1.12 2.50 1.41 0.29 1.09
0.53 1.17 0.66 0.14 0.51
0.25 0.55 0.31 0.06 0.24
z/pm
60
p+ x 10 7 / pm  3 p_ x 10 7 / pm 3 p x 107 / pm 3 op+ x 107 / pm  3 op_ x 107 / pm  3
3.73 0.25 3.01 0.70 2.76
80 4.71 4.02 4.58 0.13 0.56
The densities are plotted in Fig. 11.4(a) and the difference densities are plotted in Fig. 11.4(b). P11.7
P
=
11fr1 2dr ~ 11fr1 2 or ,
(a) From Problem 11 .5
or
=
1.oopm 3
MOLECULAR STRUCTUR E
20
15 ME Q.
'0 10 X
<:>.
5 0  100
100
0
200
z/pm
Figure 11.4(a)
10 8 6
1:
Q.
.0
x <:>.
"0
4
2 0 2 4 6 100
200
100
0 z/pm
Figure ll.4(b)
Therefore, the probability of finding the electron in the volume
P = 8.6
X
or at nucleus A is
10 7 pm  3 x 1.00pm 3 = 18.6 x 10 7 1.
(b) By symmetry (or by taking (c) From Fig. 11.4(a),
z=
1/Ii (~R) =
106pm) P
=
18.6 x 10 7 1
3.7 X 10 7 pm  3 , so
P=
rI3 .7  xI0 7'1
(d ) From Fig. 11 .5, the point referred to lies at 22.4 pm from A and 86.6 pm from B.
0.0 pm
Figure 11.5
A _'''..,. B
e 22.4/ 52.9
Therefore,
1/1
=
1/1 2 = 4.9
+ e  86.6/ 52.9
12 16pm X
10 7 pm  3 ,
3/ 2
so
P
_0._65_ +_0:;.1=9 _ 4 3/ 2 12 16pm 3 / 2  6.98 x 10 pm .
= 14.9 x 10 7 1.
229
230
STUDENT'S SOLUTIONS MANUAL
For the antibonding orbital, we proceed similarly. (a) 1/I~ (z
= 0) = 19.6 X 10 7 pm 3 [Problem 1l.5], so
(b) By symmetry, P
(c)
GR) =
1/1~
(d) We evaluate
=
0,
12.0 x 106 1.
so
=
P
_ 0.65  0.19 _
1/1~ = 5.49 EH
= El =
[QJ.
1/1 at the point specified in Fig. 1l.5.
1/1  621 pm 3/ 2
P11.9
12.0 x 10 6 1.
P

