Student Solutions Manual to accompany Principles of General Chemistry

Student Solutions Manual

to accompany

Prepared by Patricia Amateis and Martin Silberberg

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Student Solutions Manual to accompany

Principles of General Chemistry Martin S. Silberberg

Prepared by

Patricia Amateis Virginia Tech and

Martin S. Silberberg

_ Higher Education Boston

Burr Ridge, IL

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The McGraw-Hili Companies

•

Student Solutions Manual to accompany PRINCIPLES

OF GENERAL CHEMISTRY

PATRICIA AMATEIS AND MARTIN S. SILBERBERG Published by McGraw-Hili Higher Education, an imprint of The McGraw-Hili Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright

© 2007 by The McGraw-Hili Companies, Inc. All rights reserved.

No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hili Companies, Inc., including, but not limited to, network or other electronic storage or transmission, or broadcast for distance learning.

Recycled/acid free paper This book is printed on recycled, acid-free paper containing 10% postconsumer waste. 34

5 6 7 8 9 0 QPD/QPD 0 9 8 7

ISBN-13: 978-0-07-310721-9 ISBN-I 0: 0-07-310721-2

www.mhhe.com

Contents 1

1

Keys to the Study of Chemistry

2

The Components of Matter

4

The Major Classes of Chemical Reactions

3

11

5

Gases and the Kinetic-Molecular Theory

7

Quantum Theory and Atomic Structure

9

Models of Chemical Bonding

11

Theories of Covalent Bonding

6

8

10

12

26 52

Stoichiometry of Formulas and Equations

66

Thermochemistry: Energy Flow and Chemical Change

93

Electron Configuration and Chemical Periodicity The Shapes of Molecules

119

111

83

102

142

The Properties of Solutions

15

Organic Compounds and the Atomic Properties of Carbon

14 16

17

152

Intermolecular Forces: Liquids, Solids, and Phase Changes

13

161

Periodic Patterns in the Main-Group Elements: Bonding, Structure, and Reactivity Kinetics: Rates and Mechanisms of Chemical Reactions Equilibrium: The Extent of Chemical Reactions

221

184

207

18

Acid-Base Equilibria

20

Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions

19 21

22

23

239

Ionic Equilibria in Aqueous Systems

261

Electrochemistry: Chemical Change and Electrical Work

301

The Transition Elements and Their Coordination Compounds Nuclear Reactions and Their Applications

335

176

325

289

PREFACE Welcome To Your Student Solutions Manual

Your Student Solutions Manual (SSM ) includes detailed solutions for the Follow-up Problems and highlighted End­ of-Chapter Problems in Martin Silberberg's Principles of General Chemistry. You should use the SSM in your study of chemi stry as a study tool to: • better understand the reasoning behind problem solutions. The • •

plan-solution-check format illustrates the problem-solving thought process for the Follow-up problems and for selected End-of-Chapter problems. better understand the concepts through their applications in problems. Explanations and hints for problems are in response to questions from students in general chemistry courses. check your problem solutions. Solutions provide comments on the solution process as well as the answer.

To succeed in general chemistry you must develop skills in problem solving. Not only does this mean being able to fol low a solution path and reproduce it on your own, but also to analyze problems you have never seen before and develop a solution strategy. Chemistry problems are story problems that bring together chemistry concepts and mathematical reasoning. The analysis of new problems is the most challenging step for general chemistry students. You may face this in the initial few chapters or not until later in the year when the material is less familiar. When you find that you are having difficulty starting problems, do not be discouraged. This is an opportunity to learn new skills that will benefit you in future courses and your future career. The following two strategies, tested and found successful by many students, may help you develop the skills you need. The first strategy is to become aware of your own thought processes as you solve problems. As you solve a problem, make notes in the margin concerning your thoughts. Why are you doing each step and what questions do you ask yourself during the solution? After you complete the problem, review your notes and make an outline of the process you used in the solution whi le reviewing the reasoning behind the solution. It may be useful to write a paragraph describing the solution process you used. The second strategy helps you develop the ability to transfer a solution process to a new problem. After solving a problem, rewrite the problem to ask a different question. One way to rewrite the question is to ask the question backwards-find what has been given in the problem from the answer to the problem. Another approach is to change the conditions-for instance, ask yourself what if the temperature is higher or there is twice as much carbon dioxide present? A third method is to change the reaction or process taking place-what if the substance is melting instead of boiling? At times, you may find slight differences between your answer and the one in the SSM. Two reasons may account for the differences. First, SSM calculations do not round answers until the final step so this may impact the exact numerical answer. Note that in preliminary calculations extra significant figures are retained and shown in the intermediate answers. The second reason for discrepancies may be that your solution route was different from the one given in the SSM. Valid alternate paths exist for many problems, but the SSM does not have space to show all alternate solutions. So, trust your solution as long as the discrepancy with the SSM answer is small, and use the different sol ution route to understand the concepts used in the problem. Some problems have more than one correct answer. For example, if you are asked to name a metal, there are over 80 correct answers. You may also find slight differences between the answer in the textbook appendix and the one i.n the S S M . These differences occur when questions are open-ended, such as, "Give an example of an acid." The example given in the SSM may differ from the one in the Appendix, but both answers are correct. Patricia Amateis

Chapter 1 Keys to the Study of Chemistry Most numerical problems wil l have two answers, a "calculator" answer and a "true" answer. The calculator answer, as seen on a calculator, will have one or more additional significant figures. These extra digits are retained in all subsequent calculations to avoid intermediate rounding error. Rounding of a calculator answer to the correct number of significant figures gives the true ( final) answer. FOLLOW-UP PROBLEMS

1.1

Plan: The real question is "Does the substance change composition or just change form?" Solution: a) Both the solid and the vapor are iodine, so this must be a physical change. b) The burning of the gasoline fumes produces energy and products that are different gases. This is a chemical change. c) The scab forms due to a chemical change.

1 .2

Plan: We need to know the total area supplied by the bolts of fabric, and the area required for each chair. These two areas need to be in the same units. The area units can be either ft2 or m2 . The conversion of one to the other will use the conversion given in the problem : 1 m 3 .28 1 ft. Solution : 200 m 2 (3 .28 1 ft) 2 I chair 205 .047 205 chairs (3 bOltS) I bolt 3 1 .5 ft 2 (l m) 2 Check: 200 chairs require about 200 x 32 / 1 0 640 m2 of fabric. (Round 3 1 .5 to 32, and 1 / (3.28 1 ) 2 to 1 / 1 0) Three bolts contain 3 x 200. = 600. m2 .

1 .3

(J

J(

(

=

=

]=

=

P lan: The volume of the ribosome must be determined using the equation given in the problem. This volume may then be converted to the other units requested in the problem. Solution : 3 V i 1tf 3 i ( 3 . 1 4 1 59 ) 2 1 .4 nm 5 1 3 1 .45 nm3 3 3 2 ( 1 0- 9 m) 3 ( 1 dm) 3 (5 1 3 1 .45 nm 3 ) 5. 1 3 1 45 X 1 0-2 1 = 5.13 X 1 0-2 1 dm3 3 3 ( I nm) (0. 1 m) =

=

]= ( )( ) = ( )( 6 ] =

(

l ,l1L 1L 5 . 1 3 1 45 X 1 0- 15 5.13 X ( I dm) 3 1 0 L Check: The magnitudes of the answers are reasonable. The units also agree.

(5 . 1 3 1 45 x 1 0- 21 dm 3 )

-

1 .4

=

10-15 ilL

Plan: The time is given in hours and the rate of delivery is in drops per second. A conversion(s) relating seconds to hours is needed. This will give the total number of drops, which may be combi ned with their mass to get the total mass. The mg of drops will then be changed to kilograms. The other conversions are given in the inside back cover of the book. Solution : 60 � I .5 droPS 65 mg I O-3 g 1 kg 60 min 2.808 2.8 kg 8.0 h 1h 1 mm 1s 1 drop 1 mg 1 0 3 g Check: Estimating the answer - (8 h) (3600 s / h) (2 drops / s) (0.070 kg/ I 0 3 drop) 4 kg. The final units in the calculation are the desired units.

(

J( J(

J( J[ )[ ) = =

=

1 .5

Plan: The volume unit may be factored away by multiplying by the density. Then it is simply a matter of changing grams to kilograms. Solution: 7.5 g 4.6 cm 3 � 0.0345 0.034 kg cm 3 1 000 g Check: 5 x 7 1 1 000 0.035, and the calculated units are correct.

( J(

=

1 .6

J

=

=

Plan: Using the relationship between the Kelvin and Celsius scales, change the Kelvin temperature to the Celsius temperature. Then convert the Celsius temperature to the Fahrenheit value using the relationship between these two scales. T (in°C) T (in K) - 273 . 1 5 =

Solution: T (in°C) =234 K - 273 . 1 5 -39. 1 5 =

=

-39°C

T (in O F) � (-39. 1 5°C) + 32 -38 .47 -38°F 5 Check: S ince the Kelvin temperature is below 273, the Celsius temperature must be negative. The low Celsius value gives a negative Fahrenheit value. =

=

=

1 .7

P lan: Determine the significant figures by determining the digits present, and accounting for the zeros. Only zeros between non-zero digits and zeros on the right, if there is a decimal point, are significant. The units make no difference. Solution: a) 3 1 .Q7Q mg; five significant figures b) 0.06Q6Q g; four significant figures c) 85Q.oC; three significant figures - note the decimal point that makes the zero significant. d) 2.000 x 1 02 mL; four significant figures e) 3 .9 x 1 0-{i m; two significant figures - note that none of the zeros are significant. f) 4.Q l x 1 0-4 L; three significant figures Check: All significant zeros must come after a significant digit.

1 .8

P lan: Use the rules presented in the text. Add the two values in the numerator before dividing. The time conversion is an exact conversion, and, therefore, does not affect the significant figures in the answer. The addition of 25 .65 and 37.4 gives an answer where the last significant figure is the one after the decimal point (giving three significant figures total). When a four significant figure number divides a three significant figure number, the answer must round to three significant figures. An exact number ( l min 1 60 s) wil l have no bearing on the number of significant figures. Solution: 25.65 mL + 37.4 mL 5 1 .434 5 1.4 mL/min 1 min 73.55 s 60 s Check: (25 + 35) 1 (5 1 4) (60)4/5 48. The approximated value checks well.

(

=

J

=

=

END-OF-CHAPTER PROBLEMS

1 .2

Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples. Solution: Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have a definite volume. The volume of the container does not affect the volume of a solid or liquid. a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of a balloon. Helium is a gas. b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury indicates the temperature.

2

c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is possible that solid particles of food will be present. 1 .3

Plan: Define the terms and apply these definitions to the examples. Solution: Physical property A characteristic shown by a substance itself, without interacting with or changing into other substances. Chemical property A characteristic of a substance that appears as it interacts with, or transforms into, other substances. a) The change in color (yellow-green and silvery to white), and the change in physical state (gas and metal to crystals) are examples of physical properties. The change in the physical properties indicates that a chemical change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an example of a chemical property. b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with magnetism are physical properties. No chemical changes took place, so there are no chemical properties to observe.

1 .5

Plan: Apply the definitions of chemical and physical changes to the examples. Solution: a) Not a chemical change, but a physical change simply cooling returns the soup to its original form. b) There is a chemical change cooling the toast will not "un-toast" the bread. c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus this is a physical change, and not a chemical change. d) This is a chemical change converting the wood (and air) into different substances with different compositions. The wood cannot be "unburned."

-

-

-

1 .7

Plan: A system has a higher potential energy before the energy is released (used). Solution: a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel bums. The fuel has a higher potential energy. b) Wood, like the fuel, is higher in energy by the amount released as the wood bums.

1.11

P lan: Re-read the section in the Chapter on experimental design (Experiment step of the scientific method). Solution: A well-designed experiment must have the following essential features. I) There must be two variables that are expected to be related. 2) There must be a way to control all the variables, so that only one at a time may be changed while keeping all others constant. 3) The results must be reproducible.

1 . 14

P lan: Review the table of conversions in the Chapter or inside the back cover of the book. Solution: a) To convert from .In 2 to cm 2 , use

(2.54 cm) 2 (lin) 2

.

(1000 m) 2

b) To convert from km2 to m2 , use. -'----'--2 (1 km)

c) This problem requires two conversion factors: one for distance and one for time. It does not matter which conversion is done first. Alternate methods may be used. To convert distance, mi to m, use: 1 .609 1 000 m 1.609 X 103 mimi l km 1 m)

( �)(

)

=

3

(

)( )

To convert time, h to s, use: 1h 1 min __ _ 60 min 60 s

=

1 h / 3600 s

(

. f: . 1 .609 X l O m Therefore, the complete conversion actor IS 1 ml when you start with a measurement of mi/h? .

1000 g 2.20 51b

d) To convert from pounds (lb) to grams (g), use use

(

(1 ft) 3 ( 1 2 in) 3

)(

( 1 in) 3 (2.54 cm) 3

)

=

)( ) Ih 3600 s

--

=

0.4469 m h ' . Do the UnIts cance I mi s

; to convert volume from ft3 to cm3

3.531 x 10-5 fe/cm3

1.16

Plan: Review the definitions of extensive and intensive properties. Solution: An extensive property depends on the amount of material present. An intensive property i s the same regardless of how much material is present. a) Mass is an extensive property. Changing the amount of material will change the mass. b) Density is an intensive property. Changing the amount of material changes both the mass and the volume, but the ratio (density) remains fixed. c) Volume is an extensive property. Changing the amount of material will change the size (volume). d) The melting point is an intensive property. The melting point depends on the identity of the substance, not on the amount of substance.

1.18

Plan: Anything that increases the mass or decreases the volume will increase the density ( density

=

mass ) volume

Solution: a) Density increases. The mass of the chlorine gas is not changed, but its volume is smaller. b) Density remains the same. Neither the mass nor the volume of the solid has changed. c) Density decreases. Water is one of the few substances that expands on freezing. The mass is constant, but the volume increases. d) Density increases. Iron, like most materials, contracts on cooling, thus the volume decreases whi le the mass does not change. e) Density remains the same. The water does not alter either the mass or the volume of the diamond. 1 .2 1

Plan : Use conversion factors from the inside back cover: 1 0-12 m Solution: 2 1 O- 1 m I : 1.43 nm Radius 1 430 pm 1 pm 10 m

[

=

1 .23

)( )

=

1 pm; 1 0-9 m

=

a) Plan: Use conversion factors: (0.0 1 m? ( 1 cm) 2 ; ( I km) 2 ( 1 000 m)2 Solution: (I km) 2 (0.0 I m) 2 1.77 X 10-9 km2 1 7 .7 cm 2 ( l cm) 2 (J000 m) 2 b) Plan: Use conversion factor: ( l inch)2 (2.54 cm) 2 Solution: $3.2 5 (J in) 2 = 8.9 1 639 =$8.92 1 7.7 cm 2 (2.54 cm) 2 l in 2 =

(

=

)

)(

=

=

[

)( )

4

=

1 nm

1.25

Plan: Use the relationships from the inside back cover of the book. The conversions may be performed in any order. Solution: 5.52 g (l cm)3 a) = 5.52 X 103 kg/m3 cm3 (0.0] m) ' 1 000 g

( )(

b) l .27

( )( 5.52 g cm3

)(� )

(2.54 Cm)3 (l in)3

)(

)( J(

J

(1 2 in)3 � 2.205 Ib = 344.66 1 = 3451b/ft3 I kg ( 1 ft)3 1 000 g

Plan: The conversions may be done in any order. Use the definitions of the various SI prefixes. Solution: (l mm)3 1 . 72f.1Jl13 ( I X l O-6m)3 9 a) Volume = , = 1.72 X 10- mm3 /eell ' cell (l f.1Jl1)3 (I X 1 0 3 m) l .72f.1Jl13 ( 1 x 1 0 -6m)3 1L = l .72 X 1 0-10 = 10-10 L b) Volume = (l Os cell i \ cell ( I X 1 0 -3 m (l f.1Jl1)3

(

l .29

,

)(

)(

)(

)

)(

)

Plan: The mass of the contents is the mass of the full container minus the mass of the empty container. The volume comes from the mass density relationship. Reversing this process gives the answer to part b. Solution: a) Mass of mercury = 1 85 . 5 6 g - 5 5 .32 g = 1 30.24 g

( )

Volume of mercury = volume of vial = (1 30.24 g ) � = 9.6260]6 = 9.626 em3 1 3 .53 g

(

)

0.997 g = 9.597 1 4 g water l cm3 Mass of vial filled with water 5 5 .32 g + 9.597 1 4 g 64.9 1 7 1 4

b) Mass of water = (9.6260 1 6 cm3 ) =

1 .3 1

=

=

64.92 g

Plan: Volume of a cube = (length of side) 3 Solution: The value 1 5.6 means this is a three significant figure problem. The final answer, and no other, is rounded to the correct number of significant figures. 1 m 1 O -3 m 1 5.6 mm � = 1 .56 cm (convert to cm to match density unit) I mm 1 0 2 m Al cube volume = ( 1 .56 cm) 3 = 3 .7964 cm3 1 0.25 g mass = 2.69993 = 2.70 g/em3 density volume 3 .7964 cm3

[ )( )

=

l .33

=

Plan: Use the equations given in the text for converting between the three temperature scales. Solution: a) T (in 0c) = [T (in O F ) - 32] � = (72°F - 32) � = 22.222 = 22°C 9 9 T (in K) = T (in 0c) + 273. 1 5 = 22.222°C + 273 . 1 5 = 295 .372 = 295 K b) T (in K) = T (in 0c) + 273 . 1 5 = - I 64°C + 273 . 1 5 = 1 09. 1 5 = 109 K T (in O F) = 2. T (in 0c) + 32 = 2. (- 1 64°C) + 32 = -263 .2 =-263°F 5 5 c) T (in 0c) = T (in K) -273 . 1 5 = 0 K 273 . 1 5 = -273 . 1 5 = -273°C -

T (in O F)

=

2. T (in 0c) + 32 = 2. (-273 . 1 5°C) + 32 = - 459.67 = 5

5

5

-

460.oF

1 .36

2 Plan: Use 1 0-9 m = I nm to convert wavelength. Use 0.0 1 A = 1 pm, 1 0- 1 m = 1 pm, and 1 0-9 m = 1 nm Solution: 7 1 0-9 m a) 255 nm = 2.55 x 10- m I nm 0.0 1 A 1 pm I O-9 m = 6830 A b) 683 nm 1 pm 1 nm 1. 0- 12 m

[ ) [ )[ --

)( )

1 .4 1

Plan: Review the rules for significant figures. Solution: Initial or leading zeros are never significant; internal zeros (occurring between non-zero digits) are always significant; terminal zeros to the right of a decimal point are significant; terminal zeros to the left of a decimal point are significant only if they were measured.

1 .42

Plan: Review the rules for significant zeros. Solution: a) No significant zeros (leading zeros are not significant) b) No significant zeros (leading zeros are not significant) c) 0.039Q (terminal zeros to the right of the decimal point are significant) d) 3 .Q900 x 1 04 (zeros between nonzero digits are significant; terminal zeros to the right of the decimal point are significant)

1 .44

Plan: Use a calculator to obtain an initial value. Use the rules for significant figures and rounding to get the final answer. Solution: ( 2.795 m )( 3 . 1 0 m ) = 1 .337 1 = 1.34 m (maximum of 3 significant figures allowed) a) 6.48 m b) V = c)

(�)

1t

( 9.282 cm )3 = 3.34976 x 1 0 3 = 3.350

X

103 em3

(maximum of 4 significant figures allowed)

1 . 1 1 0 cm + 1 7.3 cm + 1 08.2 cm + 3 1 6 cm = 442.61 = 443 em (no digits allowed to the right of the decimal since 3 1 6 has no digits to the right of the decimal point)

1 .46

Plan: Review the procedure for changing a number to scientific notation. Solution: a) 1.310000 x 105 (Note that all zeros are significant.) b) 4.7 x 10-4 (No zeros are significant.) c) 2.10006 x 105 d) 2.1605 X 103

1 .48

Plan: Review the examples for changing a number from scientific notation to standard notation. Solution: a) 5550 (Do not use terminal decimal point since the zero is not significant.) b) 10070. (Use terminal decimal point since final zero is significant.) c) 0.000000885 d) 0.003004

1 .5 0

Plan: Calculate a temporary by simply entering the numbers into a calculator. Then you will need to round the value to the appropriate number of significant figures. Cancel units as you would cancel numbers, and place the remaining units after your numerical answer.

6

Solution : ( 6.626 x 1 0-34 JS) ( 2 .9979 x 1 08 mls) 4.062 1 8 x 1 0-19 a) 9 X 489 1 0- m 9 4.06 x 10-19 J (489 X 1 0- m limits the answer to 3 significant figures; units of m and s cancel)

( 6.022

2 2 1 0 3 molecules/mol ) ( 1 . 1 9 x 1 0 g) = 1 .5555 X 1 024 b) 46.07 g/mol 24 2 1.56 x 10 molecules ( 1 . 1 9 x 1 0 g limits answer to 3 significant figures; units of mol and g cancel) 2 1 1 c) ( 6.022 x 1 0 3 atoms/mol)( 2. 1 8 x 1 0- 1 8 J/atom ) 2 2 = 1 . 82333 x 1 0 5 2 3 1.82 x lOs J/mol (2. 1 8 x 1 0-18 J/atom limits answer to 3 significant figures; unit of atoms cancels) X

( - )

1 .52

Plan: Exact numbers are those that have no uncertainty. Unit definitions and number counts of items in a group are examples of exact numbers. Solution : a) The height of Angel Fal ls is a measured quantity. This is not an exact number. b) The number of planets in the solar system is a number count. This � an exact number. c) The number of grams in a pound is not a unit definition. This is not an exact number. d) The number of millimeters in a meter is a definition of the prefix "milli-." This � an exact number.

1 .54

P lan: Observe the figure, and estimate a reading the best you can. Solution : The scale markings are 0.2 cm apart. The end of the metal strip falls between the mark for 7.4 cm and 7.6 cm. If we assume that one can divide the space between markings into fourths, the uncertainty is one-fourth the separation between the marks. Thus, since the end of the metal strip fal ls between 7.45 and 7.55 we can report its length as 7.50 ± 0.05 em. (Note: If the assumption is that one can divide the space between markings into halves only, then the result is 7.5 ± 0. 1 cm.)

1 .56

P lan: Calculate the average of each data set. Remember that accuracy refers to how close a measurement is to the actual or true value while precision refers to how close multiple measurements are to each other. Solution : 8.72 g + 8 . 7 4 g + 8.70 g = 8.7200 = 8.72 g Iavg = a) 3 8 .56 g + 8 .77 g + 8.83 g = 8.7200 = 8.72 g Ilavg = 3 8.50 g + 8.48 g + 8.5 1 g lIIavg = = 8.4967 = 8.50 g 3 8.4 1 g + 8.72 g + 8.55 g = 8.5600 = 8.56 g I Vavg = 3 Sets I and II are most accurate since their average value, 8.72 g, is closest to the true value, 8.72 g. b) To get an idea of precision, calculate the range of each set of values: largest value - smallest value. A small range is an indication of good precision since the values are close to each other. lrange = 8.74 g - 8.70 g = 0.040 g IIrange = 8.83 g - 8.56 g = 0.27 g IIIrange = 8.5 1 g - 8.48 g = 0.030 g IVrange= 8.72 g - 8.4 1 g = 0.3 1 g Set III is the most precise (smallest range), but is the least accurate (the average is the farthest from the actual value). c) Set I has the best combination of high accuracy (average value = actual value) and high precision (relatively small range).

7

d) largest range). has both low accuracy (average value differs from actual value) and low precision (has the 1. 5 9 SolPlan:utiIfon:it is necessary to force something to happen, the potential energy wil be higher. b) a) Set

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=

=

Note: The densities are additive because the volume of the ball and the volume of the H2 gas are the same: 0.0899 + 0.2 1 glLb = 0.30 giL. d) Because the density of O 2 is greater than that of air, a ball filled with O2 will sink. e) Density of ball filled with nitrogen: 0.2 1 giL + 1 . 1 65 giL = 1 .3 8 giL. Ball will sink because density of ball fi lled with nitrogen is greater than density of air. g .0560 L 0.665 84 g For ball filled with f) To sink, the total mass of the ball and gas must weigh

· �� )(

)

C

=

hydrogen: 0.665 84 - 0. 1 7 g = 0.4958 = 0.50 g. More than 0.50 g would have to be added to make the ball sink. 1 .67

P lan: Use the surface area and the depth to determine the volume. The volume may then be converted to liters, and finally to the mass using the giL. Once the mass of the gold is known, other conversions stem from it. Solution: 1 000 IL 5 . 8 X I O - 9g a) (3800 m)(3.63 x 1 08 km 2 ) = 8.00052 X 1 0 12 = 8 .0 x 1 0 1 2 g 3 3 (l km) 1 0- m b) Use the density of gold to convert mass of gold to volume: 0.0 1 m) 3 1 cm 3 (8.00052 x 1 01 2 g) = 4. 1 4535 x 1 0 5 = 4.1 X 1 0 5 m 3 1 9.3 g ( l cm) 3

(

c) ( 8.00052 1 .69

x

m( )(

( )( ) )( )( )

2 1 01 g

1 tr. O Z . 31.1 g

)( L )

$3 70.00 = 9.5 1 830 x 1 0 13 = $9.5 X I tr. oz.

10

13

Plan: Use the equations for temperature conversion given in the chapter. The mass and density wil l then give the volume. Solution : a) T (in 0c) = T (in K) - 273 . 1 5 = 77.36 K - 273 . 1 5 = -1 95.79°C

b) T (in OF ) = � T (i n 0c) + 32 = � (- I 95.79 °C) + 32 = -320.422 = -320.42°F 5 5 c) The mass of nitrogen is conserved, meaning that the mass of the nitrogen gas equals the mass of the liquid nitrogen. We first find the mass of liquid nitrogen and then convert that quantity to volume using the density of the liquid. g 4. M ass of liquid nitrogen = mass of gaseous nitrogen = (895 .0 L = 4086.57 g N 2

{ �� )

( )

Volume of liquid N 2 = (4086.57 g) � = 5 .05 1 4 = 5.05 L 809 g 1 .72

Plan: Determine the volume of a particle, and then convert the volume to a convenient unit (cm 3 in this case). Use the density and volume of the particles to determine the mass. In a separate set of calculations, determine the mass of all the particles in the room or in one breath. Use the total mass of particles and the individual mass of the particles to determine the number of particles. Solution: 3 4 4 2.5/lm V (�m 3 ) = 31rr 3 = 3 1t -- = 8. 1 8 1 2 = 8.2 /lm 3 2 ( l cm) 3 V (cm3 ) = 8 . 1 8 1 2 �m 3 = 8. 1 8 1 2 X 1O-12 cm 3 ( 1 0 4 ,lJ111) 3

()

() ( ) ( ) ( )

Mass (g) = 8. 1 8 1 2 X 1 0- 1 2 cm 3

2.5 g = 2.045 cm 3

Calculate the volume of the room in m3 : Volumeroom = 1 0.0 ft x 8.25 ft x 1 2.5 ft = 1 .03 1 9

x

x

1 0-11 = 2.0 x 1 0- 1 1 g each microparticle 1 0 3 fe

0-2 m)3)=2. 9 1 95 x 101 m3 3 )( 0 (1 .031X 103 ft3)( (112in)ft)33)( 2. 5 4cm) in)3 cm)3 (2. 9 195 x 101 m3)( 50.1 m3Jig )( 10I --{)Jigg)= 1.460 x 10-3= 1. 5 x 10-3 g for all the micropartic1es in the room icle ) Number of microparticles In room = 1.460 x 10- g [2.I mi045cropart x 10- g =7.1394X 107= -6 ( 50 10 m ( 3 ( ) ) Mass 0. 5 00 I L m --Il gg) = 2. 5 x 10-8 g in one O. 5 00- L breath c 1 e Number of micropartic1es I.n one breath = 2. 5 x 10-8 g ( 12.mi045croparti x 10- g J = 1.222X 103= 1.73 andPlan:densiDetteyrmialonnge thwie tthotconversi al mass ofonsthfrom e Eartthhe'sincrustsideinbackmetcover. ric tonsThe(t) bymasscombiof each ningitnhdie vdeptidualh, elsurface area e ment comes from t h e concent r at i o n of t h at element mul t i p l i e d by t h e mass of t h e crust. Solution: 2. 8 g )(� J( It J=4. 998X 1019t Ooo m)J 3)( (0.(t0cm)Im)33)( Icm3 (35 km)(5.10x 108 m )( ( (lkm) 1000g 1000 kg (4.(4.(4.999998988 xxx 101101101999t)t)t) (2.(4.x752510-Xx 4105105g Rutheni gg Oxygen Siliconum t)t)=t)==12..4.32595974198 Xx 1025101025=== 1. 7 5 SolPlan:utiIonn:visualizing the problem, the two scales can be set next to each other. There are 50 divisions between the freezing point and boiling point of benzene on the ox scale and 74.6 divisions (80.1°e - 5SC) on the °e scale. So ox= ( --74.5060°eX J °e This does not account for the offset of5. 5 divisions in the °e scale from the zero point on the ox scale. So ox= ( 74.5060°eX J cae - 5. 5°C) eheck: Plug in 80.1 °e and see if result agrees with expected value of 50oX. So oX= ( 74.50°6°eX J (80.1 °e - 5. 5 °C)=500X Use this formula to find the freezing and boiling points of water on the oX scale. = ( 74.5060°eX J (o.oooe - 5. 5°C)= oX = ( 500X ) (lOo.ooe 5. 5 °C) = 74.6°e (l

(I

3

.