 4
10 7 pm  3 ,
X
 3/ 2
7.41 x 10 so
pm
.
= 15.5 x 10 7 1.
P
hcRH [Section 1O.2(b)].
Draw up the following table using the data in question and using
?
?
?
~
~
=x= x4nEoR 4nEoao R 4nEo x (4nEon2 / mee2) R
so that
(e2 / 4nEoR) Eh
ao
=. R
Draw up the following table. R / ao
0
2 (e / 4n EoR) / Eh
00
(VI  V2) / Eh
0
(E  EH) / Eh
00
0.007 l.049
2
3
4
0.500 0.031 0.425
0.333 0.131 0.132
0.250 0.158 0.055
00
0 0 0
The points are plotted in Fig. 11.6. The contribution V2 decreases rapidly because it depends on the overlap of the two orbitals. P11.11
The internuclear distance (r )n ~ n2ao , would be about twice the average distance (~ l.06 x 106 pm) of a hydro genic electron from the nucleus when in the state n = 100. This distance is so large that each of the following estimates is applicable. Resonance integral,
f3
Overlap integral, S
~ E
~
8 (where 8
(where
E ~
~
0).
0).
MOLECULAR STRUCTURE
231
0.5
0.4
..c:
0.3
... .. .!
~
:c L<J I
~
0.2
0.1
o
2
4
3
5
Figure 11.6
Coulomb integral , et Binding energy
£,,=100 for atomic hydrogen .
R::
= 2{£+ = 2{
 £,,=100)
~ ~;
 £,,= 100 }
= 2{et  £11= 100)
Vibrational force constant, k Rotational constant, B
R::
0, because of the weak binding energy.
= h2 /2hcl = h2 / 2hc/JrIB
R::
0 because
rIB is so large.
The binding energy is so small that thermal energies would easily cause the Rydberg molecule to break apart. It is not likely to exist for much longer than a vibrational period. P11.13
In the simple Hiickel approximation
10] [0]0/~~0
~~~0 o 2
eto  £
0
0 0
eto  £
f3
f3
0
++
3
0 0 eto  £ f3
f3 f3 f3 etN  £
=0.
(£  etO)2 x {(£  etO) x (£  etN)  3(3 2) =
o.
Therefore, the roots are £  eto
= 0 (twice)
and
(£  eto) x (£  etN)  3f32
= o.
232
STUDENT'S SOLUTIONS MANUAL
Each equation is easily solved (Fig. 11.7(a)) for the permitted values of E in terms of ao , aN , and f3. The quadratic equation is applicable in the second case.
8C
ao+aN+ {( ao  aN)2 + 12,82 11 /2
E+
2
OJ
c 0
'".... 5
II)
ao
.~
>.
OJ) .... cII) II)
a o +aN { (aoaN)2 + 12,82 }1/2
I<
2 E_
Figure 11.7(a)
In contrast, the IT energies in the absence of resonance are derived for just one of the three structures, i.e. for a structure containing a single localized IT bond.
l
ao  E
f3
f3 aN  E
I = o.
Expanding the determinant and solving for E gives the result in Fig.l1 .7(b). Delocalization energy
= 2 (E_ (with resonance) 
E_ (without resonance)}
II)
u
C
OJ
c
ao+aN+ { (aoC>N )2+ 4,82} 1/2
E+
2
~ ~
~ 0
5 .~ >.
......c ...
OJ)
aO+aN  { (ao  aN)2 +4fJ2} 1/2
I<
2
Figure 11.7(b)
E_
4f32
Delocalization energy :::::: '(ao  aN)
P11.1S
(a) The transitions occur for photons whose energies are equal to the difference in energy between the highest occupied and lowest unoccupied orbital energies: EphO(OD
= ELUMO 
EHOMO·
If N is the number of carbon atoms in these species, then the number of IT electrons is also N . These N electrons occupy the first 12 orbitals, so orbital number NI2 is the HOMO and orbital number
MOLECULAR STRUCTURE
233
I + N/2 is the LUMO. Writing the photon energy in terms of the wavenumber, substituting the given energy expressions with this identification of the HOMO and LUMO, gives
= (a + 2f3 cos (! N + l)rr)
hev
N+I
= 2f3
(
cos
_ (a
+ 2f3 cos
!Nrr ) N+ I
(!N + I)rr !Nrr )  cos   . N+ I N+I
Solving for f3 yields hev f3  ;,:;( (!N + I)rr !Nrr ). 2 cos  cos  N+ I N+ I Draw up the following table Species
N
v/ cm I
estimated f3/eV
C2H4 C4H6 C6 HS CsHIO
2 4 6 8
61500 46080 39750 32900
 3.813 4.623 5.538 5.873
(b) The total energy of the Jr electron system is the sum of the energies of occupied orbitals weighted by the number of electrons that occupy them. In CSHIO , each of the first four orbitals is doubly
occupied, so Err
4
4 (
= 2 LEk = 2 L k= 1
a
k)
k
4
+ 2f3COS; = 8a +4f3 L
k=1
Cos ;
= 8a +9.518f3 .
k= 1
The delocalization energy is the difference between this quantity and that of four isolated double bonds, Edeloc
= E rr

8(a
+ f3) = 8a + 9.518f3 
8(a
Using the estimate of f3 from part (a) yields Edeloc
+ f3 ) = 11.158 f3 1·
=
18.913 eV
I.
(c) Draw up the following table, in which the orbital energy decreases as we go down. For the purpose of comparison, we express orbital energies as (Ek  a)/f3 . Recall that f3 is negative (as is a for that matter), so the orbital with the greatest value of (Ek  a)/f3 has the lowest energy. Energy Orbital
(Ek a) / f3
6 5 4 3 2
1.8019 1.2470 0.4450 0.4450 1.2470 1.8019
Coefficients
0.2319 0.4179 0.5211 0.5211 0.4179 0.2319
2
3
4
5
6
0.4179 0.5211 0.2319 0.2319 0.5211 0.4 179
0.5211 0.2319 0.4 179 0.4179 0.2319 0.52 11
0.5211 0.23 19 0.4179 0.4179 0.2319 0.5211
0.4179 0.5211 0.2319 0.23 19 0.5211 0.4179
0.23 19 0.4179 0.5211 0.5211 0.4179 0.23 19
234
STUDENT'S SOLUTIONS MANUAL
The orbitals are shown schematically in Fig. 11.8, with each vertical pair of lobes representing a p orbital on one of the carbons in hexatriene. Shaded lobes represent one sign of the wavefunction (say positive) and unshaded lobes the other sign. Where adjacent atoms have atomic orbitals of the same sign, the resulting molecular orbital is bonding with respect to those atoms ; where adjacent atoms have different sign, there is a node between the atoms and the resulting molecular orbital is anti bonding with respect to them. The lowest energy orbital is totally bonding (no nodes between atoms) and the highestenergy orbital totally anti bonding (nodes between each adjacent pair). Note that the orbitals have increasing antibonding character as their energy increases. The size of each atomic
6
5
4
3
2
Figure 11.8
MO LE CULAR STRU CTURE
235
p orbital is proporti onal to the magnitude of the coefficient of that orbital in the molecular orbital.
So, for example, in orbitals 1 and 6, the largest lobes are in the center of the molecule, so electrons that occupy those orbitals are more likely to be fo und near the center of the molecule than on the ends. In the ground state of the molecule, there are two electrons in each of orbitals 1, 2, and 3, with the result that the probability of fi nding a 7T electron in hexatriene is uniform over the enti re molecule. P11.17
Mathcad may be used to demonstrate the power of matrix diago nali zation techniq ues. The function eigenvalsO performs diagonalization to find the eigenvalues of a matri x while the function eigenvecsO fi nds the eigenvectors. To show use of the techn ique, the 7T conjugated systems of both benzene and hexatriene are analyzed below wi th the Huckel approx imation. The energy levels are reported as x values where x = (E  ex)/ (J . The molecular orbital for a level is identified by counting nodal planes. Energy increases with the number of nodal planes in the wavefunction. Modem mathematic software makes it very easy to check both normalization of a wavefunction and the orthogonality of two wavefunctions so this is demonstrated fo r benzene. Writing the Huckel secular matrix as M(x) , the functions eigenvalsO and eigenvecsO use M (O) as their arguments. The transpose operator (T ) is used to place the coefficients of each eigenfunction along a row. Eigenvalues are sorted with the sortO function. The sum of the squares of the p orbital coefficients equals I when the eigenfunction is normalized. When two eigenfunctions are orthogonal, the sum of their multiplied coefficients equals zero. For benzene:
0
x
M(x) :=
1 0 0 0
x
0 0
x
0 0 0
1 0 0
eigenvecs(M(O))T =
0 0 0
x
1 0
1 0 0 0
sort(eigenvals(M(O))) =
 2 x4 1 x3 1 x3 x2
X
0.569 0.096 0.508  0.274 0.408  0.408
x2 x
2
0.201 0.541  0.491  0.303 0.408 0.408
Checking the normalization of l{!2a :
0.368 0.445  0.017 0.577 0.408 0.408
0.569 0.096 0.508  0.274 0.408 0.408
Xl
0.201 0.541  0.491 0.303 0.408 0.408
0.368 0.445 0.017 0.577 0.408 0.408
l{!2a
l{!2b l{!3a
l{!3b l{!l l{!4
L5 [ ( eigenvecS(M(O))T) .J2 = 1 i= O
0,1
5
Checking the orthogonality of l{!2a and l{!3a :
L [(eigenveCS(M(O))T)
i= O
. ' ( eigenVeCS(M(O))T) 0,1
.J= 0
2,1
236
STUDENT' S SOLUTI ONS MAN UAL
Placing a total of 6 7r electrons of benzene into the three lowest energy levels gives a total x tenn energy: 5
2·
L sort(eigenvals (M(O»); = 8 ;=3
= 2{E I + E2a + E2b} = 2{(a + (3xl) + (a + (3x2a) + (a + (3x2b)}
Err
= 6a + 2(xl + X2a + X2b){3 = 6a + 8{3. Edeloc = Err  6(a + (3) = 2{3. For hexatriene
M(x)
=
x
1
1 0 0 0 0
x
1 0 0 0
eigenvecs(M(O))T
0 1 x
1 0 0
=
0 0 1 x
0 0 0 1 x
0
0.418 0.521 0.232  0.521 0.418 0.232
0 0 0 0 1
sort(eigenvals(M(O)))
x
0.521  0.232 0.418 0.232 0.521 0.418
0.232 0.418  0.521 0.418  0.232 0.521
0.232 0.418  0.521  0.418  0.232 0.521
=
1 .802  1.247 0.445 0.445 1.247 1.802
0.521  0.232 0.418  0.232 0.521 0.418
Es E5 E4 E3 E2 E1
0.418  0.521 0.232 0.521  0.418 0.232
1/12 1/13 1/11 1/14 1/15 1/16
5
2.
L sort(eigenvals(M(O)))i = 6.988 i=3
Err
= 2{El + £2 + E3 } = 2{(a + (3x l ) + (a + (3x2) + (a + (3x3) } = 6a + 2(xl + x2 + X3){3 = 16a + 6.988{3 1.
Edeloc
= Err
 6(a
+ (3 ) = I O.988{31·
Remembering that the resonance integral is negative, we see that the delocalization energy stabilizes the 7r orbitals of the closed ring conj ugated system (benzene) to a greater extent than what is observed in the open chain conjugated system (hexatriene). The unusually large stabilization energy of benzene also demonstrates the validity of the HUckel 4n + 2 rule for planar, cyclic conjugated 7r systems. P11.19
. In all of the molecules considered, the HOMO is bonding with respect to the carbon atoms connected by double bonds, but antibonding with respect to the carbon atoms connected by single bonds. (The bond lengths returned by the modeling software suggest that it makes sense to talk about double bonds and single bonds. Despite the electron delocalization, the nominal double bonds are consistently shorter than the nominal single bonds.) The LUMO had just the opposite character, tending to weaken the
MOLECULAR STRUCTURE
237
c=c bonds but strengthen the CC bonds. To arrive at this conclusion, examine the nodal surfaces of the orbitals. An orbital has an anti bonding effect on atoms between which nodes occur, and it has a binding effect on atoms that lie within regions in which the orbital does not change sign. The rr* + rr transition, then, would lengthen and weaken the double bonds and shorten and strengthen the single bonds, bringing the different kinds of polyene bonds closer to each other in length and strength. Since each molecule has more double bonds than single bonds, there is an overall weakening of bonds. (See Fig. 11.9.)
HOMO
LUMO
Figure 11.9
238
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems P11.21
Since
( I ) (Z
=~
1/12p, '
1/ 2
2rr
x
I
= ..;2 r" R21(YII '
) 3/2
aD
 YII ) [SectionlO.2] ,
I (Z ) 3/2 _eP
=.j2
.j2
1/12p,. .
4( 8rr
/
4( 8rr
3)1/ 2 sinOcosl/J
(Z )3/2 !:..eP/4 sinOcos l/J,
2rr
2
aD
I
= 2.R21( YI ,1 + I
YI ,I ) [Section 10.2]
= I (  I ) 1/2 x
(Z ) 3/2 !:..eP/ 4 sinOsinl/J [Tables 10.1 , 9.3] .
2rr
4
p
3)1/ 2 sinO(e"P+e'Q»[Tables . . 10.1, 9.3]
/
2
aD
= 1(  I ) 1/2 x 4
p
2
aD
I (Z ) 3/ 2 _eP
=
~)eP/4,
x (2
2
aD
Therefore,
1/1 = _1_ x .J3
X
~
(~) 1/2 x (~)3/2
x
4
rr
I (2 ( .j2
aD
P) 2



1 P sm . u cos 'I' A. + .J3 P sm. uCJ sm. 'I'A.) e  p/4 22 2 2 f)
 
I ( I ) (Z
=4
6rr
1/2 x
pip IIp  ) 3/2 x { 2sinecosl/J+   sine sini/J } e p / 4 aD 2 .j2 2 22
I ( I ) (Z
=4
6rr
1/2
x
= 1(1)1  /2 x 4
6rr
aD
)3/2
x
{
(3) }
P () 22' 1+ .j2sinOcosl/JY'2sinOsinl/J
p
/
4
p( ) + [cosl/J.J3sinl/J].sm 0 )} e  p/4.
(Z) 3/2 x { 2  aD
e
2
.j2
The maximum value of 1/1 occurs when sin 0 has its maximum value (+ I, when e = 90°; that is, in the xy plane), and the term multiplying p /2 has its maximum negative value (  ), when I/J = 120°).
MOLECULAR STRUCTURE
P11.23
The normalization constants are obtained from
f N
2 1/I dr
2
f
?
H1/I
± l/rs)2dr
(1/IA
Therefore, N
H
= I, 1/1 = N(1/IA ± 1/IB),
f
= N 2 (1/11 ± 1/1~ ± 21/IA l/rs)dr = N 2(l + 1 ± 2S) = 1.
1
= 2( 1 ± S) , e2 1 e2 1  .   . 
fi? = __ V2 2m
41[60
rs
41[60
R
41[60
= E1/I implies that
Multiply through by
1/1*(= 1/1) and integrate using
Then for
1/1 = N(1/IA + 1/IB)
hence EH
f
and
rA
2
e 1 + '.
f
2
1/1 2 dr + e  . 1 41[60
1/IA ~1/IAdr
rs
=
f
41[60
rA
+ e  . I 41[60
2
e N
1/1 2dr 
f (
1/IB ~1/IBdr [by symmetry]
2
which gives EH
R
f
R
(
I ) 1+ S
X
(VI
1/1 1/IB + 1/1A) dr rA rs
=
+ V2)
VI
= E
/(e 2 /41[ 60)
= E
239
240
or
STUDENT'S SOLUTIONS MANUAL
E
+ V2 + e 2 . 1 1+ S 4nso R
VI
=
EH    
as in Problem 11.8. The analogous expression for E_ is obtained by starting from
1/1 = N(1/IA  1f;s) with N2
= __1__ 2(1  S)
and following through the stepwise procedure above. The result is
as in Problem 11.9. P11.25
(a) Expanding the determinant yields
This is a quadratic equation in E where a
E±
=
=
b ± .Jb2  4ac 2a (aA
+ aB) 2
=
(aA
=
1, b
+ aB)
2
[(aA  aB)2
=
±
(aA +aB) , and c
= aAaB 
f32 . The solution is
(ai + 2aAaB + a~  4aAaB + 4f3 2) 1/2 '''''=2:
+ 4f32JI / 2
±2
(b) By hypothesis, (aA  aB)>> f32, so the term in parentheses can be expanded:
MOLE CU LAR STRU CTURE
241
Solutions to applications P11.27
The secular determi nant for a cyclic species 2 x
1 0 0
3
N I
N
0
0 0 0 0
0 0 0
x
I
x
0
x
=
1
o
000 where x
HNhas the form
x
(aE) / fJ or E = a  fJx.
Expandi ng the determinant, fi ndi ng the roots of the polynomial, and solving for the total binding energy yields the fo llowing table. Note that a < 0 and fJ < O. Species
Permitted x (roots)
Number ofe
Total binding energy
2, 0, 0, 2
4a
6
+4f3 2, ~ ( I  ~) , ~ ( I  ~) , ~ ( I +vrs) , ~ ( I +~) 4a + (3 +~) f3  2, ~ ( I  ~) , ~ ( I  ~) , ~ ( I + vrs) , ~ ( I +~) Sa + ~ (5 +3~) f3  2, ~ ( I  ~),~, ( I ~), ~ ( 1 +~) , ~ ( I +~) 6a + (2 +2~) f3
6
 2,  I, 1 , 1, 1, 2
6a
6
2,  1.248,  1.248, 0.445 , 0.445, 1.802, 1.802
4 4 5
H4 ~ 2H2
6. r U
Ht ~ H2 + Hj
6. r U
+ 8f3 6a + 8.992f3
= 4(a + fJ)  (4a + 4fJ) = O. = 2(a + fJ) + (2a + 2fJ)  (4a + 5.236fJ) = 0.764f3 < O.
The above 6. r U values indicate that H4 and Ht can fall apart without an energy penalty. Hs ~ H2
+ H}
6. r U
= 2(a + fJ)
 (4a
= 2.472fJ >
H6 ~ 3H2
6. r U
= 6(a + fJ)
 (6a
+ 2fJ)
 (6a
+ 6.472fJ)
O.
+ 8fJ)
= 2f3 > O.
Hj ~ 2 H2 + Hj
6. r U = 4(a
+ fJ) + (2a + 4f3) 
=  0.992fJ > O.
(00
+ 8.992fJ)
242
STUDENT'S SOLUTIONS MANUAL
The 6. r U values for Hs, H6 , and Hj suggest that they are stable. Satisfies Hiickel 4n + 2 low energy rule Species
Correct number e
H4 , 4eHt , 4eHs , 6eH6 , 6eHj,6e
Stable
No No Yes Yes Yes
No No Yes Yes Yes
Hiickel's 4n + 2 rule successfully predicts the stability of hydrogen rings. P11 .29
This question refers to six 1,4benzoquinones: the unsubstituted, four methylsubstituted, and a dimethyldimethoxy species. The table below defines the molecules and displays reduction potentials and computed LUMO energies. 0
R6v~ R5
R3 0
Species
2 3 4 5 6
R2
R3
R5
H CH 3 CH3 CH3 CH3 CH 3
H H H H H CH3 CH 3 C H 3 CH3 CH3 CH3 CH30
R6
r/v
H H H H CH3 CH30
0.078 0.023 0.067 0.165 0.260
ELUMo / eV* 1.706 1.651 1.583 1.371 1.233 1.446
* Semiempirical, PM3 level, PC Spartan pro™. (a) The calculations for the species 15 are plotted in Fig. 11.10. The figure shows that a linear relationship between the reduction potential and the LUMO energy is consistent with these calculations. (b) The linear leastsquares fit from the plot of ELUMO vs. r is
I
I
ELUMo / eV = 1.621  1.435r / V Solving for r
yields
r / V = (ELUMo / eV
+
1.621)/ 1.435.
(r2
= 0.927) .
MOLECULAR STRUCTURE
243
1.2 ......,,:,...::,.,....,,
1.3
> ~
 1.4
o
~  1.5
...J
l<.l  1.6
1.7  1.8 L_'_'_ _'_'_'_ _'_'_'
D.3
D.2
D. I
D.O
E!'/ V
0.1
Figure 11.10
Substituting the computed LUMO energy for compound 6 (a model of ubiquinone) yields
~ = [(1.446 + 1.621) / 1.435] = 10.122 V I. (c) The model of plastoquinone defined in the problem is compound 4 in the table above. Its experimental reducing potential is known ; however, a comparison to the ubiquinone analog based on ELUMO ought to use a computed reducing potential:
~ = [(1.371
+ 1.621)/1.435] = 10.174 V I.
The better oxidizing agent is the one that is more easily reduced, the one with the less negative reduction potential. Thus we would expect compound 6 to be a better oxidizing agent than compound 4, and ubiquinone a better oxidizing agent than plastiquinone . (d) Impact 17.2 states that coenzyme Q (which is another name for ubiquinone) acts as an oxidizing agent (oxidizing NADH and FADH2) in respiration (i.e. in the overall oxidation of glucose by oxygen). Plastoquinone, on the other hand, acts as a reducing agent (reducing oxidized plastocyanin) in photosynthesis (Impact 123.2). Respiration involves the oxidation of glucose by oxygen, while photosynthesis involves a reduction to glucose and oxygen. It stands to reason that the better oxidizing agent, ubiquinone, is employed in oxidizing glucose (i.e. in respiration), while the better reducing agent (that is, the poorer oxidizing agent) is used in reduction, i.e. in photosynthesis. (Note, however, that both species are recycled to their original forms: reduced ubiquinone is oxidized by iron (III) and oxidized plastoquinone is reduced by water.)
12
Molecular symmetry
Answers to discussion questions 012.1
The point group to which a molecule belongs is determined by the symmetry elements it possesses . Therefore the first step is to examine a model (which can be a mental picture) of the molecule for all its symmetry elements. All possi ble symmetry elements are descri bed in Section 12.1. We list all that apply to the molecule of interest and then follow the assignment procedure summarized by the flow diagram in Figure 12.7 of the text.
012.3
The dipole moment is a fixed property of a molecule and as a result it must remain unchanged through any symmetry operation of the molecule. Recall that the dipole moment is a vector quantity ; therefore both its magnitude and direction must be unaffected by the operation. That can only be the case if the dipole moment is coincident with all of the symmetry elements of the molecule. Hence molecules belonging to point groups containing symmetry elements that do not satisfy this criterion can be eliminated. Molecules with a center of symmetry cannot possess a dipole moment because any vector is changed through inversion. Molecules with more than one Cn axis cannot be polar since a vector cannot be coincident with more than one axis simultaneously. If the molecule has a plane of symmetry, the dipole moment must lie in the plane; if it has more than one plane of symmetry, the dipole moment must lie in the axis of intersection of these planes. A molecule can also be polar if it has one plane of symmetry and no Cn . Examination of the character tables at the end of the Data section shows that the only point groups that satisfy these restrictions are C s , Cn , and Cnv .
012.5
A representative is a mathematical operator (usually a matrix ) that represents the physical symmetry operation. The set of aU these mathematical operators corresponding to all the operations of the group is called a representation. See Section 12.4(a) for examples.
012.7
Selection rules tell us which transition probabilities between energy levels are nonzero, namely, which spectroscopic transitions will have a nonzero intensity. The intensities are given by the transition moment integral, eqn 12.9, which has the form of the integral of the product of three functions as described by eqn 12.8. Without actually having to perform the integrations involved, group theory can tell us which of these integrals wiU be nonzero, and hence tell us which are the allowed transitions. Such integrals will be nonzero only if the representation of the triple product in the point group of the molecule spans A I or contains a component that spans AI. In practice, it is usually sufficient to work with the product of the characters of the representations, rather than the matrix representatives themselves. See Examples 12.6 and 12.7.
MOLECULAR SYMMETRY
245
Solutions to exercises E12.1(b)
CCI4 has 14 C3 axes 1(each CCl axis), 13 C2 axes 1(bisecting CICCI angles), 13S4 axes 1(the same as the C2 axes), and
E12.2(b)
16 dihedral mirror planes I(each CICCI plane).
Only molecules belonging to Cs , Cn, and Cnv groups may be polar, so ... (a) CH3CI (C3v) 1 may be polar I along the CCI bond; (b) HW2 (CO) 10 (D4h) I may not be polar I (c) SnCI4 (Td ) 1 may not be polar I
E12.3(b)
The factors of the integrand have the following characters under the operations of D6h
E Px Z
2
pz Integrand
1 2
2C6
2C3
C2
3C~
3C~
I
2
0 I I 0
0 I \ 0
I I
2
2 I 1 2
2S3
2S6
I 1
I
\
\
I
I
(ih 2 I I 2
3(id
3(iv
0
0
I 0
I 0
The integrand has the same set of characters as species Elu , so it does not include AI g; therefore the integral 1vanishes I. E12.4(b)
We need to evaluate the character sets for the product A 1g E2uq , where q E
2C6
2C3
C2
3C~
3C~
I I
I 2 2 4
I 0 0 0
0 0 0
AI g
I
E2u
2
I
(x ,y)
2 4
I I
Integrand
2S3
I
2  2 4
= x , y, or z
2S6
(ill
3(id
3(iv
I 2
I 0 0 0
1 0 0 0
1
2
I
4
To see whether the totally symmetric species AI g is present, we form the sum over classes of the number of operations times the character of the integrand c(A Ig)
= (4) + 2( I) + 2(1) + (4) + 3(0) + 3(0) + (4) + 2( 1 ) + 2( 1) + ( 4) + 3(0) + 3(0) = 0
Since the species A I g is absent, the transition is 1 forbidden 1 for x or ypolarized light. A similar analysis leads to the conclusion that AI g is absent from the product AI g E2uz; therefore the transition is forbidden. E12.S(b)
The classes of operations for D2 are: E, C2(X), C2 (Y). and C2(Z). How does the function xyz behave under each kind of operation? E leaves it unchanged. C2 (x) leaves x unchanged and takes y to  y and z to  z, leaving the product xyz unchanged. C2(Y) and C2(Z) have similar effects, leaving one
246
STUD ENT' S SOLUTIONS MANUAL
axis unchanged and taking the other two into their negatives. These observations are summarized as follows E
C2(X)
C2(Y)
C2(Z)
xyZ
A look at the character table shows that thi s set of characters belong to symmetry species ~. E12.6(b)
A molecule cannot be chiral if it has an ax is of improper rotation. The point group Td has 5 4 axes 1 and 1mirror planes (= 51) I, which preclude chirality. The Th group has, in addition, a
1
1center of inversion (= 52) I· E12.7(b)
The group mUltiplication table of group C4v is E
C+ 4
C4
C2
av(x)
av(y)
ad (xy )
E
E
av(y)
ad (xy)
E
C2 C4 C+ 4
av(x)
C+
C+ 4 C2
C4
C+ 4 C
ad(XY) ad ( xy)
ad (xy) av(y) av(x) ad (xy)
E
av(y)
av(x)
ad (xy)
ad (xy)
av(Y)
E
C2 E
C4 C+ 4 E
C+ 4 C4 C2
C2
E
4
C4 C2
4
C2 av(x)
avCx)
C4 ad (xy)
C2 C+ 4 ad(XY)
av(Y)
av(y)
ad (xy)
ad (  xy)
av(x)
ad(xy)
ad (xy)
av(x)
a v(y)
ad (xy)
C2 C+
av(Y)
av(x)
ad (xy)
C4
ad ( xy ) ad (xy)
E12.8(b)
E
4
C4 C+ 4
ad (  xy) ad (xy) av(x) av(Y)
See Figure 12.1. (a) Sharpened pencil: E, Coo, a v; therefore 1 Coov 1
(b) Propellor: E , C3, 3C2; therefore 1 D31
(c) Square table: E, C4 , 4av; therefore 1 C4v
I;
(d) Person: E, a v (approximately); therefore E12.9(b)
Rectangular table: E, C2, 2av; therefore 1 C2v 1
[SJ.
We follow the flow chart in the text (Figure 12.7). The symmetry elements found in order as we proceed down the chart and the point groups are (a) Naphthalene: E, C2, C~ , C~, 3ah, i; 1 D2h 1 (b) Anthracene: E, C2 , C~ , C~, 3ah, i; 1 D2h 1 (c) Dichlorobenzenes:
(i) 1,2dichlorobenzene: E, C2, a y, a~; 1 C2y
1
(ii) 1,3dichlorobenzene: E, C2, a y, a~; 1 C2y
1
(iii) lAdichlorobenzene: E, C2, C; ,
q, 3ah,i; ID2h I·
MO LE CULA R SYMM ETRY
247
(b)
(a)
C3
Figure 12.1
E12.10(b) (a) HF C ",V (b )
(e)
F
F
Ccp,F F
43:
D lh
F F
e2v
F
(e)
(f) Td
F F I I I I
F ~L
,
__ _ ___ _
............ F
The fo llowing responses refer to the tex t flow chart (Figure 12.7) for assignin g poi nt groups. (a) HF : li near, no i, so I Coov
I
(b ) IF7 : nonlinear, fewer than 2Cn with n > 2, Cs , 5C~ perpendicular to Cs , O"h , so I DSh I (c) Xe02F2: nonlinear, fewe r than 2Cn with n > 2, C2, no C~ perpendicular to C2, no O"h , 20"v, so I C2v
I
(d) Fe2 (CO)9: nonli near, fewer than 2Cn with n > 2, C3, 3C2 perpend ic ular to C3, O"h , so D3h (e) cubane (CgHg): nonlinear, more than 2CII with n > 2, i, no Cs, so@;] (f) tetrafl uoroc ubane (23): nonlinear, more than 2CII with n > 2, no i, so~.
I
I
248
STUDENT'S SOLUTIONS MANUAL
E12.11(b) (a) Only molecules belonging to Cs , Cn , and C Il V groups may be polar. In Exercise 12.9(b)
Iarrhadichlorobenzene Iand Imetadichlorobenzene Ibelong to C2v and so may be polar; in Exercise 12.6(b), IHF and Xe02F21 belong to C groups, so they may be polar. IlV
(b) A molecule cannot be chiral if it has an axis of improper rotation  including disguised or degenerate axes such as an inversion centre (S2) or a mirror plane (S , ). In Exercises 12.S(b) and 12.6(b), all the molecules have mirror planes, so I none Ican be chiral. E12.12(b) In order to have nonzero overlap with a combination of orbitals that spans E, an orbital on the central atom must itself have some E character, for only E can multiply E to give an overlap integral with a totally
symmetric part. A glance at the character table shows that IPx and Py Iorbitals available to a bonding N atom have the proper symmetry. If d orbitals are available (as in S03), I all d orbitals except dZ2 1cou ld have nonzero overlap. E12.13(b) The product rf x r(J1.) x Ii must contain Al (Example 12.7). Then, since Ii = BI , r(J1.) = r (y) = B2 (C2v character table), we can draw up the following table of characters
E B2 BI BIB2
C2
frv
a v'
I
I I I
I I I =A2
1
Hence, the upper state is I A21, because A2 x A2 = A I . E12.14(b) (a) Anthracene
H
H
H
:@: H
H
D.
H
The components of J1. span B3u(X), B2u(Y) , and Blu(z). The totally symmetric ground state is Ag . Since Ag x r
I
I
= r in this group, the accessible upper terms are B3u (xpolarized), ~ (y
polarized), and ~ (zpolarized). (b) Coronene, like benzene, belongs to the D6h group. The integrand of the transition dipole moment must be or contain the Ai g symmetry species. That integrand for transitions from the ground state is AIgqf, where q is x,y, or z and! is the symmetry species of the upper state. Since the ground state is already totally symmetric, the product qf must also have Ai g symmetry for the entire integrand to have Aig symmetry. Since the different symmetry species are orthogonal , the only way qfcan have Ai g symmetry is if q and! have the same symmetry. Such combinations include
I
ZA2u, xElu , and yElu , Therefore, we conclude thattransitions are allowed to states with A2u or Elu symmetry.
I
MOLECULAR SYMMETRY
249
E12.15(b)
E AI A2 E sine
2C3
3uv
 I
I
1 Linear combinations of sin e and cos e
2
case Product
0 I I
The product does not contain AI , so I yes Ithe integral vanishes.
Solutions to problems P12.1
I
I
(a) Staggered CH3CH3 : E, C3, C2, 3Ud ; D3d [see Fig . 12.6(b) of the text].
I
(b) Chair C6H1 2: E , C3, C2, 3Ud ; D3d I· Boat C6 H 12: E, C2, u V , (c) B2H6: E , C2,
2C~ ,
Uh ;
u~; I C2v I·
ID2h I·
(d) [Co(enh] 3+ : E, 2C3, 3C2; I D3 1·
I
(e) Crown Ss : E, C4 , C2, 4C~ , 4Ud, 2Ss ; D4d I· Only boat C6H 12 may be polar, since all the others are D point groups. Only [Co (enh ]3+ belongs to a group without an improper rotation axis (SI = u) , and hence is chiral. P12.3
Consider Fig. 12.2. The effect of Uh on a point P is to generate UhP , and the effect of C2 on UhP is to generate the point C2UhP. The same point is generated from P by the inversion i, so C2UhP = iP for all points P. Hence, 1CZUh = i I, and i must be a member of the group. p
Figure 12.2
P12.5
We examine how the operations of the C3v group affect lz = XPy  YPx when applied to it. The transformations of x,y, and z, and by analogy Px, Py, and pz, are as follow s (see Fig. 12.3)
250
STUDENT'S SOLUTIONS MANUAL
E(x,y,z) + (x,y,z), av(x, y,z)
+
(x,y,z),
a~(x,y,z) +
(x,  y,z),
a;(x ,y, z) + (x ,y,  z), cj(x,y,z) + (!x + !J3Y, !5x  !y,z) , C3(X,y,Z) + (!X  !J3Y,!5x  !y, z) .
y
Figure 12.3
MOLECULAR SYMMETRY
251
The characters of all a operations are the same, as are those of both C3 operations (see the C3v character table); hence we need consider only one operation in each class. El z = XPy  YPx = 'z, avl z
= XPy + YPx = Iz
[(x,y,z) + (x,y,z)],
Ct lz = ( !x + !.J3y ) x (!.J3Px  !py)  (!.J3x  !y) x (!Px [(x, y , z) + (!x
= ~ (.J3xPr + XPy = xPv 
YPx
+ !.J3y , 
+ !.J3py)
!.J3x  !y, z) ]
3ypx  .J3ypy  .J3xPx
+ 3xpy 
YPx
+ .J3ypy )
= lz·
ct
The representatives of E , a v, and are therefore all onedimensional matrices with characters I, I , I respectively. It follows that I z is a basis for A2 (see the C3v character table). P12.7
The multiplication table is
ay
ax ax ax ay az
ax ay az
iaz iay
az az iay iax
ay jax iaz
The matrices 1 do not form a group 1 since the products iaz , jay, jax and their negatives are not among the four given matrices. P12.9
(a) In C3v symmetry the His orbitals span the same irreducible representations as in NH3, which is Al + AI + E. There is an additional Al orbital because a fourth H atom lies on the C3 axis. In C3v, the d orbitals span A I + E + E [see the final column of the C3v character table]. Therefore, all five d orbitals may contribute to the bonding.
I
I
(b) In C2v symmetry the His orbitals span the same irreducible representations as in H20 , but one 'H20' fragment is rotated by 90° with respect to the other. Therefore, whereas in H20 the HIs orbitals span AI + B2 [HI + H2, HI  H2], in the distorted C14 molecule they span Al + B2 + Al + BI [HI + H2 , HI  H2, H3 + H4 , H3 14]. In C2v the d orbitals span 2AI + BI + B2 + A2 [C2v
I
I
character table] ; therefore, all except A2(dxy ) may participate in bonding. P12.11
(a) We work through the flow diagram in the text (Fig. 12.4). We note that this complex with freely rotating CF3 groups is not linear, it has no Cn axes with n > 2, but it does have C2 axes; in fact it has two C2 axes perpendicular to whichever C2 we call principal, and it has a ah. Therefore, the point group is 1D2h I· (b) The plane shown in Fig. 12.4 below is a mirror plane so long as each of the CF3 groups has a CF bond in the plane. (i) If the CF3 groups are staggered, then the AgCN axis is still a C2 axis; however, there are no other C2 axes. The AgCF3 axis is an S2 axis, though, which means that the Ag atom
252
STUDENT'S SOLUTIONS MANUAL
is at an inversion center. Continuing with the flow diagram, there is a Uh (the plane shown in the figure). So the point group is I C2h I. (ii) If the CF3 groups are eclipsed, then the axis through the Ag and perpendicular to the plane of the Ag bonds is still a C2 axis ; however, neither of the Ag bond axes is a C2 axis. There is no Uh but there are two Uv planes (the plane shown and the plane perpendicular to it and the Ag bond plane) . So the point group is I C2v I.
NC
Figure 12.4
P12.13
i,
(a) C2v. The functions x 2 , and Z2 are invariant under all operations of the group, and so z(5 z2  3r 2) transforms as Z(AI) , y(5i  3r2) as y (B2) , x(5x 2  3r2) as x(B I), and likewise for z(x 2  i ), y(x2  Z2 ), and x(z2  i ) . The function xyz transforms as BI x B2 X AI = A2 . Therefore, in group C2v, f ~ 12A I + A2 +2BI + 2B21.
i,
(b) C3v. In C3v, z transforms as AI , and hence so does z3. From the C3v character table, (x 2 xy) is a basis for E, and so (xyz, z(x 2  i)) is a basis for AI x E = E. The linear combinations y(5i  3r2) + 5y (x2  Z2) <X Y and x(5x 2  3r2) + 5x(Z2  y2) <X X are a basis for E. Likewise, the two linear combinations orthogonal to these are another basis for E. Hence, in the group C3v, f ~ IAI +3EI· (e) Td. Make the inspired guess that thef orbitals are a basis of dimension 3 + 3 + I, suggesting the decomposition T + T + A. Is the A representation A I or A2? We see from the character table that the effect of S4 discriminates between AI and A2. Under S4, x ~ y, y ~  x, Z ~  z, and so xyz ~ xyz. The character is X = I, and so xyz spans A I. Likewise, (x 3, y3, z3) ~ (y3,  x 3,  z3) and X = 0 + 0  I = I . Hence, this trio spans T 2. Finally,
resulting in X
= I, indicating T I· Therefore, in Td , f ~ 1A I + T I + T 21·
(d) Oh . Anticipate an A + T + T decomposition as in the other cubic group. Since x ,y, and z all have odd parity, all the irreducible representatives will be u. Under S4, xyz ~ xyz (as in (e)), and so the representation is X = I (see the character table). Under S4, (x 3,y3, Z3 ) ~ (y3, _ x 3,  Z3) , as before, and X =  I, indicating Tlu . In the same way, the remaining three function s span T 2u. Hence, in Oh , f ~ I A2u + Tlu + T2u I· (The shapes of the orbitals are shown in Inorganic Chemistry, 3rd edn, D. F. Shriver, and P. W. Atkins, Oxford University Press and W. H. Freeman & Co (1999).) The f orbitals will cluster into sets according to their irreducible representations. Thus (a) f ~ A I + T I + T 2 in Td symmetry, and there is one nondegenerate orbital and two sets of triply degenerate orbitals. (b) f ~ A2u + Tlu + T2u, and the pattern of splitting (but not the order of energies) is the same.
MOLECULAR SYMMETRY
P12.15
253
We begin by drawing up the fo llowing table.
E C2
av a v'
N2s
N2px
N2py
N2pz
02px
02py
02pz
O' 2px
O' 2py
O' 2pz
X
N2s N2s N2s N2s
N2px  N2px N21Jx N2px
N2py N2py N2py N2py
N2pz N2pz N2pz N2pz
0 2px O' 2px O'2px 02p"
0 2py O'2py O'2py 02py
02p: O' 2pz O' 2pz 02pz
O' 2px 02px 02px O'2p.t
O' 2py 02py 02py O'2py
O' 2p: 02pz 02p: O' 2pz
10 0
The character set (10, 0, 2, 4) decomposes into 14AI + 2BI adapted linear combinations as described in Section 12.5. l{f(A I) l{f(A I) l{f(A I) l{f(AI) l{f(BI)
= = = = =
(column (column (column (column (column
N2s N2p: 02p: + O' 2pz 02py + O' 2py N2p.t
I) 4) 7) 9) 2)
l{f( BI ) l{f(B2) l{f(B2) l{f(A2) l{f(A2)
= = = = =
+ 3 B2 + A21. We then form
02px + O' 2px N2py 02py + O' 2py 0 2p: + O'2P: 02px + O' 2IJx
2 4
sy mmetry
(column 5) (column 3) (colum n 6) (column 7) (colu mn 5)
(The other columns yield the same combinations.) P12.17
Consider phenanthrene with carbon atoms as labeled in the stucture below. a
a'
e
f'
e'
(a ) The 2p orbitals involved in the 7T system are the basis we are interested in. To find the irreproducible representat ions spanned by thi s basis, consider how each basis is transformed under the sy mmetry operations of the C2v group. To find the character of an operation in this basis, sum the coefficients of the basis terms th at are unchanged by the operation.
E C2
av a v'
a
a'
b
b'
c
a a' a'  a
a' a a a'
b  b' b'  b
b'  b b b'
c c' c' c
d
d'
e
e'
f
f'
d d' d' d
d'
e  e' e' e
e' e
f f' f' f
f'
c' c' c c c'
d d d'
e  e'
f f f'
0
'" 0
g'
X
0'
14
'"  0'
g'"
0'
g
to

'"0 to
0'
'"
0 0 1 4
To find the irreproducible representations that these orbitals span, multiply the characters in the representation of the orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 12.5(a». The table below illustrates the procedure, beginning at left with the C2v character tab le.
E AI A2 BI B2
C2
I  I
av
a vI
I
 I I I
I 1
product
The orbitals span 17 A2 + B21.
E
C2
av
a v'
sumlh
14 14 14 14
0 0 0 0
0 0 0 0
14 14 14 14
0 7 7 0
254
STUDENT'S SOLUTIONS MANUAL
To find symmetryadapted linear combinations (SALCs), follow the procedure described in Section 12.5(c). Refer to the table above that displays the transformations of the original basis orbitals. To find SALCs of a given symmetry species, take a column of the table, multiply each entry by the character of the species' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species A I are I, I, I, I, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALCs of A I symmetry. (No surprise here : the orbitals span only A2 and B I .) An A2 SALC is obtained by multiplying the characters I, I, I ,  I by the first column : I( , , ) I (  a. ') 4aaa+a=2a
The A2 combination from the second column is the same. There are seven distinct A2 combinations in all : 1!(a  a' ), !(b  b'), . . . , !(g  g') I. The BI combination from the first column is ±(a + a' + a' + a) = !(a + a'). The B 1 combination from the second column is the same. There are seven distinct B I combinations in all: 1!(a + a'), !(b + b' ), ... , !(g + g' ) I. There are no B2 combinations, as the columns sum to zero. (b) The structure is labeled to match the row and column numbers shown in the determinant. The HUckel secular determinant of phenanthrene is:
a
b
c
d
g
e
g'
f'
e'
d'
c'
b'
a'
a
0
0
0
0
0
0
0
0
fJ
b
0
f3
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0 a£ fJ 0 0 a£ fJ 0 fJ a£ 0 c 0 fJ f3 a  £ fJ d 0 0 fJ a£ 0 0 e 0 f3 0 0 0 0 f3 0 0 0 g 0 f3
f3 0 0 a£ f3 0 a£ f3 f3 a£ f3 0 f3 a £ 0 0 f3
,/
0
0
0
0
0
f'
0
0
0
0
0
e'
0
0
0
0
0
0
0
0
fJ
0
0
0
0
0
0
0
0
0
0
0
0
0
0
'"
d'
c'
0
0
0
0
b'
0
0
0
0
0
0
0
f3
0
a'
fJ
0
0
0
0
0
0
0
0
0 0 0 f3 0 0 a £ f3 0 a£ f3 f3 a  £ f3 0 f3 a£ 0 0 f3 0 0 0 f3
0 0 0 0 0
fJ a £
This determinant has the same eigenvalues as in exercise 11 .13b(b). (c) The ground state of the molecule has A l symmetry by virtue of the fact that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has A I character. If a transition is to be allowed, the transition dipole must be nonzero, which in tum can JL \IIj includes the totally symmetric species A I · only happen if the representation of the product Consider first transitions to another A I wavefunction, in which case we need the product A I JLA I .
\II;
MOLECULAR SYMMETRY
255
Now AlAI = AI, and the only character that returns AI when mUltiplied by AI is AI itself. The + A I transitions are allowed. (Note: transitions from the AI ground state to an AI excited state are transitions from an orbital occupied in the ground state to an excitedstate orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (A2 or B I) to the other; in that case, the excitedstate wavefunction will have symmetry of A 18 I = 8 2 from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A 1P.82 = P.B2, and the only species that yields A I when multiplied by B2 is B2 itself. Now the y component of the dipole operator belongs to species 8 2, so these transitions are also allowed (ypolarized).
z component of the dipole operator belongs to species A I, so zpolarized A I
Solutions to applications P12.19
The shape of this molecule is shown in Fig. 12.5. A
Q
OO
60
0
Figure 12.5
COB
(a) Symmetry elements 1E, 2C3, 3C2, ah , 253, 3av I· Point group 1 D3h I· I (b)
D (E)
=
D (C3)
(
=
0
0
1
o
0
(
0
0
I
0 1
o
5;
C~ and are counter clockwise rotations. a v is through A and perpendicular to 8C a~ is through 8 and perpendicular to AC. a~' is through C and perpendicular to AB.
D(av )
=
0 0
G D· = G ~) 1
D(a~/)
0 0
~ G0I
0
D (" ;)
~).
256
STUDENT'S SOLUTIONS MANUAL
(c) Example of elements of group multiplication table
D(C,)D(C,)
D(a; )D(a. )
~ G~
DG~ 1)
~ G~
D~
~
D(a:).
G ~)(: 0
0
~G
0 0
D~
0 0
D
D(c;).
D3h
E
C3
C2
ay
a y'
ah
E C3 C2 ay a y' ah
E C3 C2 ay a y' ah
C3 C'3 a y' a y' a y"
C2 a" y E £ C3 C2
ay a" y E £ C3 ay
a y' ay C3 C3 £ a y'
ah C3 C2 ay , ay £
C3
(d) First, determine the number of s orbitals (the basis has three s orbitals) that have unchanged positions after application of each symmetry species of the D3h point group.
Unchanged basis members
3
o
3
o
This is not one of the irreducible representations reported in the D3h character table but inspection shows that it is identical to A') + £ ' . This allows us to conclude that the three s orbitals span 1A;
+ £ ' I.
COMMENT. The multiplication table in part (c) is not strictly speaking the group multiplication ; it is instead
the multiplication table for the matrix representations of the group in the basis under consideration.
P12.21
(a) Following the flow chart in Fig. 12.4 of the text, note that the molecule is not linear (at least not in the mathematical sense); there is only one Cn axis (a C2), and there is a ah . The point group, then,
isl C2h I.
MOLECULAR SYMM ETR Y
b
d
~
a
.0
0
c
e
f
h
o
k'
oj
j'
k
g
e'
g'
j'
00.0 h'
c'
E C2
ail
a
a'
b
b'
c
c'
a a' ,  a a
a' a a a'
b b b' b
b' b b b'
c c' c' c
c' c c c'
a'
.0
.0
~
f'
d'
b'
(b) The 2pz orbitals are transformed under the symmetry operations of the
j'  j' j
C2h
group as follows .
j'
k
k'
j'
k k' k' k
k' k k k'
j' j'
257
X
22 0 0 22.
To find the irreproducible representations that these orbitals span, we multiply the characters of orbitals by the characters of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section 12.5(a». The table below illustrates the procedure, beginning at left with the C2h character table. E
C2
Ag All Bg BII
ah
I I
1 1
product
I  I
1
E
C2
22 22 22 22
0 0 0 0
ah
0 0 0 0
22 22 22 22
sum / h 0 II II 0
The orbitals span I II Au + II Bg I. To find symmetryadapted linear combinations (SALes), follow the procedure described in Section 12.5(c) . Refer to the table above that displays the transformations of the original basis orbitals. To find SALes of a given symmetry species, take a column of the table, multiply each entry by the character of the species ' irreproducible representation, sum the terms in the column, and divide by the order of the group. For example, the characters of species Au are I, I, I, I, so the columns to be summed are identical to the columns in the table above. Each column sums to zero, so we conclude that there are no SALes of Ag symmetry. (No surprise: the orbitals span only Au and Bg.) An Au SALe is obtained by multiplying the characters 1, I, 1 , I by the first column: ±(a
+ a' + a' + a)
= !(a + a').
The Au combination from the second column is the same. There are I I distinct Au combinations in
!
!
all : 1 (a + a'), (b + b') , .. . ! (k 4I ( a  a,  a,
+ a)
+ k' ) I. The Bg combination from the first column is
I ( = 2"a a ') .
The Bg combination from the second column is the same. There are I I distinct Bg combinations
!
!
in all: 1 (a  a'), (b  b'), ... !(k  k') I. There are no Bu combinations, as the columns sum to zero.
258
STUDEN T'S SOLUTIONS MANUAL
(C) The structure is labeled to match the row and column numbers shown in the determinant. The HUckel
secular determinant is: a a b
c
b
c
aE fJ 0 aE fJ fJ 0 a E fJ 0
0
0
0
0
0
k
0
0
0
k
k'
j'
I
'1
c'
b'
a'
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0
.. . aE fJ 0 0 0 0 aE 0 0 0 fJ fJ ... 0 a  E fJ 0 0 fJ 0 0 a  E fJ 0 fJ 0 0 0 aE fJ fJ .. . 0 0 0 0 a / fJ
k'
0
0
0
j'
0
0
0
'1
0
0
0
c'
0
0
0
0
0
0
0
0
0
b'
0
0
0
0
0
0
0
0
0
a'
0
0
0
0
0
0
0
0
0
I
.. .
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
a  E fJ 0 a  E fJ fJ 0 a E fJ
The energies of the fi lled orbitals are a + 1.98137,8 , a + 1.92583,8, a + 1.83442,8, a + 1.70884,8, + 1.55142,8 , a + 1.36511,8, a + 1.15336,8, a + 0.92013,8 , a + 0.66976,8 , a + 0.40691,8 , and a + 0.1 3648,8. The 7r energy is 27.30729,8 .
a
(d ) The ground state of the molecule has Ag symmetry by virtue of the fac t that its wavefunction is the product of doubly occupied orbitals, and the product of any two orbitals of the same symmetry has Ag character. If a transition is to be allowed, the transition dipole must be nonzero, which in tum can only happen if the representation of the product Ilit ILllij includes the totally symmetric species Ag. Consider first transitions to another Ag wavefunction, in which case we need the product AgILAg. Now AgAg = A g, and the only character that returns Ag when mUltiplied by Ag is Ag itself. No component of the dipole operator belongs to species A g, so no Ag __ Ag transitions are allowed. (Note: such transitions are transitions from an orbital occupied in the ground state to an excitedstate orbital of the same symmetry.) The other possibility is a transition from an orbital of one symmetry (Au or Bg) to the other; in that case, the excitedstate wavefunction will have symmetry of AuBg = Bu from the two singly occupied orbitals in the excited state. The symmetry of the transition dipole, then, is A glLBu = ILBu , and the onl y species that yields Ag when multiplied by Bu is Bu itself. The x and y components of the di pole operator belongs to species B u, so these transitions are allowed.
13
Molecular spectroscopy 1. rotational and vibrational spectra
Answers to discussion questions 013.1
(1) Doppler broadening. This contribution to the linewidth is due to the Doppler effect, which shifts the frequency of the radiation emitted or absorbed when the atoms or molecules involved are moving towards or away from the detecting device. Molecules have a wide range of speeds in all directions in a gas and the detected spectral line is the absorption or emission profile arising from all the resulting Doppler shifts. As shown in Justification 13.3, the profile reflects the distribution of molecular velocities parallel to the line of sight which is a bellshaped Gaussian curve.
(2) Lifetime broadening. The Doppler broadening is significant in gas phase samples, but lifetime broadening occurs in all states of matter. This kind of broadening is a quantum mechanical effect related to the uncertainty principle in the form of eqn 13.18 and is due to the finite lifetimes of the states involved in the transition. When r is finite, the energy of the states is smeared out and hence the transition frequency is broadened as shown in eqn 13.19.
(3) Pressure broadening or collisional broadening. The actual mechanism affecting the lifetime of energy states depends on various processes, one of which is collisional deactivation and another of which is spontaneous emission. Lowering the pressure can reduce the first of these contributions; the second cannot be changed and results in a natural linewidth. 013.3
(1) Rotational Raman spectroscopy. The gross selection rule is that the molecule must be anisotropically polarizable, which is to say that its polarizability, a, depends upon the direction of the electric field relative to the molecule. Nonspherical rotors satisfy this condition. Therefore, linear and symmetric rotors are rotationally Raman active.
(2) Vibrational Raman spectroscopy. The gross selection rule is that the polarizability of the molecule must change as the molecule vibrates. All diatomic molecules satisfy this condition as the molecules swell and contract during a vibration, the control of the nuclei over the electrons varies, and the molecular polarizability changes. Hence both homonuclear and heteronuclear diatomics are vibrationally Raman active. In polyatomic molecules it is usually quite difficult to judge by inspection whether or not the molecule is anisotropically polarizable; hence group theoretical methods are relied on for judging the Raman activity of the various normal modes of vibration. The procedure is discussed in Section 13. 17(b) and demonstrated in Illustration 13.6. 013.5
The exclusion rule applies to the benzene molecule because it has a center of symmetry. Consequently, none of the normal modes of vibration of benzene can be both infrared and Raman active. If we wish to characterize all the normal modes we must obtain both kinds of spectra. See the solutions to Exercises 13.2S(a) and 13.2S(b) for specific illustrations of which modes are IR active and which are Raman active.
260
STUDENT'S SOLUTIONS MANUAL
Solutions to exercises E13.1 (b)
The ratio of coefficients Al B is (a)
A
8JThv3
B
c3
x 1O J s) x (500 x 10 SI) 3 I 8JT(6.626 ',..:,,',;';;;'= 7.73 8 34
6
(2.998 x 10 m s I ) 3 .
10 32 J m3 s
X
(b) The frequency is 34 c A 8JT h 8JT (6.626 X 10 J s) 1 v =  so  =  3 = = 6.2 x 10 2 A B A (3 .0 x 10 m) 3
E13.2(b)
28
Jm
 3
s
1
A source approaching an observer appears to be emitting light of frequency Vapproaching
v
=  s
.
[13.15 , Section 13.3]
1 c 1 Since vex , AObs = (1  sic) A A
For the light to appear green the speed would have to be
s=
(I 
AObS) c = (2.998 x 108 m s I) x (1  520 nm) = 16.36 x 107 m A 660 nm
SI
or about 1.4 x 108 m.p.h. (Since s
~
Vb o s 
c, the relativistic expression 1 + (5 I C»)1 /2 ( 1 _ (s ic) v
should really be used. It gives s E13.3(b)
= 7.02 X
107 m
1 5 .)
The linewidth is related to the lifetime r by
oiJ =
5.31 cm
I
[13.19] so r
r i ps
5.31 cm
ov
=
I
ps
(a) We are given a frequency rather than a wavenumber
(5 .31 cm
iJ = vic so r = or 11.59 ns
E13.4(b)
=
x (2.998 x 10 100 x 106 S I
)
I I
(b) r
I
5.31 cm2.14 cm I ps
=
I
2.48 ps
1
The linewidth is related to the lifetime r by
oiJ =
5.31 cm I r i ps
(5 .31 em I)c so 0 v = ','r i ps
10
em
S I)
s P
= 1.59 X
103 s P
1
I .
261
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
(a) If every collision is effective, then the lifetime is 1/( 1.0 x 109 S I) 
8v
10 I (5.31 cm ) x (2.998 x 10 cm sI) 3 1.0 x 10
=
= 1.6 x
i
08  I s
= 1.0 x
109 s
= 1.0 x
103 pS
= I 160 MHz I
(b) If only one collision in 10 is effective, then the lifetime is a factor of 10 greater, 1.0 x 104 ps _
8v E13.5(b)
=
(5.31 cm I) x (2.998 x 10 10 cm SI) 1.0 x 104
= 1. 6 x
0 I i 7s
= I16 MHz I
The frequency of the transition is related to the rotational constant by hv
= f+..E = hef+..F = heB[J(J + I) 
where J refers to the upper state (J
(J  I)J]
= 2heBJ
= 3). The rotational constant is related to molecular structure by
Ii Ii  4ncl  4n emeffR2
B_=_ where I is moment of inertia, meff is effective mass, and R is the bond length. Putting these expressions together yields v
liJ
= 2eBJ = 2nmeffR 2
The reciprocal of the effective mass is
= m I + m I =
m I
C
eff
0
(12 U)I + (l5.9949u)  1 = 8.78348 x 1025 k  I 1.66054 x lO 27 kgu 1 g
_I
So v _ (8.78348 x 1025 kgI) x (1.0546 x 10 34 J s) x (3) 2n (1128 . I x 10  12 m) 2 E13.6(b)
_I
11
3.4754 x 10
s
1
(a) The wavenumber of the transition is related to the rotational constant by
hev
= f+..E = hef+..F = hcB[J(J + 1) 
where J refers to the upper state (J
(J  I)J]
= 2heBJ
= 1). The rotational constant is related to molecular structure by
Ii B=4ncl
where I is moment of inertia. Putting these expressions together yields v
I
= 2BJ = ~
so 1= liJ
2ncl
ev
= 13.307 x
=
34 (1.0546 x 10 J s) x (I) 2n(2.998 x 1010 cm s I) x (16.93 cm I )
1047 kg m2 1
(b) The moment of inertia is related to the bond length by I
= meffR 2 so R = (  I
meff
) 1/2
262
STUDENT'S SO LUTIONS MANUAL
m
I
eff
I
and R
x 1026 kgI) x (3.307
= {(6.0494 =
E13.7(b)
(1.0078 U)  I + (80.9163 U)  I 1.66054 x 1027 kg u I
 I
= mH + mBf =
1.414 x 1O
lO
= 6.0494 x
10
26
kg
I
10 47 kg m 2)} 1/ 2
X
= 1141.4 pm 1
m
The wavenumber of the transition is related to the rotational constant by
= /'<"E = hc6.F = hcB[J(J + 1) 
hci)
(J  I )J]
= 2hcBJ
where J refers to the upper state. So wavenumbers of adjacent transitions (transitions whose upper states differ by 1) differ by
_ /'<" v
n
n
2rrc/
2rr c /'<"i!
= 2B =   so I =   
where I is moment of inertia, m eff is effective mass, and R is the bond length. _ So 1
34 (1.0546 X 10 J s)  15420 1046 k 21 . x gm 2rr(2.9979 x 1010 cm s I) x (1.033 cm I)
The moment of inertia is related to the bond length by
1= meffR2 soR
=(
I ) 1/2
meff
m I  m I eff 
and R
(18.9984 u)  I + (34.9688 u)  I 1.66054 x 10 27 kg u I

CI 
= {(4.89196 x =
E13.8(b)
+ mI
F
= 4.89196 X
1025 kgI
1025 kg  I) x (5.420 x 10 46 kg m 2)}1 / 2
1.628 x 10 10 m
= 1162.8 pm I
The rotational constant is
B_ _
n_ _
n
so R
n ) 1/2
=(
 4rrc/  4rr c(2moR2)
8rrcmoB
where I is moment of inertia, meff is effective mass, and R is the bond length.
R_ ( 
(1.0546 8rr(2.9979 x
= 1.1621
x 10
10
IOlO
m
X
10 34 J s)
cm sI) x ( 15.9949 u) x (1.66054 x 10 27 kg u 1)(0.39021)
= 1116.21
pm
I
) 1/2
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
E13.9(b)
263
This exercise is analogous to Exercise 13.9(a), but here our solution will employ a slightly different algebraic technique. Let R = Roc, R' = Re s , 0 = 16 0 , C =1 2 C. Ii 1=   [Comment 13.4] 4JrB
1.05457 x 10 34 J s = 1.3799 x 10 45 kg m2 = 8.3101 x 10 19 u m2 (4Jr ) x (6.0815 x 109 s I) 1.05457 x 10 34 J s I (OC 34 S) = = 1.4145 X 10 45 kg m2 = 8.5184 x 10 19 u m2 (4Jr ) x (5.9328 x 109 S I)
I (OC 32 S) =
The expression for the moment of inertia given in Table 13.1 may be rearranged as follows.
? ? 2 , 2 '2 = mAmR? + me mR, 2  m t.. R + mAme RR  me R
= mA (mB + mc) R2 + 111c(I11 A + I11B)R'2 + 2I11Al11e RR' Let me = 11132s and 111~ = 11134s 1m

me
mA
= 
me
(mB
I'm'
mA
me
me
 , =,
+ mc) R 2 + (m A + mB )R t? + 2mARR,
(mB
,2
,
+ me )R + (m A + mB )R 2 + 2m ARR
(a) ,
(b)
Subtracting
1m  I'm', = me
me
[(lilA  ) l11e
(mB +mc)  (m  ,A ) (mB +m~) ] R2
me
Solving for R2 m~ /m  me I'm' I11BI11 A
Substituting the masses, with 111
I1l A
(m~  mc)
= mo, mB = me, me = m32s' and m~ = m34s
+ 12.0000 + 31.9721 ) u = 59.9670 u (15.9949 + 12.0000 + 33.9679) u = 61.9628 u
= ( 15 .9949
m' =
R2 =
(33.9679 u) x (8.3 101 x 10 19 u m2 ) x (59.9670 u) ( 12.0000 u) x ( 15.9949 u) x (33.9679 u  3 1.9721 u) (33.9721 u) x (8.51 84 X 10 19 u m2) x (61.9628 u) ( 12.0000 u) x ( 15 .9949 u) x (33.9679 u  31.9721 u) 20 2 51.6446 x 10 19 m2) _ 383.071  1.3482 x 10 m
R = 1.1611
X
10 10 m = 1116.1 pm 1= Roe
264
STUDENT'S SOLUTIONS MANUAL
Because the numerator of the expression for R2 involves the difference between two rather large numbers of nearly the same magnitude, the number of significant figures in the answer for R is certainly no greater than 4. Having solved for R, either equation (a) or (b) above can be solved for R'. The result is R'
=
1.559
X
10 10 m
= 1155.9 pm 1= Res
E13.10(b) The wavenumber of a Stokes line in rotational Raman is VStokes
= Vi 
2B(21
+ 3) [13.42a)
where J is the initial (lower) rotational state. So VStokes
= 20623 em  I 
2(1.4457 emI) x [2(2)
E13.11(b) The separation oflines is 4B, so B
Then we use R withmeff
=
=(
Ii
=
+ 3) = 120603 emII
~ x (3.5312 emI)
= 0.88280 cm I
) 1/2 [Exercise 13.8(a»)
4Jl"meff CB
1m(19F)
=
1 x (I8.9984u) x (1.6605 x 1O 27 kgu 
l
)
=
1.577342 x 10 26 kg
1.0546 x 10 34 J S ( R = 4Jl"(1.577342 x 10 26 kg) x (2.998 x IO I0 cm s l ) x (0.88280cm
)1 / 2 l)
= 1.41785 x 10 10 m = 1141.78 pm 1 E13.12(b) Polar molecules show a pure rotational absorption spectrum . Therefore, select the polar molecules based
on their wellknown structures. Alternatively, determine the point groups of the molecules and use the rule that only molecules belonging to CIl , CIlV , and Cs may be polar, and in the case of c" and CIlV, that dipole must lie along the rotation axis. Hence all are polar molecules. Their point group symmetries are (a) H20 , C2v , (b) H202 , Cz , (c) NH3, C3v, (d) NzO, Coov
~ show a pure rotational spectrum. E13.13(b) A molecule must be anisotropically polarizable to show a rotational Raman spectrum; all molecules
except spherical rotors have this property. So 1CHzCI21, 1CH3CH3 Raman spectra; SF6 cannot. E13.14(b) The angular frequency is
W
k)
=;;; (
1/ 2
= 2Jl"V
k=10.7INm 1 1
E13.15(b)
W =
, ) 1/2 (~)1 /2 ~,./ = (_k meff
meff
[prime = 2H 37 CI)
~ and 1N20 1can display rotational
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
265
The force constant, k, is assumed to be the same for both molecules. The fractional difference is
(~) ( ~) meff meff ~)1 /2 ( meff 1/ 2 _
w' w w
Wi 
W
(m~ff) 1/ 2 _
=
w
1= {
men
1/ 2
(+) (_1 ) (_1_) I~eff 1/ 2 _
meff
1/ 2
=
(mm~: )' /2 _ I
meff
mHmCI mH + mC!
(1.0078 u) x (34.9688 u) (2.0 140 u) + (36.9651 u) } 1/ 2 '''.,,,,...:... x  I  { ( 1.0078 u) + (34.9688 u) (2.0140 u) x (36.965 1 u) = 0.284 Thus the difference is 128.4 percent 1 E13.16(b) The fundamental vibrational frequency is
k ) 1/ 2 w = (= 2n v = 2n cii meff We need the effective mass
m;;nl = mil + mi l = (78.9183 u)I + (80.9 163 u) I = 0.0250298 u I [2n(2.998 x 1010 em sI) x (323 .2 cm I)]2 x ( 1.66054 x 10 27 kg u I) k =0.0250298u71~=1245.9Nm
1
1
E13.17(b) The ratio of the population of the ground state (No) to the first excited state (NI ) is No  = exp NI
(a)
(hV)   = exp (hCii) 
No
 = exp
kT
kT
( (6.626 x
NI
(b )
No
=exp NI
( (6.626
34
=~
J s) x (2.998 x IO IOcms l ) x (32 1 em I») (1.38 1 x 10 23 J K I) x (800K)
=~
X 10
34
E13.18(b) The relation between vibrational frequency and wavenumber is
w =
(~) 1/ 2 =
2nv = 2n cii
meff
so
ii = _1_
2n C
(~. ) 1/ 2 m eff
(km;;~) 1/ 2
2nc
The reduced masses of the hydrogen halides are very similar, but not identical I meff
= mn I + /Jl x I
~
J s) x (2.998 X 1010 em s I) x (321 em I») (1.38 1 x 1O 23 JK I) x (298 K)
10
~
266
STUDENT'S SOLUTIONS MANUAL
We assume that the force constants as calculated in Exercise 13.18(a) are identical for the deuterium halide and the hydrogen halide. ForDF _I m eff
=
v
=
(2.0140 u)I + (18 .9984 u)I 1.66054 x 10 27 kg u I
= 3.3071
x 10
26
kg
I
{(3.3071 x 1026 kg l ) x (967.04kgs 2)}1 /2 _I  II 27r(2.9979 x 10Iocms l)  . 3002.3 cm .
For Del _I l1leff
=
(2.0140 u)  I + (34.9688 u)  I 1.66054 x 10 27 kg u I
= 3.1624 x
10
26
kg
I
{(3.1624 x 10 26 kgI) x (515.59kgs 2)}1 /2 _I _I I v= 27r(2.9979 x lO lo cms l) .2143 .7cm. ForDBr _I l1leff
=
jj
(2.0140 u)  I + (80.9163 u)  I = 3.0646 x 1026 kg I 1.66054 x 10 27 kg u I
=
{(3.0646 x 1026 kg l ) x (411.75kgs2)}1 /2 27r(2.9979 x IO lo cms l )
=
I
1885.8 cm
_ I I
ForD! m
_I eff
v =
=
(2.0140 U)  I + (126.9045 U)I 1.66054 X 10 27 kg u I
= 3.0376 x
1026 kg I
{(3.0376 x 1026 kg I) x (314.21 kgs2)}1 /2 I _ I = . 1640.1 cm I. 27r(2.9979 x 10Iocms l )
E13.19(b) Data on three transitions are provided. Only two are necessary to obtain the value of
datum can then be used to check the accuracy of the calculated values. I1G(v
=
I1G(v
=2
I + 0) + 0)
= v  2vxe = 2345.15 cm I [13 .57] = 2v  6vxe = 4661.40 cm I [13 .58]
Multiply the first equation by 3, then subtract the second.
v = (3)
x (2345 .15 cm I)  (4661.40 cm I)
= 12374.05 cm I 1
Then from the first equation
Xe
=
v
2345.15 cm  I
2v
(2374.05  2345.15)cm(2) x (2374.05 cm I)
1
= 16.087 .
x 103 1
v and Xe . The third
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
Xe
267
data are usually reported as Xe i! which is
=
xei!
14.45 cm
I
= 3 + 0) = 3i! 
tlG(v
= (3) x (2374.05 emI) = 6948.74 cm I
12vxe
( 12) x (14.45 emI)
which is close to the experimental value. E13.20(b)
tlG v+ I/2
= i! 
2(v + l )xe i! [13.57)
where tlG v+I/2
= G(v + I)
 G (v)
Therefore, since
a plot of tlG v + I/2 agai nst v should give a straight line which gives ( I  2xe )i! from the intercept at v = 0 and  2xe i! from the slope. We draw up the following table
o
v G(v) / cm
I
tlG v+ 1/ 2/ cm 
1
2
3
4
1144.83
3374.90
5525.51
7596.66
2230.07
2150.61
2071.15
1991.69
9588.35
The points are plotted in Figure 13.1.
2200
., E
u ....... ..... '"
2 100
+
c.5
2000
0
3
2
v
Figure 13.1
The intercept lies at 2230.51 and the slope Since i!  2xe i!
= 2230.51
= 76.65 emI; hence xei! = 39.83 emI.
em I it follows that i!
= 2310.16 emI
The dissociation energy may be obtained by assuming that a Morse potential describes the molecule and that the constant De in the expression for the potential is an adequate first approximation for it. Then i!
De
= 4xe
i!2
[13 .55)
= 4xe i! =
(23 10.16 cm I )2 (4) X (39.83 em I)
= 33 .50 x
3
10 emI
= 4.15 eV
268
STUDENT'S SOLUTIONS MANUAL
However, the depth of the potential well De differs from Do, the dissociation energy of the bond, by the zeropoint energy; hence _ 3 cm _ 1) (1) x (231O.16cm l ) Do=De  1 v=(33.50x10
2
= 13.235 x
104 cm I 1= 14.01 eV 1
E13.21(b) The wavenumber of an Rbranch IR transition is
iiR = v + 2B(J + 1) [13.62c] where J is the initial (lower) rotational state. So
VR
= 2308 .09 cm I + 2(6.511 cm I )
x (2
+ 1) = 12347.16 cm I
1
E13.22(b) See Section 13.10. Select those molecules in which a vibration gives rise to a change in dipole moment.
It is helpful to write down the structural formulas of the compounds. The infrared active compounds are
COMMENT. A more powerful method for determining infrared activity based on symmetry considerations is
described in Section 13.15.
E13.23(b) A nonlinear molecule has 3N  6 normal modes of vibration, where N is the number of atoms in the
molecule; a linear molecule has 3N  5. (a) C6H6 has 3(12)  6
= ~ normal modes.
= 1421 normal modes. HC=CCCH is linear; it has 3(6)  5 = [II] normal modes.
(b) C6H6CH3 has 3(16)  6 (c)
E13.24(b) (a) A planar AB3 molecule belongs to the D3h group. Its four atoms have a total of 12 displacements,
of which 6 are vibrations. We determine the symmetry species of the vibrations by first determining the characters of the reducible representation of the molecule formed from all 12 displacements and then subtracting from these characters the characters corresponding to translation and rotation. This latter information is directly available in the character table for the group D3h. The resulting set of characters are the characters of the reducible representation of the vibrations. This representation can be reduced to the symmetry species of the vibrations by inspection or by use of the little orthogonality theorem.
D3h
E
ah
X (translation) Unmoved atoms X (total, product) X (rotation) X (vibration)
3 4 12 3 6
1 4 4 1 4
2C3
2S3
3C~
3av
0 I 0 0 0
2 1 2 2 2
1 2 2 1 0
I 2 2 1 2
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
269
+ A~ + 2£' .
X (vibration) corresponds to A'I
Again referring to the character table of D3h, we see that E' corresponds to x and y, A~ to z; hence 1A ~ and £ ' are IR active I. We also see from the character table that £' and A; correspond to the
quadratic terms; hence 1A; and E' are Raman active I· (b) A trigonal pyramidal AB 3 molecule belongs to the group C3v. In a manner similar to the analysis
in part (a) we obtain
X (total) X (vi bration)
0 2
12 6
2 2
X (vibration) corresponds to 2AI
+E
active and that 1 AI the Raman spectra.
1
+ 2E . We see from
the character table that 1 AI and E 1 are IR
are also Raman active. Thus all modes are observable in both the IR and
E13.25(b) (b) The boatlike bending of a benzene ring clearly changes the dipole moment of the ring, for the
moving of the CH bonds out of the plane will give rise to a noncancelling component of their dipole moments . So the vibration is 1IR active I. (a) Since benzene has a centre of inversion, the exclusion rule applies: a mode which is IR active (such as this one) must be 1Raman inactive I.
+ Alu + A2g + 2E lu + Eig . The rotations Rx and Ry span El g , and the + Alu · So the vibrations span 1 A Ig + A2g + £Iu 1
E13.26(b) The displacements span A]g
translations span E]u
Solutions to problems Solutions to numerical problems P13.1
Use the energy density expression in terms of wavelengths (eqn 8.5) [ = pd)"
where p
= ),. 5 (e I
8nhc
lC
j AkT
I)

.
Evaluate 9
8nhc
700 x 1O m
[=
1
d)"
400 x 10 9 m
I)
),.5(ehcjAkT 
at three different temperatures. Compare those results to the classical, RayleighJeans expression (eqn 8.3). [class
so [class
= Pclass d)"
=
700 x 10
1
400 x 10 9
where 9
m
OJ
8nkT
Pclass
8nkT d)" 4 ),.
= ~' 8nkT 1700 x
=  3 ),.3
10 9 m
400 x 10 9 m
.
270
STUDENT'S SOLUTIONS MANUAL
EI 1 m 3
TIK
Eclassl1 m 3
2.136 x 10 6 9.884 x 10 4 3.151 x 10 1
(a) 1500 (b) 2500 (c) 5800
2.206 3.676 8.528
The classical values are very different from the accurate Planck values! Try integrating the expressions over 400  700 ~m or mm to see that the expressions agree reasonably well at longer wavelengths. P13.3
On the assumption that every collision deactivates the molecule we may write T
= ~ = !!... (JTm)I /2 z
4ap
For Hel, with m T ~
~
kT
36 u,
) (1.381 x 1O 23 1K I ) x (298K) ( (4) x (0.30 x 10 18 m 2 ) x (1.013 x 105 Pa)
x ( JT x (36) x (1.66 1 x 10 27 kg) ) 1/2 (1.381 x 1O 23 1K I ) x (298K) "'" 2.3 x 10 10 s. 8£ "'" h8\!
fi
= T
[13.18] .
The width of the collisionbroadened line is therefore approximately
O\!
~
1 hT

=
I (h) x (2.3 x 10 10 s)
~
I700MHz I.
The Doppler width is approximately 1.3 MHz (Problem 13.2). Since the collision width is proportional 1.3 to p [o\! ex I I T and T ex l i p], the pressure must be reduced by a factor of about 700 = 0.002 before Doppler broadening begins to dominate collision broadening. Hence, the pressure must be reduced to below (0.002) x (760 Torr) =
P13.5
B=

fi
4JTcl
l1Ieff =
II Torr I.
R2 =
[13.24] ;
memo = (12.0000U) x (I5 .9949U») x ( 1.66054 x 1O 27 kgu l ) me + mo (I2.0000u) + (15 .9949u)
= 1.13852 x  fi = 2.79932 X 4JTC
__fi__ 4JTCmeff B
10 44 kg m.
10 26 kg.
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
x 10 10 m
Ro
= 1.1283
R~
=
RI
= 1.2352 X
271
= 1112.83 pm I. 44
2.79932 X 10 kg m = 1.52565 x 10 20 m2, ( 1.13852 x 10 26 kg) x (1.6 116 x 102 m I ) 10
10
m
= 1123.52 pm I.
The change in internuclear distance is roughly 10 per cent, indicating that the rotations and
COMMENT,
vibrations of molecules are strongly coupled and that it is an oversimplification to consider them independently of each other.
P13.7
The separations between neighboring lines are 20.81,20.60, 20.64, 20.52,20.34,20.37,20.26 Henee B =
(!) x (20.51 em  I) =
Ii 1=4rrcB
I
R= (meff
=
34 1.05457 x 10 J s (4rr) x (2 ,99793 x 10 10 ems  1 x (I0.26em 
) 1/2
[Table 13. 1]
with
meff
= 1.6266 x
_ (2.728 x 10 47 kgm2) 1/ 2 _ 129 5  1 . pm 1.6266 x 10 27 kg COMMENT,
mean: 20.51 emI.
1O.26em 1 and
l)
= I2.728 x
10
47
21 kgm.
10 27 kg [Exercise 13.6(a)]
I·
Ascribing the variation of the separations to centrifugal distortion, and not just taking a simple
average wou ld result in a more accurate value, Alternatively, the effect of centrifugal distortion could be minimized by plotting the observed separations against J , fitting them to a smooth curve, and extrapolating that curve to J = 0, Since B ex at a factor
~ I
and I ex melt, B ex _1_, Hence, the corresponding lines in 2H35 C1 will lie melt
to low frequency of 1 H 35Cllines. Hence, we expect lines at 110.56, 21,1 1, 31.67 " .. cm 1
P13.9
R
(_li_) 4rrjl.cB
1/2
and
v
= 2cB(J + 1)
(63.55) x (79.9 1) We use jl.(CuBr) ~ (63.55) + (79.91) u
[13.37, with v
= 35.40 u
and draw up the following table. J
13
14
15
v/ MHz
84421.34
90449.25
96476,72
0.10057
0.10057
0.10057
= cii].
I.
272
STUDENT'S SOLUTIONS MANUAL
= (_:_::_
H ence, R
_
_
1.05457 x 10 34==_:_::___:__) Js ~~~...:..::...1/ 2 l l
(4rr) x (35 .40) x (1.6605 x 10 27 kg) x (2.9979 x 10Iocm s
) x
(0.10057cm
)
= 1218 pm I· P13.11
Plot frequency against J as in Fig. 13.2. 26~
r~~~~~~~~~~~~~~
= 8300.2 + 8603.2x. R2 = 1.000
... J y

~
.
... .. .. _J. ....... . ... ..... ..... ! ..•..
25~
~
~
~
..:.....:..... :..... .:.
N
:r:
~ 240000 ..... :......:..... : .. ... :.
.
u
" & e u.. 0)
.
23~
22~
23
24
26
25
27
28
29
30
J
Figure 13.2
The rotational constant is related to the wavenumbers of observed transitions by _ V
(
v
= 2B J + I) = e
so v
= 2Bc(J + I) .
A plot of v versus J , then, has a slope of 2Be. From Fig.13 .3, the slope is 8603 MHz, so
The most highly populated energy level is roughly 1
kT ) 1/ 2
J max ( 2heB so J max
_ _
and J max
P13.13

2"
IP (1.38 1 x 1O 23 JK  I ) x (298K) ( (6.626 x 10 34 J s) x (8603 x 106 s I) )
=
(1.38 1 x 10 23 J K I ) x (lOOK)
( (6.626 x 10 34 J s) x (8603 x 106 s I) )

IP
1 _ 6 298K 2 at 2
~

1
 2 = [ill 15 at lOOK.
The Lewis structure is
[9=N=9rl: VSEPR indicates that the ion is \ linear \ and has a center of symmetry. The activity of the modes is consistent with the rule of mutual exclusion; none is both infrared and Raman active. These transitions
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
273
may be compared to those for C02 (Fig. 13.40 of the text) and are consistent with them. The Raman active mode at 1400 cm I is due to a symmetric stretch (vd, that at 2360 cm I to the antisymmetric stretch (V3) , and that at 540 cm I to the two perpendicular bending modes (V2). There is a combination band, VI + V3 = 3760 cm I ::::;, 3735 cm I , which shows a weak intensity in the infrared. P13.15
Do
(a)
= De
= 1v 
 v' with v'
IHCI :
v' =
{(1494.9) 
~xev [Section 13.11].
0) x (52.05)}cm 1 = 1481.8cm
Hence, Do = 5.33  0.18 = 15.15eV (b)
2
HCI :
2meffWXe
= a [13.55] , so vXe ex 
I
meff
13.20(a)], so v ex 2
I
, implying v
,
or 0.184eV.
I.
2
n
l
v
2
as a is a constant. We also have De = _ [Exercise 4xev
I ex ~. Reduced masses were calculated in Exercises 13.18(a)
meff
meff
and 13.18(b), and we can write m ( IHCI») 1/2 eff 2 ( meff ( HCI)
=
veHCI)
= (meffc'2HCI»)
_2
XeV( HCI)
v' e HCI)
meff( HCI)
=
(1) x (2 144.2) 
x v(IHCI) = (0.7172) x (2989.7cm') = 2144.2cm' ,
X Xe v(
(~)
IHCl)
=
(0.5144) x (52.05cm')
x (26.77cm')
Hence, Do e HCI) = (5.33  0.132) eV = 15.20 eV P13.17
=
= 26.77cm l •
1065.4cm' , 0.132eV.
I.
(a) In the harmonic approximation D e =DO+1V so v=2(DeDo).
_ v=
2( 1.51 x 10 23 J  2 x 10 26 J) (6.626 x
1O 34
Js) x (2.998 x
108
=
,,,, 152m I
ms I) '
The force constant is related to the vibrational frequency by
W
= (
k ) 1/2
meff
= 2n:v = 2n:cv
The effective mass is
k
= [2n: (2.998 = 12.72
x 10 8 ms
l
)
x (l52m 
l
)f
x (3 .32 x 10 27 kg)
x 10 4 kg s 2 1.
The moment of inertia is I
= meffR~ =
(3.32 x 10 27 kg) x (297
X
10 12 m) 2 = 12.93 x 10 46 kg m2 1.
274
STUDENT'S SOLUTIONS MANUAL
The rotational constant is
(b) In the Morse potential V
Xe=
4De
De = Do
and
+~
2
(I 
~xe) ii = 2
Do
+~
2
~) ii. 8~
(I 
This rearranges to a quadratic equation in ii I
I ii 2 16D  2" ii e
+ De
 Do = 0 so
4(1.51
V
=
4(De  Do) 16De
2
~
2(l6De )1
1O 23 J)
X
(
~~1(6.626 X 10 34 J s) X (2.998 X 10 8 m s I)
10 26 J 1.51 X 10 23 J 2
X
= 1293m 1 I. and P13.19
x =
(293m I)
X
(6.626
X
1O 34 Js)
4( 1.51
e
X
X 10 23
(2.998 J)
X
10 8 ms
l
)
~ = 0.96.
(a) Follow the flow chart in Fig. 12.7 of the text. CH3Cl is not linear, it has a C3 axis (only one), it does not have C2 axes perpendicular to C3 , it has no ah, but does have 3 a v planes; so it belongs to 1C3v I. (b) The number of normal modes of a nonlinear molecule is 3N  6, where N is the number of atoms. So CH3CI has nine normal modes.
I
I
(c) To determine the symmetry of the normal modes, consider how the Cartesian axes of each atom are transformed under the symmetry operations of the C3v group; the 15 Cartesian displacements constitute the basis here. AllIS Cartesian axes are left unchanged under the identity, so the character of this operation is 15. Under a C3 operation, the H atoms are taken into each other, so they do not contribute to the character of C3. The z axes of the C and CI atoms, are unchanged, so they contribute 2 to the character of C3; for these two atoms x+
x 2
3 1/ 2 Y 2
+
so there is a contribution of 1 / 2 to the character from each of these coordinates in each of these atoms. In total , then X = 0 for C3 . To find the character of a v, call one of the a v planes the yz plane; it contains C, CI, and one H atom. The y and z coordinates of these three atoms are unchanged, but the x coordinates are taken into their negatives, contributing 6  3 = 3 to the character for this operation ; the other two atoms are interchanged, so they contribute nothing to the character. To find the irreproducible representations that this basis spans, we multiply its characters by the characters
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
275
of the irreproducible representations, sum those products, and divide the sum by the order h of the group (as in Section l3 .S(a». The table below illustrates the procedure.
basis AI A2 E
E
2C2
3av
15 I I 2
0
3 I
 I
0
basis x AI basis x A2 basis x E
E
2C2
3av
sum/ h
15 15 30
0 0 0
3 3 0
3 2 5
Of these 15 modes of motion, 3 are translations (an A I and an E) and 3 rotations (an A2 and an E) ; we subtract these to leave the vibrations, which span 12AI +A2+3E (two AI modes, one A2 mode, and 3 doubly degenerate E modes).
I
(d) Any mode whose symmetry species is the same as that of Iall but the A2 mode are infrared active I.
x, y, or z is infrared active. Thus
(e) Only modes whose symmetry species is the same as a quadratic form may be Raman active. Thus 1all but the A 2 mode may be Raman active I.
Solutions to theoretical problems P13.21
The center of mass of a diatomic molecule lies at a distance x from atom A and is such that the masses on either side of it balance
and hence it is at mB x = R m
m = m A +mB.
The moment of inertia of the molecule is
P13.23
Refer to the flow chart in Fig. 12.7 in the text. Yes at the first question (linear?) leads to linear point groups and therefore linear rotors. If the molecule is not linear, then yes at the next question (two or more CII with n > 2?) leads to cubic and icosahedral groups and therefore spherical rotors. Lf the molecule is not a spherical rotor, yes at the next question leads to symmetric rotors if the highest CII has n > 2; if not, the molecule is an asymmetric rotor. (3) CH4 : not linear, but more than two CII (/1 > 2), so 1spherical rotor
(b) CH3CN: not linear, C3 (only one of them), so 1symmetric rotor I. (c) C02: linear, so I linear rotor I.
I.
276
STUDENT'S SOLUTIONS MANUAL
(d) CH30H: not linear, no CIl, so I asymmetric rotor I. (e) Benzene: not linear, C6 , but only one highorder axis, so I symmetric rotor
I.
(0 Pyridine: not linear, C2 , is highest rotational axi s, so I asymmetric rotor I. P13.25
S(v, J)
= (v +
6.S?
6.SJ
Dv
+ BJ(J + I ) [13 .61].
= v  2B(21 = v + 2B(21 +
= I , M = 2] . 3) [6.v = I , M = +2]. I ) [6.v
The transition of maximum intensity corresponds, approximately, to the transition with the most probable value of J, which was calculated in Problem 13.24,
kT )1 /2
Jrnax
I
= ( 2hcB
2
The peaktopeak separation is then 6.S
= 6.SLx  6.S~,ax = 2B(2Jrnax +3) 
(2B(21rnax  I)}
= 8B (Jrnax +!)
= 8B (~) 1/ 2 = (32BkT) 1/ 2 2hcB
hc
To analyze the data we rearrange the relation to
hC(6.S)2 B=32kT Ii and convert to a bond length using B =   , with I = 2mxR2 (Table 13.1) for a linear rotor. This gives 4J[c/ Ii
)1 / 2
R = ( 8J[cmx B
(
=
I ) (2kT)I /2 J[c6.S x ;;;;
We can now draw up the following table
T/K
mx / u 6.S/cm 1 R/pm
HgCl 2
HgBr2
HgI2
555 35.45 23 .8 227.6
565 79.1 15.2 240.7
565 126.90 11.4 253.4
Hence, the three bond lengths are approximately 1230, 240, and 250 pm
I.
Solutions to applications P13.27
ctroscopy for studying the 00 ,.'::..:..:..:.:::..::..:.cc=.....:..L.C..,,! stretching mode because such a mode would be infrared inactive or at best only weakly active.
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
277
(The mode is sure to be inactive in free 02, because it would not change the molecule's dipole moment. In a complex in which 02 is bound, the 00 stretch may change the dipole moment, but it is not certain to do so at all, let alone strongly enough to provide a good signal .) (b) The vibrational wavenumber is proportional to the frequency, and it depends on the effective mass as follows,
_ (k) 1/ 2 ,
vex
so
meff
and JJe802) = (0.943)(844cm l ) = I 796cm  1 I. Note the assumption that the effective masses are proportional to the isotopic masses. This assumption is valid in the free molecule, where the effective mass of 02 is equal to half the mass of the 0 atom; it is also valid if the 0 2 is strongly bound at one end, such that one atom is free and the other is essentially fixed to a very massive unit. (c) The vibrational wavenumber is proportional to the square root of the force constant. The force constant is itself a measure of the strength of the bond (technically of its stiffness, which correlates with strength), which in tum is characterized by bond order. Simple molecular orbital analysis of 0 2, O2, and O~ results in bond orders ofl2, 1.5, and I respectively I. Given decreasing bond order, one would expect decreasing vibrational wavenumbers (and vice versa). (d) The wavenumber of the 00 stretch is very similar to that of the peroxide anion, suggesting
IFe~+o~ I·
(e) The detection of two bands due to 16 0 18 0 implies that the two 0 atoms occupy nonequivalent positions in the complex. Structures 7 and 8 are consistent with this observation, but structures 5 and 6 are not. P13.29
According to Problem 1O.27(a), the Doppler effect obeys I  SIC) 1/ 2
Vrecedin g
= vf where f = (   I + sic
This can be rearranged to yield 1 f2 s=  2c, I +f
We are given wavelength data, so we use
f=
VSlar V
=~ . Aslar
The ratio is: 654.2 nm = 09260 706.5 nm . ,
f
=
S
= I
2
so
I
I
I  0.9260 + 0.92602 C = 0.0768c = 2.30 x 107 m s  1.
278
STUDENT'S SOLUTIONS MANUAL
The broadening of the line is due to local events (collisions) in the distant star. It is temperature dependent and hence yields the surface temperature of the star. Eqn 13.17 relates the observed Iinewidth to temperature: OAobs
= 2A (2kTln2)1 / 2 so
T =
(C 0,A )2
m
C
m
2kln2'
2A
T = ((2.998 x 10 ms )(61.8 x 1O 12 m)2 [(47.95U)(1.661 x lO 27 kg U I )] 2(654.2 x 10 9 ) 2(1.381 x 1O 23 JK I )ln2 ' 8
T = 18.34
P13.31
EJ
X
105 K
I.
= J(J + l)hcB ,
EI  Eo
I
gJ
= 2hcB = hc
= 2J + 1.
(_1___1_) . I) =:21( 1 Ashoner
1
B_l(
:2
Ashorter 
Alonger
Alonger
I( 1 ) x (

Ashoner 
1 1
1
+ (/':;.A/Ashorter)
1) +
Ashoner
/':;.A
)
2
A shorter
1(
1 ) ( 1 ) ( 10 nm) 387.5 nm x 1  1 + (0.061 /3 87.5) x 10 2 cm '
9
= 2" B
= 12.031 cm i I. 2(6.626
2hcB
X
10 34 J S) x (3.00
1010 cm SI) x (2.031 cm I ) 1.381 x 10 23 J K I
k
X
= 5.847K. Intensity of J f + J absorption line IJ ex gJe  E} l kT
Solve for T
T
P13.33
EI  Eo) =(
k
x (
1 In(glhshonerlgohlonger)
)
 ( = 5.847K
I In(3 x 4)
)
= I2.35 K.I
Temperature effects. At extremely low temperatures (10K) only the lowest rotational states are populated. No emission spectrum is expected for the cloud and star light microwave absorptions by the cloud are by the lowest rotational states. At higher temperatures additional highenergy lines appear because higher energy rotational states are populated. Circumstellar clouds may exhibit infrared absorptions due to vibrational excitation as well as electronic transitions in the ultraviolet. Ultraviolet absorptions may indicate the photodissocation of carbon monoxide. High temperature clouds exhibit emissions. Density effects. The density of an interstellar cloud may range from one particle to a billion particles per cm 3. This is still very much a vacuum compared to the laboratory high vacuum of a trillion particles
MOLECULAR SPECTROSCOPY 1: ROTATIONAL AND VIBRATIONAL SPECTRA
279
per cm 3 . Under such extreme vacuum conditions the halflife of any quantum state is expected to be extremely long and absorption lines should be very narrow. At the higher densities the vast size of nebulae obscures distant stars. High densities and high temperatures may create conditions in which emissions stimulate emissions of the same wavelength by molecules. A cascade of stimulated emissions greatly amplifies normally weak linesthe maser phenomena of Microwave Amplification by Stimulated Emission of Radiation. Particle velocity effects. Particle velocity can cause Doppler broadening of spectral lines. The effect is extremely small for interstellar clouds at 10K but is appreciable for clouds near high temperature stars. Outflows of gas from pulsing stars exhibit a red Doppler shift when moving away at high speed and a blue shift when moving toward us. There will be many more transitions observable in circumstellar gas than in interstellar gas, because many more rotational states will be accessible at the higher temperatures. Higher velocity and density of particles in circumstellar material can be expected to broaden spectral lines compared to those of interstellar material by shortening collisional lifetimes. (Doppler broadening is not likely to be significantly different between circumstellar and interstellar material in the same astronomical neighborhood. The relativistic speeds involved are due to largescale motions of the expanding universe, compared to which local thermal variations are insignificant.) A temperature of 1000 K is not high enough to significantly populate electronically excited states of CO; such states would have different bond lengths, thereby producing transitions with different rotational constants. Excited vibrational states would be accessible, though, and rotationalvibrational transitions with P and R branches as detailed later in this chapter would be observable in circumstellar but not interstellar material. The rotational constant B for 12CI60 is 1.691 cm  I . The first excited rotational energy level, J = 1, with energy J(J + l)hcB = 2hcB , is thermally accessible at about 6 K (based on the rough equation of the rotational energy to thermal energy kT). In interstellar space, only two or three rotational lines would be observable; in circumstellar space (at about 1000 K) the number of transitions would be more like 20.
Molecular spectroscopy 2: electronic transitions
Answers to discussion questions 014.1
The process of the detennination of the term symbol for dioxygen, 3 ~;, is described in Section 14.1 (b) and will not be repeated here. The interpretation of the symbol follows~ the letter ~ means that the magnitude of the total orbital angular momentum about the internuclear axis is 0; the left superscript 3 means that the component of the total spin angu lar momentum about the internuclear axis is I (2 x I + I = 3) ; the subscript g means that the parity of the term is even; and the superscript  means that the molecular wavefunction for 0 2 changes sign upon reflection in the plane containing the nuclei.
014.3
A band head is the convergence of the frequencies of electronic transitions with increasing rotational quantum number, 1. They result from the rotational structure superimposed on the vibrational structure of the electronic energy levels of the diatomic molecule. See Figs 14.8 and 14.11. To understand how a band head arises, one must examine the equations describing the transition frequencies (eqns 14.5). As seen from the analysis in Section 14. 1(e), convergence can onl y arise when ternlS in both (B'  B) and (B' + B) occur in the eq uation . Since only a term in (B'  B) occurs for the Q branch, no band head can arise for that branch.
014.5
The overall process associated with fluorescence involves the following steps. The molecule is first promoted from the vibrational ground state of a lower e lectronic level to a higher vibrationalelectronic energy level by absorption of energy from a radiation fie ld. Because of the requirements of the FranckCondon principle, the transition is to excited vibrational levels of the upper electronic state. See Fig. 14.22. Therefore, the absorption spectrum shows a vibrational structure characteristic of the upper state. The excited state molecule can now lose energy to the surroundings through radiationless transitions and decay to the lowest vibrational level of the upper state. A spontaneous radiative transition now occurs to the lower e lectronic level and this fluorescence spectrum has a vibrational structure characteristic of the lower state. The fluorescence spectrum is not the mirror image of the absorption spectrum because the vibrational frequencies of the upper and lower states are different due to the difference in their potentia l energy curves.
014.7
See Section 14.5 for a detailed description of both the theory and experiment involved in laser action. Here we restrict our discussion to only the most fundamental concepts. The basic requirement for a laser is that it has at least three energy levels. Of these levels, the highest lying state must be capable of being efficiently populated above its thermal equilibrium value by a pulse of radiation. A second state, lower in energy, must be a metastable state with a long enough lifetime for it to accumu late a popUlation greater than its thermal eq uilibrium value by spontaneous transitions from the higher overpopulated state.
MOLECULAR SPECTROSCOP Y 2: ELECTRONIC TRANSITIONS
281
The metastable state must than be capable of undergoing stimulated transitions to a third lower lying state. This last requirement implies not only that the metastable state must have more than its thermal equilibrium population, but also that it must have a higher population than the third lower lying state, namely, that it achieve population inversion. See Figs 14.28 and 14.29 for a description of the three and fourlevel lasers. The amplification process occurs when low intensity radiation of frequency equal to the transition frequency between the metastable state and the lower lying state stimulates the transition to the lower lying state and many more photons (higher intensity of the radiation) of that frequency are created. Examples of practical lasers are listed and discussed in Further information 14.1 .
Solutions to exercises E14.1 (b)
According to Hund's rule, we expect one I JTu electron and one 2JT g electron to be unpaired. Hence S = I
[1].
and the mUltiplicity of the spectroscopic term is The overall parity is u x g = the complete core), one electron occupies a u orbital another occupies a g orbital. E14.2(b)
0
since (apart from
Use the BeerLambert law I log  = &[J]I = (327 dm 3 mol  I cm  I) x (2.22 X 10 3 mol dm  3) x (0.15 cm) [0
= 0.10889
I = 10 0.10889 = 0.778 II
The reduction in intensity is 122.2 percent 1 E14.3(b)
&
I [J]/
I
= 100 to
10
[13 .2 133] ,.  I 4
(6.67 x 10 mol dm
E14.4(b)
) x (0.35 cm) mol  I cm  I [I dm = 10cm]
= 787
x
= 17.9
x 105 cm 2 mol  I I
= 787 dm 3 mol  I cm I
The BeerLambert law is I log 10 [J]
E14.S(b)
103 cm)
_
log 0.655
3
=
= &[J]I
so
[J]
 I
I
I
t;/
10
= Iog
(323dm 3 mol  1 cm  I x (0.750cm)
10g(1  0.523)
= 11.33 .
x 10 3 mol dm 3
1
.
Note: a parabolic Iineshape is symmetrical, extending an equal distance on either side of its peale. The given data are not consistent with a parabolic lineshape when plotted as a function of either wavelength or wavenumber, for the peak does not fall at the center of either the wavelength or the wavenumber range. The exercise will be solved with the given data assuming a triangular Iineshape as a function of wavenumber.
282
STUDENT'S SOLUTIONS MANUAL
The integrated absorption coefficient is the area under an absorption peak A
=
f
edii
If the peak is triangular, this area is A
= ! (base) x (height) = ![(l99 x 1O9 m)  1 6
010
=15 . x 1
E14.6(b)
(275 x 1O9 m )I] x (2.25 x 104 dm 3 mol 1cm I )
3 I I I (1.56 x 109dm3mImolIcml) x (lOOcmm l ) dm m mol cm = 3 103 dm m 3
Modeling the 7r electrons of 1,3,5hexatriene as free electrons in a linear box yields nondegenerate energy levels of n2 h 2 Ell
=
8m L2 e
The molecule has six 7r electrons, so the lowestenergy transition is from n the box is 5 times the CC bond distance R. So
= 3 to n = 4. The length of
Modelling the 7r electrons of benzene as free electrons on a ring of radius R yields energy levels of m I2 1i2
Em/
=u
where I is the moment of inertia: I = meR2 . These energy levels are doubly degenerate, except for the nondegenerate ml = O. The six 7r electrons fill the m[ = 0 and 1 levels, so the lowestenergy transition is from m[ = 1 to ml = 2 (2 2 _ 12)h 2

87r 2meR2
Comparing the two shows
I
Therefore, the lowestenergy absorption willi rise in energy. E14.7(b)
The BeerLambert law is I
log  = e[J]l = log T
10
so a plot (Figure 14.1) of log T versus [J] should give a straight line through the origin with a slope m of d. So e = mil.
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
The data follow [dye] / (mol dm 3 ) 0.0010 0.0050 0.0100 0.0500
o __
T
10gT
0.73 0.21 0.042 1.33 x 10 7
0.1367 0.6778 1.3768 6.8761
~~~~~~~~~CC,
~ 3.5384 X10 3 : 137.60:X • R2 h 1:000. : : : Y
.. ...
 2
6 8
om
0.00
0.02
0.03
0.04
0.05
0.06
3
Figure 14.1
[dye]/(mol dm )
The molar absorptivity is £
E14.8(b)
I
3 =  138dm mol  I = 522dm 3 molI emI
0.250cm·
I .
The BeerLambert law is
=
log T £
£[J]I
I
so
£
= [J]/log T
~
=
(0.0155 mol dm  ) x (0.250 em)
logO.32 = 1128dm 3 mol  I emI I
Now that we have £, we can compute T of this solution with any size of cell T
E14.9(b)
= lO e[J]1 = 1O{ (128dm
3
mol  I cm l ) x(0.0155 mol dm  3 ) x(0.450cm) }
= @J.TI
The BeerLambert law is I
log 
10
=
(a)
1= 
(b)
1= 
£[J]I
(30dm 3
so
I
I
£[1]
10
1 = Iog
I 1 ,, x log  = 0.0 10 em 3 mol  I em I) x (1.0 mol dm  ) 2
I
I
(30dm 3 mol  I emI) x (1.0 moldm  3 )
x logO. 10
I
= I0.033em I
283
284
STUDENT'S SOLUTIONS MANUAL
E14.10(b) The integrated absorption coefficient is the area under an absorption peak
A
=
f
s dv
We are told that S is a Gaussian function, i.e. a function of the form S = smaxexp
(
_X2)
~
where x = v  vmax and a is a parameter related to the width of the peak. The integrated absorption coefficient, then, is
A
=
f
oo
00
Smax exp
(_X2 ) ~ dx = smaxa.jJi
We must relate a to the halfwidth at halfheight, XI 12
So
A
( 2) X
I
2 Smax
1/ 2 = Smax exp ;;r
In!
so
2
= SmaxXl/2 (~) 1/ 2 = (1.54 X
= 1l.39 X 108
dm 3 molI cm 2
=
2 X 1/ 2 a2
and
104 dm 3 mol 1 em I) x (4233 emI) x
C:
2) 1/ 2
1
In SI base units (1.39 x 108 dm 3 molI cm 2) x (lOOOcm 3 dm 3 ) A=~~1 lOOcmm
9
= 11.39 x 10 m molII E14.11 (b)
Fi is formed when F2 loses an antibonding electron, so we would expect Fi to have a shorter bond than F2. The difference in equilibrium bond length between the ground state (F2) and excited state (Fi + e ) of the photoionization experiment leads us to expect some vibrational excitation in the upper state. The vertical transition of the photoionization will leave the molecular ion with a stretched bond relative to its equilibrium bond length. A stretched bond means a vibrationally excited molecular ion, hence a stronger transition to a vibrationally excited state than to the vibrational ground state of the cation.
I
I
Solutions to problems Solutions to numerical problems P14.1
The potential energy curves for the X3 1:; and B3 1:; electronic states of 0 2 are represented schematically in Fig. 14.2 along with the notation used to represent the energy separation of this problem. Curves for
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
285
the other electronic state of 02 are not shown. Ignoring rotational structure and anharmonicity we may write
voo ~
Te
I + (v'  v) = 6.175 eV x
2
(8065.5 cmI eV
= 149 364 cm I
I
)
+ I (700 2
1580) cm
_I
I.
\7.v' = 4 ii'
L 00 emT, = 6. 175eV
. __ .
T
v' = 3
v, ~=; 2
v' = 0 '    , . .          1
state 5.1147 eV
\rv = 3 v
~
1580 em l
o
TR
COMMENT. Note that the selection rule
Figure 14.2
tov = ±1 does not apply to vibrational transitions between different
electronic states.
Question. What is the percentage change in voo if the anharmonicity constants Xe V (Section 13.11), 12.0730 cm  I and 8.002 cm I for the ground and excited states, respectively, are included in the analysis? P14.3
Initially we cannot decide whether the dissociation products are produced in their ground atomic states or excited states. But we note that the two convergence limits are separated by an amount of energy exactly equal to the excitation energy of the bromine atom : 18345 cm I  14660 cm I = 3685 cm I . Consequently, dissociation at 14660cm 1 must yield bromine atoms in their ground state. Therefore, the possibilities for the dissociation energy are 14660cm 1 or 14660cm 1  7598cm 1 = 7062 cm I depending upon whether the iodine atoms produced are in their ground or excited electronic state.
286
STUDENT'S SOLUTIONS MANUAL
in order to decide which of these two possibilities is correct we can set up the following BornHaber cycle. (I)
rBr(g)
>
~h(g)
+ ~Br2(1)
(2)
~ 12 (s)
>
~12(g)
(3)
~ Br2(I)
>
~ Br2 (g)
(4)
~12(g)
>
I(g)
(5)
~ Br2 (s)
>
Br(g)
IBr(g)
>
I(g)
+ Br(g)
= !::,.fH" (IBr, g) !::"H~ = ~!::"subH" ( 12' s) !::"Hr = ~!::"vapH" ( Br2' I) !::,.H: = ~!::"H(Il ) !::"H~ = ~!::"H(BrBr)
!::"Hf
!::"H"
= !::,.fH" (lBr, g) + ~ !::" subH" (I 2'S) + ~!::"vapH"(Br2 , 1) + ~!::"H(I I ) + ~!::"H(BrBr)
!::"H"
= {40.79 + ~
x 62.44 +
~
x 30.907
+~
x 151.24
+~
x 192.85 } kJ mol 
I
[Table 2.7 and data provided) = 177.93 kJ mol  I = 1 14874 cm  I
I.
Comparison to the possibilities 114660 cm I 1 and 7062 cm  I shows that it is the former that is the correct di ssociation energy. P14.S
We write
S
= smaxex2 = s rnaxeii2 /2r the variable being vand r being a constant. v is measured from
the band center, at which
v = o. S = ~Smax when v2 = 2r In 2. Therefore, the width at halfheight is A
2
uV /
implying that
I 2 r=. 81n2
Now we carry out the integration
f = 1
00
=
A
sdv
= Smax A
Smax
00
2 e  ii /2r dv
2n!::,.v2I/ 2 )1 / 2 (
8 1n 2
n
= s max(2 rn) 1/ 2
= () 41n 2
1/2
= 1.0645s max !::,. VI /2, with v centered on va·
Since
I/? v = I , !::,.VI /2 ~ !::,.A 2 [A ~ AO). A
A
= 1.0645s max
AO
(
!::" AI/ 2 ) A6 .
_
Smax!::,.VI /2
[1
00
00
2
e _x ill•  n 1/ 2J _
= 1.0645s max !::,.VI /2,
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
From Fig. 14.6 ofthe text, we find 6.A 1/2 = 38 nm with AD
287
= 290 nm and E"max ~ 235 dm3 mol  I cm I ;
hence 7 3 I 1.0645 x (235dm mol  I cm ) x (38 x 10 cm) 11 I 106 d 3 I I 21 . x m mo cm . 7 (290 X 10 cm)2
A 
Since the dipole moment components transform as A I (z) , 8 I (x), and 82 (y), excitations from AI to AI , 8 I , and 82 terms are allowed. P14.7
We use the technique described in Example 13.5, the BirgeSponer extrapolation method, and plot the 1 difference 6. vu against v + 2' We then draw up the following table.
6.i\
688.0665.1641.5617 .6591.8561.2534.0 I
1
2
3 2
5 2
7 
9 
11
2
2
2
13 2
502.1
465 .5
428.9
388.2
343.1
300.9
255 .0
15 2
17 2
19 2
21
23 2
25 2
27 2
v+
6.v v 1
v+2
2
2
The data are plotted in Fig. 14.3. Each square corresponds to 25 cm I . The area under the nonlinear extrapolated line is 295 squares; therefore the dissociation energy is 7375 cm I . The 3 I:;; + X excitation energy (where X denotes the grou nd state) to v = 0 is 49357.6 cm I which corresponds to 6.12 eY. The 3 I:;; dissociation energy for
is 7375 cm I , or 0.91 eV. Therefore, the energy of
is 6.12 eV energy of
+ 0.91
eV
= 7.03 eV.
Since 0 * + 0 is 190 kJ mol  I, corresponding to 1.97 eV, the
0 2(X) + 20
is 7.03 eV  1.97 eV COMMENT.
= 15.06 eV I.
This value of the dissociation energy is close to the experimental value of 5.08 eV quoted
by Herzberg [Further reading , Chapters 13 and 14], but differs somewhat from the value obtained in Problem 14.2. The difficulty arises from the BirgeSponer extrapolation, which works best when the experimental data fit a linear extrapolation curve as in Example 13.5. A glance at Figure 14.3 shows that the plot
288
STUDENT'S SO LUTI ONS MANUAL . .
700 : . •
.
:
:
.... ..... .......... .... .. .... ...... ............... :
:
:
:
:
:
:
:
:
:
':"':"':"':" ':"'7"':" ':" ': "':" • ." '7" ':" ' ~' " '7 " ':"':'" "7" : ••
:
..
··;···~· · · ·~ · ··;···i ... ~ ... ~ ... ~ ....~ ... ~ ... ! . ...:: . . . : .. . :.... i· ··~ .. .: ... j ... ( ....:.... .i··· (... ...:.... .j •. . {.. •• ..;••.• .i··· . . .. ..~ ...: ... ~ .. ..~ " '! .. .~ .. ..~ ... i ... . . . . :.. . . ... ~ ...~ ... ~ ... ~ ... ~ ... ~ ... ~ ... "":'" .. . : ... ... ':" .... ..... :
600
~
.; ..i~t~~.
:;:::;.:. .,. ::~>~k:,.,." ...,. ,..;. ~ . ;.. .,.
.,' .,'
·· · f· ·~·~·~· ·
..': .. " ...•. .,_ ......... . :
. .•..; •. j .. { .•. : . . • ! ... { .
300 ;..
.... ... .: ... ...... ...:. ~
~
~
.
.
:
::
...
~ .
:
· ·1· ··'; . . ·j. · ·· 1 . . ~.
. . 0 :. ......... ........ . ......... ... . ~
'
o
4
:
:
:
.~ .. ·l··)·· .. ~ .. ·~ .. .
... ;. ... :... ~.... : ... :.. . ... ~ ... ;... ~ ....;... ;... .. ~ ~ ::~ ~ ~ ~ !... ~ ....~ ... !.. . , .).. .. ; ... .,........ ; ... '.. ~, : ~: . . . . . . ~
j ... ;
~
.. . . ...
.. +... i... ~....;... !. . ... ... ... ...
xtrianpola tion.;';"+" :; ·· .....;;'..... ebeg s . . ; . . here "1".,. . ...!...
.. .. ., ... ..
... ~ .. ~ ... ~ .. !...~ ..
. . . ~ . . . ~ . .. ~ . . . . ~ .. . " .. !
· ... ~ ... Inaccurate .. ., .... .. ~ ... nonlinear
.., .. i .. ~ ...,
...;... ~
. . ....:... ~....~ ... :... ~ .~ .. :._. ~'.. . ... ~ ... ~....~ ;... ~....~ ... ~ ... ~.. ~~ .. ~ ... ~ . . .... : : : : ,: :
.
200
~ .. .~ 100 i....;
.• . j •.•
: : l·J{m·;~~:'~latiOn::TT .j.,...! .. ;.. ,.....::L'>~·T ·L:: :'>; ::':L::::LL:::::
... ~ . . . ! . " :
. ..., ....
:.... ~ ... ;... ~....~ ...;.. .~
:, :
~
) ... r::j:·····: :
~
500
, ...:....,...
.. .,... ...: ... ..
8
.
.
.!.
.~
~
...
... :....:....:....:....: 12
" ~~i"'~''' ~'' . ..~ .
":' .. ~
...~ ... ~ ...j ...:. .
•. • j . •
~
.. ~ ..
.. ~ "<..: : '\
. :
. :
'~"' l"'~""~
....
.. ! ..
..! ... ~ ...}... i ··1·"J··j··
..... . .:
.... ..... : ....:.. . .:.... :....:...... : ... : .... :... ~:'\ . 20 24 28
32
Figure 14.3
of the data is far from linear; hence. it is not surprising that the extrapolation here does not compare well to the extrapolation quoted in Problem 14.2. The extrapolation can be improved by using a quadratic or higher terms in the formula for e.G [Chapter 13].
P14.9
Draw up a table like the following . Hydrocarbon
hVrnax/eV
Benzene Biphenyl Naphthalene Phenanthrene Pyrene Anthracene
4.184 3.654 3.452 3.288 2.989 2.890
 9.7506 8.9169 8.8352  8.7397 8.2489 8.2477
*Semiempirical. PM3 level . PC Spartan PrOTM. Figure 14.4 shows a good correlation:
P14.11
r2
= 0.972.
(a) The molar concentration corresponding to 1 molecule per cubic 11m is: n
v
6.022 x 1023 mol I
i.e. nanomolar concentrations.
x
I 6 (10 11 m m )3 1 (1.0 11 m3 ) (10 dm m )3
= 11.7 x
109 mol dm 3 1.
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS 8.0
289
...,.......~,~.~,
8.5 9.0 9.5
10.0
L_~_'_~_'._~
2.5
3.0
_ _'_~~
3.5
4.0
4.5
Figure 14.4
(b) An impurity of a compound of molar mass 100 g mol l present at 1.0 x 10 7 kg per 1.00 kg water can be expected to be present at a level of N molecules per cubic Ilm where N is
N=
1.0 X 10 7 kg impurity 1.00 kg water
6.022 x 1023 mol 1
x
~~
100 x 10 3 kg impurity mol l
x (1.0 x 103 kg water m 3) x (lO6 m )3,
= 16.0
N
1.
x 10 2
Pure as it seems, the solvent is much too contaminated for singlemolecule spectroscopy.
Solutions to theoretical problems P14.13
We need to establish whether the transition dipole moments
JLfi =
f
VrfJL1fi dr [13 .13]
connecting the states I and 2 and the states I and 3 are zero or nonzero. The particle in a box wavefunctions are 1fn = (2IL)I /2sin (nrexIL) [9.5] . Thus
M,l
f f
ex
and it3, 1 ex
sin
sin
C~x ) x sin (rc;) dx ex
havi ng used sin a sin fJ the standard form
f x(cos ax)
dx
= ! cos(a I
x
a
a
1 dx = Ioo x cos (rex) L (re IL)
2
L 1xcos (3rex) 0
dx
=
I
L
2
[cos
(rc;)  cos C~x) ] dx
C~x) 
cos
C~x ) ] dx
sin ax.
cos (rex) 
(3rel L)
[cos
! cos(a + fJ). Both of these integrals can be evaluated using
fJ) 
= 2' cos ax + 
L
L
fx fx
C~x ) x sin (rc;) dx ex
cos
IL + x sin (rex) IL= 2 (L)  2 1= 0, 0
(3rex) L
(reIL)
L
ore
IL +  x  sin ( 3rex ) IL = 2 ( L )21= o. 0
(3rel L)
L
0
3re
290
STUDENT'S SOLUTIONS MANUAL
=f O.
Thus M,I
In a similar manner,
/L3,1
= O.
COMMENT. A general formula for Pfi applicable to all possible particle in a box transitions may be derived.
The result is (n = f, m = i)
__ eL [cos(n  m)n  1 _ cos(n JLnm 
(n _ m)2
n2
(n
+ m)n + m)2
1] .
For m and n both even or both odd numbers, JLnm = 0; if one is even and the other odd, JLnm
=f O.
See also
Problem 14.17.
Question. Can you establish the general relation for /Lnm above? P14.15
We need to determine how the oscillator strength (Problem 14.16) depends on the length of the chain. We assume that wavefunctions of the conjugated electrons in the linear polyene can be approximated by the wavefunctions of a particle in a onedimensional box. Then
f = /Lx
8rr 2mev 3he
2
2I#Lfd [Problem 14.16].
= e foL l/fn, (x)xl/fn(x)dx,
2)1 /2
= (L
l/fn
sin
(nrrx)
L
2e fL =L 10 xsin (n'L rrx) sin (nJU) L dx
=
I:
(8eL) n(n + I) rr2 (2n + 1)2
if n'
= n + 2.
if n'
= n + 1.
The integral is standard, but may also be evaluated using 2 sin A sin B Problem 14.13. hv
= En+1
 En
=
+ B) as in
h2 e
~
n,
8rr2)(me )( h ) he2 8meL2 (2n ( 3
Therefore,f ex
B)  cos(A
= (2n + 1)2· 8m L
Therefore, for the transition n + 1
f
= cos(A 
n2 (n
+
2 (8eL)2 n (n+ 1)2 1) ~ (2n + 1)4
=
(64)[n2(n+I)2] 3Jl"2 (2n + 1)3 .
+ 1)2 3 .
(2n + 1) The value of n depends on the number of bonds: each Jl" bond supplies two Jl" electrons and so n increases by 1. For large n, n4 8n 3
f ex 
~
n 
8
and
f ex n.
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
291
Therefore, for the longest wavelength transitionsf increases as the chain length is increased. The energy of the transition is proportional to (2n + I) j L 2 , but, as n ex L, this energy is proportional to I j L. I:!.E=
but L
(2n + l)h 2 2 [l:!.n=+IJ 8me L
= 2nd is the length of the chain (Exercise 14.6(a)), with d the carbon<:arbon interatomic distance.
Hence
Therefore, the transition moves toward the red as 1 L is increased 1 and the apparent color of the dye 1shifts towards blue I· P14.17
Jt
=
eSR [givenJ.
S= [I + ~ao + ~3 (~)2] eao
Rj ao
[Problem I1.3J.
RS)2 ( ao fa· We then draw up the following table.
R j ao fifo
°°
I
0.737
2 1.376
345 1.093 0.573 0.233
678 0.080 0.024 0.007
These points are plotted in Fig. 14.5. The maximum inf occurs at the maximum of RS, d
(RS) dR
3] 3 ( )
dS = S + R= 1+ R  I R
dR
[
ao
ao
e
Ra j 0
=
°
at R
= R*.
That is, I + R* _ ~ (R*) 3 = 0. ao 3 ao This equation may be solved either numerically or analytically (see Abramowitz and Stegun, Handbook of mathematical functions, Section 3.8.2), and R* = 2.1 0380ao. As R + 0, the transition becomes s + s, which is forbidden . As R + single atom because its wavefunction does not extend to the other. P14.19
00,
the electron is confined to a
The fluorescence spectrum gives the vibrational splitting of the lower state. The wavelengths stated correspond to the wavenumbers 22 730, 24390, 25 640, 27030 cm I , indicating spacings of 1660, 1250, and 1390 cm I . The absorption spectrum spacing gives the separation of the vibrational levels of the upper state. The wavenumbers of the absorption peaks are 27800,29000,30300, and 32 800 cm I . The vibrational spacings are therefore 1200, 1300, and 2500 cm I .
292
STUDENT'S SOLUTIONS MANUAL
o
2
6
4
8
R/ ao
P14.21
Figure 14.5
Use the ClebschGordan series [Chapter 10] to compound the two resultant angular momenta, and impose the conservation of angular momentum on the composite system. (a) 02 has S = 1 [it is a spin triplet]. The configuration of an 0 atom is [He]2s22p4, which is equivalent to a Ne atom with two electronlike 'holes'. The atom may therefore exist as a spin singlet or as a spin triplet. Since SI = I and S2 = 0 or SI = I and 52 = I may each combine to give a resultant and ~ may with 5 = 1, both may be the products of the reaction. Hence multiplicities be expected.
lED
!.
= O. The configuration of an N atom is [He]2s22p3. The atoms may have S = ~ or Then we note that 5 I = ~ and 5 I = ~ can combine to give 5 = 0; S I = and 52 = can also combine
(b) N2, S
to give 5 = 0 (but SI be expected.
= ~ and S2 =
!
!
! cannot). Hence, the mUltiplicities 14 + 41 and 12 + 2 1may
Solutions to applications P14.23
Fraction transmitted to the retina is (I  0.30) x (I  0.25) x (I  0.09) x 0.57
= 0.272.
The number of photons focused on the retina in 0.1 s is 0.272 x 40mm 2 x 0.1 s x 4 x J0 3 mm 2 sI
= 14.4 x
10 3 1. More than what one might have guessed.
MOLECULAR SPECTROSCOPY 2: ELECTRONIC TRANSITIONS
P14.25
293
The integrated absorption coefficient is
A
f
=
£(i.i)di.i [13.5].
Ifwe can express £ as an analytical function of i.i, we can carry out the integration analytically. Following the hjnt in the problem, we seek to fit £ to an exponential function , which means that a plot of In £ versus i.i ought to be a straight line (Fig. 14.6). So if In £ = mi.i
+ b,
then
£ = exp(mi.i) exp(b)
and A = (e b / m) exp(mi.i) (evaluated at the limits of integration). We draw up the following table and find the bestfit line. )Jnm
£/ (dm 3 mol  I em I)
i.i/em  I
In £/ ( dm 3 mol  I emI)
292.0 296.3 300.8 305.4 310. 1 315.0 320.0
1512 865 477 257 135.9 69.5 34.5
34248 33748 33248 32748 32248 31746 31250
4.69 4.13 3.54 2.92 2.28
1.61 0.912
5
4
3 2
o 31000
32000
33000
ii / (em I)
So A
=
34000
35000
Figure 14.6
3 3 e38.383 [ (1.26 X I0 em) (1.26XIO Cm) ] 3 I I ex  ex dm mol cm1.26 x 10 3 em p 290 x 10 7 em p 320 x 10 7 em
294
P14.27
STUDENT'S SOLUTIONS MANUAL
(a) The integrated absorption coefficient is (specializing to a triangular lineshape) A
=
f
e(v) dv
= (l/2) A
=
(l/2)e max !lv
x (l50dm 3 mol I cm I ) x (34483  31250) cm I ,
= 12.42 X 105 dm3 mo1 1 cm 2 1
(b) The concentration of gas under these conditions is e
= n = p = V
RT
2.4 Torr 3 I (62.364Torrdm mol K I ) x (373K)
= 1.03 x
10
4
mol dm
3
.
Over 99 per cent of these gas molecules are monomers, so we take this concentration to be that of CH3I (If 1 of every 100 of the original monomers turned to dimers, each produces 0.5 dimers; remaining monomers represent 99 of99.5 molecules.) Beer's law states A
= eel = (l50dm3 mol I cm I )
x (1.03 x 104 moldm 3) x (12.0cm)
= 10.1851.
(c) The concentration of gas under these conditions is n p e    
V  RT 
100 Torr (62.364 Torr dm 3 mol I K I ) x (373 K)
= 4.30
3 3 x 10 mol dm
Since 18 per cent of these CH3I units are in dimers (forming 9 per cent as many molecules as were originally present as monomers), the monomer concentration is only 82/91 of this value or 3.87 x 10 3 mol dm 3 . Beer's law is
= eel = (l50dm3 mol I cm I )
A
x (3.87 x 10 3 moldm 3) x (12.0cm)
= 16.971.
If this absorbance were measured, the molar absorption coefficient inferred from it without
consideration of the dimerization would be
e
= A/el = 6.97/«4.30 = 1135 dm 3 mol I
x 10 1 moldm 3 ) x (12.0cm»
cm I
l
an apparent drop of 10 per cent compared to the lowpressure value. P14.29
In Fig.14.7
!lEI I
= ~ = All
386.4nm
!lEoo
= he =
he
he
= 5.1409 x
10
= 5.1250 x
10
19
J
= 3.2087eV
J
= 3.1987eV.
and Aoo
387.6nm
= VI + (u + 1/2)vlhe + J(J + l)Blhe. J) = Vo + (u + 1/2)vohe + J(J + l)Bohe.
Energy of excited singlet, SI : EI CU, J) Energy of excited singlet, So : Eo(u,
19
MOLECULAR SPECTROSCOPY 2: ELECTRON IC TRANSITIONS
295
v=oo
Mil Moo 4I+I~. 386.4 nm 387.6 nm
The midpoint of the 00 band corresponds to the forbidden v = 0 + O. 6. Eoo = E, (0, 0)  Eo (0, 0) = (V,  Vo)
+ ! (v,
Q branch
A Figure 14.7
(6.J = 0) with J = 0 and
 vo)hc.
The midpoint of the II band corresponds to the forbidden v=I+1.
( I)
Q branch
(6.J
0) with J = 0 and
(2)
MUltiplying eqn I by three and subtracting eqn 2 gives
= 2(V,
36.Eoo  6.E" V,  Vo
 Vo).
= !(36.Eoo 
6.E,d
= !(3(S.12S0)
 (S . 1409) }lO'9 J
=S.I171 x 1O ' 9 J= 13. 193eV I.
(3)
This is the potential energy difference between So and S, . Equations ( I) and (3) may be solved for v,  vo . v,  vo = 2(6.Eoo  (V,  Vo)}
= 2(S.12S0 
S. I I7I}lO'9 J/ hc 2
= 1.5800 x 10 ' J = 0.009861SeV = 179 .S38 cm' I. The
v,
value can be determined by analyzing the band head data for which J
6.EIQ(1)
= E, (O , J ) 
= V, 6. E oo(1) = V,
+ !(v, Vo + ! (v , 
 Vo 
Eo(l , J
+I
+ J .
+ I)
+ J (J + I )B,hc  (1 + I ) x (J + 2)Bohc. vo)he + J(1 + l)B, he  (J + I) x (1 + 2)Bohe.
3vo)hc
296
STUDENT'S SOLUTIONS MANUAL
Therefore, t::..Eoo(J)  t::..ElO(J) t::..EOO(Jhead)
=
t::..ElO(Jhead)
=
=
vohe.
he 388.3 nm
19
= 5.1158
x 10
= 4.7117
x 10 19 1.
he
421.6 nm
1.
(5.1158  4.7117) x 10 19 1 he
4.0410 x 10 20 1 = he VI
= 0.25222 eV =
I
2034.3 em I I·
= vo+79.538cm 1 =
(2034.3
(III)
In  100
+ 79.538) cm I =
2113 .8cm1
=
4.1990 x 1Ohe
2o
J
hev i =kTeff ·
Teff = kin (10  0 )
4.1990 x 10 20 J .,,~(1.38066 x 1O 23 JK J )ln(l0)
= 11321
K
I.
11  1
= 0 and v = 1 vibrational states is the inverse of the relative intensities of the transitions from those states; hence ~ = ITQJ.
The relative population of the v
0.1
It would seem that with such a high effective temperature more than eight of the rotational levels of the S J state should have a significant population. But the spectra of molecules in comets are never as clearly resolved as those obtained in the laboratory and that is most probably the reason why additional rotational structure does not appear in these spectra.
Molecular spectroscopy 3: magnetic resonance
Answers to discussion questions 015.1
Detailed discussions of the origins of the local, neighboring group, and solvent contributions to the shielding constant can be found in Sections 15.5(d), (e), and (f) as well as the books listed under Further reading . Here we will merely summarize the major features. The local contribution is essentially the contribution of the electrons in the atom that contains the nucleus being observed. It can be expressed as a sum of a diamagnetic and paramagnetic part, that is, a(local) = ad + a p . The diamagnetic part arises because the applied field generates a circulation of charge in the ground state of the atom. In tum, the circulating charge generates a magnetic field. The direction of this field can be found through Lenz's law, which states that the induced magnetic field must be opposite in direction to the field producing it. Thus it shields the nucleus. The diamagnetic contribution is roughly proportional to the electron density on the atom and it is the only contribution for closed shell free atoms and for distributions of charge that have spherical or cylindrical symmetry. The local paramagnetic contribution is somewhat harder to visualize since there is no simple and basic principle analogous to Lenz's law that can be used to explain the effect. The applied field adds a term to the hamiltonian of the atom that mixes excited electronic states into the ground state and any theoretical calculation of the effect requires detailed knowledge of the excited state wave functions. It is to be noted that the paramagnetic contribution does not require that the atom or molecule be paramagnetic. It is paramagnetic only in the sense that it results in an induced field in the same direction as the applied field. The neighboring group contributions arise in a manner similar to the local contributions. Both diamagnetic and paramagnetic currents are induced in the neighboring atoms and these currents result in shielding contributions to the nucleus of the atom being observed. However, there are some differences: The magnitude of the effect is much smaller because the induced currents in neighboring atoms are much farther away. It also depends on the anisotropy of the magnetic susceptibility (see Chapter 20) of the neighboring group as shown in eqn 15.23 . Only anisotropic susceptibilities result in a contribution. Solvents can influence the local field in many different ways. Detailed theoretical calculations of the effect are difficult due to the complex nature of the solutesolvent interaction. Polar solventpolar solute interactions are an electric field effect that usually causes deshielding of the solute protons.
298
STU DE NT'S SOLUTI ONS MANUAL
Solvent mag netic antisotropy can cause shielding or deshielding, for example, for solutes in benzene solution. In addition, there are a variety of specific chemical interactions between solvent and solute that can affect the chemical shi ft. See the references listed under Fu rther reading for more details. 015.3
Both spinlattice and spinspin relaxation are caused by flu ctuating magnetic and electric fie lds at the nucleus in question and these fie lds result from the rando m thermal motions present in the solution or other form of matter. These random motions can be a res ult of a number of processes and it is hard to summarize all that could be important. In theory every known nuclear interaction coupled with every type of motion can contribute to re laxation and detailed treatments can be exceedingly complex. However, they all depend on the magnetogyric ratio of the atom in question and the magnetogyric ratio of the proton is much larger than that of J3c. Hence the interaction of the proton with flu ctuating local magnetic fi elds caused by the presence of neighboring magnetic nuclei wi ll be greater, and the relaxation will be quicker, corresponding to a shorter relaxation time for protons. Another consideration is the structure of compounds containing carbon and hydrogen. Typicall y the C atoms are in the interior of the molecule bonded to other C atoms, 99% of which are nonmagnetic, so the primary relaxation effects are due to bonded protons. Protons are on the outside of the molecule and are subject to many more interactions and hence faster re laxation.
015.5
Spinspin couplings in NMR are due to a polarization mechanism that is transmitted through bonds. The fo llowing description applies to the coupling between the protons in a HxC Hy group as is typical ly fo und in organic compounds. See Figs. 15.201 5.22 of the text. On Hx , the Fermi contact interaction causes the spins of its proton and electron to be ali gned anti parallel. The spin of the electron fro m C in the HxC bond is then aligned antipara llel to the electron fro m Hx due to the Pauli exclusion principle. The spin of the C electron in the bond Hy is then aligned parallel to the electron from Hx because of Hund 's rule. F inally, the ali gnment is transmitted through the second bond in the same manner as the fi rst. This progression of alignments (antiparallel x antiparallel x parallel x antiparallel x antiparallel) yields an overall energetically favorable parallel alignment of the two proton nuclear spins. Therefore, in thi s case the coupling constant, 2 iHH is negati ve in sign. The hyperfine structure in the ESR spectrum of an atomic or molecular system is a result of two interactions: an ani sotropic dipolar coupling between the net spin of the unpaired electrons and the nuclear spins and also an isotropic coupling due to the Fermi contact interaction. In solution, onl y the Fermi contact interaction contributes to the spl itting as the dipolar contribution averages to zero in a rapidly tumbling system. In the case of rrelectron radicals, such as C6H6', no hyperfi ne interaction between the unpaired electron and the ring protons might have been expected. The protons lie in the nodal plane of the molecular orbital occup ied by the unpai red electron, so any hyperfine structure cannot be ex plai ned by a simple Fermi contact interaction. which requires an unpaired electron density at the proton. However, an indirect spin polarization mechani sm, similar to th'at used to ex plai n spinspin couplings in NMR, can account for the ex istence of proton hyperfi ne interactions in the ESR spectra of these systems. Refer to Fig. 15.6 of the text. Because of Hund's rule, the unpaired electron and the fi rst electron in the CH bond (the one fro m the C atom), will tend to align parallel to each other. The second electron in the CH bond (the one from H) will then al ign anti parallel to the fi rst by the Pauli principle, and fi nally the Fenni contact interaction will align the proton and electron on H anti parallel. The net res ult (parallel x antiparal lel x anti parallel) is that the spins of the unpaired electron and the proton are aligned parallel and effectively they have detected each other.
MOLECULAR SPECTRO SCOPY 3: MAGNETIC RESONAN CE
299
Solutions to exercises E15.1(b)
For 19F,
~ = 2.62835,
g
= 5.2567
iLN
v
= VL = y2rr~
.
wIth y
gl iLN ~
Hence, v =  h = = 6.49 E15.2(b)
£1/1/
X
= g liLIiN
(5.2567) x (5.0508 x 10 27 JT I) x ( 16.2 T) (6.626 x 10 34 J s)
10 8 s I = 1649 MHz 1
= yt0a1111 = gl iL N!J81111
= 1, 0, 1 £1/1/ =  (0.404) x
1111
( 2.3466 x 10 26
=12.35 E15.3(b)
X
(5.0508 x 10 27 J T I )
J)
( 11.50 T )1111
1111
10 26 J, 0, +2.35 x 10 26 J
I
The energy separation between the two levels is 1";.£
= hv
y~ ( 1.93 X 107 T  I sI) x ( 15.4 T ) where v         , :      2rr 2rr
= 4.73 E15.4(b)
X
x 10 7 s l
= 147.3 MHz 1
A 600 MHz NMR spectrometer means 600 MHz is the resonance fie ld for protons for which the magnetic field is 14. 1 T as shown in Exercise 15.1 (a). In highfield NMR it is the field not the frequency that is fi xed. (a) A 14N nucleus has three energy states in a magnetic field corresponding to 1111
=
+ 1, 0,  I . But
1";.£(+ I + 0) = 1";.£(0 +  I )
The allowed transitions correspond to 1";.1/1/ 1";.£
= ± I; hence
= h v = yt0a = gliLN ~ = (0.4036)
x (5.051 x 1O 27 Jr l ) x (l4.IT)
= 12.88 x 10 26 J
I
(b) We assume that the electron gvalue in the radical is equal to the free electron gvalue, ge Then 1";.£
= h v = geiLB ~ [37) = (2.0023) = 15.57
x (9.274 x 10 24 Jr' ) x (0.3OOT)
x 10 24 J
I
=20023.
300
STUDENT'S SOLUTIONS MANUAL
COMMENT. The energy level separation for the electron in a free radical in an ESR spectrometer is far greater
than that of nuclei in an NMR spectrometer, despite the fact that NMR spectrometers normally operate at much higher magnetic fields .
E15.5(b)
6.£
= hv =
Hence !!1J ,
E15.6(b)
= gliLN~
y!i!$
hv
=  = gliLN
[Solution to Exercise 15.1(a)]
(6.626 x 1O 34 JHz l ) x (150.0 x 106 Hz) (5.586) x (5 .051 x 10 27 JTI)
In all cases the selection rule 6.ml
= ±I
I
=3
523 T
I
~'~
is applied; hence (Exercise 15.4(b)(a»
hv 6.626 x 10 34 J HzI v Bx I 27  gliLN  5.0508 x 10 JTgl
= (1.3119 x 10 7 ) x
(~J T gl
= (0.13119) x
(~) T gl
We can draw up the following table 31p
!!1JjT
0.40356 97.5 244
gl
300 MHz 750 MHz
(a)
(b)
5.2567 7.49 18.7
2.2634 17.4 43.5
Magnetic fields above 20 T have not yet been obtained for use in NMR spectrometers. As
COMMENT.
discussed in the solution to Exercise 15.4(b), it is the field, not the frequency, that is fixed in highfield NMR spectrometers . Thus an NMR spectrometer that is called a 300 MHz spectrometer refers to the resonance frequency for protons and has a magnetic field fixed at 7.05 T.
E15.7(b)
The relative population difference for spin
! nuclei is given by
Na  NfJ ~ yh~ Na + NfJ ~ 2kT
[Justification 15.1]
oN
Ii =
gliLN!!1J
=~
27 I 1.405 (5 .05 x 1O JT ) ~ = 8.62 x 10 7 (!!1J j T) 2(1.381 x 1O 23 JK I) x (298K)
(a) For 0.50 T (b) For 2.5 T (c) For 15.5 T
oN
N
0:
oN N
=
(8 .62 x 10 7 ) x (0.50)
= 14.3
x 10 7 1
= (8.62 x 10 7 ) x (2.5) = 12.2 x 106 1 = (8.62 x 10 7 ) x (15.5) = 11.34 x 10 5 1
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
E15.8(b)
301
The ground state has I
.
= + 2: = ex spm,
ml
mI
I
=   = f3 2
.
spm
Hence, with
8N
Net  NfJ
Net  Nete6E /kT
Ii = Net + NfJ = Net + Net e 6E / kT 1 e 6E / kT I + e 6E / kT
8N=
NglJAN flg 2kT
~
[Justification 15 .1]
I  (I  6.E / kT) 1+ I
~
gIJAN ~ 6.E = 
2kT
2kT
[for 6.E
« kT]
Nhv =2kT
Thus, 8N ex v
8N(SOOMHz) 8N(60MHz)
=
(SOOMHz) (60 MHz)
= ITTI
This ratio is not dependent on the nuclide as long as the approximation 6.E v  VO (a) 8 =   x 106 [15 .S] VO
« kT holds.
Since both v and VO depend upon the magnetic field in the same manner, namely
V=
I
gIJAN ~ 
and VO
h
=
gIJAN~O.
[Exercise 15 . I(a)]
h
I
8 is independent of both ~ and v. (b) Rearranging [15 . IS], v  VO
v  VO(SOO MHz) v  vO(60 MHz)
=
= v08
x 10 6 and we see that the relative chemical shift is
(SOO MHz) (60 MHz)
= ITTI
COMMENT. This direct proportionality between v 
and
VO
VO
is one of the major reasons for operating an
NMR spectrometer at the highest frequencies possible.
E15.9(b)
~I oc
=
(l  a) ~
16. flgloc 1 = 1(6.a)l flg
= 11.16 
~
1[8(CH3) 
3.361 x 1O 6 flg
(a) flg
= 1.9T, 16.~locl = (2.20
(b) flg
= 16.5 T, l6. flgloc 1 = (2.20 x
8(CH2)]I ~
= 2.20 x
1O6 ~
x 10 6 ) x (1.9T)
= 14.2
10 6 ) x (16.5 T)
x 10 6 T I
= 13.63
x 10 5 T
I
302
STUDENT'S SOLUTIONS MANUAL
v  Vo
E15.10(b)
I~vl
= v08 == (v
x lO6
VO)(CH2)  (v  VO)(CH3) = V(CH2)  V(CH3)