7. 1

=

1

L

0 -3

X

107

II

microparticIes in the room

� 3

11

1 .2

k

1 0 3 microparticIes in a breath

l

2

/

(I

FPwater OX

BPwate r

X

/

X

/

-3.7°X

-

63.3°X

10

IS

2.3 X 1 025 g Oxygen 1 .4 X 1 025 g Silicon 1 5 X 1 0 5 g Ruthenium (and Rhodium)

Chapter 2 The Components of Matter FOLLOW-UP PROBLEMS

2. 1

Plan: The mass fraction of uranium in pitchblende was determined in the problem as (71.4/84.2 = 0.84798). The remainder of the sample must be oxygen (1.00000 - 0.84798 = 0.15202). The mass fraction of the uranium will be used to determine the total mass of pitchblende. The total mass of pitchblende and the mass fraction of oxygen may then be used to determine the mass of oxygen. Solution: . Mass 0 f PltC · hblende = 2 . 3 t Uramum

(

1.00 t Pitchblende

0.84798 t uranium

(

)

. hblende = 2 . 7 1 23 = 2.7 t PltC

)

. 0. 1 5202 t oxygen = 0.4 1 23 = 0.41 t oxygen Mass of oxygen = 2.7123 t PItchblende 1 .00 t pitchblende Check: Adding the amount of oxygen calculated to the mass of uranium gives the calculated 2.7 t of pitchblende. 2.2

Plan: The subscript (Atomic Number = Z) gives the number of protons, and for an atom, the number of electrons. The Atomic Number identifies the element. The superscript gives the mass number (A) which is the total of the protons plus neutrons. The number of neutrons is simply the Mass Number minus the Atomic Number (A - Z). Solution: a) Z= 5 and A = II, there are 5 p+ and 5 e- and II - 5 = 6 nO Atomic number = 5 = B. b) Z= 20 and A = 41, there are 20 p+ and 20 e- and 41 - 20 = 2 1 nO Atomic number = 20 = Ca. c) Z= 53 and A = 131, there are 53 p+ and 53 e- and 1 3 1 - 53 = 78 nO Atomic number = 53 = I.

2.3

Plan: To find the percent abundance of each B isotope, let x equal the fractional abundance of lOB and (1 - x) equal the fractional abundance of liB. Remember that atomic mass = isotopic mass of lOB x fractional abundance) + (isotopic mass of 11B x fractional abundance). Solution: 1 Atomic Mass = (loB mass) (fractional abundance of lOB) + ( 18 mass) (fractional abundance of liB) Amount of lOB + Amount liB = I (setting lOB = x gives liB = I - x) 1 0. 8 1 amu = ( 1 0.0129 amu)(x) + ( 1 1 .0093 amu) (1 - x) 10. 8 1 amu = 11.0093 - 11.0093x + 10.0 1 29 x 10.8 1 amu = 1 1 .0093 - 0.9964 x -0.1993 = - 0.9964x x = 0.20; 1 - x = 0.80 ( 1 0.8 1 - 1 1 .0093 limits the answer to 2 significant figures) Fraction x 1 00% = percent abundance. 1 l % abundance of OB = 20. %; % abundance of 1B = 80. % 11 Check: The atomic mass of B is 1 0.81 (close to the mass of B, thus the sample must be mostly this isotope.)

2.4

Plan: Locate these elements on the periodic table and predict what ions they will form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Or, relate the element's position t o the nearest noble gas. Elements after a noble gas lose electrons t o become positive ions, while those before a noble gas gain electrons to become negative ions. Solution: a) 16S2- [Group 6A(l6); 6 - 8 = -2]; sulfur needs to gain 2 electrons to match the number of electrons in 18Ar. b) 37Rb+ [Group IA(I)]; rubidium needs to lose I electron to match the number of electrons in 36Kr. c) 56Ba2+ ]Group 2A(2)]; barium needs to lose 2 electrons to match the number of electrons in 54Xe.

11

2.5

e second namewhile the l anditshunchanged, thtoe tfihrestperinameodibelc taoblngse. Theto thmete metal aname ing with Iifonithcerebinisaryanycompounds, Plaon:ngsWhento thdeale nonmetal. refer , doubt belnonmet to the nonmetal root. Sola) ution:ails hasin an -ide suffix added andand fromfromoxibromi de, is dine, is in b)c) is iins in and and fromfromchlsulorifidde,e, iissiinn d) is in PlSolan:utiUseon: the charges to predict the lowest ratio leading to a neutral compound. a) Zinc shouldsoform Zns an2+ accept and oxygen shoulda.form 02-, these wil combine to give The charges cancel t h i s i a bl e formul b) Silversoshoulthisdisforman accept Ag+ andablebromi ne should form Br-, these wil combine to give AgBr. The charges cancel formula. c) Lithiumso should formacceptLi+aandble chlformula. orine should form cr, these will combine to give The charges cancel t h i s i s an d)ThiAluminum should form A13+ and sulfur should formsoSt2h-,istoisproduce al combia. nation the formula is s way the charges wil cancel an accepta neut ablerformul Plnamean: Detor formula. ermine thThee names ororsymbol seofionseachalwofaysthegospecifirste.sRevi presentew .thThen combi ne the piaetcesuretcovered o producein athe met a l posi t i v e rul e s for nomencl For metofalsthliekmete manyal iotnransiis intdiiocnatmeted abyls,athRoman at can form eachimwimedi th a aditefferent tchapt hmete iaoleni'r.scname. charge numeralmorewitthhanin oneparentionheses ly follocharge, wing the SolTheution:Roman numerals mean that the lead is Pb4+ and oxygen produces the usual 02-. The neutral a)combi a istwo copper ions, each of which must be I. This is one of the b)twoSulcommon finatde,ionikicharges es oxide, foris copper Thissoiisons.spltheTheitformul betwIeencharge on thcuprous e coppersulisfiide).ndicated with a Roman numeral. Thic) Bromide, s gives thleikname ( c ommon name eofothertoelements in thout.e sameThus,coltuhmne irofonthmuste peribeodic (tianbldiec, atforms ath-Ia Roman ion. Twonumeral of these). Thiionss requi r es a t o t a l cancel them e d wi is onedofe).the two common charges on iron ions. This gives the name (or ferrous bromi d) The mercuric ion is Hg2+, and two ions (Cn are needed to cancel the charge. This gives the formula n: DetermiThenemettheanames ortisymbol salofweach offithrste .species present. Then combine the pieces to produce a name orSolPlaformula. l or posi v e i o ns ays go utiocupri n: c ion, Cu2+, requires two nitrate ions, N0 -, to cancel the charges. Trihydrate means three water The 3 mol e cul e s. These combi n e t o gi v e: The zinconlion,y forms Zn2+, requi res+ itown,osohydroxi denumeral ions, s aretounnecessary. cancel the charges. Thesede icombi ne thaso gitvhe:e appropriate c)b)charge. Lithium t h e Li Roman The cyani o n, These combine to give n: DetermiThene mettheanames ortisymbol salofweach offithrst.e speci escorrect presentio. nsThenaccordi combinglney.the pieces to produce a name orSolPlaformula. l or posi v e i o ns ays go Make n: ion is NH/ and the phosphate ion is P0 3-. To give a neutral compound they should combine a) Theutioammonium 4 t o gi v e t h e correct formul a b) Aluminum givtoesgiA1ve3+thande correct the hydroxi dea ion is Parent To gihvesese a neutare rrequi al compound theytheshoul d ocombi nn.e formul r ed around pol y at mi c i o c)numeral Manganese is Mn, and Mg,Theinotthhere formul a, is magnesium. Magnesi uhydrogen m only formscarbonat the Mge (or2+ ibion,carbonat so Romane) ion. s are unnecessary. i o n i s HC0 -, whi c h i s cal l e d t h e 3 or The correct name is Zinc Group 2B(1 2) oxygen, Silver Group lB(ll) bromine, Lithium Group IA(I) chlorine, Aluminum Group 3A(13) sulfur,

2.6

Group 6A(16). Group 7A(17). Group 7A(17). Group 6A(16).

ZnO.

(+2 -2 = 0),

(+1

-1 = 0),

LiCl.

-I

(+1

= 0),

AhS3 '

[2(+3) + 3(-2) = 0],

2.7

I

[+4 + 2(-2) = 0], -2.

Pb02.

+

+

copper(I) sulfide

=

+2

+2

iron(II) bromide

-1

2.8

HgCh.

a)

CU(N03)2·3H20.

OH-,

Zn(OHh

CN",

lithium cyanide.

2.9

[3(+1) + (-3) = 0]

(NH4 )3 P04 .

OH-.

[+3 + 3(-1) =0]

AI(OHh

magnesium hydrogen carbonate

12

magnesium bicarbonate.

d) Either use the "-ic" suffix or the "(1Il)" but not both. Nitride is N 3-, and nitrate is N0 3 -. This gives the correct name: chromium(III) nitrate (the common name is chromic nitrate). e) Cadmium is Cd, and Ca, in the formula, is calcium. Nitrate is N0 3 -, and nitrite is N02-. The correct name is

calcium nitrite. 2.10

Plan: Determine the names or symbols of each of the species present. The number of hydrogen atoms equals the charge on the anion. Then combine the pieces to produce a name or formula. The hydrogen always goes first. For the oxoanions, the -ate suffix changes to -ic acid and the -ite suffix changes to -ous acid. Solution: a) Chloric acid is derived from the chlorate ion, CI0 3-. The -I charge on the ion requires one hydrogen. These combine to give the formula: HC103. b) As a binary acid, HF requires a "hydro-" prefix and an "ic" suffix on the "fluor" root. These combine to give the name: hydrofluoric acid. c) Acetic acid is derived from the acetate ion, which may be written as CH 3 COO- or as C2H 3 02-. The - I charge means that one H is needed. These combine to give the formula: CH3 COOH or H C 2 H 3 0 2• d) Sulfurous acid is derived from the sulfite ion, SO/-. The -2 charge on the ion requires two hydrogen atoms. These combine to give the formula: H 2 S0 3• e) H BrO is an oxoacid containing the BrO- ion (hypobromite ion). To name the acid, the "-ite" must be replaced with "-ous." This gives the name: hypobromous acid.

2. 11

Plan: Determine the names or symbols of each of the species present. The number of atoms leads to prefixes, and the prefixes lead to the number of that type of atom. Solution: a) Sulfur trioxide one sulfur and three (tri) oxygens, as oxide, are present. b) Silicon dioxide one silicon and two (di) oxygens, as oxide, are present. c) N 2 0 Nitrogen has the prefix "di" 2, and oxygen has the prefix "mono" 1 (understood in the formula). d) SeF6 Selenium has no prefix (understood as 1), and the fluoride has the prefix "hexa" 6. -

-

=

=

=

=

2.12

Plan: Determine the names or symbols of each of the species present. For compounds between nonmetals, the number of atoms of each type is indicated by a prefix. Solution : a) Suffixes are not used in the common names of the nonmetal listed first in the formula. Sulfur does not qualify for the use of a suffix. Chlorine correctly has an "ide" suffix. There are two of each nonmetal atom, so both names require a "di" prefix. This gives the name : disulfur dichloride. b) Both elements are nonmetals, and there is more than one nitrogen. Two nitrogens require a "di" prefix; the one oxygen could use an optional "mono" prefix. These combine to give the name dinitrogen monoxide or dinitrogen oxide.

c) The heavier Br should be named first. The three chlorides are correctly named. The correct name is bromine trichloride. 2.13

Plan: First, write a formula to match the name. Next, multiply the number of each type of atom by the atomic mass of that atom. Sum all the masses to get an overall mass. Solution : 2 a) The peroxide ion is 02 -, which requires two hydrogen atoms to cancel the charge: H 2 0 2. Molecular mass (2 x 1. 008 amu) + (2 x 16.00 amu) 34. 016 34.02 amu. formula mass 132.9 amu +35.45 amu 168. 35 1 68.4 amu. b) A Cs + 1 ion requires one C I- 1 , ion to give: CsCI; 2 c) Sulfuric acid contains the sulfate ion, S04 -, which requires two hydrogen atoms to cancel the charge: H 2 S04; molecular mass (2 x 1.008 amu) + 32 . 07 amu + (4 x 16.00 amu) 98.086 98.09 amu. d) The sulfate ion, SO/-, requires two + I potassi urn ions, K+, to give K 2 S04; formula mass (2 x 39.10 amu) + 32 . 07 amu + (4 x 16.00 amu) 1 74.27 amu. =

=

=

=

=

=

=

=

=

=

=

13

2. 1 4

Plan: Since the compounds only contain two elements, finding the formulas involve simply counting each type of atom and developing a ratio. Solution: a)There are two brown atoms (sodium) for every red (oxygen). The compound contains a metal with a nonmetal. Thus, the compound is sodium oxide, with the formula Na20. The formula mass is twice the mass of sodium plus the mass of oxygen: 2 (22.99 amu) + ( 1 6.00 amu) 61 .98 amu. b) There is one blue (nitrogen) and two reds (oxygen) in each molecule. The compound only contains nonmetals. Thus, the compound is nitrogen dioxide, with the formula N02. The molecular mass is the mass of nitrogen plus twice the mass of oxygen: ( 1 4.0 I amu) + 2 ( 1 6.00 amu) 46.01 amu. =

=

END-OF-CHAPTER PROBLEMS

2. 1

Plan: Refer to the definitions of an element and a compound. Solution: Unlike compounds, elements cannot be broken down by chemical changes into simpler materials. Compounds contain different types of atoms; there is only one type of atom in an element.

2.4

Plan: Review the definitions of elements, compounds, and mixtures. Solution: a) The presence of more than one element (calcium and chlorine) makes this pure substance a compound. b) There are only atoms from one element, sulfur, so this pure substance is an element. c) The presence of more than one compound makes this a mixture. d) The presence of more than one type of atom means it cannot be an element. The specific, not variable, arrangement means it is a compound.

2.6

Plan: Restate the three laws in your own words. Solution: a) The law of mass conservation applies to all substances - elements, compounds, and mixtures. Matter can neither be created nor destroyed, whether it is an element, compound, or mixture. b) The law of definite composition applies to compounds only, because it refers to a constant, or definite, composition of elements within a compound. c) The law of multiple proportions applies to compounds only, because it refers to the combination of elements to form compounds.

2.7

Plan: Review the three laws: Law of Mass Conservation, Law of Definite Composition, and Law of Multiple Proportions. Solution: a) Law of Definite Composition - The compound potassium chloride, KCl, is composed of the same elements and same fraction by mass, regardless of its source (Chile or Poland). b) Law of Mass Conservation - The mass of the substance inside the flashbulb did not change during the chemical reaction (formation of magnesium oxide from magnesium and oxygen). c) Law of Multiple Proportions - Two elements, 0 and As, can combine to form two different compounds that have different proportions of As present.

2.8

Plan: Review the definition of percent by mass. Solution: a) No, the mass percent of each element in a compound is fixed. The percentage of Na in the compound NaCI is 39.34% (22.99 amu / 5 8 .44 amu), whether the sample is 0.5000 g or 50.00 g. b) Yes, the mass of each element in a compound depends on the mass of the compound. A 0.5000 g sample of NaCI contains 0. 1 967 g ofNa (39.34% of 0.5000 g), whereas a 50.00 g sample of NaCI contains 1 9.67 g of Na (39.34% of 50.00 g). c) No, the composition of a compound is determined by the elements used, not their amounts. ]f too much of one element is used, the excess will remain as unreacted element when the reaction is over.

14

2.9

•

2. 1 1

Plan: Review the mass laws: Law of Mass Conservation, Law of Definite Composition, and Law of Multiple Proportions. Solution: Experiments 1 and 2 together demonstrate the Law of Definite Composition. When 3 .25 times the amount of blue compound in experiment I is used in experiment 2, then 3 .25 times the amount of products were made and the relative amounts of each product are the same in both experiments. In experiment I , the ratio of white compound to colorless gas is 0.64:0.36 or 1 . 78: I and in experiment 2, the ratio is 2.08: 1 . 1 7 or 1 .78: 1 . The two experiments also demonstrate the Law of Conservation of Mass since the total mass before reaction equals the total mass after reaction. Plan: The difference between the mass of fluorite and the mass of calcium gives the mass of fluorine. The masses of calcium, fluorine, and fl uorite combine to give the other values. Solution: Fluorite is a mineral containing only calcium and fluorine. a) Mass of fluorine = mass of fluorite - mass of calcium = 2.76 g 1 .42 g = 1.34 g F b) To find the mass fraction of each element, divide the mass of each element by the mass of fluorite: 1 .42 g Ca = 0.5 1 449 = 0.5 1 4 Mass fraction of Ca = 2.76 g fluorite 1 .34 g F Mass fraction of F = 0.4855 1 = 0.486 2.76 g fluorite c) To find the mass percent of each element, multiply the mass fraction by 1 00: Mass % Ca = (0.5 1 4) ( l 00) = 5 1 .449 = 5 1 .4% M ass % F = (0.486) ( l 00) = 48 .55 1 = 48.6% -

2. 1 3

Plan: Since copper is a metal and sulfur is a nonmetal, the sample contains 88.39 g Cu and 44.61 g S. Calculate the mass fraction of each element in the sample. Solution : Mass of compound = 88.39 g copper + 44.6 1 g sulfur = 1 33 .00 g compound 1 0 3 g compound 88.39 g copper = 3 . 4983 8 x 1 0 6 M ass 0 f copper = (5264 k g compoun d) . 1 kg compound 1 3 3 .00 g compound

(

Mass of sulfur = (5264 kg comp O Und)

(

I 0 3 g compound 1 kg compound

)( )(

44.6 1 g sulfur l 33 .00 g compound

)

)

=

=

=

2. 1 5

3.498 x 1 0 6 g copper

1 .76562 X 1 0 6 6 1. 766 X 1 0 g sulfur

Plan: The law of multiple proportions states that if two elements form two different compounds, the relative amounts of the elements in the two compounds form a whole number ratio. To illustrate the law we must calculate the mass of one element to one gram of the other element for each compound and then compare this mass for the two compounds. The law states that the ratio of the two masses should be a small whole number ratio such as 1 :2, 3 :2, 4:3, etc. Solution: 47.5 mass % S = 0.90476 = 0.904 Compound 1 : 52.5 mass % CI 3 1. 1 mass % S = 0.45 1 379 = 0.45 1 Compound 2 : 68.9 mass % CI 0.904 = 2 .0044 = 2.00 / 1 .00 Ratio: 0.45 1 Thus, the ratio of the mass of sulfur per gram of chlorine in the two compounds is a small whole number ratio of 2 to 1 , which agrees with the law of multiple proportions. -----

15

2. 1 8

Plan: Determine the mass percent of sulfur in each sample by dividing the grams of sulfur i n the sample by the total mass of the sample. The coal type with the smallest mass percent of sulfur has the smallest environmental impact. Solution: 1 1 .3 g sulfur Mass % In Coal A (1 00 %) 2.9894 2.99% S (by mass) 378 g sample ·

=

Mass % In Coal 8 ·

=

( ( (

J J )

=

1 9.0 g sulfur ( 1 00 %) 3 . 8384 3 . 84% S (by mass) 495 g sample =

20.6 g sUlfur (1 00 %) 675 g sample Coal A has the smallest environmental impact. M ass % In Coal C ·

=

=

=

=

3.05 1 9 3 .05% S (by mass) =

2. 1 9

Plan: This question is based on the Law of Definite Composition. If the compound contains the same types of atoms, they should combine in the same way to give the same mass percentages of each of the elements. Solution: Potassium nitrate is a compound composed of three elements - potassium, nitrogen, and oxygen - in a specific ratio. If the ratio of elements changed, then the compound would be different, for example to potassium nitrite, with different physical and chemical properties. Dalton postulated that atoms of an element are identical, regardless of whether that element is found in India or Italy. Dalton also postulated that compounds result from the chemical combination of specific ratios of different elements. Thus, Dalton ' s theory explains why potassium nitrate, a compound comprised of three different elements in a specific ratio, has the same chemical composition regardless of where it is mined or how it is synthesized.

2.20

Plan: Review the discussion of the experiments in this chapter. Solution : a) Millikan determined the minimum charge on an oil drop and that the minimum charge was equal to the charge on one electron. Using Thomson 's value for the mass-ta-charge ratio of the electron and the detennined value for the charge on one electron, Millikan calculated the mass of an electron (charge/( charge/mass» to be 2 9. I 09 x 1 0- 8 g. b) The value -1 .602 x 1 0- 1 9 C is a common factor, detennined as follows: -3 .204 x 1 0- 1 9 C / -1 .602 X 1 0- 1 9 C 2.000 19 19 -4. 806 X 1 0- 1 9 C / -1 .602 X 1 0- 1 9 C 3 .000 -8 .0 1 0 X 1 0- C / -1 .602 X 1 0- C 5.000 - 1 .442 X 1 0- 1 8 C / - 1 .602 X 1 0- 1 9 C 9.000 =

=

=

=

2.23

Plan: The superscript is the mass number. Consult the periodic table to get the atomic number (the number of protons). The mass number - the number of protons the number of neutrons. For atoms, the number of protons and electrons are equal. Solution: # of Protons # of Neutrons # of Electrons Mass Number Isotope 3 6 Ar 18 18 18 36 3 8 Ar 18 20 18 38 18 22 18 40 Ar 40 =

2.25

Plan: The superscript is the mass number (A) and the subscript is the atomic number (Z, number of protons). The mass number - the number of protons the number of neutrons. For atoms, the number of protons the number of electrons. =

=

16

Solution: a) ' : 0 and '� O have the same number of protons and electrons (8), but different numbers of neutrons. ' : 0 and '� O are isotopes of oxygen, and ' : 0 has 1 6 - 8 = 8 neutrons whereas '� O has 1 7 - 8 = 9 neutrons. Same Z value

b) �� Ar and :� K have the same number of neutrons (Ar: 40 - 1 8 = 22; K: 41 - 1 9 = 22) but different numbers of protons and electrons (Ar = 1 8 protons and 1 8 electrons; K = 1 9 protons and 1 9 electrons). Same N value c) � Co and �� Ni have different numbers of protons, neutrons, and electrons. Co: 27 protons, 27 electrons and 60 - 27 = 33 neutrons; Ni : 28 protons, 28 electrons and 60 - 28 = 32 neutrons. However, both have a mass number of 60. Same A value 2.27

Plan: Combine the particles in the nucleus (protons + neutrons) to give the mass number (superscript, A). The number of protons gives the atomic number (subscript, Z) and identifies the element. Solution:

2.29

Plan: The subscript (Z) is the atomic number and gives the number of protons and the number of electrons. The superscript (A) is the mass number and represents the number of protons + the number of neutrons. Therefore, mass number - number of protons = number of neutrons. Solution: a) �82Ti c) ' ; B b) ;: Se 48 - 22 = 26 79 - 34 = 45 11 -5=6

2.3 1

Plan: To calculate the atomic mass of an element, take a weighted average based on the natural abundance of the isotopes: (isotopic mass of isotope 1 x fractional abundance) + (isotopic mass of isotope 2 x fractional abundance) Solution: 39.89% 60. 1 1 % . . = 69 .7 230 = 69 . 72 amu AtomiC mass of gallium = (68.9256 amu) + (70.9247 amu) 1 00% 1 00%

(

2.33

)

(

)

Plan: To find the percent abundance of each CI isotope, let x equal the fractional abundance of 3 5 CI and (1 equal the fractional abundance of 37 Cl. Remember that atomic mass = isotopic mass of 3 5 CI x fractional abundance) + (isotopic mass of 37 CI x fractional abundance). Solution: Atomic mass of CI = 3 5 .4527 amu 35 .4527 = 34.9689x + 36.9659( 1 - x) 35 .4527 = 34.9689x + 36.9659 - 36.9659x 35 .4527 = 36.9659 - 1 .9970x 1 .9970x = 1 .5 1 32 x = 0.75774 and 1 x = 0.24226 % abundance 3 5C I 75.774 % % abundance 3 7CI 24.226% -

=

=

17

-

x)

2.35

Plan: Review the section in the chapter on the periodic table. Solution : a) In the modern periodic table, the elements are arranged in order of increasing atomic number. b) Elements in a column or group (or family) have simi lar chemical properties, not those in the same period or row. c) Elements can be classified as metals, metal loids, or nonmetals.

2.38

Plan: Locate each element on the periodic table. The Z value is the atomic number of the element. Metals are to the left of the "staircase", nonmetals are to the right of the "staircase" and the metalloids are the elements that lie along the "staircase" l ine. Solution : 4A( 1 4) metalloid a) Germanium Ge 6A( l 6) nonmetal S b) Sulfur 8A( 1 8) nonmetal c) Helium He 1 A( I ) metal Li d) Lithium 6B(6) metal e) Molybdenum Mo

2.40

Plan: Review the section in the chapter on the periodic table. Remember that alkaline earth metals are in Group 2A(2) and the halogens are in Group 7A( 1 7); periods are horizontal rows. Solution : a) The symbol and atomic number of the heaviest alkaline earth metal are Ra and 88. b) The symbol and atomic number of the lightest metalloid in Group 5A( 1 5 ) are As and 33 . c) The symbol and atomic mass of the coinage metal whose atoms have the fewest electrons are Cu and 63.55 amu.

d) The symbol and atomic mass of the halogen in Period 4 are

Br

and 79.90 amu.

2.42

Plan: Review the section of the chapter on the formation of ionic compounds. Solution : These atoms wil l form ionic bonds, in which one or more electrons are transferred from the metal atom to the nonmetal atom to form a cation and an anion, respectively. The oppositely charged ions attract, forming the ionic bond.

2.44

Plan: Assign charges to each of the ions. Since the sizes are similar, there are no differences due to the sizes. Solution: Coulomb 's law states the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. The product of charges in MgO (+2 x -2 = -4) is greater than the product of charges in LiF (+ 1 x - 1 = - 1 ). Thus, MgO has stronger ionic bonding.

2.47

P lan: Locate these elements on the periodic table and predict what ions they wil l form. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Solution : Potassium (K) is in Group l A( I ) and forms the K+ ion. Iodine (1) is in Group 7A( 1 7) and forms the r ion (7 - 8 = - 1 ).

2 .49

Plan: Use the number of protons (atomic number) to identify the element. Add the number of protons and neutrons together to get the mass number. Locate the element on the periodic table and assign its group and period number. Solution : a) Oxygen (atomic number = 8) mass number = 8p + 9n = 1 7 Period 2 Group 6A( l 6) Period 2 Group 7 A( 1 7) b) Fluorine (atomic number = 9) mass number = 9p + I O n = 1 9 c) Calcium (atomic number = 20) mass number = 20p + 20n = 40 Group 2A(2) Period 4

18

2.5 1

Plan: Determine the charges of the ions based on their position on the periodic table. Next, determine the ratio of the charges to get the ratio of the ions. Solution: Lithium [Group I A( 1 )] forms the Li+ ion; oxygen [Group 6A( 1 6)] forms the 0 2- ion (6 8 -2). The ionic compound that forms from the combination of these two ions must be electrically neutral, so two Li+ ions combine with one 0 2- ion to form the compound Li 2 0. There are twice as many Li+ ions as 02- ions in a sample of Li 2 0. 1 0 2. ion = 2.65 X 1 0 2 ° = 2.6 x I 020 02- ions Number of 0 2- ions = (5 .3 x I 0 2 0 Li + iOnS) 2 Li + ions

-=

]

(

2.53

Plan: The key is the size of the two alkali ions. The charges on the sodium and potassium ions are the same as both are in Group I A( I ), so there wil l be no difference due to the charge. The chloride ions are the same, so there wil l be no difference due to the chloride. Solution: Coulomb ' s law states that the energy of attraction in an ionic bond is directly proportional to the product of charges and inversely proportional to the distance between charges. (See also problem 2 .44.) The product of the charges is the same in both compounds because both sodium and potassium ions have a + I charge. Attraction increases as distance decreases, so the ion with the smaller radius, Na+, wil l form a stronger ionic interaction (NaCl).

2.55

Plan: Review the definitions of empirical and molecular formulas. Solution: An empirical formula describes the type and simplest ratio of the atoms of each element present in a compound whereas a molecular formula describes the type and actual number of atoms of each element in a molecule of the compound. The empirical formula and the molecular formula can be the same. For example, the compound formaldehyde has the molecular formula, CH 2 0. The carbon, hydrogen, and oxygen atoms are present in the ratio of I :2: 1 . The ratio of elements cannot be further reduced, so formaldehyde ' s empirical formula and molecular formula are the same. Acetic acid has the molecular formula, C 2 H 4 02 . The carbon, hydrogen, and oxygen atoms are present in the ratio of 2 :4:2, which can be reduced to I :2: I . Therefore, acetic acid ' s empirical formula is CH 2 0, which is different from its molecular formula. Note that the empirical formula does not uniquely identify a compound, because acetic acid and formaldehyde share the same empirical formula but are not the same compound.

2.56

Plan: Review the concepts of atoms and molecules. Solution: The mixture is similar to the sample of hydrogen peroxide in that both contain 20 billion oxygen atoms and 20 billion hydrogen atoms since both O 2 and H 2 02 contain 2 oxygen atoms per molecule and both H 2 and H 2 0 2 contain 2 hydrogen atoms per molecule. They differ in that they contain different types of molecules: H 2 0 2 molecules in the hydrogen peroxide sample and H 2 and O 2 molecules in the mixture. In addition, the mixture contains 20 billion molecules ( 1 0 billion H 2 and 1 0 billion O2 ) while the hydrogen peroxide sample contains 1 0 billion molecules.

2.57

Plan: Examine the subscripts and see if there is a common divisor. If one exists, divide all subscripts by this value. Solution: a) To find the empirical formula for N 2 H 4 , divide the subscripts by the highest common divisor, 2 : N 2 H 4 becomes NH2 b) To find the empirical formula for C 6 H ' 2 06 , divide the subscripts by the highest common divisor, 6: C 6 H I 2 06 becomes CH20

2.59

Plan: Locate each of the individual elements on the periodic table, and assign charges to each of the ions. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound.

19

Solution : a) Lithium is a metal that forms a + I (group I A) ion and nitrogen is a nonmetal that forms a -3 ion (group +3 -3 SA, 5 - 8 = -3). +I + 1 -3 The compound is Li3 N, lithium nitride. Li N Li3N b) Oxygen is a nonmetal that forms a -2 ion (group 6A, 6 - 8 = -2) and strontium is a metal that forms a +2 ion (group 2A). +2 -2 Sr 0 The compound is SrO, strontium oxide. c) Aluminum is a metal that forms a +3 ion (group 3A) and chlorine is a nonmetal that forms a -1 ion (group 7A, 7 - 8 = -1 +3 -3 +3 - 1 +3 - 1 The compound is AICI3 , alu minum chloride. AICI3 AI CI 2.6 1

Plan : Based on the atomic numbers (the subscripts) locate the elements on the periodic table. Once the atomic numbers are located, identify the element and based on its position, assign a charge. For A-group cations (metals), ion charge = group number; for anions (nonmetals), ion charge = group number minus 8. Find the smallest number of each ion that gives a neutral compound. Solution: a) 1 2 L is the element Mg (Z = 1 2). Magnesium [group 2A(2)] forms the Mg2 + ion. 9M is the element F (Z = 9). Fluorine [group 7 A( 1 7)] forms the F- ion (7-8 = 1 ) The compound formed by the combination of these two elements is MgF2, magnesiu m fluoride. b) 1 1 L is the element Na (Z = 1 1 ). Sodium [group IA( I )] forms the Na+ ion. 1 6M is the element S (Z = 1 6). Sulfur[group 6A( 1 6)] will form the S 2- ion (6-8 = -2). The compound formed by the combination of these two elements is Na 2 S, sodiu m sulfide. c) 1 7 L is the element CI (Z = 1 7). Chlorine [group 7 A( 1 7)] forms the Cl- ion (7-8 = - 1 ). 3 8 M is the element Sr 2 (Z = 38). Strontium [group 2A(2)] forms the Sr + ion. The compound formed by the combination of these two elements is SrC h , stronti u m chloride. -

.

2.63

P lan: Review the rules for nomenclature covered in the chapter. For metals, like many transition metals, that can form more than one ion each with a different charge, the ionic charge of the metal ion is indicated by a Roman numeral within parentheses immediately fol lowing the metal ' s name. Solution: a) Tin (IV) chloride = SnCl4 The (IV) indicates that the metal ion is Sn4+ which requires 4 cr ions for a neutral compound. b) FeBr) = iron(III) bromide (common name is ferric bromide); the charge on the iron ion is +3 to match the -3 charge of 3 Br- ions. The +3 charge of the Fe is indicated by ( I I I ) + 6 -6 +3 -2 c) cuprous bromide = CuBr (cuprous is + 1 copper ion, cupric is +2 copper ion) d) Mn 2 0) = manganese(III) oxide Use ( I I I ) to indicate the +3 ionic charge of Mn: Mn 2 03

2.65

Plan : Review the rules for nomenclature covered i n the chapter. Compounds must b e neutral. Solution: a) Barium [Group 2a(2)] forms Ba2+ and oxygen [Group 6A( 1 6)] forms 02- (6 - 8 = -2) so the neutral compound forms from one barium ion and one oxygen ion. Correct formula is BaO. b) Iron(Il) indicates Fe 2+ and nitrate is NO)- so the neutral compound forms from one iron(II) ion and two nitrate ions. Correct formula is Fe(N03h c) Mn is the symbol for manganese. Mg is the correct symbol for magnesium. Correct formula is MgS. Sulfide is the S 2- ion and sulfite is the SO)2- ion.

2.67

Plan : Acids donate H+ ion t o solution, s o the acid i s a combination o f H+ and a negatively charged ion. Binary acids (H plus one other nonmetal) are named hydro + nonmetal root + ic acid Oxoacids (H + an oxoanion) are named by changing the suffix of the oxoanion: -ate becomes -ic acid and -ite becomes -ous acid. Solution: a) Hydrogen sulfate is H S0 4-, so its source acid is H 2 S04. Name of acid is sulfu ric acid. b) HI0 3 , iodic acid (I03- is the iodate ion)

20

c) Cyanide is C� ; its source acid is d) H z S, hydrosulfuric acid

HCN hydrocyanic acid .

2.69

Plan: This compound is composed of two nonmetals. Rule I ("Names and Formulas of B inary Covalent Compounds") indicates that the element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. Solution : Disulfur tetrafluoride SZF4 di- indicates two S atoms and tetra- indicates four F atoms.

2.7 1

Plan: Break down each formula to the individual elements and count the number of each. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: a) There are 1 2 atoms of oxygen in AI 2 (S0 4 h The molecular mass is: Al 2(26.98 amu) 53 .96 amu S 3(32.07 amu) 96.2 1 amu o 1 2( 1 6.00 amu) 1 92.0 amu 342.2 amu

b) There are 9 atoms of hydrogen in (N H 4 ) 2 H P04 . The molecular mass is: N 2( 1 4.0 1 amu) 28.02 amu H 9( 1 .008 amu) 9.072 amu P 1 (30.97 amu) 30.97 amu o 4( 1 6.00 amu) 64.00 amu 1 3 2 .06 amu

c) There are 8 atoms of oxygen in CU3(OHMC03) 2 ' The molecular mass is: Cu 3(63.55 amu) 1 90.6 amu o 8( 1 6.00 amu) 1 28.0 amu H 2( 1 .008 amu) 2.0 1 6 amu C 2( 1 2.0 1 amu) 24.02 amu 344.6 amu

2.73

P lan: Review the rules of nomenclature and then assign a name. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: a) (NH4) z S04 28.02 amu 2( 1 4.0 1 amu) N 8.064 amu 8( 1 .008 amu) H 32.07 amu 1 (32.07 amu) S 64.00 amu 4( 1 6.00 amu) 0 1 3 2 . 1 5 amu

b) NaH z P04 Na H P 0 c) KHC03 K

H C 0

22.99 amu 2.0 1 6 amu 30.97 amu 64.00 amu

1 (22.99 amu) 2( 1 .008 amu) 1 (30.97 amu) 4( 1 6.00 amu)

1 1 9.98 amu

39. 1 0 amu 1 .008 amu 1 2.0 1 amu 48.00 amu

1 (39. 1 0 amu) 1 ( l .008 amu) 1 ( 1 2.0 1 amu) 3 ( 1 6.00 amu)

1 00. 1 2 amu

21

2. 7 5

2. 7 7

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. Divide the molecular formula by the largest factor to give the empirical formula. Use the nomenclature rules in the chapter to derive the name. This compound is composed of two nonmetals. Rule ("Names and Formulas of Binary Covalent Compounds") indicates that the element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution: The compound's name is disulfur dichloride . (Note: Are you unsure when and when not to use a prefix? If you leave off a prefix, can you definitively identify the compound? The name s u lfur dichloride would not exclusively identify the molecule in the diagram because sulfur dichloride could be any combination of sulfur atoms with two chlorine atoms. Use prefixes when the name may not uniquely identify a compound.) The empirical formula is S 2I2 CI 2/2 or S C I . The molecular mass is amu) + amu) 1 35.04 amu .

1

2(32. 07

2(35.45

=

Plan: Use the chemical symbols and count the atoms of each type to give a molecular formula. D ivide the molecular formula by the largest factor to give the empirical formula. Use the nomenclature rules in the chapter to derive the name. This compound is composed of two nonmetals. Rule I ("Names and Formulas of Binary Covalent Compounds") indicates that the element with the lower group number is named first. Greek numerical prefixes are used to indicate the number of atoms of each element in the compound. The molecular (formula) mass is the sum of the atomic masses of all of the atoms. Solution : a) Formula is S03. Name is sulfur trioxide . ( 1 S atom and three 0 atoms) Molecular mass amu) + amu) = 80.07 amu b) Formula is N 2 0. Name is dinitrogen monoxide . N atoms and 1 0 atom) Molecular mass amu) + amu) 44.02 amu =

=

(32. 0 7 2(14. 0 1

3(16.00 (16.00

=

(2

2. 79

Plan: Review the discussion on separations. Solution: Separating the components of a mixture requires physical methods only; that is, no chemical changes (no changes in composition) take place and the components maintain their chemical identities and properties throughout. Separating the components of a compound requires a chemical change (change in composition).

2. 8 2

Plan: Review the definitions in the chapter. Solution : a) Distilled water is a compound that consists of H 2 0 molecules only. How would you classify tap water? b) Gasol ine is a homogeneous mixture of hydrocarbon compounds of uniform composition that can be separated by physical means (distillation). c) Beach sand is a heterogeneous mixture of different size particles of minerals and broken bits of shells. d) Wine is a homogenous mixture of water, alcohol, and other compounds that can be separated by physical means (distillation). e) Air is a homogeneous mixture of different gases, mainly N 2 , O2 , and Ar.

2. 84

Plan: Use the equation for the volume of a sphere in Part a) to find the volume of the nucleus and the volume of the atom. Calculate the fraction of the atom volume that is occupied by the nucleus. For Part b), calculate the total mass of the two electrons; subtract the electron mass from the mass of the atom to find the mass of the nucleus. Then calculate the fraction of the atom's mass contributed by the mass of the nucleus. Solution: a) Fraction of volume

=

Volume of Nucleus Volume of Atom

-------­

( �) 1t(2. 5 10-1 5 ( �) 1t(3.1 10-11 X

x

22

m

t

m)

3

=

5.2449 10-1 3 = X

5.2

X

1 0-1

3

b) Mass of nucleus = mass of atom - mass of electrons 6.64648 x 1 0-24 g - 2(9. 1 0939 X 1 0-2 8 g) 6.64466 X 1 0-24 g 2 6.64466 x 1 0 - 4 g ) ( Mas s of Nuc leus = 0.999726 1 7 = 0.999726 = . . Fraction of mass = M ass of Atom ( 6.64648 x 1 0-24 g ) As expected, the volume of the nucleus relative to the volume of the atom is small while its relative mass is large. =

2.86

=

Plan: Determine the percent oxygen in each oxide by subtracting the percent nitrogen from 1 00%. Express the percentage in grams and divide by the atomic mass of the appropriate elements. Then divide by the smaller ratio and convert to a whole number. Solution: I a) ( 1 00.00 - 46.69 N)% = 53.3 1 % 0 I mol 0 1 mol N ( 53 . 3 1 g 0 ) = 3.3326 mol N ( 46.69 g N ) = 3 .33 1 9 mol 0 1 6.00 g 0 1 4.0 1 g N

)

(

3 . 3 3 1 9 mol 0 ----- = 1 .0000 mol 0 3.33 1 9

3 .3326 mol N ---- = 1 .0002 mol N 3 .3 3 1 9

NO

1 : 1 N :O gives the empirical formula: II

( 1 00.00 - 36.85 N)% = 63 . 1 5% 0 1 mol N = 2.6303 mol N ( 36.85 g N ) 1 4.0 1 g N

)

(

2.6303 mol N 2.6303

=

)

(

( 63 . 1 5 g 0)

(

)

1 mol 0 = 3 .9469 mol 0 1 6.00 g 0

3 . 9469 mol 0 = 1 .500 1 mol O 2 .6303

1 .0000 mol N

I : 1 .5 N :O gives the empirical formula: III

)

( 1 00.00 - 25 .94 N)% = 74.06% 0 I mol N = 1 .85 1 5 moI N ( 2 5 .94 g N ) 1 4.0 1 g N

(

) ) J

1 :2.5 N :O gives the empirical formula: N 2 0s 5 3 .3 I g 0 ( 1 .00 g N ) = 1. 1 4 1 8 = 1 .14 g 0 46.69 g N II

( 1 .00 g N )

III

( 1 .00 g N )

( ( (

(

)

1 mol 0 = 4.6288 mol 0 1 6.00 g 0

4.6288 mol 0 = 2.5000 mol 0 1 . 85 1 5

1 . 85 1 5 mol N = 1 .0000 mol N 1 . 85 1 5

b)

( 74.06 g O )

63. 1 5 g 0 = 1 .7 1 37 36.85 g N

=

1 .7 1 g 0

74.06 g 0 = 2.8550 = 2 .86 g 0 25.94 g N

23

2.88

Plan: The mass percent comes from determining the kilograms of each substance in a kilogram of seawater. The percent of an ion is simply the mass of that ion divided by the total mass of ions. Solution: a) For chloride ions: 1 8980. mg C I - 0.00 1 g � 00% = .898% (1 cr Mass% cr = ) 1 I kg seawater I mg 1 000 g cr: 1 .898% 1 .056% Na+: 80/-: 0.265% 2 Mg +: 0.127% 2 Ca +: 0.04% 0.038% K+: HC03-: 0.014% Comment: Should the mass percent add up to I OO? No, the maj ority of seawater is H 2 0. b) Total mass of ions in 1 kg of seawater = 1 8980 mg + 1 0560 mg + 2650 mg + 1 270 mg + 400 mg + 380 mg + 1 40 mg = 34380 mg % Na+ = ( 1 0560 mg Na +/34380 mg total ions) ( 1 00) = 30.7 1 553 = 30.72% c) Alkaline earth metal ions are Mg2 + and Ca2 +. Total mass% = 0. 1 27 + 0.04 = 0. 1 67 = 0.17% Alkali metal ions are K+ and Na+. Total mass% = 1 .056 + 0.03 8 = 1 .094% Total mass percent for alkali metal ions is 6.6 times greater than the total mass percent for alkaline earth metal ions. Sodium ions (alkali metal ions) are dominant in seawater. d) Anions are cr, SO/-, and HC03-. Total mass% = 1 .898 + 0.265 + 0.0 1 4 = 2.177% Cations are Na+, Mg2 +, Ca2 +, and K+. Total mass% = 1 .056 + 0. 1 27 + 0.04 + 0.03 8 = 1 .26 1 0 = 1 .26% The mass fraction of anions is larger than the mass fraction of cations. Is the solution neutral since the mass of anions exceeds the mass of cations? Yes, although the mass is larger the number of positive charges equals the number of negative charges.

)(

[

2 .90

)( )

Plan: First, count each type of atom present to produce a molecular formula. Divide the molecular formula b y the largest divisor to produce the empirical formula. The molecular mass comes from the sum of each of the atomic masses times the number of each atom. The atomic mass times the number of each type of atom divided by the molecular mass times 1 00 percent gives the mass percent of each element. Solution: The molecular formula of succinic acid is C4H604• Dividing the subscripts by 2 yields the empirical formula C 2 H30 2 . The molecular mass of succinic acid is 4( 1 2.0 1 amu) + 6( 1 .008 amu) + 4( 1 6.00 amu) = 1 1 8.088 = 1 1 8.09 a m u . 4 ( 1 2 . 0 1 amu C ) 1 00% = 40.68 1 5 = 40.68% C %C= 1 1 8 .088 amu %H=

[ [( [(

6 1 .008 amu H 1 1 8 .088 amu

) )) ))

1 00% = 5 . 1 2 1 6 = 5. 122% H

4 1 6.00 amu 0 1 00% = 54. 1 969 = 54.20% 0 1 1 8.088 amu Check: Total = (40.68 + 5 . 1 22 + 54.20)% = 1 00.00% The answer checks. %0=

2.92

P lan: To find the formula mass of potassium fluoride, add the atomic masses of potassium and fluorine. Fluorine has only one naturally occurring isotope, so the mass of this isotope equals the atomic mass of fluorine. The atomic mass of potassium is the weighted average of the two isotopic masses: (isotopic mass of isotope 1 x fractional abundance) + (isotopic mass of isotope 2 x fractional abundance) Solution: 93 .258% . 6.730% Average atomIc mass of K = (3 8.9637 amu) = 39.093 amu + (40.96 1 8 amu 1 00% 1 00% The formula for potassium fluoride is KF, so its molecular mass is (39.093 + 1 8.9984) = 58.091 amu

)

(

24

{

)

2.94

)

(

Plan: First, count each type of atom present to produce a molecular formula. Determine the mass percent of each element. Mass percent = total mass of the element 1 00 . The mass of TNT times the mass percent of molecular mass of TNT each element gives the mass of that element. Solution: The molecular formula for TNT is C 7 Hs06N 3 . (What is its empirical formula?) The molecular mass of TNT is: C 7( 1 2.0 1 amu) 84.07 amu H 5( 1 .008 amu) 5.040 amu o 6( 1 6.00 amu) 96.00 amu N 3 ( 1 4.0 1 amu) 42.03 amu 227. 1 4 amu

( (

) )

The mass percent of each element is: 84.07 amu C= 1 00 = 37.0 1 % C 227. 1 4 aum 0=

H=

96.00 amu 1 00 = 42.26% 0 227. 1 4 aum

( ( ( (

N=

( (

)

5.040 amu 1 00 2.2 1 9% H 227 . 1 4 aum

)

=

42.03 amu 1 00 = 1 8.50% N 227. 1 4 aum

Masses: 37.0 1 % C kg C = ( I .00 Ibs ) = 0.370 Ib C 1 00% TNT kg H = kg 0 =

2.2 1 9% H 1 00% TNT 42.26% 0 1 00% TNT

J J( J( J (I .OO

1 .00 Ibs ) = 0.0222

Ib H

1 .00 Ibs ) = 0.4223

Ib 0

1 8.50% N Ibs ) = 0. 1 85 1b N 1 00% TNT Note: The percent ratio yields the mass of a substance in the compound. kg N =

2. 1 04

Plan: Remember that a change is physical when there has been a change in physical form but not a change in composition. In a chemical change, a substance is converted into a different substance Solution: 1 ) Initially, all the molecules are present in blue-blue or red-red pairs. After the change, there are no red-red pairs, and there are now red-blue pairs. Changing some of the pairs means there has been a chemical change. 2) There are two blue-blue pairs and four red-blue pairs both before and after the change, thus no chemical change occurred. The different types of molecules are separated into different boxes. This is a physical change. 3) The identity of the box contents has changed from pairs to individuals. This requires a chemical change. 4) The contents have changed from all pairs to all triplets. This is a change in the identity of the particles, thus, this is a chemical change. 5) There are four red-blue pairs both before and after, thus there has been no change in the identity of the individual units. There has been a physical change.

25

Chapter 3 Stoichiometry of Formulas and Equations FO LLOW-UP PROBL E M S

3. 1

a) Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles. Solution: I O -3 g 1 mol C 2.6228 X 1 0-2 2 . 62 X 1 0-2 mol C 3 1 5 mg C 1 mg 1 2.0 1 g C Check: 3 1 5 mg is less than a gram, which is less than 1 1 1 2 mol of C. b) Plan: Avogadro's number is needed to convert the number of atoms to moles. The molar mass of manganese can then be used to determine the number of grams. Solution: 54.94 g n 1 mol Mn 2 2 .9377 x 1 0-2 3 .22 x 1 0 0 Mn Atoms 2 3 6.022 X 1 0 Mn Atoms 1 mol Mn

[ )( =

=

)=

[

2.94

x

)(

2 1 0- g M n

M )=

Check: The exponent, 20, from the Mn atoms is much smaller than that for Avogadro's number, 23, thus the mass is much smal ler than the molar mass of Mn. 3.2

a) P lan: Avogadro's number is used to change the number of molecules to moles. Moles may be changed to mass using the molar mass. The molar mass of tetraphosphorus decaoxide requires the chemical formula. Solution : Tetra 4, and deca 1 0 to give P 4 0 1 0 . The molar mass, .M., is the sum of the atomic weights, expressed in glmol : 1 23 .88 glmol P 4(30.97) 1 60.00 g/mol ° 1 O( 1 6.00) 283.88 glmol of P 4 0 l o 1 ol 22 283.88 g 4.65 x 1 0 molecules P40l O 2 1 .9203 2 1 .9 g P40 1 0 � 1 mol 6.022 x L O molecules Check: The exponents indicate there is less than 1 1 1 0 mole of P 4 0 l o . Thus the grams must be less than 1 1 1 0 the molar mass. b) Plan: Each molecule has four phosphorus atoms, so the total number of atoms is four times the number of molecules. Solution : P 23 1 .86 X 1 0 P atoms (4.65 x 1 0 22 P4 0 l0 molecule J\ 1 P 04 atoms 4 l0 molecule Check: There are more phosphorus atoms than molecules so the answer should be larger than the original number.

=

= =

== =

=

)(

[

)=

=

)=

3.3

Plan: The formula of ammonium nitrate and the molar mass are needed. The total mass of nitrogen over the molar mass times 1 00% gives the answer. Solution : The formula for ammonium nitrate is NH 4N0 3 . There are 2 atoms of N per each formula. a) Molar mass N H 4N0 3 (2 x 1 4 .0 1 ) + (4 x 1 .008) + (3 x 1 6.00) 80.05 glmol 2 x 1 4.0 1 glmol total mass of N (1 00 ) 3 5 .003 1 35.00% N ( 1 00 ) 80.05 glmol molar mass of N H 4 N0 3

= =

------ = =

26

=

b) Plan: Convert kg to grams. Use the mass percent found in (a) to find the mass ofN in the sample. Solution: (3 5 . 8 kg -1 03 g 35 . 00 g N = 1 .2530 x 1 04 = 1 .2 5 x 1 04 g Nitrogen I kg 1 00 g NH 4 N0 3 Note: The percent ratio yields the mass of nitrogen in the compound. Check: Nitrogen and oxygen are about the same mass. The difference is very slight and is neglected. Now 40% of the atoms remaining are N, so the answer should be about 40%.