= vO[8(CH2)  8(CH3)] x lO6 = (3.36  1.16) x 1O 6 vo = 2.20 x 1O6 vo (a)
VO
= 350 MHz
I~ v I = (2.20 x 10 6 ) x (350 MHz) = 770 Hz [Figure 15.1]
(b)
VO
= 650 MHz
I~vl = (2.20 x 10 6 ) x (650MHz) = 1.43 kHz
at 350MHz
Figure 15.1
At 650 MHz, the spin spin splitting remains the same at 6.97 Hz, but as the splitting appears narrower on the 8 scale.
~v
has increased to 1.43 kHz,
E15.11(b) The difference in resonance frequencies is
~v =
(vo x 10
6
)
~8 = (350s l ) x (6.8  5.5) = 4.6 x 102 SI
The signals will be resolvable as long as the conformations have lifetimes greater than T
= (2rr ~8)1
The interconversion rate is the reciprocal of the lifetime, so a resolvable signal requires an interconversion rate less than rate = (2rr ~8) = 2rr (4.6 x 102 SI) = \2.9 x 103 sI \
E15.12(b)
v=
gIItN~ h
[Solution to exercise 15.I(a)]
ve1p) ge1p) Hence, v(IH) = g(IH) 2.2634 orve1p) =   x 500MHz=1203MHzl 5.5857 The proton resonance consists of 2 lines (2 x [2 x (4 x
~) + I].
~ + I)
and the 31 P resonance of 5 lines
The intensities are in the ratio 1:4:6:4: I (Pascal's triangle for four equivalent
I 5.5857 . . h h h . spin :1 nuclei, Section 15.6). The lines are spaced 2.2634 = 2.47 times greater m t e p osp orus regIOn than the proton region. The spectrum is sketched in Figure 15.2.
MOLECULAR SPECTROSCOPY 3 : MAGNETIC RESONANCE
303
Proton resonance
Phosphorus resonance
Figure 15.2
E15.13(b) Look first at A and M, since they have the largest splitting. The A resonance will be split into a widely
spaced triplet (by the two M protons); each peak of that triplet will be split into a less widely spaced sextet (by the five X protons). The M resonance will be split into a widely spaced triplet (by the two A protons); each peak of that triplet will be split into a narrowly spaced sextet (by the fi ve X protons). The X resonance will be split into a less widely spaced triplet (by the two A protons); each peak of that triplet will be split into a narrowl y spaced triplet (by the two M protons). (See Figure 15 .3.) lAM > l AX> lM X
M protons (a)
d
X protons
j 
l AX
I I
(b)
Figure 15.3
Only the splitting of the central peak of Figure IS.3 (a) is shown in Figure IS.3(b).
E15.14(b) (a) Since all JHF are equal in this molecule (the CH2 group is perpendicular to the CF2 group), the H
and F nuclei are both chemicall y and magnetically equivalent. (b) Rapid rotation of the PH3 groups about the MoP axes makes the P and H nuclei chemically and magnetically equivalent in both the cis and trailsform s. E15.15(b) Precession in the rotating frame follows
304
STUDENT'S SO LUTIONS MANUAL
Since w is an angular frequency, the angle through which the magnetization vector rotates is
"" _ S0 ""' 1
en _
 
g] flNt
34
(n) x (1.0546 x 10 J s) 1 4 1 940 x 10 T (5.586) x (5 .0508 x 10 27 JTI) x (12 .5 x 10 6 s) .
a 90° pulse requires! x 12.511 s = 16.25 I1S
=
I
(6.626 X 10 34 J s) x (2.998 x 108 m SI) (2) x (9.274 x 10 24 J TI ) x (8 x 10 3 m) 
[iliJ 13T .
E15.17(b) The g factor is given by
hv
h
flB fA
flB
g =;
= g
6.62608 x 10 34 J s 9.2740 x 10 24 JT 1
~,~
1
_I
71.448mTGHz x 9.2482GHz 330.02 mT 
= 7.1448
X
10
11
THz 1
= 71.448mTGHz 1
1
2.0022
E15.18(b) The hyperfine coupling constant for each proton is 12.2 mT
I, the difference between adjacent lines in
the spectrum. The g value is given by
=~= g
flB fA
(71.448mTGHz 1) x (9.332GHz) _~ 334.7mT ~
E15.19(b) If the spectrometer has sufficient resolution, it will see a signal spilt into eight equal parts at
± 1.445 ±
1.435 ± 1.055 mT from the center, namely 1328.865,330.975,331.735, 331.755,333.845,333.865, 334.625 , and 336.735 mT
I
If the spectrometer can only resolve to the nearest 0.1 mT, then the spectrum will appear as a sextet with intensity ratios of 1: I :2:2: I : I. The four central peaks of the more highly resolved spectrum would be the two central peaks of the less resolved spectrum. E15.20(b) (a) If the CH2 protons have the larger splitting there will be a triplet (1:2:1) of quartets (1:3:3:1).
Altogether there will be 12 lines with relative intensities 1(4 lines), 2(2 lines), 3(4 lines), and 6(2 lines). Their positions in the spectrum will be determined by the magnitudes of the two proton splittings which are not given. (b) If the CD2 deuterons have the larger splitting there will be a quintet (1:2:3:2:1) of septets (1 :3:6:7:6:3:1). Altogether there will be 35 Lines with relative intensities 1(4 lines), 2(4 lines), 3(6 lines), 6(8 lines), 7(2 lines), 9(2 lines), 12(4 lines), 14(2 lines), 18(2 Iines),and 21(1 line). Their positions in the spectrum will determined by the magnitude of the two deuteron splittings which are not given.
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
305
E15.21(b) The hyperfine coupling constant for each proton is 12.2mT I, the difference between adjacent lines in the spectrum. The g value is given by hv
h
hv
 = 71.448mTGHz
g=   s o f$ =  , J1B f$ J1Bg
_ I
J1B l
(a)
(b)
f$ = (71.448mTGHz ) x (9.312GHz) =1332.3mTI
2.0024 f$
=
l
(71.448 mT GHz ) x (33.88 GHz) = 11209 mT 2.0024
E15.22(b) Two nuclei of spin
II = 1 Igive five lines in the intensity ratio 1:2:3:2: 1 (Figure 15.4).
I I I II III
I II
2
2
3
I
First nucleus with I = I second nucleus with 1= 1
Figure 15.4
E15.23(b) The X nucleus produces four lines of equal intensity. Three H nuclei split each into a 1:3:3: 1 quartet. The three D nuclei split each line into a septet with relative intensities 1:3:6:7:6:3: 1 (see Exercise 15.20(a». (See Figure 15.5.)
II
Figure 15.5
Solutions to problems Solutions to numerical problems P15.1
g/ Bo
= 3.8260
(Table 15.2) . (6.626 x 10 34 JHz l ) x v
hv
=  = g/J1N
()(3 .8260) x (5.0508 x
Therefore, with v
= 300 MHz,
10 27 JTI)
= 3.429 x
8
10 (v/Hz) T.
306
STUDENT'S SOLUTIONS MANUAL
8N Ii
gliLNBO
~ ~
[Exercise 15.4(a)]
(3.8260) x (5.0508 x 1027 J r ' ) x (I0.3T)
=
(2) x (1.381 x 1O23 JK ') x (298K)
I
51
= . 2.42 x 1 0 .
Since gl < 0 (as for an electron , the magnetic moment is anti parallel to its spin), the [ ] state (ml lies lower. P15.3
= !)
The envelopes of maxima and minima of the curve are determined by T2 through eqn 15.30, but the time interval between the maxima of this decaying curve corresponds to the reciprocal of the frequency difference t. v between the pulse frequency vo and the Larmor frequency VL, that is, t. v = Ivo  VL I:
t.v
I
,
=   = 10s = 10Hz. O. IOs
Therefore the Larmor frequency is 1300 x 106 Hz ± 10Hz. 1 According to eqns 15.30 and 15.32 the intensity of the maxima in the flO curve decays exponentially as e r/ T2 . Therefore T2 corresponds to the time at which the intensity has been reduced to I Ie of the original value. In the text figure, this corresponds to a time slightly before the fourth maximum has occurred, or about I 0.29 s I. P15.5
It seems reasonable to assume that only staggered conformations can occur. Therefore the equilibria are as shown in Fig. 15.6.
When R3 =
~
= H, all three of the conformations in Fig. 15.6 occur with equal probability ; hence
31HH (methyl)
=
*e + 1(
2 319)
[t
= trans, g = gauche;
CHR3 ~
= methyl].
Additional methyl groups will avoid being staggered between both R, and R2. Therefore
= !(J( + 19) 31HH (isopropyl) = 1(
31HH(ethyl)
[R3 = H, R4 [R3 = ~
= CH3],
= C H3]·
We then have three simultaneous equations in two unknowns 1, and 19 .
te1( + 2 1g) = 7.3 Hz, 3
!cJ1( 31(
+ 31g) = 8.0Hz,
= 11.2 Hz.
(I)
(2)
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONAN CE
307
The two unknowns are overdetermined. The first two equations yield 3it = 10.1,3 i g = 5.9. However, if we assume that 3i t = 11.2 as measured directl y in the ethyl case then 3i g = 5.4 (eqn I) or 4.8 (eqn 2), with an average value of 5.1. Using the original form of the Karplus equation,
or 11.2
= A + B,
5.1 = 0.25A
+ B.
These simultaneous equations yield A = 6.8 Hz and B = 4.8 Hz. With these values of A and B, the original form of the Karplu s equation fits the data exactly (at least to within the error in the values of 3it and 3i g and in the measured values reported). From the form of the Karplus equation in the text [15 .27] we see that those values of A, B, and C cannot be determined from the data given, as there are three constants to be determined from only two values of i . However, if we use the values of A, B, and C given in the text, then
+ 5 Hz(cos 360°) = II Hz, I Hz(cos 60°) + 5 Hz(cos 120°) = 5 Hz.
it = 7 Hz  I Hz(cos 180°) i g = 7 Hz 
The agreement with the modern form of the Karplus equation is excellent, but not better than the original version. Both fit the data equally well. But the modern version is preferred as it is more generally applicable.
I
I
P15.7
The proton COSY spectrum of Initropropane shows that (a) the CaH resonance with 8 = 4 .3 shares a crosspeak with the Cb H resonance at 8 = 2.1 and (b) the Cb H resonance with 8 = 2.1 shares a crosspeak with the Cc  H resonance at 8 = 1.1. Off diagonal peaks indicate coupling between H's on various carbons. Thus peaks at (4,2) and (2,4) indicate that the H 's on the adjacent CH2 units are coupled. The peaks at ( 1,2) and (2, I) indicate that the H's on CH3 and central CH2 units are coupled. See Fig. 15.7.
P15.9
Refer to Fig. 15.4 in the solution to Exercise 15.20(a). The width of the CH3 spectrum is 3aH = 16.9 mT I. The width of the CD3 spectrum is 6ao . It seems reasonable to assume, since the hyperfine interaction is an interaction of the magnetic moments of the nuclei with the magnetic moment of the electron, that the strength of the interactions is proportional to the nuclear moments.
and thus nuclear magnetic moments are proportional to the nuclear gvalues; hence
ao
~
0.85745
  x aH = 0. 1535a H = 0.35 mT. 5.5857
Therefore, the overall width is 6ao
= 12.1
mT
I.
308
STUDENT'S SOLUTIONS MANUAL abc N0 2CH 2CH 2CH 3
~ L'=A =® 2
0
=cw
@
=@
=@
3
4
5 4
5
P1S.11
3
2
0
. 5.7mT WewnteP(N2s) =   55.2mT
Figure 15.7
~ =~(lOpercentofitstime) ;
~
1.3 mT
P(N2pz) = 3.4 mT = ~ (38 per cent of its time).
The total probability is
I
(a) peN) = 0.10 + 0.38 = 0.481 (48 per cent of its time) . (b) P(O) = I  peN) = j 0.52j (52 per cent of its time). The hybridization ratio is P(N2p) = 0.38 = P(B2s ) 0.10
13.8l. ~
The unpaired electron therefore occupies an orbital that resembles an sp3 hybrid on N, in accord with the radical 's nonlinear shape. From the discussion in Section 11.3 we can write 2 1+ cos¢ a =  1 cos¢
b2 = 1 _ a2 = 2 cos ¢
1  cos¢ A=
b'2 ~ a
=
1 cos¢ A , implying that cos ¢ =  1 + cos¢ 2+ e
Then, since A = 3.8, cos ¢ = 0.66, so ¢ = ~
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
309
Solutions to theoretical problems P15.13
Use eqn 15.22 and Illustration 15.2. For hydrogen itself, we have:
The only difference in wavefunction (and therefore in the expectation value of \ / r) between hydrogen and a more general hydrogenic ion is that the latter has ao/Z where the former has ao, so: e 2 /J Z _....:."',0,  = 12n:m e ao
P15.15
ynJ.LOml
=
Bnuc
rearranges to R
=
4 "~R3
1.78 x 10 5 Z .
2
(1  3 cos e)[ 15.28]
( gl J.LNJ.LO ) 1/3 = (5 .5857) x (5.0508 x 10
4n:Bnuc
[ml
= +~ , e = 0,
yn
l
= gfJ.L N
27 J T I) X (4n: x 10 7 T2 r 1 m3»)
which
1/ 3
(4n:) x (0.715 x 1O3 T)
= (3.946 x 10 30 m 3 ) 1/ 3 = 1158 pm P15.17
glJ.LNJ.Lo =4n:R3
I.
The shape of spectra l line I(w) is related to the free induction decay signal G(t) by
10
I(w) = a Re
00
iw G(t)e (dt
where a is a constant and Re means take the real part of what follows. Calculate the lineshape corresponding to an oscillating, decaying function G(t) = coswote  (IT
I(w)
= aRe 10
1 Re = a 2
= ~a Re 2 =
1°
00
00
(e ,.~( .~
+ elL\) ( )e(I r + ',w( dt
I
I
roo {ei(lL\)+w+i/ r )( + ei (lL\)  wi IT)(\dt
10
~ aRe [ 2
iw G(t)e ( dt
I
i(WO
+ w + i/ r)
_ _ __ I __ ] iCwo  w  i/ r) .
310
STUDENT'S SOLUTIONS MANUAL
When wand
wo
are similar to magnetic resonance frequencies (or higher), only the second term in
brackets is significant (because
I(w)
~
I
(wo + w)
I 2
I i(WOw)2+I / r
I
I
«
I but
I
(wo 
w)
may be large if w
~ wo).Therefore,
a Re::
=  a Re ...,,,;,;;2 (wo  w) 2 + 1/ r 2 ar
2 which is a Lorentzian line centered on wo, of amplitude ~A r and width  at halfheight. r
P15.19
For nonweak fields in which the external magnetic field is comparable to the spinorbit coupling field of an unpaired electron it is necessary to include a spinorbit coupling term with coupling constant A [10.41] and apply the secondorder perturbation equation [9.65b]. The Hamiltonian is H = geYeBOsz YeBOlz + Al . s = YeB . (ges + I) + Al . s where vector notation is used for the electron orbital and spin angular momentum. The first order perturbation equation [9.65a] gives E ( I)
=
YeB . (ge(s ) + (I )) + A (I . s )
=
YegeB . (s )  YeB · (I ) + A (s ) . (I).
The expectation value of orbital angular momentum, (I ), equals zero for real states and (sz) gives
= ms which
The secondorder perturbation term is written using the ground state '0' and 'n' excited states with the energy difference !lEno = En  Eo [9.65b], which is positive.
The numerator may be expanded and simplified by discarding secondorder terms in Bo and negligibly small.
E(2)
= _L n",O
(OI{YeBolzlln )(nIAI · s lO) + (OIAI · sln)(nl{YeBOlzl IO) !lEno
_ " AYeBo L n",O
(Ollzln )(nl/ · slO) + (011· sln)(nllzIO ) !lE nO
Sz
as
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
311
The last manipulation uses the assumption that the local field is parallel to the applied field (i.e., I . S = lzsz). Combination of the first and secondorder perturbation estimates gives
E
(spin) 

E(I)
+ E(2)


_
~&
8
"+ 2A ~ 8 O~I'~ , , " (Ollzln)(nllzIO)
o~u
n;loO
Comparison with the effective spin Hamiltonian, _ gge
"
2A~
(Ollzln)(nllzIO)
n;loO
I':.Eno
= gYeBosz, indicates that
E(spin)
.
I':.E"o
g increases with increasing strength of the spinorbit coupling (A) and with decreasing excitation energy
(I':.E"o). This analysis is presented on p. 434 of P.w. Atkins and R.S. Friedman, Molecular quantum mechanics, 3rd edn, Oxford University Press, 1997.
Solutions to applications
P15.21
(8 nucl)
=
g/J1NJ10m/
4nR
3
J~max (l
 3 cos2 8) sin 8d8 J;maxsin 8 d8
=;Q,.
The denominator is the normalization constant, and ensures that the total probability of being between
o and 8max is 1.
Xmax (I X
Xmax 
If 8max
=
n (complete rotation), cos 8max
x~ax) 1
= I
=
g/J1NJ10m/
4
nR
and (8nu cl)
3
= O.
2
(cos 8max
+ cos8max )
.
312
STUDENT'S SOLUTIONS MANUAL
If emax = 30°, cos 2 emax (B nuel )
+ cos emax
= 1.616, and
_ (5 .5857) x (5 .0508 x 10 271 T I) x (4n x 10 7 T2 r l m3 ) x (1.616) 
~~~~~~~~n_~~~~~~
(4n) x (1.58 x 10
10 m)3
x (2)
= 10.58 mT I· P15.23
The desired result is the linear equation [110 =
[E~ov~v
 K,
so the first task is to express quantities in terms of [110 , [Elo, ~V , 8v , and K , eliminating terms such as [I], [EI], [E], VI , VEl , and v. (Note: symbolic mathematical software is helpful here.) Begin with v: [11
V
= [11
[Ell
+ [Ell VI + [ll + [Ell VEl
=
[110  [Ell [110 VI
+
[Ell [110 VEl ,
where we have used the fact that total I (i.e. free I plus bound I) is the same as initial 1. Solve this so it must also be much greater than [Ell: [Ell
=
[I]o(v  VI) VEl 
VI
[Il08v ~'
where in the second equality we notice that the frequency differences that appear are the ones defined in the problem. Now take the equilibrium constant
K _ [E][Il _ ([Elo  [EI]) ([110  [EI])  [Ell [Ell
~
([Elo  [EI]) [110 [Ell
We have used the fact that total I is much greater than total E (from the condition that [110 it must also be much greater than [EI], even if all E binds 1. Now solve this for [Elo: E = K + [110 EI = (K + (110) ([Il08V) = (K [110 [1 [110 ~v [ 10
»
[Elo), so
+ [Ilo)8v . ~v
The expression contains the desired terms and only those terms. Solving for [110 yields:
[110=
~ K ,
which would result in a straight line with slope P15.25
[Elo~ v
and yintercept K if one plots [llo against 1/ 8v.
When spin label molecules approach to within 800 pm, orbital overlap of the unpaired electrons and dipolar interactions between magnetic moments cause an exchange coupling interaction between the spins. The electron exchange process occurs at a rate that increases as concentration increases . Thus the process has a lifetime that is too long at low concentrations to affect the 'pure' ESR signal. As the concentration increases, the linewidths increase until the triplet coalesces into a broad singlet. Further increase of the concentration decreases the exchange lifetime and therefore the linewidth of the singlet.
MOLECULAR SPECTROSCOPY 3: MAGNETIC RESONANCE
313
When spin labels within biological membranes are highly mobile, they may approach closely and the exchange interaction may provide the ESR spectra with information that mimics the moderate and high concentration signals below.
ESR spectrum of ditertbutyl nilroxide
Low concentration
Moderate concentration
Hi gher concentration
Hi gh concentration
Figure 15.8
P15.27
Assume that the radius of the disk is I unit. The volume of each slice is proportional to (length of slice x8x ) , Fig. IS.9(a). Length of slice at x
= cos e, e = arcosx.
x
= 2 sin e.
314
STUDENT'S SOLUTIONS MANUAL
x ranges from 1 to
+ l.
Length of slice at x = 2 sin(arcosx) .
•
Ox Ox ___ x
Figure 15.9(a)
2
1.5 MRI absorption
intensity f(xl
0.5
0 1
 0.5
0 x
0.5
Figure 15.9(b)
= 2 sin (arcos x) against x between the limits 1 and + l. The plot is shown above. The volume at each value of x is proportional tof(x) and the intensity of the MRI signal is proportional to the volume, so Fig. 15. 9(b) represents the absorption intensity for the MRI image of the disk.
Plotf(x)
Statistical thermodynamics 1. the concepts
16
Answers to discussion questions 016.1
Consider the value of the partition function at the extremes of temperature. The limit of q as T approaches zero, is simply gO, the degeneracy of the ground state . As T approaches infinity, each term in the sum is simply the degeneracy of the energy level. If the number of levels is infinite, the partition function is infinite as well. In some special cases where we can effectively limit the number of states, the upper limit of the partition function is just the number of states. In general, we see that the molecular partition function gives an indication of the average number of states thermally accessible to a molecule at the temperature of the system.
016.3
We evaluate {3 by comparing calculated and experimental values for thermodynamic properties. The calculated values are obtained from the theoretical formulas for these properties, all of which are expressed in terms of the parameter {3 . So there can be many ways of identifying {3, as many as there are thermodynamic properties. One way is through the energy as shown in Section 16.3(b). Another is through the pressure as demonstrated in Example 17.1. Yet another is through the entropy, and this approach to the identification may be the most fundamental. See Further reading for elaboration of this method.
016.5
An ensemble is a set of a large number of imaginary replications of the actual system. These replications are identical in some respects, but not in all respects. For example, in the canonical ensemble, all replications have the same number of particles, the same volume, and the same temperature, but not the same energy. Ensembles are useful in statistical thermodynamics because it is mathematically more tractable to perform an ensemble average to determine the (time averaged) thermodynamic properties than it is to perform an average over time to determine these properties. Recall that macroscopic thermodynamic properties are averages over the time dependent properties of the particles that compose the macroscopic system. In fact, it is taken as a fundamental principle of statistical thermodynamics that the (sufficiently long) time average of every physical observable is equal to its ensemble average. This principle is connected to a famous assumption of Boltzmann's called the ergodic hypothesis. A thorough discussion of these topics would take us far beyond what we need here. See the references under Further reading.
Solutions to exercises E16.1(b)
Ne  f3 ej 11;=
q
whereq
=L
e f3ej
316
STUDENT'S SOLUTIONS MANUAL
Thus
.
n2
1
= , 11£0 = 300cm 1 2
GIven 
nl
k = (1.38066
In
X
lcmI ) = 0.69506cm 1 K I 10 23 JK I) x ( 1.9864 x 10 23 J
(:~) = M/kT
T=
11£0
11£0
kln(n2/nl)
kln(nl/n2)
· 300cm l _ ~ = (0.69506cm I KI)ln(2) =622.7K~~
E16.2(b)
f3
A = h( 2rrm )
(a)
= (6.626
1/ 2
(
[16.19] = h _1_)
1/ 2
2rrmkT
10 34 J s)
X
x C2rr) x (39.95) x (1.6605 x 1O2;kg) x (1.381 x 1O23JKI) x T
)'/2
276pm (T / K) 1/2
V
q= 
(b)
A3
[16.19] =
(i) T = 300 K,
(1 00 x 10 6 m 3) x (T /K)3/2
.
A = 1.59
(ii) T = 3000K,
(2.76 x 10 10 m)3 10 11 m =
X
A = 15.04 pm
I,
'115.9pm'~
= 4.76 x lO 22 (T /K)3/2 q = 12.47 x 1026 1,
q = 17.82 x 1027 1
Question. At what temperature does the thermal wavelength of an argon atom become comparable to its diameter? E16.3(b)
The translational partition function is qtr
so
E16.4(b)
= hV3 (2kTrrm)3 /2
qXe qHe
q=
= (mxe)3 /2 = (131.3U)3/2 =1187.91 mHe 4.003 u
L gje/3e = 2 + 3e/3e + 2e/3€2 1
j
levels
f3£O
=
hey kT
1.4388(Y / cm I)
=
T/K
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
317
Thus q = 2 + 3e(1.4388 x I250/2000) + 2e(1.4388 x I300/2000) = 2 + 1.2207 + 0.7850 = 14.0061 E16.5(b)
E = U  U(O) = _'!.... dq =
q df3
=
'!....~(2 + q df3
_'!.. (3f:lefJE1 2f:2efJE2) = q
=
3e fJE1 + 2e fJE2 )
Nheq (3yefJhCVI +2yefJhCVz )
(NAhe) x {3( 1250 cm I) x (e(1.4388X 1250/2(00») 4.006 +2(l300cm l ) x (e(1.4388X I300/2000» )}
= (NAhe)
x (2546cm l )
4.006
= (6.022 x 1023 molI) x (6.626 x 10 34 Js) x (2.9979 x IO lO cms l ) x (2546cm I)/4.006 = 17.605 kJ molII E16.6(b)
In fact there are two upper states, but one upper level. And of course the answer is different if the question asks when 15 percent of the molecules are in the upper level, or if it asks when 15 percent of the molecules are in upper state. The solution below assumes the fonner.
each
The relative population of states is given by the Boltzmann distribution
(heY)
n 2 = exp (~E)   = exp  nl kT kT
n2
hey
so In =  nl kT
hey
ThusT=    kln(n2 / n l) Having 15 percent of the molecules in the upper level means 0.15 10.15
so
n2
= 0.088
nl
(6.626 x 1O 34 J s) x (2.998 x 1OIOcmsl) x (360cm l ) and T =           = 23   : :           (1.381 x 1O JKI) x (in 0.088) =1213KI E16.7(b)
The energies of the states relative to the energy of the state with mt = 0 are YN~, 0, + YN Ii = 2.04 X 10 27 J T I . With respect to the lowest level they are 0, YN Ii, 2YN Ii. The partition function is
q
=L
eE"ale / kT
states
where the energies are measured with respect to the lowest energy. So in this case
q = 1 + exp
YN~ ) + exp (2YN kT ~ ) ( ;;:r
YN ~,
where
318
STUDENT'S SOLUTIONS MANUAL
As ~ is increased at any given T , q decays from q = 3 toward q = I as shown in Figure 16.1 (a).
2
Figure 16.1(3) The average energy (measured with respect to the lowest state) is (£ )
=
'"
L.. slales
£
Slale e
E"ate / kT
I
+ YN ~ exp (YN ~/ kT) + 2YNrJ!$ exp (2YN ~/ kT) 1+ exp (YN ~/ kT ) + exp ( 2YN ~/ kT)
q
The expression for the mean energy measured based on zero spin having zero energy becomes YN ~
(I  exp (2YN ~/ kT»
1+ exp (YN ~/ kT) + exp (2YN ~/ kT) As
~
is increased at constant T, the mean energy varies as shown in Figure 16.1(b).
Figure 16.1(b) The relative populations (with respect to that of the lowest state) are given by the Boltzmann factor
exp
6.£) = (u
exp
(YN ~) kT
or
YN ~ _ (2.04 x 10 JT i ) x (20.0 T) Note that k l.381 X 10 23 J K  i 27
= 2 95 .
x 10 3 K
so the populations are 3
(3)
exp (2.95 x 101.0 K
(b)
exp (2.95 x 10298
K) K) =
= 10.9971
3
and
10.999991 and
K)) = K)) =
2(  2.95 x 10exp ( 1.0 K exp (2 (2.95 x 10298
3
3
10.9941
10.999981
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
E16.8(b)
(a) The ratio of populations is given by the Boltzmann factor n2 nl
("""E) = e25.0K/ T kT
= exp
n3 = e  50.0 K/T
and
nl
(I) At LOOK n2 nl
= exp (2S.0K) = 11.39 x 10 11 1 1.00 K · .
and n3 = exp (  SO.OK) = 11.93 x 10 22 1 nl 1.00 K . (2) At 2S .0 K n2
=exp nl
(SO.OK) ~ =~ 25.0K
( 2S.0K) ~ =~ 2S .0 K
and

(25 .0K) ~ =~ lOOK
and
~ n3 =exp (50.0K) =~ nl lOOK
n3
nl
=exp
(3) At 100 K n2
=exp nl
(b) The molecular partition function is q =
L
eESla,elkT
= 1+ e 25.0K/T + e 50.0K/T
states
At 25.0 K, we note that e2S.0K/T = e I and eSO.OK/ T = e 2 q = I +e I +e 2 =ll.s031
(c) The molar internal energy is Urn = Urn (0)
_:A G;)
where/3 = (kT)  1
So Um = Urn(O)  NA (2S.0K)k (e2S.0K/T + 2eSO.O K/T) q At 25.0 K 1023 mol  I) x (25 .0K) x (1.381 x 10 23 J K I) 1.503 I 2 x (e +2e ) (6.022
X
= 188.3 J molI I (d) The molar heat capacity is CV.rn =
( aa~m ) v
= NA(25.0 K )k aa T
=NA(25.0K)k x
_
(2~~2K
~
(e25.0K/T + 2e50.0 K/T)
(e 2S.0 K/T +4eSO.O K/T)
~ ( e  2S.0 K/T + 2e SO.OK/ T) aq ) ~ aT
319
320
STUDENT'S SOLUTIONS MANUAL
where aq = _25_.0_K aT
so Cv
=
.m
(e 2S.0 K / T
+
2eSO.OK / T)
T2 NA(25 .0K) 2k ( (e2S.0K/T + 2e SO.OK/T)2) e 2S.0K / T +4e SO.O K / T _    _ _ _ __ T 2q q
At 25 .0K Cv m ,
=
(6.022 x 1023 mol  I) x (25.0K)2 x (1.381 x 1O 23 JK  I) (25.0K)2 x (1.503) I 2 x (e I + 4e2 _ (e + 2e )2) 1.503
;::::::~~;::_::_:::':::.:..
= 13.53 J K I molI I (e) The molar entropy is
At 25.0K Sm =
88.3Jmol 1 25.0 K + (6.022
X
1023 molI ) x (1.38 1 x 10 23 J K I) In 1.503
=16.92JK l mol  1 1
E16.9(b)
no nl
Set 
no
I e
=  and solve for T.
In
G)
T
heB = ,".,.
= In3 +
(:;B)
k( I+ln3)
6.626
X
10 34 Js x 2.998 x lO iO cms 1 x 1O.593cm 1 +1.38 1 x 1O 23 JK  1 x (1 + 1.0986)
= 17.26 K 1 E16.10(b) The SackurTetrode equation gives the entropy of a monatomic gas as
whereA=
h
~
v2kTrrm
(a) At 100 K
6.626
X
10 34 J s
A~=
 12(1.381 x 1O 23 JKI) x (lOOK) x rr(131.3u) x (1.66054 x 1O 27 kgu  l )jl /2 = 1.52 x 10 11 m
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
_I
= (8 .3145J K
and Sm
_I mol
) In
321
I 2 23 (eS/ ( 1.381 x 10 J K ) x ( lOOK») (1.013 x lOs Pa) x (1.52 x 10 II m)3
I
= 1147 J K I mol  I
(b) At298.15K
1\
6.626 x 10 34 J s
= =
/2 (1.38 1 x 10
23
1/2 JK  I) x (298.15K) x JT(l31.3u) x (1.66054 x 1O 27 kgu  I
)1
8.822 x 10 12 m
and
Sm
= (8.3 145JK
_I
_I mol
)In
(e S / 2( 1.381 x 10 23 J K I) x (298.15 K») (1.013 x IOsPa) x (8.822 x 1O 12 m) 3
= 1169.6J K I mol  I I E16.11(b)
q
=
1
e{Je
I  e hc{Jv
_ (1.4388cmK) x (321 cm I) 600 K hcf3v =
Thus q
=
I  eO.76976
=

= 0.76976
1.863
The internal energy due to vibrational excitation is
U  U(O)
=
N s e{Je I _ e{Je Nheve  hcv{J I  ehcv{J
Sm
and hence NAk
=
NhcV ehcv{J  1
::::; = (0.863) x (Nhc) x (32 1 cm I)
U  U (0) + Inq NAkT
= (0.863)
x
( he )
kT
I

x (321 cm  ) + In(1.863)
_ (0.863) x ( 1.4388 Kcm) x (32Icm l ) 600K +In(1.863)
= 0.664 +
0.62199
=
1.286
and Sm = 1.286R = 110.7 J K I mol  I I E16.12(b) Inclusion of a factor of (N!)  I is necessary when considering indistinguishable particles. Because of
their translational freedom , gases are collections of indistinguishable particles. The factor, then , must be included in calculations on I (a) C02 gas I.
322
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P16.1
Number of configurations of combined system, W
= (1020)
W
x (2 x 1020)
s = kin W[16.34]; S
= k In (2
x 1040)
= 92.8 x
S,
= 12 X
1040
= kIn WI ;
=
W, W2.
I.
S2
= kin W2 .
= k{ln 2 + 40 In 10) = 92.8k
(1.381 x 10 23 J K' )
= 11.282 x
10 2' J K  '
I.
s, = k In(1020) = k{20 In 10) = 46.lk = 46.1 S2
x ( 1.381 x 10 23 J K  ' )
= k 1n(2
= 46.7 x
x 1020)
= I 0.637
X
10 2' J K'
I.
= k{ln 2 + 20 In 10) = 46.7k
( 1.381 x 10 23 J K' )
= I 0.645
X
10 2' J K  '
I.
These results are significant in that they show that the statistical mechanical entropy is an additive property consistent with the thermodynamic result. That is, S = S, + S2 = (0.637 X 10 2' + 0.645 x 10 21) J K' = 1.282 X 10 2 ' J K'.
S = kin W [16.34].
P16.3
Therefore,
) (auas) v = Wk (aw au v or
But from eqn 3.45
( au) as v = T . So,
Then
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
323
Therefore, ~w
~U
W
kT
 ~ 
 (1.381 x 10 23 J K
= 2.4
1
P16.5
=
q
x 10
v
25
.1\ =
.1\3'
I)
x 298 K
1
h
(21tmkT) 1/
2
[16.19, f3 = kiT]'
and hence T
= (~)
x
21tmk
(i)2/3 V
10 34 J s)2 ) = ( (21t) x (39.95) x ( 1.6605 x 1027 kg) x (1.381 x 10 23 J K I) (6.626
X
10 )2/3 x ( 1.0 x 10 6 m3 = 13.5 x 10 15 K 1[a very low temperature].
The exact partition function in one dimension is 00
q
= Le(1I2I) h2P /8mL2. 11=1
For an Ar atom in a cubic box of side 1.0 em, h 2 f3
(6.626
X
10 34 J s)2
(8) x (39.95) x (1.6605 x 10 27 kg) x (1.381 x 10 23 J K 1) x (3.5 x 10 15 K) x (1.0 x 10 2 m)'
8mL2
= 0.171. 00
Then q
=L

e O. 171 (1I
2
I)
= 1.00 + 0.60 + 0.25 + 0.08 + 0.02 + ... = 1.95.
11=1
The partition function for motion in three dimensions is therefore q
= (1.95)3 = [iliJ.
COMMENT. Temperatures as low as 3.5 x 10 15 K have never been achieved. However, a temperature of
2 x 10 8 K has been attained by adiabatic nuclear demagnetization (Chapter 3).
Question. Does the integral approximation apply at 2 x 10 8 K? P16.7(b)
(a) q
= Lgje P' j [16.9] = L j
We use he f3 = (i) q
gjehcPvj.
j
I I 1 at 298 K and I at 5000 K. Therefore, 207 cm3475 cm
= 5 + e4707/207 + 3e 475 1/207 + 5e10559/207
= (5) + (1.3
x 10 10)
+ (3.2 x
10 10 ) + (3.5 x 10 22 )
= 15.00 I.
324
STUDENT'S SOLUTIONS MANUAL
(ii) q
= 5 + e 4707/3475 + 3e475 1/ 3475 + 5e10559/ 3475 = (5)
+ (0.26) + (0.76) + (0.24)
g'e fJsj
= _1_ _ =
(b) Pj
1
Srn
[16.7, with degeneracy gj included]
q
= ~ = [QQJ at 298 K and !0.80 !at 5000 K .
Therefore, Po
(c)
!.
g 'e  hcf3vj
q
P2 =
= ! 6.26
q
3e475 1/ 207 5.00 3e475 1/3475
P2
=
=
Urn  Urn (0)
6.26
T
= 16.5 x
I
10 1I at 298 K .
= [QJ3J at 5000 K.
+Nklnq [16.35].
We need Urn  Urn (0), and evaluate it by explicit summation Urn  Urn (0) = E = N A q
L gjCje{3£j
[16.28 with degeneracy gj included] .
j
In terms of wavenumber units (i) Urn  Urn (0) = _I_tO NAhc
(ii) UmUrn(O) NAhc
+ 4707 emI
x e  4707/207
+ ... } =
4.32 x 1O 7cm l ,
5.00
= _I_tO + 4707 emI
x e  4707/ 3475
+ ... } = 1178 emI .
6.26
Hence, at 298 K Urn  Urn(O) = 5.17 x 10 6 J mol  I
and at 5000 K Urn  Um(O) = 14. IOkJ mol  I.
It follows that (i) Srn =
5. 17 x 10 6 J mOl  I) ( 298 k
I + (8.314 J K Imol ) x
(In 5.00)
= 113.38 J K  I mol I I[essentially R In 5]. (ii) Srn
P16.9
q
=
3 (14.09 x 10 J mOl  I) 5OO0K
= Lgje{3£j j
Pi
gi e {3£i
[16.9]
+ (8.314 J K I mol  I)
= Lgje hc{3 Vj. 1
gie  hcf3vi
= q [16.7] = =q
x (In 6.26)
= 1 18.07 J K I mol  I I.
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
325
We measure energies from the lower states, and write q
= 2 + 2e hc {3v = 2 + 2e( 1.4388 x I21.1 )/(T/ K) = 2 + 2e 174.2/ (T/ K).
This function is plotted in Fig. 16.2. (a) At 300 K,
Po
= ~ = I + e:74.2/300 = I0.641,
PI = I  Po = I 0.361·
4
q 3
2
o
200
400
600
800
1000
T/ K
Figure 16.2
(b) The electronic contribution to U m in wavenumber units is U
I dq
 Um(O)
m _:"''= NAhc
hcq dfJ
=
[16.3 Ial
2ve hc {3v
= q
( 121.1 cm  I) x (e174.2/300) I + e 174.2/300
I
= 43.45 cm
which corresponds to 10.52 kJ mol  I I. For the electronic contribution to the molar entropy, we need q and U m as at 300 K. These are 300 K U m  Um(O)
q
0.518 kJ mol  I 3.120
500K 0.599 kJ mol I 3.412

U m (0) at 500 K as well
326
STUDENT'S SOLUTIONS MANUAL
Then we form Sm
=
Urn  Um(O)
+ R In q [16.35].
T
At300K
Sm=
At1500K
I 518JmOI ) 300K +(8.3 141JK l mol l )x(ln3.120)=11.2JK l mol l
I
(
Sm=
I 518JmOI ) 500K +(8.3141JK l mol 1) x (In 3.412) = 11.4JK 1 mo1 1 .
I
(
P16.11
]
1
At lOOK, hcf3 = 69.50cm 1 and, at 298 K, hcf3 = 207.22cm I ' Therefore, at 100 K, (a) q = 1 + e 213 .30/69.50 + e 435 .39/69.50 + e636.27/69.50 + e845.93/69.50 = 11.0491 and at 298 K (b) q = 1 + e 213 .30/207.22 + e 425.39/207.22 + e636.27/207.22 + e845.93/207.22 = ehc{3v;
In each case, Pi =   [16.7] . q
Po =
~q
= (a) I 0.9531,
(b) I 0.6451·
e hc{3V\ PI =   = (a)1 0.0441, q
(b) I 0.230 I·
e  hc {3V2
P2 =   = (a)1 0.0021, q
(b)10.083
I.
For the molar entropy we need to form Um  Um(O) by explicit summation: Urn  Urn (0)  NA q
'~ " i
. {3s; _ NA fie q
"'hcv,e. ~ i
hc{3v;
[1629 . , 1630] .
= 1123 J mol I (at 100 K) 1,11348 J molI (at 298 K) Srn =
Urn  Urn(O) T + R In q [16.35].
(a) Sm =
1 123 Jmo1I I 1I lOOK + R In 1.049 = 1.631 K mo1.
(b)Srn=
1348Jmo]1 I 1 1I 298K +Rln1.55 = 8.17JK mo1.
Solutions to theoretical problems P16.13
(a) W =
N! nl !n2! . . .
[16.1] =
I.
5! = 0!5!0!0!0!
OJ.
I.
[ill.
I
STATISTI CAL THERMODYNAMICS 1: THE CONCEPTS
327
(b) We draw up the following table. 0
& 2&
4 3 3 2 2 1 0
0 I 0 2 1 3 5
w=
3& 4& 5&
0 0 1 0 2 1 0
0 0 1 I 0 0 0
0 I 0 0 0 0 0
N!
111 !112!··· 5 20 20 30 30 20
I 0 0 0 0 0 0
I
I I
I
The most probab le configurations are (2, 2, 0, 1, 0, 0 I and (2, 1, 2, 0, 0, 0 I jointly.
P16.15
(a)
I1j
~
= e fJ(£j  £o) = e  fJj£, which implies that jl3& = In I1j 
In 110 and therefore that In I1j
= In 110 
j
k& .
T
&
Therefore, a plot of In I1j agai nst j should be a straight line with slope  kT . Alternatively, plot In Pj againstj, since Inpj
j&
= const 
 . kT
We draw up the fo llowing table using the information in Problem 16.8.
o
j
2 0.69
4 1.39
2
3
2 0.69
o
I
[most probable configuration]
&
These are points plotted in Fig. 16.3 (full line). The slope is  0.46 and, since he corresponds to a temperature
T
=
(50cm
l)
x (2 .998 x 10 10cms l ) x (6.626 x 1O 34 J s) (0.46) x ( 1.381 x 1O 23 JK I)
=
= 50 cm I, the slope
~
160K.
(A better estimate, 104 K represented by the dashed line in Fig. 16.3, is fo und in Problem 16. 17.) (b) Choose one of the weight 2520 configurations and one of the weight 504 configurations, and draw up the fo llowing table.
W
= 2520
j
0
I1j
4 1.39 6 1.79
Inl1j
W=504
I1j
In I1j
2 3 1.10 0  00
I 0 I 0
Inspection confirms that these data give very crooked lines.
3
4
0
I 0 1 0
 00
1 0
328
STUDENT'S SOLUTIONS MANUAL 1.6
1.2
0.8 ;,,~
.E 0.4
0
{).4
0
3
2
4 I
00 .+ P16.17
dj3 =
=
Hence, efJ'
N   [16.3Ib],
withq=
I P [16 . 12] . 1 e ,
 ee P' qdfJ = 1 e P" 1 dq
dlnq
ae
dlnq dfJ
=
(a) U  U(O)
Figure 16.3
U  U(O)
ee(fJ')
N
1 e fJ '
1+a, . . that, fJ = In 1 (1 + =Implymg e
a
I)
 . a
1 For a mean energy of e, a = I, fJ =  In 2, implying that e
T = _ e _ 1n2 = (50cm
l
k~2
1 (b) q = 1  e fJ , =
1
(
) =
a
1
x
)
(~) k~2
= 1104 K
I.
~
1+ a .

I+a S U  U(O) (c)  = + ln q[16.35]=afJe +ln q Nk NkT
= a ~ (I +
~) +
In(I + a)
= a In(l
+ a)  a In a + In ( 1 + a)
=1 ( I +a) ln(I +a ) a lnal· . When the mean energy IS e, a P16.19
P=kT(alnQ)
av
= 1 and then ~ ~.
[17 .3] T .N
= kT (a In(qNIN!») av
[16.45b] T,N
= NkT (a In q )
= kT (a[N Inq InN!])
av
T .N
av
T.N
STATISTICAL THERMODYNAMICS 1: THE CONCEPTS
=NkT
aIn(V /A3») av T.N
(
3
=NkT
_
329
a[lnV  InA ]) (
av
N~T
=NkT
(alnV) av
T .N
T .N
IPV = NkT = nRT I.
or
Solutions to applications P16.21
= e1V(r)  V(ro)} / kT
At equilibrium N(r)/V N(ro)/V
=
Since V(r)
N(oo)/V N(ro)/V
GMm /r, V(oo)
[16 .00]
= 0 and [Note:
V(r) is potential energy, V is volume]
= eV (ro) / kT
which says that N(oo)/V <X e V(ro) / kT = constant. This is obviously not the current distribution for planetary atmospheres where lim N (r) / V = O. Consequently, we may conclude thatthe earth's atmosphere, r> oo
or any other planetary atmosphere, cannot be at equilibrium. P16.23
Each protein binding site can be represented as a distinct box into which a ligand, L, may bind. All possible configurations are shown in the following table and the configuration count of i indistinguishable ligands n! being placed in n distinguishable sites is seen to be given by the combinatorial: C(n, i) = . . (n I)!I! C(4, 0)
L
L
P16.25
1
L L
L L L L
C(4, 2)
L
=
4! (4 _ 2)!2!
= 6 conformations
L L
L L
L L L L
L L L
(n  i
Pi=
I conformation
L
L L
L
=
C(4,1)= (4 ~ ) !I! = 4 conformations
L
L L L
4! (4  O)!O!
4'
L
L L L
=
4! C(4,3) = (4 _ 3)!3! = 4 conformations
1
+ I )asi q
C(4,0)
=
4! (4 _ O)!O!
= I conformation
(n i+ l )as i
(i)
= L;~o ipi.
(a) The fraction distribution of molec ules with i coiled residues depends dramatically upon the value of the stability parameter s. When s < I , low values of i are observed but large i values are observed
330
STUDENT'S SOLUTIONS MANUAL
when s > I. Thus, when s < I, the polypeptide is largely helical and, when s > I, it is more of a random coil. See Fig. 16.4(a) . Fraction of molecules with i coiled residues when
0.25
(J
=
0.050
.r..,,
02
0.1.5
s= 1.5 0.1
OD.5
10
1.5
L
Figure 16.4(a)
(b) The (i) plot, Fig . 16.4(b), shows that for s < 0.5 the polypeptide model is helical and little changed as s is varied. At the other extreme, for s > 1.5 the polypeptide is largely a random coil which changes only slightly with variance of s. The mean number of coiled residues changes rapidly in the middle range of 0.5 < s < 1.5 giving an overall sigmoidal dependence upon s. Dependence of the mean number of coiled residues upon the stability parameters. From left to right: s=O.OOO 1, 0.001 , 0.0 I, and 0.05
~r~~~~
1.5
(i )
10
s
Figure 16.4(b)
Statistical thermodynamics 2: applications
17
Answers to discussion questions 017.1
An approximation involved in the derivation of all of these expressions is the assumption that the contributions from the different modes of motion are separable. The expression qR = kT / heB is the high temperature approximation to the rotational partition function for nonsyrnmetrical linear rotors. The expression q v = kT / hev is the high temperature form of the partition function for one vibrational mode of the molecule in the harmonic approximation. The expression qE = gE for the electronic partition function applies at normal temperatures to atoms and molecules with no low lying excited electronic energy levels.
017.3
Residual entropy is due to the presence of some disorder in the system even at T = O. It is observed in systems where there is very little energy differenceor nonebetween alternative arrangements of the molecules at very low temperatures. Consequently, the molecules cannot lock into a preferred orderly arrangement and some disorder persists.
017.5
Equations of state can be thought of as expressions for the pressure of a gas in terms of the state functions, n, V, and T . They are obtained from the expression for the pressure in terms of the canonical partition function given in eqn 17.3. p
= kT
(aavQ) . In
T
Partition functions for perfect and imperfect gases are different. That for the perfect gas is given by Q = qN / N! with q = V / A 3 . There is no one form for imperfect gases. One example is shown in Selftest 17.1 . Another which can be shown to lead to the van der Waals equation of state is I 2rrmkT  2 ) N! ( h
Q= 
3N / 2
(VNb)e
aN2/ kTV
.
For the case of the perfect gas there are no molecular features in the partition function, but for imperfect gases there are repulsive and attractive features in the partition function that are related to the structure of the molecules. 017.7
See Justification 17.4 for a derivation of the general expression (eqn 17 .54b) for the equilibrium constant in terms of the partition functions and difference in molar energy, t::;,rEO , of the products and reactants in a chemical reaction. The partition functions are functions of temperature and the ratio of partition functions
332
STUDENT'S SOLUTIONS MANUAL
in eqn 17.54b will therefore vary with temperature. However, the most direct effect of temperature on the equilibrium constant is through the exponential term e I':. ,Eo/ RT. The manner in which both factors affect the magnitudes of the equilibrium constant and its variation with temperature is described in detail for a simple R ;= P gas phase equilibrium in Section 17.8(c) and justification 17.5.
Solutions to exercises E17.1(b)
CV.m = !(3 + vR+ 2vt)R [17.35] with a mode active if T >
11M .
= !(3 + 3 + O)R = 3R [experimental = 3.7R] CV,m = !(3 + 3 + 2 x l)R = 4R [experimental = 6.3R] CV ,m = !(3 + 2 + O)R = ~ R [experimental = 4.5R]
(a) 0 3: CV,m
(b) C2H6: (c) C02:
Consultation of the Herzberg references in Further reading , Chapters 13 and 14, turns up only one vibrational mode among these molecules whose frequency is low enough to have a vibrational temperature near room temperature. That mode was in C2H6 , corresponding to the "internal rotation" of CH3 groups. The discrepancies between the estimates and the experimental values suggest that there are vibrational modes in each molecule that contribute to the heat capacityalbeit not to the full equipartition valuethat our estimates have classified as inactive. E17.2(b)
The equipartition theorem would predict a contribution to molar heat capacity of !R for every translational and rotational degree of freedom and R for each vibrational mode. For an ideal gas,
Cp,m = R + CV,m' So for C02 With vibrations
Cv,m/ R =3(!)+2(!)=(3 X 46)=6.5
Without vibrations CV,m/ R
Experimental y =
and
= 3 (!) + 2 (!) = 2.5
37. 11 Jmol  1K 1 (37 . 11  8.3145) J mol  1K1
=
~
7.5 6.5
y==~
and
3.5 2.5
~
y==~
[ill 1.29
The experimental result is closer to that obtained by neglecting vibrations, but not so close that vibrations can be neglected entirely. E17.3(b)
The rotational partition function of a linear molecule is [Table 17.3]
qR
= 0.6950
x
a
T/ K (B / cm 
= I)
(0.6950) x (T / K) 2 x 1.4457
(a) At 25 °C:
qR = (0.2404) x (298) = ~
(b) At 250°C:
qR
= (0.2404)
x (523)
= @]
= 0.2404(T / K)
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
E17.4(b)
333
The symmetry number is the order of the rotational subgroup of the group to which a molecule belongs (except for linear molecules, for which a = 2 if the molecule has inversion symmetry and 1 otherwi se). (a) C02 : full group Dooh; subgroup C2 ; hence a =
[I]
(b) 0 3: full group C2v; subgroup C2 ; a =
[I]
(c) S03: full group D3h; subgroup {E , C3,
C~ , 3C2}; a =
GJ
(d) SF6: full group Oh; subgroup 0; a = 1241
(e) A12Cl6 : full group D2d ; subgroup D2; a = E17.S(b)
[I]
The rotational partition function of a nonlinear molecule is [Table 17.3] R 1.0270 (T / K) 3/2 1.0270 x 298 3/ 2 q =  a (ABC / cm  3) 1/2 = (2) x (2.02736 x 0.34417 x 0.293535) 1/2 [a = 2] = 15837 1
The hightemperature approximation is valid if T >
~,
where
hc (ABC) 1/3
eR=k
(6.626 x 10 34 J s) x (2.998 X 1010 cm SI) x [(2.027 36) x (0.344 17) x (0.293535) cm 3 jl / 3 1.381 x 10 23 J K I = 10.8479KI E17.6(b)
qR = 5837 [Exercise 17.5(b)] All rotational modes of S02 are active at 25 °C; therefore R
R
Urn  Urn (O)
ER
SR
rn
= T
+R
= ~R + E17.7(b)
3 = E R = '2RT InqR
Rln(5837) = 184.57 J K I mol I I
(a) The partition function is
states
leve ls
where g is the degeneracy of the level. For rotations of a symmetric rotor such as CH3CN, the energy levels are EJ = hc[BJ(J + I ) + (A  B)K2] and the degeneracies are gJ.K = 2(21 + 1) if K t= 0 and 2J + I if K = O. The partition function , then, is q = 1+ t(21 + l)e  IhcBJ (J+I )/ kTl J= I
(I t + 2
K= I
e  IhC(AB)K2/kTl)
334
STUDENT'S SOLUTIONS MANUAL
To evaluate this sum explicitly, we set up the following columns in a spreadsheet (values for A 5.28 cm I , B = 5.2412 cm  I , and T = 298.15 K) J
J(J + I )
2J + I
e  hcBJ (J+I )/ kT
0 2 3
0 2 6 12
I 3 5 7
I 0.997 0.99 1 0.982
82 83
6806 6972
165 167
4. 18 x 10 5 3.27 x 10 5
e  {hc(A B)K 2/ kTi
K sum
J sum
8.832 23.64 43.88
I 0.976 0.908 0.808
2.953 4.770 6.381
9.832 33.47 77 .35
0.079 0.062
8 x 10 71 2 x 10 72
11 .442 11 .442
7498 .95 7499.01
Jterm
=
The column labeled K sum is the term in large parentheses, which includes the inner summation. The J sum converges (to 4 significant figures) only at about J = 80; the K sum converges much more quickly. But the sum fai ls to take into account nuclear stati stics, so it must be divided by the symmetry number (a
= 3).
= 12.50 x
At 298 K, qR
103 1. A similar computation at T
= 500 K
yields qR = 15.43 x 103 1.
(b) The rotational partition functio n of a nonlinear molec ule is [Table 17.3 with B R
q
=
1.0270 (T / K)3/2  a  (A BC/c m  3 ) 1/ 2
1.0270 (T / K )3/2 3 (5.28 x 0.307 x 0.307)1 /2
At 298 K, qR
= 0.485 x 298 3/ 2 = 12.50 x
At 500 K , qR
= 0.485
x 5003 / 2
= 15 .43
= C]
= 0.485
x (T / K)3/ 2
103 1
x 103 1
The hightemperature approximation is certainly valid here. E17.8(b)
The rotational partition function of a nonlinear molecule is [Table 17.3] R
1.0270
(T / K)3/2
q 
a
(ABC/cm3) 1/2
(a) At 25 °C,
qR
qR
(b) At 100 °C ,
E17.9(b)
= 1.549 X =
1.0270 x (T / K)3/ 2 (3. 1752 x 0.3951 x 0.3505) 1/2
= 1.549 x
(T / K)3/2
(298)3/ 2 = 17.97 x 1031
1.549 x (373)3/ 2 = 11.12 x 104 1
The molar entropy of a collection of oscillators is given by Srn
=
Urn  Urn(O) +klnQ [17. 1]
T
= efJhcv _
T
Ov
hcv
where (&)
= NA(& ) +Rlnq
I
= k eOv / T
_
1 [17.28], q
and Ov is the vibrational temperature hcv/k. Thus
Srn
=
R(Ov/ T) _ RI ( I _ Ov / T) n e  I
vv./ T eA
=
I I_
e fJhcv
1 I _ e Ov / T [17 .19]
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
335
A plot of Sm / R versus T lev is shown in Figure 17.1.
2.5 ~
'i;
2
""' 1.5
0.5
o
o
4
2
6
10
8
TIO v
Figure 17.1
The vibrational entropy of ethyne is the sum of contributions of this form from each of its seven normal modes. The table below shows results from a spreadsheet programmed to compute Sm / R at a given temperature for the normalmode wave numbers of ethyne. T
= 298 K
T
= 500 K
ii/em  I
ev/K T /8v
Sm/ R
T/ev
Sm/ R
612 729 1974 3287 3374
880 1049 2839 4728 4853
0.216 0.138 0.000 766 0.000 00217 0.000 00146
0.568 0.479 0.176 0.106 0.103
0.554 0.425 0.0229 0.000 818 0.000 652
0.336 0.284 0.105 0.0630 0.0614
The total vibrational heat capacity is obtained by summing the last column (twice for the first two entries, since they represent doubly degenerate modes). (a) At 298 K, Sm
= 0.708R = 15.88 J molI
(b) At 500 K, Sm
=
1.982R
K 1 1
= 116.48 J mol  I K I 1
E17.10(b) The contributions of rotational and vibrational modes of motion to the molar Gibbs energy depend on
the molecular partition functions G m  Gm(O) = RT In q [17.9; also see Comment to Exercise 17.6(a)]
The rotational partition function of a nonlinear molecule is given by
qR =
~ a
(kT)3 /2 (~)1 /2 = 1.0270 ( (T/K)3 )1 /2 he ABC a ABC/ cm 3
and the vibrational partition function for each vibrational mode is given by
v q
=
I
1 e(J / T
where 8
hcii
=
k
=
1.4388 (ii/em I)
:=::::
(T /K)
336
STUDENT'S SOLUTIONS MANUAL
R _
At 298 K
q

1.0270 ( 298 3 )  2  (3.553) x (0.4452) x (0.3948)
1/ 2
3
= 3.35 x 10
and C~,  C~(O) = (8.3 145 J mol I K I) x (298 K) In 3.35 x 10 3
= 20.1 x 103 J mol  I = 120.1 kJ molI
I
The vibrational partition functions are so small that we are better off taking
Inq; "'" e II .4388(IIIO)/298} = 4.70 x 10 3 In
qi "'" e 1 1.4388(705 )/ 298} = 3.32 x 10 2
Inqj "'" e 11.4388(1042)/298} = 6.53 x 10 3
so
C~  C~(O) =  (8.3 145 J mol  IK  I) x (298 K)
x (4.70 x 10 3 + 3.32 x 10 2 + 6.53 x 10 3) =IIOJmol  1 =1O. llOkJmol1 1
where g = (2S + 1) x
E17.11(b)
I {2
for 2: states for
n, ~,
...
states
The 32: term is triply degenerate (from spin), and the I ~ term is doubly (orbitally) degenerate. Hence
q=3+2e fJ e At 400 K {3e =
( 1.4388 cm K) x (7918.1 cm I) = 28 .48 400K
Therefore, the contribution to C m is COl  COl (0) =  RT In q [Table 17.4 for one mole 1
RT Inq =  (8.3 14J K I mol  I) x (400 K) x In (3 +2 x e 2848 )
= (8.3 14J K I molI ) x (400 K) x (In 3) = 13.65 kJmol  1
COMMENT. The contribution of the excited state is negligible at this temperature.
I
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
=
E17.12(b) The degeneracy of a species with S
Srn
=
Urn  Um(O)
T
+R
In q
337
~ is 6. The electronic contribution to molar entropy is
= R In q
(The term involving the internal energy is proportional to a temperaturederivative of the partition function, which in tum depends on excited state contributions to the partition function ; those contributions are negligible.)
Srn E17.13(b) Use Sm
= (8.3145 Jmol 1 K 1) In6 =[ 14.9Jmol 1 K 1 [ =Rlns [17.52bl
Draw up the following table 0
n:
s
1
2
6
Sm / R 0 1.8
3
4
5
0
m
p
a
b
c
0
m
p
6
6
6
2
6
6
1.8
3 1.1
6
1.8
1.8
1.8 0.7
1.8
1.8
3 1.1
6
6
1.8 0
where a is the 1,2,3 isomer, b the 1,2,4 isomer, and c the 1,3,5 isomer. E17.14(b) We need to calculate
Each of these partition functions is a product
with all qE
= 1.
The ratio of the translational partition functions is virtually 1 (because the masses nearly cancel; explicit calculation gives 0.999). The same is true of the vibrational partition functions. Although the moments of inertia cancel in the rotational partition functions , the two homonuclear species each have a = 2, so q
R 9 Br2)qR(SIBr2) qR (19BrS1 Br)2
C
= 0.25
The value of I'1Eo is also very small compared with RT, so
338
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P17.1
qE (a ) qE af3 v
Urn  Urn CO) = NA CV,rn = kf32

C~rn ) v
[17 .31a].
Let x = f3 £, then df3 = (1 /£) dx.
Therefore,
We then draw up the following table. TIK
(kT/hc)/mo\1
x CV,m/ R CV.rn / (J K  I mol I)
50
298
500
34.8 3.48
207 0.585
348 0.348
[Q;iliJ 2.9 1
10.0791 10.0291 0.654
0.244
Note that the double degeneracies do not affect the results because the two factors of 2 in
COMMENT.
q cancel when U is formed. In the range of temperature specified , the electronic contribution to the heat capacity decreases with increasing temperature. P17.3
The energy expression for a particle on a ring is
Therefore 00
00
m= OO
m= OO
STATISTICAL THERM ODYNAMI CS 2: APPLICATIONS
339
The summation may be approximated by an integration
q
~ .!.. / 00 e
00
a
U  U(O)
m
=
fn 2 / 2IkT dml
SOl
=
/ 00 e00
x2
d.x
~ .!.. ( 2n /;T) 1/2 = .!.. (2;/) 1/ 2 a
a
ti
= _N aInq = N ~ In.!.. ( 27r2 /)1 / 2 = !!. = ~NkT = ~RT af3
C V. ol =
.!.. ( 2IkT) 1/ 2 a ti 2
aU m ) ( aT
af3
ti f3
a
2f3
2
2
(N
ti f3
= NA) .
[  I I [ v = '2I R =4.2JK mol .
Urn  UOl(O) +Rlnq T
1 I (2nlkT) = R + Rln  22 a ti
1/ 2
1 1 ( 2n) x (5.341 x 1O 47 kgm 2) x (1.38 1 x 1O 23 JK I ) x (298»)1 /2 =R+Rln2 3 ( 1.055 X 10 34 J s)2
= ~R + 1.31R = 1.81R, orllS J K 1 molI 2
P17.5
I.
The absorption lines are the values of differences in adjacent rotational terms. Using eqns 13.25, 13.26, and 13.27, we have
F(J+l)F(J)=
E(J
+ I) 
E(J)
he
=2B(J+I )
for J = 0 , I , .... Therefore, we can find the rotational constant and recon struct the energy levels from the data. To make use of all of the data, one would plot the wavenumbers, which represent F(J + I )  F(J) , vs. J ; from the above equation, the slope of that linear plot is 2B. Inspection of the data show that the lines in the spectrum are equally spaced with a separation of 2 1. 19 cm I , so that is the slope: slope
= 21.19cm 1 =2B
so
B= 10.S95cm
l
.
The partition function is
00
q
= L (2J + l)efJE(J)
where
E(J)
=
heBJ(J
+ I ) [13 .25]
1=0
+ I is the degeneracy of the energy levels.
and the factor of 2J At 25°C , heBf3
00
q
=L
heB
=
(2 J
kT
=
6.626 x 10 34 J s x 2.998 X 10 10 cm S I x IO.S95cm  1 1.381 x 10 23 J K 1 x 298. 15 K
+ l )eO.OSII 2J(J+ I)
1=0
= 1 + 3e0.OSlI 2x 1x2 + Se 0.OS I1 2x2x3 + 7e 0.OS I1 2x3 x4 + .. . = 1 + 2.708 + 3.679 + 3.791 + 3.238 + ... = [ 19.90 [.
= 0 .05112.
340 P17.7
STUDENT'S SOLUTIONS MANUAL
The molar entropy is given by
Sm = Urn 
Um(O) +R(ln qm _ T NA
where Urn  Um(O) T
= NA (aln q ) a{3
I) and
v
qm NA
= q~qRqVqE NA
The energy term Urn  Urn (0) works out to be
Translation: Te
~ = 2.561 x 1O 2(T / K)5 /2 x (M/g mol I)3/2 [Table 17.3) NA
= 2.561
x 10 2 x (298)5/2 x (38.00)3/2 = 9.20 x 106
= ikT .
and (s T)
Rotation of a linear molecule:
0.6950 = __
qR
a
T/ K x    I [Table 17.3].
B/cm
The rotational constant is
Ii Ii B     ::.; 4rrcl  4rrCf.,l,R2 (1.0546 x 10 34 1 s) x (6.022 x 1023 mol  I) 4rr(2.998 x JOlOcms l) x x 19.00 x 103 kgmol l ) x (190.0 x 1O 12 m)2
(!
= 0.4915cm  1 so qR = 0.6950 x ~ = 210.7. 2 0.4915
= kT.
Also (sR) Vibration :
lexp(1.4388(ii/cm I) /( T / K»
= v
1.129. (6.626 x 10 34 ls) x (2.998 x IO lO cms l ) x (450.0 cm  l )
hciJ
(s )
= ehcv /kT _ = 1.149 X
I
exp(1.4388(450.0)/298)  I
10 211.
The Boltzmann factor for the lowestlying excited electronic state is
~p
(
I  exp( 1.4388(450.0) /298)
(1.60geY) x (1.602 x 10 19 leV I») (1.381 x 10 23 1 K I) x (298K)
=6 x
l
0 28
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
341
so we may take qE to equal the degeneracy of the ground state, namel y, 2 and (e E ) to be zero. Putting it all together yields
um 
U (0)
T
m
=
N ~ (lkT
T
)
+ kT + 1.149 x
10 21 J
=
2.2R +
N (1.149 X 10 2 1 J) _A_ _ _ _ __
2
= (2.5)
_I x (8.3 1451 mol
_I K
)
+
T
(6.022 x 1023 mol I) x (1.149 x 10 2 1 J) 298 K
= 23.11 J molI K I. R (In
~:
 1)
= (8.3 145 J mol I K I) =
P17.9
176.31 mol  I K I
x (In[(9.20 x 106 ) x (2 10.7) x (1.129) x (2)]  I}
and
S~ = 1199.4 J mol I
K I \.
(a) The probability distribution of rotational energy levels is the Boltzmann factor of each level, weighted
by the degeneracy, over the partition function (21
L
+ l)e hcBJ (1+I) / kT + l)ehcBJ(J+I) / kT
=;;::;;:::;=
(2J
[17 .13] .
J=O
It is conveniently plotted agai nst J at several temperatures usi ng mathematical software. This distribution at 100 K is shown in Fig. 17.2(a) as both a bar plot and a line plot.
Rotational distributions 0. 15 r    ,     ,      ,     ,      ,       ,    r       ,
lOOK 0.1 rI}(T)
Figure 17.2(a)
The plots show that higher rotational states become more heavily populated at higher temperature. Even at lOOK the most populated state has 4 quanta of rotational energy; it is elevated to 13 quanta at 1000 K. Values of the vibrational state probability distribution, y
p" (T)
e <} / kT
=  qy  =
e  IIhcv / kT
1  e I,e ll / kT [17.19]
are conveniently tabulated against v at several temperatures. Computations may be discontinued when values drop below some small number like 10 7 .
342
STUDENT'S SOLUTIONS MANUAL
p~(T)
v
lOOK
300K
600K
1000K
3.02 X 10 5 9.15 x 10 10
0.095 5.47 x 10 3 3.01 x 10 5 1.65 x 10 7
0.956 0.042 1.86 x 8.19 x 3.61 x 1.59 x
0 2.77 x 10
14
2 3 4 5
10 3 10 5 10 6 10 7
Only the state v = 0 is appreciably populated below 1000 K and even at 1000 K only 4 % of the molecules have I quanta of vibrational energy. (b) The classical (equipartition) rotational partition function is
kT 1eB
R
T
qclassical(T) = I  = 
where
~
8R
( I)
[17.15b].
is the rotational temperature. We would expect the partitIOn function to be well
approximated by this expression for temperatures much greater than the rotational temperature. 8R
= heB =
(6.626 x 10 34 J s) x (2.998
k
X
10 10 cm
1.381 x 1O
23 J
S I )
K I
x ( 1.931 cm
I
)
'
8R = 2.779 K.
«
In fact ~ T for all temperatures of interest in this problem (100 K or more). Agreement between the classical expression and the explicit sum is indeed good, as Fig. 17 .2(b) confirms. The figure di splays the percentage deviation (q~assical  qR ) 100/ qR . The maximum deviation is about 0.9% at 100 K and the magnitude decreases with increasing temperature. (c) The translational , rotational, and vibrational contributions to the total energy are specified by eqns 17.25b, 17.26b, and 17 .28 respectively. As molar quantities, they are
uT = ~ RT,
U
R
= RT ,
The contributions to the difference in energy from its 100 K value are !':..UT (T) = UT(T) U T (100 K), etc. Fig. 17.2(c) shows the individual contributions to !':..U(T) . Translational motion contributes 50% more than the rotational motion because it has 3 quadratic degrees of freedom compared to 2 quadratic degrees of freedom for rotation. Very little change occurs in the vibrational energy because very high temperatures are required to populate v = 1,2 • . ... states (see Part a ).
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
343
C lassical partition function error
0
0.2 c
.!2 OJ .;;  0.4
"
"0
"'"
OIl
= " 0.6 ~
"
I>
0.8
I
200
0
400
600
Temperature/ K
800
1000
Figure 17.2(b)
Energy cbange contributions
T /K
Figure 17.2(c)
The derivative dU v / dT may be evaluated numerically with numerical software (we advise exploration of the technique) or it may be evaluated analytically using eqn 17.34:
Cv
(}y
v
{(}y (
dT 
T
 dU R
V,m 
eOv /2T )} 2 leOv/ T
where = hev / k = 3 122 K. Fig. 17 .2(d) shows the ratio of the vibrational contribution to the sum of translational and rotational contributions. Below 300 K, vibrational motion makes a small , perhaps negligible, contribution to the heat capacity. The contribution is about 10% at 600 K and grows with increasing temperature.
344
STUDENT'S SOLUTIONS MANUAL Relative contributions to the heat capacity 0.2 ,      ,       ,        ,       ,       ,
C~. m
0I .
C~. m+ C~. m
o
L _ _ _ _ _ _
~
o
____
~~~
200
_ _ _ _ _ _ _ _L __ _ _ _ _ __ i_ _ _ _ _ _
400
600
~
800
1000
TI K
Figure 17.2(d)
The change with temperature of molar entropy may be evaluated by numerical integration with mathematical software. 6.S(T)
= SeT) 
S(100 K)
=
I
T
lOOK
= (T ilOOK
=
6.S(T)
I
T
CV,m(T) T 27
R
C (T)dT p,m [3.18] T
+ R dT [2.48]
+ C~.meT) dT. T
lOOK
= 2R In (_T_) + (T C~,m(T) dT 2
lOOK
T
ilOOK
~ /),.ST +R (T)
'~~~ /),.SV (T)
Fig. 17 .2( e) shows the ratio of the vibrational contribution to the sum of translational and rotational contributions. Even at the highest temperature the vibrational contribution to the entropy change is less than 2.5% of the contributions from translational and rotational motion. The vibrational contribution is negligible at low temperature.
Relati ve contributions to the entropy change 0.03 ,       .      ,        ,        ,       ,
0.02
0.01
o L _ _ _
o
~
_____
200
~~
____
400
~
_ _ _ _ _ L_ _ _ _
600
TI K
800
~
1000
Figure 17.2(e)
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
P17.11
+ DCI ;=! HDO + HCI.
H20 K
345
= q"(HDO) q" (HCI) e{3Mo [17.54; NA factors cancel] . q" (H20 )q" (DCI)
Use partition function expressions from Table 17.3. The ratio of translational partition functions is
q~(HDO)q~(HCI) = (M (HDO)M (HCI ) ) q~(H20)q~(DCI)
3/ 2 = (19.02
X 36.46) 3/2 18.02 x 37.46
M(H20)M(DCI)
= 1.041 .
The ratio of rotational partition functions is qR (HDO)qR(HCI)
=
qR(H20)qR(DCI)
a (H20 ) (A(H20 )B(H 20 )C(H20 ) / cm3) 1/2B(DCI)/cm 1  1  (A(HDO)B(HDO)C(HDO)/cm 3) 1/2B(HCI) /c m 1
=2 x
(27.88 x 14.51 x 9.29)1 /2 x 5.449 (23.38 x 9.102 x 6.417)1 /2 x 10.59
= 1.7
07
(a = 2 for H20 ; a = I for the other molecules).
The ratio of vibrational partition functions (call it Q) is q v (HDO)q v (HC!)
q(2726. 7)q( 1402.2)q(3707.5 )q(299 I )
=
q v (H20 )q v (DCI)
where q(x) I:!.Eo he
q(3656.7)q(l594.8)q(3755 .8)q(2 145 )
=Q
I
=
I_
e 1.4388x/ (T / K ) ·
= ~ {(2726.7 + 1402.2 + 3707 .5 + 2991) 2
= 162cm
l
(3656. 7 + 1594.8
+ 3755 .8 + 2 145)}cm I
.
So the exponent in the energy term is f3I:!.Eo
Therefore, K
I:!. Eo
he
kT
k
=  =  =
I:!. Eo
x 
1.041 x 1.707 x
Q
he
I x 
T
=
x e 233 / (T/ K)
=
1.4388 x ( 162) T/K
233
= +. T/K
1.777 Qe 233/ (T / K).
We then draw up the following table (using a computer)
T/ K K
100 18.3
200 5.70
and specifically K
300 3.87
400 3.19
500 2.85
600 2.65
700 2.51
800 2.41
900 2.34
1000 2.29
= 13.891 at (a) 298 K and [IiI] at (b) 800 K.
Solutions to theoretical problems P17.13
(a) Bv and BR are the constant factors in the numerators of the negative exponents in the sums that are the partition functions for vibration and rotation. They have the dimensions of temperature, which
346
STUDENT'S SOLUTIONS MANUAL
occurs in the denominator of the exponents. So high temperature means T » (}y or OR and only then does the exponential become substantial. Thus (}y and ~ are measures of the temperature at which higher vibrational and rotational states, respectively, become significantly populated:
~ _ hcf3 _ (2.998 x IOlO cm 
k