{ )(

3 .4

Plan: The moles of sulfur may be calculated from the mass of sulfur and the molar mass of sulfur. The moles of sulfur and the chemical formula will give the moles af M. The mass of M divided by the moles of M will give the molar mass of M. Solution: ) ( 2.88 g S I mol S 2 mol M = 0.0599 mol M 32.07 g S 3 mol S (3 . 1 2 g M) / (0.0599 mol M) = 52. 1 = 52. 1 g M / mol M The element is Cr (52.00 glmo!); M is Chromiu m and M 2 S 3 is chromiu m(III) sulfide. Check: Given that, the starting masses are similar, but the final formula has 1 .5 times as many S ions as M ions. M should have a molar mass near 1 .5 times the molar mass of sulfur: 32 x 1 .5 = 48

(

3.5

)

)(

)

Plan: Two calculations are needed - one for carbon and one for hydrogen. This is because these are the only elements present in benzo[a]pyrene. If we assume there are 1 00 grams of this compound, then the masses of carbon and hydrogen, in grams, are numerically equivalent to the percentages. Using the atomic masses of these two elements, the moles of each may be calculated. Dividing each of the moles by the smaller value gives the simplest ratio of C and H. The smallest multiplier to convert the ratios to whole numbers gives the empirical formula. Comparing the molar mass of the empirical formula to the molar mass given in the problem allows the molecular formula to be determined. Solution: Assuming 1 00 g of compound gives 95.2 1 g C and 4.79 g H. Then find the moles of each element using their molar masses. I mol C = 95.2 1 g 7.92756 mol C 1 2.0 1 g C

c( (

) )

I mol = 4.75 1 98 mol H H 1 .008 g H Divide each of the moles by 4.75 1 98, the smaller value. 4.75 1 98 mol H = 7 .92756 mol C = 1 .6683 mol C '. 1 .000 mol H 4.75 1 98 4.75 1 98 The value 1 .668 is 5/3 , so the moles of C and H must each be multiplied by 3 . If it is not obvious that the value is near 5/3 , use a trial and error procedure whereby the value is multiplied by the successively larger integer until a value near an integer results. This gives C S H 3 as the empirical formula. The molar mass of this formula is: (5 x 1 2.0 1 glmol) + (3 x 1 .008 glmol) = 63 .074 glmol Dividing 252.30 glmol by 63.074 glmol gives 4.000. Thus, the empirical formula must be multiplied by 4 to give C 2oH1 2 as the molecular formula of benzo[a]pyrene. Check: Determine the molar mass of the formula given and compare it to the value given in the problem: .At (20 x 1 2.0 1 glmol) + ( 1 2 x 1 .008 glmol) = 252.296 glmo!. This is very close to the given value. 4.79 g H

=

27

3.6

Plan: The carbon in the sample is converted to carbon dioxide, the hydrogen is converted to water, and the remaining material is chlorine. The grams of carbon dioxide and the grams of water are both converted to moles. One mole of carbon dioxide gives one mole of carbon, while one mole of water gives two moles of hydrogen. Using the molar masses of carbon and hydrogen, the grams of each of these elements in the original sample may be determined. The original mass of sample minus the masses of carbon and hydrogen gives the mass of chlorine. The mass of chlorine and the molar mass of chlorine will give the moles of chlorine. Once the moles of each of the elements have been calculated, divide by the smallest value, and, if necessary, multiply by the smallest number required to give a set of whole numbers for the empirical formula. Compare the molar mass of the empirical formula to the molar mass given in the problem to find the molecular formula. Solution : Determine the moles and the masses of carbon and hydrogen produced by combustion of the sample. 1 2.0 I g C I mol CO 2 I mol C = = 0. 1 2307 g C 0.0 1 0248 mOl c 0.45 1 g C0 2 I mol C 44. 0 1 g CO 2 I mol CO 2

(

J(

(

J

J(

J

(

(

J

J

1 .008 g H 1 mol H 2 0 2 mol H 0.006904 g H 0.0068495 mol H 1 mol H 1 8.0 1 6 g H 2 0 I mol H 2 0 The mass of chlorine is given by: 0.250 g sample - (0. 1 2307 g C + 0.006904 g H ) 0. 1 20 g CI The moles of chlorine are: 1 mol CI 0. 1 20 g C I 0.0033850 mol CI. This is the smallest number of moles. 3 5 .45 g CI Dividing each mole value by the lowest value, 0.0033850: 0.0033850 mol CI 0.0068495 mol H 0.0 1 0248 mol C 2 .02 mol H ' 1 .00 mol CI 3 .03 mol C' ' 0.0033850 0.0033850 0.0033850 These values are all close to whole numbers, thus the empirical formula is C 3 H 2 CI. The empirical formula has the following molar mass: (3 x 1 2.0 1 glmol) + (2 x 1 .008 g/mol) + (3 5 .45 glmol) 73 .496 glmol C 3 H 2 CI Dividing the given molar mass by the empirical formula mass: ( 1 46.99 glmol) / (73 .496 glmol) 2.00 Thus, the molecular formula is two times the empirical formula, C 6 H4C12 . Check: Carbon is about one-fourth the mass of carbon dioxide, or 0. 1 1 grams of C in the compound. The mass of hydrogen is very small, thus, most of the remaining mass of the compound (0.250 - 0. 1 1 0. 1 4) must be chlorine. Chlorine is about three times as heavy as carbon, thus, there must be three carbons for each chlorine. Also, check that the molar mass of the molecular formula agrees with the given molar mass. (6 x 1 2.0 1 glmol) + (4 x 1 .008 glmol) + (2 x 35 .45 glmol) 1 46.992 glmol 0.06 1 7 g H 2 0

(

J

=

=

=

=

=

=

=

=

=

=

=

3.7

Plan: In each part it is necessary to determine the chemical formulas, including the physical states, for both the reactants and products. The formulas are then placed on the appropriate sides of the reaction arrow. The equation is then balanced. a) Solution: Sodium is a metal (solid) that reacts with water (liquid) to produce hydrogen (gas) and a solution of sodium hydroxide (aqueous). Sodium is Na; water is H 2 0; hydrogen is H 2 ; and sodium hydroxide is NaOH . Na(s) + H 2 0(l) � H 2 (g) + NaOH(aq) is the equation. Balancing wil l precede one element at a time. One way to balance hydrogen gives: Na(s) + 2 H 2 0(l) � H 2 (g) + 2 NaOH(aq) Next, the sodium will be balanced: 2 Na(s) + 2 H 2 0(l) � H 2 (g) + 2 NaOH(aq) On inspection, we see that the oxygens are already balanced. Check: Reactants (2 Na, 4 H , 2 0) Products (2 Na, 4 H, 2 0) =

28

b ) Solution: Aqueous nitric acid reacts with calcium carbonate (solid) to produce carbon dioxide (gas), water (liquid), and aqueous calcium nitrate. Nitric acid is HN03 ; calcium carbonate is CaC0 3 ; carbon dioxide is CO2 ; water is H 2 0; and calcium nitrate is Ca(N0 3 ) 2 ' The starting equation is HN0 3 (aq) + CaC03 (s) � CO 2 (g) + H 2 0(l) + Ca(N0 3 Maq) Initially, Ca and C are balanced. Proceeding to another element, such as N , or better yet the group of elements in N03 - gives the following partially balanced equation: 2 HN03 (aq ) + CaC0 3 (s) � CO 2 (g) + H 2 0(l) + Ca(N0 3 )z(aq) Now, all the elements are balanced. Check: Reactants (2 H, 2 N, 9 0, I Ca, I C ) Products (2 H , 2 N, 9 0, 1 Ca, I C) c) Solution: We are told all the substances involved are gases. The reactants are phosphorus trichloride and hydrogen chloride, while the products are phosphorus trifluoride and hydrogen chloride. Phosphorus trifluoride is PF 3 ; phosphorus trichloride is PCI 3 ; hydrogen fluoride is HF; and hydrogen chloride is HC!. The initial equation is: P CI 3 (g) + HF(g) � PF 3 (g) + HCI(g) Initially, P and H are balanced. Proceed to another element (either F or CI); if we wi ll choose CI, it balances as: PCI 3 (g) + H F(g) � PF 3 (g) + 3 HCI(g) The balancing of the Cl unbalances the H, this should be corrected by balancing the H as: P C h (g) + 3 HF(g) � PF 3 (g) + 3 HCl(g) Now, all the elements are balanced. Check: Reactants (1 P, 3 Cl, 3 H , 3 F) Products ( I P, 3 Cl, 3 H , 3 F) d) Solution: We are told that nitroglycerine is a liquid reactant, and that all the products are gases. The fonnula for nitroglycerine is given. Carbon dioxide is CO2 ; water is H 2 0; nitrogen is N 2 ; and oxygen is O 2 • The initial equation is: C 3 HSN 3 0 9(l) � CO 2 (g) + H 2 0(g) + N 2 (g) + 0 2 (g) Counting the atoms shows no atoms are balanced. One element should be picked and balanced. Any element except oxygen wil l work. Oxygen wil l not work in this case because it appears more than once on one side of the reaction arrow. We wil l start with carbon. Balancing C gives: C 3 HSN 3 0 9 (l) � 3 CO 2 (g) + H 2 0(g) + N 2 (g ) + 0 2 (g) Now balancing the hydrogen gives: C 3 HsN 3 0 9 (l) � 3 CO2 (g) + 5 /2 H 2 0(g) + N 2 (g) + 0 2 (g) Similarly, if we balance N we get: C 3 HsN 3 0 9 ( l) � 3 CO2 (g) + 5 /2 H 2 0(g) + 3/2 N 2 (g) + 0 2 (g) Clearing the fractions by multiplying everything except the unbalanced oxygen by 2 gives: 2 C3 HSN 3 0 9 (l) � 6 CO 2 (g) + 5 H 2 0(g) + 3 N lCg) + 0 2 (g) This leaves oxygen to balance. Balancing oxygen gives: 2 C 3 HSN 3 0 9( l) � 6 CO 2 (g) + 5 H 2 0(g) + 3 N 2 (g) + 1 12 0 2 (g ) Again clearing fractions by multiplying everything by 2 gives: 4 C 3 HSN 3 0 9 ( l) � 12 CO 2 (g) + 10 H2 0(g) + 6 N 2 (g) + 0 2 (g) Now all the elements are balanced. Check: Reactants ( 1 2 C 20 H 1 2 N 36 0 ) Products ( 1 2 C 20 H 1 2 N 36 0) =

=

=

3.8

a) P lan: The reaction, like all reactions, needs a balanced chemical equation. The atomic mass of aluminum is used to determine the moles of aluminum. The mole ratio, from the balanced chemical equation, converts the moles of aluminum to moles of iron. Finally, the atomic mass of iron is used to change the moles of iron to the grams of iron.

29

Solution: The names and formulas of the substances involved are : iron(I1I) oxide, Fe203 , and aluminum, AI, as reactants, and aluminum oxide, AIz03 , and iron, Fe. The iron is formed as a liquid; all other substances are solids. The equation begins as : Fe203 (s) + AI(s) � AIz03 (s) + Fe(l) There are 2 Fe, 3 0, and 1 AI on the reactant side and 1 Fe, 3 0, and 2 AI on the product side. Balancing aluminum: Fe203 (s) + 2 Al(s) � AI20 3 (s) + Fe(l) Fe203 (s) + 2 AI(s) � AIz03 (s) + 2 Fe(l) Balancing iron: Check: reactants: (2 Fe, 3 0, 2 AI) products: (2 F e, 3 0, 2 AI) Using the balanced equation and the atomic masses, calculate the grams of iron: =

( 1 3 5 g AI )

(

(J

1 mol AI 26.98 g AI

2 mol Fe 2 mol AI

(J

5 5 . 85 g F e 1 mol Fe

J

=

279.457

=

27 9 g Fe

Check: The number of moles of aluminum produces an equal number of moles of iron. Iron has about double the mass of Al (2 x 27 54). Thus, the initial mass of aluminum should give approximately 2 x 1 3 5 g ( 270) of Iron. b) P lan: The grams of aluminum oxide must be converted to moles. The formula shows there are two moles of aluminum for every mole of aluminum oxide. Avogadro ' s number will then convert the moles of aluminum to the number of atoms. Solution: 23 6.022 x 1 0 atoms AI 1 mol AI2 03 2 mol Al 22 22 1 . 1 8 X 1 0 a toms Al l . l 8 1 25 x 1 0 ( 1 . 00 g AI2 03 ) 1 0 1 .96 g Al203 I mol Al203 I mol Al =

=

(

)(

)(

)

=

=

Check: The molar mass of aluminum oxide is about 1 00, so one mole of aluminum oxide is about 1 1 1 00 the mass of aluminum oxide. Doubling the moles of aluminum oxide gives the moles of aluminum atoms. Multiplying 22 Avogadro ' s number by 21 1 00 gives about 1 .2 x 1 0 atoms. 3.9

Plan: (a) Count the molecules of each type, and find the simplest ratio. The simplest ratio leads to a balanced chemical equation. The substance with no remaining particles is the limiting reagent. (b) Use the balanced chemical equation to determine the mole ratio for the reaction. Solution: (a) The balanced chemical equation is B2(g) + 2 AB(g) � 2 AB2(g) The limiting reagent is AB since there is a B2 molecule left over (excess). (b) Moles of AB2 from AB : (1 . 5 mol AB Moles of AB2 from B 2 : (1 . 5 mol B 2

{ 2 mol AB 2 \ 2 moi AB

{

2 mo1 AB 1 mol B 2

2

J

)

=

=

1 .5 mol AB2

3 . 0 mol AB2

Thus A B i s the Limiting reagent and only 1.5 mol of AB 2 will form. Check: The balanced chemical equation says twice as many moles of AB are needed for every mole of B2. Equal numbers of moles means AB is not twice as great as B2. Thus, AB should be limiting. 3.10

Plan: First, determine the formulas of the materials in the reaction and begin a chemical equation. Balance the equation. Using the molar mass of each reactant, determine the moles of each reactant. Use mole ratios from the balanced equation to determine the moles of aluminum sulfide that may be produced from each reactant. The reactant that generates the smaller number of moles is limiting. Change the moles of aluminum sulfide from the limiting reactant to the grams of product using the molar mass of aluminum sulfide. To find the excess reactant amount, find the amount of excess reactant required to react with the limiting reagent and subtract that amount from the amount given in the problem.

30

Solution: The balanced equation is: 2AI(s) + 3 S(s) � AI 2 S3(s) Determining the moles of product from each reactant: I mol A1 2 S3 I mol AI ( 1 0.0 g AI ) = 0. 1 8532 mol A I 2 S3 26.98 g AI 2 mol AI

(

(

)( )(

)

)

1 mol S 1 mol A1 2 S3 = 0. 1 55909 mol AI 2 S 3 32.07 g S 3 mol AI Sulfur produces less product, thus it is limiting. The moles of product formed are calculated from the moles and the molar mass. 1 50. 1 7 g AI 2 S3 = 23.4 1 3 = 23.4 g AIzS3 (0. 1 55909 mol A1 2 S3 ) 1 mol AI 2 S3 The mass of aluminum used in the reaction is now determined : ( 1 5 .0 g S { I mol S 2 moI AI 26.98 g AI = 8 .4 1 258 g Al used \ 32.07 g S 3 mol S I mol AI Subtracting the mass of aluminum used from the initial aluminum gives the mass remaining. Un-reacted aluminum = 1 0.0 g - 8.4 1 285 g = 1 .5872 = 1 .6 g A I e x c e s s Check: The reactant ratios in the balanced chemical equation are the same as the mass ratio given in the problem. Thus, the substance with the larger molar mass should be limiting. This is the sulfur. An additional check is based on the aluminum calculation. If the sulfur were not the limiting reactant, the "remaining" aluminum would have been a negative number. ( 1 5.0 g S)

J

(

)(

3. 1 1

Plan : Determine the formulas, and then balance the chemical equation. The grams o f marble are converted to moles, a mole ratio (from the balanced equation) gives the moles of CO2 , and finally the theoretical yield of CO 2 is determined from the moles of CO2 and its molar mass. Solution : The balanced equation: CaC03(s) + 2 HCI(aq) � CaCI 2 (aq) + H 2 0(l) + CO 2 (g) The theoretical yield of carbon dioxide: I moi CaC03 44.0 1 g C0 2 I mol CO 2 = 4.39704 g C0 2 ( 1 0.O g Cac o3 ) 1 mol CO 2 1 00.09 g CaC03 1 mol CaC03 The percent yield: 3 .65 g CO 2 actual yield ( 1 00% ) = ( 1 00%) = 83.0 1 04 = 83.0 % 4.39704 g CO 2 theoretical yield Check: Initially, there are about 1 0/ 1 00 moles of marble (mass of marble/molar mass). Theoretically, about 1 1 1 0 the molar mass of carbon dioxide should form.

)

(

J

J(

J(

(

3. 1 2

)

)(

(

J

Plan: Multiply the molarity by the liters to determine the moles. Solution : 0.50 Ol 84 m L ) = 0.042 mol KI

(

�

KJ) [ \o:t }

Check: The molarity is 0.5, this ratio impl ies the moles should be 0.5 (84 mL) = 42 mmole (0.042) 3. 1 3

Plan: Use the molar mass to change the grams to moles. The molarity may then be used to convert to volume. Solution : 1 L 1 mol C 1 2 H 22 0 1 1 = 0. 1 1 95 1 = 0.1 20 L ( 1 35 g C 1 2 H 22 0 l d 342 .30 g C 1 2 H 22 0 1 1 3.30 mol C 1 2 H 22 0 I I Check: All units cancel except the volume units requested. The mass given is about 1 13 mole and the molarity term gives another 1 /3 ratio, thus the answer should be about ( 1 13 ) ( 1 13 ) = 1 19 L .

(

J

(

J

31

3.14

Plan: Determine the new concentration from the dilution equation ( Mconc)(Vconc) = ( Mdil)(Vdi l ) . Convert the molarity (moUL) to glm L in two steps (one-step is moles to grams, and the other step is L to mL.) Solution : ( m3 ) = M eo nc Vc o n e M M dil Vd m3 =

( 7. 5 0 M) 25. 0 500. ( 0. 3 75 L J ( 98. 09 (8) 0.10 0.10 3 (0. 0 )( 78.00 i1

=

0.375 J [ ) 0.036784 25/500 1120. 3. 1 5 3.4 3 3 ) 3. 84615 1 0 3. 8 )(

g H 2 S0 4 1 O -3 L mol H 2 S0 4 = I mL I mol H 2 S0 4 I = Check: The dilution wi ll give a concentration lower by calculation gives ( 1 /20) ( 1 00) (0.00 1 ) = 0.04.

3.15

3.16

= 3.68 x

1 0-

2

g/mL solution

Checking the calculation by doing a rough

P lan: The problem should be solved like Sample Problem to facilitate the comparison. A new balanced chemical equation is needed. A g sample of aluminum hydroxide, for direct comparison, is converted to moles. The mole ratio from the balanced equation converts from moles of AI(OH )3 to moles of HCI. F inally, the moles of acid reacting with the AI(OH ) 3 are compared to the x 1 0-3 mol of HCI that was shown in the Sample Problem to react with g of Mg(OH) 2 . Solution : Balanced equation: HCI(aq) + AI(OH Ms) � AICI 3 (aq) + H 2 0(l) I mol AI(OH) 3 mol H CI 3 X X 1 0- mo l HCI = -3 = 1 g AI(OH h g AI(OH ) 3 1 mol AI(OH ) 3 Magnesiu m hydroxide is less effective than aluminum hydroxide because it reacts with fewer x 1 0-3 ) moles of HCI. Check: The greater molar mass of AI(OH ) 3 is about and this compound has a larger mole ratio from the balanced equations. This gives x -3 ) = x 1 0-3 .

(3.4 10 (6/8) (1. 56/8,) 3. 8

3/2

(3. 4

a) Plan: The formula of lead( U ) acetate is needed. The moles of compound are converted to moles of lead; this value times the inverse of the molarity gives the volume of the solution. Solution : The formula is Pb(C 2 H 3 0 2 ) 2 which may be written as Pb(CH3COO) 2 . 1 mol Pb(C 2 H 3 0 2 h I 2 mol Pb + ) = = 0.267 L ( 2 mol Pb(C 2 H 3 0 2 h 1 mol Pb + Check: A reverse calculation may serve as a check (complete units are unnecessary): . m OI 1 m O I L = 0.4005 = mol L I mol b) Plan: The formulas are used for the balanced chemical equation. The molarity and the volume of the sodium chloride solution are used to determine the moles ofNaCI. Either the lead or the chloride ion is limiting - the one producing the fewer moles of product is limiting. Convert the lower number of moles of product to grams of product. Solution: Balanced reaction: Pb(C 2 H 3 0 2 Maq) + N aCI(aq) � PbCI 2 (s) + NaC 2 H 3 0 2 (aq) Because of the l : l ratio, there will be mol of PbCI 2 produced from mol of Pb 2+. The moles of PbCI 2 produced from the NaCI are: 1 O -3 mol NaCI --I mol PbCI 2 = mol PbCI 2 mol NaCI Im L The NaCI is limiting. The mass of PbCI 2 may now be determined using the molar mass. g PbCI 2 mol PbCI 2 ) = = 5 9.1 g PbC h mol PbCI 2

0.400

[

(0.267 ) ( 1 5 ) ( J

) ( 1. 5 0

(0. 2 125

) 0. 2 66667

0.400

2

( 3.40 1

L

0.400 2 J [ LL ) (125 mL ) ( 2 J 0.2 125 ( 278.1 J 59.0962 1 32

0.400

Check: There are 0.400 mol of lead ions; they wi ll require 2 x 0.400 0.800 mol of chloride ion. The volume of NaCI solution needed to supply the chloride ions is (0. 800/3 .4) 0.235 L (235 mL). The NaCI must be limiting because less than 235 mL are suppl ied. =

=

END-OF-C HAPTER PROBLEM S

3.2

Plan: The formulas are based on the mole ratios of the constituents. Avogadro's number al lows the change from moles to atoms. Solution : 1 2 mol C a) Moles of C atoms ( 1 mol Sucrose ) 1 2 mol C I mol C l2 H 22 0 " 2 6.022 x 1 0 3 C atoms 12 mol C 24 7.226 x 1 0 C atoms b) C atoms ( I mol C ' 2 H 22 0 , , ) 1 mol C ' 2 H 22 0" I mol C Plan: It is possible to relate the relative atomic masses by counting the number of atoms. Solution : a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yel low balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of six red balls means that they are that much I ighter. Because the red ball is l ighter, more red atoms are required to make 1 gram. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 gram. d) Neither element has more atoms per mole. Both the left and right elements have the same number of atoms per mole. The number of atoms per mole (6.022 x 1 0 23 ) is constant and so is the same for every element.

(

=

=

3 .6

3.7

(

J[

J

=

)

=

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element times the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance. Solution : a) The molar mass, M, is the sum of the atomic weights, expressed in g/mol : Sr 87.62 g Sr/mol Sr(OH ) 2 0 = (2 mol 0) ( 1 6.00 g O/mol 0) 32.00 g O/mol Sr(OH) 2 2.0 1 6 g H/mol Sr(0H)2 H (2 mol H ) ( 1 .008 g Hlmol H ) =

=

=

=

=

=

1 2 1 .64 g/mol of S r(OH) 2

b) M (2 mol N) ( 1 4.0 1 g N/mol N ) + ( 1 mol 0) ( 1 6.00 g O/mol 0) 44.02 g/mol of N 2 0 c) M ( I mol Na) (22.99 g Nalmol Na) + ( 1 mol CI) (35.45 g Cllmol Cl) + (3 mol 0) ( 1 6.00 g O/mol 0) 1 06.44 g/mol of NaCI0 3 d) M (2 mol Cr) (52.00 g Cr/mol Cr) + (3 mol 0) ( 1 6.00 g O/mol 0) 1 52.00 g/mol of C r2 0 3 =

=

=

=

=

=

3.9

Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element times the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance. Solution: a) M ( 1 mol Sn) (I 1 8.7 g Sn/mol Sn) + (2 mol 0) ( 1 6.00 g O/mol 0) 1 50.7 g/mol of S n0 2 b) M ( l mol Ba) ( 1 37.3 g Balmol Ba) + (2 mol F) ( 1 9.00 g Flmol F) 1 75.3 g/mol of BaF 2 c) M = (2 mol AI) (26.98 g Allmol AI) + (3 mol S) (32.07 g Simol S) + ( 1 2 mol 0) ( 1 6.00 g O/mol 0) =

=

=

=

=

342 . 1 7g/mol of AI z ( S 04) 3

d) M ( I mol Mn) (54.94 g M n/mol Mn) + (2 mol CI) (35 .45 g Cllmol C I) =

33

=

1 25.84 g/mol of M nCI z

3. 1 1

Plan: The mass of a substance and its number of moles are re lated through the conversion factor of M, the mol ar mass expressed i n g/mol . The moles of a substance and the number of entities per mole are rel ated by the conversion factor, Avogadro ' s number. Solution : a) M of K M n04 = 3 9 . 1 0 + 54.94 + (4 x 1 6 . 00) = 1 5 8 .04 gimol of KMn04 Mass of KM n04 =

( 0 . 5 7 mol KMn04 )

(

1 5 8 . 04 g KMnO 4 1 mol K M n 0 4

)

= 90.08 = 9.0

x

I

1 0 g KM n0 4

b) M of Mg(NO) 2 = 2 4 . 3 1 + (2 x 1 4 . 0 I ) + (6 x 1 6 . 00) = 1 48 . 3 3 g/mol Mg(N03)2 Moles 0 f ° atoms =

(8. 1 8 g

)

Mg ( N O 3 ) 2

(

1 m o l Mg(N03 ) 2 1 48 . 3 3 g M g ( N 0 3 h

)(

6 mol ° atoms 1 mol Mg( N03 h

= 0 . 3 3 0 8 8 = 0.33 1 mol 0 atoms c ) M of C u S 0405 H20 = 63 . 5 5 + 3 2 . 07 + (9 x 1 6 .00) + ( l O x 1 . 008) = 249 .70 g/mol (Note that the waters of hydration are incl uded in the molar mass . )

(

3

o atoms = 8 . 1 x 1 O - g C u c mPd = 1 .758 1 x 1 0 3. 1 3

20

= 1 .8

X

)

10

(

20

1 mol Cu C mp d 249 .70 g Cu C mpd

)(

9 mol ° atoms 1 mol Cu Cmpd

0 atoms

)[

6 . 022 x 1 0

23

0 atoms

1 mol ° atoms

]

P lan: Determi n e the molar mass of each substance, then perform the appropri ate molar conversions. Solution : a) M of M nS04 = ( 5 4 . 94 g M n/mol M n ) + ( 3 2 . 0 7 g S/mo l S ) + [(4 mol 0) ( 1 6 . 0 0 g O/mo l 0)] = 1 5 1 . 0 1 gimol of M n S 04 M ass of M n S 04 = ( 0 . 64 mol MnSO 4 )

(

1 5 1 .0 \ g MnSO 4 I moi MnS0 4

)

= 96.65 = 97 g M n S 04

b) M of Fe(C l04)3 = ( 5 5 . 8 5 g Fe/mol Fe) + [(3 mol C I ) ( 3 5 .45 g C l/mol C I ) ] + [( 1 2 mol = 3 54 . 2 0 gimol of Fe(C I 04)3 Moles Fe( C I 04») = ( 1 5 . 8 g Fe(CI0 4 ) 3 )

(

1 mol Fe(C I 0 4 ) 3 3 54.20 g Fe(CIO 4 ) 3

0)

= 1 . 74 1 26 x 1 0

24

(

)[

)(

= 1 .74

X

10

24

( 1 6.00 g O/mol

0)]

)

2 = 0 . 044608 = 4.46 x 1 0- mol Fe(CI0 4h c) M of N H4N02 =[(2 mol N ) ( 1 4 . 0 1 g N/mol N)] + [(4 mol H ) ( 1 .008 g H/mol H ) ] + [ ( 2 m o l 0) ( 1 6 . 00 g O/mol 0 ) ] = 64 .05 gimol of N H 4N 02 23 1 mOI 2 mol N 6 . 022 x 1 0 N atoms N atoms = 92 . 6 g N H 4 N O ? - 64.05 g 1 mol N H 4 N 0 2 1 mol N

3. 1 5

)

]

N atoms

P l an : The formula o f each compound must b e determined from its name. The molar mass for each formula comes from the formula and atomic masses from the peri odic table. Avogadro ' s number i s also necessary to find the number of particles. S o l uti o n : 2 a) C arbonate i s a polyatomic a n i o n with t h e formu la, C O ) -. C opper (J) indicates C u +. The correct formula for this ionic compound is C U2CO). M of C u2C 03 = (2 x 63 . 5 5 ) + 1 2 . 0 1 + (3 x 1 6 . 00) = 1 8 7 . 1 1 gimo l Mass C U2CO) =

(8.4 1

mol C U 2 C03

)

(

1 8 7 . 1 1 g Cu 2 CO 3

I

mol C U 2 C03

)

= 1 5 7 3 . 5 9 5 = 1 .57

x

3 1 0 g CU 2 C03

b ) D i n i trogen pentaox ide has the formula N20S ' Di- indicates 2 N atoms and penta- i ndicates 5 ° atoms. M of N20s = ( 2 x 1 4 . 0 \ ) + (5 x 1 6 .00) = 1 08 . 02 gimol M ass N20S =

( 2 . 04

X

10

21

N 2 0S mOlecu les

= 0 . 3 65926 = 0.366 g N 2 0S

)

[

34

1

�3O l N 2 0S

6 . 02 2 x 1 0

N 2 0S molecules

J(

1 08 . 02 g N 2 0S 1 mol N 2 0S

)

c) The correct formula for this ionic compound is NaCI04. There are Avogadro ' s number of entities (in this case, formula units) in a mole of this compound . .M ofNaCI04 = 22.99 + 3 5 .45 + (4 x 1 6.00) = 1 22 .44 glmol Moles NaCI04 = ( 5 7 . 9 g NaCI02 ) FU = formula units FU NaCI04 = ( 5 7 . 9 g NaCI02 )

(

(

1 mol NaCIO 4

1 22 .44 g NaCI04

1 mol NaCI0 4

1 22 .44 g NaCI04

J[

J

= 0.47288 = 0.473 mol NaCl04

3 6.022 x 1 02 FU NaC I04

I

mol NaCl04

]

=

23 2 . 8477 1 1 5 x 1 0 = 2.85 X 10 23 F U NaCl04 d) The number of ions or atoms is calculated from the formula units given in part c. Note the unrounded initially calculated value is used to avoid intermediate rounding. 2 . 8477 1 1 5 x 1 0

2 . 8477 1 1 5 x 1 0

2 . 8477 1 1 5 x 1 0 2 . 8477 1 1 5 x 1 0

3. 17

23 23 23 23

mol NaCI04

mol NaCI04

mol NaCl04 mol NaCl04

[ [ ( (

1 Na+ lon 1 FU NaCI04

I CIO; Ion

I

FU NaCl04 1 Cl atom

1 FU NaCI04 4 0 atoms 1 FU NaCI0 4

]

]

J J

+ .

= 2.85 x 10 23 Na Ions = 2.85

X

1 0 23 Cl04- ions

= 2.85 x 10 23 CI atoms = 1 . 1 4 x 10

24

0 atoms

P lan: Determine the formula and the molar mass of each compound. The formula gives the number of atoms of each type of element present. Masses come from the periodic table. Mass percent = (total mass of element in the substancelmolar mass of substance) x 1 00. Solution: a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH/ and bicarbonate ions, HC0 3The formula of the compound is NH4HC0 3 .M of NH4HC03 = ( 1 4. 0 1 glmol) + (5 x 1 .008 glmol) + ( 1 2 .0 1 glmo!) + (3 x 1 6. 00 glmo!) = 79.06 glmol NH4HC0 3 In 1 mole of ammonium bicarbonate, with a mass of 79.06 g, there are 5 H atoms with a mass of 5 .040 g. (5 mol H ) (1 .008 g / mol H) x 1 00% = 6.374905 = 6.375% H 79.06 g / mol NH4 HC03 •

b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na+, dihydrogen phosphate ions, H2P04-, and seven waters of hydration. The formula is NaH2P04·7H20. Note that the waters of hydration are included in the molar mass . .M of NaH2P04·7Hp = (22.99 glmol) + ( 1 6 x 1 .008 glmol) + (30.97g1mol) + ( 1 1 x 1 6.00 glmol) = 246.09 glmol NaH2P04°7H20 In each mole of NaH2P04·7H20 (with mass of 246.09 g), there are 1 1 x 1 6.00 glmol or 1 76.00 g of oxygen.

( 1 1- mol 0 ) ( 1 6.00 g / mol 0 ) -'-------'--'-------'-x 1 00 % = 7 1 . 5 1 855 = 71 .52% 0 246.09 g l mol

35

3. 1 9

Plan : Determine the formula of c i splat i n from the figure, and then calculate the molar mass from the formula. The molar mass i s necessary for the subsequent calculati ons. Solution : The form ula for c i splatin i s Pt(C IMNH3) 2 .M of Pt( C I M N H 3 h = 1 95 . 1 + (2 x 3 5 .45) + (2 x 1 4 .0 1 ) + (6 x 1 . 008) = 3 0 0 . 1 glmol

. . latm ( 2 8 5 . 3 g clsp

.

a) Moles cisplatm =

b)

H

a

to ms =

3.2 1

=

(0 . 98 m o l c l.

3 . 540936

x

10

spla

24

.

tm

= 3.5

) X

(

(

I mol ci splatin

.

. J(

.

300. 1 g clsp latm

6 mol H . 1 mol clsplatm 1 0 24 H atoms

J

6 . 02 2

. · I atm = 0 . 950683 1 = 0 . 950 7 mo I CISP

x

10

23 H

1 mol H

atoms

)

Plan: Determine the formulas for the compounds where needed. Determine the molar mass of each formula. Calculate the percent nitrogen by dividing the mass of nitrogen i n a mo l e of compound by the molar mass of the compound, and multiply the result by 1 00%. Then rank the values. Soluti o n : Mo lar Mass Cg/mol) Name Formu la 101.1 1 KN03 Potassium n itrate 80.05 N H4N03 Ammonium n itrate 1 32. 1 5 (NH4)2S04 Ammon i u m sulfate 60.06 U rea CO(NH2)2 Calcul ating the nitrogen percentages :

( 1 mol N ) ( 1 4 . 0 1 g / mol N )

Potassium nitrate

1 0 1 .l 1 g / mol

( 2 mol N ) ( 1 4 . 0 1 g / mol N )

Ammon i u m n itrate

80.05 g / mo l

)

(2 m o l N ( 1 4 . 0 1 g/mol N

Ammon i u m sulfate

1 3 2 . 1 5 g/mol

)

x

x

1 00 = 1 3 . 8 5 6 1 96 = 13.86% N

x

1 00 = 3 5 . 003 1 2 3 = 35.00% N

1 00

( 2 mol N ) ( 1 4 . 0 1 g / mol N )

-'-----'--'------"x

U rea

3 .22

)

60.06 g/ mol

= 2 1 .203 1 8 = 2 1 .20% N

1 00 = 4 6 . 65 3 3 = 46.65 % N

Plan: The volume must be converted from cubic feet to cubic centi meters (or vice versa) . The vol ume and the density w i l l give you mass, and the mass with the molar mass gives you mo les. P art (b) requi res a conversion fro m cubic dec i meters, i n stead of cubic feet, to cubic centi m eters. The density al lows you to change these cubic centimeters to mass, the molar mass al lows you to find moles, and finally Avogadro ' s number al lows you to make the last step. Solution : The molar mass of galena is 2 3 9 . 3 g/mol . 3 3 ( 1 2 in) ( 2 . 54 Cm) 7 46 g PbS 1 mol PbS 3 a ) Moles P b S = l . 00 ft PbS 3 3 3 2 3 9 . 3 g PbS ( 1 ft) ( I in) 1 cm 8 8 2 . 7 5 66886 b ) Lead atoms 3 (0. 1 m) 3 1 . 00 dm PbS 3 ( 1 dm) =

=

(

{ \

= 1 . 8773 1

)(

{\

(

x

10

=

883 mol PbS

)[

25

3 ( l cm) 2 3 ( 1 0 - m)

= 1 .88

X

)(

7 .46 g PbS 3 I cm

)( .

)(

)(

J(

1 mol PbS 2 3 9 . 3 g Pb S

1 0 25 Pb atoms

36

I mol Pb I mol PbS

)(

J

6 . 02 2

x

10

23

Pb atoms

I mo l Pb

)

3 . 24

P lan: Remember that the molecular formula te l l s the actual number of moles of each e l e ment i n one mo le of compound. Soluti o n : a) No, y o u c a n obtain t h e empirical formula from the number of m o l e s of e a c h type of atom i n a compound, but not the mo lecular formula. b ) Yes, you can obtain the mo lecular formu la from the mass percentages and the total number of atoms. Solution P l an� 1 ) Assume a 1 00 . 0 g samp le and convert masses (from the mass% of each element) to moles using molar mass. 2 ) I dentity the e lement with the lowest number of moles and use this number to divide i nto the number of moles for each e l e ment. You now have at least one elemental mole ratio (the one with the sma l lest number of moles) equal to 1 . 00 and the remai n i ng mole ratios that are l arger than one. 3) Examine the numbers to determine i f they are whole numbers. I f not, multi p l y each number by a whole n u mber factor to get whole numbers for each el ement. You w i l l have to use some j udgment to decide when to round. 4) Write the empirical formula usi ng the who l e numbers from step 3 . 5 ) Chec k the total n umber o f atoms i n the empirical formula. I f i t equals the total n umber o f atoms given then the empirical formula i s also the molecular formu la. I f not, then divide the total number of atoms given by the total n umber of atoms in the empirical formula. This should give a whole number. M ultiply the n u mber of atoms of each el ement in the empirical formul a by this whole number to get the mo lecular formula. I f you do not get a whole number when you divide, return to step 3 and revise how you multipl ied and rounded to get whole numbers for each element. c) Yes, you can determine the molecular formula from the mass percent and the n u mber of atoms of one element in a compound. S o lution plan : I ) Fol low steps 1 -4 i n part b. 2 ) C ompare the n u mber of atoms given for the one element to the number in the emp i rical formula. Determi ne the factor the number i n the emp irical formula must be multi p l i ed by to obtain the given n u mber of atoms for that e lement. M ultip ly the empirical formula by this number to get the molecular formula. d) No, the mass % will only lead to the empirical formula. e) Yes, a structural formula shows all the atoms i n the compound. S o l ution plan: Count the number of atoms of each type of e lement and record as the number for the molecular formula.

3 .25

Plan: Exami n e the number of atoms of each type in the compound. Divide al l atom n umbers by any common factor. The final answers must be the l owest whole-number values. S o l ution: a) C2H4 has a ratio of 2 carbon atoms to 4 hydrogen atoms, or 2 :4 . This ratio can be reduced to I :2, so that the empirical formula is CH 2 . The empirical formula mass i s 1 2 . 0 1 + 2( 1 . 0 0 8 ) 1 4.03 g/mol. b ) The ratio of atoms i s 2 : 6 : 2 , or 1 : 3 : I . T h e empirical formula i s CH 3 0 a n d i t s empirical formula mass i s 1 2 . 0 1 + 3 ( 1 .008) + 1 6 .00 3 1 .03 g/mot. c) S ince, the ratio of e lements cannot be further reduced, the molecular formula and empirical formula are the same, N 2 0S' The formula mass is 2 ( 1 4 . 0 1 ) + 5 ( 1 6.00) 1 08.02 g/mot. d) The ratio of e lements i s 3 atoms of barium t o 2 atoms of phosphorus t o 8 atoms of oxygen, or 3 : 2 : 8 . This ratio cannot be further reduced, so the empirical formula is also Ba 3 (P0 4b with a formula mass of 3 ( 1 3 7 . 3 ) + 2(30.97) + 8 ( 1 6 . 0 0 ) 60 1 . 8 g/mol. e) The ratio of atoms i s 4: 1 6, or 1 :4. The empirical formula i s TeI4, and the formula mass is 1 27 . 6 + 4( 1 2 6 . 9 ) 635.2 g/mol. =

=

=

=

=

3.27

P lan: Determine t h e molar mass o f each empirical formula. T h e molar mass of each compound divided b y its empirical formula mass gives the number of ti mes the empirical formula i s within the mo lecule. M ultiply the emp irical formula b y the number of times the empirical formula appears to get the molecular formula.

37

Solution : Only approximate whole number values are needed. a) CH2 has empirical mass equal to 1 4.03 glmol

(

42.08 g/mol 1 4.03 g / mol

)

=3

Multiplying the subscripts in CH2 by 3 gives C3H6 b) NH2 has empirical mass equal to 1 6 .03 glmol

(

32.05 g/mol 1 6 .03 g / mol

)

=2

Multiplying the subscripts in NH2 by 2 gives N 2 H4 c) N02 has empirical mass equal to 46.0 1 glmol

(

92.02 g/mol 46. 0 1 g / mol

)

=2

Multiplying the subscripts in N02 by 2 gives N 2 04 d) CHN has empirical mass equal to 27.03 glmol

(

)

1 3 5 . 1 4 g/mol = 5 27.03 g/mol

Multiplying the subscripts in CHN by 5 gives CsHsNs 3 .29

Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to moles of an element. Using the smallest number of moles present, convert the mole ratios to whole numbers. Solution : a)

(

0.063 mol CI 0.063 mol C l

J

(

= I

0.22 mol 0 0.063 mol Cl

J

= 3.5

The formula is Cl , 0 3 .5, which in whole numbers (x 2) is CIz07 b) ( 2 .45 g Si )

(

(

1 mol Si 28 .09 g Si

0.08722 mol Si 0.08722 mol Si

J

J

( 1 2.4 g CI )

= 0.08722 mol Si

(

= 1

0.349788 mol Cl 0.08722 mol Si

The empirical formula is SiCI4• c) Assume a 1 00 g sample and convert the masses to moles.

( 1 00 g )

(

(

27.3% C 1 00%

2.273 1 mol C 2.273 1 mol C

J

J(

I mol C 1 2. 0 1 g C

J

(

4.5438 mol O 2.273 1 mol C

The empirical formula is CO 2 • 3.3 1

1 mol Cl 3 5 .45 g CI

J

)

= 0.349788 mol Cl

=4

( 1 00 g )

= 2.273 l moI C

= I

J

(

(

72.7% 0 1 00%

J(

1 mol 0 1 6.00 g 0

)

= 4.5438 mol 0

=2

P lan: The balanced equation for this reaction is: M(s) + F2(g) � MF2(s) since fluorine, like other halogens, exists as a diatomic molecule. The moles of the metal are known, and the moles of everything else may be found from these moles using the balanced chemical equation. Solution : a) Determine the moles of fluorine. Moles

F = ( 0.600 mol M )

(

2 mol

F

1 mol M

J

= 1 .20 mol F

38

F) ( 19.1 00 F ) = 40.=0 40.0

b) The grams o f M are the grams of M F2 m i n us the grams of F present. Grams M

=

46. 8

(1.20 = 24.0 0. 600

g (M + F)

-

mol

g

mol F

c) The molar mass i s needed to identify the el ement. M o l ar mass of M gMI mol M The metal with the c losest m o l ar mass to

3. 3 3

100

24.0 g M

g I mol g/mol i s calci u m .

P l an : Assume grams o f cortisol s o the percentages are n umerically equivalent t o the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. D i v i de all moles by the lowest number of moles and convert to whole numbers to determ ine the empirical formula. The empirical formula mass and the given molar mass w i l l then re late the empirical formula to the molecular formula. Solution :

(69. 6 ) ( 12.1 0 1 ) = 5. 7952 = (8. 3 4 ) ( 1.1008 . ) = 8.2738 = (22.1 ) ( 16.1 00 ) 1. 3 8125 738 ) 6. 00 ( 1.1.33 8125 952 J = 4.20 ( 1.8.328125 8125 ) = 1.00 ( 1.5.378125 5. 21 (12. 0 1 30 (1. 008 5 (16. 00 362.45

M o les C

mol C

g C

=

Moles H

g H

Moles °

g 0

mol

mol C

g C

mol H

mol

g H

mol °

mol °

=

g °

C

H

mol

H

mol 0

=

mal O

mol °

maI O

4.20

The carbon value i s not close enough to a whole n umber to round the value. The smallest number that may be multiplied by to get c l ose to a whole number i s (You may wish to prove thi s to yoursel f. ) A l l three ratios need to be multi p l i e d by five to get the empirical formula Of C2 1 H3 0 0S . The empirical formula mass i s : g C/mo l ) + g H/mol ) + g O/mo l ) glmol The empirical formula mass and the molar mass given are the same, so the empirical and the molecular formulas are the same . The molec u l ar formula is C 2 1 H 30 0S. =

3. 34

P l an : I n combustion analysis, fi nding the moles o f carbon and hydrogen i s rel atively s i mp l e because all o f the carbon present in the sampl e is found in the carbon of CO2, and all of the hydrogen present in the samp l e i s found i n the hydrogen of H 20 . The moles of oxygen are more difficult to fi nd, because additional O2 was added to cause the combustion reacti o n . The masses of CO2 and H20 are used to find both the mass of C and H and the moles of C and H . S ubtracting the masses of C and H from the mass of the sampl e gives the mass of O . Convert the mass of ° to moles of 0. Take the moles of C , H, and ° and divide by the smallest value, and convert to a who l e number t o g e t the empirical formula. Determ ine t h e empirical formula m a s s and compare i t to t h e mol ar mass given i n the prob lem to see how the empirical and mo lecular form ulas are rel ated. F i nally, determine the molecular formula. Solution: ( There i s n o intermediate ro unding. ) I n i tial mole determinati o n : Moles C

M o les H

( 44. 0 1 1 )( 18.02 = (0.0 10202 C ) ( 12.1 0 1 = (0.020422 ) ( 1. 008

= (0.449 = (0. 184

g CO2

)

g H 20

I

mol CO 2 g CO2

mo l H 2 0 g H20

Now determi ne the masses of C and H : Grams C

Grams H

mol

mol H

g C

mol C

I

) 0.0 10202 J( J ( 1 2 J = 0. 020422 ) = 0.122526 ) = 0.020585 I

I

mol C

mo l H

mol

H20

gC

g H

mol H

=

mol CO2

gH

39

mol C

mol H

- 0.122526 - 0.020585 0.0 16389 (0.0 16389 ( 16.1 00 J 0.0010243 0.0.00010243 20422 J 19.9 20 ( 0.0. 00010243 ( 9. 9 6 10 010243 J J 10 (12. 0I 20 (1.008 1 (16. 00 156.26

Determine the mass and then the moles of 0: g (C, H, and 0) gC gH= gO mol 0 = mol 0 g 0) Moles 0 = g0 Divide by the smal lest number of moles: (Rounding is acceptable for these answers.) mol O mol H mol C =I = mol ° mol ° mol ° Empirical formula = C , oH 2 0 0 Empirical formula mass = g C/mol) + g H/mol) + g O/mol) = glmol The empirical formula mass matches the given molar mass so the empirical and molecular formulas are the same. The molecular formula is C IO H 20 0 .

0.1595

10202 ( 0.0.00010243

3. 3 6 3. 3 7

=

=

=

Plan: Examine the diagram and label each formula. We wil l use A for red atoms and 8 for green atoms. Solution : The reaction shows A 2 and 8 2 molecules forming AB molecules. Equal numbers of A 2 and 8 2 combine to give twice as many molecules of AB. Thus, the reaction is A 2 + 8 2 4 AB. This is the answer to part b.

2

Plan: Balancing is a trial and error procedure. Do one blank/one element at a time. Solution : a) lQ Cu(s) + _ Ss(s) 4 � Cu 2 S(s) Hint: Balance the S first, because there is an obvious deficiency of S on the right side of the equation. Then balance the Cu. b) P 4 0 , o (s) + Q H 2 0( 1) 4 :! H 3 P04 ( 1) Hint: Balance the P first, because there is an obvious deficiency of P on the right side of the equation. Balance the H next, because H is present in only one reactant and only one product. Balance the ° last, because it appears in both reactants and is harder to balance. c) _B 2 0 3 (s) + Q NaOH(aq) 4 2. Na3 B0 3 (aq) + J. H 2 0( 1) Hint: Oxygen is again the hardest element to balance because it is present in more than one place on each side of the reaction. If you balance the easier elements first ( 8 , Na, H), the oxygen will automatically be balanced. d) 2. CH 3 N H 2 (g) + 0 2 (g) 4 2. CO 2 (g) + l H 2 0(g) + _N 2 (g) ) CH -± 3 N H 2 (g + 2. 0 2 (g) 4 :! CO 2 (g) + lQ H 2 0(g) + 2. N 2 (g ) Hint: You should balance odd/even numbers of oxygen using the "half' method, and then multiply al l coefficients by two. _

9/2

3. 3 9

Plan: The names must first be converted to chemical formulas. The balancing is a trial and error procedure. Do one blank/one element at a time. Remember that oxygen is diatomic. Solution : a) :! Ga(s) + J. 0 2 (g) 4 2. Ga2 0 3 (s) b) 2. C 6H ' 4 ( 1) + l2. 0 2 (g) 4 .ll CO 2 (g) + H 2 0(g) c) J. CaCI 2 (aq) + 2. Na3 P0 4 (aq) 4 Ca3 (P04 Ms) + Q NaCI(aq)

14

3.4 1

Plan : Convert the ki lograms o f oxygen to the moles o f oxygen. Use the moles o f oxygen and the mole ratios from the balanced chemical equation to determine the moles of KN0 3 . The moles of KN0 3 and its molar mass will give the grams. Solution : 3 mol KN0 3 mol 02 g 3 a) Moles KN0 3 = kg 02 ) = 2.22 x 1 0 mol KN 03 I kg g O2 mol O2 1 03 g mol O2 mol KN0 3 g KN0 3 kg O2 ) b) Grams KN0 3 = I kg g O2 mol O2 I mol KN0 3 = = 2.24 x 1 0 5 g KN0 3

(88.6 (88. 6 223958.65

[ 10 )( 32.I 00 )( 4 5

[ )( 32.1 00 ) ( 4 5 40

) 2215 )( 101.11

=

)

The beginning of the calculation is repeated to emphasize that the second part of the prob lem is s i mply an extension of the fi rst part. T here is no need to repeat the entire calculation, as only the fi nal step ti mes the answer of the fi rst part w i l l give the final answer to thi s part. 3 .43

P lan : F i rst, balance the equati on. Convert the grams of di borane to moles of diborane using its mo lar mass. Use mole ratios fro m the balanced chemical equation to determine the moles of the products. U se the moles and molar mass of each product to determi ne the mass formed. Solution : The balanced equation i s : B 2 H 6 (g ) + 6 H 2 0(l) � 2 H3B03(S) + 6 H 2 (g). M ass H 3 B03 =

M ass H 2

=

=

3 .45

(33.6 1

1 50.206

(33.6 1

=

(

I mol B 2 H 6

2 7 . 67 g B 2 H 6

1 50.2 g H 3 B0 3

g B2 H 6

1 4 . 69268

)

g B2 H 6

=

)

(

I

mol B2 H 6

2 7 . 6 7 g B2 H 6

1 4.69 g H 2

J(

J(

2 mol H 3 B03 I mol B2 H 6

6 mol

H2

1 mol B2 H 6

)(

J(

6 1 . 8 3 g H 3 B03 I mol H 3 B03

2.0 1 6 g H 2 I

mol H 2

)

J

Plan: Write the balanced equation by first writing the formulas for the reactants and products . Reactants : formula for phosphorus is given as P4 and formula for chlorine gas i s CI2 (chlorine occurs as a d i atomi c molecule). Products : formula for phosphorus pentachloride - the name indicates one phosphorus atom and five chlorine atoms to give the formula P C l s . Convert the mass of phosphorus to moles, use the mole ratio from the balanced chemical equati on, and fi n a l l y use the molar mass of chlorine to get the mass o f chlori ne. S o l ution : Formulas give the equati o n : P 4 + C I 2 � PCls Balancing the equati o n : P4 + 1 0 C I 2 � 4 PC I s Grams C h

3 .47

=

=

(355 g

P4

)

(

I

mol P4

1 23 . 8 8 g P4

J

(

1 0 mol C I 2 I mol P4

J(

70. 90 g C I 2 I mol C I 2

J

=

203 1 . 76

=

2.03

x

3 1 0 g Ch

Plan: Convert the given masses to moles and use the mole ratio from the balanced chemical equation to find the moles of CaO that w i ll form. The reactant that produces the least moles of CaO i s the I i m iting reactant . Convert the moles of CaO fro m the l i m iting reactant to grams using the molar mass. S o l ution : a) Moles CaO fro m Ca

b) Moles CaO from O 2

( o? ) (

I mol Ca

=

( 4 . 2 0 g ca )

40.08 g Ca

=

( 2 . 80 g

3 2 . 00 g 02

-

1 mol ° 2

)( )(

2

mol ca

) o J

o

=

2 mol Ca 2 mol c a

=

I mol 0 2

0 . 1 04790

0 . 1 7 5 00

=

=

0. 1 05 mol CaO

0. 1 75 mol CaO

c) Calciu m i s the l i m iting reactant si nce it wi l l form less calcium oxide. d ) The mass of C aO formed i s determined by the l i miting reactant, Ca. Grams CaO

3 .49

=

(4.20 g ca )

(

I mol Ca 4 0 . 0 8 g Ca

)(

2 mol CaO 2 mol Ca

)(

5 6 . 0 8 g ca

o

1 mol CaO

)

=

5 . 8 766

=

5.88 g CaO

P l an : First, balance the chemi cal equation. Determine which o f the reactants i s the l i miting reagent. Use the l i m iting reagent and the mo l e ratio from the bal anced chemical equation to determi ne the amount of materi al formed and the amount o f the other reactant used. The difference between the amount of reactant used and the i n itial reactant s upp l ied gives the amount of excess reactant remai ning. S o l ution : The balanced chemical equation for this reaction i s : 2 l C I 3 + 3 H 2 0 � r C I + H I03 + 5 H C I H i nt : Balance the equation by starting with oxygen. The other el ements are in multiple reactants and/or products and are harder to balance initially.

41

N ext, fi n d the l i m iting reactant by using the molar ratio to find the smaller number of mo les of H I 03 that can be produced from each reactant gi ven and excess of the other: Moles H I 03 from I C I 3 =

( 685 g

M o l e s H I 03 fro m H 2 0 =

( 1 1 7 .4 g

ICI3

)

(

H )- O

I

(

mol T C I 3

233.2 g ICI3

)

I

)(

mol H 2 °

1 8 .02 g H 2 0

1 mol H I03

)(

2 m o l IC I 3

)

= 1 .468696 = 1 .4 7 mol H I 03

1 mol H I03 . 3 mol H 2 0

)

= 2 . 1 7 1 66 = 2 . 1 7 mol H I 03

I C I 3 is the l i miting reagent and w i l l produce 1 .47 mol HI0 3 . Use the l i m iting reagent to fi n d the grams of H I03 formed. G rams H I 03 =

( 685 g

ICI3

)

(

I mol I C I 3 2 3 3 . 2 g lC I 3

J(

I mol H I 03 2 mol l C I 3

)(

1 7 5 . 9 g H I03 I mol H I03

J

= 2 5 8 . 3 5 5 = 258 g H I 0 3

The remai ning mass of the excess reagent can be cal culated from the amount of H20 comb i n i ng with the l i miting reagent.

( 685 g

Grams H20 required to react with 685 g I C I 3 :

ICI3

. = 7 9 . 3 9 8 = 79.4 g H20 reacted Remain i n g H 20 = 1 1 7 .4 g 79.4 g = 38.0 g H z O

)

(

1 mol I C I 3 233.2 g ICI3

)(

3 mol H 2 0 2 mol ICI3

)(

1 8 .02 g H 2 0 1 mol H 20

)

-

3.5 1

P l a n : Write the bal anced equation: formula for carbon is C, formu l a for oxygen i s O2 and formula for carbon dioxide is CO2, Determ ine the l i miting reagent by see ing which reactant wi l l yi e l d the smal ler amount of product. The l i miting reactant is used for a l l subsequent calculations. S o l ution : C(s) + 02(g) � CO2(g) Moles CO2 from C =

( 0 . 1 00 mo l C )

M o l e s CO2 from O2 =

( 8 . 00 g ° 2 )

(

(

(

) J(

1 mol CO 2 1 mol C I mol °2

3 2 . 00 g 02

)(

= 0 . 1 00 mo I C02 I mol CO2 I mol 02

)

)

= 0 . 2 5 000 = 0 . 2 5 0 mol CO2

The C is the l i m iting reactant and wi l l be used to determine the amount of CO2 that wi l l form . Grams CO2 =

c l ( 0 . 1 00 mol )

mol CO 2 1 mol C

44 . 0 1 g CO 2 I

mol CO2

= 4 .40 1 = 4.40 g CO z

Since the C is l i miti ng, the O z is in excess. The amount remai n i ng depends on how much combines with the l i miting reagent. . . O2 req U i red to react wIth 0 . 1 00 mo l of C = Remaining O2 = 8 . 00 g 3.53

-

( 0 . 1 00 mol C )

3 . 2 0 g = 4.80 g O z

(

I mol 02 I

mol C

)(

3 2 . 00 g O2 1 mol 0 2

)

= 3 .2 0 g O2

P l a n : The question asks for the mass of each substance present a t t h e end o f t h e reaction. "S ubstance" refers to both reactants and products. Solve this problem using multip l e steps. Recognizing that this is a l i miting reactant pro b l em, first write a bal anced chemical equation. U sing the mo l ar re lationsh ips from the balanced equation, determine whi c h reactant i s l i m iti ng. Any product can be used to pred ict the l i miting reactant; i n this case, AICI3 is used. Additional significant figures are retained unti l the last step . S o l ution : The balanced chemical equation i s :

AI(N 02Maq) + 3 N H4C I(aq) � A I C I3(aq ) + 3 N2(g) + 6 H 20(l) Now determi n e the l i miting reagent. We w i l l use the moles of A I C I 3 produced to determine which is l i m iting. Mole A I C l 3 from AI(N02)3 =

( 62 . 5 g A I ( N 02 )3 )

= 0 . 3 7 8 7 6 = 0 . 3 7 9 mol A I C I 3

(

l mol A l ( N 0 2 h

1 6 5 . 0 I g A I ( N 02 )3

42

)(

I

I

mol A I C I 3

mol Al ( N 0 2 ) 3

)

Mole AICI) from NH4CI = ( 54.6 g NH4CI )

(

1 mol NH 4 CI 53 .49 g NH4CI

)(

1 mol AICI 3 3 mol NH4CI

) 0. 34025 =

= 0.340 mol AICI)

Ammonium chloride is the limiting reactant, and it is important for all subsequent calculations. Mass of substances after the reaction: AI(N02») : AI(N02) 3 required to react with 54.6 g of NH4CI: ( 54.6 g NH 4 C I ) 1 mol NH4CI 1 mol AI(N02 h l 65 .0 1 g AI(N02 )3 5 3 .49 g NH4CI 3 mol NH4CI I mol AI(N02 )3 = 5 6 . 1 4474 = 56. 1 g AI(N02») AI(N02) 3 remaining: 62 . 5 g - 56. 1 g = 6.4 g A I(N 02 h

)(

(

)(

)

NH4CI: none left since it is the limiting reagent.