hcv k
=
x (6.626 x 10 34 ) s) x (60.864cm l ) ( 1.38I x lO 23 )K  I )
S I)
= 187.55 K 1
and (}y
=
(6.626
X
10 34 )
s)
x (4400.39 cm  I ) x (2.998 x 10 10 cm (1.381 x 1O 23 )K I )
sI)
=
1 1 6330 K .
(b) and (e) These parts of the solution were performed with Mathcad 7.0 and are reproduced on the following pages.
Objective: To calculate the equilibrium constant K(T) and Cp(T) for dihydrogen at bigh temperature for a system made with n mol H 2 at I bar. H 2(g) ;=: 2H(g)
At equilibrium the degree of dissociation, a, and the equilibrium amounts of H 2 and atomic hydrogen are related by the expressions nH2 = ( I  a)n
and
nH
= 2a n.
The equilibrium mole fractions are XH2 XH
= (l  a)nl{(l  a)1I + 2an) = (l  a)/(l + a), = 2an/{(I  a)n + 2an} = 2a/(I + a) .
The partial pressures are
PH 2
= (l 
a)p/(1
+ a)
and
PH
= 2ap/(1 + a).
The equilibrium constant is
4a 2     ;?,
(I  a)
where p
= p" = I
bar.
The above equation is easily solved for a :
1a = (KI(K + 4»1 /21· The heat capacity at constant volume for the equilibrium mixture is
The heat capacity at constant volume per mole of dihydrogen used to prepare the equilibrium mixture is Cv
= Cv(rnixture)/n = {nHCv,m(H) + nH2CV,m(H2)}/n = I2aCv,m(H) + (l  a)CV,m(H2) ,.
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
347
The formula for the heat capacity at constant pressure per mole of dihydrogen used to prepare the equilibrium mixture (Cp ) can be deduced from the molar relationship Cp,m Cp
= CV,m + R .
= (nHC".m(H) + nH2Cp,m (H2) lin = nH (Cv.m(H) + RI + nH2 ( CV,m(H2) + RI n
n
= nHCv.m(H) +nnH2C v.m(H 2)
+R(nH:nH2)
=Cv +R( I+a ).
Calculations J = joule mol = mole h = 6.62608 x 10 34 J s R = 8.31451 J K  1 mol I
s = second = gram e = 2.9979 x 10 8 m s I NA = 6.022 14 X 10 23 mol I g
kJ = 1000 J bar=l x lO5 Pa k = 1.38066 X 10 23 J KpG = I bar
I
Molecular properties of H2: v = 4400.39 cm I ,
B = 60.864 cm I ,
D = 432.1 kJ molI.
I g mol I he\!
ev = k '
heB
~  k .
Computation of K (T ) and a (T ) N = 200,
i = 0, ... , N
T; = 500 K
+
i x 5500 K N
qv ; = I _ e (&v I T;) ' Keq; a;
=(
Keq;
) 1/ 2
+4
See Fig. 17.3(a) and (b). Heat capacity at constant volume per mole of dihydrogen used to prepare the equilibrium mixture is (see Fig. 17.4(a» C v (H) =
[ill,
C V(H2') =
,
2 .5R +
ev [ T;
e<&v /2T;) ]2R x  ;:,.;;;1  e&v I T;
CV;
= 2a;Cv(H ) + (\ 
a; )C v(H2) .
348
STUDENT' S SOLUTIONS MANUAL
(a)
i5 0.5
0 0
1000
2000
3000
4000
5000
6000
T;/ K
Figure 17.3(a)
(b) 100
80
60 0
:.(
40
20
0 0
1000
2000
3000
4000
5000
T;/ K
6000
Figure 17.3(b)
The heat capacity at constant pressure per mole of dihydrogen used to prepare the equilibrium mixture is (see Fig. 17.4(b» Cp ; = Cv; P17.1S
+ R(l + ai)·
Eqn. 17.42 relates the second virial coefficient to the pairwise intermolecular potential energy:
In order to relate the pairwise potential to the van der Waals equation, we must express that equation as a virial series. (See Table 1.7.) The equations are RT a van der Waals p =     2 ' Vrn  b Vrn
virial p = RT ( 1 + B Vrn Vrn
+ ... )
STATISTICAL THERM ODYNAMICS 2: APPLI CATI ONS
27 26 25 I
"0
,.:.:E
24
.., 23
U
;;:
22 21 20 0
1000
2000
4000
3000 T;/K
5000
6000
Figure 17.4(a)
42
40
38
I
"0 E
,.:.:
36
uo.
34
::;
32
30
28 ~~~~~~~ o 1000 2000 3000 4000 5000 6000 Ti / K
Figure 17.4(b)
349
350
STUDENT'S SOLUTIONS MANUAL
Expand the van der Waals equation as a power series in 1/ Vm:
Thus, the second virial coefficient in terms of van der Waals parameters is B=b
~. RT
The pairwise potential and Mayer Ifunction are: for 0.:::: r < r, for
rl .::::
e/3Ep
=0
Ep + E
e/3Ep
= e/3£
1= I 1 = e /3£ 
Ep + 0
e /3Ep
=
1=0
Ep +
r < r2
for r2 .:::: r
00
«
Expand the exponential, because E B
~
2rr.NA
1
kT, so {3E
«
1> 0
I.
r3 (3 r 3) } 2rr.NA { { 3 + (1 + (3E  1) 3  3 = 3 r r2
1
1
3 1

( 3 r 3) }.
E r2 
1
kT
Comparing this result to the virial coefficient from the van der Waals equation, we identify
a=
and
2rr.NAEm(r~  rf) 3 '
~~..:....=.~
where Em is E expressed as a molar quantity. Thus the van der Waals b is proportional to the volume of the hardsphere (repulsive) part of the potential. The a parameter is more complicated, but it is where the attractive part of the potential appears, including both the depth of the attractive well and the range of distances over which it operates. Use eqn 17.45 to compute the limiting isothermal JouleThomson coefficient .
dB
hm /1T = B  T dT
p>o
=
2rr.NA {r3 _ 3 I
= 2rr.NA {r3 _
3
1
E(r~
 r
i)} _T 2 rr.NA
{ E(r~
3
kT
2E(r~ 
ri) } = b _
kT
~. RT
The JouleThomson coefficient itself is [2.55] /1
= _ /1T = 2rr.NA {2E(r~ Cp
3Cp
kT
ri) _ ri}
 r
kT2
= b  ~~ . Cp
i)}
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
P17.17
(a) Ethene belongs to the D 2h point group, whose rotational subgroup includes E and 3 C2 elements around different axes. So a = 4. The rotational partition function of a nonlinear molecule is [Table 17.3] R
=
q
1.0270 (T / K) 3/2  a  (ABC/c m3) 1/2
(b) Pyridine belongs to the R
q P17.19
351
1.0270
e2v
=
1.0270 x 298 .15 3/ 2 (4) x (4.828 x 1.0012 x 0.8282)1 /2
group, the same as water, so a
(T /K)3/2
= a (ABC/cm 3 ) 1/ 2 =
~
=~.
= 2.
1.0270 x 298.15 3/ 2 (2) x (0.2014 x 0.1936 x 0.0987)1 /2
=
1
41
4 26 x 10 . .
The partition function of a system with energy levels e (J) and degeneracies g(J) is q
= Lg(J)e  P£(J) J
The contribution of the heat capacity from this system of states is Cv
= kfJ2 (au) afJ
[17.3Ia] v
= N C~;q)v = ~
where U  U(O)
G;)v·
Express these quantities in terms of sums over energy levels
u
_'!.. ( L
U(O) =
q
g(J)e (J)eP£(J») =
'!.. Lg(J)e (J)eP£(J) q
J
J
and Cv ) kfJ
= (au) = '!.. (_ L afJ
v
= _'!.. L q
q
J
q g(J )e 2 (J)e  P£(J») _ N '"' g(J)e(J)e P£(J) ( a ) q2 afJ
g(J)e2(J)e  P£(J)
7
+~ q
J
(I)
L g(J)e(J)eP£(J) L g(J')e(J')eP£(J'). J'
J
Finally a double sum appears, one that has some resemblance to the terms in l;(fJ). The fact that l;(fJ) is a double sum encourages us to try to express the single sum in Cv as a double sum. We can do so by multiplying it by one in the form (L: g(J ')e  Pd J')/q, so J'
~ = _ q2 N kfJ 2
'"' g(J)e2(J)eP£(J) '"' g(J')eP£(J' ) ~ ~ J
+ q2 N
J'
'"' g(J)e(J)eP£(J) '"' g(J')e(J')eP£(J' ). ~ ~ J
J'
Now collect terms within each double sum and divide both sides by N :
~ kNfJ2
=
~ '"' g(J)g(J') e 2(J )e  P[£(J)+£(J' )1  ~ '"' g(J)g(J')e(J)e(J')eP[£(J)+£(J') ]. q ~ q2 ~ J,./'
J,./'
352
STUDENT' S SO LUTIONS MANUAL
Clearly the two sums could be combined, but it pays to make one observation before doing so. The first sum contains a term £2(1 ), but all the other factors in that sum are related to 1 and l ' in the same way. Thus, the first sum would not be changed by writing £2 (1') instead of £2(1) ; furthermore, if we add the sum with £2 (1' ) to the sum with 8 2(1), we would have twice the original sum. Therefore, we can write (fi nall y combining the sums)
Recogni zing that £2(1)
+ 82 (1 ' ) 
28 (1 )£ (1' )
= [£(1 ) 
8(1 ' )f , we arrive at
kNfJ 2
= 2 ~(fJ ).
Cv
For a linear rotor, the degenerac ies are g(1) 8(1)
so fJ £(1 )
= hcBl(1 +
I)
= ()R kJ (1 +
= 2J + I. The energies are
I)
= ()R l(1 + I ) / T .
The total heat capacity and the contributions of several transitions are plotted in Fig. 17.5 . One can evaluate CV.m / R using the following ex pression, deri vable fro m eqn (I) above. It has the advantage of using single sums rather than double sums.
(fJ) is defined in such a way that J and J' each run independently from 0 to infinity. Thus, identical terms appear twice. (For example, both (0, 1) and (1,0) terms appear with identical value in (fJ ). In
COMMENT.
the plot, though, the (0,1) curve represents both terms.) One could redefine the double sum with an inner sum over J ' running from 0 to J  1 and an outer sum over J running from 0 to infinity. In that case, each term appears only once, and the overall factor of 1/2 in Cv would have to be removed .
P17.21
All partition fun ctions other than the electronic partition function of atomic I are unaffected by a magnetic field; hence the relative change in K is due to the relati ve change in qE . E _ '"'
q L e
 g/ls fJ l3Mj
,
M __ 1 _ ! +! + 1 .  ~ J 2 ' 2' 2 ' 2, g  3'
MJ
Since gft BfJ B qE
=L
«
1 for normall y attainable fields, we can expand the exponentials
{ 1  gftBfJBMJ
+ ~(gftBfJBMJ)2 + ... }
MJ
~ 4+ ~(gft BfJB)2
L MJ
MJ [LMJ
0
= 0] = 4(1 + 1 9(ftBfJ B )2 )
[g= ~l
MJ
This partition function appears squ ared in the numerator of the equil ibrium constant expression. (See solution to E 17 .14(a).) Therefore, if K is the actual equilibrium constant and KOis its value when B 0,
=
STATISTICAL THERMODYNAM ICS 2: APPLICATIONS
353
1.2
0.8
u'" c:.:::
E
0.6
0.4 1,3 0.2
=::::::: 
0,3
0 2
0
4
3
5
Temperature , T IeR
Figure 17.5
we write
For a shift of I per cent, we req uire
Hence B ~
0.067kT
=
(0.067) x ( 1.381 x 1O 23 JK
J.i.B
9.274 x 10 24 J T
I)
x (lOOOK)
I
~
~
lOOT .
Solutions to applications P17.23
S
= k In W [16 .34] .
so S = k In 4N = Nk In 4
=
(5 x 10 8 ) x ( 1.38 x 1O 23 J K
I
)
x In4
= 19.57 x
10
15
JK
 I
I.
Question. Is this a large residual entropy? The answer depends on what comparison is made. Mu lti ply the answer by Avogadro's number to obtain the molar residual entropy, 5.76 x 10 9 J K I mol I, surely
354
STUDENT'S SOLUTIONS MANUAL
a large numberbut then DNA is a macromolecule. The residual entropy per mole of base pairs may be a more reasonable quantity to compare to molar residual entropies of small molecules. To obtain that answer, divide the molecule's entropy by the number of base pairs before multiplying by NA. The result is 11.5 J K 1 mol I, a quantity more in line with examples discussed in Section 17.7. P17.2S
The standard molar Gibbs energy is given by
c:  c:
G
(0) = RTln qrn NA
Translation (see table 17.3 for all partition functions) :
=
(2.561 x 10 2) x (2000)5/2 x (38.90)3/2 = 1.111
X
109 .
Rotation of a linear molecule: q
R
kT 0.6950 ==x ahcB
a
T /K B / cm  I
.
The rotational constant is Ii B   
Ii ;=:;;
 4ncl  4ncmeffR2
where meff
mBmSi
= mB
B _ so qR
=
+mSi
4n(2.998 x
IO lO
0.6950 x 2000 0.5952 1
(l0.81) x (28.09) 1O 3 kg mol I :::.,...:::: x 10.81 + 28 .09 6.022 x 1023 mol 1
cms
l)
= 1.296 x
1O 26 kg.
34 1.0546 x 1O J s I _ 05952 . cm x (1.296 x 10 26 kg) x (190.5 x 1O 12 m)2
= 2335.
Vibration: qv
= __;::;;;;:1
ehcv/kT
1.4388(ii / cm I») 1 exp ( T/ K
1  exp (
1.4388(772) ) 2000
= 2.467. The Boltzmann factor for the lowestlying electronic excited state is exp (
 (1.4388) x (8000») _ 3 2 103 2000  . x .
The degeneracy of the ground level is 4 (spin degeneracy = 4, orbital degeneracy excited level is also 4 (spin degeneracy = 2, orbital degeneracy = 2), so qE
= 4(1 + 3.2 x
10 3)
= 4.013 .
= 1), and that of the
STATISTICAL THERMODYNAMICS 2: APPLICATIONS
355
Putting it all together yields
= (8.3145 J mol I K I) x
cr:  cr: (0)
x In[(1.111 x
= 5.135 x P17.27
109 )
(2000 K)
x (2335) x (2.467) x (4.013)]
105 J mol  I
= 1513.5 kJ mol I I·
The standard molar Gibbs energy is given by <>
<>
NA
where qm NA
= RTln qm
cr:  cr: (0)
T<>
= ~qRqV qE [17.53] NA
See Table 17.3 for partition function expressions. First, at 10.00 K T<>
Translation : ~ NA
x 1O 2(T / K)5 /2(M / g mol  I )3/2
= 2.561 = (2.561
x 10 2) x (10.00)5/2 x (36.033)3/2
= 1752.
Rotation of a nonlinear molecule:
qR
= ~ (kT) 3/2 (~)1 /2 = 1.0270 a
hc
ABC
X
a
(T/K)3 /2 . (ABC / cm3) 1/2
The rotational constants are
B= Ii
so
4ncJ
ABC= (  Ii 4nc
)3  , IA/B/C
1.0546 x 10 34 J s )3 ABC = ( 4n(2.998 x IO lO cm sI) (loIDA m 1)6
x
= R
so q
~~
(39.340) x (39.032) x (0.3082) x (u A2) 3 x (1.66054 101.2cm  3
1.0270
= 2
X
( 10.00)3/2 (101.2)1 /2
X
10 27 kg u I )3
= 1.614.
Vibration : for each mode
v q
I
=
I 
e  hcv/ kT
1  exp (
1.4388(ii/ cm  I)) T/ K
1  exp (
1.4388(63.4) ) 10.00
= 1.0001 Even the lowestfrequency mode has a vibrational partition function of 1; so the stiffer vibrations have q v even closer to I. The degeneracy of the electronic ground state is 1, so qE = 1. Putting it all together yields cr:  cr:(0)
= (8 .3145J mol I K I) x = 1660.8 J mol  I I.
(lO.OOK) In[(1752) x (1.614) x (I) x (l)]
356
STUDENT'S SOLUTIONS MANUAL
Now at 1000 K q<>
Translation: ~ = (2.561 x 10 2) x (1000)5/ 2 x (36.033)3/2 = 1.752 X 10 8 .
NA
1.0270
( 1000)3/2
Rotation:
qR   2  X (101.2) 1/2  1614.
Vibration:
q'(
I = ,,,,,,..,...., = 1 I 47
Iexp ( 
qi =
.
,
1
:(:(:l.4::3:::8:::8) X:(1:::2=24:.:::5)....,) = 1.207, 1  exp     I0'00
v q3 =
( 1.4388) x (63.4») 1000
1
(
(1.4388) x (2040») = 1.056, 1  exp  1000
qV = (11.47) x ( 1.207) x (1.056) = 14.62 .
Putting it all together yields
c:,  c:,(0) =
(8.3 1451 mol I K I ) x ( IOOOK) x 10[( 1.752 x J08 ) x ( 1614) x (14.62) x (I )]
=2.415 x 105 1 mol  I =1 241.5 k1 mol I I.
18
Molecular interactions
Answers to discussion questions 018.1
Molecules with a permanent separation of electric charge have a permanent dipole moment. In molecules containing atoms of differing electronegativity, the bonding electrons may be displaced in such a way as to produce a net separation of charge in the molecule. Separation of charge may also arise from a difference in atomic radii of the bonded atoms. The separation of charges in the bonds is usually, though not always, in the direction of the more electronegative atom but depends on the precise bonding situation in the molecule as described in Section 18.1 (a). A heteronuc1ear diatomic molecule necessarily has a dipole moment if there is a difference in electronegativity between the atoms, but the situation in polyatomic molecules is more complex . A polyatomic molecule has a permanent dipole moment only if it fulfills certain symmetry requirements as discussed in Section 12.3(a). An external electric field can distort the electron density in both polar and nonpolar molecules and this results in an induced dipole moment that is proportional to the field. The constant of proportionality is called the polarizability.
018.3
Dipole moments are not measured directly, but are calculated from a measurement of the relative permittivity, Sr (dielectric constant) of the medium. Equation 18.15 implies that the dipole moment can be determined from a measurement of Sr as a function of temperature. This approach is illustrated in Example 18.2. In another method, the relative permittivity of a solution of the polar molecule is measured as a function of concentration. The calculation is again based on the Debye equation, but in a modified form. The values obtained by this method are accurate only to about 10%. See the references listed under Further reading for the details of this approach. A third method is based on the relation between relative permittivity and refractive index , eqn 18.17, and thus reduces to a measurement of the refractive index. Accurate values of the dipole moments of gaseous molecules can be obtained from the Stark effect in their microwave spectra.
018.5
If the AH bond in the AH · .. B arrangement is regarded as formed from the overlap of an orbital on A, l/IA , and a hydrogen Is orbitall/lH , and if the lone pair on B occupies an orbital on B, 1/1'8 , then, when the two molecules are close together, we can build three molecular orbitals from the three basis orbitals:
One of the molecular orbitals is bonding, one almost nonbonding, and the third antibonding. These three orbitals need to accommodate four electrons, two from the AH bond and two from the lone pair on B.
358
STUDENT'S SOLUTIONS MANUAL
Two enter the bonding orbital and two the nonbonding orbital, so the net effect is a lowering of the energy, that is, a bond has formed. 018.7
A molecular beam is a narrow stream of molecules with a narrow spread of velocities and, in some cases, in specific internal states or orientations. Molecular beam studies of nonreactive collisions are used to explore the details of intermolecular interactions with a view to detennining the shape of the intermolecular potential. The primary experimental information from a molecular beam experiment is the fraction of the molecules in the incident beam that is scattered into a particular direction. The fraction is normally expressed in terms of dI , the rate at which molecules are scattered into a cone that represents the area covered by the 'eye' of the detector (Fig. 18.14 of the text). This rate is reported as the differential scattering crosssection, a , the constant of proportionality between the value of dI and the intensity, I, of the incident beam, the number density of target molecules, N , and the infinitesimal path length dx through the sample: dI
= alNdx.
The value of a (which has the dimensions of area) depends on the impact parameter, b, the initial perpendicular separation of the paths of the colliding molecules (Fig. 18.15), and the details of the intermolecular potential. The scattering pattern of real molecules, which are not hard spheres, depends on the details of the intermolecular potential, including the anisotropy that is present when the molecules are nonspherical. The scattering also depends on the relative speed of approach of the two particles: a very fast particle might pass through the interaction region without much deflection, whereas a slower one on the same path might be temporarily captured and undergo considerable deflection (Fig. 18.17). The variation of the scattering crosssection with the relative speed of approach therefore gives information about the strength and range of the intermolecular potential. Another phenomenon that can occur in certain beams is the capturing of one species by another. The vibrational temperature in supersonic beams is so low that van der Waals molecules may be formed, which are complexes of the form AB in which A and B are held together by van der Waals forces or hydrogen bonds. Large numbers of such molecules have been studied spectroscopically, including ArHCI, (HClh, ArC02 , and (H20h. More recently, van der Waals clusters of water molecules have been pursued as far as (H20k The study of their spectroscopic properties gives detailed information about the intermolecular potentials involved.
Solutions to exercises E18.1(b)
A molecule that has a center of symmetry cannot be polar. S03(D3h) and XeF4(D4h) cannot be polar.
ISF41 (seesaw, e 2v ) may be polar. E18.2(b)
+ M~ + 2MiM2 cose)i /2 [18.2a] = [(1.5)2 + (0.80)2 + (2) x (1.5) x (0.80)
M = (Mf
x (cos 109S)]i /2
D= II.4D I
MOLECULAR INTERACTION S
E18.3(b)
359
The components of the dipole moment vector are J.Lx = I>iXi = (4e) x (0)
+ (2e) and JLy =
L
+ (
2e) x ( J62 pm)
x (l43pm) x (cos 30°) = ( 572 pm )e
+ (2e)
qiYi = (4e) x (0)
x (0)
+ (2e)
x (l43 pm ) x (sin 30°) = ( 143 pm )e
i
The magnitude is JL = (JL~
+ JL;') 1/ 2 =
« 570)2
+ (143)2) 1/2 pm e =
(590 pm )e
= (590 x 10 12 m) x ( 1.602 x 10 19 C) = 19.45 x 10 29 C m 1 and the direction is
e=
tan  I JLy = tan  I  143 pm e = 1194.0° 1from the xaxis (i.e. 14.0° below the JLx
572pm e
negative xax is). E18.4(b)
The molar polarizati on depends on the polari zability through
Thi s is a linear eq uati o n in T 
so
J.L =
I
with slope
9E:okm) 1/2 = ( ;:;;
? (4.275 x 1O 9 Cm) x (m/(m 3 moJ  i K»i /2
and with yintercept
Since the molar polarization is linearly dependent on T  I , we can obtain the slope m and the intercept b
m=
P m .2  Pm.1
TI
and
I 
T
I
(75.74  71.43) cm 3 mol  I 3 3 I (320.0K) I _(42 1.7K) 1 =5 .72 x 10 cm mol K
2
b = Pm  mT 1 = 75.74cm 3 mol I  (5.72 x 10 3 cm 3 mol I K) x (320.0 K)  1 = 57.9cm 3 mol  I
It follows that JL = (4.275 x 10 29 C m) x (5.72 x 10 3 ) 1/ 2 = 13.23 x 10 30 C m 1
and
360
E18.5(b)
STUDENT'S SOLUTIONS MANUAL
The relative permittivity is related to the molar polarization through
er  I pPm ==C er +2 M
C
=
( 1.92gcm 3 )
r
X
= 0.726
I
(0.726) + I = 18.971 10.726
The induced dipole moment is
= ae = 4JTeoa ' e
J.L*
= 4JT(8.854
12
10
X
rl
C2 m  ' )
X
(2.22
X
10 30 m 3 )
X
(15 .0
X
10 3 V m')
x 1O 36 C m 1
= 13.71 E18.7(b)
(32. l6cm 3 mol')
X
85 .0gmol
e = 2
E18.6(b)
2C + 1 er =   , IC
so
If the permanent dipole moment is negligible, the polarizability can be computed from the molar polarization
3eoP NA
m a=
so
and the molar polarization from the refractive index
pPm
M
a
=
er  l 6 r +2
3 X (8.854

(6.022
= 13.40 E18.8(b)
X
3eOM(n;I)
n;I
= 1l~+2 X
X
a=
so
10 12 r l C2 m I )
1023 mol  ')
X
n} +2
NAP
(2.99
X
(65.5gmol  ' )
X 1Q6
gm  3)
X
(1.622 2
10 40 C 2 m2 J ' 1
The solution to Exercise 18.7(a) showed that
M)
= (3 eO
a
pNA
X
(~)
or
+2
n~
a
= (3M) 4JTpNA
I
X
which may be solved for nr to yield
n. __ r
1
(f3 + 2a. ' ) f3'  a '
' /2
,
(3)
f3 = (4JT) nr
with
X _
= (
(0.865
X
33.14+2 x 2.2 33.14  2.2
X
f3 = 3M I
4JTpNA
(72.3 g mol I)
106 g m 3 ) ) 1/ 2
X
(6.022
=~
X
1023 mol

I)
1.6222 +2
(11;1)
n}+2
MOLECULAR INTERACTIONS
E18.9(b)
361
The relative permittivity is related to the molar polarization through I
Cr 
pPm
  =   == C so Cr
+2
M
Cr
2C + I 1 C
= 
The molar polarization depends on the polarizability through so
C
pNA =
3eoM
2 ( 4n coa , +1L )
3kT
(1491 kgm  3) x (6.022 x 1023 mol  I) C = ~:::~~;==:::~::;::=.:3(8.854 x 1O 12 J I C2 m I) x ( 157.01 x 10 3 kg mol I) x (4n(8.854 x 10 12 r I C2 rn I) x (1.5 x 10 29 m 3) (5.17 x 1O 30 Cm)2 ) +3 (1.381 x 1O 23 JK I) x (298K) C = 0.83
E18.10(b)
M
p
Vm =
and
er =
2(0.83) + 1 I."Zl I _ 0.83 = L..!iJ
18.02gmol 1 5 3 _I = 999.4 x 10 3 gm 3 = 1.803 x 10 m mol 2(7.275 x 1O 2 Nm l ) x (1.803 x 105 m 3 molI) (20.0 x 10 9 m) x (8.314J K I mol I) x (308.2K)
2yVm
rRT

= 5.119 x 10
2
P = (5.623 kPa) e00511 9 = 1 5.92 !cPa E18.11(b)
y =
~pghr= ~ (0.9956 g cm 3 )
1
x (9.807ms 2) x (9.11 x 1O 2 m)
3 x (0.16 x 1O m) x
C~c~_~3)
=17.12 x 1O 2 Nm  1 1 E18.12(b)
Pin  Pout
2y (2) x (22.39 x 1O 3 N m I) I 5 = ~ [18.38] = 2.20 x 10 7 m = 2.04 x 10 p~
I
Solutions to problems Solutions to numerical problems P18.1
The positive (H) end of the dipole will lie closer to the (negative) anion. The electric field generated by a dipole is g =
(~) x (~)[18.21] 4n cO r (2) x (1.85) x (3.34 x 10 30 Cm) (411:) x (8.854 x 10 12 J I C2 ml) x r3
l.ll x 10 19 Vm I
l.ll x 108 Vrn I
(rlm)3
(rlnm)3
362
STUDENT'S SO LUTIONS MANUAL
(a) rff =
11.1
(b) rff = (c) rff = P18.3
X
1. I I
X
1. I I
X
10 8 Y m'
1
when r = 1.0nm.
I
108 Y m  , 0.3 3 =4x 109 Ym  '
I forr=0.3nm .
10 8 Y m  \ = 14kYm'1 forr=30nm. 303 . .
The equations relating dipole moment and polarizability volume to the experimental quantities fr and pare P
m 
x (frl) (M) P fr + 2
411 , NAJL 2 [\8.14] and Pm = NACf +  3 9fokT
[18.15, with a = 411foO"].
Therefore, we draw up the following table (with M = 1I9.4 g mol  ').
e;oe
80
70
60
40
20
0
20
TIK
193 5.18 3.1
203 4.93 3.1
213 4.69 7.0
233 4.29 6.5
253 3.95 6.0
273 3.66 5.5
293 3.41 5.0
0.41
0.41
0.67
0.65
0.63
0.60
0.57
1.65 29.8
1.64 29.9
1.64 48.5
1.61 48.0
1.57 47.5
1.53 56.8
1.50 45.4
1000 / (T / K)
Cr cr I f r +2 p i g cm  3 Pm / (ern 3 molI) Pm
is plotted against 1IT in Fig. 18.1. 50
o
2
3 4 5 103/(T /K) m.p!.
6
Figure IS.1
The (dangerously unreliable) intercept is ~ 30 and the slope is ~ 4.5
X
103 . It follows that
MOLECULAR INTERACTIONS
363
To determine fJ we need fJ
Ok) 1/2 = ( 9& NA x
=
_ 2 (slope x cm 3 mol I K)I /
I
(9) x (8.854 x 10 12 r l C2 m I ) x ( 1.381 x 10 23 J K 6.022 x 10 23 mol I
x (,lope x om' mol' K)'I'
I
mol )1 / 2
= (4.275
x 10 29 C) x ( K m
= (4.275
X
I
)
1/2
)
x (slope x cm3 molI K) I/ 2
10 29 C) x (slope x cm 3 m I )I /2
= (4.275 x 10 32 C m) x (slope) 1/ 2 = (1.282 x 10 2 D) x (slope) 1/ 2 = (1.282 x 10 2 D) x (4.5 x 103)1 / 2 =
I 0.86 D I.
The sharp decrease in Pm occurs at the freezing point of chloroform (63°C), indicating that the dipole reorientation term no longer contributes. Note that Pm for the solid corresponds to the extrapolated, dipolefree, value of Pm , so the extrapolation is less hazardous than it looks.
P18.5
4n 3
Pm = NM:t
,
2
NAfJ+ [18 .15, with a
9&okT
,
= 4n&oa].
Therefore, draw up the following table.
T/ K
292.2
309.0
333.0
387.0
413 .0
446.0
1000/( T / K) Pm /(cm 3 molI )
3.42 57.57
3.24 55.01
3.00 51.22
2.58 44.99
2.42 42.51
2.24 39.59
The points are plotted in Fig. 18.2. The extrapolated (least squares) intercept lies at 5.65 cm 3 mol  I (not shown in the figure), and the least squares slope is 1.52 x 104 cm 3 K  I mol  I. It follows that
a'
3 x 5.65 cm 3 mol  I 4n x 6.022 x 1023 molI
3P m (at intercept) 4nNA
fJ
= \ 2.24 x
10 24 cm 3
= 1.282 x
10 2 D x ( 1.52 x 104 ) 1/ 2 [from Problem 18.3]
\.
The highfrequency contribution to the molar polarization, refractive index:
p:n = (M) p
X
p:n, at 273 K may be calculated from the
(&+ 2I) [18 . 14] = (M)  I) .  x (n2 ;n, + 2  ' 
&,
p
= 11 .58 D I.
364
STUDENT'S SOLUTIONS MANUAL
3.0 1000 KJT
2.0
4.0
Figure 18.2
Assuming that ammonia under these conditions ( 1.00 atm pressure assumed) can be considered a perfect gas, we have
p M
and p
=
pM RT
RT
=
p
=
Then P~, = 2.24
82.06 cm 3 atm K I molI x 273 K 1.00 atm X
= 2.24
4 3 I x 10 cm mol .
I
I
( 1.000379)2  I } 104 cm 3 mol  I x { 2 = 5.66 cm 3 mol  I . (1.000379) + 2
If we assume that the highfrequency contribution to Pm remains the same at 292.2 K then we have
N fL2
= Pm  Pm' = (57.57  5.66) cm 3 molI
_A_
9t:okT
Solving for fL we have
The factor
(
Therefore fL
9~:k )
1/2
has been calculated in Problem 18.3 and is 4.275 x 10 29 C x (mol / K m) 1/2.
= 4.275
x 10 29 C x
= 5.26
10 30 C m
X
The agreement is exact'
(~I)
1(2
= 11.58 D I.
x (292 .2 K) 1/2 x (5.191 x 10 5 ) 1/2(m 3 / mo!) 1/2
MOLECULAR INTERACTIONS
P18.7
365
(a) The depth of the well in energy units is f: = hcD e = 11.51 x 10
23
J
I·
The distance at which the potential is zero is given by Re =2 1/ 6ro
so
ro =ReT I/6 = T I/ 6 (297 pm) = 1265pm l.
(b) In Fig. 18.3 both potentials were plotted with respect to the bottom of the well, so the LennardJones
potential is the usual LJ potential plus f:. 10
8
....
u
~
;:J
6
6

Morse
''
~ 4
2
0 200
400
300
500
r/pm
Figure 18.3
Note that the LennardJones potential has a much softer repulsive branch than the Morse. P18.9
Neglecting the permanent dipole moment contribution, N A fY.
Pm = 
3f:o
[18 .15]
.
(6.022 X 1023 mol  I) x (3.59 x 10 40 r I C2 m 2 ) 3(8.854 x 10 12 r l C 2 m I) = 8.14 x 10 6 m 3 mol  I = 18. 14 cm) mol  I I. f:  I
_r_ _
=
f: r +2

[18.16]
M =
f: r
pP
~
(0.7914 gcm 3 ) x (8. 14cm 3 mol32.04gmol 
I = 0.201 f: r
nr = f: : /
2
[18.17]
+ 0.402 ;
1
l)
= 0.201.
1f: r = 1.76 1.
= ( 1.76) 1/ 2 = [Lill.
The neglect of the permanent dipole moment contribution means that the results are applicable only to the case for which the applied field has a much larger frequenc y than the rotational frequency. Since red light has a frequency of 4.3 x 10 14 and a typical rotational frequency is about I x 10 12 Hz, the results apply in the visible.
366
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems P18.11
Exercise 18.7 showed
a= (3epNAoM) x (n;nt + 2I) n;  I Therefore,  2 nr + 2
=
nt+2'
4npNA
) 1/ 2
(
=
( I _ 4na pNA 3M
8na'p) 1/ 2 1+3kT I _ 4na'p 3kT
[( 1+8na'p) x ( 1+4na'p)] 1/ 2 3kT
~ (
= I + const.
PNA + 8na' 3M ,
~ ~
3kT
12na'p
1++ ··· 3kT
x p
)1 / 2
A
2na'p
~ I+
kT
~ with constant =
3M) (n;  I) a, = (4rr.N P x n; + 2 = P19.13
, = (3M   ) x (n;I) 
3M
I
I
a
4na'NA P
Solving for n r , nr =
Hence, nr
or
.E]
M M [for a gas, P = = Vm RT
~ I+X] [ _I Ix
[(l+X) I /2~1+~X].
2n~
IT' From the first line above,
(~ (n;  I) ;;r+2 . 3kT)
x
Consider a single molecule surrounded by N  1(~ N) others in a container of volume V. The number of molecules in a spherical shell of thickness dr at a distance r is 4nr2 x (N IV) dr . Therefore, the interaction energy is
u=
l
a
R4 rr.r 2 x (N) x (C6)   dr V
r6
4nNc6 1R dr
=
V
a
r4
where R is the radius of the container and d the molecular diameter (the di stance of closest approach). Therefore,
u
4n) =(3
x
(N) V
(C6) x
(
I
I)
R3  d3
~
4nNC6 3Vd 3
because d « R. The mutual pairwise interaction energy of all N molecules is U = !Nu (the! appears because each pair must be counted only once, i.e. A with B but not A with Band B with A). Therefore,
U= 2
For a van der Waals gas, /l 2a = V
(au) av
2nN2C6

2 3
T
3V d
and therefore a
=
MOLECULAR INTERACTIONS
P18.15
367
The number of molecules in a volume element dr isN dr IV = Ndr. Tbeenergy of interaction of these molecules with one at a distance r is V N dr . The total interaction energy, taking into account the entire sample volume, is therefore
u
f
=
VN dr = N
f
V dr
[Vis the interaction energy, not the volume].
The total interaction energy of a sample of N molecules is ~Nu (the ~ is included to avoid double counting), and so the cohesive energy density is
U
2
 V = 2rrN
1
00
C6
a
dr r4
Nlc 6
2rr
=3
X
~.
However, N = NAP 1M, where M is the molar mass; therefore
P18.17
Once again (as in Problem 18.16) we can write
8(v) =
I
. ( rr  2 arCSIn 0 R,
b
+ R2(V)
)
b ::: R,
+ R2(V)
b > R, +R2(V)
but R2 depends on v
Therefore, with R, (a)
8(v)
= ~R2 and b = ~R2'
= rr2arcsin( 1 + 2e1 v/ . ) . v
(The restriction b ::: R, + R2(V) transforms into This function is plotted as curve a in Fig. 18.4. The kinetic energy of approach is E (b)
8(E)
~R2
:::
= ~mv2 , and so
= rr  2 arcsin (1 + 2 e~(E/E*)1/2 ) with E* =
in Fig. 18.4.
~R2 + R2e v/ v', which is valid for all v.)
~mv*2. This function is plotted as curve b
368
STUDENT'S SOLUTIONS MANUAL
Solutions to applications P18.19
(a) The energy of induceddipoleinduceddipole interactions can be approximated by the London formula (eqn 18.25): V
C = = 3a;a;  Ilh   = 3a'21 6 6 r6
2r
II
+h
4r
160
120
40
2
4 (a) v/ v*
and
6 8 (b) £ / £*
10
Figure 18.4
where the second equality uses the fact that the interaction is between two of the same molecule. For two phenyl groups, we have
V=

3(1.04
X
10 29 m 3 )2(5 .0eY)(1.602 x 10 19 JeyI) 4(1.0 x 10 9 m )6
= 6.6 x
?3
10  J
or l39 Jmol  l l· (b) The potential energy is everywhere negative. We can obtain the distance dependence of the force by
taking
dV
6C
F==dr r7 .
This force is everywhere attractive (i.e. it works against increasing the distance between interacting groups). The force approaches zero as the distance becomes very large ; there is no finite distance at which the dispersion force is zero. (Of course, if one takes into account repulsive forces , then the net force is zero at a distance at which the attractive and repulsive forces balance. ) P18.21
(a) The dipole moment computed for transNmethylacetamide is iL = (3.092 D) x (3 .336 X 10 30 C m D I) =
I 1.03
X
10 29 C m I
(semiempirical, PM3 level, PC Spartan Pro™). The dipole is oriented mainly along the carbonyl group. The interaction energy of two parallel dipoles is given by eqn 18.22: V
=
iL liL"zf (e)
4rreor
3
where/eel
= 1
3 cos
2
e
MOLECULAR INTERACTIONS
369
and r is the distance between the dipoles and () the angle between the direction of the dipoles and the line that joins them . The angular dependence is shown in Fig. 18.5. Note that V (() is at a minimum for () = 0 0 and 1800 while it is at a maximum for 90 0 and 270 0 .
20
r
0
o
E
~
:::: 20
40
o
50
150
100
200
250
350
300
O/ deg
Figure 18.5
(b) If the dipoles are separated by 3.0 nm, then the maximum energy of interaction is: 29
Vrnax =
(1.03 1 x 10 C m) 2 4rr(8.854 x 10 12 r I C2 m  I ) x (3 .0 x 10 9
I 
= 3.55 x 10
m )3
 23
I
J.
In molar units Vrnax = (3.55
X
10 23 J) x (6.022
X
1023 molI) = 21 Jmol I = 2.1 x 10 2 kJmol 
l.
Thus, dipoledipo1e interactions at this distance are dwarfed by hydrogen bonding interactions. However, the typical hydrogen bond length is much shorter, so this may not be a fair comparison. P18.23
Here is a solution using MathCad.
(8)
Data:=
7.36 3.53 ( 1.00
logA := (DataT )(0)
8.37 8.~ 4.24 4.09 1.80 1.70
S:= (DataT )(1)
info := regress(Mxy, togA,1)
(b)
W:= 1.5 S := 4.84 Given
7.47 3.45 1.35
7.25 2.96 1.60
6.73 8.52 7.87 2.89 4 .39 4.03 1.60 1.95 1.60
W:= (DataT )(2)
Mxy:= augment(S,W) b=
0.957) bo 0.362 bl ( 3.59 b2
W := Find(W)
W = 1.362
b := submatrix(info, 3,5,0,0)
Estimate for Given/Find Sotve Bank togA:= 7.60 togA = bo + b1 . S + b2 . W
7.53) 3.80 1.60
Materials 1: macromolecules and aggregates
19
Answers to discussion questions 019.1
Number average is the value obtained by weighting each molar mass by the number of molecules with that mass (eqn 19.1 )
 = IVI"
~N;M;.
Mn
;
In this expression, N; is the number of molecules of molar mass M; and N is the total number of molecules. Measurements of the osmotic pressures of macromolecular solutions yield the number average molar mass. Weight average is the value obtained by weighting each molar mass by the mass of each one present (eqn 19.2)
LN;M;
 = I" Mw
m
~m;M; i
= =,i = , L.,N;M;
[19.3].
In this expression, m; is the total mass of molecules with molar mass M; and m is the total mass of the sample. Light scattering experiments give the weight average molar mass. Zaverage molar mass is defined through the formu la (eqn 19.4)
LN;M; Mz
=
2'
LN;M;
The Zaverage molar mass is obtained from sedimentation equilibria experiments . 019.3
Contour length: the length of the macromolecule measured along its backbone, the length of all its monomer units placed end to end. This is the stretchedout length of the macromolecule, but with bond angles maintained withi n the monomer units. It is proportional to the number of monomer units, N , and to the length of each unit (eqn 19.30). Root mean square separation: one measure of tbe average separation of tbe ends of a random coil. It is the square root of the mean value of R2 , where R is the separation of the two ends of the coil. This mean val ue is calculated by weighting each possible value of R2 with the probability, f (eqn 19.27), of that value of R occurring. It is proportional to N 1/ 2 and the length of each unit (eqn 19.31).
MATERIALS 1: MACROMOLECULES AND AGGREGATES
371
Radius of gyration: the radius of a thin hollow spherical shell of the same mass and moment of inertia as the macromolecule. In general , it is not easy to visualize this distance geometrically. However, for the simple case of a molecule consisting of a chain of identical atoms this quantity can be visualized as the root mean square distance of the atoms from the center of mass. It also depends on N I / 2, but is smaller than the root mean square separation by a factor of (1/6) I / 2 (eqn 19.33). 019.5
For a molecular mechanics calculation, potential energy functions are chosen for all the interactions between the atoms in the molecule; the calculation itself is a mathematical procedure that locates the energy minima (local and global) of the molecule as a function of bond distances and bond angles. Because only the potential energy is included in the calculation, contributions to the total energy from the kinetic energy are excluded in the result. The global minimum of a molecular mechanics calculation is a snapshot of the molecular structure at T = O. No equations of motion are solved in a molecular mechanics calculation. The structure of a macromolecule (or any molecule, for that matter) can, in principle, be determined by solving the time independent Schr6dinger equation for the molecule with methods similar to those described in Chapter II. But, due to the very large size of macromolecules, these methods may be impractical and, due to approximations to make them tractable, inaccurate. In a molecular dynamics calculation, equations of motion are integrated to determine the trajectories of all atoms in the molecule. The equations of motion can, in principle, be either classical (Newton's laws of motion) or quantum mechanical. But, in practice, due to the very large number of atoms in a macromolecule, Newton's equations of motion are used. Quantum mechanical methods are too time consuming, complicated, and at this stage too inaccurate to be popular in the field of polymer chemistry.
019.7
A surfactant is a species that is active at the interface of two phases or substances, such as the interface between hydrophilic and hydrophobic phases. A surfactant accumulates at the interface and modifies the properties of the surface, in particular, decreasing its surface tension. A typical surfactant consists of a long hydrocarbon tail and other nonpolar materials, and a hydrophilic head group, such as the carboxylate group, C0 2, that dissolves in a polar solvent, typically water. In other words, a surfactant is an amphipathic substance, meaning th at it has both hydrophobic and hydrophilic regions. How does the surfactant decrease the surface tension? Surface tension is a result of cohesive forces and the solute molecules must weaken the attractive forces between solvent molecules. Thus molecules with bulky hydrophobic regions such as fatty acids can decrease the surface tension because they attract solvent molecules less strongly than solvent molecules attract each other. See Section 19. IS(b) for an analysis of the thermodynamics involved in this process .
019.9
A Langmuir Blodgett (LB) film is a monolayer or multilayer film that has been placed upon a substrate by transferring a surface film from a liquid to the substrate. A Langmuir trough, shown in Fig. 19.1 (a), is designed to perform the transfer. A surface film of waterinsoluble, fi lmforming molecules is assembled upon the water by mechanical compression. Dipping and withdrawing the substrate affects monolayer transfer. Repeated dipping produces multi layers. Weak van der Waals forces hold the monolayers together. Selfassembled monolayers (SAMs) do not require assembl y by mechanical compression. SAMs form from charged materials that have adsorptiondesorption properties that promote selfassembly as shown in Fig . 19.1(b). The substrate is si mply immersed in a dispersion of the charged materials, withdrawn, and rinsed. Films are held together with either strong ionic bonds or covalent bonds.
372
STUDENT'S SOLUTIONS MANUAL Substrate
Figure 19.1(a)
selfassembly ~

F;oo",19.1(b)
Both methods yield wellorganized monolayers but LB films upon water provide better organizational control than is possible with spontaneous selfassembled films. However, not requiring mechanical compression, SAMs are much more versatile. The strong bonding of SAMs gives longlasting, stable films in contrast to the less stable van der Waals LB films.
Solutions to exercises E19.1(b)
The numberaverage molar mass is (eqn 19.1)
Mn =
~ LN;M; =
[3 x (62)
+ 2 ~ (78)] kg mol
I
I
= 68 kg mol  I
I
The massaverage molar mass is (eqn 19.3)
_ LN;Ml _ 3 x (62)2 Mw LN;M; 3 x (62) E19.2(b)
+2 X +2 x
(78)2 k II  169k I I g mo g mo (78)
I
For a random coil , the radius of gyration is ( 19.33)
Rg = L(N / 6) 1/ 2 so N = 6(Rg / L)2 = 6 x (18.9 nm / 0.450 nm )2 = 11 .06 x 104 1 E19.3(b)
(a) Osmometry gives the numberaverage molar mass, so
(m l / MI)MI (ml / Md
+ (m2/ M2) M2 + (m2/ M 2)
MATERIALS 1: MACROMOLECULES AND AGGREGATES
373
(b) Lightscattering gives the massaverage molar mass, so
Mw = mIMI +m2 M 2 = (25) x (22) +(75) x (22/3) kgmol I =lllkg molII ml+m2 25+75 E19.4(b)
The formula for the rotational correlation time is
r =
4rra 31/ 3kT
1/(H20, 20 0c) = 1.00 x 10 3 kgm I sI[CRC Handbook] I 3 r = 4rr x (4.5 x 1O 9 m)3 x 1.00 x 10 kgm sI = 19.4 x 10 8 s I 3 x 1.381 X 10 23 J K I x 293 K E19.5(b)
The effective mass of the particles is meff = bm = (I  pl!s)m [19.14] = m  pl!sm = l!pp  l!p = l!(pp  p) where l! is the particle volume and Pp is the particle density. Equating the forces meffrw2 =/s = 6rr1/as [19.15, 19.12] or l!(pp  p)rw2 = 1rra3(pp  p)rw2 = 6rr1/as Solving for s yields 2a 2(pp  p)rw2 s=   '     91/
. .. S2 a~(pp  ph (a2)2 (pp  ph Thus, the relatIve rates of sedimentatIOn are  = 2 = . Sl a l (pp  P)I al (pp  P)I The value of this ratio depends on the density of the solution. For example, in a dilute aqueous solution with p = 1.0 I g cm 3 , the difference in polymer densities matters in that the factor involving densities is significantly different than I: S2 2 ( 1.10  0.794)  = (8.4) SI ( 1.18  0.794)
~ =~
In a less dense organic solution, for example a dilute solution in octane with p = 0.71 gcm  3, the density difference has a smaller effect, for the factor involving densities is closer to 1: S2 2 (1.10  0.71h  = (8.4) SI (1.18  0.71)1
~ =~
In both cases, the larger particle sediments faster. E19.6(b)
The molar mass is related to the sedimentation constant through eqns 19.19 and 19.14: 
SRT
SRT
M = bD = :(I pl!s:)D
374
STUDENT'S SOLUTIONS MANUAL
where we have assumed the data refer to aqueous solution at 298 K. (7.46 x 1O 13 s) x (8.314SJK 1 molI) x (298K)

~~~~~~~~~~ [I  (lOOOkgm 3 ) x (8.01 x 1O4 m 3 kg I)] x (7.72 x 10 11 m 2 s l )
Mn=
= 1120kgmOI 1 1 E19.7(b)
See the solution to Exercise 19.5(b). In place of the centrifugal force meff r 2 we have the gravitational force meffg. The rest of the analysis is similar, leading to
2a2 (pp  p)g ~
• 
(2) x (lS.S x 
9T]

=11.47 x E19.8(b)
m ) 2 x (12S0  1000) kgm 3 x (9.81 m s 2 ) (9) x (8.9 x 1O 4 kgm IS I)
1O6
~~~~
1O4
m s11
The molar mass is related to the sedimentation constant through eqns 19.19 and 19.14: 
SRT
SRT
M          bD  (I  pvs)D
Assuming that the data refer to an aqueous solution, (S.l x 1O 13 s) x (8.314SJ K I molI) x (293K) 1 6 _I 1 = S kg mol [ 1  (0.997 gcm 3 ) x (0.72 1 cm 3 g I)] x (7 .9 x 10 11 m 2 s I)
M = E19.9(b)
In a sedimentation experiment, the weightaverage molar mass is given by (eqn 19.20) Mw
=
2RT (ri 
C2
rf)bw 2
In 
so
CI
C2
In CI
=
Mw(r~  r? )bw 2 2RT
='
This implies that M r 2 bw2
~RT
In C =
so the plot of In
C
Mwbw2
In
+ constant
versus r2 has a slope m equal to 
=  and M w = 2RT
2RTm   2
bw
(8.3 14SJK 1 mol I) x (293K) x (821 cm  2 ) x ( IOOcmm  I)2 Mw = [I _ (lOOOkgm 3 ) x (7.2 x 1O 4 m 3 kg I)] x [(1080 s l ) x (2JT)]2
_
2
X
=13 . 1 x 10 3 kg mol  I 1 E19.10(b) The centrifugal acceleration is
2 2 a = rw so a / g = rw / g
a/
3 = (S. SO cm) x [2JT X ( 1.32 x 10 s I) ] 2 = 13.86 x 105 1 I g ( IOOcm m ) x (9.81 m s 2) . .
MATERIALS 1: MACROMOLECULES AND AGGREGATES
375
E19.11(b) For a random coil, the rms separation is [19.31)
Rrms = N 1/ 2 l = ( 1200) 1/2 x (1.1 25 nm ) = 138.97 nm 1 E19.12(b) Polypropylene is  (CH (CH3)CH2)  N, where N is given by
N
=
174kgmol 1 0 ..::::... :I = 4.13 x I 3 3 42.1 x 10 kg mol
Mpolymer Mmonomer
The repeat length l is the length of two CC bonds. The contour length is [19.30) Rc = Nl = (4.13 x 103 ) x (2 x 1.53 x IO Io m) = 11.26 x 10 6 m 1
The rms seperation is [19.31) Rrms = INI /2 = (2 x 1.53 x 10 10 m) x (4.13 x 103 ) 1/2 = 11.97
X
10 8 m 1= 19.7 nm
Solutions to problems Solutions to numerical problems P19.1
s S =  2 [19.16) . rev
. dr s I dr dIn r Since s =   =   =  dt ' r
r dt
dt
and, if we plot In r against t , the slope gives S through 1 dIn r
S= ev 2 dt .
The data are as follows t/ min
15.5
29.1
36.4
58.2
r/cm
5.05 1.619
5.09 1.627
5.12 1.633
5.19 1.647
In(rlcm )
The points are plotted in Fig. 19.2. The leastsquares slope is 6.62 x 10 4 min  I, so S=
P19.3
6.62 x 10 4 min  I (6.62 x 10 4 minI ) x (I min / 60 s) 1 3 ~ 2 = 2 =4.97 x 10 s or~. 4 ev (2IT x 4.5 x 10 / 60 s)
[1)) = lim (1) / 1)0  I) [19 .23). c+o
C
376
STUDENT'S SOLUTIONS MANUAL
1.64
c ..... , ... ... , ... .
Eu
.E ~
1.62
1.60
o
40
20
60
I/ min
Figure 19.2
We see that the yintercept of a plot of the righthand side against c, extrapolated to c begin by constructing the following table using 1)0 = 0.985 g m I sI.
1)/1)0 c
I)
3
/ (dm gI)
1.32
2.89
5.73
9.17
0.0731
0.0755
0.0771
0.0825
= 0, gives [1)]. We
(
The points are plotted in Fig. 19.3 . The leastsquares intercept is at 0.0716, so [1)]
8.2
,.
8.0
OIl
Ma ~
,.
7.8
):!..
7.6
>?

S 7.4 00 7.2
Figure 19.3
= 10.0716 dm3 gI
I.
MATERIALS 1: MAC ROMOLECULES AND AGGREGATES
P19.5
377
We follow the procedure of Example 19.5. Also compare to Problems 19.3 and 19.4. [17]
= lim
co (
17 / 170 
I)
[ 19.23]
c
and
[17]
= K~
[19.25]
with K and a from Table 19.4. We draw up the following table using 170
17 / ( 10 3 kg m 
I S I )
= 0.647
x 10 3 kg m I s I.
o
0.2
0.4
0.6
0.8
1.0
0.647
0.690 0.332
0.733 0.332
0.777 0.335
0.821 0.336
0.865 0.337
« 17 / 170  1)/c) /(100cm 3 g l )
The values are plotted in Fig 19.4, and the yintercept is 0.330 .
.,
0.336
eo
~E
u
8 0.334
~ Ie
~
0.332
0.330 0
0.2
0.4
0.6
0.8
cI(g1 I00 em 3)
Hence [17]
and
=
M
y
g mol I
(0.330) x ( 100cm 3 gI)
=
(33 0
3  I
. cm g 8.3 x 10 2 cm 3 g I
That is, M = G58 kg mol  I P19.7
Figure 19.4
= 33.0cm 3 g I ) 1/ 0.50
= 158 x 10 3
.
I.
The empirical MarkKuhnHouwinkSakurada equation [19.25] is
As the constant a may be nonintegral the molar mass here is to be interpreted as unitiess, that is, as M y I(g mol  I). The units of K are then the same as those of [17]. We fit the data to the above equation and obtain K and a from the fitting procedure. The plot is shown in Fig 19.5.
1
K = 0.0117 cm 3 g I
1
and
1a = 0.7171·
378
STUDENT'S SOLUTIONS MANUAL
y
= 0.01167xo.7 1661 ,
R
= 0.99983
250
200
,. Oil
ISO
M e ~
.£
100
50.0
0.00
'2L.J,,'L.......L..L.......,'''..L......L..I....L.......1,L,,L.....J
o
0.4
0.2
0.8
0.6
1.0
Figure 19.5
(Many plotting programs can fit a power series directly. If not, the equation can be transformed into a linear one In[1)]
= InK +alnMy
so a plot of In [1)] versus In M v will have a slope of a and a yintercept of In K.) COMMENT. This value for a is not much different from that for polystyrene in benzene listed in Table 19.4.
This is somewhat surprising as one would expect both the K and a values to be solventdependent. THF is not chemically similar to benzene . On the other hand, benzene and cyclohexane are very much alike, yet the values of K and a as determined in Example 19.5 are markedly different from those in Table 19.4 for polystyrene in cyclohexane.
P19.9
See Section 5.5(e) and Example 5.4.
h
RT
BRT
e
pgMn
pgMn
 = = +
2'
e [Example 5.4].
We plot hie against e. Draw up the following table. c/(g/lOO cm 3 )
0.200
0.400
0.600
0.800
1.00
hlcm h  / (100 cm 4 g l) e
0.48
1.12
1.86
2.76
3.88
2.4
2.80
3.10
3.45
3.88
The points are plotted in Fig. 19.6, and give a leastsquares intercept at 2.043 and a slope 1.805. Therefore, RT I pgMn  _ Mn 
=
(2.043) x (100 cm 4 g I)
=
2.043
X
10 3 m4 kgI and hence
(8 .314 J K  1 molI) x (298 K) _ (0.798 x 10 3 kg m 3 ) x (9.81 m s2) x (2 .043 x 10 3 m4 kg I)
=
1
k II 1 . 155 g mo
MATER IALS 1: MACROMOLE CU LES AN D AGGR EG ATES
379
4
,.
01)
"uE
8
3
~ ~
0.2
0.4
0.6
0.8
cI(gli ()() em 3)
Figure 19.6
From the slope, 4
BRT? = ( 1.805) x
pgM ~
(~OO cm g~ l)
= 1.805
X
104 cm 7 g 2 = 1.805
X
10 4 m7 kg 2
",/(100 cm )
and hence
B=
(p!~n)
x Mn x ( 1.805 x 10 4 m 7 kg 2 )
(155 kg mol  I) x ( 1.805 x 10 4m7 kg 2)
2.043 x 10 3 m4 kg I = 113.7 m 3 mol I I.
Solutions to theoretical problems P19.11
See the discussion of radius of gyration in Section 19.8(a). For a random coi l Rg ex N I / 2 ex MI / 2. For a rigid rod, the radius of gyration is proportional to the length of the rod, which is in turn proportional to the number of polymer units, N , and therefore also proportional to M. Therefore, poly(ybenzylLglutamate) is rodlike whereas polystyrene is a random coil (in butanol ).
P19.13
Call the constant of proportionality K , and evaluate it by requiring that L etMM = (2y)I / 2X
and N =
10
00
so
1dN =
N.
dM=(2y)I /2dt
Ke (M A1)l / 2y dM = K (2y) 1/2
1: e
x2
dx where a = M / (2y)I /2.
Note that the point x = 0 represents M = M, and x =  a represents M = O. In a narrow distribution, the number of molecules with masses much different than the mean falls off rapidly as one moves away
380
STUDENT'S SOLUTIONS MANUAL
from the mean; therefore, dN
Hence, K =
Mn
N
(2ny)
= 1 N
=
f
~
0 at M ::: 0 (that is, at x ::: a) . Therefore
1/2 . It then follows from turning eqn 19.1 into an integral that
MdN
I
1 1
(2ny) I/2
00
1
(2ny)I /2 2y
= () n
1/2
1
00
=
[(2y)I /2x
0
00
(
a
_x2
xe
 ? / 2Y dM Me(MM)
0
+ M]e
X2
(2y)I /2dM
M _x2 )
+1 /2 e (2y)
dx.
Once again extending the lower limit of integration to 
P19.15
00
adds negligibly to the integral, so
(a) Following Justification 19.4, we have
.
a
3
2 _ a2R2
wlthf=4n(n l / 2 ) R e 2
Therefore, Rrms =4n
= Hence, Rrms =

3
2a
(
a )3
n l/ 2
3
,a= ( 2Nl2 )
{oo Re4
Jo
_ a2 R2
1/2 [19 .27].
(
a )3
dR=4n n l / 2
x
(
3)
"8 x
(n) 1/2 aiD
2 2 =Nl .
11N I / 2 1.
(b) The mean separation is
(c) The most probable separation is the value of R for which! is a maximum, so set d! / dR = 0 and solve for R.
MATERIALS 1: MACROMOLECULES AND AGGREGATES
381
Therefore, the most probable separation is
= 4000 and 1 = 154 pm, R rm s = 19.74 nm I; (b) Rmean = 18.97 nm I;
When N (a) P19.17
(c) R*
= 17.95 nm I·
We use the definition of the radius of gyration given in Problem 19.19, namely,
R~ = ~ LRJ. j
(a) For a sphere of uniform density, the center of mass is at the center of the sphere. We may visualize
the sphere as a collection of a very large number, N, of small particles distributed with equal number density throughout the sphere. Then the summation above may be replaced with an integration. 2
Rg =
I N
J; r 2P(r)dr
N=~::7.~':P(r)dr
P(r) is the probability per unit distance that a small particle will be found at distance r from the center, that is, within a spherical shell of volume 4Jtr 2dr. Hence, P(r) = 4Jtr2dr. If P(r) were normalized, the integral in the numerator would represent the average value of r2 , so N times that integral replaces the sum. The denominator enforces normalization. Hence
Rg
_(3)
"5
1/ 2
a.
(b) For a long straight rod of uniform density the center of mass is at the center of the rod and P( z) is constant for a rod of uniform radius; hence,
2
Rg
COMMENT.
rl / 2 Z2d Z 2 Jo
13 (1/) 2 3
I
2
= 2Jci/2 dz = ~ = 12 1 ,
~ 1 Rg
= 2.J3
.
The radius of the rod does not enter into the result. In fact, the distribution function is P(r ,z),
the probability that a small particle will be found at a distance r from the central axis of the rod and Z along that axis from the center, that is, within a squat cylindrical shell of volume 2nrdrdz. Integration radially outward from the axis is the same in numerator and denominator.
For a spherical macromolecule, the specific volume is
so
3vsM ) a= (  4JtNA
1/ 3
382
STUDENT'S SOLUTIONS MANUAL
and
= (~) 1/2 x (3 Vs M) 1/ 3
R
5
g
4rrNA
( ~ ) 1/2 x
=
(3 vs/ em 3 gI) x em 3 gI x (MIg mol I) x g mOl  I) 1/ 3 (4 rr ) x (6.022 x 1023 mo l l )
5
=
(5.690 x 10 9 ) x (v s/e m 3 g I) 1/ 3 x (M /g mol I) 1/ 3 em
=
(5.690 x 1O
ll
m) x {(vs/em 3 gI) x (M/g mol I)}1 /3 .
That is Rg/ nm = [ 0.05690 x When M
=
1(v s/e m 3 g l ) x (M /g mol  I )1 13) [.
100 kg mol I and Vs
= 0.750 em3 gI ,
R g/ nm = (0 .05690) x {O.750 x 1.00 x 1O5} 1/3 = 12.40 I.
For a rod, Vmol
Rg
_

= rra 2l, so
Vmo l
2rra 2y'3
_ vsM  x
NA
I 2rra 2y'3
(0.750 em 3 g I) x ( 1.00 x 105 g mol  I) (6.022 x 10 23 mol  I) x (2 rr ) x (0.5 x 10 7 em)2 x y'3
= 4.6 COMMENT.
~=
x 10 6 em
= 146 nm I.
Rg may also be defined through the relation
L:m;rl iL,mi '
Question. Does this definition lead to the same formulas for the radii of gyration of the sphere and the rod as those derived above? P19.19
Refer to Fig. 19.7.
....
~=
j
Figure 19.7 The defi nition in the text (eqn 19.32) is
so
R 2g
=
1~hij " 2 2N2 ij
=
I " " 2' 2N2 ~~hij . j
MATERIALS 1: MACROMOLECULES AND AGGREGATES
383
The scalar quantity hij can be written as the dot product hij . hij . If we refer all our measurements to a common origin (which we will later specify as the center of mass), the interatomic vectors hij can be expressed in terms of vectors from the origin: hij = Rj  R; . (If this is not apparent, note that R; + h ij = Rj .) Therefore
R2 = _1 " "(R  R ) . (R  R) g 2N2 L L } I } I ;
j
=~ " " ( R j . Rj + R;· R; 2N L L I
2R;· Rj)
}
=~ L L(Rj + R;  2R;· Rj ). 2N . . I
}
Look at the sums over the squared terms:
j
j
If we choose the origin of our coordinate system to be the center of mass, then
" L R; " = L Rj
=0
and
1LRj' "2 Rg2 = N
j
j
for the center of mass is the point in the center of the distribution such that all vectors from that point to identical individual masses sum to zero. P19.21
Write I
= aT, then
( ~) aT
=a
(au) at
and, using PI9.20,
I
T
=1aT=O
.
Thus the internal energy is independent of the extension. Therefore
t
= aT =
T(~) aT = IT (as) at I[p19.20] I
T
and the tension is proportional to the variation of entropy with extension. Extension reduces the disorder of the chains, and they tend to revert to their disorderly (nonextended) state.
Solutions to applications P19.23
The center of the sphere cannot approach more closely than 2a; hence the excluded volume is
vp
where
4
(4
= "3rc(2a) 3= 8 "3rca 3) = ~ ~ Vrnol
is a molecular volume.
384
STUDENT'S SOLUTIONS MANUAL
The osmotic virial coefficient, B (see eqn 5.41), arises largely from the effect of excluded volume. If we imagine a solution of a macromolecule being built by the successive addition of macromolecules to the solvent, each one being excluded by the ones that preceded it, then the value of B turns out to be (PI9.18)
where
vp
is the excluded volume due to a single molecule. 1 = NA x 2
B(BSV)
= B(Hb) =
32 rca 3 3
(l~rc)
(I~rc)
16 3 = rca NA 3
x (6.022 x 1023 molI) x (14.0 x
x (6.022 x
10
1023 molI) x (3 .2 x 1O
9
m)3 = 128 m3 mol II.
9 m)3
= I 0.33 m3 mol I I.
.
Smce n = RT [J] +BRT[Jf + .. . [5.41], if we write n ° = RT[J] then
,
n  n°
no
~
BRT[J]2 RT[J]
= B[J] .
For BSV, g [J]=(1.0 ) x (lOdm 3) = M
3 IOgdm=9.35 x 1O 7moldm3 1.07 x 107 g mol I = 9.35 x 1O 4 molm 3
and
nn°
no
For Hb, [J] = and P19.25
nn°
n0
= (28 m3 mol I) x (9.35 10 g dm 3
66.5 x 103 g mo]
X
10 mol m3) = 2.6 x 10 2 corresponding to 12.6 per cent I. 4
I = 0.15 mol m 3
= (0.15 mol m 3) x (0.33 m3 molI) = 5.0 x 10 2 which corresponds tol5 percent I.
(a) We seek an expression for a ratio of scattering intensities of a macromolecule in two different conformations, a rigid rod or a closed circle. The dependence on scattering angle e is contained in the Rayleigh ratio Re. The definition of this quantity, in eqn 19.7, may be inverted to give an expression for the scattering intensity at scattering angle
e
Ie
sin 2 ¢
= ReIo2, r
where ¢ is an angle related to the polarization of the incident light and r is the distance between sample and detector. Thus, for any given scattering angle, the ratio of scattered intensity of two conformations is the same as the ratio of their Rayleigh ratios: Prod
P ee
The last equality stems from eqn 19.8, which related the Rayleigh ratios to a number of angleindependent factors that would be the same for both conformations, and the structure factor (Pe)
MATERIALS 1: MACROMOLECULES AND AGGREGATES
385
that depends on both conformation and scattering angle. Finally, eqn 19.9 gives an approximate value of the structure factor as a function of the macromolecule's radius of gyration Rg , the wavelength of light, and the scattering angle: 3)..2 161[2R~ sin2 (~e) 3)..2 The radius of gyration of a rod of length 1 is Rrod =
1/ (12)1 /2 [Section 19.8(a)].
For a closed circle, the radius of gyration, which is the rms distance from the center of mass [PI9.19], is simply the radius of a circle whose circumference is I:
[ = 21[Ree
[
so
Ree =  . 21[
The intensity ratio is: Irod
3)..2 _11[2[2 sin 2 (~e)
Icc
3)..2  4[2 sin 2 (~e)
Putting the numbers in yields:
er lrod / lee
20 0.976
45 0.876
90 0.514
(b) I would work at a detection angle at which the ratio is smallest, i.e. most different from unity, provided I had sufficient intensity to make accurate measurements. Of the angles considered in part (a), 1900 1is the best choice. With the help of a spreadsheet or symbolic mathematical program, the ratio can be computed for a large range of scattering angles and plotted (Fig. 19.8).
u
0.5
::::u
J 0.0
0.5
L.~~'~~'~~'~~'
o
45
90 (JIO
135
180
Figure 19.8
A look at the results of such a calculation shows that both the intensity ratio and the intensities themselves decrease with increasing scattering angle from 0° through 180°, that of the closed circle conformation changing much more slowly than that of the rod. Note: the approximation used above yields negative numbers for Prod at large scattering angles; this is because the approximation, which depends on the molecule being much smaller than the wavelength, is shaky at best, particularly at large angles.
386
P19.27
STUDENT'S SOLUTIONS MANUAL
The molar mass is given by eqn 19.19 Mn
=
SRT bD
=
(1
SRT D [19.14, for b]  pV s )
_ (4.5 x 1O 13 s) x (8.3 141 K I mol I) x (293 K) _I _ II (1  0.75 x 0.998) x (6.3 x 10 11 m2 SI) . 69 kg mol . Now combine!
= 6nary [19.12] with! = kT / D [19.11]:
kT
a P19.29
= 6nryD =
(1.381 x 10 23 1 K  I) x (293 K) I (6n) x (1.00 x 10 3 kgm S I) x (6.3 x 10 11 m 2 s l )
= 13.4 nm I·
The isoelectric point is the pH at which the protein has no charge. At that point, then, its drift speed under electrophoresis, s, vanishes. Plot the drift speed against pH and extrapolate the line to s = O. The plot is shown in Fig. 19.9. o~~~~
D. I
D.3
D.4+++++~
4
3.5
4.5
5.5
5
pH
6
Figure 19.9
Isoelectric pH is the xintercept on the graph, that is, the value of x at which y solving the fit equation: s/(l1m/s) so pH
=
 0.17pH + 0.655
= O. One can find this by
=0
= 13.851.
COMMENT.
One could obtain the result to about ±O.05 pH by reading the value directly from the
graph. P19.31
(a) The data are plotted in Fig. 19.10. Both samples give rise to tolerably linear curves, so we estimate the melting point by interpolation using the bestfit straight line. The bestfit equation has the form Tm/ K = m! + b, and we want Tm when! = 0.40: 10 2 mol dm  3 :
Csalt
= 1.0 x
Csalt
= 0.15 mol dm 3 :
Tm
Tm
= (39.7 x 0.40 + 324) K = 1340 K I·
= (39.7 x 0.40 + 344) K = 1360 K I·
MATERIALS 1: MACROMOLECULES AND AGGREGATES
375
365
::.:
f.=
•
..
370
387
• 0.01 • 0.15 Linear (0.01)
···· ··· ···f ~· · ·· · ···
360 355 350 345 340 335 0.3
0.4
0.6
0.5
0.7
f
0.8
Figure 19.10
(b) The slopes are the same for both samples. The different concentrations of dissolved salt simply offset the melting temperatures by a constant amount. The greater the concentration, the higher the melting point. This behavior is not what is typically observed with small molecules, where the presence of dissolved impurities disrupts freezing and depresses the freezing point. The dissolved ions can interact with charged regions of the macromolecule that might otherwise experience unfavorable intramolecular interactions. For example, if two regions bearing negative charge would have to approach each other in the absence of dissolved salts, the incorporation of a cation very close to each region and an anion in between tbem would turn an unfavorable interaction into a favorable one. (See Fig. 19. 11).
Figure 19.11
The melting points are greater at both larger fractions of Gc base pairs and at larger salt concentrations. Tm increases with the number of Gc base pairs because this pair is held togethar with the three hydrogen bonds in the double helical structure, whereas the A T pair is held with two hydrogen bonds (see Section 19.11). The f:!"Hm contribution is greater for the Gc pair. Low salt concentrations destabilize the double helix by inadequately contributing to the attractive forces between the solution and the sugarphosphate backbone of the double helix. This makes it easier for a base to rotate out from the center of the double helix. P19.33
The peaks are separated by 104 g mol I , so this is the molar mass of the repeating unit of the polymer. This peak separation is consistent with the identification of the polymer as polystyrene, for the repeating group of CH2CH(C6Hs) (8 C atoms and 8 H atoms) has a molar mass of 8 x (12 + I) g molI = 104 g moli . A consistent difference between peaks suggests a pure system and points away from different numbers of subunits of different molecular weight (s uch as the Ibutyl initiators) being incorporated into the polymer molecules . The most intense peak has a molar mass equal to that of n repeating groups plus
388
STUDENT'S SOLUTIONS MANUAL
that of a silver cation plus that of terminal groups: M(peak)
= nM(repeat) + M(Ag+) + M(terrninal) .
If both ends of the polymer have terminal tbutyl groups, then
= 2M(tbutyl) = 2(4
M(terminal) and n
P19.35
=
x 12 + 9) g mol 1
M(peak)  M(Ag+)  M(terrninal) M(repeat)
=
=
114 g molI
25598  108  114 104
= 12441 .
The procedure is that described in Problem 19.7. The data are fitted to the MarkKuhnHouwinkSakurada equation. [1)]
= KM".,
[19.25].
The values obtained for the parameters are K
= 12.38
X
10 3 cm 3 gI 1 and
a
= I0.9551·
This K value is smaller than any in Table 19.4 or that in Problem 19.7. The value for a is quite close to 1. When a = 1 exactly, the molar mass, M v corresponds to the weight average molar mass, M w · COMMENT.
The magnitude of the constant a reflects the stiffness of the polymer chain as a result of
rrorbital interactions between heterocyclic rings.
20
Materials 2: the solid state
Answers to discussion questions 020.1
Lattice planes are labeled by their Miller indices h, k, and I , where h, k, and 1 refer respectively to the reciprocals of the smallest intersection di stances (in units of the lengths of the unit cell , a, b, and c) of the plane along the x , y, and z axes.
020.3
If the overall amplitude of a wave diffracted by planes (hk/ ) is zero, that plane is said to be absent in the diffraction pattern. When the phase difference between adjacent planes in the set of planes (hkl) is n , destructive interference between the waves diffracted from the planes can occur and this wi ll diminish the intensity of the diffracted wave. This is illustrated in Fig. 20.21 in the text. The overall intensity of a diffracted wave from a plane (hkl) is determined from a calculation of the structure factor, Fhkl , which is a function of the positions (hence, of the Miller indices) and of the scattering factors of the atoms in the crystal (see eqn 20.7). If Fhkl is zero for the plane (h kl) , that plane is absent. See Example 20.3.
020.5
The majority of metals crystalli ze in structures that can be interpreted as the closest packing arrangements of hard spheres. These are the cubic closepacked (ccp) and hexagonal closepacked (hcp) structures. In these models, 74% of the volume of the unit cell is occupied by the atoms (packing fraction = 0.74). Most of the remaining metallic elements crystallize in the bodycentered cubic (bcc) arrangement, which is not too much different from the closepacked structures in terms of the efficiency of the use of space (packing fraction 0.68 in the hard sphere model). Polonium is an exception; it crystallizes in the simple cubic structure, which has a packing fraction of 0.52. See the solution to Problem 20.24 for a derivation of all the packing fraction s in cubic systems. If atoms were trul y hard spheres, we would expect that all metals would crystalli ze in either the ccp or hcp closepacked structures. The fact that a significant number crystallize in other structures is proof that a simple hard sphere model is an inaccurate representation of the interactions between the atoms. Covalent bonding between the atoms may influence the structure.
020.7
Because enantiomers give almost identical diffraction patterns it is difficult to distinguish between them. But absolute configurations can be obtained from an analysis of small differences in diffraction intensities by a method developed by J.M . Bijvoet. The method makes use of extra phase shifts that occur when the frequency of the Xrays approaches an absorption frequency of atoms in the compound . The phase shifts are called anomalous scattering and result in different intensities in the diffraction patterns of different enantiomers. See Section 23.7(b) of the 7th edition of this text for an explanation of the origin of this anomalous phase shift. The incorporation of heavy atoms into the compound makes the observation of the extra phase shift easier to observe, but with very sensitive modern diffractometers this is no longer strictly necessary.
390
020.9
STUDENT'S SOLUTI ONS MANUAL
The FenniDirac distribution is a version of the Boltzmann distribution that takes into account the effect of the Pauli exclusion principle. It can therefore be used to calculate the population, P, of a state of given energy in a manyelectron system at a temperature T: P
=
1
+I'
';:''C7,':;:
e (E/.L )/ kT
In this expression, /l is the Fermi energy, or chemical potential, the energy of the level for which P = 1/ 2. The Fenni energy should be distinguished from the Fermi level, which is the energy of the highest occupied state at T = O. See Fig. 20.54 of the text. From thermodynamics (Chapter 3) we know that dU = pdV + TdS + /ldn for a onecomponent system. This may also be written dU = pdV + TdS + /ldN, and this /l is the chemical potential per particle that appears in the FD distribution law. The term in dU containing /l is the chemical work and gives the change in internal energy with change in the number of particles. Thus, /l has a wider significance than its interpretation as a partial molar Gibbs energy and it is not surprising that it occurs in the FD expression in comparison to the energy of the particle. The Helmholtz energy, A, and /l are related through dA =  p dV  S dT + /l dN, and so /l also gives the change in the Helmholtz energy with change in number of particles. To fully understand how the chemical potential /l enters into the FD expression for P, we must examine its derivation (see Further reading) which makes use of the relation between /l and A and of that between A and the partition function for F D particles.
Solutions to exercises E20.1(b)
(1 , 0,1) is the midpoint of a face. All face midpoints are alike, including
G,!, 0) and (0, !,!) .
There are six faces to each cube, but each face is shared by two cubes. So other face midpoints can be described by one of these three sets of coordinates on an adjacent unit cell. E20.2(b)
Taking reciprocals of the coordinates yields ( I, ~, I) and
I
yields the Miller indices (3 13) and (643) E20.3(b)
(!, ~, ! ) respectively. Clearing the fractions
I
The distance between planes in a cubic lattice is a dhkl
=
(h2
+ k2 + [2) 1/ 2
This is the distance between the origin and the plane which intersects coordinate axes at (h la, k i a , [I a). d

121 
(I
523pm
+ 22 + I) 1/ 2
 \ 214 m\ P
MATERIALS 2: THE SOLID STATE
E20.4(b)
391
The Bragg law is /1A
= 2d sine
Assuming the angle given is for a firstorder reflection, the wavelength must be A = 2(128.2 pm) sin 19.76° = 186.7 pm 1
E20.S(b)
Combining the Bragg law with Miller indices yields, for a cubic cell sin ehkl
= ~ (h2 + k 2 + p)I / 2 2a
In a facecentered cubic lattice, h, k, and l must be all odd or all even. So the first three reflections would be from the ( I I I ), (2 0 0), and (2 2 0) planes. In an fcc cell , the face diagonal of the cube is 4R, where R is the atomic radius. The relationship of the side of the unit cell to R is therefore
4R
so
a=
../2
Now we evaluate
A A  =  = 2a 4../2R
154pm
4../2 (144 pm)
=0.189
We set up the following table
E20.6(b)
e
hkl
sin
III 200 220
0.327 0.378 0.535
19.1 22.2 32.3
38.2 44.4 64.6
In a circular camera, the di stance between adjacent lines is D = R6. (2e), where R is the radius of the camera (distance from sample to film) and is the diffraction angle. Combining these quantities with the Bragg law (A = 2d sin relating the glancing angle to the wavelength and separation of planes), we get
e
e,
d)
1 A D = 2R6.e = 2Rt{sin 2
= 2(5 .74cm ) x ( sinE20.7(b)
I
I
96.035 95.401 pm )  sin  I = 0.054cm 2(82.3 pm) 2(82.3 pm )
I
The volume of a hexagonal unit cell is the area of the base times the height c. The base is equivalent to two equilateral triangles of side a. The altitude of such a triangle is (/ sin 60°. So the volume is
v=
2
Ua x asin 60° ) c = (/2 c sin 60° = (1692.9 pm)2 x (506.96 pm) x sin 60°
= 1.2582 x 109 pm 3 = 11.2582 nm 3 1
392
E20.8(b)
STUDENT'S SOLUTIONS MANUAL
The volume of an orthorhombic unit cell is V = abe= (589 pm) x (822 pm) x (798 pm) =
3.86 x 108 pm 3 10 3 = 3.86 x 1O 22 cm 3 (10 pmcm I)
The mass per formula unit is m=
135.01 gmol I = 2.24 x 10 22 g 6.022 x 1023 molI
The density is related to the mass m per formula unit, the volume V of the unit cell, and the number N of formula units per unit cell as foHows
Nm
3 22 N = P V = (2.9 g cm ) x (3.86 x 10 cm 3) = Isl m 2.24 x 10 22 g L::J
so
p=
V
A more accurate density, then, is p =
E20.9(b)
22 5(2.24 x 10 g) 1 31 = 2.90gcm 3.86 x 10 22 cm 3
The di stance between the origin and the plane which intersects coordinate axes at (hla, klb, lie) is given by h2
k2
l2 )1 /2
dhkl = (  2 +  2 +  Z a b c d322
= 1182pm
=
(32 (679 pm)2
+
22 (879 pm)z
22 )1 /2 + ;:(860 pm)z
I
E20.10(b) The fact that the III reflection is the third one implies that the cubic lattice is simple, where all indices give reflections. The III reflection would be the first reflection in a facecentered cubic cell and would
be absent from a bodycentered cubic. The Bragg law
can be used to compute the cell length a=
A 2sinehkl
(h2+k2+p)I /2=
137pm (I2+12+12)1 /2=390pm 2sin 17.7°
With the cell length, we can predict the glancing angles for the other reflections expected from a simple cubic ehkl = sinI elOO
(~ (h 2 + k Z + P)I / Z)
= sin I (0.176(h 2 + k Z + p)I /Z)
= sinI (0.176(1 2 +0+0)1 /2) = 10.1° (checks)
ello = sinI (0.176(12 + 12 +O)I / Z) = 14.4° (checks)
ezoo = sin  I (0.176(2 2 + 0 + O)I /Z) = 20.6° (checks)
MATERIALS 2: THE SOLID STATE
393
These angles predicted for a si mple cubic fit those observed, confirming the hypothesis of a simple lattice; the reflections are due to the ( 100), ( 110), (III), and (200) planes.
I
I
E20.11(b) The Bragg law relates the glancing angle to the separation of planes and the wavelength of radiation
A = 2d sin
e
so
e=
A sin  \ 2d
The distance between the origin and plane which intersects coordinate axes at (hla, klb, lie) is given by
So we can draw up the following table
hkl 574.1 796.8 339.5
100 010 III
4. 166 3.000 7.057
E20.12(b) All of the reflections present have h + k + I even, and all of the even h + k + I are present. The unit cell,
I
then, is bodycentered cubic
I
E20.13(b) The structure factor is given by
All eight of the vertices of the cube are shared by eight cubes, so each vertex has a scattering factor off/8. The coordinates of all vertices are integers, so the phase
When h + k + I is even,
= I; when h + k + I is odd,
= 8(f/ 8)(I) +f(_I)h+k+1 = 12f for h + k + I even
and
0 for h + k
+ I odd I
E20.14(b) There are two smaller (white) triangles to each larger (gray) triangle. Let the area of the larger triangle be
A and the area of the smaller triangle be a. Since b
= ~B(base) and h = ~H(height), a = !A . The white
394
STUDENT'S SOLUTIONS MANUAL
space is then 2N A/4, for N of the larger triangles. The total space is then (NA Therefore the fraction filled is NA/(3NA/2) = 12/31
+ (NA/2» =
3NA/2.
E20.15(b) See Figure 20.1.
Figure 20.1 The body diagonal of a cube is a.J3. Hence
a.fi
=
2R + 2r
.fiR
or
= R+r
[a
= 2R]
~ = 10.7321 R
E20.16(b) The ionic radius of K+ is 138 pm when it is 6fold coordinated, 151 pm when it is 8fold coordinated.
(a) The smallest ion that can have 6fold coordination with it has a radius of
(../2 1) x (138 pm) =
IS7pm I·
(b) The smallest ion that can have 8fold coordination with it has a radius of ( .J3  1) x (151 pm) = 1lllpml· E20.17(b) The diagonal of the face that has a lattice point in its center is equal to 4r, where r is the radius of the
atom. The relationship between this diagonal and the edge length a is 4r
= a../2
so
a
= 2hr
The volume of the unit cell is a 3 , and each cell contains 2 atoms. (Each of the 8 vertices is shared among 8 cells; each of the 2 face points is shared by 2 cells.) So the packing fraction is 3
2Vatom _ 2(4/3)7l"r _ _ 7l"_ 10 1 370 VeeU  (2../2r) 3  3(2)3/2 .
E20.18(b) The volume of an atomic crystal is proportional to the cube of the atomic radius divided by the packing
fraction. The packing fraction for hcp, a closepacked structure, is 0.740; for bcc, it is 0.680. So for titanium 0.740(122 p m)3 Vhcp = 0.680 126 pm = 0.99
Vbce
MATERIALS 2: THE SOLID STATE
395
The bcc structure has a smaller volume, so the transition involves a 1contraction I. (Actually, the data are not precise enough to be sure of this. 122 could mean 122.49 and 126 could mean 125.51, in which case an expansion would occur.) E20.19(b) Draw points corresponding to the vectors joining each pair of atoms. Heavier atoms give more
intense contributions than light atoms. Remember that there are two vectors joining any pair of atoms
(AB and As); don ' t forget the AA zero vectors for the center point of the diagram . See Figure 20.2 for C6H6 ·
Figure 20.2
E20.20(b) Combine E
=
!kT and E
=
!mv 2
=
2::,:2 ' to obtain
6.626 X 10 34 J s  (mkT) I/ 2  [(1.675 x 10 27 kg) x ( 1.381 x 10 23 J K 1) x (300 K)]1 /2 h
A

 I252 pm 1
E20.21 (b) The lattice enthalpy is the difference in enthalpy between an ionic solid and the corresponding isolated
ions. In this exercise, it is the enthalpy corresponding to the process MgBr2(s)
+
Mg2+ (g)
+ 2Br  (g)
The standard lattice enthalpy can be computed from the standard enthalpies given in the exercise by considering the formation of MgBr2(s) from its elements as occuring through the follow ing steps: sublimation of Mg(s), removing two electrons from Mg(g), vaporization of Br2(I) , atomization of Br2(g), electron attachment to Br(g) , and formation of the solid MgBr2 lattice from gaseous ions
So the lattice enthalpy is f..LW(MgBr2 , s)
=
f.. SUbH G (Mg, s)
+ f.. aIH G f..LH
(MgBr2' s)
G
+ f..ionHG(Mg, g) + f.. vapH
(Br2, g)
+ 2f.. eg H
= [148 + 2 187 + 3 1 + 193 
G
G (Br2, I)
(Br, g)  f..rW(MgBr2 ' s)
2(331)
+ 524] kJ
mol  I
= 12421 kJ mol I I
396
STUDENT'S SOLUTIONS MANUAL
E20.22(b) Tension reduces the disorder in the rubber chains; hence, if the rubber is sufficiently stretched, crystallization may occur at temperatures above the normal crystallization temperature. In un stretched rubber the random thermal motion of the chain segments prevents crystallization. In stretched rubber these random thermal motions are drastically reduced. At higher temperatures the random motions may still have been sufficient to prevent crystallization even in the stretched rubber, but lowering the temperature to 0 °C may have resulted in a transition to the crystalline form. Since it is random motion of the chains that resists the stretching force and allows the rubber to respond to forced dimensional changes, this ability ceases when the motion ceases. Hence, the seals failed . COMMENT. The solution to the problem of the cause of the Challenger disaster was the final achievement, just before his death , of Richard Feynman , a Nobel prize winner in physics and a person who loved to solve problems. He was an outspoken person who abhorred sham, especially in science and technology. Feynman concluded his personal report on the disaster by saying, 'For a successful technology, reality must take precedence over public relations, for nature cannot be fooled ' (James Gleick, Genius: The Life and Science of Richard Feynman . Pantheon Books, New York (1992) .)
E20.23(b) Young's modulus is defined as:
E
normal stress normal strain
= ,
where stress is deforming force per unit area and strain is a fractional deformation. Here the deforming force is gravitational, mg, acting across the crosssectional area of the wire, Jr r2. So the strain induced in the exercise is . stram
stress
=~=
mg Jr(d/2)2E
=
4mg Jrd2E
=
4(1O.0kg)(9.8ms 2) Jr(O.1O x 103 m) 2(2 15 x 109 Pa)
1
=
5.8 x I
0 2 1
The wire would stretch by 5.8%.
E20.24(b) Poisson's ratio is defined as: Vp
=
transverse strain normalstra"'in
where normal strain is the fractional deformation along the direction of the deforming force and transverse strain is the fractional deformation in the directions transverse to the deforming force. Here the length of a cube of lead is stretched by 2.0 percent, resulting in a contraction by 0.41 x 2.0 percent, or 0.82 percent, in the width and height of the cube. The relative change in volume is:
v + 6V = ( 1.020) (0.9918) (0.9918) = 1.003 V
and the absolute change is: 6V
=
(1.003  1)(1.0dm 3 )
= 1 0.003 dm 3 1
E20.2S(b) ptype; the dopant, gallium, belongs to Group 13, whereas germanium belongs to Group 14. E20.26(b)
Eg
= hVmin
and
Vmin
= Eg / h =
1.I2eV (1.602 X IO6.626 x 10 34 J s leV
19
J) = I
2.71 x 10
14
Hz
I
MATERIALS 2: THE SOLID STATE
E20.27(b)
m = ge lS(S + I)} 1/2/18 Therefore, si nce m
[20.34, with S in place of s1
= 4.oo/1B
+ I) = (£)
S(S
397
X
(4.00)2
= 4.00,
S
impl yi ng that
= 1.56
Thus S ~ ~, implying three unpaired spins. In actuality most Mn2+ compounds have [ ] unpaired spins. E20.28(b)
XM
Xm
= XVm = p =
(7 .9 x 10 6 ) x (84.15 gmol l ) 0.811 gcm 3
I
I
= 18 .2 x 10 4 cm3 mol  I = '18.2 xl0 IOm3mo1, 1 E20.29(b) The molar susceptibility is given by
N02 is an oddelectron species, so it must contain at least one unpaired spin ; in its ground state it has one unpaired spin, so S = Therefore,
!.
Xm = (6.022 x 10 23 mol  I) x (2.0023)2 x (9.274 x 10 24
x
10 7 T2 r l m 3 )
(41T X
1T1)2 x (!) x (! + I)
~~~~~~~~~~
3(1.381 x 1O 23 JK  I) x (298K)
= 11.58 x
10 8 m 3 mol I I
The expression above does not indicate any pressuredependence in the molar susceptibility. However, the observed decrease in susceptibility with increased press ure is consistent with the fact that N02 has a tendency to dimerize, and that dimerization is favored by higher pressure. The dimer has no unpaired electrons, so the dimerization reaction effectively reduced the number of paramagnetic species.
E20.30(b) The molar susceptibility is given by
so
S (S
+ I) =
3kTXm 2
2
NA ge /10/1B
3(1.38 1 x 1O 23 JK I ) x (298 K)
S(S
+ I) = (6:'.O'22xI0"""'23;moI;1) 'x(2:'.0023)"""2
= 2.84
so
S
=
I
+ ..II + 4(2.84) 2
= 1.26
398
STUDENT'S SOLUTIONS MANUAL
[TI.
corresponding to 12.521 effective unpaired spins. The theoretical number is The magnetic moments in a crystal are close together, and they interact rather strongly. The discrepancy is most likely due to an interaction among the magnetic moments.
E20.31 (b) The molar susceptibility is given by
Xm
=
NAg~J.LOJ.L~S(S
+ I)
3kT
Mn 2+ has five unpaired spins, so S
Xm =
= 2.5 and
(6.022 x 1023 mol  I) x (2.0023)2 x (4JT x 1O 7 T 2 r 3(1.381 x 1O 23 JKI)
l
m 3)
~~~~~~~~
+ I) (298 K) ~~~~ (9.274 x 10 24 JTI) 2 x (2.5) x (2.5
x
10 7 m 3 mol  II
= 11.85 X
E20.32(b) The orientational energy of an electron spin system in a magnetic field is
£ =geJ.LBMS~
The Boltzmann distribution says that the population ratio r of the various states is proportional to
r=exp
( 1:1£) ~
where 1:1£ is the difference between them . For a system with S = I, the Ms states are 0 and between adjacent states
r
= ex (geJ.LBMs~ ) = ex kT
p
p
± I. So
24 ((2.0023) x (9.274 x 10 JTI) X (I) x (l5.0T») (1.381 x 1O 23 JKI) x (298K)
= 10.9351 The population of the highestenergy state is r2 times that of the lowest; r2
= 10.8731
Solutions to problems Solutions to numerical problems P20.1
A = 2dhkl sin Bllkl =
?
(h
2a sin Bhkl ., 2 2 1/2 [eqn 20.5, msertmg eqn 20.2] +k +l )
= 2a sin 6.0 = 0.209a. 0
In an NaCI unit cell (Fig. 20.3) the number of formula units is 4 (each comer ion is shared by 8 cells, each edge ion by 4, and each face ion by 2).
MATERIALS 2: THE SOLID STATE
399
Figure 20.3
Therefore, implying that
_ a
P20.3
(
a
4M
= ( 
)1/3
pNA
[Exercise 20.8(a)] .
I ) 1/3_ 5635. pm
(4) x (58.44 g mol ) (2.17 x 10 16 g m  3 ) x (6.022 x 1023 molI )
and hence).. = (0.209) x (563.5 pm) = 1118 pm I. See Fig. 20.23 of the text or Fig. 20.1 of this manual. The length of an edge in the fcc lattice of these compounds is
Then (I) a(NaCl)
= 2(rNa+ + rCl  ) = 562.8 pm ;
(2) a(KCI )
= 2(rK+ + rCl  ) = 627.7 pm;
(3) a(NaBr)
= 2(rNa+ + rBr ) = 596.2 pm ;
(4) a(KBr)
= 2(rK + + rBr ) = 658.6 pm .
If the ionic radii of all the ions are constant then
+ (4) = (2) + (3). ( I) + (4) = (562.8 + 658 .6) pm = (2) + (3) = (627 .7 + 596.2) pm =
( I)
1221.4 pm. 1223 .9 pm.
The difference is slight; I hence the data support Ithe constancy of the radii of the ions. P20.5
For the three given reflections sin 19.076°
= 0.32682,
For cubic lattices sin ehkl
=
sin 22.171 ° )"(h2
= 0.37737,
+ k 2 + 12) 1/ 2 2a
sin 32.256°
= 0.53370.
[20.5 with 20.2].
First consider the possibility of simple cubic; the first three reflections are (100), (110), and (Ill). (See Fig. 20.22 of the text.) sin e( IOO) sine(lIO)
=
I 0.32682 M f.    [not simple cubic]. ,,2 0.37737 ·
400
STUDENT'S SOLUTIONS MANUAL
Consider next the possibility of bodycentered cubic; the first three reflections are (110). (200). and (211).
.J2 I = J4 = .J2 i=
sin 9(110) sin 9(200)
0.32682 0.37737 (not bcc).
Consider finally facecentered cubic; the first three reflections are (III). (200). and (220). sin 9(111) :::::=:
sin 9(200)
J3 J4 = 0.86603
=
which compares very favorably to 0.32682 / 0.37737
0.86605.
Therefore. the lattice is
1facecentered cubic I·
This conclusion may easily be confirmed in the same manner using the second and third reflection.
a
= _A_(h 2 + k 2 + p)' /2 = (
p
= 
2 sin 9 NM
[Exercise 20.8(a)]
NAV
154.18pm ) x (2) x (0.32682)
J3 = 1408.55 pm I. .
.
(4) x (107.87 gmol  ')
= ,'',,,23 (6.0221 x 10
mol') x (4.0855 x 1O8 cm)3
= 110.507 g cm  ' I· This compares favorably to the value listed in the Data section. P20.7
A = 2a sin 8100
as dlOo = a.
A
Therefore. a =
2 sin 9100
and
a(KCI)
sin 8100 (NaCI)
a(NaCl)
sin 9IOO(KCI)
Therefore. a(KCI)
= ( 1.l14)
sin 6°0'    = 1.114. sin 5°23'
x (564pm)
= 1628 pm I.
The relative densities calculated from these unit cell dimensions are p(KCI) p(NaCI)
= (M(KCI)
) x (a (NaCI))3 M(NaCI) a(KCI)
Experimentally p(KCI) p(NaCI)
=
3 1.99gcm2.17 gcm 3
= 0.917.
and the measurements 1are broadly consistent I·
= (74.55) 58.44
x (564 p m)3 628 pm
= 0.924.
MATERIALS 2: THE SOLID STATE
P20.9
401
As demonstrated in Justification 20.3 of the text, closepacked spheres fill 0.7404 of the total volume of the crystal. Therefore 1 cm 3 of closepacked carbon atoms would contain 0.74040 cm 3 ;: = 3.838 x 1023 atoms
(1 nr3 )
154.45) 25 x 10 10 cm. ) ( r = (  2  pm = 77.225 pm = 77.2
Hence the closepacked density would be (3.838 x 1023 atom) x (12.01 u/atom) x (1.6605 x 10 24 gu I )
mass in 1 cm 3 P= I cm3
1 cm 3
3 = 17 .654 g cm 1·
The diamond structure (solution to Exercise 20.17(a» is a very open structure, which is dictated by the tetrahedral bonding of the carbon atoms. As a result many atoms that would be touching each other in a normal fcc structure do not in diamond; for example, the C atom in the center of a face does not touch the C atoms at the corners of the face.
m (unit cell) P20.11
(2) x (M(CH2CH2»/NA
P = V (unit cell) = ab=c   (2) x (28.05gmol l ) (6.022 x ]023 mol I ) x [(740 x 493 x 253) x 10 39 ] m 3 = 1.0] x 106 gm 3 =11.01 gcm 3 1.
P20.13
(a) When there is only one pair of identical atoms, the Wierl equation reduces to 2 sinsR
lee ) =/  
where
sR
4n
I
A
2
s=sine.
Extrema occur at sR = sin sR/cos sR = tan sR and this equation may be solved either graphically or numerically to give the extrema values shown in Fig. 20.4(a). The angles of extrema are calculated using the Br2 bond length of228.3 pm (Table 13.2), the equation
e=
2 sin 
I
SRA) ( 4nR 
Neutron diffraction:
,
and sR extrema values shown in Fig. 20.4(a).
el st max
. _I
elst min
= 2 Sin
. _ I
e2nd max
= 2 Sin
= 0,
( (0.9534 x 3n/2)(78 pm») 4n(229.0 pm) ((0.9836 x 5n/2)(78 pm») 4n(229.0 pm)
~
= ~, ~
= ~.
402
STUDENT'S SOLUTIONS MANUAL
Extrema in units of nl 2
0.5 /lf2
0.9986
o
* 17
0.9989
* 19
0.9534 • 3 ~.5~0~5~10~~15~i 20~J25 sRI(rc/2)
Figure 20.4(a)
Electron diffraction: 81SI max = 0, . =2sinI(0.9534x3n/2)(4.Opm»)_~ 4n(229.0 pm) ~,
8
lsI nun
82 nd max
(b)
1=

. I (0.9836 x 5n/2)(4 pm»)  ~ 2 Sin 23° 4n(229.0 pm)  1. .
sinsRiJ
LiJ j J jsR . , IJ
= 4fdCl
sin SRCCl R s CCI
4n 1 s =  sin8 A
2
+ 6fci2 sin SRCICl [4CClpairs,
SinX) = (4) x (6) x (17) x (f2) x ( x I
sinx ? = (408) x jx
6CIC1 pairs]
SRClCI
+ (6)
x (17) x (f2)
sin(!!)1 3 / 2x 8
(3")1 /2x
[x = SRCCi].
. (8) 1/ 2
sm 3" x + (1062)....:.......: x
This function is plotted in Fig. 20.4(b). We find X max and Xmin from the graph, and Smax and Smin from the data. Then, since x = SRCCl, we can take the ratio x/s to find the bond length RcCI. The calculation of s requires the wavelength of the electron beam.
From the de Broglie relation [8.12], h
A   P 
h 7"""7
(2m eeV) 1/ 2
6.626 X 10 34 J s (2 x (9.109 x 10 31 kg)(1.609 x 1O 19 C)(1.00 x 104 y»)
=
12.2 pm.
MATERIALS 2: THE SOLID STATE
403
400
200 IIf2
0
 200
400
10
5
0
25
20
15
30
40
35
x
Figure 20.4(b)
We draw up the following table. Minima
Maxima 8(expt.)
3°0'
5°22'
7°54'
1°46'
4°6'
6°40'
9° 10'
s/prn I x(calc.)
0.0270 4 .77 177
0.0482 8.52 177
0.0710 12.6 177
0.0159 2.89 177
0.0368 6.52 177
0 .0599 10.6 177
0.0819 14.5 177
(x / s)/pm
I
Hence, ReCi geometry. P20.15
=
I
177 pm and the experimental diffraction pattern is consistent with tetrahedral
The volume per unit cell is
v=
abc = (3.688 1 nm) x (0.9402nm) x (1.7652 nm ) = 6.121 nrn 3 = 6.121 x 10 21 cm 3.
The mass per unit cell is 8 times the mass of the formula unit, RuN2C28HMS4, for which the molar mass is
M
= {I 0 1.07 + 2(14.007) + 28(12.011) + 44(1.008) + 4(32.066)} g molI = 638.01 g mol  I.
The density is m 8M P =  =  =
V
NA V
l
8(638.0Igmol ) 1 31 = 1.385 g cm . (6.022 x 1023 mol  I) x (6.121 x 10 21 cm 3 )
The osmium analog has a molar mass of 727. I g molI. If the volume of the crystal changes negligibly with the substitution, then the densities of the complexes are in proportion to their molar masses, Pos
=
3 727.1 638.01 (1.385 gcm  )
= 1 1.578 g crn 3 1.
P20.17
In (G/S)
= In(Go / S)