( )(

( 54.6 g NH4CI ) ( 54.6 g NH4CI

3.55

3.57

1 mol NH4CI 5 3 .49 g NH4CI 1 mol NH4CI 5 3 .49 g NH4CI

)( J(

1 mol AICI3 3 mol NH4CI 3 mol N 2 3 mol NH4CI

)( ) ( 28.02 J J 1 33 . 3 3 g AlCI3 1 mol AICI3 g N2

1 mol N2

= 45 .3656 = 45.4 g A ICI3

= 2 8 . 60 1 = 28.6 g N 2

Plan: Multiply the yield of the first step by that of the second step to get the overall yield. Solution : It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 1 00%. (0.65) ( 1 00%) = 5 3 . 3 = 53 %

(0. 82)

Plan: Balance the chemical equation using the formulas of the substances. Determine the yield (theoretical yield) for the reaction from the mass of tungsten(VI) oxide. Use the density of water to determine the actual yield of water in grams. The actual yield divided by the yield j ust calculated (with the result multiplied by 1 00%) gives the percent yield. Solution: (Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: W0 3 (s) + 3 H2(g) � W(s) + 3 H20(l) Theoretical yield of H20 :

( 4 1 .5 g W03 )

(

1 mol W03 23 1 .9 g W03

( 00

J )

(

3 mol H 2 0 1 mol W03

Actual yield, in grams, of H20 :

( 9.5

(

0

mL H 2 0 )

1.

Calculate the percent yield: Theoretical Yield

J

g H20

I mol H 2 0

J

= 9.6743 9 g H20

g H20 = 9.50 g H 2 °

1 mL H 2 0

Actual Yield

J

( 18.02

x 1 00% =

(

9.50 g H 2 0 9.67439 g H20

J

x 1 00% = 98. 1 974 = 98.2%

43

3 . 59

quantitieswiofl l form. ( o d on. Sinceof products tyhe balancedof thechemicalcculalaequati of amounts d e t g. Onltefigures lsiPlimganin:itifinWricant ) . t resul final the l unti postponed be l wi Sol u(tigo)n: CI2(g) C H3C I (g) HCI (g) CH4 Mole HCI from CH4 = g CH4)( molgCH4CH4 ) ( molmol CH4 ) = Mole HCI from C I2 = g C1 2 ) ( molgCICI2 2 )( molmol CI 2 ) = mol Chlorine is the l i miting reactant. Grams CH3CI = ( g CI 2 ) ( molgCICI2 ) ( molmol ) ( mol ) ( ) =ning of the= calculation i s repeated to emphasize that the The begi n extensi to repeatto thithsepartenti. re calc answeroofn ofthethfirste firstpartpart.wil lThere give theis nofinalneedanswer first stemust p is tobedetermi nneed.theFichemi c, altheformul as soCF4a balis anced lSolPliman:uittiinTheogn:reactant determi n al l y mass ng toonthise: correct number of significant figures ( Roundi The bal a nced chemi c al equati (CN)2(g) F2(g) CF4(g) N F3(g) mol ( CNh mo l C 4 F4 Mol e CF4 from ( CN)2 = ( g(CN)2) ( g (CNh )( F ) = Mole CF4 from F2 = ( g F ) ( I molg F2 )( molmolCF4F2 ) = mol CF4 F2 is the l i miting reactant, and wi l be used to calculate the yield. Grams CF4 = g F )( molg F )( molmolCF4F )( g CF4F4 ) = = n: The spheres represent partimolcelessofofsolsoluutete/volandumethe(amount solute solution iSolPltsaconcentrat i o n. Mol a ri t y = L) a)b) ution: hashas more more solsolvutente added because it contmoleaculins es because sol v ent c)d) hashas aa hil ogwerherconcentrati mol arity, because moles it has fewer moles ty), because on (and imolt hasarimore P n aI cases, the de I n ltl. on f mo lant' y Lmolofesols solutIuOten ) b . molar mass i s used to convert moles to grams. theSolliters.umoltioThen:ar mass. a) Grams Ca( C2H302)2 = mL)[ mLL J( mol aL C 0 )( J = =

reactants are present, we must determine which is R un ing to the correct number

80.0% �

+

+

1 1 6.04

( 1 8.5

1

1

2

�

7

2

+

1 52.04

( 80.0

2

F2

2

2 7

38.00

I

I

80.0% 1 00%

I F2 38.00 2

2 7

3 .074558 mol C

0.60 1 50

88.0 1

I mol

2

52.9383

C

of of solution.

--..J!!J..:.

CH 3 CI

chemical equation can be written. The determined from the limiting reactant. will be postponed until the final result.)

2 I mol (CN) 2

Box C Box B Box C Box B

3 . 65

CH3CI

2

80.0

80.0

0.606488

mol

second part of the problem is simply an ulation as only the final step(s) times the

of

+

1 . 1 53367

50.48 g CH 3 CI 1 CH 3 CI

CH 3 CI CI 2

24.5 g C H 3 CI

24.4924

3.64

CH 3 CI

1

1

I 70.90

43.0

3 .6 1

1

1 70.90

(43.0

CH 3 C I

I

52.9

g

C

F4

per given volume o f

determines

2 more spheres than Box A contains.

have displaced two solute molecules. of solute per volume of solution. of solute per volume of solution.

fi

I

.

0

(M

=

.

.

.

WI'11 e Important. Volume must be expressed In

The chemical formulas must be written to determine

( 1 75 .8

5 .7559

1 0- 3

0.207

I

5.76 g Ca(C 2 H 3 0 2 ) 2

44

C ( 2 H3 2 )2 1

1 58. 1 7

1 mol

g

Ca(C 2 H 3 0 2 h Ca(C 2 H 3 0 2 h

(2 1 . 1 g KI f

)

1 mol KI \ 1 66.0 g KI 0. 1 27 1 08 moles Kl 1 O -3 L Volume (SOO. mL -O.SOO L I mL 0. 1 27 1 08 mol KI = 0.2S42 1 6 = 0.254 M KI Molarity KI

b ) Moles KI

=

{ J

=

c ) Moles NaCN 3 .67

=

=

O . S OO L

=

(

=

J

0.8S0 mol NaCN ( 1 4S.6 L ) 1L

1 23 . 76

=

=

1 24 mol NaCN

Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new volume, however, it is much easier to use MI V I M2 V2. Part (c ) may be done as two dilution problems or as a mole problem. The dilution equation does not require a volume in liters; it only requires that the volume units match. Solution: a) MI 0.2S0 M KCI V I 3 7.00 mL M2 ? V2 I SO.OO mL (0.2S0 M)(37.00 mL ) (M2)( I SO.0 mL ) MI VI = M2V2 (0.2S0 M)(37.00 mL ) M2 M2 0.06 1 667 0.06 1 7 M KCI I SO.O mL b ) MI 0.0706 M (NH 4 hS0 4 V I 2S . 7 1 mL M2 ? V2 SOO.OO mL = I (M2)( I SOO.0 mL 2V2 (0.0706 M)(2S .7 1 mL M VI M ) ) (0.0706 M)(2S .7 I mL ) M2 0.003630 0.00363 M (NH4) 2 S04 M2 SOO.O mL =

=

=

=

=

=

_

=

=

=

=

=

=

=

-

_

=

=

c ) When working this as a mole problem it is necessary to find the individual number of moles of sodium ions in each separate solution. (Rounding to the proper number of significant figures will only be done for the final answer. ) 3 1 mol Na + 0.288 mol NaCl 1 O- L (3.S8 mL ) Moles Na+ from NaCl solution I mol NaCl I mL IL 0.00 1 03 1 04 mol N a+ 3 2 mol Na + 6.S 1 x 1 0 -3 mol Na 2 S0 4 1O- L Moles Na+ from Na2 S0 4 solution (soo. mL ) 1 mol Na 2 S0 4 I mL 1L 0.006S I O mol Na+ ( 0.00 1 03 1 04 + 0.006S 1 0) mol Na + I mL --3- 0.0 1 497486 0.0 1 50 M Na+ ions Molarity of Na+ ( 3 . S 8 + SOO. ) mL 1 0- L =

J[ ]

(

=

=

][ ]

[

]( )

=

=

3 .69

]

[

[

=

]

[

=

Plan: Use the density of the solution to find the mass of I L of solution. The 70.0% by mass translates to 70.0 g solutell OO g solution and is used to find the mass of HN0 3 in I L of solution. Convert mass of HN0 3 to moles to obtain moles/L, molarity. Solution: 70.0 g HN?3 1 .4 1 g Solution I mL 987 g HN03 / L a) Mass HN0 3 per liter 3 I mL 1 0- L 1 00 g SolutI O n =

(

(

1 .4 1 g Solution . b ) Molanty of HN0 3 = I mL I S .66 1 7 1 5.7 M HN03 =

=

(J )( J (--)( I mL 1 0 -3 L

J 3 ( J

70.0 g HN 0 . 1 00 g SolutI On

4S

=

I mol HN0 3 63 .02 g HN0 3

)

3.7 1

cal aciequatid wioln togivfienthed the e toes molof acies,dandalongusewiththe tbalheamolncedarichemi ofdcalrequiciumred.carbonat the mass Plmolan:esConvert the of y t mol The aci c ri o hydrochl of volSoluumetion:required. The molarity of the solution is given in the calculation as mollL. HCI(aq) CaC03(s) CaCI2(aq) CO2(g) H20(l) Volume required = g cac0 { �0��1 �����J [ �:o�:c�J ( 0.383 �I JUO�) volumeUseof the reactininogn.whiUsechtheis themolliamriittyinandg reactant. on tofordetermi calaequati write andne thebalamolnceetshofe chemi step i stotodetermi Pleachan: ofThethefirstreactants de u prel as each tSolhe ulitimoin:ting reactant to determine the mass of barium sulfate that wil form. The bal2(aq)ancedNa2S04(aq) chemical equatiBaS04(s) on i s : NaCI(aq) BaCI The mole and l imiting reactant calculations are: [-mol Moles BaS04 from BaCI2 = mL ) ( mol BaCI J [ molmol a J Moles BaS04 from Na2S04 = mL) [--) [ molLNa )( molmol ) mol Sodium sulfate is the limiting reactant. [ mL ) ( moI l J( molmol )[233.mol4 g ) Grams BaS04 = = = firstof thepartHCIofthaleoprobl etmh theis amolsimaplriety.dilution problem (MI VI = M2V2). The second part requires the molSolPlan:uatirTheomass ng wi a) MI =n:M VIM= M2V2 VI = M)(VIM2 =) = M V2gal=) gal gal) V VI = gallons (unrounded) Instructiner.ons:AddBeslsureowlytoandwearwitgoggl contai h mixeisntgo protectgalloyour n of eyes!MPourHCIapproxi into thematwatelyer. Digallutelotnso of water gallonsintowitthhewater. Volume needed = g H I ( molgHCIHCI J( mol HCI J ( ) = ne andthis themassmoltoedetermi s of watneer theare mass requirofed.watWeercanpresentassumeandanyconvert masstheof mass narceitonemolhydrates ofe wiThel usemolThees ofmass g),narcei andofusewater (tPlhweean:hydrate. wil bepresent. converted to moles. Finally, the ratio of the moles of hydrate to moles of watSoluetir owin:l give the amount of water ) n e hydrat e Moles narceme. hydrate = ( g narceme. hydrate ) [ mol gnarcei narceine hydrate = mol narceine hydrate 2

�

+

+

( 1 6.2

=

3 .73

845 . 1 923

=

3

+

�

+

+

(25.0 mL) ( 68.0

( 68.0 mL)

1 1 .7

?

(3.5 M)(5.0 I 1 .7 M

-

_

1 O-3 L 1

2

1 O-3 L 0. 1 60 1L 1 1 O-3 L 0.055 1 1 mL

0.055

3.5 ( 3 . 5 M)(5 .0

( 1 1 .7

3 .76

BaS04

I

B CI2

Na2S04

BaS04

Na2S04

0.00400

BaSO4

1

1

1

=

=

0.00374

BaS04

BaS04

1

=

5.0

l .4957

I

C

(9.55

22.38725

1

2 SO4

I

L

1 .5

b)

Na2S04

2

0.87 g BaS04

0.8729 1 6

3 .75

HCI

1

1

845 mL HCl solution

)

1 36.46

3.0

1 1 .7

1L

1 l .7

22.4 mL muriatic acid solution

1 00

1 00

1 499.52

0.200 1 9

46

1 mL 1 O-3 L

5 .0

BaS04

BaS04

Moles H 2 0 =

( 1 00 g n arcei. n e hydrate )

(

1 0 . 8% H ?- O 1 00% narceine hydrate

J

(

1 mol H 2 ° 1 8 . 02 g H 2 0

)

= 0 . 5 99 3 3 mol H 20 The rati o of water to hydrate i s : ( 0 . 5 99 3 3 mol) / ( 0 . 200 1 9 mo l ) = 3 Thus, there are three water molecules per mole of hydrate. The formula for narceine hydrate is narceineo3 H 2 0 3.77

Plan : Determi n e the formula, then the molar mass o f each compo und. Determ ine the mass o f hydrogen i n each formula. The mass of hydrogen divided by the molar mass of the compound (with the result multi p l i ed by 1 00%) w i l l give the mass percent hydrogen . Ranki ng, based on the percents, i s easy. Sol ution : N ame C h e m ical M o l ar mass Mass percent H formula (glmol ) [( mass H ) / ( molar mass)] x 1 00% Ethane C2H6 30.07 [ ( 6 x 1 . 008) / ( 3 0 . 07)] x 1 00% = 2 0 . 1 1 % H Propane C3Hg 44.09 [ ( 8 x 1 . 008) / (44 . 09)] x 1 00% = 1 8 .29% H Cetyl pal mitate C32 H 6 402 480.83 [ ( 64 x 1 . 008) / (48 0 . 83 )] x 1 00% = 1 3 .42% H Ethano l C 2 H sO H 46.07 [ ( 6 x 1 . 008) / (46 . 0 7 )] x 1 00% = 1 3 . 1 3 % H B enzene C6H 6 78 . 1 1 [ ( 6 x 1 . 008) / ( 7 8 . 1 1 )] x 1 00% = 7 . 743% H The hydrogen percentage decreases i n the fo l l owing order: Ethane > Propane > Cetyl palmitate > Ethanol > Benzene

3.8 1

P l an : I f 1 00 . 0 g of d i nitrogen tetroxide reacts with 1 00 . 0 g of hydrazine ( N 2 H 4) , what i s the theoretical yi eld of ni trogen if no side reaction takes p l ace? F i rst, we need to ident i fy the l i miting reactant. The l i miting reactant can be used to calcu late the theoretical yie l d . Determ ine the amount of l i miting reactant required to produce 1 0 . 0 grams of N O . Reduce the amount of l i miting reactant by the amount used to produce NO. The reduced amount of l i miting reactant is then used to calculate an "actual yield." The "actual" and theoretical yields w i l l give the max i m u m percent yield. Solution : Determ i n i n g the l im i t i n g reactant : Nz from N z04 =

( 1 00 . 0 g

N ? 04 -

)

(

I mo I N 2 04 9 2 . 02 g N 2 0 4

NZ04 is the l i m it i ng reactant. . . Theoretical Yield of Nz =

( 1 00 . 0 g N 2 04 )

(

)(

3 mol N 2 1 mol N 2 0 4

I mol N 2 ° 4 92 . 02 g N 2 04

)(

( 1 0. 0 g NO )

(

I mol N O 30.0 1 g NO

)(

= 3 . 2 60 1 6 mo I N z

3 mol N 2 1 mol N 2 04

= 9 1 .3497 g Nz ( unrounded) H o w much l i miting reactant used to produce 1 00 . 0 g NO? G rams N ?04 used = -

)

2 mol N 2 ° 4 6 mol N O

)(

)(

2 8 . 02 g N 2 I mol N 2

92.02 g N 2 0 4 1 mol N 2 0 4

)

)

= 1 0 . 22 1 g Nz04 ( unrounded)

Determi n e the "actual y i e l d . " A m o u n t of N z04 avai lable t o produce N 2 = 1 00 . 0 g NZ04 - m a s s of N 204 required t o produce 1 0 . 0 g N O 1 00 . 0 g - 1 0 .22 1 g = 8 9 . 7 7 9 g NZ04 ( unrounded)

(

"Actual yie ld" of N -? = 8 9 . 7 7 9 g N 2 0 4

{

I mol N 2 ° 4 92.02 g N 2 0 4

)(

3 mol N 2 1 mo I N 2 0 4

)(

Theoretical yield = ["Actual yield" / theoreti cal yield] x 1 00% [ ( 8 2 . 0 1 28 5 g N2) / ( 9 1 . 3497 g N2)] x 1 00% = 8 9 . 7 790 = 89.8%

47

2 8 . 02 g N 2 I mol N 2

)

3 . 82

Plan: Count the number of each type of molecule i n the reactant box and i n the product box. S ubtract any molecules of excess reagent ( molecules appearing in both boxes) . The remain i ng material is the overal l equation. This will need to be s i mp l i fi ed i f there i s a common factor amon g the substances i n the equation. The balanced chemical equation is necessary for the remainder of the problem. Solution: a) The contents of the boxes give: B 2 -7 AB3 2 B2 B2 i s i n excess, so two molecules need to be removed from each side. This gives:

6 AB2 + 5 6 + 6 +3 6 + (5. 0 { J 5. 0 (3 . 0 { J 6. 0 (5 . 0 {\ 5. 0 3.0

AB2 B 2 -7 AB3 Three i s a common factor among the coeffic ients, and a l l coeffi c i ents need to be divided by thi s value to give the fi nal balanced equati o n : 2 AB2 B2 -7 2 AB 3 b) B2 was i n excess, thus AB2 is the l i miting reactant. c) Moles of A B 3 from

M o les of AB3 from

AB2

82

2 mol A B 3 2 mol A B 2

mol A B 2

=

2 mo1 A B 3 I mol B 2

mol B 2

=

=

mol A83

mol A83

=

AB2 is the l i miting reagent and maximum is 5.0 mol A B3 . d)

Moles of

82 that reacts with

The un-reacted B2 is

3. 8 5

mol

mol - 2 . 5 mol

=

AB 2

mol A B 2

=

I mol B 2

2 mol A B 2

I)

=

2 . 5 mol B2

0.5 mol B2 .

P l an : Count t h e total number of spheres i n each box. T h e number i n b o x A d ivided by the volume change in each part wi l l give the number we are looking for and allow us to match boxes. S o l ution : The number i n each box i s : A 1 2, B C and D a ) When the volume i s tri pled, there should b e spheres i n a box. This i s box C . b ) When the volume i s doubled, there should be 1 212 spheres i n a box. This i s box B . c) When the volume i s q uadrupled, there should b e spheres i n a box. This i s box D . =

=

6, 12/3 4, 4 3. 12/46 3 =

=

=

=

=

3.89

Plan: T h i s prob lem may be done as two d i l ution problems with the two fi nal molarities added, or, as done here, it may be done by calculating, then adding the moles and dividing by the total volume. S o l ution : M KBr

=

Total M o l e s K B r

M o les K B r from Solution

( 0.053 �t }0.200 L ) + ( 0. 078 t 0.200 L + 0. 5 50 L m

KBr

M KBr

3. 92

KBr

I

=

1 ++ } 0. 5 50 L )

Volume Solution I

Total Volume

M oles K B r from S olution 2

Volume S olution 2

=

0.071333

0.071 M KBr

=

Plan: Deal with the methane and propane separate ly, and combine the results. Balanced equations are needed for each hydrocarbon. The total mass and the percentages w i l l give the mass of each hydrocarbon . The mass of each hydrocarbon i s changed to moles, and through the balanced chemical equation the amount of CO2 produced by each gas may be found. S u m m i ng the amounts of CO2 gives the total from the m ixture . S o l ution : The balanced chemical equations are : M ethane : CH 4 (g) 2 02(g) -7 CO2(g) 2 H20(l) C 3 H g (g) 02(g) -7 CO2(g) H20(l) Propan e : M ass of CO2 from eac h : M et h ane:

(200

++ 5 3 + + 4 ( 25.0% ) ( ) ( -J 100% 16.04

· . g M Ixture

I mol C H 4

g CH4

48

I mol

CO 2

I mol C H 4

J ( 44. 0 1

CO 2 CO 2

g

I mol

J

=

1 3 7 1 88 g C 02 .

(--)(

I mol C 3 Hg 75.0% P ropane: ( 200 .g M·Ixture) 1 00% 44.09 g C 3 Hg Total CO2 = 1 37. 1 88 g + 449. 1 83 g = 586.3 1 8 = 586 g CO2 3 .93

)(

3 mol CO2 I mol C 3 Hg

)(

)

44.01 g CO2 = 449. 183 g CO2 1 mol CO2

Plan: lfwe assume a 1 00-gram sample of fertilizer, then the 30: 1 0: 1 0 percentages become the masses, in grams, ofN, P20 5 , and K20. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y:1.0 ratio, divide the moles of each element by the moles of potassium. Solution: A 1 00-gram sample of 30: I 0: I 0 fertilizer contains 30 g N, 1 09 P20 5 , and 109 K20. I mol N = 2. 1 4 1 3 mol N (unrounded) Moles N = ( 30 g N) 1 4. 0 1 g N

(

( (

Moles P = (l O g P20 5 )

)

)( )(

I mol P20 S 141.94 g P20 S

)

2 mol P = 0. 1 4090 mol P (unrounded) I mol P2 0S

)

1 mol K20 2 mol K = 0.2 1 23 1 mol K (unrounded) 94.20 g K20 I mol K20 This gives a ratio of 2 . 14 1 3:0.14090:0.2 1 23 1 The ratio must be divided by the moles ofK and rounded. (2. 1 4 1 3/0.2 123 1 ):(0. 14090/0.2123 1 ):(0.2 1 23 1 10.2 1 23 I ) 1 0.086:0.66365: 1.000 1 0:0.66: 1.0 Moles K = ( 1 0 g K20)

3 .95

Plan: Assume 1 00 grams of mixture. This means the mass of each compound, in grams, is the same as its percentage. Find the mass of C from CO and from CO2 and add these masses together. Solution: 1 00 g of mixture = 35 g CO and 65 g CO2. 1 mol C 1 2.01 g C 1 mol CO = 1 5.007 g C (unrounded) C from CO = ( 35.0 g CO) 28.01 g CO I mol CO I mol C

(

C from CO2 = ( 65.0 g C02) Mass percent C = 3 .97

[

(

)(

I mol CO2 44.0 I g CO2

)

)(

)(

I mol C I mol CO2

)(

)

)

12.0 I g C = 17 . 738 gC (UnrOUnded) I mol C

( 1 5.007 + 1 7.738)g x 100% = 32.745 = 32.7% C 1 00 g Sample

Plan: Determine the molecular formula from the figure. Once the molecular formula is known, use the periodic table to determine the molar mass. Convert the volume in (b) from quarts to mL and use the density to convert from mL to mass in grams. Take 6.82% of that mass and use the molar mass to convert to moles. Solution: a) The formula of citric acid obtained by counting the number of carbon atoms, oxygen atoms, and hydrogen atoms is C6Hs07. Molar mass = (6 x 12.01 ) + (8 x 1 .008) + (7 x 1 6.00) = 192.12 g/mol b) Determine the mass of citric acid in the lemon juice, and then use the molar mass to find the moles. l m L 1 .09 g 6.82% l moI C 6Hg�7 IL Moles C6H g0 7 = (1.50 qt) 1 00% 1 92. 1 2 g aCId mL 1 .057 qt 1O-3 L = 0.549 1 04 = 0.549 mol C6Hs07

(

)( )( )( ) (

49

)

3 .99

Plan: Use the mass percent to find the mass of heme in the sample; use the molar mass to convert the mass of heme to moles. Then find the mass of Fe in the sample. Solution : 6.0% heme . a) Grams of heme ( 0.45 g hemoglobIn ) 0.0270 = 0.027 g heme 1 00% hemoglobin =

(

(

)

=

I mol heme I 4.3 7963 X 1 0-s 4.4 x 10-5 mol heme 616.49 g heme ) 1 mol Fe 55.85 g Fe c) Grams of Fe = (4.3 7963 x 1 0 - 5 mol heme . 1 mol heme I mol Fe 2.44602 x 1 0-3 2.4 X 1 0-3 g Fe l mol hemin 65 1 .94 g hemin d) Grams of hemin = (4.37963 x 1 0 -s mol heme 1 mol heme 1 mol hemin 2. 85526 x 1 0-2 2.9 X 10-2 g hemin b) Mole of heme = ( 0.027 g heme )

3 . 1 02

=

=

=

=

{

{

)(

)

)(

=

)

Plan: Determine the molecular formula and the molar mass of each of the compounds. From the amount of nitrogen present and the molar mass, the percent nitrogen may be determined. The moles need to be determined for part (b). Solution: a) To find mass percent of nitrogen, first determine molecular formula, then the molar mass of each compound. Mass percent is then calculated from the mass of nitrogen in the compound divided by the molar mass of the compound, and multiply by 1 00%. Urea: CH4N 2 0, oM = 60.06 g/mol 2(1 4.0 1 glmol 1 00% 46.6533 46.65% N in urea %N 60.06 g/mol CH 4 N 2 0 Arginine: C6H I SN402, oM 1 75 .22 g/mol 4(1 4.0 1 g/mol N) %N = I 00% 3 1 9 8265 3 1 98% N in a r g inin e 1 75.22 glmol C6 H IsN 4 02 Ornithine: C s H 1 3N202, oM 1 33 . 1 7 glmol 2(1 4.0 I glmol 1 00% 2 1 .04077 = 2 1.04% N in ornithine %N 1 33 . 1 7 g/mol C sH 13 N 2 02 =

=

b) Grams ofN

=

=

(

(

( (

=

N) J N)

J(

=

=

J

=

I mol C SH I3 N 2 0 2 1 33 . 1 7 g C S H I 3 N 2 0 2 30. 1 30390 = 30.13 g N

( 1 43 . 2 g C 5 H 1 3 N 2 0 2 )

3 . 1 06

=

J

=

=

.

.

=

1 mol C H 4N 2 0 1 mol C S H I3 N 2 0 2

J(

2 mol N I mol C H 4N 2 0

J(

1 4.0 1 g N 1 mol N

J

Plan: The balanced chemical equation is needed. From the balanced chemical equation and the masses we can calculate the limiting reagent. The limiting reagent wi ll be used to calculate the theoretical yield, and finally the percent yield. Solution : Determine the balanced chemical equation: ZrOCI2"8H20(s) + 4 H2C204"2H20(s) + 4 KOH(aq) � K2Zr(C204)3( H2C204)"H20(S) + 2 KCI(aq) + 20 H20(l) Determine the limiting reactant: I mol product Moles product from ZrOCI2"8 H20 ( 1 .60 g Zr cmP d ) I mol Zr cmpd 322.25 g Zr cmpd 1 mol Zr cmpd 0.004965 1 mol product (unrounded)

(

=

=

50

J(

J

(

Moles product from H2C204"2H20 =

)(

)

1 mol H2C204 " 2H20 1 mol Product 1 26.07 g H 2C204 " 2H 20 4 mol H2C204 " 2H20 = 0.0 1 03 1 1 7 mol product (unrounded) Moles product from KOH = irrelevant because KOH is stated to be in excess. The ZrOCIz"8H20 is limiting, and will be used to calculate the theoretical yield: 1 mol product 54 1 .53 g Product 1 mol Zr cmpd G rams pro d uct = ( 1 . 60 g Zr cmp d ) 322.25 g Zr cmpd 1 mol Zr cmpd 1 mol product = 2.68874 g product (unrounded) Finally, calculating the percent yield: 1 .20 g Actual Yield . Percent YIeld = x 1 00% = 44.63 1 = 44.6% Yield x 1 00% = 2.68874 g Theoretical Yield

( 5 .20 g H 2 C204 " 2H20 )

(

(

)(

)

(

51

)

)(

)

.

Chapter 4 The Major Classes of Chemical Reactions FOLLOW-UP PROBLEMS

4. 1

Plan: We must write an equation showing the dissociation of one mole of compound into its ions. The number of moles of compound times the total number of ions formed gives the moles of ions in solution. If the moles of compound are not given directly, they must be calculated from the information given. Solution: a) One mole of KCI04 dissociates to form one mole of potassium ions and one mole of perchlorate ions. KCI04(s) � K+(aq) + CI04-(aq) Therefore, 2 moles of solid KCI04 produce 2 mol of K+ ions and 2 mol of CI04- ions. b) Mg(C2H 3 02)z(S) �Mg2 +(aq) + 2 C 2H 302-(aq) First convert grams ofMg(C 2 H 302)2 to moles of Mg(C 2 H 302)2 and then use molar ratios to determine the moles of each ion produced. 1 mol Mg(C 2H 02 ( 354 g Mg(C 2 H 3 0 2 h) 1 42.40 g Mg(C H3 0h ) = 2.48596 mol (unrounded) 2 3 2 2) The dissolution of 2.48596 molMg(C 2 H 302)z(S) produces 2.49 mol Mg2+ and (2 x 2.48596) =

4.97 mol C2 H302-

I

(

(NH4)2Cr04(S) � 2 NH/(aq) + CrO/-(aq) c) First convert formula units to moles and then use molar ratios as in part b). (�H 4h Cr0 4 3 . 1 2 1 886 mol (unrounded) ( 1 .88 x I 0 2 4 Formula Units ) 6.022I mol x 1 0 Formula Units The dissolution of 3 . 1 2 1 886 mol (NH 4) 2 Cr04(S) produces (2 x 3. 1 2 1 886) 6.2 4 mol NH/ and 3.1 2 mol Cr042 -.