(E g /2k)
x
liT.
404
STUDENT'S SOLUTIONS MANUAL
Thus, the slope of a In (G) against l i T plot equals Eg/ 2k. The data has minimal uncertainty so the slope can be calculated by the twopoint difference method. Altematively, a linear regression fit of (I I T, In (G/ S» data points gives the slope. slo e = t;.ln (G/ S) = In (.0847)  In(2.86) _ _ p t;. ( l i T ) (1/312 K) _ ( 1/ 420 K) 4270 K, Eg = 2k x (slope) = 2 x (1.381 x 10 23 J K) x (4270) = 1.18 x 10 19 J.
This is equivalent to 71.0 kJ/mol or l 0.736 eV
P20.19
I.
The molar magnetic susceptibility is given by
NAg~f1,Q/.L~S(S+ I)
Xm =
3kT
.
ForS=2
' Xm
= (6.3001 x 10
6
) x
S(S+ I)
DK
6 = (6.300 1 x lO ) x (2)x(2+ 1) 3 298 m mol
m 3 molI [Illustration 20.3] .
1_1 =63 I I . 0. 127 x 10 m mol.
6 F S 3 (6.300I x lO )x(3)x(3+1 ) 3 I I 63  II or = , Xm = 298 m mol = . 0.254 x 10 m mol . F
or
S
=
4
, Xm =
6 (6.3001 x 10 ) x (4) x (4 + I ) 3 I I 6 3 II 298 m mol = . 0.423 x 10 m mol  :
Instead of a single value of S we use an average weighted by the Boltzmann factor SO x 10 3 J mol  I ) 9 exp ( (8.3 145 J molI K I) x (298 K) = 1.7 x 10 .
Thu s the S = 2 and S = 4 form s are present in negligible quantities compared to the S = 3 form . The compound 's susceptibility, then, is that of the S = 3 form , namely I 0.254 x 10 6 m 3 molI
P20.21
I.
If the unit cell vo lume does not change upon substitution of Ca for Y, then, the density of the superconductor and that of the Yonly compound will be proportional to their molar masses. Msuper = [2 (200.59) + 2( 137.327) + (I  x) x (88 .906) + x(40.078 ) + 2(63.546) + 7.55(15 .999) ] g mol  I, Msuper /(g mol  I) = 1012.6  48.828x. The molar mass of the Yonly compound is 1012.6 g mol  I, and the ratio of their densities is Psuper = 1012.6  48.828x = I _ 0.04822x 101 2.6
PYonl y
sox = __1_ (I _ psuper) .
0.04822
PYonly
The density of the Yonly compound is its mass over its volume. The volume is 21 cm 3, 2 3 VYonly = a e = (0.38606 nm) 2 x (2.89 15 nm) = 0.43096 nm = 0.43096 x 10so the density
PYonly
2M 2( 101 2.6 g mol  I) 804 3 = NAV = (6.022 x 1023 mol  I) x (0.43096 x 10 21 cm3) = 7. gcm .
MATERIALS 2: THE SOLID STATE
405
The extent of Ca substitution is
x
=
I ( 7.651) ~ 0.04822 I  7.804 = ~.
COMMENT.
The precision of this method depends strongly on just how constant the lattice volume
really is.
Solutions to theoretical problems P20.23
If the sides of the unit cell define the vectors a , b, and c, then its volume is V Introduce the orthogonal set of unit vectors i, ] , k so that
= axi + ay] + azk , b = bxi + b r ] + bzk , c = cxi + cy] + czk . a
Then V
=a .b x c =
ay az by b: . cy c z
ax bx Cx
Therefore
V2
=
ax bx Cx
ay by cy
a z ax b z bx C z Cx
ay by cy
az bz
ax bx Cx
Gy
a z ax b z bx Cz Cx
ay by
a: bz
cy
C
by cy
C
z
z
[interchange rows and columns, no change in value]
axa x + ayay + aza z bxax + byay + bza z
axbx + ayby + azb z bybx + byby + bzbz
axcx + aycy + azcz bxcx + byc)' + bzc z
G~ + 0~ + ~~
G~+0~ + ~~
GG + 00 + ~~
a2
b .a c .a
a2
a .b a ·c b2 c .b
b .c c2
=
ab cos y (l C cos f3
= (l 2 b 2 c2 ( I  cos 2 ex  cos 2
Hence V
= abc( I 
V
= abc(l
 cos 2 Y + 2 cos ex cos f3 cos y) I / 2
= y = 90°
 cos 2 f3) I / 2
ac cos f3 bc cos ex c2
f3  cos 2 Y + 2 cos ex cos f3 cos y) 1/ 2.
cos 2 ex  cos2 f3
For a monoclinic cell, ex
ab cos y b2 bc cos ex
= Iabc sin f31.
.
= a .b
x c [given] .
406
STUDENT'S SOLUTIONS MANUAL
For an orthorhombic cell, ex
= f3 = y = 90°, and
V=labcl· P20.25
The four values of hx and 7/2 , and so Fhkl
P20.27
ex: I
+ ky + lz that occur in the exponential functions of F have the values 0, 5/2, 3,
+ e5irr + e6irr + e7irr = I
_ I
+I
 I
= [QJ.
According to eqn 20.18, G=
E 2(1
+ vp)
E
and K=    
3(1  2vp)'
Substituting the Lameconstant expressions for E and yields
Vp
JL(3A G
=
+ 2JL»/(A + JL) 2(1 + __A __ ) 2(A + JL)
(JL(3A
and
K=
into the righthand side of these relationships
+ 2JL)
A+JL
A ). 3(1 __ A+JL
Expanding leads to:
as the problem asks us to prove. P20.29
Permitted states at the low energy edge of the band must have a relatively long characteristic wavelength while the pennitted states at the high energy edge of the band must have a relatively short characteristic wavelength. There are few wavefunctions that have these characteristics so the density of states is lowest at the edges. This is analogous to the MO picture that shows a few bonding MOs that lack nodes and few antibonding MOs that have the maximum number of nodes. Another insightful view is provided by consideration of the spatially periodic potential that the electron experiences within a crystal. The periodicity demands that the electron wavefunction be a periodic function of the position vector r. We can approximate it with a Bloch wave: 1/f ex: e ik .r where k = kxi + kyJ + ki, is called the wavenumber vector. This is a bold, 'free' electron approximation and in the spirit of searching for a conceptual explanation, not an accurate solution, suppose that the wavefunction satisfies a Hamiltonian in which the potential can be neglected: if = (/i2/2m)V2. The eigenvalues of the Bloch wave are: E = /i 2 IkI 2 /2m. The Bloch wave is periodic when the components of the wave number vector are multiples of a basic repeating unit. Writing the repeating unit as 2Jl' I L where L is a length that depends upon the structure of the unit cell, we find: kx = 2nxJl' I L where nx = 0, ± I, ±2, ... . Similar equations can be written for ky and kz and with substitution the eigenvalues become:
E = 112m (2rr/ij L) 2 (n; + n; + n~). This equation suggests that the density of states for energy level E can be visually evaluated by looking at a plot of permitted nx , ny, nz values as shown in Fig. 20.5. The
MATERIALS 2: THE SOLID STATE
407
number of nx , ny, n z values within a thin, spherical shell around the origin equals the density of states that have energy E. Three shells, labeled 1, 2, and 3, are shown in the graph. All have the same width but their energies increase with their distance from the origin. It is obvious that the low energy shell 1 has a much lower density of states than the intermediate energy shell 2. The sphere of shell 3 has been cut into the shape determined by the periodic potential pattern of the crystal and, because of this phenomenon, it also has a lower density of states than the intermediate energy shell 2. The general concept is that the low energy and high energy edges of a band have lower density of states than that of the band center.
Figure 20.5
P20.31 1/ 2
with 1/f = (
Then, since Xm =
P20.33
NA f.1.0~
_1_ 3) na o
er/ao
[20.32, m =0],
If the proportion of molecules in the upper level is P, where they have a magnetic moment of2f.1.B (which replaces (S (S + I)} 1/ 2 f.1.B in eqn 20.35), the molar susceptibility Xm =
(6.3001 x 10 6 ) x [S (S + I)] 3 TjK m molI [Illustration 20.3]
408
STUDENT'S SOLUTIONS MANUAL
is changed to (6.300 1 x 10 6 ) x (4) x P 3 _l 2 25.2P T/K m mol [2 replacesS(S+ I)] = T / K x lO 6 m 3 mol 
Xm=
l
.
The proportion of molecules in the upper state is ehcv/kT
P = ,::: /C7"=T [Boltzmann distribution] = :::~
I
+ e
I CV
I
<
+ ehcv / kT
hcv ( 1.4388cmK) x (121 cm I) 174 and= =kT T T/ K 25.2 x 10 6 m 3 mol  I Th ere f ore Xm = c=,:,,:;;:~~ , (T / K) x (I +e I74 / (T/ K) · This function is plotted in Fig. 20.6.
~4 I
(5
E
"'E
t
~2 E
~ o
100
200
300
T/ K
400
500
Figure 20.6
COMMENT. The explanation of the magnetic properties of NO is more complicated and subtle than indicated
by the solution here. In fact the fuJI solution for th is case was one of the important triumphs of the quantum theory of magnetism which was developed about 1930. See J.H. van Vleck, The theory of electric and magnetic susceptibilities. Oxford University Press (1932).
Solutions to applications P20.35
The Xray diffraction pattern of fibrous BDNA (Figure 20.26) is discussed in ImpacT120.1 . Figures 20.27 and 20.28 provide definitions of the helical tilt angle ex and the baselayer spacing h. The helical pitch p is the vertical rise per turn of the helix . The characteristic Xshape of the diffraction pattern is that of a helix with incident radiation (Cu Kex 0.1542 nm) perpendicular to the cylindrical axis. An angle () = 2.6° between the line of the incident radiation and the Line from sample to the first spot on the X gives p = A/ sin () = 0. 1542 nm / sin (2.6°) = 3.4 nm. 10 spots (counting two ' missing fourth ' spots) along the X diagonal indicate that there 10 baseplanes per turn of the helix with each accounting for a turn of 36°. The very large spot is at a distance (I / h) which is 10 times the distance I / p shown in
MATERIALS 2: THE SOLID STATE
409
Fig. 20.7(a). Consequently, h = 0.34 nm. The missing fourth spots on the X diagonals indicate two coaxial sugarphosphate backbones that are separated by 3p/8 along the axis. The periodic h spacing of the large, very electrondense phosphorus atoms causes the 1/ h spots to be very intense. The fact that the fibrous Xray sample was saturated with water suggests that the phosphates are to the outside.
Fourth spot in each direction is missing. Counting the missing sputs, there are a total of 10 spots along the diagonal.
Figure 20.7(a) Figure 20.7(b) shows the twodimensional zigzag projection of the helical sugarphosphate backbone. It serves to define the projection length l , perpendicular distance d between backbone planes, and the heli x radius r . Examination of the ri ght triangle that shows the definition of Cl yields tan (Cl)
= !!...
or
4r
r
p
=  = 4 tan(Cl)
3.4 nm 4 tan (40°)
=
1.0 nm.
2a( i 1'/2
_______~_?L __ ___ .
;;\.2a( Figure 20.7(b) Examination of the right triangle containing the angle Cl also shows that l sin (Cl) = p /2 while the right triangle containing the angle 2Cl shows that l sin (2Cl) = d. Dividing these two equations yields sin (2Cl) sin (Cl) d
2d p
or
2 sin(Cl) COS(Cl) sin(Cl)
p
= p COS(Cl) = (3.5 nm) cos( 40°) = 2.6 nm.
Finishing, l
2d
p
=  = 2 sin (Cl)
3.4 nm 2 sin (400)
= 2.6 nm.
or
COS(Cl)
d
=. p
410 P20.37
STUDENT'S SOLUTIONS MANUAL
Refer to Fig. 20.8.
,.. B~~~rL~~'
r
c
l~\~.~
Figure 20.8
Evaluate the sum of ±(II ri), where ri is the distance from the ion i to the ion of interest, taking +(llr) for ions of like charge and (l / r) for ions of opposite charge. The array has been divided into five zones. Zones Band D can be summed analytically to give  In 2 = 0.69. The summation over the other zones, each of which gives the same result, is tedious because of the very slow convergence of the sum. Unless you make a very clever choice of the sequence of ions (grouping them so that their contributions almost cancel), you will find the following values for arrays of different sizes lO x 10
20 x 20
50x50
100 x 100
200 x 200
0.259
0.273
0.283
0.286
0.289
The final figure is in good agreement with the analytical value, 0.289 259 7 ... For a cation above a flat surface, the energy (relative to the energy at infinity, and in multiples of e2 / (4n ero) where ro is the lattice spacing (200 pm» , is Zone C + 0
+ E = 0.29 
0.69 + 0.29
which implies an attractive state.
= [3[ill
PART 3 Change
Molecules in motion
Answers to discussion questions 021.1
(a) See Section 24. 1 which discusses the collision theory of gas phase reactions. Their rate depends on the number of colli sions having a relative kinetic energy above a certain critical value Ba. The relative kinetic energy, in turn, depends on the relative velocities of the colliding molecules. The rate of reaction also depends on the number of collisions per unit volume per unit time, or collision density, ZAB , the formula for which is derived in Justification 24.1. ZAB also depends on the average relative velocity of the colliding molecules. (b) A complete analysis of the composition of planetary atmospheres is a complicated process. See the solution to Problem 21 .35 for detailed calculations on the depletion of the Earth 's atmosphere and on the atmosphere of planets in general. Also see the solution to Problem 16.21 which deals with the inherent instability of planetary atmospheres. The simple answer to this question, though, is that light molecules are more likely to have velocities in excess of the escape velocity than are heavy molecules. Therefore, heavy molecules will remain in the atmosphere much longer than light molecules, though all will eventually escape, unless there is a source of replenishment.
021.3
Gases are very dilute systems and on average the molecules are very far apart from each other except when they collide. So what little resistance there is to flow in a gaseous fluid is almost entirely due to the collisions between molecules. The frequency of collisions increases with increasing temperature (see eqns 21.1Ib and 24.8) ; hence the viscosity of gases increases with temperature. In liquids, on the other hand, the molecules are very close to each other, which results in there being strong forces of attraction between them that resist their movement relative to each other. However, as the temperature increases, more and more molecules are likely to have sufficient kinetic energy to overcome the forces of attraction, resulting in decreased viscosity.
021.5
(a) This is Fick 's first law of diffusion in one dimension written in terms of concentrations rather than activities ; hence, it applies strictly only to ideal solutions. (b) In addition to the restriction to ideal solutions as in (a), the derivation of this expression uses the additional approximation that the frictional retarding force on a moving particle is proportional to the first power of the speed of the particle (as opposed to a more general functional relation). (c) The restrictions of parts (a) and (b) still apply, as well as a third, which is the assumption that the particle is spherical.
021.7
Because the drift speed governs the rate at which charge is transported, we might expect the conductivity to decrease with increasing solution viscosity and ion size. Experiments confirm these predictions for bulky ions, but not for small ions. For example, the molar conductivities of the alkali metal ions increase
414
STUDENT'S SOLUTIONS MANUAL
from Li+ to Cs+ (Table 21.6) even though the ionic radii increase. The paradox is resolved when we realize that the radius a in the Stokes formula is the hydrodynamic radius (or 'Stokes radius') of the ion, its effective radius in the solution taking into account all the H20 molecules it carries in its hydration sphere. Small ions give rise to stronger electric fields than large ones, so small ions are more extensively solvated than big ions. Thus, an ion of small ionic radius may have a large hydrodynamic radius because it drags many solvent molecules through the solution as it migrates. The hydrating H20 molecules are often very labile, however, and NMR and isotope studies have shown that the exchange between the coordination sphere of the ion and the bulk solvent is very rapid. The proton, although it is very small , has a very high molar conductivity (Table 21.6)! Proton and I7ONMR show that the times characteri stic of protons hopping from one molecule to the next are about 1.5 ps, which is comparable to the time that inelastic neutron scattering shows it takes a water molecule to reorientate through about I rad (I  2 ps).
Solutions to exercises E21.1 (b)
(a) The mean speed of a gas molecule is
c = (~R:) 1/2 so
~(He) e(Hg)
= (M(Hg») 1/ 2 = (200.59) 1/2 = 17.0791 M(He) 4.003
(b) The mean kinetic energy of a gas molecule is ~ me 2 , where e is the root mean square speed
So ~ me 2 is independent of mass, and the ratio of mean kinetic energies of He and Hg is E21.2(b)
(a)
OJ
The mean speed can be calculated from the formula derived in Example 21.1. l 1 c= (8RT)I /2 = (8 x (8.314JK mol ) x (298K») 1/ 2 =14.75 x 102 m s 1 1 Jr M Jr x (28.02 X 10 3 kg moll )
(b) The mean free path is calculated from A = kTj(21 /2ap) [21.13] with a = Jrd 2 = 10 10 m)2 = 4.90 x 10 19 m 2 (1.381 Then , A =
X
21/2 x (4.90xlO 19 m 2) x
10 23 J K
(I
I
)
Jr
x (3 .95
x (298 K)
x 10 9 Torr) x
(7c:oa~~rr
)
X
(1.01 3XI05 I aI m
X
pa)
=14 x 104 m l (c) The collision frequency could be calculated from eqn 21.11 , but is most easily obtained from
eqn 21 .12, since A and c have already been calculated 1 2 z  c  4.75 x 10 ms =  A
4.46
X
104 m
II x 10.
2 SI
I
MOLECULES IN MOTION
415
Thus there are 100 s between collisions, which is a very long time compared to the usual timescale of molecular events. The mean free path is much larger than the dimensions of the pumping apparatus used to generate the very low press ure. kT p = ~/2 [21.1 3]
E21.3(b)
2
a
p
=n
aA
2
d , d
a)I/2
=;; (
=
(0.36 nm2 )1/2 n
= 0.34 nm
1 (1.381 x 1O 23 JK  ) x (298K) 18 2 (21/2) x (0.36 x 1O m ) x (0.34 x 1O 9 m)
=
=
I
7
2.4 x 10 Pa
I
This pressure corresponds to about 240 atm, which is comparable to the pressure in a compressed gas cylinder in which argon gas is normally stored. E21.4(b)
The mean free path is
A=
E21.S(b)
kT
(1.381 x 10 23 1 K  1) x (217 K )
21/2ap
21/2 [0.43 x (1O 9 m) 2] x (12. 1 x 103 Paatm  l )
 =
I41 x
10
7
m
= 0.43 nm 2,
Substituting a
p
=
12. 1 x 103 Pa, m
= [16 / (nmkT)]'/2 ap
= (28.02 u), and T = 2 17K we obtain
4 x (0.43 x 10 18 m 2) x (12. 1 x 10 3 Pa)
z=
~~~~~
[n x (28.02) x (1.6605 x 10 27 kg) x (1.381 x 1O 23 JK  I ) x (2 17K)]1/2
= 19.9 x
108 S I
1
The mean free path is kT (1.381 x 10 23 1 K I ) x (25 + 273) K 5.50 x 10 3 m Pa A == 2 1/2ap 2 1/ 2 [0.52 x (1O 9 m)2]p p
E21.7(b)
I
.
Obtain data from Exercise 21.4(b) The expression for z obtained in Exercise 2 1.5(a) is z
E21.6(b)
=
(a)
A=
3 5.50 x 10 m Pa ( 15atm) x (1.013 x 105Pa atm 
(b)
A=
3 5.50 x 10 m Pa ( LObar) x (105 Pa bar  I)
(c)
A=
3 5.50 x 10 m Pa ( 1.0Torr) x (1.013 x 105 Pa atm  1/760 Torr atm  I)
= I5.5
I
l)
9
= 3.7 x 10 m
8 x 10 m
I
I = I 4. 1 x
The fraction F of molecules in the speed range from 200 to 250 m s I is
F
=
25om s '
1
200m s 1
f(v)dv
5 10 m
I
416
STUDENT'S SOLUTIONS MANUAL
where! (v) is the Maxwell di stribution. This can be approximated by
F ~ ! (v) ~v
= 4][ ( M  )3 / 2 v 2 exp (MV2)  ~ v, 2][RT
2RT
with! (v) evaluated in the middle of the range
F
~
4][ (
X
I
F
exp
~ 9.6
44.0 x 10 3 kgmol I 2][(8.314SJK  l mol  l ) x (300K)  (44.0
)
3~
x 22S m s(
I
2
)
X103kgmol I) x (22Sms I)2) x (SOm s _I) ,
2(8.3 14SJK 1 mol I) x (300K)
(
x 10 2
1
COMMENT. The approximation we have employed, taking f (u) to be nearly constant over a narrow range of
speeds, might not be accurate enough, for that range of speeds includes about 10 percent of the molecules. You may wish to do the integration without this approximation (a considerably more complicated process) to see how much difference there is.
E21.8(b)
The number of collisions is
N
pAt
= Zw At =    '   c  = (2][ mk T) 1/2
/2][
( III Pal x (3.5 x 1O 3 m) x (4.0 x 1O2 m) x (lOs) x (4.00 u) x (1.66 x 1O 27 kgu l ) x (1.381 x 1O 23 JK I) x (IS00K)jl /2
=ls.3 x E21.9(b)
10
21
1
The mass of the sample in the effusion cell decreases by the mass of the gas which effuses out of it. That mass is the molecular mass times the number of molecules that effuse out
6.m
= mN = mZwAt =
mpAt (2][mkT) 1/2
m )1 /2 )1/2 = pAt ( = pAt (M 2][kT 2][RT
= (0.224 Pa) X ][ x
(~x 3.00 x
l 1O 3 m)2 x (24.00 h) x (3600Sh )
X
300 x 10 3 kg molI } 1/2 { 2][ x (8.3l4SJK  l mol I) x (4S0K )
= 14.98
x 10 4 kg
I
E21.10(b) The time dependence of the pressure of a gas effusing without replenishment is p
= POe l / r
where
T
ex
.Jfii
MOLECULES IN MOTION
417
The time t it takes for the pressure to go from any injtial pressure Po to a prescribed fraction of that pressurefpo is
fpo
t = r In = r Inf Po
so the time is proportional to r and therefore also to .;m. Therefore, the ratio of times it takes two different gases to go from the same initial pressure to the same final pressure is related to their molar masses as follow s
~
= (MI )1 /2
12
M2
and
I
Mnuorocarbon = (28.01 gmol _ I) x (82.3  S)2 = 554 g mol I 18.5 s
So
I
E21.11(b) The time dependence of the pressure of a gas effusion without replenishment is
p
= Po e
where r =
I r /
so
t
=r
Inpo / p
1/ 2 = ~ (2rrM) 1/2 AoV (2rrm) IT Ao RT
=( so t = (8.6
22.0 m 3
2rr x (28.0 x 10 3 kg mol  I) X
rr x (0.50 x 1O3 m )2 ) X
122kPa 105 s) In  = 105 kPa
I1.5
(
X
(8.3145JK 1 molI) x (293K) ) 104 s
I
E21.12(b) The flux is
J
dT
I
dz
3
dT
=  K  =   ACV me [X]'
dz
where the minus sign indicates flow toward lower temperature and 1
A= v'2Na' e =
So J
(8kT) rrm
1/ 2
=
(8RT) 1/2 rrM , and[M]=n / V=N / NA
= _ 2Cv .m (RT) 1/ 2 dT 3aNA
rrM
dz
2 x (28 .832  8.3145) J K  I mol  I
)
 ( 3 x [0.27 x (I0 9 m)2] x (6.022 x 1023 molI)
x
(8.3145JK 1 molI) x (260K»)i /2 ( rr x (2.016 x 1O 3 kgmol  l )
=10.17Jm 2 s 1 1
I
x (3.5 Km )
1/2
= 2.4 x
5
LO s
418
STUDENT'S SOLUTIONS MANUAL
E21.13(b) The thermal conductivity is
= ~ACv,mC [Xl = 2CV,m(RT) 1/ 2
K
3aNA
3
so a
a= 2CV.m(RT) 3KNA
(0.240mJCm 2 s l ) x (KcmI)I
=
K
so
TrM
1/ 2
TrM
x 10 1Jm  I sI K I
= 0.240
2x(29.12S8.314S)JKlmol1 ) I 1 23 3 x (0.240 X 10 J m sI KI) x (6.022 x 10 molI)
=(
x ((8.314SJK 1 molI) x (298K))1 /2 Tr
= 11.61
x (28.013 x 10 3 kg mol  I)
x 10 19 m 2
1
E21.14(b) Assuming the space between sheets is filled with air, the flux is
J
=
k
~: = [(0.241 = 1.45 x
3
2
x 10 J cm SI) x (KcmIfl] x
C ~O:~:~ SO
10 3 J cm 2 sI.
So the rate of energy transfer and energy loss is
E21.1S(b) The coefficient of viscosity is
1]
= ~AmNc = ~ (mkT)1 /2 3a
3 T)
so (J
= 1.66 tiP = 166 x
(

3x

(166
so
a = ~ (mkT)1 /2 31]
Tr
Tr
10 7 kgm I sI
2)
x 1O 7 kgm 1 SI)
1/ 2
x ((28.01 x JO 3 kgmol l ) x (1.381 x 1O 23 JK I) x (273K) Tr
= 13.00 x
10 19 m2
)
x (6.022 x 1023 molI)
1
E21.16(b) The rate of fluid flow through a tube is described by
dV _ (pln  p~u() Trr4 so . _ (16l1]PO dV 16l1]PO Pm Trr4 dt
dt 
+
2 )
Pout
Several of the parameters need to be converted to SI units r
=
!(IS x 10 3 m)
= 7.S x
10 3 m
1/2
Kl)
MOLECULES IN MOTION
419
and dV = 8.70cm 3 x (i02 m cm  I) 3 s I = 8.70 x 1O 6 m 3 s l . dt
Also, we have the viscosity at 293 K from the table. According to the TI /2 temperature dependence, the viscosity at 300 K ought to be 300 K ) 1/2 ( 300 ) 1/2 l l 7 11 (300K)=11 (293 K)x ( 293 K =(l76 x lO kgm s )x 293 = 1.78 x 10 7 kgm  I S I l l 5 7 . _ {(16 ( 10.5m) x (178 x 1O kgm  s ) x (1.00 x 10 pa))
Pm
lr
= 11.00
X
X
(7.5
x 10 3 m)
4
105 Pa I
For the exercise as stated the answer is not sensitive to the viscosity. The flow rate is so low
COMMENT.
that the inlet pressure would equal the outlet pressure (to the precision of the data) whether the viscosity were that of N2 at 300 K or 293 K, or even liquid water at 293 K!
E21.17(b) The coefficient of viscosity is
11 =
~ AmNc = ~ (mkT) 1/2 3
= (
3a
lr
) x ((78.12 x 1O 3 kgmol l ) x (1.381 x 1O23 JK I)T)I /2 l 3[0.88 x (lO9 m)2] lr x (6.022 x 1Q23 mo l ) 2
= 5fi x
10 7 x (T / K ) I/2 kg m I sI
(a) At273K
11 = (5.72 x 10 7) x (273)1/2 kgm I sI =10.95 x lO 5 kgm 1 S I 1
(b) At298 K
11 = (5 .72 x 10 7) x (298) 1/2 kg m I S I = 10.99 x 10 5 kgm I s I 1
(c) AtlOOOK
11 = (5. 72 x 10 7 ) x (1000)1/2 kg m I s I = 11.81
E21.18(b) The thermal conductivity is
k
(RT )1 /2
2C = 1ACV.mC [Xl = ~ 3aNA
(a) K
lrM
2x [(20. 786 8.3 145)JKlmoll] ) ( 9 23 = 3 [0.24 x (10 m) 2] x (6.022 x 10 mol I) x ((8.3 145 JK 1 mol I) x (300K») 1/2 I lr (20. l8 x 10 3 kg mol ) = 10.0114 Jm I s I K I I
X
10 5 kg m I s ' I
420
STUDENT'S SOLUTIONS MANUAL
The flux is
so the rate of energy loss is
2 x [(29. 125  8.3145) J K I mol I]
(b)
=
K
)
( 3 [0.43 x (10 9 m) 2] x (6.022 x 1023 molI) x (8.3145JK l mol 
l
x 10 3
7r (28.013
)
x (300K) )
1/ 2
kg molI)
The flux is
so the rate of energy loss is
E21.19(b) The rate of fluid flow through a tube is described by
dV
(P~l  P~UI) 7r r 4
dt
16l'7PO
so the rate is inversely proportional to the viscosity, and the time required for a given volume of gas to flow through the same tube under identical pressure conditions is directl y proportional to the viscosity
'7CFC
=
(208j1.P) x (I8.0s)
n.os
= I52.0j1.P I= 52.0
x 10
7
kgm
The coefficient of viscosity is
'7
= ~ AmNc = (~) 3
30"
x
(mkT) 7r
1/ 2
=
(_2 ) (mkT) x
37r cf2
7r
1/ 2
I
s
I
MOLECULES IN MOTION
421
so the molecular d.iameter is
d
=
C~ry) 1/2 x
= (3n (52.0 x
(m;T)
1O~7
1/ 4
kg mI S I)) 1/2 x (1.381 x 1O 23 JK I) x (298K»)1 /4 n x (6.022 x 1023 mol I)
x ((200 x 1O 3 kgmol 
l
)
= 9.23 x IO Io m = / 923pm / E21.20(b)
K
I 2C =  ACV. mC [Xl = 3 V.m 3 · aNA
(RT) 
1/ 2
nM
2 x (29.125  8.3145) 1 K I mol  I
)
= ( 3[0.43 x (1O 9 m)2J x (6 .022 x 1023 molI) = 19.0 x 10 3 Jm I S I K
1
((8.31451 K 1 molI) x (3OOK») 1/2 l 3 x 7r x (28 .013 x 1O kgmol  )
I
E21.21 (b) The diffusion constant is
D
I
2(RT) 3/2
3
3apNA(nM) I/2
=  AC = c,:,.___
,,;c
2[(8 .31451K 1 molI) x (298K)] 3/2 3 [0.43 x (10 9 m)2J p(6.022 x 1023 mol  I) x
In (28.013 x
10 3 kg mol  I) 11 /2
1.07m 2 s 1
p / Pa The flux due to diffusion is
D~
J = _Dd[Xl = dx
dx
(!!.) V=
_
(~)
RT
dp dx
where the minus sign indicates flow from high pressure to low. So for a pressure gradient of O. IOatmcm  1 2
J = (
D / (m s I)
) x (0.20
X
105 Pam  I)
(8.3145 J KI mol  I) x (298 K) = (8.1 mol m 2 S I) x (D /( m 2 5 1» 2
(a)
I
I
1.07 m SI ? = 0.107 m 5 1 10.0 ,_ __ _ , and} = (8. 1 molm 2 s l ) x (0.107) =1 0.87 molm 2 s 1 I D =
422
STUDENT'S SOLUTIONS MANUAL
I1.07 x 10 5 m2 S I I
2
1.07 m s I = 100 X 103
D=
(b)
and] = (8.lmolm  2 s l ) x (1.07 x 10 5 ) = 18.7 x 1O 5 molm 2 s 1 1 (c)
E21.22(b) Molar ionic conductivity is related to mobility by
A = zuF = (\) x (4.24 x 1O 8 m 2 s 1 V I) x (96485CmOI  I ) = 14.09 x 10 3 S m2 molI I E21.23(b) The drift speed is given by u!'!.¢
S
= us =   = I
(4.01 x 10 8 m 2 S I V I) x (l2 .OV) I 5 _I I = 4.81 x 10 m s 1.00 x 10 2 m
E21.24(b) The limiting transport number for Cl in aqueous NaCI at 25°C is
u_
t':. = u+ + u_ =
~
7.91 5. 19
+ 7.91
=~
(The mobilities are in 10 8 m 2 S I VI .) E21.2S(b) The limiting molar conductivity of a dissolved salt is the sum of that of its ions, so
J\~
(MgI 2 ) = A (Mg2+) = (18.78
+
2A
(1)
+ 2(\2.69)
J\~ (Mg (C2H30 zh) + 2J\~ (NaI)
=
so
u=
A
u=
zF
2 5.54 x 10 3 S m mol (\) x (96485Cmol
it
2J\~ (NaC2H302)
 2(9.10» mSm 2mol 1 =125.96mSm z mol  II
E21.26(b) Molar ionic conductivity is related to mobility by
A = z/LF

l
1
I
= 5.74 x 10 8 m 2 VI s I
I
)
2 3 = 7.635 x 10 S m mol  I = 17.913 x 10 8 m 2 VI S I I (I) x (96485 C molI) 7.81 x 10 3 S m 2 mol  I
,,c: =
( I) x (96485Cmol 
l
)
I8.09 x 10
8
m 2 VI s ~
I
MOLECULES IN MOTION
423
E21.27(b) The diffusion constant is related to the mobility by
D=
uRT
(4.24 x 10 8 m 2 s I VI) x (8.3145J K I molI) x (298K)
zF
(1) x (96485Cmol I)
= 11.09 x
10 9 m2 sI 1
E21.28(b) The mean square displacement for diffusion in one dimension is
In fact, this is also the mean square displacement in any direction in two or threedimensional diffusion from a concentrated source. In three dimensions
So the time it takes to travel a distance
/PI
is
E21.29(b) The diffusion constant is related to the viscosity of the medium and the size of the diffusing molecule
as follows kT D=67rl'/a
a
= 2.07
(1.381 x 1O 23 JK  I ) x (298K)  67rI'/D  67r (1.00 x 10 3 kgm I S I) x (1.055 x 10 9 m 2 sI)
kT a   
so
x 10
10
.,''::::';==.,.,
= 1207 pm 1
m
E21.30(b) The EinsteinSmoluchowski equation related the diffusion constant to the unit jump distance and time )...2
D=2r
)...2
so
r= 
2D
If the jump distance is about one molecular diameter, or two effective molecular radii , then the jump distance can be obtained by use of the StokesEinstein equation kT kT D =   = 67rl'/a 37r1'/)...
so
kT
)...=
37rI'/D
= 1200 x 10 I l si = 20 ps E21.31(b) The mean square displacement is (from Exercise 21.28(b))
424
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P21.1
The time in seconds for a disk to rotate 360° is the inverse of the frequency. The time for it to advance 2° is (2° / 360 0 )/v. This is the time required for slots in neighboring disks to coincide. For an atom to pass 1.0cm through all neighbouring slots it must have the speed Vx = = 180 v cm = 180(v/ Hz) cm sI . (2/ 360 )/ v Hence, the distributions of the xcomponent of velocity are
v/Hz vx/ (cm s I) I (40 K) I (lOOK)
20
40
80
100
120
3600 0.846 0.592
7200 0.513 0.485
14400 0.069 0.217
18000 0.015 0.119
21600 0.002 0.057
Theoretically, the velocity distribution in the xdirection is m ) 1/2 mv2 . j(vx ) = ( 2rrkT e , / 2kT [21.6, with M / R = m/ k].
Therefore, as I o:j, 10: mv 2
?
83.8 x (1.6605
Smce  '
10 27 kg) x {1.80 ( v / Hz) m s I }2
X
(2) x (1.38\ x 10 23 J K I ) x (T)
2kT
write J 0:
,)1/2 (T e mv; / 2kT.
(T ;K) 1/2 e 1.63 x
2
1O (v/ Hd / (T / K)
JI (40 K) 1/ ( \ 00 K)
20
40
SO
100
120
0.80 0.56
0.49 0.46
(0.069) 0.209
0.016 0.116
0.003 0.057
in fair agreement with the experimental data. I
(X ) = 
P21.3
N
(a) (h) =
L. NiXi [see Problem 21.2].
~ {I.SOm + 2 x 53
(b)
Ifh2i = 11.89 mi·
(1.S2m)
X
10 2( v / Hz)2 T/ K
, we can
and draw up the following table, obtaining the constant
of proportionality by fitting I to the value at T = 40 K , v = SO Hz v/ Hz
1.63
+ .. . + 1.98m} =11.89m I·
MOLECULES IN MOTION
P21.S
425
The number of molecules that escape in unit time is the number per unit time that would have collided with a wall section of area A equal to the area of the small hole. That is, dN
dt = Zw A =
 Ap
(2nmkT)
1/2 [21.14]
where p is the (constant) vapor pressure of the solid. The change in the number of molecules inside the cell in an interval !'>.t is therefore !'>.N = ZwA!'>.t, and so the mass loss is
!'>.w
( m ) 1/2 m ) 1/2 = !'>.Nm = Ap ( !'>.t = Ap  !'>.t . 2nkT . 2nRT
Therefore, the vapor pressure of the substance in the cell is
For the vapor pressure of germanium
p
=(
4.3 x 10 8 kg ) x (21T ) n x (5.0 x 1O4 m)2 x (7200 s)
= 7.3 x 10 3 Pa, or 17.3 mPa
P21.7
X
(8.3 l4JK 1 molI ) x ( 1273 K»)1 /2 72.6 x 10 3 kg mol  I
I.
The atomic current is the number of atoms emerging from the slit per second, which is ZwA with A = 1 x 10 7 m2 . We use
Zw
= (2m~T) 1 /2
[21.14]
p/ Pa [(2n) x (M l gmol  I ) x (1.6605 x 10 27 kg) x (1.381 x 1O 23 JK
= (1.35
(a) Cadmium:
(b) Mercury:
x
1023 m  2 s I) x (
p/ Pa
(M i g mol I )1 /2
) .
I)
x (380K)]1 /2
426
STUDENT'S SOLUTIONS MANUAL o  Xe 1/2 [21. 29 ], Am = C [21.2 8] Am = Am
P21 .9
eR
where C = 20.63 m I (C = K* R *, where K* and R* are the conductivity and resistance of a standard solution, respectively). Therefore, we draw up the fo llowing table. e/ M
0.0005
0.001
0.005
O.OlO
0.020
0.050
(e/ M)I/2
0.224 3314 12.45
0.032 1669 12.36
0.071 342.1 12.06
0.1 00 174.1 11 .85
0. 141 89.08 11.58
0.224 37. 14 11. 1I
R/ Q
Am / (mS m2 mol I)
The values of Am are plotted against e l / 2 in Fig. 21.1.
12.6
12.4 I
0 E
12.2
~E t/l
~ 12.0 ....... E
.:::
11 .8
Figure 21.1
The limiting val ue is
x
A~
= 112.6 mS m2 mol 1 I. The slope is  7.30; hence
= 17 .30mSm 2 mol I M I/ 2 1. Am = (5 .01 + 7.68) mS m2mol 1
(a)
= 111.96 mS m2 mol 
(b)
K
1

(+7.30 mS m2mol 
l
) X
(0.0 10)1 /2
I.
= eArn = ( lOmol m 3 ) x (11.96 mS m2 molI) = 119.6mS m2 m 3 = 1119.6mS mI
C (c) R =  = K
20.63m 1 119.6mSm
1
=1 172.5Q I· .
.
I.
MOLECULES IN MOTION
P21.11
S
IOV I = utf: [21.42] wi th tf: =    = lOY cm . l .oocm
s(Li+) = (4.0 1 x 1O4 cm 2 s 1 VI) x (lOVcm
l
)
= 14.0 x 1O 3 cms 1 I.
s(Na+) = (5. 19 x 1O 4 cm2 s 1 V I) x (IOVcm l ) = 15.2 x 1O 3 cmss(K+) = (7 .62 x 10 4 cm 2 s I V I)
X
(lOV cm I) = 17.6
X
l
l·
10 3 cms I I·
t = d i s with d = 1.0cm:
+ t(Li )
=
4.0
X
1.0cm ~ 10 3 cm s I = ~,
(a) For the distance moved during a halfcycle, write
r' / 2v
d = Jo
r
r' / 2v
1/ 2v
s dt = J o
utf: dt = u£o J o
[tf: = tf:o sin(2nvt)]
sin(2nvt) dt
u X (IOV cm  I) nx(l'.0'xI""'073  'Ic) [assume tf:o = IOV]=3. 18 s
utf:o nv
That is, d l cm = (3 .18 d(Li+) = (3. 18 d(Na+) = 11.7
X
X
X
10 3 )
10 3 )
X
I,
10 6 cm
I
(u l cm2 V  1 s I). Hence,
(4.0
X
3
10 uVscm .
X
X
10 4 cm) = 11.3
d(K+) = 12.4
X
X
10 6 cm
10 6 cm
~
I·
(b) These correspond to about @],~, and []I] solvent molecule diameters respectively.
ze VF
P21.13
zeAFl
t =   =   [2 1.52] 1M 1M
= (21mOlm  3 ) x (n ) x (2.073 x 1O 3 m)2 x (9.6485 x 104 C mOII» ) x 18.2 X 10 3 A = (1.50 x 103 m I s) x
(~) i:!.t
= (1.50) x (l l mm) . i:!.t Is
Then we draw up the following table. 200
400
600
800
1000
64 0.48 = 1 t+ 0.52
128 0.48 0.52
192 0.48 0.52
254 0.48 0.52
318 0.48 0.52
i:!.tl s ll mm t+ L
(~) i:!.t
427
428
STUDENT'S SOLUTIONS MANUAL
Hence, we conclude that t+ =10.481 and t_ = 10.52 1. For the mobility of K+ we use A+ u+F t+ = [21 .50] = [21.44]
A.:;'
A.:;'
to obtain _ t~ A.:;' _ (0.48) x ( 149.9 Scm mol u+  y 9.6485 x 104Cmol 1 2
t+ A~ [21.50]
A+ = P21.15
§
l
)
_I7.5 x 10
4
.
= (0.48) x ( 149.9 S cm 2 mol  I) =
InS cm
2
I
I
2
cm s
V
molI
I,
I.
RT de =   x  [21.58] . dx e
de
(0.05  0. 10) M 1 O. IOm =  0.50Mm [linear gradation] .
dx =
RT = 2.48 x 103 J molI = 2.48 x 103 N mmol  I . (a) § =
(b) § =
(c) § =
P21.17
I 2.48kNmmOI ) O. IOM x (0.50Mm(
(
(
I 2.48kNmmOI  ) 0.075M x (O.50Mm 
l
)
1 )
I
II
l .
I
II
1
I
II
= 12kNmol  I , 2.1 x 1O 20 Nmolecule= 17kNmol  I , 2.8 x 1O 20 Nmolecule
I .
I
I 2.48kNmmOI) x (0.50 M m I) = 25 kN mol  I , 4.1 x 10? 0 N molecule I. 0.05 M
If diffusion is analogous to viscosity [Section 21.5 , eqn 2 1.26] in that it is also an activation energy
controlled process, then we expect
Therefore, if the diffusion constant is D at T and D' at T',
R In Ea =
(;,
(~) 
1
(8 .3 14JK molI) x In I I
~)
G:~~)
_I = 9.3kJmol

298 K
273 K
That is, the activation energy for diffusion is 19.3 kJ mol  I I. P21.19
I
(X2 }= 2Dt [21.83],
kT
D =   [21.67]. 6rrary
Hence, kT
kTt
ry=== 6nDa 3na (x2)
1.381
X
1O 23 JK
1
x (298.15K) x t
(3 n ) x (2. 12 x 10 7 m) x
(x2)
.
MOLECULES IN MOTION
_ I)
_I
and therefore 1] / ( kg m
s
=
2.06
X
429
10 11 (t/s)
((x2) / cm2 )
.
We draw up the following table. t/ s
108(x 2)/cm2 103 1] / (kg m Is I)
30
60
90
120
88.2 0.701
113.4 1.09
128 1.45
144 1.72
Hence, the mean value is 11.2 x 10 3 kg m I sI P21.21
I.
The viscosity of a perfect gas is I]
= 1. JV
mAc = mc
=
3a.J2
3
~ (mkT) 1/ 2 3a
so a =
~ (mkT) 1/ 2 31]
IT
IT
The mass is
m
=
17.03 x 10 3 kg mol  I
=="';
6.022 x 1023 mol  I
= 2.828
x 10 26 kg.
(a)
x ((2.828 x 10 26 kg) x (1.38: x 10 23 J K
= 4.25 x 10(b)
19
m2 = ITd 2 so d = (
4.25 x
I
)
1O 19 IT
x (270 K) )
m2 )1 / 2
1/ 2
= 13.68 x 10 10 m i·
2
a=:,,,,3(17.49 x 1O 6 kgm 1 s I)
x
C
2 .828 x 10 26 kg) x (1.38: x 1O 23 JK
=2.97xlO
COMMENT.
19
m 2 =ITd2
so
297
d= ( .
1 )
10X
IT
x (490K»), /2
19 2 )1 / 2
m
=13 .07 x lO IOm
l.
The change in diameter with temperature can be interpreted in two ways. First, it shows the
approximate nature of the concept of molecular diameter, with different values resulting from measurements of different quantities. Second , it is consistent with the idea that, at higher temperatures, more forceful collisions contract a molecule's perimeter.
430
STUDENT'S SOLUTIONS MANUAL
Solutions to theoretical problems P21.23
The most probable speed of a gas molecule corresponds to the condition that the Maxwell distribution be a max imum (it has no minimum); hence we find it by setting the first derivative of the function to zero and solve for the value of v for which this condition holds.
= 4n (  m
fey)
2nkT
)3/2v2e mv /2kT = const x v2e mv /2kT 2
2
df (v) ) = o.  = 0 when (/11.V2 2 kT
ds
So, v (most probable)
2kT)I /2 ( 2RT)I /2 = c* = ( ;;;= M
The average kinetic energy corresponds to the average of ~mv2. The average is obtained by determining (v 2) = fooo v2f(v)dv = 4n(m/ 2n)3/2 x ( l / kT)3 /2 fooo v4e  mv2 /2kT dv. The integral evaluates to (3/8)n 1/2(m / 2kT)5/2. Then 3kT
m
thus (£)
P21.2S
= ~m(v2) = ~kT .
f;
Write the mean velocity initially as a; then in the emerging beam (v x ) = K vxf(vx)dvx where K is a constant that ensures that the distribution in the emergent beam is also normalized. That is, I = K f; j(vx)dvx = K (m/2nkT ) 1/ 2 f~' e mv;J 2kT dvx . This integral cannot be evaluated analytically but it can be related to the error function by defining 2 my; x =2kT
which gives dvx
=
(2kT /m) 1/ 2 dx. Then b
2kT I=K ( m )1 / 2 (  )' /2 Io e_2 X dx 2nkT m 0
=
K
10r e _x b
n l/ 2
2
[b = (m / 2kT) 1/ 2 x a]
I
dx
= "2 Kerf (b)
where erf (z) is the error function [Table 9.2]: erf (z) = (2 / n 1/ 2) f~ eTherefore, K
=
x2
dx.
2 rf . e (b)
The mean velocity of the emerging beam is (v x ) = K (~) 1/ 2 lo a vxe mv; /2kT dvx 2nkT 0
a m )' /2( kT)lo d ( emvx2/2kT d vx) 2nkT m 0 d vx
=K (
MOLECULES IN MOTIO N
431
= (Vx )initial = (2kT / mn) 1/ 2.
Now use a
Thi s expression for the average magnitude of the onedimensional velocity in the x direction may be obtained from (Vr )
=2 =
00 1o
vxf(vx) d vx
=2
100 ( 0
m )1 /2 e mvx2 / 2kT dv Vx  x
2nkT
(~)1 /2 ( 2kT) = ( 2kT)1 /2 2nkT
m
Inn
It may also be obtained very quickly by setting a = with erf(b) = erf(oo) = I.
Substituting a erf(I / JTI / 2).
=
Therefore, (vx )
=
00
in the expression for (vx ) in the emergent beam
(2kT /m n) 1/2 into (v x ) in the emergent beam,
ema2/ 2kT
=
e l / lT and erf (b)
=
2kT) 1/2 x ;,,( mn erf ( n :/2)"
From tables of the error function (expanded version of Table 9.2), or from readily available software, or by interpolating Table 9.2, I ) _ erf(0.56) _ 0.57 and e I I " _ 0.73 . erf ( n l/2
Therefore, (v x) P21.27
= 10.47 (V x )initial I·
The most probable speed, c* , was evaluated in Problem 21.23 and is
c*
=
v( most probable)
2kT ) 1/2
= ( ;;;
Consider a range of speeds 6. v around c* and nc* ; then, with v
Therefore, f(3c*) f( c*)
P21.29
= 9 x e 8 = I3.02 x
10 3 ~
= c* ,
f(4c*) = 16 x e 15 = f(c*)
14.9x 10
6 1.
The current Ij carried by an ion) is proportional to its concentration Cj, mobility Uj, and charge number IZj I (Section 2 1.7). Therefore
where A is a constant. The total current passing through a solution is 1= L lj
=A L j
CjUj Zj .
432
STUDENT'S SOLUTIONS MANUAL
The transport number of the ionj is therefore tj
=
!l =
ACjUj lj
I
CjUjZj
.
A L j CjUjZj L j CjUjlj
If there are two cations in the mixture
P21.31
t'
C'U'
t"
c"u"
p(x)
p(6d)
II
~(N+s)! ~(Ns) 12N
=
[Justification 21.7] ,
NI
I~
.
(N
+ 6)
I!I~
= 4, p(6J..) = [Q] (m! =
(b) N
=
6, p(6J..)
(c) N
=
12,p(6J..)
P
z".
I
N!
(a) N
(NB O! P21.33
I
=
if z' =
=
61 6!0;2 6
=
.
I
(N  6) !2N
=
12! 9!3!2 12
x s J...
00
for m < 0).
1 26
=
=
I 64
= 10.06161·
12 x 11 x 10 ~ 3 x 2 x 212 =~.
= I). N! = ,,::
{~ (N
+ n)}I{ ~(N 
n)}!2N
The intermediate mathematical manipulations of Justification 21.7 begin with the above expression. Simpbfication of the expression proceeds by taking the natural logarithm of the expression, applying Stirling 's approximation to each term that has the In(x!) form , checking for term cancellations, and simplification using basic logarithm properties.
= In(2rr)1 / 2 + (x +~) In x Basic logarithm properties: In(x x y) = In x + In y Stirling's approximation: Inx!
In (x/y) In (xY)
=
x.
In x In y
= y Inx
Taking the natural logarithm and applying Stirling's formula gives
In P
=
In {
N!
}
+ n))!{ ~(N  n)}!2N = InN! In({~(N + n)}!) In({~(N {~(N
n)}!) ln2N
=~+ (N + ~)lnNpI
 ~ + {~(N + n) +~} In{~(N + n)}  [In (2 TC) 112 + {~ (N  n)
+ ~} In {~ (N

~(pI +A)]
 n)}  ~ (pi ;tl)]  In 2N .
MOLECULES IN MOTION
InP
= (N + ~)lnN In(2Jt)1 / 2 ln2N  {~ (N + n) + ~I In{~ (N + n)1 
I
~ In ~::+l (N
1 liN
I ~ I(N~::+ll_IIN
{~(N 
+ n + IJ In
n)
+ ~ }ln{~ (N 
433
n)1
1~ (1 + ~ ) 1 liN 
n + l)1n
1~ (1  ~) 1
~ In IN~:;'+II_IIN +n+ IJ H~) +In(1+~) 1 ~ {N 

n + I I { In
(~) + In (I  ~ ) }
In
+11'+
l)lnm lIN+n+
l)1n(1
+~)
~{N Ii' + I I In (~)  ~{N  n + I I In (I  ~)

I
~ In (N~:;'+l l _ IN + l)1n G) lIN + n+ l)1n (1 +~) ~ {N 

= In
+ I I In (1
/2)N+~
(N (N / 2)N+l n l /2
= In
In P
n
I

1
~)
I  {N 2
+ n + I I In (n 1+  )N
(n)
I {N  n + II In 1   . 2 N
(n2N )' /2  ~ {N + n + 1I In (1 + ~ )  ~ IN  n+ I I In (1  ~ ) .
Solutions to applications P21.35
The work required for a mass, m, to go from a distance r from the center of a planet of mass m' to infinity is
1
00
=
w
Fdr
where F is the force of gravity and is given by Newton's law of universal gravitation, which is
Gmm'
F= 2. r
G is the gravitational constant (not to be confused with g) . Then
,= 1
00
w
r
Gmm' 2dr r
Gmm' =. r
Since, according to Newton 's second law of motion, F
Gm' g=7·
= mg, we may make the identification
434
STUDENT'S SOLUTIONS MANUAL
Thus, w = grm. This is the kinetic energy that the particle must have in order to escape the planet 's gravitational attraction at a distance r from the planet's center; hence w = ~ m v2 = mg/" . Ve = (2g Rp) 1/2 [Rp = radius of planet]
which is the escape velocity. (a) ve = [(2) x (9.8 1 ms  2) x (6.37 x 10 6 m)]1 /2 = 111.2 km s I 1
(b)
m(Mars) R (Earth) 2 (6.37)2 x 2 x g(Earth) = (0.108) x x (9.8 1 m s2) m(Earth) R(Mars) 3.38
g(Mars) =
= 3.76ms 2 Hence, Ve = [(2) x (3.76ms 2) x (3.38 x 106 m)]1 /2 =
Is.o km s I I.
Since c = (8RT / nM)I /2, T = nMc 2/8 R and we can draw up the following table.
Earth Mars
11.9 2.4
He
02
23.7 4.8
190 38
[c = 11.2 km SI] [c=S.Okm s l ]
In order to calculate the proportion of molecules that have speeds exceeding the escape velocity, Ve, we must integrate the Maxwell di stribution [2l.4] from Ve to infinity.
P=
00 1
f(v)dv =
Ve
100 Ve
(m
4n   ) 3/ 2 v2 e mv2/ 2kT dv 2nkT
m]
[M
= k
R
This integral cannot be evaluated analytically and must be expressed in terms of the error function . We proceed as follow s. Defining f3
= ml 2kT and y2 = f3 v2 gives v = /
P
3 2  lfl  I/21 00 = 4 n (fl) fl n
f31 /2ve
frl /2y, v 2 = fr l y2 , Ve =
4100
y 2 e Y d y = l 2 ?
n /
,
2 Y d y ye
f3 1/ 2Ve
The first integral can be evaluated analytically ; the second cannot.
00
1 o
P
2
n l/ 2
i e Y dy =   ; hence
= 1
4
2
1 f3 1/ 2ve
2
2
ye Y (2ydy) = 1  172 n o n
~ /2
1 f31 / 2ve 0
fl I/ 2Ye,
2
yd(e Y
).
MOLECULES IN MOTION
This integral may be evaluated by parts
[erfc(z)
From f3
f3
=
I  erf(z» ).
= !!!.... = !!. and Ve = (2gRp) 1/ 2, 2kT
1/ 2
ve
=
2RT
MgRp
(
1/ 2
~)
For H2 on Earth at 240 K l 2 1/2V = (0.002016k g mol  ) x (9.807ms ) x (6.37 x 106 m »)1 / 2 = 7.94, f3 e (8.3141Klmol l) x (240K)
P
= erfc(7 .94) + 2 (:.~~ ) e(7.94)2 = (2 .9 x
10 29 )
+ (3.7
x 10 27 )
= 13.7 x
10 27 1.
At 1500 K
f3
1/2
Ve=
(0.002016 kg mol  I) x (9.807ms 2 ) x (6.37 x 106 m »)1 /2 =3.18, (8.3141K 1 molI) x (I500K)
P= erfc(3 .18) + 2 (~'II/~) e(3 18)2 = (6.9 x 10
6)
+ ( 1.46 x
10 4) = 11.5 x 10 4 1.
For H2 on Mars at 240 K
f3
_ (0.002016 kg mol  I) x (3.76m s2) x (3.38 x 106 m») 1/2 _ ve  3.58 (8.3141K 1 molI) x (240K) ,
1/2
P
= erfc(3.58) + 2 (~'~/~) e (358)2 = (4.13
x 10 7) + (1.10 x 10 5 )
P
= erfc(1.43) + (1.128)
= 0.0431 + 0.209 = I 0.251.
x ( 1.43) x e (IA3)2
= 11.1
x 10 5
For He on Earth at 240 K
f3
P
1/2
ve
=
(0.004003kg mol  l ) x (9.807ms 2) x (6.37 x 106 m»)1 /2 _ = 11.19 (8.3 141 K I molI ) x (240K) ,
= erfc(l1.2) + ( 1.128) x
(11.2) x e _ (112)2
= 0 + (4
X
10 54 )
= 14
x 10 54 1.
I.
435
436
STUDENT'S SOLUTIONS MANUAL
At ISOO K, f31 / 2ve
P
= 4.48,
= erfc(4.48) + ( 1.128)
= 11.0 X
x (4.48) x e(4.48)2
=
(2.36 x 10 10 ) + (9.71 x 10 9 )
10 8 1.
For He on Mars at 240 K 1'11 / 2 p
P
~=
((0.004003 kg molI) X (3.76ms 2) X (3.38 (8.314JK 1 mol I) X (240K)
= erfc(S.OS) + (1.128)
= 14.9 x
10 11
At lS00 K, f31 /2ve
P
x (2.02) x e(2.02)2 = (4.28 x 10 3)
For 0 2 on Mars at 240 K, f31 / 2ve
= 14.3,
= erfc(14.3) + (J.J28)
x 10 11)
+ (0.0401) = 10.4441.
0 at both temperatures.
x (14.3) x e(14.3)2 = 0
+ (2.S
X
10 88 )
= 12.s
x 10 88 1~ O.
= S.71,
= erfc(S.71) + (J.J28) = 14.s x
+ (4.79
,
= 2.02,
~
p
=S.~
I.
= erfc(2.02) + (1.J28)
At lS00 K, f31 / 2ve
106 m») 1/ 2
x (S .OS) x e(5.05)2 = (9.21 x 10 13)
For 02 on Earth it is clear that P
P
X
x (S.71) x e(5.71 )2
= (6.7
x 10 6 )
+ (4.46
x 10 14 )
10 14 1.
Based on these numbers alone, it would appear that H2 and He would be depleted from the atmosphere of both Earth and Mars only after many (millions?) years; that the rate on Mars, though still slow, would be many orders of magnitude larger than on Earth; that 0 2 would be retained on Earth indefinitely ; and that the rate of 02 depletion on Mars would be very slow (billions of years?), though not totally negligible. The temperatures of both planets may have been higher in past times than they are now.
In the analysis of the data, we must remember that the proportions, P, are not rates of depletion, though the rates should be roughly proportional to P. The results of the calculations are summarized in the following table. 240K
ISOO K He
He P(Earth) P(Mars)
3.7 x 10 27 l.l x 10 5
4 X 10 54 4.9 X 10 11
o o
I.S
X
10 4
0.2S
1.0 x 10 8 0.044 4.S
o X
10 14
MOLECULES IN MOTION
P21.37
437
Dry atmospheric air is 78.08% N2, 20.95% 02, 0.93% AI, 0.03% C02, plus traces of other gases. Nitrogen, oxygen, and carbon diox.ide contribute 99.06% of the molecules in a volume with each molecule contributing an average rotational energy equal to kT. The rotational energy density is given by
ER PR=V=
0.9906N (e R ) V
0.9906kT pNA RT
= 0 .9906 P
0.9906(1.013 x 105 Pa)
=
= 0.1004 J cm I .
The total energy density (translational plus rotational) is
P21.39
In
= PK + PR = 0. 15 1 cm  3 + 0.101 cm 3
In
= 0.25 J cm 3.
For order of magnitude calculations we restrict our assumed values to powers of 10 of the base units. Thus
P
=
Igcm 3 = I x 103 kgm  3,
'1 (air)
=1x
10 5 kg m I s I [see comment and question below].
We need the diffusion constant kT D= . 6Jt'1a
a is calculated from the volume of the virus which is assumed to be spherical m
( I x 105 u) x (I X 10 27 kg u I )
P
I x 103 kg m 3
V="'"
V
= 1Jta 3 .
a"'"
V)1 /3"'"
(4 
D "'" ( (6Jt)
(I
x 1O 25 m 3 )1 / 3
4
""' l x lO
25
3
m.
8
"'" I x 1O m.
( I x 1O 23 JK I )x (300K) ) 9 2 \ "'" I x 10 m s . x 1O 5 kgm  l s I )(1 x 1O 5 m)
X (1
For threedimensional diffusion,
Therefore it does not seem likel y that a cold could be caught by the process of diffusion. COMMENT. In a Fermi calculation only those values of physical quantities that can be determined by scientific
common sense should be used . Perhaps the value for 1) (air) used above does not fit that description .
Question. Can you obtain the value of '1 (air) by a Fermi calculation based on the relation in Table 21.3 ?
438
P21.41
STUDENT'S SOLUTIONS MANUAL
c(x, t ) = Co
+ (cs 
co){ I  erf(~)} where ~(x, t) =
x 1/2' (4Dt )
In order for c(x, t ) to be the correct solution of this diffusion problem it must satisfy the boundary condition , the initial condition, and the diffusion equation (eqn 2 1.68). Accordi ng to Justification 9.4,
erf(O
= 1
2 [ 00 2 rr 1/2 J~ eJ' dy.
= 0, ~ = 0, and erf(O) = 1 (2/ rrl /2) fooo ei dy = I  (2 / rrl / 2) x (rrl /2/2) = O. Thus, c(O, t ) = Co + (c s  co){ I  O} = cs . The boundary condition is satisfied. At the initial time (t = 0), ~(x , 0) = 00 and erf(oo) = I. Thus, c(x, 0) = Co + (c s  co ){l  I) = Co. The initial condition At the boundary x
is satisfied. We must find the analytical forms for 'dc/'d t and 'd 2 c/ 'dx 2 . If they are proportional with a constant of proportionality equal to D, c(x , t ) satisfies the diffusion equation.
The constant of proportionaljty between the partials equals D and we conclude that the suggested solution sati sfies the diffusion equation. Diffusion through alveoli sites (about I cell thick) of oxygen and carbon dioxide between lungs and blood capi llaries (also about I cell thjck) occurs through about 0.075 mm (the diameter of a red blood cell). So we will examine diffusion profiles for 0 ::: x ::: 0. 1 mm . The largest distance suggests that the
3 .r~r_~_rr_,
xl mm
Figure 21.2
MOLECULES IN MOTION
439
longest time that must be examined is estimated with eqn 2l.82.
Figure 2l.2 shows oxygen concentration distributions for times between 0.01 sand 4.0 s. Illustration 5.2 uses Henry's law to show that the equilibrium concentration of oxygen in water equals 2.9 x 104 mol dm 3. We use this as an estimate for Cs and take Co to equal zero.
22
The rates of chemical reactions
Answers to discussion questions 022.1
The timescales of atomic processes are rapid indeed : according to the following table, a nanosecond is an eternity. Note that the limes given here are in some way typical values for times that may vary over two or three orders of magnitude. For example, vibrational wavenumbers can range from about 4400 cm I (for H 2) to 100 cm I (for 12) and even lower, wi th a corresponding range of associated times. Radiative decay rates of electronic states can vary even more widely: Times associated with phosphorescence can be in the millisecond and even second range. A large number of timescales for physical, chemical, and biological processes on the atomic and molecular scale are reported in Figure 2 of A.H. Zewail, Femtochemi stry: atomicscale dynamics of the chemical bond. 1. Phys. Chel11. A 104, 5660 (2000) . Process
tIns
Reference
Radiative decay of electron ic excited state Rotational motion
I X 10 1 3 X 10 2 3 x 10 5 2 x 10 5 1 X 10 4 I x 10 3 3 x 10 3
Section 13.3b
Vibrational motion Proton transfer (in water) Initial chemical reaction of vision* Energy transfer in photosynthesist Electron transfer in photosynthesis Polypeptide helixcoil transition Collision frequency in liquids
2 x 102 4 X 10 4
B ~ lcm  1
ii ~ 1000 cm 
I
Section 2 1.7a Impact 1I4.1 Impact 123 .2 Impact 123.2 Impact 122.1 Section 21. I bt
*Photoisomerization ofretinal from IIcis to alltrans. t Time from absorption until electron transfer to adjacent pigment. t Use formula for gas collision frequ ency at 300 K, parameters for benzene from Data section, and density of liquid benzene. 4
Radiative decay of excited electronic states can range from about 10 9 s to 10 seven longer for phosphorescence involving ' forbidden ' decay paths. Molecular rotational motion takes place on a scale of 10 12 to 10 9 s. Molecular vibrations are faster still , about 10 14 to 10 12 s. The mean time between collisions in liquids is similarly short, 10 14 to 10 13 s. Proton transfer reactions occur on a timescale of about 10 10 to 10 9 S. Impact 114.1 describes several events in vision, including the 200fs photoisomerization that gets the process started. Impact 123.2 lists timescales of several energytransfer and electrontransfer steps in photosynthesis. Initial energy transfer (to a nearby pigment) has a timescale of 10 around 10 13 to 10 11 s, with longerrange transfer (to the reaction center) taking about 10 s. Immediate electron transfer is also very fast (about 3 pS), with ultimate transfer (leading to oxidation of water and
THE RATES OF CHEMICAL REACTIONS
441
reduction of plastoquinone) taking from 10 10 to 10 3 s. lmpact 122.1 discusses helixcoil transitions, including experimental measurements of times cales of tens or hundreds of microseconds (10 5 to 10 4 s) for formation of tightly packed cores. The ratedetermining step for the helixcoil transition of small polypeptides has a relaxation time of about 160 ns in contrast to the faster 50 ns relaxation time oflarge protein. 022.3
The determination of a rate law is simplified by the isolation method in which the concentrations of all the reactants except one are in large excess. If B is in large excess, for example, then to a good approximation its concentration is constant throughout the reaction. Although the true rate law might be v = k[A][B], we can approximate [B] by [B]o and write v = k' [A],
where k' = k[B]o [22.10]
which has the form of a firstorder rate law. Because the true rate law has been forced into firstorder form by assuming that the concentration of B is constant, it is called a pseudo firstorder rate law. The dependence of the rate on the concentration of each of the reactants may be found by isolating them in turn (by having all the other substances present in large excess) and so constructing a picture of the overall rate law. In the method of initial rates, which is often used in conjunction with the isolation method, the rate is measured at the beginning of the reaction for several different initial concentrations of reactants. We shall suppose that the rate law for a reaction with A isolated is v = k[A]G; then its initial rate, Vo is given by the initial values of the concentration of A, and we write Vo = k[A]g . Taking logarithms gives log vo = log k
+ a 10g[A]0. [22.11]
For a series of initial concentrations, a plot of the logarithms of the initial rates against the logarithms of the initial concentrations of A should be a straight line with slope a. The method of initial rates might not reveal the full rate law, for the products may participate in the reaction and affect the rate. For example, products participate in the synthesis of HBr, where the full rate law depends on the concentration of HBr. To avoid thi s difficulty, the rate law should be fitted to the data throughout the reaction. The fitting may be done, in simple cases at least, by using a proposed rate law to predict the concentration of any component at any time, and comparing it with the data. Because rate laws are differential equations, we must integrate them if we want to find the concentrations as a function of time. Even the most complex rate laws may be integrated numerically. However, in a number of simple cases analytical solutions are easily obtained, and prove to be very usefu l. These are summarized in Table 22.3. In order to determine the rate law, one plots the righthand side of the integrated rate laws shown in the table against t in order to see which of them results in a straight line through the origin. The one that does is the correct rate law.
022.5
The ratedetermining step is not just the slowest step: it must be slow and be a crucial gateway for the formation of products . If a faster reaction can also lead to products, then the slowest step is irrelevant because the slow reaction can then be sidestepped. The ratedetermining step is like a slow ferry crossing between two fast highways: the overall rate at which traffic can reach its destination is determined by the rate at which it can make the ferry crossing. If the first step in a mechanism is the slowest step with the highest activation energy, then it is ratedetermining, and the overall reaction rate is equal to the rate of the first step because all subsequent steps are so fast that once the first intermediate is formed it results immediately in the formation of products. Once over the initial barrier, the intermediates cascade into products . However, a ratedetermining step
442
STUDENT'S SOLUTIONS MANUAL
may also stem from the low concentration of a crucial reactant or catalyst and need not correspond to the step with highest activation barrier. A ratedetermining step arising from the low activity of a crucial enzyme can sometimes be identified by determining whether or not the reactants and products for that step are in equilibrium: if the reaction is not at equilibrium it suggests that the step may be slow enough to be ratedetermining. 022.7
Case II
Case I
G
G
P2 Reaction coordinate
!\ Reaction coordinate
Case III
G
R~ P2 Reaction coordinate
Figure 22.1
Simple diagrams of Gibbs energy agai nst reaction coordinate are useful for distinguishing between kinetic and thermodynamic control of a reaction. For the simple parallel reactions R + PI and R + P2, shown in Fig. 22. 1 cases I and II, the product PI is thermodynamicall y favored because the Gibbs energy decreases to a greater extent for its formation. However, the rate at which each product appears does not depend upon thermodynamic favorability. Rate constants depend upon activation energy. In case I the activation energy for the formation of P I is much larger than that for form ation of P2. At low and moderate temperature the large activation energy may not be readily available and PI either cannot form or forms at a slow rate. The much smaller activation energy for P2 formation is available and, consequentl y, P2 is produced even though it is not the thermodynamically favored product. Thi s is kinetic control. In this case, [P2l/ [PIl = k2 / kl > I [22.46]. The activation energies for the parallel reactions are equal in case n and, consequently, the two products appear at identical rates. If the reactions are irreversi ble, [P2l/ [Pll = k2 / kl = I at all times. The results are very different for reversible reactions. The activation energy for PI + R is much larger than that for P2 + R and PI accumulates as the more rapid P2 + R + PI occurs. Eventually the ratio [P2l/ [P ll approaches the equilibrium value for which
[P2l ) ( [PIl
= e(6~6GIl/RT
< I.
eq
This is thermodynamic control. Case III represents an interesting consecutive reaction series R + PI + P2. The first step has relatively low activation energy and PI rapidly appears. However, the relatively large activation energy for the second step is not available at low and moderate temperatures. By usi ng low or moderate temperatures and short reaction times it is possible to produce more of the thermodynamically less favorable PI . This is kinetic control. High temperatures and long reaction times will yield the thermodynamically favored P2. The ratio of reaction products is determined by relative reaction rates in kinetic controlled reactions. Favorable conditions include short reaction times, lower temperatures, and irreversible reactions. Thermodynamic control is favored by long reaction times, higher temperatures, and reversible reactions. The ratio of products depends on the relative stability of products for thermodynamically controlled reactions.
THE RATES OF CHEMICAL REACTIONS
022.9
443
The primary isotope effect is the change in rate constant of a reaction in which the breaking of a bond involving the isotope occurs. The reaction coordinate in a CH bond breaking process corresponds to the stretching of that bond. The vibrational energy of the stretching depends upon the effective mass of the C and H atoms. See eqn 13.50. Upon deuteration, the zero point energy of the bond is lowered due to the greater mass of the deuterium atom. However, the height of the energy barrier is not much changed because the relevant vibration in the activated complex has a very low force constant (bonding in the complex is very weak), so there is little zero point energy associated with the complex and little change in its zero point energy upon deuteration. The net effect is an increase in the activation energy of the reaction. We then expect that the rate constant for the reaction will be lowered in the deuterated molecule and that is what is observed. See the derivation leading to eqns 22.5122.53 for a quantitative description of the effect. A secondary kinetic isotope effect is the reduction in the rate of a reaction involving the bonded isotope even though the bond is not broken in the reaction. The cause is again related to the change in zero point energy that occurs upon replacement of an atom with its isotope, but in this case it arises from the differences in zero point energies between reactants and an activated complex with significantly different structure. See Illustration 22. 3 for an example of the estimation of the magnitude of the effect in a heterolytic dissociation reaction. If the rate of a reaction is altered by isotopic substitution it implies that the substituted site plays an important role in the mechanism of the reaction. For example, an observed effect on the rate can identify bond breaking events in the ratedetermining step of the mechanism. On the other hand, if no isotope effect is observed, the site of the isotopic substitution may play no critical role in the mechanism of the reaction .
Solutions to exercises E22.1(b)
E22.2(b)
v
= _ d[A] = _~ d[8] = d[C] = ~ d[D] = l.00moldm 3 s1 so dt
3 dt
dt
2 dt
Rate of consumption of A
= 11.0 mol dm 3 S I 1
Rate of consumption of B
= 13.0 mol dm  3 S I 1
Rate of formation of C
= 11.0 mol dm 3 S I 1
Rate of form ation of D
= 12.0 mol dm 3 s I 1
. of 8 Rate of consumptIOn Rate of reaction
'
d[8] = I 1.00 mol dm  3 s  ~ =  dr
I
=  ~ d[8] = 10.33 mol dm 3 s I 1= d[C] = ~ d[D] = _ d[A] 3 dt
Rate of formation of C
= 10.33 mol dm  3 S I I
Rate of formation ofD
= 10.66 mol dm 3 sI I
Rate of consumption of A = 10.33 mol dm  3 S I I
.
dt
2 dt
dt
444
E22.3(b)
STUDENT'S SOLUTIONS MANUAL
The dimensions of k are dim of v
amount x length 3 x time I
(dim of [AJ) x (dim of [BJ)2
(amount x length  3 )3
= length 6
x amount 2 x time I
In mol, dm, s units, the units of k are Idm 6 mol  2 S I I
E22.4(b)
(a) v
=  d[A] = k[A][Bj2 so
(b) v
=
d[A]
dt
d[C]
dt
d[C] so 
dt
dt
= k[A][Bj2
= k[A][B] 2
The dimensions of k are amount x length 3 x time I
dim of v
   ==   : ;  
dim of [A] x dim of [B] x (dim of [CJ)I
amount x length  3
= time I
The units of k are ~
v E22.S(b)
= d~~] = I k[A][B][C]  1 I
The rate law is
v = k[AY' ex p"
= {Po(1 
f)}a
where a is the reaction order, andl the fraction reacted (so that I lis the fraction remaining). Thus VI
{Po(1  !J)}(I {Po(l  h))"
=
(1 /1)a h
and
a
1
= _In;(v,.I_I_v2:): = In
(_I_II) 1
E22.6(b)
(I  0.100)
= 12 00 I
.
The halflife changes with concentration, so we know the reaction order is not 1. That the halflife increases with decreasing concentration indicates a reaction order < 1. Inspection of the data shows the halflife roughly proportional to concentration, which would indicate a reaction order of 0 according to Table 22.3. More quantitatively, if the reaction order is 0, then (I)
tl / 2
ex p
and
t 1/2
;m 1/ 2
PI P2
We check to see if this relationship holds ( I) tl / ?
340 s
t (2)
178 s
 =   = 1/ 2
1.91
so the reaction order is E22.7(b)
h
In (9.71 /7 .67) In 1  0.200
@] .
The rate law is 1 d[A] v =    =k[A]
2 dt
and
~ P2
= 55 .5 kPa = 1.92 28.9kPa
THE RATES OF CHEMICAL REACTIO NS
445
The halflife formula in eqn 22.13 is based on the assumption that
= k[A].
_ d[A] dt
That is, it would be accurate to take the halflife from the table and say
In 2
tl / 2
= k'
where k' = 2k . Thus t
1/ 2
=
In 2
2(2.78 x 10 7 S I)
=
I1.80
X
106 s
I
Likewise, we modify the integrated rate law (eqn 22. 12b), noting that pressure is proportional to concentration:
(a) Therefore, after 10 h, we have 4 p = (32. 1 kPa) exp[2 x (2.78 x 1O 7 s l ) x (3.6 x \0 s) ] = 13 1.5 kPal
(b) After 50 h, p = (32.1 kPa)exp[2 x (2.78 x 10 7 SI) x ( 1.8 x 105 s) ] = 129.0 kPa E22.8(b)
From Table 22 .3, we see that for A
kt
+ 2B
+
I
P the integrated rate law is
 2[P])] = [B]o  I 2[A]0 In [[A]O([B]O ([A]o  [P])[B]o
(a) Substituting the data after solving for k
k =
I [ (0.075 x (0.080  0 .060) ] x In (3.6 x 103 s) x (0.080  2 x 0.075) x (mol dm 3 ) (0.075  0.030) x 0.080
= \ 3.47 x 1O 3 dm 3 mol  l s l \ (b) The halflife in terms of A is the time when [A) = [A]0/2 = [P], so
(A) tl / 2
I I [[A]O([B]0(2[A]0 / 2»] = k([B]o _ 2[A]0) n ([A]0[B]0/2)
which reduces to (A) 
t 1/ 2

=
I I (2 2[A]0) k([B]o _ 2 [A]0) n  [B]o I x ln ( 2  0.150) (3.47 x 10 3 dm 3 mol I S I) x (0.070 mol dm 3 ) 0.080
=856fs=~
446
STUDENT'S SOLUTIONS MANUAL
The halflife in terms of B is the time when [8] [A]o
1
(1 /2(B)
=
k([B]o _ 2[A]o) In
= [B]o/2 and [P] = [B]o/4:
([B]o  2[B]O)]
[ ( [A]o 
[B]o) [B]o 4
which reduces to
1/2
E22.9(b)
1
(B)
t
I (
= k([B]o _ 2[A]o) n
[A]o/2 ) [A]o  [B]o/4
=
I 0.075/2 ) x In ( 0.Q75  (0.080/4) (3.47 x 10 3 dm 3 mol  I s I) x (0.070 mol dm 3 )
=
1576 s
= I0.44 h I
(a) The dimensions of a secondorder constant are
dim ofv (dim of [A))2
amount x length 3 x time I
  
"'=
(amount x length 3 )2
= length 3 x
amount I x timeI
I
In molecule, m, s units, the units of k are m3 moleculeI s I
I
The dimensions of a thirdorder rate constant are dim of v (dim of [A])3
amount x length 3 x time I



3,,
(amount x length )3
= length
6
x amount
I
In molecule, m, s units, the units of k are m6 molecule 2 s I
 2.  1 x lIme
I
COMMENT. Technically, "molecule" is not a unit , so a number of molecules is simply a number of individual
objects, that is, a pure number. In the chemical kinetics literature, it is common to see rate constants given in molecular units reported in units of m 3 S  1 , m 6 S  1 , cm3 S 1 , etc.
(b) The dimensions of a secondorder rate constant in pressure units are dim of v (dim of p )2
press ure x time(pressure)
i
~::;;2
= pressure  Ix .lIme _ I
In SI units, the press ure unit is N m 2
=
I
Pa, so the units of k are Pa I S I
The dimensions of a thirdorder rate constant in pressure units are dim of v (dim of p)3
press ure x timeI
=: ,:;3
(pressure)
= pressure
2.
x tIme
In SI pressure units, the units of k are I Pa 2 s I I. E22.10(b) The integrated rate law is
k( =
I In [A]o( [B]o  2[C)) [Table 22.3] [Blo  2[A]o ([A]o  [C))[B]o
_I
I
THE RATES OF CHEMICAL REACTIONS
447
Solving for [C] yields. after some rearranging [A]o[B]o{exp[kt([B]o  2[A]0)]  I} [C] = [B]'oexp=[k:t(C:: [B==]:o::2=[A::]'o)=]::2==[A:"C]=o [C] so mol dm  3 =
(0.025) x (0.150) x (eO. 21x (0.lOO)x r/ s  I) (3.75 x 10 3 ) x (eO.021x r/s  I) (0.150) x e O.2Ix (0.lOO)x r/s  2 x (0.025) = (0.150) x eO.02I xr/s  (0.050)
03 (0.21 I) 3 ) x e [C] = (3 .75 x I rna I d m 3 =.165 ' x 10 rna I d m 31. (0. 150) x e0 21  (0.050) 3 126 [C] = (3.75 x 10 ) x (e  I) mol dm  3 = 1 0.025 mol dm 3 1 (0. 150) x e 12.6  (0.050)
(a)
(b)
E22.11 (b) The rate law is
v
= _~ d[A] =
k[A] 3
2 dt
which integrates to
2kt
t
I(I I)
= 2'
= (4(3 .50 X = 11.5
[A]~
[A]2 
~m6 moI 2 sI Jx CO.02I m~1 dm 
10 4
x 106 s
3 )2
 (0.077
m~I dm )2 ) 3
I
E22.12(b) A reaction nthorder in A has the following rate law
_ d[A] = k[Al"
so
dt
d[A] [A]"
= k dt = [Ar" d[A]
Integration yields [A]I  n  [A]I  II _ _ _ _'0'
= kt
ln Let tl /3 be the time at which [A]
d
(~ [A]O)III  [A]b II
[A]b  "[(Vn  I]
In
1n
= ~,~
so kt1 /3
an
= [A]0 / 3.
tl / 3 =
3" 1  I [A]In ken _ I)
°
E22.13(b) The equilibrium constant of the reaction is the ratio ofrate constants of the forward and reverse reactions:
K
= kr kr
so
kr
= Kk r .
The relaxation time for the temperature jump is (Example 22.4): r = {kr
+ kr([B] + [C])}I
so
kr = r  I  kr([B]
+ [CD
448
STUDENT'S SOLUTIONS MANUAL
Setting these two expressions for kr equal yields so
k r 
T(K
I
+ [B] + [CD
Hence I
kr = ~~~~~~~ (3.0 X 10 6 s) x (2.0 X 10 16 + 2.0 X 10 4 + 2.0 x 10 4 ) mol dm 3 = 18.3 x 10 8 dm 3 mol  I sI
I
E22.14(b) The rate constant is given by
k
= A exp (
Ea ) RT
[22.31]
so at 24°C it is 1.70
X
10 2 dm 3 mol  I SI
= A exp (
Ea
(8.31451K I molI) x [(24 + 273) K]
)
and at 37 °C it is 2.01 x 10 2 dm 3 mol  I S I
= A exp (
Ea
(8.31451 KI mol  I) x [(37
+ 273) K]
)
Dividing the two rate constants yields 1.70 2.01 so
2
X X
1010 2
= exp
2 (1.70 X 10 ) In 2.01 x 10 2
=
[(
(
Ea ) 8.31451K  I mol  I x
(1
1)]
297K  310K
Ea ) (1 I) 8.31451K Imol  1 x 297K31OK
and Ea = _ ( __1_ _ _1_)1 In (1.70 x IO~ ) x (8.31451 K I mol  I) 297 K 310 K 2.0 I x 10
= 9.9
x 10 3 1 mo]I
= 19.9 kJ mol  I I
With the activation energy in hand, the prefactor can be computed from either rate constant value A = kexp
Ea )
( RT
2 3 I I ( = (1.70 x 10 dm mol s ) x exp
= I0.94 dm 3 mol I sI
I 3 9.9 x 10 J mol) I (8.3145JKI mol ) x (297K)
I
E22.15(b) (a) Assuming that the ratedetermining step is the scission of a CH bond, the ratio of rate constants
for the tritiated versus protonated reactant should be [22.53 with hciJ
= fiw = Ii(kj /1) 1/ 2]
THE RATES OF CHEMICAL REACTIONS
449
The reduced masses will be roughly I u and 3 u respectively, for the protons and 3H nuclei are far lighter than the rest of the molecule to which they are attached. So A~
(1.0546
2 x (1.381 x 1O 23 JK I) x (298K)
(~ /2  ~ /2 ) (l u) (3 u)
x ~
10 34 I s) x (450Nm I)I / 2
X
~:.,,=:~;C"":"""":"::::c
27
I kgu  ) 1/2
2.8
~ e 2S = 10.06 ~
so ::
x ( 1.66 x 1O
1/ 16 1
(b) The analogous expression for 16 0 and ISO requires reduced masses for C 16 0 and CI SO bonds. These reduced masses could vary rather widely depending on the size of the whole molecule, but in no case will they be terribly different for the two isotopes. Take 12CO, for example:
/L 16 = _(1_6_.0_u_)_x_(1:2_.0_u_) (16.0+ 12.0) u
( 1.0546
= 6.86 u
and
/LI S =
( 18.0u) x (12.0u) (18.0 + 12.0) u
= 7.20 u
10 34 J s) x (1750 N m I) 1/ 2
X
A ==::::::::::;;;;:=;:=:::::=:23
2 x ( 1.381 x 1O
1K I) x (298K)
1 1/2 1 1/2 ) x ( 1.66 x 1O 27 kgu  I)  1/2 (6.86 u) (7.20 u)
x (
= 0.12 so kl s kl 6
= e O.12 = 10.891
At the other extreme, the 0 atoms could be attached to heavy fragments such that the effective mass of the relevant vibration approximates the mass of the oxygen isotope. That is, /L16 ~ 16 u and /L IS ~ I 8 u so A ~ 0. 19 1
E22.16(b)
k'a
k = k k a b
kl s = e O.19 = 10.831
so
kl6
1
+ kaPA
[analogous to 22.67]
Therefore, for two different pressures we have
so
ka =
(~  ~) (~  ~)  I P
P'
k
k'
1
2.2
X
10 4 sI
)1
450
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems P22.1
A simple but practical approach is to make an initial guess at the order by observing whether the halflife of the reaction appears to depend on concentration. If it does not, the reaction is firstorder; if it does, it may be secondorder. Examination of the data shows that the first halflife is roughly 45 minutes, but that the second is about double the first. (Compare the 0 ~ 50.0 minute data to the 50.0 ~ 150 minute data.) Therefore, assume secondorder and confirm by plotting I/[Al against time. If the reaction is secondorder, it will obey
1 [Al
I
= kt + [Ala
[22.15bl .
We draw up the following table (A
tlmin m(urea)/g m(A)/g
[AJ /(mol dm  3 ) [Ar l /(dm 3 mol I)
0 0 22.9 0.381 2.62
= NH4CNO).
20.0 7.0 15.9 0.265 3.78
50.0 12.1 10.8 0.180 5.56
65.0 13.8 9.1 0.152 6.60
150 17.7 5.2 0.0866 11.5
The data are plotted in Fig. 22.2 and fit closely to a straight line. Hence, the reaction is / secondorder /. The rate constant is the slope: 1k
= 0.0594 dm 3 mol  I min I I.
12
10 t
0 E
8
E
~

~
I ~
6
'"'
4
2
0
20
40
60
80 t/ min
100
120
140
Figure 22.2
To find [AJ at 300 min, use eqn 22.15c: [Ala [Al
= I + kt[Ala
0.382 mol dm  3 = 0.0489 mol dm 3. 1 + (0.0594) x (300) x (0.382)
THE RATES OF CHEMICAL REACTIONS
451
The mass of NH4CNO left after 300 minutes is
P22.3
The procedure adopted in the solutions to Problems 22. 1 and 22.2 is employed here. Examination of the data indicates the halfLife is independent of concentration and that the reaction is therefore 1firstorder I. That is confirmed by a plot of In (A = nitrile). t/( 103 s) [Al /(mol dm
3
)
0
2.00
4.00
6.00
8.00
10.00
12.00
1.10
0.86
0.67
0.52
0.41
0.32
0.25
0.78
0.6 1
0.47
0.37
0.29
0.23
0.246
0.496
0.749
0.987
l.235
1.482
ill [Alo
In
(l&)
(l&) against time (eqn 22. 12b). We draw up the following table
0
slope
A leastsquares fit to a linear equation gives k coefficient of 1.000. P22.5
11.23
X
10 4
S i
1 with a correlation
As described in Example 22.5, if the rate constant obeys the Arrhenius equation [22.29] , a plot of In k against I / T should yield a straight line with slope Ea/ R . However, since data are available only at three temperatures, we use the twopoint method, that is,
R In (576/ 2.46)
41
Ea= «(1 / 313K)(i / 273K)) =9.69 x 10 Jmol
.
For the pair () = 20°C and 40°C, Rln (576/45. 1)
4
1
Ea= ((l/313K)(1/293K)) =9.71 x 10 Jmol.
The agreement of these values of Ea indicates that the rate constant data fits the Arrhenius equation and that the activation energy is 19.70 x 104 J molI I. P22.7
The data for this experiment do not extend much beyond one halflife. Therefore the halflife method of predicting the order of the reaction as described in the solutions to Problems 22.1 and 22.2 cannot be used here. However, a si milar method based on threequarters lives will work. For a firstorder reaction, we may write (analogous to the derivation of eqn 22.13)
~[Alo
3 4
4 3
kt3 /4 =  In   =  In  = In  = 0.288
[Alo
or
0.288
t3 /4
= k'
452
STUDENT'S SOLUTIONS MANUAL
T hus the threequarters life (or any given frac tional life) is also independent of concentration for a firstorder reaction. Examination of the data shows that the first threequarters life (time to [A] = 0.237 mol dm  3) is about 80 min and by interpolation the second (time to [A] = 0.178 mol dm 3) is also about 80 min. Therefore the reaction is firstorder and the rate constant is approximately
k
0.288 0.288 ~ t3/4 80 min
= 
=
36 x 10 3 minI. .
A leastsquares fit of the data to the firstorde r integrated rate law [22.12b] gives the slightly more accurate result, k = 13.65 x 10 3 min  I I. The halflife is
rl /2
In 2
In 2
=  k = 3.65
x 10 3 min  I
= 1190 min I..
The average lifetime is calculated form [A] [A]o
= e  kt
[22.12b].
which has the form of a distribution functio n. The ratio ~ is the fraction of sucrose molecules that have lived to time t. The average lifetime is then
The denominator ensures normalization of the distribution function . COMMENT.
The average lifetime is also called the relaxation time. Compare to eqn 22.28. Note that the
average lifetime is not the halflife. The latter is 190 minutes. Also note that 2 x 13/ 4
P22.9
I
11 / 2 .
The data do not extend much beyond one halflife; therefore, we cannot see whether the halflife is constant over the course of the reaction as a preliminary step in guessing a reaction order. In a first order reaction, however, not only the halflife but any other similarly defined fractional lifetime remains constant. (T hat is a property of the exponential funct ion .) In this problem, we can see that the ~ life is not constant. (It takes less than 1.6 ms for [CIO] to drop from the first recorded val ue (8.49 /Lmol dm  3) by more than of that value (to 5.79 !Lmol dm  3 ) ; it takes more than 4 .0 more ms for the concentration to drop by not even of that value (to 3.95 !LmOI dm 3 ). So our working assumption is that the reaction is not firstorder but secondorder. Draw up the following table.
1
1
rims
[CIO]/(!Lmol dm 3)
( II[CIO])/(dm3!Lmoll)
0. 12 0.62 0.96 1.60 3.20 4.00 5.75
8.49 8.09 7.10 5.79 5.20 4 .77 3.95
0. 11 8 0. 124 0.141 0. 173 0. 192 0.210 0.253
THE RATES OF CHEM ICAL REACTIONS
453
The plot of [CIO] vs. t in Fig. 22.3 yields a reasonable straight line; the linear least squares fit is: (I / [CIO)) / (dm\lmol  I)
= 0.118 + 0 .0237(t / ms)
R2 = 0.974.
The rate constant is equal to the slope
The halflife depends on the initial concentration (eqn 22.16): '1 / 2
= __1  = k[ClO]o
I (2.37 x 10 7 dm 3 mol I sI )(8.47 x 10 6 mol dm 3 )
= 14.98
x 10 3 s I.
0.30
,.
0.25
"0
E
:::l
ME
0.20
"0
:::::'
I~
0.15 0.10 2
0
P22.11
A+B
~
d[P]
P,
3 rIm s
4
5
6
Figure 22.3
= k[A]'"[BJ"
dt
and, for a short intervall3t, 8[P]
~
k[A]'"[B]"8t
Therefore, since 8[P]
= [P]f
 [P]o
=
[P]I>
[Chloropropanel · . d d f [P ] . I· h [Propene] IS III epen en! 0 ropene , Imp ylllg t at m [Chloropropane] [HCI]
= {P(HCI)
10 0.05
7.5 0.03
= I.
5.0 0.01
These results suggest that the ratio is roughly proportional to p (HCI)2, and therefore that m A is identified with HC!. The rate law is therefore d[Chloropropane]
"d"":t'"'::'
I
= k[Propane][HCI]
I
3
I
I
and the reaction is firstorder in propene and thirdorder in HC!.
= 3 when
454 P22.13
STUDENT'S SOLUTIONS MANUAL
2HCI ;=; (HCI)z,
[(HClh ] = KI [HClf
KI
HCI + CH3CH=CH2 ;=! complex (HClh + complex rate
+
[complex]
= K2[HCI][CH3CH =CH2]
CH3CHCICH3 + 2HCI k
d[CH3CHCICH3]
=
K2
dt
= k[(HClh][complex].
Both (HClh and the complex are intermediates, so substitute for them using equilibrium expressions: rate
= k[(HClh][complex] = k(KI [HClf)(K2[HCI][CH3CH=CH2]) = 1 kKIK2[HCI] 3[CH 3CH=CH2] 1
which is thirdorder in HCI and firstorder in propene. One approach to experimental verification is to look for evidence of proposed intermediates, using infrared spectroscopy to search for (HClh , for example. P22.1S
We can estimate the activation energy of the overall reaction by proceeding as in P22 .5: R In (keff /k~ff)
I
R In 3
I
I
E  18 kJ mol a,eff ((l / T)  (l / T'»  (I / 292K)(l / 343K )  '
To relate this quantity to the rate constants and equilibrium constants of the mechanism (P22.13), we identify the effective rate constant as keff = kKI K2 and apply the general definition of activation energy (eqn 22.30): ?
d In keff
= RT~ = RT
Ea,eff
keff d( l i T) d(l I T) ~
2 din
din keff
= R d(l I T)'
This form is useful because rate constants and equilibrium constants are often more readily differentiated when considered as functions of liT rather than functions of T, as in this case: In keff
=
In k + In KI + In K2
__ Rdlnkeff R dink _RdlnKI _RdlnK2 =E. +t:..H +t:..H so Ea,eff d(l / T) d(l I T) d(l / T) d(l l T) a r 1 r 2 dinK since  d(l / T)
t:..rH . = [van't Hoff equation, 7.23b]. R
Hence Ea = Ea,eff  t:..rHI  t:..rH2 = (18 + 14 + 14) kJ mol  I P22.17
I = _ k'a k
kakb
= 1+10 kJ mol I I·
I + _ [analogous to 22.67] . kaP
We expect a straight line when
~ k
is plotted against
~ . We draw up the following table.
P
plTorr
84. 1
11 .0
2.89
0.569
0.120
0.067
1/(plTorr)
0.012 0.336
0.091 0.448
0.346 0.629
1.76 1.17
8.33 2.55
14.9 3.30
1O 4 /(kl s I)
THE RATES OF CHEM ICAL REACTIONS
455
These points are plotted in Fig. 22.4. There are marked deviations at low pressures, indicating that the Lindemann theory is deficient in that region.
4
.,...... '"1;
~
..,.......
3
2
I
0
4
8
12
l/ (p/ Torr)
P22.19
Figure 22.4
The reasoning that led to eqn 22.46 holds as long as the rate laws for the two products have the same reaction orders:
Then, si nce Ea . 1 > Ea.2, the exponent in the exponential function is negative, and it gets less negati ve as the temperature increases. Therefore, the exponential function itself increases and the product concentration ratio also increases. COMMENT. A qualitative argument can be made that leads to the same conclusion, provided one under
stands that the activation energy is a measurement of the strength of a reaction's temperature dependence. (See eqn. 22 .30.) Since Ea.1 > Ea.2, the rate of reaction 1 increases faster with increasing temperature than does the rate of reaction 2.
Solutions to theoretical problems P22.21
A ;=" B d[A] dt
= k[A] + k'[B]
At all times, [Al Therefore, [B]
d~~]
+ [Bl
= [Alo
and
=
k' [Bl
[AJ.
+ k' {[Alo + [Blo 
[All = (k
To solve, one must integrate
f
(k
d[Al k' ([Alo
+ k' )[A] 
+ k[Al .
+ [Blo·
= [Ala + [Blo 
= k [A ]
d[B] dt
+ [Blo) = 
f
dt.
+ k' )[Al + k'( [A]o + [B]o) .
456
STUDENT'S SOLUTIONS MANUAL
· . [Al Th e so IutlOn IS
k'([Alo + [Blo) + (k[Alo  k' [Blo)e(k+k')r = ::,=........:..:''k+k'
The final composition is found by setting t
=
[Aloo
and [Bloo
=
C~ k') [Alo
= 00:
x ([Alo + [Blo).
+ [Blo 
=
[Aloo
(_k_)
x ([Alo
k +k'
+ [Blo) .
Note that
P22.23
d[Al dt
= 2k[Aj2[Bl
(a) Let [Pl = x at
I,
'
2A
+ B +
then [Al = Ao  2x and [Bl = Bo  x = ~  x. Therefore,
dx d[Al = 2= 2k (Ao dr dr
dx = k(Ao dt
P.
(I
 2x) 2 x
2
2x) x (Bo  x) ,
)= I
Ao  x
2
k(Ao  2x) 3
2
'
1 = 10r (Ao dx2r) 3 = 4I x [( Ao 12x )2  (IAo )2] .
"2kt
Therefore, kt (b) Now Bo
2x(Ao  x)
= ,
Aii(Ao  2x)
2
= Ao , so
dx 2 2  = k(Ao  2x) x (Bo  x) = k(Ao  2x) x (Ao  x) , dt kt _
r ..,._:,dx_,_:x (Ao  x) .
 10 (Ao 
2x)2
We proceed by the method of partial fractions (which is employed in the general case too), and look for the values of ex, fJ, and y such that ex (Ao  2x)2
(Ao  2x)2 x (Ao  x)
+ _fJ_ + _y_ . Ao  2x
Ao  x
This requires that ex(Ao  x)
+ fJ(Ao 
2x) x (Ao  x)
+ y(Ao 
2x)2
= I.
Expand and gather terms by powers of x: (Aoex
+ A6fJ + A6Y) 
(ex
+ 3fJAo + 4yAo)x + (2fJ + 4y)x 2 = I.
THE RATES OF CHEMICAL REACTIONS
This must be true for all x; therefore
+ 3Ao,8 + 3Aoy = 0,
a
2,8
+ 4y = 0. 2
=
These solve to give a
Ao
,,8
2
I
Ao
Ao
= 2' and y = 2·
Therefore, kt
r(
=
(2/Ao) _ (2/A6) (Ao2x)2 Ao2x
10
+ (i/A6))
I
(i/Ao)
dx
Aox x
I
 +  In(Ao  2x)   In(Ao  x) ) (Ao  2x A2 A2 0 0 0 I
P22.2S
d[Al The rate law  dt kt =
n I
tl /2
kt3 /4
=
t3 /4
v
=
2n 
= k([Alo
=
1 
I
x)([Blo
I
+ x).
 x)  k([Blo
[Blo
[E22.12(a)l.
4 ),,1  (I[Alo )"1] . (n _I 1 ) [( 3[Alo
dv The extrema correspond to dx [Alo  x
[Al3
C~ I) [C~lOr1  c~lor1
(3"4),,1 
= k([Alo 
dv dx
I_I)
[Al"
ktl /2 =
= t3 /4,
Hence, 
P22.27
I integrates to
(_1_) x (_1_1 _ _
Att = tl / 2,
At t
= k[Al" for n f.
+x
or
+ x).
= 0, or 2x
=
[Alo  [Blo
or
x=
[Alo  [Blo 2
457
458
STUDENT'S SOLUTIONS MANUAL
Substitute into v to obtain
Since v and x cannot be negative in the reaction, 1[Blo::: [Alo I·
To see the variation of v with x , let [Blo v
or
=
k([Alo  x)([Alo + x)
k[:16 =
(I  [~~6)
Thus we plot vk[Aol
= (I
2
= k([A16 
(1 +
=
= [Alo . The rate equation becomes
[:10)
x )
(1
X2) = (l 
2
[Aol
=
k[A16  kx 2
[:10).
X2) against x
[Aol
= X from X = o.
x x . The plot is shown in Fig. 22.5 in which X =   .   < I corresponds to realIty . [Alo [Alo
1.0
r===~::__:_.
0.8
0.6 N
><: I ~
0.4
0.2
00 0.0
0.4
0.2
0.6
x
The integrated rate law is [14C]
= [14 Cloe kl
[22.12bl with k
1.0
Figure 22.5
Solutions to applications P22.29
0.8
In2
=
tl / 2
[22.131.
THE RATES OF CHEMICAL REACTIONS
459
Solve for t. t =
~ In k
P22.31
[14C]O = ~ In [14C]O = (5730 Y) x In (1.00) = 12720 I. [14C] In 2 [14C] In 2 0.72 Y
A simple but practical approach is to make an initial guess at the order by observing whether the halflife of the reaction appears to depend on concentration. If it does not, the reaction is firstorder; if it does, it may be secondorder. Examination of the data shows that the halflife is roughly 90 minutes, but it is not exactly constant. (Compare the 60 + 150 minute data to the 150 + 240 minute data; in both interval s the concentration drops by roughly half. Then examine the 30 + 120 minute interval, where the concentration drops by less than half.) If the reaction is firstorder, it will obey
In
(:J =
kt [22. 12b].
If it is secondorder, it will obey I
I
C
Co
 = kt + 
[22.15b].
See whether a firstorder plot of In C vs. time or a secondorder plot of II c vs. time has a substantially better fit. We draw up the following table.
tl min c/(ng cm 3 ) (ng cm 3 )/c In {c/(ng cm 3 ))
30 699 0.00143 6.550
60 622 0.00161 6.433
120 413 0.00242 6.023
150 292 0.00342 5.677
240 152 0.00658 5.024
360 60 0.0167 4.094
480 24 0.0412 3.178
7 ..... . ........ ......
6 .5
.Y.. ;;.::Q.OOJ6:ix ..t. .6...8:iQ
6 ~
'I'
5.5
E u
01)
c:
"5'
=
5 4 .5 4 3.5 3
0
100
200
300 tlmin
400
500
Figure 22.6(a)
The data are plotted in Figs . 22.6(a) and (b). The firstorder plot fits closely to a straight line with just a hint of curvature near the outset. The secondorder plot, conversely, is strongly curved throughout. Hence,
460
STUD ENT'S SOLUTIONS MANUAL 0.05
~
~ R2 = 0.8403:
0.04
•
~ 0.03
'7 E u
OIl
.5 0.02
0.0 1
0 0 tlmin
Figure 22.6(b)
the reaction is 1firstorder I. The rate constant is the slope of the firstorder plot: k 10.459 h
I
=
1
0.00765 min 
I
1
=
I·
The halflife is (eqn 22.13) In2
tl / 2
=
k
In 2 = 0.459 h I
~
=~=191 min I·
COMMENT. As noted in the problem, the drug concentration is a result of absorption and elimination of the
drug, two processes with distinct rates . Elimination is characteristically slower, so the later data points reflect elimination only, for absorption is effectively complete by then . The earlier data points, by contrast, reflect both absorption and elimination. It is, therefore, not surprising that the early points do not adhere so closely to the line so well defined by the later data.
P22.33
(a) For the mechanism k.
hhhh ...
~
hehh . . .
hehh . ..
kb ~
ecce . ..
k~
the rate equations are d[hhhh ... J dt
I
= ka[hhhh . .. J + ka[hehh . .. J,
d[hehh . .. J    = ka[hhhh .. .J dt d[ccee .. .J """:""d:t "
= kb[hchh . .. J 
I
I
ka[hehh . .. J  kb[hehh . .. J + kb[eeee . .. J, I
kb[eeee . . .J.
THE RATES OF CHEMICAL REACTIONS
461
(b) Apply the steadystate approximation to the intennediate: , ' 0 ka[hchh . . .]  kb[hchh . .. ] + kb[cccc . .. ] =
d[hchh . .. ]    = ka[hhhh . .. ] dt
so [hchh . .. ] Therefore,
ka[hhhh . . .] + k~[cccc . .. ] = ~ k~ + kb
d[hhhh . .. ] ~
kakb k~k~ [hhhh . .. ] + [cccc . . .].
=
~+~
~+~
This rate expression may be compared to that given in the text [Section 22.4] for the mechanism k
A .::= B . k' keff
Here hhhh ... .::= cccc ... with k~ff
(c) It is difficult to make conclusive inferences about intennediates from kinetic data alone. For example,
if rate measurements show fonnation of coils from helices with a single rate constant, they tell us nearly nothing about the mechanism. The rate law d[cccc . . .] dt
= k[hhhh ... ]
is consistent with a singlestep mechanism, with a twostep mechanism with a ratedetermining second step, and with a twostep mechanism with a steadystate intennediate. Even if kinetic monitoring of the product shows production with two rate constants, the rate constants could belong to competing paths or to steps of a single reaction path. The best evidence for an intennediate's participation in a reaction is detection of the intennediate, or at least detection of structural features that can belong to a proposed intennediate but not reactant or product. P22.35
We assume a preequilibrium (as the initial step is fast), and write K
=
[unstable helix] [A][B] ,
implying that [unstable helix]
= K[A][B].
The ratedetermining step then gives v
=
d[double helix] dt
,, [B) I [k
= k2[unstable helix] = k2 K [A][B] = Ik[A]
The equilibrium constant is the outcome of the two processes kl
k,
A + B .::= unstable helix,
K=k'I
k;
.
Therefore, with v
P22.37
= k[A][B],
E!J k
=  1k2 . k'I
The Arrhenius expression for the rate constant is k
= Ae Ea / RT
[22.31] so Ink
= InA  Ea/RT [22.29].
= k2K].
462
STUDENT'S SOLUTIONS MANUAL
A plot of In k versus l i T wiU have slope Ea / R and yintercept InA . The transformed data and plot (Fig. 22.6) follow. 295 3.55 15 .08 3.39
T IK 1O 6 kl(dm 3 molI sI) lnk l(dm 3 molI S I) 10 3 K I T
15.5 15 14.5
"'" .:
14
.
223 0.494 13.11 4.48
I
'"
218 0.452 13 .02 4.59
213 0.379 12.85 4.69
206 0.295 12.59 4.85
200 0.241 12.39 5.00
195 0.217 12.29 5.13
I
In k =  1642 KIT + 20.585 R'=O.9937
~
13.5
~
13
"; ~
12.5
~
12
0.003
0.0035
0.004
0.0045
0.005
KJT
0.0055
Figure 22.7
SoEa = (8.3145JK 1 mol  I) x (1642K) = 1.37 x 104 Jmol 1 =1 13.7 kJ molI 1 and A = e20 .585 dm 3 mol I SI =18.7 x 108dm 3 mOI l s 1 1 P22.39
The rate constants are:
k=Aexp(~~a) 9
[22.31]. 3
I
I
kl = ( 1.13 x 10 dm s mol ) exp
(
1 ) 14.1 x 103JmolI I (8.3145J K mol ) x (298 K)
Compared to reaction I, reaction 2 shows a significant kinetic isotope effect whereas reaction 3 shows practically none. This difference should not be surprising: in reaction 2 a CD bond is broken, whereas
THE RATES OF CHEMICAL REACTIONS
463
in reaction 3 the D atom is simply along for the ride already attached to the 0 atom. Compare the measured isotope effect of 0.13 to that expected in reaction 2.
(1 1))
l2 I72  I72 k2 = exp (hkf kl
We take
k2 kl
2kBT
Ji,CH ~ mH
= ex
/I
""CH
and Ji,CD
[E22.15(a»).
/I
""CD
~ mD ~ 2mH,
so
34 (( (1.0546 x 10 J s) x (500 kg s2) P 2(1.381 x 1O23JKI) x (298K)
23 x (6.022 x 10 mOl I) 1 x 10 3 kg mol I
1/2)
=~ in agreement with the experimental value.
1/2) ( x
1) 1 21 /2
23
The kinetics of complex reactions
Answers to discussion questions 023.1
(a)
(l)AH~A·+H·
(2) A· ~ B· +C (3) AH + B· ~ A· + D (4) A · + B·
(b)
~
P
(l)A 2 ~A·+A·
(2) A· ~ B· + C (3) A· +P ~ B· (4)A · +B·~P
Initiation [radicals formed] Propagation [new radicals formed] Propagation [new radicals formed] Termination [nonradical product formed] Initiation [radicals formed] Propagation [radicals formed] Retardation [product destroyed, but chain not terminated] Termination [nonradical product formed]
023.3
The MichaelisMenten mechanism of enzyme activity models the enzyme with one active site that, weakly and reversibly, binds a substrate in homogeneous solution . It is a threestep mechanism . The first and second steps are the reversible formation of the enzymesubstrate complex (ES). The third step is the decay of the complex into the product. The steadystate approximation is applied to the concentration of the intermediate (ES) and its use simplifies the derivation of the final rate expression. However, the justification for the use of the approximation with this mechanism is suspect, in that both rate constants for the reversible steps may not be as large, in comparison to the rate constant for the decay to products, as they need to be for the approximation to be valid. The simplest form of the mechanism applies only when kb » k~ . Nevertheless, the form of the rate equation obtained does seem to match the principal experimental features of enzymecatalyzed reactions; it explains why there is a maximum in the reaction rate and provides a mechanistic understanding of the turnover number. The model may be expanded to include multi substrate reaction rate and provides a mechanistic understanding of the turnover number. The model may be expanded to include multisubstrate reactions and inhibition.
023.5
The primary quantum yield is associated with the primary photochemical event in the overall photochemical process which may involve secondary events as well. An example that illustrates both kinds of events is the photolysis of HI described in Section 23.8(a). The primary quantum yield is defined as the ratio of the number of primary events to the number of photons absorbed (eqn 23.28) and its value can never exceed one. However, in reactions described by complex mechanisms, the overall quantum yield, which is the number of reactant molecules consumed in both primary and secondary processes per photon absorbed, can easily exceed one. Experimental procedures for the determination of the overall
THE KINETICS OF COMPLEX REACTIONS
465
quantum yield involve measurements of the intensity of the radiation used, defined here as the number of photons generated and directed at the reacting sample, and of the amount of product formed . This ratio is the overall quantum yield. See Example 23 .5. In addition to chemical reactions, the concept of the quantum yield enters into the description of other kinds of photochemical processes, such as fluorescence and phosphorescence, and in each case there are techniques specific to the process for the determination of the quantum yield. 023.7
The Forster theory of resonance energy transfer examines the interaction between an induced oscillating dipole moment in chromophore S, the energy donor, with a second chromophore Q, the energy acceptor. The oscillating dipole moment of S is induced by incident electromagnetic radiation and the chromophores are separated by distance R. S transfers the excitation energy of the radiation to Q via a mechanism in which its oscillating dipole moment induces an oscillating dipole moment in Q. Resonance energy transfer can be efficient when R is short (typically less than about 9 nm) and when the absorption spectrum of the acceptor overlaps with the emission spectrum of the donor. Fluorescence resonance energy transfer (FRET) experiments commonly use the fluorescent spectrum and relaxation times of the Forster donor and acceptor chromophores to find the distances between fluorescent dyes at labeled sites in protein, DNA, RNA, etc. FRET is a type of spectroscopic ' ruler' . The computation uses either experimental quantum yields or relaxation lifetimes to calculate the efficiency of resonance energy transfer ET . ET
=
I 
r/Jf
=
T
I 
r/Jf,O
[23 .37] .
TO
ET is used to calculate R.
or
R
IE) ' /6
= Ro ( ~T
T
[23 .38].
Solutions to exercises In the following exercises and problems, it is recommended that rate constants are labeled with the number of the step in the proposed reaction mechanism and that any reverse steps are labeled similarly but with a prime. E23.1(b)
The intermediates are NO and N03 and we apply the steadystate approximation to each of their concentrations k2 [N02] [N03]  k3 [NO] [N20S]
=0
k, [N20 S]  k; [N02] [N03]  k2 [N02] [N0 3] Rate = _~ d [N 20 S] 2 dt d [N?Os] d~ = k, [N20 S]
I
=0
+ k, [ N02][N03]  k3[NO][N20S]
466
STUDENT'S SOLUTIONS MANUAL
From the steadystate equations k3 [NO] [N20S]
[N20 S] k; + k2
= kl
[NO] [NO] 2
= k2 [N02] [N03]
3
Substituting,
E23.2(b)
Apply the steadystate approximation to both equations
+ k3 [R'] k3 [R' ] = 0
2kl [R2]  k2 [R] [R2]
k2 [R] [R2] 
The second solves to [R']
=
2k4 [R]2 = 0
k2 [R][R2] k3
k ) and then the first solves to [R] = ( k~ [R2]
E23.3(b)
1/ 2
(a) The figure suggests that a chainbranching explosion I does not occur I at temperatures as low as 700 K. There may, however, be a thermal explosion regime at pressures in excess of 106 Pa. (b) The lower limit seems to occur when
There does not seem to be a pressure above which a steady reaction occurs. Rather the chainbranching explosion range seems to run into the thermal explosion range around log (pjPa) E23.4(b)
= 4.5
so
p
= 104.s Pa = 13
The rate of production of the product is
x 104 pa 1
THE KINETICS OF COMPLEX REACTIONS
467
HAH + is an intermediate involved in a rapid preequilibrium
= ~ so [HAH+ ] =
[HAH+ ] [HAl [H+ ] and d[BH + ] dt
k;
= kl~2
kl [HA1[H+ ]
k;
[HAl [H+ ]rBl
kl
L _ __ _ __
'
This rate law can be made independent of [H+l if the source of H+ is the acid HA, for then H+ is given by another equilibrium 2 [H+ )[A  l = K = [H+ l so [H+l = (K [HAl) 1/2 a [HAl a [HAl and d[BH + l dt E23.5(b)
= klk2K~ /2 [HAl3/ 2[Bl k;
L _ __ _ __
'
A2 appears in the initiation step only.
Consequently, the rate of consumption of [A2l is first order in A2 and the rate is independent of intermediate concentrations. E23.6(b)
The maximum velocity is kb [Elo and the velocity in general is
v
V
E23.7(b)
= k [Elo =
Olax
=
kb [Sl [Elo KM + [Sl
so
V max
= kb [Elo =
KM
+ [Sl
[Sl
v
3 (0.042 + 0.890) mol dm  (2.45 x 10 4 mol dm 3 s l) 0.890 mol dm  3
= 12.57
X 10 4 mol dm  3 sI 1
The quantum yield tells us that each mole of photons absorbed causes 1.2 x 102 moles of A to react; the stoi chiometry tells us that I mole of B is formed for every mole of A which reacts. From the yield of 1.77 mmol B, we infer that 1.77 mmol A reacted, caused by the absorption of 1.77 x 10 3 mol / ( 1.2 x 102 mol EinsteinI) = 11.5 x 10 5 moles of photons
I
E23.8(b)
The quantum efficiency is defined as the amount of reacting molecules nA divided by the amount of photons absorbed nabs. The fraction of photons absorbed Jabs is one minus the fraction transmittedflrans; and the amount of photons emitted nphoton can be inferred from the energy of the light source (power P times time t ) and the energy of the photons (he/A) . nAheNA <1>=( I  flrans) APt
(0.324mol) x (6.626 x JO34JS) x (2.998 x J08 m sl) x (6.022 x 1023 molI) ( I 0.257) x (320 x JO9 m) x (87.5 W) x (28.0 min ) x (60 s min  I)
=[D]
468
STUDENT'S SOLUTIONS MANUAL
Solutions to problems Solutions to numerical problems
These equations serve to show how even a simple sequence of reactions leads to a complicated set of nonlinear differential equations. Since we are interested in the time behavior of the composition we may not invoke the steadystate assumption. The only thing left is to use a computer and to integrate the equations numerically. The outcome of this is the set of curves shown in Fig. 23.1 (they have been sketched from the original reference). The similarity to an A + B + C scheme should be noticed (and expected), and the general features can be analysed quite simply in terms of the underlying reactions.
3.0
~
2.0
u
"0
E........ ~
1.0
0.0
o
2
4
6 t/ ms
8
10
Figure 23.1
THE KINETICS OF COMPLEX REACTIONS
P23.3
The roles are
+0
(1)
N20 + N2
(2)
o + SiH4 +
(3)
(4) (5)
(6)
SiH3 + OH OH + SiH4 + SiH3 + H2O SiH3 + N20 + SiH30 + N2 SiH30 + Si~ + SiH30H + SiH3 SiH3 + SiH30 + (H3 SihO
initiation, propagation [or transfer], propagation [or transfer], propagation, propagation, termination.
The rate of silane consumption is
Steadystate approximation (SSA) for 0
SSA forOH
so [OH]
=
kl [N20 ] . k3[SiH4]
SSA for SiH30 and SiH3
+ ks[SiH30][SiH4] = 2kl [N20] 
k6[SiH30][SiH3]
k4[SiH3][N20]
+ ks[SiH30HSiH4] 
k6[SiH30HSiH3] ~
Adding these expressions together yields
and subtracting them gives
Solve for [SiH30]:
0= 2k [N 0]  k 1k4[N2 0 ]2 I
2
k6[ SiH 30 ]
= 2klk6[SiH30][N20] 
+ ks[SiH3 O][S1.'4 ·u 0]
klk4[N20]2
+ ksk6[SiH30f[SiH4].
o.
469
470
STUD ENT'S SOLUTIONS MANUAL
If kl is small, then
[SiH 30 ] "" kl [N 20] (k4kS[SiH4]) 1/ 2 kS[S IH4] k, k6
= [N2 0 ] (
klk4 ) ksk6[SIH4]
1/ 2
Putting it all together yields
P23.5
In the steadystate approximation for [I ·],
Substitution of (2) into ( I) gives 2kbka[h][H2]
d[HI]
k~
dl
+ kb[H 2]
This simple rate law is observed when step (b) is ratedetermining so that step (a) is a rapid equilibrium and [I·] is in an approximate steady state. This is equivalent to kb[H2] « k~ and hence,
P23.7
(a)
!.i~ = e
I
/
ro [23.3 1]
or
In
(!.i) = _!.... ~ ~
A plot of In(lr 110) against t should be linear with a slope equal to  l i To (i.e. TO = l/slope) and an intercept equal to zero. Consequently, we make the plot to determine whether it is linear. If it is linear (it is), we do a linear regression fit with a zero intercept and use the regression slope to calculate TO . See Fig. 23.2. Alternatively, average the experimental values of (l I t ) In Urllo) and check that the standard deviation is a small fraction of the average (it is). The average equals  l i TO (i.e. TO =  I1average). Slope
= 0.150 ns l ,  (0.150 ns I ) I ,
TO
=
TO
= 16.67 ns I·
THE KINETICS OF COMPLEX REACTIONS
(b)
kf
= rpflTo [23.34] = 0.70/(6.67 ns)
kf
= 10.105 ns I I·
471
0 y = D.ISOlx
D.S
R2 =0.997
I ~
~
:::: =
I.S 2 2.S 3 3.S
20
IS
10
S
0
tIns
P23.9
2S
Figure 23.2
Since Jr = kf[S*]t = kf[S*]O e t / ro , we surmise that a graph of In (If / 10) against t should be linear with a slope equal to l i To in the absence of a quencher. The plot is in fact linear with a regression slope equal to 1.004 x 105 SI, TO
1
= 1.004 x 105s I = 9.96I1 s.
In the presence of a quencher, a graph of In(Jr / 10) against t is still linear but with a slope equal to l / T. This plot is found to be linear with a regression slope equal to 1.788 x lOSs I. 1
T = 1.788 x 105 sI
1 T
I TO
 =  + kQ[Q]
= 5.59I1s.
[23 .36].
(0.08206dm 3 atmK I mol')(300K)(1.788  1.004)105 sI 9.74 x 104 atm kq = 11.98 x 109 dm 3 mol' sI
I.
Solutions to theoretical problems P23.11
d[CH3CH3] dt
=
ka[CH3CH3]  kb[CH3][CH3CH3]  kd[CH3CH3][H]
+ ke[CH3CH2][H].
We apply the steadystate approximation to the three intermediates CH3, CH3CH2 , and H. d[ CH 3] dt = 2ka[CH3CH3]  kb[CH3CH3][CH3] = 0
· h Imp ' l'les th at [C H3] whIC
2ka. =kb
472
STUDENT'S SOLUTIONS MANUAL
d[ CH3CH ? ) dt 
= kb[CH3)[CH3CH3) + kd[ CH 3CH 3)[H) 
kdCH3CH2) ke[CH3CH2)[H)
= O.
These three equations give
0' [CH, CH, I  /
('';;J + [ (,~)' + (~~)
rI
[CH, CH, i
which implies that
If ka is small in the sense that only the lowest order need be retained,
The rate of production of ethene is therefore
The rate of production of ethene is equal to the rate of consumption of ethane (the intermediates all have low concentrations), so
d[
CH CH 3 3) dt
= k[CH 3 CH 3,)
Different orders may arise if the reaction is sensitized so that ka is increased.
P23.13

(M )N
= MIp

[eqn 23. 8a with (M )N
= (n)M ).
The probability Pn that a polymer consists of n monomers is equal to the probability that it has n  1 reacted end groups and one unreacted end group. The former probability is pll I; the latter I  p.
THE KINETICS OF COMPLEX REACTIONS
473
Therefore, the total probability of finding an nmer is
P"
= p"I (1
We see that (n 2 )
 p).
= (i~~2
_
and that (M2 )N  (M)N2
= M2
( 1+ P I ) = (lp)
2 
( Ip)
2
pM2 (lp)
2'
pi /2M
Hence, 8M
=  . Ip
The time dependence is obtained from p
=
kt[Alo
I I
and  
Ip
+ kt[Alo
= 1 + kt[Alo
pl / 2
Ip
P23.15
(kt[Alo(\
+ kt[Alo)) 1/ 2
= I M{kt[Alo(l + kt[Alo)) 1/ 2 I.
In termination by disproportionation, the radicals do not combine. The average number of monomers in a polymer molecule equals the number in the radical, the kinetic chain length, v. (n)
P23.17
[23.8bl.
= pl / 2(l + kt[Alo) =
Hence  and 8M
[23 .7l
= v = Ik[.M][Il  I/21 [23. 14l·
(a) A + P + P + P autocatalytic step, v = k[A][Pl. Let [Al = [Alo  x and [Pl = [Plo
+ x.
We substitute these definitions into the rate expression, simplify, and integrate.
d~~l
v= 
= k[A][Pl
d([Alo  x) dt = k([Alo  x)([Plo

dx ([Alo  x)([Plo
:::=~:::
[Alo
I
+ [Plo
(
+ x)
+ x).
= k dt.
I [Alo  x
+
I ) dx [Plo + x
= k dt.
474
STUDENT'S SOLUTIONS MANUAL
t
[Alo
+ [Plo
1
Jo
(_1_ [Alo 
[Alo
+ [Plo
1
{In (
x
+
1 ) dx [Plo +
x
+ In ([Plo
[Alo ) [Alo  x
=k
t dt.
Jo
+x)} _ . kt
[Plo
X) }= k([Alo + [Plo)t.
[Al o ) ( [Plo + In { ( [Plo [Alo _ x
In { In {
(~~i~) ([Alo + ~~~o _ [Pl) } = k([Alo + [Plo)t
G)
CAlo
+ ~~~o _
_ _ _[_P_l_ _ [Alo + [Plo  [Pl [Pl
0+ beat)[Pl = [Plo
+
I)
e
I
= at
where
a
= k([Alo + [Plo)
and
[Plo [Alo
b= .
= beal .
= ([Alo + [Plo)beal 
[Pl _ (b [Plo 
[Pl) }
(1
+
beal [Pl. [Alo) beal [Plo
= [Plo (1 + ~) beal = [Plo(b + l)eal . b
al
+ beal
(b) See Figure 23.3(a). Autocatalytic process 12 10
b=O.1
8 [P] / [P]o
b =0.2 6
4 b= 1.0 2 0
0
2
4
6 at
8
10
Figure 23.3a
The growth to [Pl reaches a maximum at very long times. As t + 00, the exponential term in the denominator of [Pl 1[Plo = (b + 1) (e al 1( I + be al )) becomes so large that the denominator becomes beal. Thus, ([Pl / [Plo)max = (b+ 1)(eal I be al ) = (b + 1) l b where b = [Plo / [Alo and this maximum occurs as t + 00. The autocatalytic curve [Pl / [Plo = (b + I )(e al 10 + beal)) has a shape that is very similar to that of the firstorder process ([Pl / [Alo) = 1  e kl . However, [Pl max = [Alo at t + 00 for the firstorder
THE KINETICS OF COMPLEX REACTIONS
475
0.8 0.6 [PJ/[AJa
4
3
2
5
kt
Figure 23.3b
process whereas [Pl max = (1 + 1/ b) [Plo for the autocatalytic mechanism. In a series of experiments at fixed [Alo and assorted [Plo, only the autocatalytic mechanism will show variation in [Pl max . Another difference is that the autocatalytic curve is initially concave up, which gives an overall sigmoidal curve, whereas the firstorder curve is concave down. See Fig. 23 .3(b). (c) Let [Pl vmax be the concentration of P at which the reaction rate is a maximum and let tmax be the corresponding time. v
+ x)
= k[A][Pl = k([Alo 
x)([Plo
= k{[Alo[Plo + ([Alo
 [Plo)x  x 2 ).
dv
 = k([Alo dt
[Plo  2x).
The reaction rate is a maximum when dv / dt
x
=
[Pl V
max
 [Pl
0
=
[Alo  [Plo 2
or
= O. This occurs when [Pl vmax [Plo
Substitution into the final equation of part (a) gives [Pl vmax [Plo
= b+1 =
(b
+ I)
2b
Solving for t max ,
atmax
(max
= In(b I) =
= I~ In(b) I.
In(b) .
ealmax 1 + beGlmax
b+l 2b
476 (d)
STUDENT'S SOLUTIONS MANUAL
d[P] dt
= k[Af[p].
= Ao x,
[A]
[P]
= Po +x, d[P] = dx = k(Ao dt dt
t
:_;;dx;_ _ Jo (Ao  X)2(PO + X)
2
X) (PO
+ X)
= kt
Solve the integral by partial fractions
__::__ = (Ao  X)2(PO
a (Ao  X)2
+ x)
a(Po
a
+ _f3_ + _y_ Ao  x Po + x
+ x) + f3(Ao
I
Poa + A oPof3 + A6Y = I + (Ao  PO)f3  2AOY = 0 f3 + y = 0
 x) (PO
(Ao  X)2(PO
+ x) + y(Ao + x)
X)2
.
This set of simultaneous equations solves to
a
I a=, Ao +PO
f3=y=. Ao +Po
Therefore,
kt=
(Ao~pJ fox
= (Ao ~ = (AO
pJ {
(Ao
~x)

+
(AO
~ po) [(Ao(A:  x»)
Therefore with y
Ao(Ao
(AO~Po) (Ao~X + Po~x) Jdx (~J + ~ [In (AoA~ J+ In ( + ~ pJ In G~~p~ ~;~)] .
[(Aolxr
x
=
Ao
+ Po )kt =
and p
(Ao
Po
= , Ao
(y) + (_1_) In ( + y) . I  Y
I  P
p p(l  y)
The maximum rate occurs at dvp
dt = 0,
Vp
= k[A]2[p]
and hence at the solution of 2k (d[A]) [A][P] dt
k[A]([A] 
+ k[Af d[P] = O. dt
+ k[A]2vp = 0 2[p])vp = o.
 2k[A][P]vp
Po)
[as VA
= vp] .
Pop: X) ] }
THE KINETICS OF COMPLEX REACTIONS
That is, the rate is a maximum when [A]
Ao x = 2Po + 2x,
or
= 2[P], which occurs at
x = !(Ao  2Po);
!(l 2p).
Y=
Substituting this condition into the integrated rate law gives
Ao(Ao + Po)ktmax = or (Ao 1
(e)
(_1_) (~(l  2p) + In~) +p 2p I
2
+ Po)2ktmax = !  p In 2p I·
d[P] dt
= k[A][P]2.
dx = k(Ao x)(Po +x) 2 [x = P  Po). dt
kt =
foX
(Ao _
x~Po + x)2·
Integrate by partial fractions (as in part (d))
kt
= (Ao
~ Po) fox {(Po ~ J + (Ao ~ Po) [Po ~ x + Ao ~ x J} dx 2
= (Ao ~ Po) = (Ao
{
(~o  Po ~
J+
~ Po) [In ( Pop: X) + In (AoA~
(Ao
~ Po) [(Po(p: +X)) + (Ao ~Po) In (~~A:~~n]·
x Po Therefore, with y = and p =  , [A]o Ao
Ao(Ao
+ Po)kt = (
p(p
y
+ y)
)
+
(_1_) + p(lp +1
p
In (
y ) . y)
The rate is maximum when dvp dt
= 2k[A][P] (d[P]) + k (d[A]) [pf dt
= 2k[A][P]vp That is, at[A]
dt
k[pfvp
= k[P](2[A] 
[P])vp
= O.
= ![P].
On substitution of this condition into the integrated rate law, we find
Ao(Ao
or (Ao
+ Po)ktmax = (
2p ) 2p(1 +p)
2 p
2
2p
P
+ Po)2ktmax =   + In 
.
+ ( 1 ) 1 +p
2 in
p
J] }
477
478
STUDENT'S SOLUTIONS MANUAL
A ~ 2R
P23.19
I.
A+ R
~
R +B
R +R
~
R2
k2.
k3.
d[A] dt = II 
k2[A][R]I,
The latter implies that [R]
d[A]

dt
d[B]

dt
=
I ) 1/2
= ( k3
' and so
( 1 ) 1/ 2 I  k2 [A] ,
k3
= k2[A][R] = k2
(/)1 /2[A]. k3
~~2
Therefore, only the combination
may be determined if the reaction attains a steady state.
k3 COMMENT. If the reaction can be monitored at short enough times so that termination is negligible compared
to initiation, then [RJ ~ 21t and ~ ~ k21t [AJ. SO monitoring B sheds light on just k2·
_d[=C_r(_C_O.;..:)s:..:.]
P23.21
dt
=1_
Hence, [Cr(CO)s]
=
d[Cr(CO)sM] dt'
k2[Cr(CO)s][CO]  k3[Cr(CO)s][M]
+ k4[Cr(CO)sM] = 0 [steady state].
1 + k4[Cr(CO)sM] k2[CO] + k3[M] .
= k3[Cr(CO)s][M] 
k4[Cr(CO)sM] .
Substituting for [Cr(COh] from above, d[Cr(CO)sM] dt . Iff
=
=
k3 / [M]  k2 k4[Cr(CO h M [CO] k2[CO] + k3[M]
= f[Cr(CO)
k2 k4[CO] k2[CO] + k3[M]
and we have taken k3/[M]
«k2~[Cr(CO)sM][CO].
Therefore,
I I k3[M] = +:::'':::::':::c f k4 k2k4[CO]
and a graph of I If against [M] should be a straight line.
Solutions to applications P23.23
(a) The mechanism considered is ka
k,
k~
k~
E + S ;=:(ES) ;=: P + E.
M] s
THE KINETICS OF COMPLEX REACTIONS
We apply the steadystate approximation to [(ES)]. d[ES] dt
= ka[E][S]
_
= [E]o 
Substituting [E]
k~[(ES)] 
kb[(ES)]
+ k~[E][P] = o.
[(ES)] we obtain
ka([E]o  [(ES)])[S]  k~[(ES)]  kb[(ES))
+ ka[E]o[S] 
(ka[S]  k~  kb  k~[P])[(ES)]
[(ES)]
Then d[P] 'dt
ka[E]o[S] ka[S] + k~
=
=k
+ k~([E]o 
[(ES)])[P] = O.
k~[E]o[P]
= O.
+ k~[E]o[P] + kb + k~[P]
[(ES)] _ k' [P][E] b b
[EJo[S] + (kVka) [E]o[P]  k' [P] b KM + [S] + (k~/ka) [P] b
=k
x ([E] _ [E]o[S] + (~/ka) [E]0[P]) o KM + [S] + (k~/ka) [P]
kb [[EJo[S]
+ (k~/ka) [E]o[P]]  k~[EJo[P]KM KM + [S] + (kVka) [P]
Substituting for KM in the numerator and rearranging d[P] dt
=
kb[E]o[S] + (k~k~/ka) [E]o[P] KM + [S] + (k~/ka) [P]
[v = d[P] ] . dt
(b) For large concentrations of substrate, such that [S]
»
KM and [S]
»
[P],
d[P]
dt = kb[E]o which is the same as for the unmodified mechanism. For [S] d[P] _ k E {[S]  (k/kb)[P] } dt  b[ ]0 [S] + (k/k~)[P]
For [S]  +0, where kp =
d[P] _ _
dt
+ kb   , . k~
kb
k~k~[E]o[P] k~
+ kb + k~[P]
_
_
k'k'
k=~
ka
»
KM, but [S]
~
[P]
479
480
STUDENT'S SOLUTIONS MANUAL
COMMENT. The negative sign in the expression for d[P]/dt for the case [S] ~ 0 is to be interpreted to
mean that the mechanism in this case is the reverse of the mechanism for the case [P] and S are interchanged.
>
O. The roles of P
Question. Can you demonstrate the last statement in the comment above? P23.25
(a)[ATPJ/(ILmol dm  3 ) v/(ILmol dm 3 S l) v/[ATP]/ Sl
0.60 0.81 1.35
0.80 1.4 0.97 1.30 1.21 0.929
2.0 l.47 0.735
3.0 1.69 0.563
[23.21]. Taking the inverse and mUltiplying by Vrnax v, we find that
Thus,
v
= Vrnax 
v . KM (EadieHofstee plot) [S]o
or
v
Vrnax
v
[S]o
(b) The regression slope and intercept of the EadieHofstee data plot of v against v/[S]o gives KM and vrnax , respectively. Alternatively, the regression slope and intercept of an alternative form of the EadieHofstee data plot of v/[S]o against v gives  1/ KM , and vrnax/ KM, respectively. The slope and intercept of the latter plot can be used in the calculation of KM and Vrnax . (c) We draw up the following table, which includes data rows required for a EadieHofstee plot (v against v/ [S]o). The linear regression fit is found for the plot as seen in Fig. 23.4.
Vrnax
= 12.30 ILmol dm  3 s l I
and
KM
= 11.10 ILmol dm 3 1·
2 1.6 I
'7'"
e
1.2
y = 1.1015x + 2.303 1
"0
"0 0.8 e
R2 =0.998
::1.
;.
0.4 0 0
0.5
1.5
Figure 23.4
THE KINETICS OF COMPLEX REACTIONS
P23.27
481
When using reaction rates v, the LineweaverBurk plot without inhibition [23 .22] has the fom :
where the intercept and slope are simple functions of Vmax and KM. When using reaction rates relative to a specific, noninhibited rate (Vrel = V/ Vreference), the LineweaverBurk plot without inhibition has the same basic form:
I
I
Vrel
= vmax,rel +
(KM)
1
vmax,rel
[S]o
The linear regression fit of the noninhibited LineweaverBurk data plot is

I
vrel
= 0.797 + (2.17)
1
[CBGP]o/l02 mol dm
3'
R2
= 0.980.
= l / intercept = 1/ 0.797 = l.25 and = slope x vmax.rel = (2.17 X 10 2 mol dm 3 ) x (l.25) = 2.71
Consequently, Vmax.rel KM
x 10 2 mol dm 3 .
The LineweaverBurk plot with inhibition has the basic form
I
vrel
=
a' vmax,rel
+
(aKM)
I
vmax,rel
[S]o '
The linear regression fit of the LineweaverBurk data plot for phenylbutyrate ion inhibition is

I
vrel
=
1.02 + (6.01)
I
[CBGP]o/l02 mol dm
3'
R2
= 0.972.
Therefore, a' = intercept x vmax,rel = 1.02 x 1.25 = 1.28 and a = slope x vmax ,reI/KM = (6.01 X 10 2 mol dm 3 ) x (l.25) / (2.71 x 10 2 mol dm 3 ) = 2.77 . Since both a > I and a' ~ 1 (see Section 23.6(c», we conclude that phenylbutyrate ion is a competitive inhibitor of carboxypeptidase. The linear regression fit of the LineweaverBurk data plot for benzoate ion inhibition is

1
vrel
= 3.75 + (3.01)
I [CBGP]o / 1O 2 mol dm 3
,
R2
= 0.999.
Therefore, a' = intercept x vmax,rel = 3.75 x l.25 = 4.69 and a = slope x vmax ,reI/KM = (3.01 x 10 2 mol dm 3 ) x (l.25)/(2.71 x 10 2 mol dm 3 ) = 1.39. Since both a ~ 1 and a' > 1, we conclude that benzoate ion is an I un competitive Iinhibitor of carboxypeptidase.
482
STUDENT'S SOLUTIONS MANUAL
4>r = 4>r,o
P23.29
T
ET = I  
R6 ET = __0_ R8 +R6
I 
[23.37].
TO
[23.38].
Equating these two expressions for ET and solving for R gives
R6
T
+ R6
TO
_ _0 _ = 1 _ _
R8
Rg +R6 ~
I(T / To)
,:_, _ I = _,T/,T...::.O_ l(T/To) I(T / TO) T/ TO
P23.31
= 10 ps/ lO3 ps = O.OlO
and R
or
R
= Ro
T/ TO ( 1  (T/TO)
)1/6
0.010 )1 / 6 = I2.6 nm.I = 5.6 nm ( 10.010
Hypothesis: The 1270 nm emission band is the emission of the first excited state of 02(a l ~
P ~ P* + 30 2 + P+ 102 porphyrin
~ 302
Test o/hypothesis: It is well known that the dioxygen state a I ~t is 0.977 e V (1270 run) above the ground state. If the hypothesis is correct the emission intensity should be proportional to both the concentration of dissolved oxygen and to the intensity of the porphyrin absorption. P23.33
(a)
k2 = 6.2 x 1O 34 cm 6 molecule 2 sI. k4
= 8.0 x
10 15 cm 3 molecule I sI.
The concentration of atomic oxygen will be very, very small making a binary collision between atomic oxygen extremely unlikely. In fact, the reaction
is ternary which makes it even less likely. Rate terms in k5 may be safely omitted from consideration. (b) For aU practical purposes d[02]/ dt = 0 because very little dioxygen reacts to form either atomic oxygen or ozone. Using ' molecules per cm 3 ', or simply cm 3 , as the concentration unit, we find that
= a2 
a2[O]
+ a3[03] 
d[03]  = k2[O](02] 2 dt
a4[O](03J.
k3[03]  k4[O](03]
THE KINETICS OF COMPLEX REACTIONS
where
109 sI cm 3,
al
= 2kl[02l = 6.478 x
a2
= k2[02f = 65.036 sI,
a3
= k3 = 0.016 SI,
a4
= k4 = 8.0
X
483
10 15 cm3 sI .
(c) We break the time period in two. The early period encompasses the first 0.05 s with the initial conditions [Olo = [03lo = O. The second period covers the remainder of the 4 hours with the initial conditions provided by the [OlO.05 s = [03lo.05 s values of the early period. Numerical integration of the coupled differential equations yields the concentrations of Figs 23.5(a), (b), and (c). Early period
2.5
2
'I' c
0 .~
c., u c 0
U
E 1.5
u
.,'"
:; u
"
0
E
00
0
0.5
0
0
0.01
0.02
0.03
0.04
Time/s
Figure 23.5a
During the early period (t < 0.05 s) UV radiation causes the formation of a small amount of atomic oxygen (less than 108 molecules cm 3). A steadystate is approached in which a near balance is established between production of atomic oxygen by dioxygen dissociation and usage of atomic oxygen to produce ozone. There is, however, a lOOfold growth in the atomic oxygen concentration over the next 4 h so it is not a perfectly steady state. After 4 h of photochemistry the percentage ozone is 0.0123%. 3.97 x 10 13 Cm 3 ) Percentage ozone ~ ( 17 3 100 1~ 0.123% I· 3.24 x 10 cmOzone production does not require low pressure in the Chapman model. Changing the oxygen pressure to 100 Torr gives 0.025 % ozone after 4 h but the increased collision rate reduces atomic oxygen to 1I5 'th the value at 10 Torr. However, a pressure increase may require the inclusion in the mechanism of the step 03 + M + 0 + 02 + M. This would reduce ozone production.
484
STUDENT'S SOLUTI ONS MANUAL
1O rr~~~~~11====c===~ 8
4 Time/ h
Figure 23.5b
Ozone
8 ~1I~~~~~====F==1
6
c 0
.~
'7 E <.)
.,'"
"5 ., .,
l: c <.)
c 0 U
4
<.)
"0
..,E "0 2
0.5
1.5
2 Time/ h
Figure 23.5c
2.5
3
3.5
4
THE KINETICS OF COMPLEX REACTIONS
P23.35
485
In this solution the notation differs slightly from that in the problem. kl to k4 are replaced by k. to kd , respectively, with k~ replaced by Ld. k7 is replaced by ke (k6 does not figure in the solution). (a)
2NO + N20
+0
k. kb kc kd Ld
o + NO + 0 2 + N N + NO + N2 + 0 20 + M + 0 2 + M 02 + M + 20 + M (b)
d[NO] dt
=  2ka[NO]2
initiation propagation propagation termination initiation
_ kb[O][NO]  kdN][NO] .
To determine the steadystate concentration of N, [N]ss , write the rate expression for d[N] / dt and set it equal to zero. d[N]
 dt = kb[O][NO] 
kc[N] ss [NO]
= O.
kb [N]ss = [0]. kc Substitution of [N]ss into the expression for d[NO] / dt indicates that, under steadystate conditions for [N], V!J = Vc and
[N] steadystate conditions: If the propagation step is much more rapid than initiation, the last term predominates d[NO]
  =  2kb[O][NO] .
dt
[N] is in steadystate and initiation is very slow: If oxygen atoms and molecules are in equilibrium
K 0 /0 2
[0]
=~= Ld
[02] [0]2 .
L [0 ]) 1/2 =(_ d_ 2 kd
Substitution into the previous rate expression yields d[NO] =
dt
2kb
(k) ~ kd
1/ 2
[02]1 /2[NO]
[N] is in steadystate; initiation is very slow ; atomic and molecular oxygen are in equilibrium.
486
STUDENT'S SOLUTIONS MANUAL
(c) Since k ex: e  Ea / RT where Ea is the activation energy, we may write the individual rate constants in the form k; ex: eE;jRT where the subscript 'a' has been dropped and 'j' represents the ith elementary step with activation energy E; . Substitution of such expressions in the last equation of part (b) yields
We conclude that the effective activation energy, Ea,eff' is given by
(d) Using the estimate that activation energy is approximately equal to the bond energies that must be broken Ea,eff
~ B(NO)
+ !B(02) 
~ 630.57 kJ molI
!B(O)
+ !(498.36 kJ molI) 
!(o)
~ 879.75 kJ molI
where this is the unimolecular bondbreakage estimate of activation energies. The previous estimate of Ea,eff may be much too high because the activation energy of step (IT) is probably being greatly overestimated. A more realistic estimate of Eb would be that difference between the energies of the NO bond that must be broken and the 02 bond that is formed. Then, Ea,eff
~ {B(NO)  B(02)}
+ !B(O)
~ B(NO)  !B(02) ~ 630.57 kJ molI  !(498.36 kJ molI)
~ 1381.39 kJ molI I. The energy of the activated complex of step (b) is the difference between bondbreakage and bondformation energies. It is interesting to compare these estimates with the value based upon E; values that have been determined by experiment E.,eff
= (161
kJ molI)
+ !(493 kJ molI) 
!(l4 kJ molI)
= 401 kJ mol I. This value is based upon experimental activation energies for the elementary steps.
THE KINETICS OF COMPLEX REACTIONS
487
(e) We now eliminate the assumption of 0 / 02 equilibrium and assume that both [N] and [0] are at steadystate value. From part (b). [Nss] = kb[O]ss/kc .
d[O] dt = ka[NO] 2 
kb[O]SS
+ kclN] + kc[N]s s [NO] 
2
2kd [O]ss [M]
+ 2Ld[02](M] = O.
+ kb[O]ss[NO]  2kdO]~s[M] + 2Ld[02](M] = O. 2kd[O]~s[M] + 2Ld[02][M] = O.
ka[NOf  kb[O]ss[NO] ka[NO] 2 
At very low values of [02] the last term is negligible so that k
[O]ss
~ ( 2kd~M]
) 1/ 2
[NO].
Substitution of the expression for [N]ss and [O]ss into the expression for d[NO] / dt (top of part (b» gives
d[~tO] =  2ka[NO]2 =
kb[O]ss[NO]  kc
(kb[~]SS) [NO]
2ka[NO]2  2kb[Olss[NO]
=  2ka[NO] 2 
ka 2kb ( 2kd[M] )
1/ 2
2
[NO] .
k ) 1/ 2 if propagation is much more rapid than initiation so that kb ( 2kd~M] becomes
d[NO] =  b __ 2k dt
(0
NO
+ 0 2 +
0
(k) __ a 
1/ 2
»
ka • this expression
[NO]2 .
2kd[M]
+ N02
ke initiation.
if the conversion has proceeded to the extent that [02] has become significant and ka [NO](02] 2ka [NOf
Applying the steadystate approximation to both [N] and [0] gives [N]ss d[O]
dt = ka[NOf
 kb[O]ss[NO] + kclN]ss[NO] 2kd [O] ~s [M]
+ ke[02](NO] =
0
and
 kb[O]ss [NO]
= kb[O]ss/kc and
+ 2Ld[02](M]
+ kc[N]s s [NO]
= 0
»
488
STUDENT'S SOLUTIONS MANUAL
Thus 2kd[O]~S[M]  2Ld[02][M]  ke [02][NO]
= O.
At high concentrations of 02 species 'M' is likely to be 02 and kd[O]~s[NO] »kb[O]ss[NO]. The value of Ld is so small that it can be neglected.
[O]ss
=
k ) 1/2 [02]1 /2[NO]1 /2. ( 2kd~]
Substitution of [N]ss and [O]ss into the expression for d[NO] I dt gives d[NO]
 = dt
kb[O]SS  kb[O]ss[NO] kc[N]ss[NO] ke [NO][02]
= 2kb[O]ss[NO] =
ke 2kb (   ) 2kd[M]
ke [NO][02]
1/2
12
32
[02] I [NO] I  ke [NO][02]'
If propagation is much more rapid than initiation, the expression becomes d[NO] _ _ =  2kb dt
(k) __ e _
1/2 [02]1 /2[NO]3 / 2 .
2kd[M]
Using the experimental values of E;, Ea .eff =
161 kJ molI
Ea .eff
is estimated to be given by
+ !(198 kJ molI) 
!(I4 kJ molI)
= 1253 kJ mol  I I. This value is consistent with the low range of the experimental values of Ea .eff , whereas the value found in part (b) is consistent with the high experimental values.
Molecular reaction dynamics
Answers to discussion questions 024.1
The harpoon mechanism accounts for the large steric factor of reactions of the kind K + Br2 + KBr+ Br in beams. It is supposed that an electron hops across from K to Br2 when they are within a certain distance, and then the two resulting ions are drawn together by their mutual Coulombic attraction.
024.3
The Eyring equation (eqn 24.53) results from activated complex theory, which is an attempt to account for the rate constants of bimolecular reactions of the form A + B ;=: C+ + P in terms of the formation of an activated complex. In the formulation of the theory, it is assumed that the activated complex and the reactants are in equilibrium, and the concentration of activated complex is calculated in terms of an equilibrium constant, which in turn is calculated from the partition functions of the reactants and a postulated form of the activated complex. It is further supposed that one normal mode of the activated complex, the one corresponding to displacement along the reaction coordinate, has a very low force constant and displacement along this normal mode leads to products provided that the complex enters a certain configuration of its atoms, which is known as the transition state. The derivation of the equilibrium constant from the partition functions leads to eqn 24.51 and in turn to eqn 24.53, the Eyring equation. See Section 24.4 for a more complete discussion of a complicated subject.
024.5
Infrared chemiluminescence. Chemical reactions may yield products in excited states. The emission of radiation as the molecules decay to lower energy states is called chemiluminescence. If the emission is from vibrationally excited states, then it is infrared chemiluminescence. The vibrationally excited product molecule in the example of Figure 24.13 in the text is CO. By studying the intensities of the infrared emission spectrum, the populations of the vibrational states in the product CO may be determined and this information allows us to determine the relative rates of formation of CO in these excited states. Multiphoton ionization (MPI). Multiphoton absorption is the absorption of two or more photons by the molecule in its transition to a higher electronic state. The frequencies of the photons satisfy the condition t:.E
= h VI + h V2 + ...
which is similar to the frequency condition for onephoton absorption. However, multiphoton selection rules are different from onephoton selection rules. Therefore, multiphoton processes allow examination of energy states that otherwise could not be reached. In multiphoton ionization, the second or third
490
STUDENT'S SOLUTIONS MANUAL
photon takes the molecule into the energy continuum above its highest lying energy state. This technique is especially useful for the study of weakly fluorescing molecules.
Resonant multiphoton ionization (REMP/). This is a variant of MPI described above, in which one or more photons promote a molecule to an electronically excited state and then additional photons generate ions from the excited state. The power of this method in the study of chemical reactions is its selectivity. In a chemically reacting system, individual reactants and products can be chosen by tuning the frequency of the laser generating the radiation to the electronic absorption band of specific molecules. Reaction product imaging. In this technique, product ions are accelerated by an electric field toward a phosphorescent screen and the light emitted from the screen is imaged by a chargecoupled device. The significance of this experiment to the study of chemical reactions is that it allows for a detailed analysis of the angular distribution of products. Femtosecond spectroscopy. See Section 24.9 for a more detailed discussion. Until recently, because of their exceedingly short lifetimes, there have been no direct observations of the activated complexes postulated to exist in the transition state of chemical reactions. But, after the development of femtosecond pulsed lasers, species resembling activated complexes can now be studied spectroscopically. Transitions to and from the activated complex have been observed and such experiments have greatly extended our knowledge of the dynamics of chemical reactions. 024.7
The Rb atom must hit the I side of CH31 in order to produce Rbi + CH3 . The orientation of CH31 can be controlled by exciting rotations about the CI axis with linearly polarized light; the optimal orientation aims the I side of CH31 at the direction of approach of the beam of Rb atoms. Two possible alignments of the reactant beams are shown in Fig. 24.1. In the top depiction, the beams are anti parallel, thereby maximizing the likelihood of collision and the volume within which collision can occur (but also putting each beam source in the path of the other beam). In the lower depiction, the beam paths are at right angles, thereby minimizing the region in which the beams collide, but facilitating the study of that welldefined collision volume by a 'probe' laser at right angles to both beams.
H 3CI
~
Hp
..
Rb
.
Rb
r Figure 24.1
Solutions to exercises E24.1(b)
The collision frequency is where a
= ][ d 2 = 4][ r2 and
(c)
SRT)I /2
= ( ][ M
MOLECULAR REACTION DYNAMICS
so
Z
16pNAr2Jr 1/2
21 /2p 1/2 2 ) (8RT) = (4Jrr kT
491
(RTM) 1/2
JrM
16 x (100 x 103 Pa) x (6.022 x 1023 molI) x (180 x 1O 12 m)2 x (Jr)I /2 [(8.314SJK l mol I) x (298K) x (28.01 x 10 3 kg mol 1)]1 /2 =16.64 X 109 s 11 The collision density is
Raising the temperature at constant volume means raising the pressure in proportion to the temperature
so the percent increase in z and Exercise 24.1 (a). E24.2(b)
ZAA
due to a 10K increase in temperature is 11.6 percent ~ same as
The appropriate fraction is given by
f = exp ( Ea) RT The values in question are (a)
( i)
(ii)
(b)
(i)
.. (II)
E24.3(b)
f=exp 
f  exp 
f  exp f=exp
I 3 (  I S X I 0 JmOI ) 124103 1 (8.314SJK l mol l ) x (300K) = . x (
(
1 3 IS x I0 Jmol (8.314SJ K I mol  I) x (800K)
)roJOl ~ 
1 3 ISO x I0 Jmol)17710271 (8.3 14SJK 1 molI) x (300K) . x
I 3 (  I S O x 10 Jmol ) 1 6 010 1 (8.314SJK l mol l ) x (800K) =1. x l
A straightforward approach would be to compute f = exp (Ea/ RT) at the new temperature and compare it to that at the old temperature. An approximate approach would be to note thatf changes from fo = exp (Eo/ RT) tof = exp (Ea / RT( I + x» , where x is the fractional increase in the temperature. If x is small, the exponent changes from Eo / RT to approximately (Ea / RT)(1  x) andf changes x from exp(Ea/RT) toexp(Ea(1x) / RT) = exp(Ea/RT) [exp(Ea/RT)r =fclo x . Thus the new Boltzmann factor is the old one times a factor of fox . The factor of increase is (a)
o:Il fox = (0.10) 10/800 = IT2Il
(i) fox = (2.4 x 10 3)10/300 = (ii)
492
STUDENT'S SOLUTIONS MANUAL
(b)
(i)
fox = (7.7 x 1027)10/300 = ~
(ii) fox E24.4(b)
=
= [DJ
(1.6 x 1010)10/800
The reaction rate is given by
so, in the absence of any estimate of the reaction probability P, the rate constant is
k
=a( =
8kn/1T) 1/2NAexp(Ea/RT) B
9
2
[0.30 x (10 m) ] x
(8(1.381 x 1O 23 JK I )X(450K»)1 /2 n(3 .930u) x (1.66 x 1O 27 kgu l )
x (6.022 x 1023 molI) exp
E24.5(b)
200 x 103 J molI ) ( (8.3145JK 1 mol I) x (450K)
The rate constant is
where D is the sum of two diffusion constants. So kd
= 4n(0.50 = 13.2
x 10 9 m) x (2 x 4.2 x 10 9 m 2 SI) x (6.022 x 1023 molI)
x 107 m 3 molI sI I
In more common units, this is
E24.6(b)
(a) A diffusioncontrolled rate constant in decylbenzene is
kd
8RT == 31]
l
8x(8.3145JK l mol ) x (298K) 3 x (3.36 x 10 3 kgm I SI)
= I 1.97x
6
3
10 m mo
II
s
II
(b) In concentrated sulfuric acid l
I
_8RT _8X(8.3145JK mOI )X(298K)_12 0 I II kd  .4 x I 5m3 mol s 31] 3 x (27 x 10 3 kgm I sI) E24.7(b)
The diffusioncontrolled rate constant is kd
8RT == 31]
l
l
8x(8.3145JK mol )x(298K) 3 x (0.60 1 x 10 3 kgm I sI)
= I 1.1 0 x
I
07 3 I II m mol s
MOLECULAR REACTION DYNAMICS
In more common units, kd = 11.10 x 1010 dm 3 mol I s I 1 The recombination reaction has a rate of with [A] = [B]
v = kdA][B]
so the halflife is given by
E24.8(b)
The reactive crosssection a* is related to the collision crosssection a by a*=Pa
so
P=a*/a.
The collision crosssection a is related to effective molecular diameters by a
= :rrd 2
so
d
= (a/:rr) 1/2
2 [ 2(dA+dB) I ]2 =41 (1 /2 a BB 1/2)2 a AA+ NowaAB=:rrdAB=:rr
so
a* P=.". !4 ( a 1/2 + a 1/2)2 AA BB 22 8.7 x 10 m ;:c::::::::±[«0.88)1/2 (0.40)1 /2) x 10 9 m]2
+
E24.9(b)
I
= 1.41 x 10 3
I
The diffusioncontrolled rate constant is 8RT 8 x (8.3145 J K I mol I) x (293 K) 6 3 I I kd =   = = 5.12 x 10 m mol s I 3 37] 3 x (1.27 x 10 kgm  s I)
In more common units, kd = 5.12 x 109 dm 3 mol  I s I. The recombination reaction has a rate of I
v = kd[A][B] = (5.12 x 109 dm 3 mol S I) x (0.200 mol dm 3 ) x (0.150 mol dm  3 )
=11.54 x10 8 moldm 3 s 1 1 E24.10(b) The enthalpy of activation for a reaction in solution is
b.+H = Ea RT = (8.3145JK 1 mol  I) x (6134K)  (8.3145JK I molI) x (298K)
= 4.852 x 104 Jmol I = 148.52kJmol 1 1
493
494
STUDENT'S SOLUTIONS MANUAL
The entropy of activation is tl +S = R ( In
B=
A) 1 B
kRT2 where B = hp'"
(1.381 X 10 23 J K I) X (8.3145JK 1 molI) x (298K)2 ~,,,';;:,.:=~.:.~
(6.626 x 10 34 J s) x (1.00 = 1.54 x lOll m 3 mol  I sI
so tl +S = (8.3145JK 1 mol  I) x
(
In
X
105 Pa)
) 8.7: X 10 12 dm 3 mol I s I _ I 3 (I000dm 3 m )x(1.54xlOllm3mollsl)
=132.2JK 1 molI I
COMMENT. In this connection, the enthalpy of activation is often referred to as "energy" of activation.
E24.11(b) The Gibbs energy of activation is related to the rate constant by
kRT2 where B=  ",
tl+G) k2 =Bexp ( ~
k2 = (6.45
X
hp
so
" k2 tl +G = RTlnB
10 13 dm 3 mol I sl)e {(5375 K)/(298 K)} = 9.47 x 105 dm 3 molI sI
= 947 m3 molI s I Using the value of B computed in Exercise 27.13(b), we obtain tl +G=(8.3145 x 1O 3 kJK l mol l ) x (298K) xln
947m3molls1 ) 3 _I I ( 1.54 X 1011 m mol s
E24.12(b) The entropy of activation for a bimolecular reaction in the gas phase is
tl+S
= R (In ~
B