]

[

=

=

NaHS04(s) � Na\aq) + H S04-(aq) d) The solution contains ( 1 .32 L solution) (0.55 mol NaHSOiL solution) = 0.726 mol ofNaHS04 and therefore 0.73 mol ofNa+ and 0.73 mol HS04-. Check: The ratio of the moles of each ion in the solution should equal the ratio in the original formula. 4.2

4.3

Plan: Each mole of acid will produce one mole of hydrogen ions. It is convenient to express molarity as its definition (moles/L). Solution: HBr(/) � H+(aq) + Br-(aq) 3 + Moles H+ = ( 45 1 mL ) 1 0- L 3 .20 mol H Br I mol H = 1 .4432 = 1.44 mol H+ L I mol H Br I mL

[ ](

J[

]

Plan: Determine the ions present in each substance on the reactant side. Use Table 4. 1 to determine if any combination of ions is not soluble. If a precipitate forms there will be a reaction and chemical equations may be written. The molecular equation simply includes the formulas of the substances, and balancing. In the total ionic equation, all soluble substances are written as separate ions. The net ionic equation comes from the total ionic equation by eliminating all substances appearing in identical form (spectator ions) on each side of the reaction arrow. Solution: a) The resulting ion combinations that are possible are iron(III) phosphate and cesium chloride. According to Table 4. 1 , iron(III) phosphate is a common phosphate and insoluble, so a reaction occurs. We see that cesium chloride is soluble.

52

Total ionic equation: Fe 3 +(aq) + 3 C reaq) + 3 Cs+(aq) + PO /-(aq) ---7 FeP04(s) + 3 cr(aq) + 3 Cs\aq) Net ionic equation: Fe 3 +(aq) + P043-(aq) ---7 FeP04(s) b) The resulting ion combinations that are possible are sodium nitrate (soluble) and cadmium hydroxide (insoluble). A reaction occurs. Total ionic equation: 2 Na+(aq) + 2 OW(aq) + Cd2+(aq) + 2 N0 3-(aq) ---7 Cd(OH)z(s) + 2 Na\aq) + 2 N0 3-(aq) Note: The coefficients for Na+ and OH- are necessary to balance the reaction and must be included. Net ionic equation: Cd2+(aq) + 2 OH-(aq) ---7 Cd(OH)z(s) c) The resulting ion combinations that are possible are magnesium acetate (soluble) and potassium bromide (soluble). No reaction occurs. d) The resulting ion combinations that are possible are silver chloride (insoluble, an exception) and barium sulfate (insoluble, an exception). A reaction occurs. Total ionic equation: 2 Ag+(aq) + SO/-(aq) + Ba2 +(aq) + 2 cr(aq) ---7 2 AgCI(s) + BaS04(s) The total and net ionic equations are identical because there are no spectator ions in this reaction. 4.4

Plan: According to Table 4.2, both reactants are strong. Thus, the key reaction is the formation of water. The other product of the reaction is soluble. Solution: Molecular equation: 2 HN0 3 (aq) + Ca(OHMaq) ---7 Ca(N0 3 Maq) + 2 H 20(l) Total ionic equation: 2 H\aq) + 2 N0 3-(aq) + Ca2 +(aq) + 2 O W(aq) ---7 Ca2+(aq) + 2 N0 3 -(aq) + 2 H20(l) Net ionic equation: 2 H\aq) + 2 OW(aq) ---7 2 H 20(l) which simplifies to: H+(aq) + OJr(aq) ---7 H20(1)

4.5

Plan: A balanced chemical equation is necessary. Determine the moles ofHCl, and, through the balanced chemical equation, determine the moles of Ba(OH)2 required for the reaction. The moles of base and the molarity may be used to determine the volume necessary. Solution: The molarity of the HCl solution is 0. 1 0 1 6 M. However, the molar ratio is not 1 : 1 as in the example problem. According to the balanced equation, the ratio is 2 moles of acid per 1 mole of base: 2 HCI(aq) + Ba(OH)z(aq) ---7 BaCI2 ( aq) + 2 H20(l) L 1 O 3 L 0. 1 0 1 6 mol HCI I mol Ba(OH h Vol ume = ( 50.00 mL ) L 2 mol HCI 0. 1 292 mol Ba(OH) 2 I mL 0.0 1 96594 = 0.0 1 966 L Ba(OH)2 solution

[ )(

=

4.6

)(

)(

)

Plan: Apply Table 4.3 to the compounds. Do not forget that the sum of the O.N . ' s (oxidation numbers) for a compound must sum to zero, and for a polyatomic ion, the sum must equal the charge on the ion. Solution: a) Sc = +3 0 -2 In most compounds, oxygen has a -2 O.N ., so oxygen is often a good starting point. If each oxygen atom has a -2 O.N., then each scandium must have a +3 oxidation state so that the sum of O.N. 's equals zero: 2(+3) + 3(-2) = o. b) Ga = +3 CI = - 1 In most compounds, chlorine has a -I O.N., so chlorine is a good starting point. If each chlorine atom has a -I O .N., then the gallium must have a +3 oxidation state so that the sum of O.N . 's equals zero: 1 (+3) + 3 (- 1 ) O. c) H + 1 P =+5 0 = -2 The hydrogen phosphate ion is HPO/-. Again, oxygen has a -2 O.N . Hydrogen has a + 1 O.N. because i t i s combined with nonmetals. The sum of the O.N. 's must equal the ionic charge, so the fol lowing algebraic equation can be solved for P: 1 (+ 1 ) + 1 (P) + 4(-2) = -2; O.N. for P +5 . d) 1=+3 F = -1 The formula o f iodine trifluoride i s IF3 . In all compounds, fluorine has a -I O.N., so fluorine is often a good starting point. If each fluorine atom has a -I O.N . , then the iodine must have a +3 oxidation state so that the sum of O.N. 's equal zero: 1 (+3) + 3(-1 ) = O. =

=

=

=

53

4.7

4.8

Plan: Use Table 4.3 and assign O.N.'s to each element. A reactant containing an element that is oxidized (increasing O.N.) is the reducing agent, and a reactant containing an element that is reduced (decreasing O.N.) is the oxidizing agent. (The O.N.'s are placed under the symbols of the appropriate atoms.) Solution: a) 2 Fe(s) + 3 Ch(g) � 2 FeCI3(s) Fe = +3 O.N . : Fe = O Cl = O Cl =- 1 Fe is oxidized from 0 to +3 ; Fe is the reducing agent. Cl is reduced from 0 to - 1 ; CI2 is the oxidizing agent. b) 2 C2H 6(g) + 7 02(g ) � 4 CO2(g) + 6 H20(l) O.N . : C = -3 0=0 C = +4 H = +I H = +1 0 = -2 0 = -2 C is oxidized from -3 to +4; C2H6 is the reducing agent. o is reduced from 0 to -2; O2 is the oxidizing agent. H remains + 1 5 CO(g) + 120s(s ) � 12(s) + 5 CO2(g) c) O.N . : C = +2 1 = +5 1 = 0 C = +4 o = -2 0 = -2 0 = -2 C is oxidized from +2 to +4; CO is the reducing agent. I is reduced from +5 to 0; 1205 is the oxidizing agent. o remains -2 Plan: To classify a reaction, compare the number of reactants used versus the number of products formed. Also examine the changes, if any, in the oxidation numbers. Recall the definitions of each type of reaction: Combination: X + Y � Z; decomposition: Z � X + Y Single displacement: X + YZ � XZ + Y double displacement: WX + YZ � WZ + YX Solution: a) Combination; Sg(s) + 1 6 F2(g) � 8 SF4(g) O.N. : S=0 F=0 S = +4 F = -1 Sulfur changes from 0 to +4 oxidation state; it is oxidized and S8 is the reducing agent. Fluorine changes from 0 to - I oxidation state; it is reduced and F2 is the oxidizing agent. b) Displacement; 2 CsI(aq) + CI2(aq) � 2 CsCl(aq) + Iz(aq) O.N . : Cs = + 1 Cl = O Cs = + I I = O I = -I C l = -1 Total ionic eqn: 2 Cs+ (aq) + 2 qaq) + Ch(aq) � 2 Cs+ (aq) + 2 Cqaq) + 12(aq) Net ionic eqn: 2 qaq) + CI2(aq) � 2 Cqaq) + IzCaq) Iodine changes from - I to 0 oxidation state; it is oxidized and CsI is the reducing agent. Chlorine changes from 0 to -I oxidation state; it is reduced and CI2 is the oxidizing agent. c) Displacement; 3 Ni(N03)2 + 2 Cr(s) � 2 Cr(N03Maq) + 3 Ni(s) O.N . : Ni = +2 Cr = O Cr = +3 Ni = O N = +5 N = +5 0 = -2 0 = -2 2 3 Total ionic eqn: 3 Ni+ (aq) + 6 N03-(aq) + 2 Cr(s) � 2 Cr+ (aq) + 6 N03-(aq) + 3 Ni(s) 2 3 Net ionic eqn: 3 Ni+ (aq) + 2 Cr(s) � 2 Cr+ (aq) + 3 Ni(s) Nickel changes from +2 to 0 oxidation state; it is reduced and Ni(N03)2 is the oxidizing agent. Chromium changes from 0 to +3 oxidation state; it is oxidized and Cr is the reducing agent.

54

END-OF-CHAPTER PROBLEMS 4.2

4.5

4.8

4. 1 0

4.12

Plan: Review the definition of electrolytes. Solution: Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds or from other electrolytes such as acids and bases. Ions must be present in an aqueous solution for it to conduct an electric current. Ions come from ionic compounds or from other electrolytes such as acids and bases. Plan: Write the formula for magnesium nitrate and note the ratio of magnesium ions to nitrate ions. Solution: The box in (2) best represents a volume of magnesium nitrate solution. Upon dissolving the salt in water, magnesium nitrate,Mg(N0 3 )2, would dissociate to form oneMg2 + ion for every two N0 3- ions, thus forming twice as many nitrate ions. Only box (2) has twice as many nitrate ions (red circles) as magnesium ions (blue circles). Plan: Compounds that are soluble in water tend to be ionic compounds or covalent compounds that have polar bonds. Solution: a) Benzene is likely to be insoluble in water because it is non-polar and water is polar. b) Sodium hydroxide, an ionic compound, is likely to be soluble in water since the ions from sodium hydroxide wil l be held in solution through ion-dipole attractions with water. c) Ethanol (CH 3 CH20 H) will likely be soluble in water because the alcohol group (-O H) wil l hydrogen bond with the water. d) Potassium acetate, an ionic compound, will likely dissolve in water to form sodium ions and acetate ions that are held in solution through ion-dipole attractions to water. Plan: Substances whose aqueous solutions conduct an electric current are electrolytes such as ionic compounds and acids and bases. Solution: a) An aqueous solution that contains ions conducts electricity. CsI is a soluble ionic compound, and a solution of this salt in water contains Cs+ and r ions. Its solution conducts electricity. b) H Br is a strong acid that dissociates completely in water. lts aqueous solution contains H + and Br- ions, so it conducts electricity. Plan: To determine the total moles of ions released, write a dissolution equation showing the correct molar ratios, and convert the given amounts to moles if necessary. Solution: a) Recall that phosphate, PO/ -, is a polyatomic anion and does not dissociate further in water. K3 P04(S) � 3 K +(aq) + PO/ -(aq) 4 moles of ions are released when one mole of K3 P04 dissolves, so the total number of moles released is (0.83 mol K3 P04) (4 mol ions/mol K3 P04) 3 3 2 3.3 mol of ions b) NiBr203 H20(s) � Ni2 +(aq) + 2 Br -(aq) Three moles of ions are released when I mole ofNiBr203 H20 dissolves. The waters of hydration become part of the larger bulk of water. Convert the grams ofNiBr203H20 to moles using the molar mass (be sure to include the mass of the water): 3 mol Ions ) 1 mol NiBr2 3H 2 0 ( 8.11 x 10- g NiBr2 3H 2 0 272.54 g NiBr2 3H 2 0 1 mol NiBr2 3 H 2 0 8.93 1 0-5 mol of ions 8.9271 x 1 0- 5 =

3

=

°

=

(

.

=

°

°

X

55

J(

°

J

c) FeCI 3(s) � Fe 3+(aq) + 3 Cr(aq) Recall that a mole contains 6.022 x 1 023 entities, so a mole of FeCI 3 contains 6.022 x 1 0 23 units of FeCl 3, (more easily expressed as formula units). Since the problem specifies only 1 .23 x 1 021 formula units, we know that the amount is some fraction of a mole. 4 moles of ions are released when one mole of FeCI 3 dissolves. 4 mo l I ons I mol FeCI 3 1.23 x 102 \ F.U. FeCI 3 2 3 6.022 x 1 0 F. u. FeC1 3 1 mol FeCI 3 3 = 8. 1 7004 x 1 0- = 8.17 x 10-3 mol of ions

)[

(

4.14

J

)(

Plan: To determine the moles of each type of ion released, write a dissolution equation showing the correct molar ratios, and convert the given amounts to moles if necessary. Solution: I mol of AIH and 3 mol of cr per mole of AICI 3 dissolved

[ )( [ I )( )[ )[ 1

-3 Moles A13+ = ( 1 00. mL) 1O L I mL 3 Moles Cr = (100. mL) 10- L mL

)[ )[ )

2.45 mol AICI 3 L

2.45 mol AICI 3 L

) )

1 mol A1 3+ = 0.2 45 mol Ae+ 1 mol AICI 3

3 mol C I= 0.735moI C r 1 mol AlCI 3

6.022 x 1 02 AI 3+ = 1.47539 X 1023 = 1.48 X 1023 Ae+ ions I mol AI + 6.022 x l O 2 3 Cl= 4.426 1 7 X 1 023 = 4.43 X 1 023 cr ions cr ions = 0.735 mol CImol CIb) Li2 S04(s) � 2 Li\aq ) + SO/-(aq) 2 mol of Lt and 1 mol of SO/- per mol of Li 2S04 dissolved

(

:

A13+ ions = 0.245 mol AI 3+

(

)

(

Moles Li+ = (1.80 L) 2.59 g Li2S0 4 I L

(

M a I es SO42- -(I . 80 L) 2.59 g Li2S04 1L

J( ( J

1 mol Li 2 S0 4 109.95 g L i2S0 4 I mol Li2 S04 109.95 g L i2S04

[J

J[

2 mol Li+ 1 mol Li 2 S04

I mol SO�-

)

= 0.08480 = 0.0848 mol Lt

)

= 0.04240 = 0.042 4 mol S0421 mol Li 2 S04 6.022 x 1O Li+ = 5. 1 06787 x 1 022 = 5.1 1 X 1 022 Li+ ions Li+ ions = 0.084802 mol Li+ I mol LI 6.022 x 1 02 3 O�= 2.5 5339 X 1 022 = 2.55 X 1022 SO/- ions S042- ions = 0.04240 I mol SO�_ 1 mol S04 + c) KBr(s) � K (aq ) + Br-(aq) I mol of L i + and I mol of Br- per mole ofKBr dissolved 1O-3L 1 .68 x 1 022 F. u.KBr I K+ . = 3.78 x 102 1 K + Ions K+ ·IOns -- ( 225 mL) I F.U.KBr L 1 mL -

(

[--)[ [ )[ )[ )[

1 O-3 L Br- ·Ions - ( 225 m L) -1 mL

(

(

X

1 02 \ K+

Moles Br- = 3.78 x 1 02 1 Br-

)[ )[

1 .68 X I 022 F.u.KBr L

1 mol K+

6.022

x

)

)

�

)[

(

Moles K+ = 3 .78

�:

)[

) )

I Br= 3.78 x 10 2 1 Br- .Ions I F.U.KBr

)

= 6.27698 X 1 0-3 = 6.28 X 10-3 mol K+ 2 3 1 0 K+

)

1 mol Br= 6.27698 X 1 0-3 = 6.28 X 1 0-3 6.022 x 1 02 3 Br-

56

mol Br-

4.16

Plan: The acids in this problem are all strong acids, so you can assume that all acid molecules dissociate completely to yield W ions and associated anions. One mole of H CI04, HN03 and H C I each produce one mole of H + upon dissociation, so moles H + moles acid. Molarity is expressed as moles/L instead of as M. Solution: 0.25 mOl 0.3 5 mol H+ a) Moles H + mol HCI04 (1.40 L) IL =

=

b) Moles H + mol HN03 =

c) Moles H + mol H C I = =

( ) 3 ) (1.8 ) (100 )( ( 7.6 ) ( 0.056 ) 0.4256 =

=

mL

=

L

L I mL

o.72 mOl IL

--

mOl IL

=

=

=

1.296 x 10-3 1.3 x 10=

3

mol H+

0.43 mol H+

4.23

P lan: Use Table 4. 1 to predict the products of this reaction. Ions not involved in the precipitate are spectator ions and are not included in the net ionic equation. Solution: Assuming that the left beaker is AgN03 (because it has gray Ag+ ion) and the right must be NaCI, then the N03- is blue, the Na+ is brown, and the cr is green. (Cr must be green since it is present with Ag+ in the precipitate in the beaker on the right.) Molecular equation: AgN03(aq) + NaCI(aq) � AgCI(s) + NaN03(aq) Total ionic equation: Ag+(aq) + N03-(aq) + Na+(aq) + C r( aq) � AgCI(s) + Na\aq) + N03-(aq) Net ionic equation: Ag\aq) + Cr(aq) � AgCI(s)

4.24

Plan: Check to see if any of the ion pairs are not soluble according to the solubil ity rules in Table 4.1 Ions not involved in the precipitate are spectator ions and are omitted from the net ionic equation. a) Molecular: Hg2(N03Maq) + 2 KI(aq) � Hg212(s) + 2 KN03(aq) Total ionic: Hg2 2+(aq) + 2 N03-(aq) + 2 K\aq) + 2 l(aq) � H g212(s) + 2 K+(aq) + 2 N03-(aq) Net ionic: H g/+(aq) + 2 l(aq) � Hg212(s) Spectator ions are K+ and N03-. b) Molecular: FeS04(aq) + Ba(OHMaq) � Fe(OHMs) + BaS04(s) Total ionic: Fe2\aq) + SO /-(aq) + Ba2+(aq) + 2 OW( aq) � Fe(OHMs) + BaS04(s) Net ionic: This is the same as the total ionic equation, because there are no spectator ions.

4.26

Plan: A precipitate forms if reactant ions can form combinations that are insoluble, as determined by the solubility rules in Table 4.1. Create cation-anion combinations other than the original reactants and determine if they are insoluble. Any ions not involved in a precipitate are spectator ions and are omitted from the net ionic equation. Solution: a) NaN03(aq) + CuS04(aq) � Na2S04(aq) + Cu(N03Maq) No precipitate will form. The ions Na+ and SO/- will not form an insoluble salt according to solubility rule #1: All common compounds of Group fA ions are soluble. The ions Cu2+ and N03- will not form an insoluble salt according to the solubil ity rule #2: All common nitrates are soluble. There is no reaction. b) A precipitate wil l form because silver ions, Ag+, and iodide ions, r, will combine to form a solid salt, silver iodide, Agl. The ammonium and nitrate ions do not form a precipitate. Molecular: NH4I(aq) + AgN03(aq) � AgI (s) + NH4N03(aq) Total ionic: NH /(aq) + l(aq) + Ag\aq) + N03-(aq) � AgI(s) + NH /(aq) + N03 -(aq) Net ionic: Ag+(aq) + l(aq) � AgI(s)

4.28

Plan: Write a balanced equation for the chemical reaction described in the problem. By applying the solubility rules to the two possible products (NaN03 and PbI2), determine that2 PbI2 is the precipitate. By using molar relationships, determine how many moles of Pb(N03)2 (and thus Pb + ion) are required to produce 0.628 g of PbI2·

57

Solution: The reaction is: Pb(N03Maq) + 2 NaJ(aq) � PbI2(s) + 2 NaN03(aq). I J mol PbI I mol .Pb 2 + 2 2 M Pb + = ( 0.628 g PbI 2 ) 46 1 .0 g Pbl 2 I mol PbI 2 35.0 mL =

4.30

(

0.03892 1 6 0.0389 M Pb2+ =

)(

][

)( ) 10 L I mL -3

--

Plan: The balanced equation for this reaction is AgN03(aq) + cr(aq) � AgCI(s) + N03-(aq). First, find the moles of AgN03 and thus the moles of Cr present in the 25 .00 mL sample by using the molar ratio in the balanced equation. Second, convert moles of CI- into grams, and convert the sample volume into grams using the given density. The mass percent of cr is found by dividing the mass of CI- by the mass of the sample volume and mUltiplying by 1 00. Solution: 1 0 3 L 0.3020 mol AgN03 I mol C I35 .45 g C I 0.467098 g C I (unrounded) (43 .63 mL ) L I mL I mol AgN0 3 1 mol CI-

[ )(

Mass of sample = ( 25 .00 mL ) Mass% CI =

Mass C I Mass Sample

x

( 1 .04 ) g

mL

1 00%

=

=

) [.

)[

26.0 g sample

0.467098 g C I 26.0 g Sample

x

1 00%

)

=

=

1 .79653 1.80% CI =

4.36

Plan: Remember that strong acids and bases can be written as ions in the total ionic equation but weak acids and bases cannot be written as ions. Omit spectator ions from the net ionic equation. Solution: a)KOH is a strong base and H I is a strong acid; both may be written in dissociated form. K1 is a soluble compound since all Group I A( I ) compounds are soluble. Molecular equation: KOH(aq) + H I (aq) � KJ(aq) + H20(!) Total ionic equation: K\aq) + OW(aq) + H\aq) + qaq) � K\ aq) + qaq) + H20(l) Net ionic equation: OW(aq) + H +(aq) � H20(!) The spectator ions are K+(aq) and qaq) b) N H3 is a weak base and is written in the molecular form. HCI is a strong acid and is written in the dissociated form (as ions). N H4C I is a soluble compound, because all ammonium compounds are soluble. Molecular equation: N H3(aq) + HCI(aq) � NH4CI(aq) Total ionic equation: N H3(aq) + H +(aq) + Creaq) � NH/ (aq) + Creaq) Net ionic equation: NH3(aq) + H+(aq) � NH / (aq) CI- is the only spectator ion.

4.38

Plan: Write an acid-base reaction between CaC03 and HC!. Solution: Calcium carbonate dissolves in HCI(aq) because the carbonate ion, a base, reacts with the acid to form CO2(g). Total ionic equation: 2 CaC03(s) + 2H +(aq) + 2 cr(aq) � Ca +(aq) + 2 cr(aq) + H20(l) + CO2(g) Net ionic equation: CaC03(s) + 2H\aq) � Ca2\aq) + H20(!) + CO2(g)

4.40

Plan: Write a balanced equation and use the molar ratios to convert the amount of KOH to the amount of CH3COOH. Solution: The reaction is: KOH(aq) + C H3COOH(aq) � KCH3COO(aq) + H20(l) 0. 1 1 80 mol KOH 1 0-3 L l mol CH 3 COOH ( I I mL M = ( 25.98 mL L I mL l I mol KOH l 52.50 mL 1 0-3 L

(

=

)[ )

]

l

0.058393 1 4 0.05839 M CH3COOH =

58

]( )

4.45

Plan: An oxidizing agent has an atom whose oxidation number decreases during the reaction. Solution: a) The S in SO/- (i.e., H2S04) has O.N. = +6, and in S02, O.N. (s) = +4, so the S has been reduced (and the r oxidized), so the H2S04 is an oxidizing agent. b) The oxidation numbers remain constant throughout; H2S04 transfers a proton to F to produce HF, so it acts as an acid.

4.46

Plan: Consult Table 4.3 for the rules for assigning oxidation numbers. Solution: a) NH20H: (O.N. for N) + 3(+ 1 for H) + 1 (-2 for 0) = 0 O.N. for N = -1 b) N2H4: 2(0.N. for N) + 4(+ 1 for H) = 0 O.N. for N = -2 c) NH/: (O.N. for N) + 4(+ 1 for H) = + 1 O.N. for N = -3 d) HN02: (O.N. for N) + 1 (+ 1 for H) + 2(-2 for 0) = 0 O.N. for N = +3

4.48

Plan: Consult Table 4.3 for the rules for assigning oxidation numbers. Solution: a) AsH). H is combined with a nonmetal, so its O.N. is + 1 (Rule 3). The O.N. for As is -3 . b) H)As04. The O.N. of H in this compound is + I , or +3 for 3 H 's. Oxygen's O.N . is -2, with total O.N. of -8 (4 times -2), so As needs to have an O.N. of +5 : +3 + (+5) + (- 8) = 0 c) AsCh . Cl has an O.N. of- 1 , total of -3, so As must have an O.N. of +3 . a)As = -3 b)As = +5 c) As = +3

4.5 0

Plan: Consult Table 4.3 for the rules for assigning oxidation numbers. Solution: a) MnO/-: (O.N. for Mn) + 4(-2 for 0) =-2 O.N. for Mn = +6 b) Mn203 : O.N. for Mn = +3 { 2(0.N. for Mn) } + 3(-2 for 0) = 0 O.N. for Mn = +7 c) KMn04: 1 (+ 1 for K) + (O.N. for Mn) + 4(-2 for 0) = 0

4.5 2

Plan: Oxidiz ing agent: substance that accepts the electrons released by the substance that is oxidized; the oxidation number of the atom accepting electrons decreases. The oxidiz ing agent undergoes reduction. Reducing agent: substance that provides the electrons accepted by the substance that is reduced; the oxidation number of the atom providing the electrons increases. The reducing agent undergoes oxidation. First, assign oxidation numbers to all atoms. Second, recognize that the agent is the compound that contains the atom that is gaining or losing electrons, not just the atom itself. Solution: 2 a) 5 H2C204(aq ) + 2 Mn04-(aq) + 6 H+ (aq) � 2 Mn + (aq ) + \ 0 CO2(g ) + 8 H20(l) H = +1 Mn = +7 H = +1 Mn = +2 C = +4 H = +1 0 = -2 0 = -2 0 = -2 C = +3 0 = -2 Hydrogen and oxygen do not change oxidation state. The Mn changes from +7 to +2 (reduction). Therefore, Mn04- is the oxidizing agent. C changes from +3 to +4 (oxidation), so H 2 C 2 04 is the reducing agent. 2 b) 3 Cu(s ) + 8 H \aq) + 2 N03-(aq) � 3 Cu + (aq) + 2 NO(g ) + 4 H20(l) Cu = O H = +1 N = +5 Cu = +2 N = +2 H = +I 0 = -2 0 = -2 o = -2 Cu changes from 0 to +2 (is oxidized) and Cu is the reducing agent. N changes from +5 (in N03-) to +2 (in NO) and is reduced, so N03- is the oxidizing agent.

4.5 4

Plan: Oxidiz ing agent: substance that accepts the electrons released by the substance that is oxidized; the oxidation number of the atom accepting electrons decreases. The oxidiz ing agent undergoes reduction. Reducing agent: substance that provides the electrons accepted by the substance that is reduced; the oxidation number of the atom providing the electrons increases. The reducing agent undergoes oxidation. First, assign oxidation numbers to all atoms. Second, recognize that the agent is the compound that contains the atom that is gaining or losing electrons, not just the atom itself.