2)
whereB
=
kR~2 hp
(1.381 x 1O 23 JK  I) x (8 .3145JK 1 molI) x [(55 (6.626 X 10 34 J s) x (1.00 X 105 Pa)
+ 273) K] 2
B=~'';;:';'''c....
= 1.86 x 1011 m3 mol I sI The rate constant is k2 = A exp
a RT ) (E
so
A = k2 exp
A = (0.23 m3 molI sI) x exp
(Ea) RT
) 49.6 x 103 Jmol  I ( (8.3145JK 1 molI) x (328K)
MOLECULAR REACTION DYNAMICS
495
= 193JK I mol  II E24.13(b) The entropy of activation for a bimolecular reaction in the gas phase is
kRT2 whereB =  ..hp
For the collision of structure less particles, the rate constant is
SkT)I /2 (t;.Eo ) k2 = N A ( rr Jk a ex p ;u:so the prefactor is A
SkT) 1/2 a rrJk
= NA ( 
= 4NA
( RT ) 1/2 a rrM
where we have used the fact that Jk = ~ m for identical particles and ki m = RI M. So A = 4 x (6.022 x 1023 mol  I) x
(S.3145J K I molI ) x (500K) I ( rr x (7S x 10 3 kg mol ) )
1/2 x (0.6S x 10 18 m 2)
= 2.13 x 108 m3 mol  I SI (1.3S1 x 10 23 J K I) X (S.3145JK I mol  I) x (500K)2 B= ~~(6.626 X 10 34 J s) x (1.00 X 105 Pa) = 4.33 x 1011 m3 mol  I S I andt;. +S = (S.3145JK I mol  I) x
(
In
I 2. 13 x 108 m3 mol sI ) ) I  2 ( 4.33 X 10" m3 mol  S I
=ISO.OJK  I mol  II E24.14(b) (a) The entropy of activation for a unimolecular gasphase reaction is
t;. +S = R (In
~
I)
where B = 1.54
X
10" m3 mol I sI [See Exercise 24. 14(a)]
sot;. +S = (S .3145JK I molI ) X
(
~
3 13 2.3 X 10 dm mol  I S I )) 1 3 ( (looo dm m 3) x (1.54 x 10" m3 mol I SI)
= 124.1 J K I mOl  II (b) The enthalpy of activation is
t;. +H = Ea  RT = 30.0 x 10 3 Jmol  I  (S.3145JK I molI) x (29SK) = 27.5 x 103 Jmo1  1 =1 27.5kJmol 1 1
496
STUDENT'S SOLUTIONS MANUAL
(c) The Gibbs energy of activation is
L'.+C
= L'.+H 
T L'.+S
= 27.5 kJ molI
 (298 K) x (24.1 x 10 3 kJ K I molI)
= I 34.7kJmol 1 I E24.1S(b) The dependence of a rate constant on ionic strength is given by
= 0 and k2 = k2, so we must find
At infinite dilution, I 10gk2
= logk2 
I 2 2A ZAZ BI /
= 0.0323
and
= log(1.55) 
2 x (0.509) x (+1) x (+1) x (0.0241)1 /2
I k2 = 1.08 dm 6 mol 2min 1 I
E24.16(b) Equation 24.84 holds for a donoracceptor pair separated by a constant distance, assuming that the
reorganization energy is constant: In ket
(L'.rG")2
=
4)"RT

L'.r G "
2RT + constant,
or equivalently In h


~t 
(L'.rC .. )2 4AkT
L'.r C " + constant, 2kT


if energies are expressed as molecular rather than molar quantities. Two sets of rate constants and reaction Gibbs energies can be used to generate two equations (eqn 24.84 applied to the two sets) in two unknowns: A and the constant. In ket I + . so
(L'.rC~)2 L'.rC~ + 4AkT 2kT
(L'.rCf)2  (L'.r C f)2
L'.AkT dA=
an
= constant = In ket .2 +
k .2
L'.rCf  L'.rC~
ket. l
2kT
4AkT
L'.rCf
+ , 2kT
= In  et + ='
(L'. C .. )2  (L'. C")2 rl r2 4 (kTln(ket .2/ket.d + (L'.rCf  L'.r C
A=
(L'.rCf)2
n
2))
(0.665 eV)2  (0.975 eV)2 4(1.381 x 1O 23 JK I)(298K) 3.33 x 106 6 '.,:'';' In  2 (0.975  0.6 5) eV 1 5 19 1.602 x 10 J eV2.02 x 10
= 11.531 eV I
If we knew the activation Gibbs energy, we could use eqn 24.81 to compute (HOA) from either rate constant, and we can compute the activation Gibbs energy from eqn 24.82: L'1 +C
=
( L'1 C ..
+ )..)2
,,r_ _'4)"
[(0.665 + l.531)eVf _ 0 22 V .Ie. 4(1.531eV)
'='" 
MOLECULAR REACTION DYNAMICS
so
(Ll+G) 2kT '
(HOA) = (hket ) 1/2 (4AkT) 1/4 exp 2 rr3 (HOA) = (
497
6.626 X 10 34 J s)(2.02 X 105 SI») 1/2 2
X (4(1.53TeV)(1.602X 10 19 JeV;:)(l.381 X 1O23 JKl)(298K») 1/4
xex (0.122eV)(1.602X 1O19 J eV l») =19.39x 1O24 JI p 2(1.381xlO 23 JKl)(298K)· . E24,17(b) Equation 24.83 applies. In Exercise 24. 17(a), we found the parameter {J to equal 12 nm 1, so:
Inketlsl = {Jr + constant
so
constant = Inketls I
+ {Jr,
and constant = In 2.02 X 105 + (12 nm 1)(1.11 nm) = 25. Taking the exponential of eqn 24.83 yields: ket = etlr+constant s 1 = e (l2/ nm )(1.48nm)+25 S1 = 11.4 X 103 s1
I.
Solutions to problems Solutions to numerical problems A = NAa*
P24.1
=
*
(~:) [Section 24.1 and Exercise 24.13(a); Jt =
a X 6.022 X 10 ( ) (
23
mol
1 )
x
!m(CH3)]
IP (8) x (1.381 X 1O 23 JK 1) x (298K) ( (n) x (1/2) x (l5.03u) x (1.6605 x lO 27 kglu) )
(a)
(b) Take a "'" nd 2 and estimate d as 2 x bond length; therefore a = (n) x (154 x 2 x 10 12 m)2 = 3.0 x 10 19 m2.
a* 4.35 x 10 20 Hence P =  = 19 = ro:I5l. a 3.0x 10~ P24.3
For radical recombination it has been found experimentally that Ea "'" O. The maximum rate of recombination is obtained when P = 1 (or more), and then k2 = A = a*NA(8kT) 1/2 = 4a*NA( kT) 1/2 [Jt = ~m]. nJt nm 2 2 12 19 2 a* "'" nd = n x (308 x 10 m)2 = 3.0 x 10 m .
498
STUDENT'S SOLUTIONS MANUAL
Hence
In
(1.381 x 1O 23 1K I) x (298K)
(
)
The rate constant is for the rate law
Therefore d[CH 3 ]/ dt
= 2k2[CH3f
I I and its solution is     [CH3] [CH3]O
= 2k2t. = 0.10 x
For 90 per cent recombination, [CH3]
[CH 3]O , which occurs when
The mole fractions of CH3 radicals in which 10 mol% of ethane is dissociated is (2) x (0.10)
'':,.' = O. 18. 1+0.10
The initial partial pressure of CH3 radicals is thus Po
= 0.18 P = 1.8 x
Therefore t
4 10 Pa
9RT
= ,,:::;:(2k2) x (1.8 x I04Pa)
(9) x (8.3141 K I mol I) x (298 K) (\.7 x 10 8 m 3 molI SI) x (3.6 x 104Pa)
= 13.6ns I· P24.S
log k2
=
logk~
+ 2AZA ZB/ I / 2 with A = 0.509 (mol dm 3) 1/2 [24.69] .
This expression suggests that we should plot log k against 11 /2 determine ZB from the slope, since we know that IZA I = I. We draw up the following table.
1/ (mol dm 3)
0.0025
0.0037
0.0045
0.0065
0.0085
(I/(mol dm3)) 1/ 2 log(k2/(dm 3 molI sI))
0.050 0.021
0.061
0.067 0.064
0.081 0.072
0.092 0.100
0.049
MOLECULAR REACTION DYNAMICS
499
These points are plotted in Fig. 24.2.
0.10
::::: I", !.... 0
E
0.05
M
E
....
~I ~
.....
OIl
.2
o 0.05
0.06
0.07
0.08
0.09
0.10
{I1(molkg ' )}' /2
Figure 24.2
The slope of the limiting line in Fig. 24.2 is "" 2.5. Since thi s slope is equal to 2 A ZAZ B x (mol dm  3 ) 1/2 = 1.018 ZA ZB , we have ZA ZB "" 2.5. But IZ A I = 1, and so IZBI = 2. Furthermore, ZA and ZB have the same sign because ZA ZB > O. (The data refer to 1 and S 2 0~ . )
~( e ;; ~ 4rr sod(l 2
)2
a*
P24.7
Taking a
[Example 24.2] .
E ea)
= rrd 2 gives
Thus, a * is predicted to increase as I  Eea decreases. The data let us construct the following table. a * I nm 2
Na K Rb Cs
Ch
0.45 0.72 0.77 0.97
Br2
0.42 0.68 0.72 0.90
12
0.56 0.97 1.05 1.34
All values of a * in the table are smaller than the experimental ones, but they do show the correct trends down the columns. The variation with Eea across the table is not so good, possibly because the electron affinities used here are poor estimates.
Question . Can you find better values of electron affinities and do they improve the horizontal trends in the table?
500
P24.9
STUDENT'S SOLUTIONS MANUAL
(a)
d[F?O] t=  kl [F20 f

k2[F][F20] ,
(1)
d[F] dt
(2)
d[OF] dt
+ k2[F][F20 ] 
= kl [F20 f
2k3[OF]2.
(3)
Applying the steadystate approximation to both [F] and [OF] and adding the resulting equations gives kl [F20f  k2[F]ss [F20 ] + 2k3 [OF]~s kl[F20f + k2[Flss [F20 ]  2k3 [OF] ~s
2k4[F] ~s [F20]
2k4 [F] ~s [F20]
=0 =0 =0
Solving for [F]ss gives k
= ( k~ [F20]
[Flss
) 1/ 2
Substituting (4) into (I ) d [F 0] _ _ 2_ dt
= kl [F20 ]2 
k2
(k.!. ) k4
1/2
[F20 ]3/2
or d[F 0]  _ _2 _ dt
= kl [F20 f + k2 (k.!. ) 1/ 2 [F20] 3/2. k4
Comparison with the experimental rate law reveals that they are consistent when we make the followin g identifications. k
= kl = 7.8
£1
k'
=
x
1O I3 e E I / RT
( 19350 K )R
= k2 (
k ) 1/ 2 k~
=
dm 3 molI s l,
160.9 kJ mol  I,
= 2.3
x
1OIOe E'/RT
dm 3 mol  I SI ,
£ ' = ( 16910 K)R = 140.6 kJ mol  I. (b)
~02
+ F2 +
F20 2F + F2 0+ !02
2F + 0 + F20
tJ.rH (F20 ) = 24.41 kJ mol  I tJ.H = D(FF) = 160.6 kJ molI tJ.H = !D(Oo) = 249. 1 kJ mol  I
MOLECULAR REACTION DYNAMICS
~H(FOF) + ~H(OF)
501
=  [ ~fH(F20)  D(FF)  !D(Oo)]
=
(24.41  160.6  249.1) kJ molI
= 385.3 kJ molI. We estimate that 1~H(FOF) Then ~H(OF)
~ EI =
= 385.3 kJ molI
160.9 kJ molI I.
 ~H(FOF)
~ (385.3  160.9) kJ molI.
1 ~H(OF)
~ 224.4kJmol I· 1
In order to determine the activation energy of reaction (2) we assume that each rate is expressed in Arrhenius form, then
or ,E' InA  RT
= InA2 
E2 
RT
I + InA, 2
I Ell
    lnA4 2 RT 2
I E4 + . 2 RT
Differentiating with respect to T, we obtain
or E2  !E4
= E'
 !E,
= (140.6  80.4) kJ molI
= 60.2 kJ moli. E4 is expected to be small since reaction (4) is termolecular, so we set E4
P24.11
~
0; then
Linear regression analysis of In(rate constant) against liT yields the following results:
where C
= 34.36,
= 23227 K, R = 10.999761
B
standard deviation
= 0.36,
standard deviation
= 252 K,
[good fit] .
502
STUDENT'S SOLUTIONS MANUAL
where C' = 28.30, B' =  21065 K,
standard deviation = 0.84, standard deviation = 582 K,
R = 10.998481
[good fit].
The regression parameters can be used in the calculation ofthe preexponential factor (A) and the activation energy (Ea) using In k = InA  Ea/RT. InA = C
+ In(22.4)
= 37.47,
Ea = RB = (8.3145 J K 1mol l ) x (23227 K) x
= 1193 kJ molI InA' = C'
(10~3 kJ)
I.
+ In(22.4) =
31.41,
3
10kJ) E~ = RB' = (8.3145 J K1mol l ) x (21065 K) x ( J=1 175kJmol 1 I. To summarize
k k'
3. 12 X 10 14 (= A) 7.29xlO ll (=A')
193 175
Both sets of data, k and k', fit the Arrhenius equation very well and hence are consistent with the collision theory of bimolecular gasphase reactions which provides an equation 24.19 compatible with the Arrhenius equation. The numerical values for k' and A may be compared to the results of Exercise 24.7(a) and are in rough agreement at 647 K, as is the value of Ea.
Solutions to theoretical problems P24.13
[J]*
=k{
a[J]* at 2
a [J]* ax 2
[J]ektdt
+ [J]e  kt [24.40],
= k[J]e kt + a[J]ekt _
k[J]e kt
at
=k
r
Jo
2
(a [J]) ektdt ax 2
= (a[J]) e kt , at
+ (a
2
[J]) e kt . ax 2
MOLECULAR REACTION DYNAMICS
503
Then, since 2
=
D a [J] ax 2
a[J] [24.39 k at '
= 0]
,
we find that
a [J]* = k (a[J]*)  e 2
1t
D 2ax
=k
 kt
at
0
1(a[J]*) t
  dt at
o
dt (a[J]) e kt at
a[J]' + = k[J] ,a[J]' +at
at
which rearranges to eqn 24.39. When t = 0, [J]' = [J], and so the same initial conditions are satisfied. (The same boundary conditions are also satisfied.) ~T
P24.1S
fu. = 2.561 NA
x 1O2(T / K)5 /2(M / gmol I )3/2 [Table 17.3]. ~T
For T "'" 300 K, M "'" 50 g mol I , fu.NA R.
q (nonlinear)
1.0270
= a
0.6950 =
a
x
10 7
(T / K) 3/2 3 1/ 2
(ABC / cm )
= 2cm l , a
For T "'" 300 K, A"'" B "'" C
qR (linear)
x
"'" 11.4 x
(T /K)
I
(B/cm  )
~
[Table 17.3].
"'" 2 [Section 13.5], qR(NL) "'" 1900 ~
[Table 17.3].
For T "'" 300 K, B "'" I cm I , a "'" I [Section 13.5] , qR(L) "'" 1200 I. q v "'"
k2 =
IT]
and
qE "'"
IT] [Table 17.3],
KkT t [24.53]
TK
=( We then use
TkT)
x
(RT) ~ ) e Mo / RT [24.51] "'" Ae Ea / RT . p x (Nq::~
S04
STUDENT'S SOLUTIONS MANUAL
[The factor of 23/ 2 comes from me
RT pG ""
= mA + mB
(8 .314JK I molI) x (300K) 105 Pa
"" 2mA
= 2.5
and qT ex: m 3/ 2 . ]
x 10
KkT ~ kT _ (1.381 x 10 23 J K I ) x (300K) ~ h 6.626 X 1034 J s
2 3 I m mol 12  I
h
= 6.25
x 10
s
Therefore, the preexponential factor
A""
(6.25 X 10 12 S I) x (2 .5 x 1O 12 m 3 molI) x (7.9 x 109 ) ~~~~ (1.4 X 107 )2
"" 6.3 x 106 m3 mol I S I
or
16.3 x 109 dm 3 mol I s I I.
If all three species are nonlinear, G
G
qA "" (1.4 X 10 7 )
NA
x (900)
=
1.3 x 1010 ""
:i!!.. NA
G
qA "" (23/ 2) x (1.4 X 10 7 ) x (900)
NA
Therefore, P
A (NL) 3.3 x = = 6.3 x A(L)
4
10 9 10
= 3.6
= I5.2 x
x 1010
10 6
I .
.
These numerical values may be compared to those given in Table 24.1 and in Example 24.1. They lie within the range found experimentally. P24.17
We consider the ydirection to be the direction of diffusion. Hence, for the activated atom the vibrational mode in this direction is lost. Therefore,
q+ = q~V q;V for the activated atom, and q
= q; q'; q,/ for an atom at the bottom of a well.
For classical vibration, q v "" kT / hv [Section 24.4]. The diffusion process described is unimolecular, hence firstorder, and therefore analogously to the secondorder case of Section 24.4 [also see Problem 24.4] we may write
+ = kl[x] d[x]  = k+[X]T+ = vK+[x] dt
Thus
[K + + = [X]+] . [x]
MOLECULAR REACTION DYNAMICS
505
where q+ and q are the (vibrational) partition functions at the top and foot of the well respectively. Therefore
= kT
k
h
I
v+
(a)
=
(kT / hV +)2 ) e /3/',.Eo (kT /hv)3
v;
=
kl
= ''
ve/3/',.Eo. Assume fl.Eo ;::';: Eo; hence
kl;::';: 1011 Hze 60 x I03/(8.314x500)
)..2 ? [ 2\.85;r ButD= ;::,;: I )..kl 2r 2
=~ (b)
P24.19
v*
I =2'
D
=
V'
x 104 s l .
I ,Problem 22.10 ] kl
x (316pm)2 x 5.4 x 104 s 1
= 12.7 x
1O 15 m 2 sI
I.
kl = 4ve/3/',.Eo = 2.2 x 105 s I.,
(4) x (2.7 x 1O 15 m 2 s I)
= 11.1
x 10 14 m 2 sI
I.
The change in intensity of the beam, dl , is proportional to the number of scatterers per unit volume, uYs, the intensity of the beam, I , and the path length dl. The constant of proportionality is defined as the collision crosssection a. Therefore,
dl
= a JVsldl
or
d In I
If the incident intensity (at I
In!...=a JVs l 10 P24.21
=
= 5.4
or
= a JVsdl.
= 0) is 10 and the emergent intensity is I , we can write
I =l/oe a A"s1 I.
A + B + C+ + P.
We assume that the only factor that changes between the atomic and molecular case is the ratio of the partition functions. (1) For collisions between atoms
q! = qr ;: ,;: 1026 , q:
= q~ ;::,;:
q~
=
1026 ,
(q~)2q~q~ ;::,;: (101.5)2 x (I)
k2(atoms) ex
1029 = 10 23 26 10 X 1026
X
(10 26 ) ;::,;: 1029 ,
506
STUDENT'S SOLUTIONS MANUAL
(2) For collisions between nonlinear molecules
k2(molecules) ex
3 x 1030 61 I x 10
=3x
Therefore k2 (atoms) l k2 (molecules) ~
10 31
10 23
31 ~ 13 x 107 1. 3 x 10· .
Solutions to applications P24.23
Collision theory gives for a rate constant with no energy barrier
8kT) 1/ 2 k=Pu ( NA IT).t
so
P
=
_k_ (
IT).t )
UNA
8kT
1/2 .
x ().t l u) x (l.66 x 10 27 kg) ) 1/2 8 x (1.381 x 1O 23 JK I ) x (298K) IT
x
(
(6.61 x 1O 13 )kl (dm 3 molI s I) (u I nm 2 ) x ().t l u) 1/ 2
The collision crosssection is
so
UAB
=
(u A1/2 + u B1/2)2 4
The collision crosssection for 02 is listed in the Data section. We would not be far wrong if we took that of the ethyl radical to equal that of ethene; similarly, we will take that of cyclohexyl to equal that of benzene. For 0 2 with ethyl
).t
so P =
=
mome mo
+ me
=
(32.0 u) x (29.1 u) (32.0 + 29.1) u
=
15.2u,
1 31 (6.61 x 10 13) x (4.7 x 109 ) = 1.6 x 10 (0.51) x (15.2)1 /2
MOLECULAR REACTION DYNAMICS
507
For 02 with cyC\ohexyl (0.40 1/ 2 + 0.88 1/ 2)2
a=
mome (32.0 u) x (77.1 u) = = 22.6u, mo+me (32.0+77.I)u
JL =
so p P24.2S
=
nm 2 =0.62nm 2,
4
(6.61 x 10 13 ) x (8.4 x 109 ) (0.62) x (22.6) 1/2
= 11.8 x
JO
31.
=
ZB
Equation 24.69 may be written in the form :
zl =
I log(k2/ kG) 2A [1 /2 2 where we have used
ZA
for the cationic protein. This equation suggests
that ZA can be determined through analysis that uses the mean value of (log(k2/k'2»/[1 /2 for several experiments over a range of various ionic strengths.
ZA
=
We draw up a table that contains data rows needed for the computation. [
0.0100 8.10 9.08
k / ko
log(k / kO )/[O.5
ZA
=
0.0150 13.30 9.18
0.0200 20.50 9.28
~ { log(k2/k2) } _ 2A mean [1 /2 
0.0250 27.80 9.13
0.0300 38.10 9.13
0.0350 52.00 9.17
9.16 ~ 2(0.509) = ~
where we have used the positive root because the protein is cationic. P24.27
Does eqn 24.83, In ket = 
f3 r + constant,
apply to these data? Draw the following table.
r/nm ketls I 0.48 0.95 0.96 1.23 1.35 2.24
1.58 3.98 1.00 1.58 3.98 6.31
x x x x x x
In ketls 10 12 28 .1 109 22.1 109 20.7 108 18.9 107 17.5 10 1 4.14
I
508
STUDENT'S SOLUTIONS MANUAL
to ................ .
rlnm
and plot In ket
Figure 24.3
r (Fig. 24.3).
YS .
The data fall on a good straight line, so the equation 1appears to apply I. The least squares linear fit equation is In ket/s
= 34.7 
so we identify 1f3 P24.29
azurin(red)
EO K
= E~ 
R2 (correlation coefficient)
13.4r / nm,
= 0.991
= 13.4 nm  1 I.
+ cytochrome c(ox) + azurin(ox) + cytochrome c(red). EE = 0.260 V  0.304 V = 0.044 V.
= e vFE" / RT [7.30] = e l (96485.3Cmol 1)(O.044V)/(8.3 145IJK 1 mol  1)(298.15K) = eJ.713 = 0.180.
kobs
= (kookAA K) 1/ 2
koo
= kobs =
2
kAAK
(
(1.5
[24.86]. 3
3
1.6 x IO dm mol
x 102
dm 3
I
s
 1)2
mol I sI) (0.180)
= . 9.5 x 1
4
3
10 dm mol
I
 1
1
s.
Processes at solid surfaces
Answers to discussion questions 025.1
(a) A terrace is a flat layer of atoms on a surface. There can be more than one terrace on a surface, each
at a different height. Steps are the joints between the terraces; the height of the step can be constant or variable. (b) The motion of one section of a crystal past another (a dislocation) results in steps and terraces. See Figures 25 .2 and 25 .3 of the text. A special kind of dislocation is the screw dislocation shown in Fig. 25.3. Imagine a cut in the crystal, with the atoms to the left of the cut pushed up through a distance of one unit cell. The surface defect formed by a screw dislocation is a step, possibly with kinks, where growth can occur. The incoming particles lie in ranks on the ramp, and successive ranks reform the step at an angle to its initial position. As deposition continues the step rotates around the screw axis, and is not eliminated. Growth may therefore continue indefinitely. Several layers of deposition may occur, and the edges of the spirals might be cliffs several atoms high (Fig. 25 .4). Propagating spiral edges can also give rise to flat terraces. Terraces are formed if growth occurs simultaneously at neighboring left and righthanded screw dislocations (Fig. 25.5). Successive tables of atoms may form as counterrotating defects collide on successive circuits, and the terraces formed may then fill up by further deposition at their edges to give flat crystal planes. 025.3
Langmuir isotherm. This isotherm applies under the following conditions: 1. Adsorption cannot proceed beyond monolayer coverage. 2. All sites are equivalent and the surface is uniform. 3. The ability of a molecule to adsorb at a given site is independent of the occupation of neighboring sites. BET isotherm. Condition number I above is removed. This isotherm applies to multilayer coverage. Temkin isotherm. Condition number 2 is removed and it is assumed that the energetically most favorable sites are occupied first. The Temkin isotherm corresponds to supposing that the adsorption enthalpy changes linearly with pressure. Freundlich isotherm. Condition 2 is again removed, but this isotherm corresponds to a logarithmic change in the adsorption enthalpy with pressure.
025.5
In the LangmuirHinshelwood mechanism of surfacecatalyzed reactions, the reaction takes place by encounters between molecular fragments and atoms already adsorbed on the surface. We therefore expect
510
STUDENT'S SOLUTIONS MANUAL
the rate law to be secondorder in the extent of surface coverage:
Insertion of the appropriate isotherms for A and B then gives the reaction rate in terms of the partial pressures of the reactants. For example, if A and B follow Langmuir isotherms (eqn 25.4), and adsorb without dissociation, then it follows that the rate law is
The parameters K in the isotherms and the rate constant k are all temperature dependent, so the overall temperature dependence of the rate may be strongly nonArrhenius (in the sense that the reaction rate is unlikely to be proportional to exp( Ea/ RT).
In the EleyRideal mechanism (ER mechanism) of a surfacecatalyzed reaction, a gasphase molecule collides with another molecule already adsorbed on the surface. The rate of formation of product is expected to be proportional to the partial pressure, PB of the nonadsorbed gas B and the extent of surface coverage, eA , of the adsorbed gas A. It follows that the rate law should be A
+ B +
P
v
= kPAes.
The rate constant, k, might be much larger than for the uncatalyzed gasphase reaction because the reaction on the surface has a low activation energy and the adsorption itself is often not activated. If we know the adsorption isotherm for A, we can express the rate law in terms of its partial pressure, PA. For example, if the adsorption of A follows a Langmuir isotherm in the pressure range of interest, then the rate law would be v
kKpAPB = ''I +KPA
If A were a diatomic molecule that adsorbed as atoms, we would substitute the isotherm given in eqn 25 .6 instead.
According to eqn 25.27, when the partial pressure of A is high (in the sense KPA » I, there is almost complete surface coverage, and the rate is equal to kpB . Now the ratedetermining step is the collision of B with the adsorbed fragments. When the pressure of A is low (KpA « I) , perhaps because of its reaction, the rate is equal to kKpAPB ; and now the extent of surface coverage is important in the determination of the rate. In the Mars van Krevelen mechanism of catalytic oxidation, for example, in the partial oxidation of propene to propenal, the first stage is the adsorption of the propene molecule with loss of a hydrogen to form the allyl radical , CH2=CHCH2. An 0 atom in the surface can now transfer to this radical, leading to the formation of acrolein (propenal, CH2=CHCHO) and its desorption from the surface. The H atom also escapes with a surface 0 atom, and goes on to form H20 , which leaves the surface. The surface is left with vacancies and metal ions in lower oxidation states. These vacancies are attacked by 02 molecules in the overlying gas, which then chemisorb as 02" ions, so reforming the catalyst. This sequence of events involves great upheavals of the surface, and some materials break up under the stress. 025.7
Zeolites are microporous aluminosilicates, in which the surface effectively extends deep inside the solid. M"+ cations and H20 molecules can bind inside the cavities, or pores, of the AIDSi framework (see
PROCESSES AT SOLID SURFACES
511
Fig. 25.29 of the text). Small neutral molecules, such as C02 , NH3, and hydrocarbons (including aromatic compounds), can also adsorb to the internal surfaces and this partially accounts for the utility of zeolites as catalysts. Like enzymes, a zeolite catalyst with a specific composition and structure is very selective toward certain reactants and products because only molecules of certain sizes can enter and exit the pores in which catalysis occurs. It is also possible that zeolites derive their selectivity from the ability to bind and to stabilize only transition states that fit properly in the pores. 025.9
The net current density at an electrode isj;jo is the exchange current density; Ct is the transfer coefficient;
J is the ratio F / RT; and T/ is the overpotential. (a) j = joJT/ is the current density in the low overpotentiallimit. (b) j = joe(lajfry applies when the overpotential is large and positive. (c) j = _joe aIry applies when the overpotential is large and negative. 025.11
The principles of operation of a fuel cell are very much the same as those of a conventional galvanic cell. Both employ a spontaneous electrochemical reaction to produce an electric current that can be used as a power source for external devices. The main difference between fuel cells and ordinary cells is that the reacting substance is a material that we normally classify as a fuel and it is continuously supplied to the cell from an external source. We wish to obtain large currents from fuel cells and in order to accomplish that goal a number of obstacles limiting the rate of reaction have to be overcome. One way to increase the rate of the reaction in the cell is to use a catalytic surface with a large effective surface area to increase the current density. Operating the cells at high temperatures can increase reaction rates and in some cases molten electrolytes and electrodes are employed.
Solutions to exercises E25.1 (b)
The number of collisions of gas molecules per unit surface area is
z _
NAP
w 
(2][ MRT) 1/ 2
(a) For N2 (i)
Zw
=
(ii)
Zw
=
(6.022 x 1023 molI) x (lO.OPa) (2][ x (28 .013 x 1O 3 kgmol l ) x (8.31451 K I molI) x (298K»I /2
(6.022 x 10 23 mol  I) x (0.150 x 1O 6 Torr) x (1.01 x 105 Paj760Torr) (2][ x (28.013 x 10 3 kg mol I) x (8.31451 KI mol I) x (298K»I /2
512
STUDENT'S SOLUTIONS MANUAL
(b) For methane Zw
(i)
=
(6.022 x 1023 mol  I) x (1O.0Pa) (27T x (16.04 x 10 3 kg mol I) x (S.3145J KI molI) x (29SK»1/2
x 1023 m 2 S I
= 3.SI
= 13.s 1 x 10 19 cm 2 sI (ii)
Zw
(6.022 x 1023 mol  I) x (0.150 x 1O 6 Torr) x (1.01 x 105 Paj760Torr) (27T x (16.04 X 10 3 kgmol I ) x (S.3145J KI mol  I) x (29SK»1 /2 = 7.60 x 10 17 m 2 S I
=
= 17.60 x 1013 cm 2 SI E25.2(b)
1
The number of collisions of gas molecules per unit surface area is NAP
Zw 
P=
so
(27TMRT)I / 2
P=
Zw A (27TMRT) 1/2
x 7T x (1/2 x 2.0 x 1O 3 m)2 x (27T x (2S.013 x 1O 3 kgmol l ) x (S.3145Jmol 1 K I) x (525K»1/2 (6.022 x 1023 mo l
= 17.3 E25.3(b)
1
l
)
2
x 10 Pa 1
The number of collisions of gas molecules per unit surface area is Zw
NAP
';=
(27TM RT) I/ 2
so the rate of colljsion per Fe atom will be ZwA where A is the area per Fe atom. The exposed surface consists of faces of the bcc unjt cell, with one atom per face. So the area per Fe is A = c2
and
rate
= Zw A =
NAPC 2 (27TM RT)I / 2
where c is the length of the unit cell. So rate
=
(6.022 x 1023 mol  I) x (24Pa) x (145 x 10 12 m)2 (27T x (4.003 x 1O 3 kgmol l ) x (S.3145JK I molI) x (I00K»1 /2
':;;;~
= 16.6 E25.4(b)
x 104 sI 1
The number of CO molecules adsorbed on the catalyst is ( 1.00atm) x (4.25 x 1O 3 dm 3) x (6.022 x 1023 molI) (0.OS206dm 3 atm K I mol  I) x (273 K)
pVNA
N=nNA=RT
=
1.14
X
1020
PROCESSES AT SOLID SURFACES
513
The area of the surface must be the same as that of the molecules spread into a monolayer, namely, the number of molecules times each one's effective area A = Na = (1.14 x 1020) x (0.165 x 10 18 m2) = 118.8 m2 1
E25.5(b)
If the adsorption follows the Langmuir isotherm, then
e=~
e
so
1 +Kp
K=::p(1  e)
v / Vmon pel  v /Vmon )
Setting this expression at one pressure equal to that at another pressure allows solution for Vmon V2 / Vmon "'.:.:.:..:...=
VI / Vmon PI(1  VI / Vmon )
Vmon = E25.6(b)
P2(l VdVmon )
P2(Vmon  V2) V2
PI  P2 (52.4  104) kPa 1 3 1 = 3 =.9.7 cm . V PI / VI  P2 / 2 (52.4/ 1.60  104/ 2.73) kPacm
The mean lifetime of a chemisorbed molecule is comparable to its halflife:
Ed)
tl /2 = Toexp ( RT E25.7(b)
PI (Vmon  VI> so :....::...c.......::::.::.::::.._"'VI
~
(10 14 s)exp (
1 3 155 x 10 Jmol) I (8.3145JK1 mol ) x (500K)
The desorption rate constant is related to the halflife by t
= (In 2) / kd
so
kd
= (In 2) / t
The desorption rate constant is related to its Arrhenius parameters by
Ed Inkd=lnA
so
and Ed =
E25.8(b)
T2
I

Ti
(In 1.35 In 1) x (8.3145JK 1 molI)
I
(600K)1  (lOOOK)1
The Langmuir isotherm is
e=
E25.9(b)
(In kl  In k2)R
RT
Kp e 1 + Kp so P = K(l _
e)
I
(a)
p =
0.20 = 0.32kPa (0.777 kPa l ) x (1  0.20)
(b)
p =
0.75 = 3.9 kPa (0.777 kPa l ) x (I  0.75)
The Langmuir isotherm is
e=~ I +Kp
I
I
I
~
=~
514
STUDENT'S SOLUTIONS MANUAL
We are looking for 0 , so we must first find K or mmon
o
m/ mmon
K=p(l  0)
Setting this expression at one pressure equal to that at another pressure allows solution for mmon
mmon =
PI  P2 (36.0  4.0) kPa = = 0.S4mg PI / ml  P2/m2 (36.0/ 0.63  4.0/ 0.21 ) kPa mg I
I
So 01 = 0.63/ 0.S4 = 10.75 1and 02 = 0.21 / 0.S4 = 0.251 E25.10(b) The mean lifetime of a chemisorbed molecule is comparable to its halflife
tl / 2
(a)
=
TO
exp
(:~ )
At 400K :
t
= 0.12 1/2 (
10 11
= 14.9 x At SOOK :
t
1/2
= 0.12 (
20 x (03 Jmol I ) 1O 12 s ex ) p ( (S.3145JK Imol l) x (400K)
X
S 1
20 x 103 Jmol1 ) 10 12 s ex ) p ( (S.3145JKI mo l l ) x (SOOK)
X
=12.4 x 1O 12 sl (b)
At400K :
t
1/2
= 0.12 (
X
= 11.6 x At SOOK :
t
= 0.12 1/2 (
200 x 10 3 J mo(I ) 1O 12 s ex ) p ( (S.3 145JKI mo l l ) x (400K)
1013 s
X
I
I ) 200 x 103 J mol1O 12 s ex l ) p ( (S.3145JKI mo l ) x (SOOK)
= [ill E25.11(b) The Langmuir isotherm is
Kp 0 0 = 1 + Kp so P = K(l  0) For constant fractional adsorption
pK = constant
KI so PIKI = P2K2 and P2 = PIK2
PROCESSES AT SOLID SURFACES
But K
RT ( ~adHe)
exp
c<
KI __ exp K2
so
515
(~adHB (~ _ ~)) R
TI
T2
= PI exp (~adHe (~  ~))
P2
R
TI
T2
3
= (8.86kPa) x exp (( 12.2 x 10 IJmOlI 8.3145 J K mol
l )
1  1 x () ) 298 K 318 K
= 16.50 kPa 1
E25.12(b) The Langmuir isotherm would be
(a)
e=~
(b)
e = ':c::
(c)
e = ':c::I + (Kp) 1/ 3
1 +Kp (Kp) 1/ 2
1 + (Kp)lf2 (Kp) 1/ 3
e versus p at low pressures (where the denominator is approximately 1) would show progressively weaker dependence on p for dissociation into two or three fragments.
A plot of
E25.13(b) The Langmuir isotherm is
e
Kp
1 + Kp so p
e=
=
e)
K (l 
For constant fractional adsorption pK
But K
= constant
c<
exp
so PIKI
(~adHe)
= P2 K 2
so P2
RT
and f'oJ.adH B =R ~adHe
=
(
and P2 PI
TI
)1
T2
KI K2
= exp (~adHe (~ _ ~))
PI
1   1
=
R
TI
T2
PI In, P2
(8.3145JK 1 molI) x (
1 1 180K  240K
)1
350kPa) x ( In l.02 x 103 kPa
= 6.40 x 104 Jmol I = 16.40kJmol 1 1 E25.14(b) The time required for a given quantity of gas to desorb is related to the activation energy for desorption by
t c< exp
and Ed
Ed
(:~)
= R ( I I TI
=
T2
so
~ = exp (Ed (~ _ ~))
)1
In 
t2
R
T2
tl
t2
(8.3145JK 1 molI) x
= 12.85
TI
x 105 Jmol I 1
1  1( 873 K 1012 K
)1 ( x
1856) In _ _s 8.44 s
516
STUDENT'S SOLUTIONS MANUAL
(a) The same desorption at 298 K would take
t = (1856 s) x exp
5 2.85X10 JmOI') x (I  I ) ) = 1.48 x 1036 s (( 8.3145 J K'mol' 298 K 873 K
I
I
(b) The same desorption at 1500 K would take
5 (844) ((2.85X I0 JmOI ' ) (1 I)) t = . S x exp 8.3145J K' mol' x 1500 K  1012 K x 104 s
= 11.38
I
E25.15(b) Disregarding signs, the electric field is the gradient of the electrical potential dL'J.rp
E:
= ~ ~
L'J.¢
d
a = ; =
a E:rE:O
0.12Cm 2 8 = (48) x (8.854 x 1O'2J1C2 m  ') = 2.8 x 10 Vm
I
,
I
E25.16(b) In the high overpotentiallimit
so
~ = e(la)!(ryl'/2) where f F  I 
 RT  25.69 mV
12 The overpotential Tl2 is
Tl2 = TIl
I + f(la)
12
In  = 105mV + )1
2 (25.69mV) ( 72mAcm ) x In 10.42 17.0rnAcm 2
=1167mvl E25.17(b) In the high overpotential Limit
) =)oe(la)!ry
so )0 =)e(a I)!ry
)0 = (17.0rnAcm 2) x e{ (0.42I) x( 105mV)/ (25.69mY)\ = 11.6 rnAcm 2 1 E25.18(b) In the high overpotential Limit
so
~ = e(la)!(ryl  ry2 )
12 So the current density at 0.60 V
12 =
(1.22 rnA cm 2) x e{ (10.50) x (0.60Y0.50Y)/(0.02569Y)\ = 18.5 rnA cm 2 1
Note: the exercise says the data refer to the same material and at the same temperature as the previous Exercise (25.18(a», yet the results for the current density at the same overpotential differ by a factor of over 5!
PROCESSES AT SOLID SURFACES
517
E25.19(b) (a) The ButlerVolmer equation gives
j =jo(e( la )!I/  e a!'1)
=
(2.5 x 10 3 Acm 2) x (e[I  0.58) X(0.30V)/ (O.02569V)} _ e[(O.58) x(0.30V)/O.02569 V)})
=
I
0.34Acm 2 1
(b) According to the Tafel equation j =joe(l a)!'1
= (2.5 x 10 3 A cm2)e[(I  0.58)x(O.30 V)/(O.02569 V)} = I 0.34Acm 2 1
The validity of the Tafel equation improves as the overpotential increases. E25.20(b) The limiting current density is
jiim
zFDc 8
=
but the diffusivity is related to the ionic conductivity (Chapter 21)
ART z2F2
D=
hm =
. so
CA
Jlim
= 8if
(1.5 mol m 3) x (10.60 X 10 3 S m 2 moli) x (0.02569 V) (0.32 X 10 3 m) x (+ I)
=11.3Am 2 1 E25.21 (b) For the iron electrode E"
= 0.44 V (Table 7.2) and the Nemst equation for this electrode (section 7.7a) is
" RT ( I ) E = E  vF In [Fe 2+]
v= 2
Since the hydrogen overpotential is 0.60 V evolution of H2 will begin when the potential of the Fe electrode reaches 0.60 V. Thus 0.60V
In[Fe
2+
[Fe 2+ ]
]
= O.44V + 0.02569V In [Fe2+] 2
=
0.16V 0.0128 V
= 14 x

= 12.5
10 6 mol dm  3 1 2
COMMENT. Essentially all Fe + has been removed by deposition before evolution of H2 begins
518
STUDENT'S SOLUTIONS MANUAL
E2S.22(b) The zerocurrent potential of the electrode is given by the Nemst equation
The ButlerVolmer equation gives
where IJ is the overpotential , defined as the working potential £ ' minus the zerocurrent potential £ . I
'7 = £ '  0.77 V + In
f
a (Fe2+) I ( 3 ) = £'  0.77 V + In r, a Fe + f
where r is the ratio of activities; so j = jo(e(0.42)E' If e{ (0.42) x (0.77 V)/ (0.025 69 V ) ) r0.42 _ e ( 0.5 8) E' If e{ ( 0.58) x (0.77 V)/ (0.025 69 V) } r0.58)
Specializing to the condition that the ions have equal activities yields j = (2.5 rnA cm  2 ) x [(e (0.42)E'lf x (3.41 x 10 6 ) _e(058)E'lf x (3.55 x 107 )] E2S.23(b)
Note. The exercise did not supply values for jo or fX . Assuming Exercise 25.22(b)
fX
= 0.5, only j po is calculated. From
j = jo(e(0.50)E'lf e (0.50)E e If r 0.50 _ e (0.50)E'lf e(0.50)Ee lf r 0.50 )
= 2jo sinh [~f £'  ~f E e + ~ In r
l
so, if the working potential is set at 0.50 V, then j = 2jo sinh[ ~ (0.91 V) / (0.02569 V)
At r = 0.1 : j /jo = 2 sinh ( 8.48 +
+ ~ In r]
~ In 0.10)
= 1.5 x 103 rnA cm 2 = 11.5 A cm 21
At r = I : j /jo = 2 sinh(8.48 + 0.0) = 4.8 x 103 rnA cm 2 = 14.8 A cm 2 1 At r = 10: j /jo = 2 sinh (8.48 +
~ In 10)
= 1.5 x 104mA cm 2 =115Acm 2 1
E2S.24(b) The potential needed to sustain a given current depends on the activities of the reactants, but the over
potential does not. The ButlerVolmer equation says
PROCESSES AT SOLID SURFACES
519
This cannot be solved analytically for 7] , but in the highoverpotentiallimit it reduces to the Tafel equation
j =joe(la )!ry
so
7]
=
1 j In (Ia)f jo
=
0.02569V 15mAcm 2 In ,.".2 10.75 4.O x 1O 2 mAcm
This is a sufficiently large overpotential to justify use of the Tafel equation. E25.25(b) The number of singly charged particles transported per unit time per unit area at equilibrium is the
exchange current density divided by the charge
N=~ e
The frequency f of participation per atom on an electrode is
f=Na where a is the effective area of an atom on the electrode surface. For the CU, H2 IH+ electrode
=14.2 x 1O 3 s 1 1 For the PtICe 4 + , Ce3+ electrode 2
I
jo 4.0 x 1O 5 AcmN =  = = 2.5 e l.602 x 1O 19 C
X
10 14
sI
cm 2
I
The frequency f of participation per atom on an electrode is
E25.26(b) The resistance R of an ohmic resistor is
R
= potential current
7]
jA
where A is the surface area of the electrode. The overpotential in the low overpotentiallimit is so
1 R=
fjoA
520
STUDENT'S SOLUTIONS MANUAL
(a) R
=
(b) R
=
0.02569V = 5.1 x 109 Q (5.0 x 10 12 A cm 2) x (1.0cm 2) 0.02569 V ~ =L.!Q£j 2 3 2 (2.5 x 10 A cm ) x (1.0cm )
= 15.1 GQ I
E25.27(b) No reduction of cations to metal will occur until the cathode potential is dropped below the zerocurrent
potential forthe reduction ofNi 2+ (0.23 Vat unit activity). Deposition ofNi will occur at an appreciable rate after the potential drops significantly below this value; however, the deposition of Fe will begin (albeit slowly) after the potential is brought below 0.44 V. If the goal is to deposit pure Ni, then the Ni will be deposited rather slowly at just above 0.44 V; then the Fe can be deposited rapidly by dropping the potential well below 0.44 V. E25.28(b) As was noted in Exercise 25 .18(a), an overpotential of 0.6 V or so is necessary to obtain significant
deposition or evolution, so H2 is evolved from acid solution at a potential of about 0.6 V. The reduction potential of Cd 2+ is more positive than this (0.40 V), so Cd will deposit (albeit slowly) from Cd 2+ before H2 evolution. E25.29(b) Zn can be deposited if the H+ discharge current is less than about I rnA cm 2. The exchange current,
according to the high negative overpotentiallimit, is
At the standard potential for reduction of Zn2+ (0.76 V) j
= (0.79 rnA cm 2) x e{ (O.5) x (O.76V)/(O.02569V)1 = 2.1 x
109 rnAcm 2
1much too large to allow deposition I. (That is, H2 would begin being evolved, and fast, long before Zn
began to deposit.) E25.30(b) Fe can be deposited if the H+ discharge current is less than about I rnA cm 2. The exchange current,
according to the high negative overpotentiallimit, is
At the standard potential for reduction of Fe2+ ( 0.44 V) j
= (l
x 10 6 Acm 2) x e  {(O.5) x (O.44V)/(O.02569V)}
= 5.2 x
10 3 Acm 2
1a bit too large to allow deposition I. (That is, H2 would begin being evolved at a moderate rate before Fe began to deposit.) E25.31 (b) The lead acid battery halfcells are Pb 4+
+ 2e
* Pb 2+
and PbS04 + 2e * Pb + SO~
1.67V  0.36 V,
PROCESSES AT SOLID SURFACES
for a total of EB
P
521
= 12.03 V I. Power is
= N = (lOO x
10 3 A) x (2.03 V)
= 10.203 wi
if the cell were operating at its zerocurrent potential yet producing lOOmA. E25.32(b) Two electrons are lost in the corrosion of each zinc atom, so the number of zinc atoms lost is half the
number of electrons which flow per unit time, i.e. half the current divided by the electron charge. The volume taken up by those zinc atoms is their number divided by number density; their number density is their mass density divided by molar mass times Avogadro's number. Dividing the volume of the corroded zinc over the surface from which they are corroded gives the linear corrosion rate; this affects the calculation by changing the current to the current density. So the rate of corrosion is
rate
jM
=  = 2epNA
(2.0Am 2) x (65.39 x 1O 3 kgmol l )
''=,2(1.602 x 1O 19 C) x (7133kgm 3) x (6.022 x 1023 molI)
= 9.5 x 10 11 ms I = (9.5 x 10 11 ms I) x = 13.0 mmyI
(I03 mmm l) x (3600 x 24 x 365s yI)
1
Solutions to problems Solutions to numerical problems P25.1
Refer to Fig. 25.1.
Figure 25.1
522
STUDENT'S SOLUTIONS MANUAL
1 Evaluate the sum of ± , where I
n
I
n is the distance from the ion i to the ion of interest, taking + 
r
for ions
of like charge and   for ions of opposite charge. The array has been divided into five zones. Zones r
Band D can be summed analytically to give in 2 = 0.69. The summation over the other zones, each of which gives the same result, is tedious because of the very slow convergence of the sum. Unless you make a very clever choice of the sequence of ions (grouping them so that their contributions almost cancel), you will find the following values for arrays of different sizes.
10 x 10
20 x 20
50 x 50
100 x 100
200 x 200
0.259
0.273
0.283
0.286
0.289
The final figure is in good agreement with the analytical value, 0.289 259 7 ... (a) For a cation above a flat surface, the energy (relative to the energy at infinity, and in mUltiples of e2j4rrEro where ro is the lattice spacing (200 pm)), is Zone C + D + E = 0.29  0.69 + 0.29
= 8m
which implies an attractive state. (b) For a cation at the foot of a high cliff, the energy is Zone A + B + C + D + E
= 3 x 0.29 + 2 x
(0.69)
= 10.51 I
which is significantly more attractive. Hence, the latter is the more likely settling point (if potential energy considerations such as these are dominant). P25.3
Refer to Fig. 25.2 .
Figure 25.2 The (IOO) and (110) faces each expose two atoms, and the (111) face exposes four. The areas of the faces of each cell are (a) (352 pm)2 = l.24 x 10 15 cm 2, (b) ./2 x (352 pm)2 = 1.75 x 10 15 cm 2, and (c) ..j3 x (352 pm)2 = 2.15 x 10 15 cm 2. The numbers of atoms exposed per square centimetre are therefore
PROCESSES AT SOLID SUR FACES
2
(a) 1.24 x 10
15
cm 2
= 11.61
2 (b ) 1.75 x 1015 cm 2 =
4 (c) 2.15 x 1O 15 cm2
x 10
11.14 x
= 11.86 x
523
21
15
cm.
21 10 15 cm. 10
15
cm
21.
For the collision freq uencies calculated in Exercise 25.1(a), the freque ncy of collision per atom is calculated by dividing the values given there by the number densities just calculated. We can therefore draw up the following table.
Propane
Hydrogen Z / (atom I sI)
100 Pa
10 7
(l00) (110)
6.8 x 105 9.6 x 105 5.9 x 105
8.7 x 10 2 1.2 x 10 1 7.5 X 10 2
(III)
P25.5
v =
Torr
100Pa
10 7 Torr
1.4 x 105 2.0 x 105 1.2 x 105
1.9 x 10 2 2.7 x 10 2 1.7 x 10 2
cz [ 25.8, BET lsotherm,Z . (I  z){l  (1  c) z }
Vrnon
= p]. p*
This rearranges to Z
1
(c  l) z
(1 z)V
cVrnon
cVrnon
=+
.
Therefore a plot of the lefthand side against isotherm. We draw up the fo llowing tables. (a) O°C, p*
z should result in a straight line if the data obeys the BET
= 3222 Torr.
p/ Torr
105
282
492
594
620
755
798
103z 103 z/ ( 1  z)(V / cm 3)
32.6 3.04
87.5 7.10
152.7 12.1
184.4 14.1
192.4 15.4
234.3 17.7
247.7 20.0
(b) 18°C, p*
= 6148 Torr.
p/ Torr
39.5
62.7
108
219
466
555
601
765
103z 103 Z/ (I  z)(V / cm 3)
6.4 0.70
10.2 1.05
17.6 1.74
35.6 3.27
75.8 6.36
90.3 7.58
97 .8 8.09
124.4 10.8
The points are plotted in Fig. 25 .3, but we analyse the data by a leastsquares procedure. The intercepts are at (a) 0.466 and (b) 0.303. Hence I
 = cVrnon
(a ) 0.466 x 10 3 cm 3,
(b) 0.303 x 10 3 cm 3.
524
STUDENT'S SOLUTIONS MANUAL
The slopes of the lines are (a) 76.10 and (b) 79.54. Hence C I
 = (a) 76.10 x cV
10 3 cm 3 ,
(b) 79.54 x 10 3 cm 3 .
mon
Solving the equations gives C 
1 = (a) 163.3, (b)262.5
and hence C
= (a) 11641,
(b)1 264
I;
Vmon
= (a) 113.1 cm 3 1,
(b) 112.5 cm 3 1.
(b)
(a)
10
20
1:u
1:u
'N' IO
'N'5
:;:
:;:I
I
. N
:::::: N 0
0 0 0
8
o
o
o
0.1
z
o
0.2
0.05
z
Figure 25.3
P25.7
()
= C1pl /c2 .
We adapt this isothenn to a liquid by noting that Wa ex: () and replacing p by [Aj, the concentration of the acid. Then Wa = ct [Ajl /c2 (with ct, C2 modified constants), and hence logwa = 10gcI
I
+
x 10g[Aj.
C2
We draw up the following table. [Aj/(mol dm 3 ) log([Aj/moldm 3 ) log(wa / g)
0.05
0.10
0.15
0.20
0.25
1.30 1.40
1.00 1.22
0.30 0.92
0.00 0.80
0.18 0.72
PROCESSES AT SOLID SURFACES
525
These points are plotted in Fig 25.4.  1.4
D.6
o
0.5
D.5
1.0
1.5
log([AJ/(mol dm 3 ))
Figure 25.4
They fall on a reasonably straight line with slope 0.42 and intercept 0.80. Therefore, C2 = 1/ 0.42 =
I 2.4 land CI = 10.161. (The units of C I are bizarre: C I = 0.16gmol o.42 dm I.26.) P25.9
Taking the log of the isotherm gives In Cads = InK
+ (In
csol) / n
so a plot of In Cads versus In Csol would have a slope of l /noo and a yintercept of In K. The transformed data and plot are shown in Fig. 25.5 . 5
y = 1.9838 + 171.06x R 2 = 0.981 . .............. .. ...... .. .. .... .... (... .
4
2
2.0
2.5
3.0
3.5
4.0
In Csol
csol/(mg gI) Cads/ (mg gI) In Csol In Cads
8.26 4.4 2.11 1.48
15.65 19.2 2.75 2.95
K=e  1.9838 mgg 1
Figure 25.5
25.43 35.2 3.24 3.56
31.74 52.0 3.46 3.95
= 10. 138 mgg l land
40.00 67.2 3.69 4.21
n = 1/ 1.71 = 10.581.
526
STUDENT'S SOLUTIONS MANUAL
In order to express this information in terms of fractional coverage, the amount of adsorbate corresponding to monolayer coverage must be known. This saturation point, however, has no special significance in the Freundlich isotherm (i.e. it does not correspond to any limiting case). P25.11
The Langmuir isotherm is n e = Kp  = 
1 + Kp
noo
so n(l
p and n
+ Kp) = nooKp
=
p
noo
1
+ . Knoo
So a plot of pin against p should be a straight line with slope Iinoo and yintercept llKn oo . The transformed data and plot (Fig. 25 .6) follow
31.00 38.22 53.03 76.38 101.97 130.47 165.06 182.41 205.75 219.91 2.90 3.22 3.30 3.35 3.36 1.00 1.17 1.54 2.04 2.49 (pln)1 (kPa mol  I kg) 31.00 32.67 34.44 37.44 40.95 44.99 51.26 55 .28 61.42 65.45
p l kPa nl(mol kgI)
70
0:0 ~
y = 24.641 +0.17313~ R2 = 0.982' .
60
I
"0
8
OJ
~
50
"
'<;: 40 30
o
100
50
noo =
150 pl kPa
200
250
300
Figure 25.6
I I = 15.78 mol kgI 0.17313 mol kg
I.
The yintercept is 1 b=   so K= Knoo bnoo K P25.13
= 7.02 x
In)
10 3 kPa
= In)o + (l
1
(24.641 kPa mol I kg) x (5.78 mol kg I)'
= 17.02pa ' I.
 Ct.)!17 [25.45] .
Draw up the following table.
I7 / mV In(jjrnA cm 2 )
50 0.98
100 2.19
150 3.40
200 4.61
250 5.81
PROCESSES AT SOLID SURFACES
527
The points are plotted in Fig. 25.7. 8 6
'I
E
u ~
4
6
~ E
2
o o
50
100
150
200
250
1J/ mV
Figure 25.7
=
The intercept is at 0.25 , and so Jo / (mA cm 2 ) (1  a)F / RT negative,
= 0.0243 m V  I. It follows that 1 
a
e 0 .25
=
10.781. The slope is 0.0243 , and so
= 0.62, and so 1a = 0.381. If TJ were large but
til "'=' Joe afq [25.46] = (0.78 mAcm 2 ) x (e 0 .38ry / 25.7mV)
= (0.78mAcm 2 ) x (e 0 .015 (ry / mV) ) and we draw up the following table.
50 1.65
P25.15
Jlim =
100 3.50
150 7.40
200 15.7
zFDc FDc [25.57aJ, and so 8 =  .  [z 8 Jlim
250 33.2
= 1]
Therefore, (9.65 x 104 Cmol I ) x (1.14 x 1O 9 m2 s l) x (0.66molm 3 ) 8=~~~~~~~ 28.9 x 10 2 A m 2 = 2.5 x 10 4 m, or 10.25 mm
P25.17
(4~T) In (~J)  IRs [25 .62].
E' = E 
P
=I
dp dl
E,
I.
= IE 
=E=
a In
10
alin (I) I 2 Rs where a
(i) 10
a  2IRs = 0 .
=
4RT F and 10
. = AJ.~ For maXImum power,
528
STUDENT'S SOLUTIONS MANUAL
which requires
This expression may be written
In
(~) = c] 
C2 / ;
CI
E
= a
I ,
2Rs
FRs
a
2RT
C2 =  =   ·
For the present calculation, use the data in Problem 25. I 6. Then
10 =A] = (5 cm 2 ) x (I mA cm 2 ) = 5 rnA,
CI = (4) C
;'<~~~;7 Y) 
=73QyI=73AI .
(3.75Q)
2
1 = 10.7,
= (2) x (0.0257 Y )
That is, In (0.201I mA)
= 10.7 
0.073(1 l mA).
We then draw up the following table.
l i mA
103
104
105
106
107
In(0.201 I mA) 10.7  0.073(1 l mA)
3.025 3. lSI
3.034 3.IOS
3.044 3.035
3.054 2.962
3.063 2.S89
The two sets of points are plotted in Fig 25.S.
3.2
§
3.1
"iii
::l C'
., '
o
:z:
3.0
....l
~
~ 2.9
2.8 102
103
104
l i mA
105
106
107
Figure 25.8
PROCESSES AT SOLID SURFACES
529
The lines intersect at I = 105 rnA, which therefore corresponds to the current at which maximum power is delivered. The power at this current is
P
= ( 105 rnA) x
( l.IOV)  (0.103 V) x (l05 rnA) x In
C~5)

( 105 rnA)2 x (3.75 n )
= 41 mW.
ro = (cRT/2pF2 Ib~) 1 /2
P2S.19
where 1=
!L
z1 (b;/b~) , b~
[19.46]
= I mol kgI .
i
For NaCI : Ib~
= bNaCI ~ [NaCI] assuming 100 per cent dissociation.
= ! ((l)2(2bNa2S04) + (2)2bNa2S04)
For Na2S04 : Ib~
= 3bNa2S04 ~ 3[Na2S04], assuming 100 per cent dissociation. ro ~
(78.54 x (8.854 x 10 12 r l C 2 m I) x (8.315 J
K
K))
I molI ) x (298.15 2 x (1.00gcm 3 ) x (10 3 kg/ g) x (106 cm 3/ m3) x (96485Cmol I)2
3.043
X
~
1/2
X
(~) 1/2 Ib
10 10 mmol l / 2 kg I/ 2 (lb~) 1 /2
304.3pmmol l / 2 kg I/ 2
~
(lb~)1 /2
These equations can be used to produce the graph of ro against bsalt shown in Fig. 25 .9. Note the contraction of the double layer with increasing ionic strength.
P2S.21
(a) The accompanying Tafel plot (Fig. 25 .10) oflnj against E shows no region of linearity so the Tafel equation cannot be used to determine jo and ex . (b)
Mads + H 2MHads
+
+e
_ K2
;=e MHads,
rate determining
HMMH.
Assuming that the dimerization is ratedetermining, two electrons are transferred per molecule of HMMH and z = 2. It is also reasonable to suppose that the first two reactions are at quasiequilibrium. According to reaction 3, the current density is proportional to the square of the functional surface coverage by MHads, e~1H , j
= zFk3e~H '
In j
= In(zFk3) + 21n ~ H .
530
STUDENT'S SO LUTION S MANUAL
Gouy Chapman diffuse double layer
5~
ITT,
4~
3~
2~
I~
o
20
40
60
100
80
bl(mmol kg I)
Figure 25.9
Tafel plot for dione in butanol
6.00
r,
• •
5.00 I
•
....I E
5« :3
•
4.00 I
•
..E
3.00 I
2.00 1.4
• I
I
I
I
I
I
1.5
1.6
 /.7
1.8
1.9
2.0
El Y
 2.1
Figure 25.10
The characteristics of this equation differ from those of the Tafel equatioin at high negative overpotentials Inj
= Injo 
airy [25.47].
PROCESSES AT SOLID SURFACES
531
At low concentration of M the value of ~H changes with the overpotential in a nonexponential manner. This makes In) nonlinear throughout the potential range.
Solutions to theoretical problems P25.23
Refer to Fig. 25.11 .
Figure 25.11 Let the number density of atoms in the solid be N. Then the number in the annulus between rand r + dr and thickness dz at a depth z below the surface is 2nN r dr dz. The interaction energy of these atoms and the single adsorbate atom at a height R above the surface is
if the individual atoms interact as C6 / d 6 , with d 2 = (R + Z)2 + r2. The total interaction energy of the atom with the semiinfinite slab of uniform density is therefore
U
= 2nNC6
We then use
and obtain
00 100 1 o
dr
0
dz
r {(R
2
+ z) + r
2 3·
}
532
STUDENT'S SOLUTIONS MANUAL
This result confinns that U ex 1/ R3. (A shorter procedure is to use a dimensional argument, but we need the explicit expression in the following.) When
v=
4€
[(~) 12 _ (~)6J R
R
= Cl 2 _ C6 RI2 R6 '
we also need the contribution from CI 2
1 1 00
U I = 2nN CI 2
o
1
00
dr
0
r {(R + z)2
dz
= 2nN CI 2 x  I + r 2}6 10 0
00
dz C1 2 = 2rr.N :' (R + Z) IO 9OR9
and therefore the total interaction energy is
We can express this result in tenns of € and CT by noting that CI 2 = 4€CT 12 and C6 = 4€CT 6 , for then
U
[I
= 8n €CT 3N
90 (CT) Ii 9 
I
T2 (CT)3 Ii J .
For the position of equilibrium, we look for the value of R for which dU / dR = 0,
Therefore, CT 9 / lOR 10 = CT 3
/4~ which implies that R
=
(~) 1/6CT
R ~ 1294 pm I. P25.25
djL I
=
 C2
(RT) ;; dVa
which implies that
~ = (  C2 RT ) dlnp
CT
x
(~) . dlnp
However, we established in Problem 25.24 that
djL'
RT Va
d Inp
CT
Therefore,
 C2
RT) x (_dV a ) __ RTVa , ( CT
Hence, din
V~2
dlnp
CT
or
c2 dlnVa =dlnp.
= dlnp , and therefore I Va = Clpl / Q I.
= 10 .858CT I. For CT = 342 pm,
PROCESSES AT SOLID SURFACES
P25.27
533
For association:
dR = konao(Req
h  R) were a
dt
= ao IS' constant,
dR
   = konaodt, Req R
loo
R
 dR  = Req  R
 In(Req  R) _ In (Req Req
lot konaodt = konaot 0
Ig = konaot
R) =
konaot ,
Req  R k ao    = e on t Req
= Req
R
1
R(t)
{ I  ekonaot} ,
= Req{l
 ekobst} where kobs
= konao I·
For dissociation: dR
 = dt
konaoR,
dR  = konaodt , R
l
R dR
Ii: = 
Req
In
P25.29
lot konaodt, 0
(~) = konaodt, Req
Let T/ osciUate between T/+ and T/ around a mean value T/O . Then T/  is large and positive (and T/+ > T/), ) ~ )oe(lct) ~f = )oe(l /2)~f
[a = 0.5]
and T/ varies as depicted in Fig. 25.12(a).
Figure 25.12a Therefore,) is a chain of increasing and decreasing exponential functions, ) =)oe(ry+ytlf!2 ex e t / r
534
STUDENT'S SOLUTIONS MANUAL
during the increasing phase of I) , where r
= 2RT Iy F ,
y is a constant, and
during the decreasing phase. This is depicted in Fig.2S.12(b).
Figure 25.12b
Solutions to applications P25.31
For the Langmuir and BET isotherm tests we draw up the following table (usingp* IS00 Torr) [Example 2S . I, Illustration 2S .3, and eqn 2S .S7b).
p i Torr
100
200
300
400
SOO
600
E. I (Torr cm 3 )
S.S9
6.06
6.38
6.S8
6.64
6.S7
67 4.01
133 4.66
200 5.32
267 5.98
333 6.64
400 7.30
V 103 z 103 zl« 1  z)( V l cm 3 »
200kPa
=
p l V is plotted against p in Fig. 25.13(a), and 103 zl ((l  z) V) is plotted against z in Fig. 2S .13(b).
'I
E u
...
0
1::
6.0
:::::; ::..
....
'"
100
200
300
400
pi Torr
500
600
Figure 25.13a
We see that the BET isotherm is a much better representation of the data than the Langmuir isotherm . The intercept in Fig. 25 . 13(b) is at 3.33 x 10 3 , and so 1l eVmon
= 3.33 x
1O 3 cm 3 . The slope of the
PROCESSES AT SOLID SURFACES
535
graph is 9.93, and so c 1 3 3   =9.93 x 10 em . cVmon
Therefore, c  1 = 2.98, and hence
Ie =
3.98
~ 1 Vmon =
75.4 cm3
1·
8
7 ~E
u
~ 6 N 5
""No 0
::
4
3
0. 1
0
0 .3
0.2
0.4
Figure 25.13b
z
P25.33
(a)    qYOC, RH=O
1 + bpyoc abcyoc
1 abcyoc
1 a
., =    + .
Parameters of regression fit:
or e
l/a
lIab
R
a
b/ppm
33.6 41.5 57.4 76.4 99
9.07 10.14 11.14 13.58 16.82
709.8 890.4 1599 2063 4012
0.9836 0.9746 0.9943 0.9981 0.9916
0.110 0.0986 0.0898 0.0736 0.0595
0.0128 0.0114 0.00697 0.00658 0.00419
I
The linear regression fit is generally good at all temperatures with 1R values in the range 0.975 to 0.991 I. (b)
Ina
= lnk. 
and lnb
!1.dH 1

= Inkb 
R
T
!1bH I
.
R T
Linear regression analysis of In a versus 1IT gives the intercept In k. and slope !1.dH IR while a similar statement can be made for a In b versus I IT plot. The temperature must be in Kelvin. For In a versus I IT lnk. = 5.605, standard deviation = 0.197, !1 .dH I R = 1043.2 K, standard deviation = 65.4 K, R = 0.9942 [good fit],
536
STUDENT'S SOLUTIONS MANUAL
= e 5 .605 = !3.68 x 1O3 !, /::;.adH = (8.31451 J K I molI)
ka
= !8.67 kJ molI
x (1043 .2K)
!.
For In b vers