59

Solution: 2 a) 8 H +(aq) + 6CI- (aq) + Sn(s) + 4N03- (aq) � SnC16 - (aq) + 4N02(g) + 4H20(l) + H = +1 N = +4 Sn = +4 H = + I cr = - 1 Sn = 0 N = +5 CI = - I 0 = -2 0 = -2 0 = -2 Oxidizing agent is N03- because nitrogen changes from +5 O.N. in N03- to +4 O.N. in N02. Reducing agent 2 is Sn because its O.N. changes from 0 as the element to +4 in SnCI6 -. 2 b) 2Mn04- (aq) + I OCr (aq) + 1 6W(aq) � 5CI2(g) + 2Mn2 +(aq) + 8H20(l) Mn = +7 Mn + = +2 H = +l H+ = + 1 CI = O CI- = -1 O=� O=� 2 Oxidizing agent is Mn04- because manganese changes from +7 O.N. in M n04- to +2 O.N. in M n +. Reducing agent is CI- because its O.N. changes from - I in cr to 0 as the element to C12.

4.56

P lan: S is in Group 6A ( 1 6), so its highest possible O.N. is +6 and its lowest possible O.N. is 6 - 8 = -2. Remember that a reducing agent has an atom whose O.N. increases while an oxidizing agent has an atom whose O.N. decreases. Solution: a) The lowest O.N. for S [(Group 6A( 1 6) ] is 6 - 8 = -2, which occurs in 2S 2-. Therefore, when S 2- reacts in an oxidation-reduction reaction, S can only increase its O.N. (oxidize), so S - can only function as a reducing agent. b) The highest O.N. for S [(Group 6A( 1 6) ] is +6, which occurs in SO/-. Therefore, when SO/- reacts in an oxidation-reduction reaction, the S can only decrease its O.N. (reduce), so SO /- can only function as an oxidizing agent. c) The O.N. of S in S02 is +4, so it can increase or decrease its O.N. Therefore, S0 2 can function as either an oxidizing or reducing agent.

4.62

Plan: Recall the definitions of each type of reaction: Combination: X + Y � Z; decomposition: Z � X + Y Single displacement: X + YZ � XZ + Y double displacement: WX + YZ � WZ + YX Solution: combination a) 2 Sb(s) + 3 C h( g) � 2 SbCI3(s) decomposition b) 2 AsH3(g) � 2 As(s) + 3 H2(g) c) 3 Mn(s) + 2 Fe(N03Maq) � 3 Mn(N03Maq) + 2 Fe(s) displacement

4.64

Plan: Two elements as reactants often results in a combination reaction while one reactant only often indicates a decomposition reaction. Review the types of reactions in Section 4.6 Solution: a) N2(g) + 3 H2(g) � 2 NH3(g) L\

b) 2 NaCI03(s) � 2 NaCI(s) + 3 02(g) c) Ba(s) + 2 H20(l) � Ba(OHMaq) + H2(g) 4.66

Plan: Review the types of reactions in Section 4.6 Solution: a) Cs, a metal, and 12, a nonmetal, react to form the binary ionic compound, CsI. 2 Cs(s) + 12(s) � 2 CsI(s) b) AI is a stronger reducing agent than Mn and is able to displace M n from solution, i.e., cause the reduction from Mn2+(aq) to Mn o(s). 2 AI(s) + 3 MnS04(aq) � AI2(S04h(aq) + 3 Mn(s) c) Sulfur dioxide, S02, is a nonmetal oxide that reacts with oxygen, O2, to form the higher oxide, S03. L\

2 S02(g) + 02(g) � 2 S03(g) It is not clear from the problem, but energy must be added to force this reaction to proceed.

60

d) Propane is a three-carbon hydrocarbon with the formula C3 HS. It burns in the presence of oxygen, Oz, to form carbon dioxide gas and water vapor. Although this is a redox reaction that cou ld be balanced using the oxidation number method, it is easier to balance by considering only atoms on either side of the equation. First balance carbon and hydrogen (because they only appear in one species on each side of the equation), and then balance oxygen . C3Hs(g) + 50z(g) .-, 3COz(g) + 4HzO(g) e) Total ionic equation: z 2 Al(s) + 3 Mn + (aq) + 3 SOl-(aq) .-, 2 AI3+ (aq ) + 3 S Ol-(aq) + 3 M n (s) Net ionic equation: z 3 2 AI(s) + 3 Mn + (aq) .-, 2 AI + (aq) + 3 Mn(s) Note that the molar coefficients are not simplified because the number of electrons lost ( 6 e-) must equal the electrons gained (6 e-). 4.68

Plan: Write a balanced equation; convert the mass ofHgO to moles and use the molar ratio from the balanced equation to find the moles and then the mass of 02 . Perform the same calculation to find the mass of the other product. Solution: The balanced chemical equation is 2 HgO(s) � 2 Hg(l) + Oz(g) I 03 g I mol ° 2 32 .00 g O 2 1 mol HgO Mass Oz = (4.27 kg HgO) 1 kg 2 1 6.6 g HgO 2 mol HgO 1 mol 02 The other product is mercury. 2 mol Hg 200.6 g Hg 1 mol HgO 1 03 g Mass Hg = (4.27 kg HgO) 1 kg 2 1 6.6 g HgO 2 mol HgO 1 mol Hg = 3 .95458 = 3 . 9 5 kg Hg

( )(

4.70

)(

( )(

)(

)(

)

)(

=

315 .420 = 31

5

g

O2

)( ) 1 kg 1 03 g

Plan: To determine the reactant in excess, write the balanced equation ( metal + Oz .-, metal oxide), convert reactant masses to moles, and use molar ratios to see which reactant mak es the smaller (" limiting") amount of product. Use the limiting reactant to calculate the amount of product formed . Solution: The balanced equation is 4 Li(s) + Oz(g) .-, 2 LizO (s)

( (

. . . . . .. a) Moles LlzO If LI IImltmg = ( 1 . 62 g LI ) Moles Liz O if Oz limiting

=

(6.00 g ° 2 )

Imol Li . 6.94 1 g LI

I mol ° 2 32.00 g 02

)( )(

2 mol Li 2 0 4 mol LI .

2 mol Li 20 1 mol 02

) )

=

. 0. 1 1 66979 mol LI2 0 (unrounded)

=

. 0.375 mol LI2 0 (unrounded)

Li is the limiting reactant; O2 is in excess. b) 0. 1 1 66979 = 0.117 mol LizO c) Li is limiting, thus there will be none remaining (0 g Li). 2 moI Li 2 0 29.88 g Li 2 0 _ . I mol Li . . Grams LlzO = ( 1 .62 g LI ) - 3 . 4869_ - 3 . 4 9 g L 12 O 6.94 1 g Li 4 mol Li 1 mol Li 2 0

(

(

)(

)( )(

)(

)

)

1 mol 0 2 32 .00 g O 2 1 mol Li 1 .867 1 66 g Oz (unrounded) 6.94 1 g Li 4 mol Li I mol 0 2 Remaining Oz = 6.00 g Oz - 1 .867 1 66 g Oz = 4. 1 3283 = 4.13 g O2

Grams Oz used = ( 1 .62 g Li )

61

=

4.73

Plan: To find the mass of Fe, write a balanced equation for the reaction, determine whether Al or Fe203 is the limiting reactant, and convert to mass. Solution: 2 AI(s) + Fe203(S) � 2 Fe(s) + AI203(s) When the masses of both reactants are given, you must determine which reactant is limiting. 1 03 g I mol Al 2 mol Fe = 37.064 mol Fe (unrounded) Mole Fe (from AI) = ( 1 .00 kg AI) I kg 26.98 g Al 2 mol Al

[ )(

)(

(

)

)

2 mol Fe Mole Fe (from Fe2 03) = (2.00 mol Fe20 3 ) = 4.00 mol Fe 1 mol Fe20 3 Fe203 is limiting, so 4.00 moles of Fe forms. 55.85 g Fe Mass = ( 4.00 mol Fe) = 223 .4 = 223 g Fe 1 mol Fe Though not required by the problem, this could be converted to 0.223 kg. 4.74

4.75

(

)

P lan: Convert the mass of Fe in a 1 25-g serving to the mass of Fe in a 737-g sample. Use molar mass to convert mass to moles and use Avogadro's number to convert moles of Fe to moles of ions. Solution: Fe(s) + 2 H+(aq) � Fe2+ (aq) + H2 (g) a) O.N .: 0 +1 +2 0 2 2 2 1 mol Fe + 6.022 x 1 0 3 Fe + ions 1 O- 3 g 1 mol Fe 49 mg Fe 2 s b) Fe + ions = ( 737 g auce) 2 1 25 g Sauce 1 mg 55.85 g Fe 1 mol Fe I mol Fe + 2 2 2 = 3 . 1 1 509 x 1 0 1 = 3.1 X 1 0 1 Fe + ions per jar of sauce

(

)[

)

P lan: Convert the mass of glucose to moles and use the molar ratios from the balanced equation to find the moles of ethanol and CO2. The amount of ethanol is converted from moles to grams using its molar mass. Solution: ° ° 1 mol C6H1 206 2 mol C2HsOH 46.07 g C2 HsOH Mass 0 f C 2HS H - ( 1 0 . 0 g C6H1 2 6 ) 1 80. 1 6 g C6H1 2 06 1 mol C6 H1 206 1 mol C2HsOH = 5 . 1 1 43 = 5.1 1 g C2HsOH 1 mol C6H1 206 2 mol CO2 22.4 L CO2 Volume CO2 = ( 1 0.0 g C6 HI 2 06 ) 1 80. 1 6 g C6H1 2 06 I mol C6HI 2 06 1 mol CO2 = 2 .4866785 = 2.49 L CO2 -

4.80

)[

)[ )(

(

(

)(

)(

)(

)(

)

)

Plan: For part (a), assign oxidation numbers to each element; the oxidizing agent has an atom whose oxidation number decreases while the reducing agent has an atom whose oxidation number increases. For part (b), use the molar ratios, beginning with Step 3, to find the moles ofN02 , then moles of NO, then moles ofNH3 required to produce the given mass of HN03. Solution: a) Step I . 4NH3(g) + 502(g) � 4NO(g) + 6H2 0(l) O.N.: N = -3 0=0 N = +2 H = +1 H = +1 0 = -2 ° 0 = -2 N oxidized from -3 to +2 by O2, and is reduced from 0 to -2. Oxidizing agent

=

O2

Reducing agent

=

O2

Reducing agent

=

NH3

2NO(g) + 02(g) � 2N02(g) N = +2 0 = 0 N° = +4 ° = -2 = -2 ° N oxidized from +2 to +4 by O2 , and is reduced from 0 to -2.

Step 2. O.N.:

Oxidizing agent

=

62

NO

Step 3 . 3 N02(g) + H20(l) � 2 HN03(l) + NO(g) O.N.: N=+4 H=+ l N=+2 H=+1 O=� O=� N=� O=� 0=-2 N oxidized from +4 to +5 by N02, and N is reduced from +4 to +2. Oxidizing agent

( )(

=

N02

Reducing agent

J(

b) Mass of N H3 : (3 .0 x 1 0 4 kg HN0 3 ) 10 3 g I mol HN0 3 3 moI N0 2 lkg 63 .02 g HN0 4 2 mol HN0 3 = J .2 1 604 X 1 0 4= 1 .2 X 104 kg NH3 4.82

N02

J(

2 mol NO 2 mol N0 2

J(

4 moI NH 3 4 moiNO

)(

17.03 g N H 3 I mol NH 3

J( ) � 3 10 g

P lan: Write a balanced equation and use the molar ratio between Na202 and CO 2 to convert the amount of Na202 given to an amount of CO 2 . Solution : The reaction is: 2 Na20 2 (s) + 2 CO 2(g) � 2 Na2C03(S) + OzCg). 1 mol Na202 44.0 1 g CO2 2 mol CO2 L Air Volume ( 80.0 g Na202 ) 77.98 g Na202 2 mol Na202 J mol CO2 0.0720 g CO2 = 627.08 = 627 L Air

)(

(

=

4.86

=

)(

)(

J

Plan: Balance the equation to obtain the correct molar ratios. Convert the mass of each reactant to moles and use the molar ratios to find the limiting reactant and the amount of CO2 produced. Solution : a) Balance the equation to obtain the correct molar ratios. This is not a redox reaction as none of the O.N. 's change. Here is a suggested method for approaching balancing the equation. - Since PO/- remains as a unit on both sides of the equation, treat it as a unit when balancing. - On first inspection, one can see that Na needs to be balanced by adding a "2" in front of NaHC03. This then affects the balance of C, so add a "2" in front of CO2. - Hydrogen is not balanced, so change the coefficient of water to "2," as this will have the least impact on the other species. - Verify that the other species are balanced. /';.

Ca( H2P04)z(S) + 2 NaHC03(s) ----7 2 CO2(g) + 2 H20(g) + CaHP04(s) + Na2 HP04(S) Determine whether Ca(H2P04)2 or NaHC03 limits the production of CO2. In each case calculate the moles of CO2 that might form. J mol NaHC0 3 2 mol CO2 31% Mole CO2 (NaH C03) = ( 1 .00 g ) 1 00% 84.0 I g NaHC0 3 2 mol NaHC0 3 3 = 3 .690 x 1 0- mol CO 2 ( unrounded) 2 mol CO2 I mol Ca( H2 P04 )2 35% Mole CO2 (Ca( H 2P04 ) 2) = ( l .00 g ) 1 00% 234.05 g Ca( H2P04 h ) 1 mol Ca( H2 P0 4 h = 2.9908 x 10-3 mol CO 2 (unrounded) S ince Ca(H2P04)2 is limiting, 3.03 x 1 0-3 mol CO2 wi ll be produced. b) Volume CO2 (2.9908 X 1 0- mol CO 2) (37.0 Llmol CO2)= 0. 1 1 06596= 0.1 1 L CO2

J(

( )(

( )(

=

4.88

i(

J

)

Plan: To determine the empirical formula, find the moles of each element present and divide by the smallest number of moles to get the smal lest ratio of atoms. To find the molecular formula, divide the molar mass by the mass of the empirical formula to find the factor by which to multiple the empirical formula. Solution: a) Determine the moles of each element present. The sample was burned in an unknown amount of O2, therefore, the moles of oxygen must be found by a different method. I mol CO2 I mol C = 4.27 1 756 X 10-3 mol C (unrounded) Moles C ( 0. 1 880 g CO2 ) 44.0 I g CO2 ) 1 mol CO2 =

(

i(

63

J

I mol H 2 0 ( 2 mol H = 3 .052 1 64 1 0-3 mol H (unrounded) ( 1 8.02 g H 2 0 J 1 mol H 2 0 J 1 mol Bi 2 0 3 )( 2 mol Bi ) = 6. 1 03004 x 1 0-4 mol Bi (unrounded) Moles Bi = ( 0. 1 422 g Bi 2 0 3) ( 466.0 g Bi 2 0 3 1 mol Bi 2 0 3 X

Moles H = ( 0.02750 g H 2 0)

Subtracting the mass of each element present from the mass of the sample will give the mass of oxygen originally present in the sample. This mass is used to find the moles of oxygen. 1 2.0I g C Mass C = (4.27 1 756 x lO- 3 mOl c ) = 0.05 1 3038 g C 1 mol C 1 .008 gH MassH = (3 .052 1 64 xl0-3 mOIH ) = 0.0030766 g H ImolH 209.0 gBi Mass B i = (6. 1 03004 x 1 0-4 mol Bi ) Imol Bi =0. 1 27553 g Bi

( (

] )

[

j

Mass ° = 0.22 1 05 g sample - (0.05 1 3038 g C + 0.0030766 g H + 0. 1 27553 g Bi) = 0.039 1 1 66 g O Moles ° = (0.039 1 1 66 g 0) Imol ° = 0.0024448 mol ° 1 6.00 g ° Divide each of the moles by the smallest value (moles Bi). C = (4.27 1 756 1 0-3 mol) / (6. 1 03004 x 1 0-4 mol) = 7 H = (3 .052 1 64 1 0-3 mol) / (6. 1 03004 x 1 0-4 mol) = 5 0 = (2.4448 1 0- 3 mol) / (6. 1 03004 x 1 0-4 mol) 4 Bi (6. 1 03004 x 1 0-4 mol) / (6. 1 03004 x 1 0-4 mol) I Empirical formula = C7Hs04Bi b) The empirical formula mass is 362 g/mol. Therefore, there are 1 086 / 362 = 3 empirical formula units per molecular formula making the molecular formula = 3 x C 7Hs04Bi = C2IHIS012Bi3. c) Bi(OH ) 3 (s) + 3 H C 7Hs0 3 (aq) � Bi(C 7Hs0 3Ms) + 3 H 20(l) 1 mol Bi(O�h 26 0 . 0 g �i(OHh 1 mg 1 00% 1 0-3 g 1 mol Active 3 mol B i d) ( 0.600 mg) . I mg 1 086 g 1 mol Active I mol BI I mol BI(OHh 1 0 3 g 88.0% = 0.48970 = 0.490 mg Bi(OH)3

(

X

X

=

X

=

=

[ )(

4.90

J

)(

)(

)[ ) ( )

)(

-

Plan: Write balanced equations and use the molar ratios to convert mass of each fuel to the mass of oxygen required for the reaction. Solution: a) Complete combustion of hydrocarbons involves heating the hydrocarbon in the presence of oxygen to produce carbon dioxide and water. Ethanol: C 2HsOH (l) + 3 0 2 (g) � 2 CO 2(g) + 3 H 20(l) Gasoline: 2 CsHls(l) + 25 0 2(g) � 1 6 CO 2 (g) + 1 8 H 20(g) b) The amounts of each fuel must be found: 90% l mL 0.742 g Gasoline = (l.OO L) = 667.8 g gasoline ( unrounded) 1 00% 1 0- 3 L ImL 0.789 g Ethanol = (I.OO L) 1 0% I mL = 78.9 g ethanol 1 00% 1 0- 3 L 1 mL I mol Cs Hls 25 mol 0 2 3 2.0 0 g 0 2 Mass O 2 (gasoline) (667.8 g CsH1s) 1 1 4.22 g CsHIS 2 mol Cs HIS I mol 0 2 = 2338.64 g O 2 (unrounded) ( 32.00 g O 2 Mass O 2 (ethanol) ( 78.9 g C 2HsOH) I mol C 2 HsOH ( 3 mol 0 2 46.07 g C 2HsOH I I mol C 2 HsOH l 1 mol 0 2 = 1 64.4 1 g O 2 (unrounded)

( )( J( ) ( J( J ( ) ( =

=

l

64

J(

)

J(

)

)

)

Total O2 = 2338.64 g O2 + 1 64.4 1 g O2 = 2503.05 = 2.50 x 103 g O2 I mo l 0 2 22.4 L c) (2503 .05 g 0 2 ) = 1 752. 1 35 = 1.75 x 103LOz 32.00 g 0 2 1 mol 0 2 d) (1 752. 1 35 L Oz

4.92

)( ) ( )( 1 00% ) = 83 83 .42 = 8.38 20.9%

x 103 L air

Plan: From the molarity and volume of the base NaOH, find the moles of NaOH and use the molar ratios from the two balanced equations to convert the moles ofNaOH to moles of HBr to moles of vitamin C. Use the molar mass of vitamin C to convert moles to grams. Solution: Mass vitamin C : 3 l moI C6Hg 06 1 1 76. 12g C6Hg 06 I mg (43 .20 mL NaOH) 1 O- L 0. 1 350 mol NaOH I mol HBr l mL lL I mol NaOH 2 mol H Br ) I mol C6Hg 06 1 0 -3 g = 5 1 3 .5659 = 5 1 3 .6 mg C6Hs 06 Yes, the tablets have the quantity advertised.

4.93

( )(

(

)(

)(

)( )

Plan: Remember that oxidation numbers change in a redox reaction. For the calculations, use the molarity and volume ofHCI to find the moles of HCI and use the molar ratios from the balanced equation to convert moles of HCI to moles and then grams of the desired substance. Solution: a) The second reaction is a redox process because the O.N . of iron changes from 0 to +2 (it oxidizes) while the O.N. of hydrogen changes from + 1 to 0 (it reduces). b) Determine the moles ofHCI present and use the balanced chemical equation to determine the appropriate quantities. 3.00 mol HCI 1 mol Fe 2 0 3 1 59.70 g Fe 2 0 3 Mass Fe203 = ( 2.50 x I 03 L ) L 6 mol HCI 1 mol Fe20 3

(

)

)(

)(

= 1 99625 = 2.00 x 105 g FeZ03 3.00 mol HCI 2 mol FeCI3 1 62.20 g FeCI3 Mass FeC I 3 = ( 2.5 0 x 1 0 3 L ) L 6 mol HCI I mol FeC I 3 = 405500 = 4.06 x 105 g FeCI3 c) Use reaction 2 like reaction 1 was used in part b. 3.00 mo l HCI 1 mol Fe 55.85 g Fe Mass Fe = ( 2.50 x 1 0 3 L ) 2 mol HCI 1 mol Fe L = 209437.5 = 2.09 x 105 g Fe . 3 .00 mol HCI I mol FeCI2 1 26.75 g FeCI2 Mass FeCI2 = ( 2.50 x 1 0 3 L ) L 2 mol HCI 1 mol FeCI2 = 4753 1 2.5 = 4.75 x 105 g FeCh d) Use 1 .00 g Fe203 to determine the mass of FeCI3 formed (reaction I ), and 0.280 g Fe to determine the mass of FeCI2 formed (reaction 2). 2 mol FeCI3 1 62.20 g FeCI 3 1 mol Fe 2 0 3 Mass FeCh = ( 1 .00 g Fe20 3) 1 mol FeCI 3 1 59.70 g Fe20 3 1 mol Fe20 3 = 2.03 1 3 g FeCI3 (unrounded) 1 mol Fe 1 mol FeCI2 1 26.75 g FeCI2 Mass FeCI2 = ( 0.280 g Fe) 55.85 g Fe I mol Fe I mol FeCI2 = 0.635452 g FeCI2 (unrounded) Ratio = (0.635452 g FeCI2) / (2.03 1 3 g FeCI3) = 0.3 1 2830 = 0.313

(

(

(

(

)(

(

)(

)(

)(

65

)(

)(

)(

)(

)(

)

)

)(

)

)

)

Chapter 5 Gases and the Kinetic-Molecular Theory FOLLOW-UP PROBLEMS

5. 1

2 Plan: Use conversion factors in Table 5 . 1 to convert pressure in torr to units of mmHg, pascals and Ib/in . Solution : Peo , = 579.6 torr

( )(

) )[

Converting from torr to mmHg:

p = ( 579.6 torr )

1 mmHg = 579.6 torr 1 torr

Converting from torr to pascals:

P = ( 579.6 torr

I atm 760 torr

4 = 7.727364 x 1 0 = 7.727 X 1 04 Pa

(

5 1 .0 1 325 x 1 0 pa latm

)[

)

)

2 1 atm 1 4.7 Ib/in = 1 1 .2 1 068 = 1 l .2 Ib/in2 760 torr 1 atm Check: For conversion to mmHg, the value of the pressure should stay the same. The order of magnitude for each conversion corresponds to the calculated answer. For conversion to pascals the order of magnitude calculation is 1 0 2 x 1 0 5 / 1 0 2 = 1 0 5 . For conversion to Ib/in 2 the order of magnitude calculation is 1 0 2 x 1 0 ' / 1 02 = 1 0 ' . Converting from torr to Ib/in 2 :

5.2

Plan: Given in the problem is an initial volume, initial pressure, and final pressure for the argon gas. The final volume can be calculated from the relationship P, V, = P2 V2 . Unit conversions for mL to L and atm to kPa must be included. Solution: 3 3 1 0- L 1 atm 1 0 pa P , = 0.87 1 atm; V, = ( 1 05 mL ) = 0. 1 05 L; P2 = ( 26.3 kPa ) 1 mL 1 kPa 1 .0 1 325 x 1 0 5 Pa = 0.25956 atm � ( 0.87 1 atm ) ( 0. 1 05 L ) V2 = ( V ) = = 0.352346 = 0.352 L I ( 0.25956 atm ) P2 Check: As the pressure goes from 0.87 1 atm to 0.25956 atm, it is decreasing by a factor of about 3 and the volume should increase by the same factor. The calculated volume of 0.352 L is approximately 3 times greater than the initial volume of 0. 1 05 L.

[ J

5.3

P = ( 579 . 6 torr )

( )

[

J

(

)

Plan: The problem asked for a temperature with given initial temperature, initial volume, and final volume. The relationship to use is V /T , = VzlT2• The units of volume in both cases are the same, but the initial temperature must be converted to Kelvin. Solution: 3 3 V, = 6.83 cm ; T , = O°C + 273 = 273 K; V2 = 9.75 cm ; T 2 = ? � = V2 T2 T, 75 cm 3 V2 T2 = T, - 389.7 1 4 390. K ( 273 K ) 6.83 �n 3 �

[ J

-

t �

-

Check: The volume increases by a factor of about 1 .4 so the temperature must have i ncreased by the same factor. The initial temperature of 273 K times 1 .4 is 380 K, so the answer 390 K appears to be correct.

66

5 .4

Plan: In this problem, the amount of gas is decreasing. Since the container is rigid, the volume of the gas will not change with the decrease in moles of gas. The temperature is also constant. So, the only change will be that the pressure of the gas will decrease since fewer moles of gas will be present after removal of the 5 .0 g of ethylene. To calculate the final pressure, use the relationship n/P I = n21P2. Since the ratio of moles of ethylene is equal to the ratio of grams of ethylene, there is no need to convert the grams to moles. (This is illustrated in the solution by listing the molar mass conversion twice.) Solution: �=� PI P2

( J

n2 --( 793 torr) P2 --( .pJ ) _ nJ

(

1 mol C2H 4 ( (35.0 - 5 .0) g C 2 H 4 ) 28.05 g C 2 H 4

(

J

J

-;-----'----�.:....:....

= 679.7 1 4 = 680. torr I mol C 2 H 4 (35.0 g C2H 4 ) 28.05 g C2H 4 Check : The amount of gas decreases by a factor of 6/7. Since pressure is proportional to the amount of gas, the pressure should decrease by the same factor, 6/7 x 793 = 680. 5.5

5.6

5.7

--

•

Plan: From Sample Problem 5 . 5 the temperature of 2 1 °C and volume of438 L are given. The pressure is 1 .37 atm and the unknown is the moles of oxygen gas. Use the ideal gas equation to calculate the number of moles of gas. Solution: ( 1 .37 atm) (438 L ) = 24.847 mol O2 n = PV / RT = 0.082 1 atm L ( (273 . 1 5 + 2 1 ) K ) mol K Mass of O2 = (24.847 mol O2) x (32 .00 glmol) = 795 . 1 04 = 7 9 5 g O2

(

)

0

0

Plan: The pressure is constant and, according to the picture, the volume approximately doubles. The volume change may be due to the temperature andlor a change in moles. Solution: The balanced chemical equation must be 2 CD ---j C2 + D2 . Thus, the number of mole of gas does not change (2 moles both before and after the reaction). Only the temperature remains as a variable to cause the volume change. Using V I / T I = V2 / T2 with V2 = 2 V ) , and T I = (-73 + 273 . 1 5) K gives: (2VI ) (-73 + 273 . 1 5 ) K V2 400.30 K - 273 . 1 5= 127. 1 5 = 127°C = T2 = TI ( VI ) VI MP Plan: Density of a gas can be calculated using a version of the ideal gas equation, d RT Two calculations are required, one with T = O°C = 273 K and P = 380 torr and the other at STP which is defined as T = 273 K and P = I atm. Solution: Density at T = 273 K and P = 380 torr ( 44.0 1 g/mol ) ( 3 80 torr) I atm = 0.981 783 = 0.982 giL d= 760 torr 0.082 1 atm L (273 K ) mol K Density at T = 273 K and P = I atm. (Note: The I atm is an exact number and does not affect the significant figures in the answer.) ( 44.0 1 g/mol ) ( 1 atm) = 1 .9638566 = 1 .96 giL d= 0.082 1 atm L (273 K ) mol K The density of a gas increases proportionally to the increase in pressure.

( J

(

(

= --

0

0

0

0

)

(

J

)

67

Check: In the two density calculations, the temperature of the gas is the same, but the pressure differs by a factor of two. The calculation of density shows that it is proportional to pressure. The pressure in the first case is half the pressure at STP, so the density at 380 torr should be half the density at STP and it is .

5.8

Plan: Use the ideal gas equation for density, d Solution: Molar mass = =

dRT -P

( l.26 g/L) =

28.95496 29.0 g/mol =

( 0.082 1

.AtP

, and solve for molar mass.

= --

RT

)

(

atm 0 L ( ( I O.O + 273. 1 5) K ) 1 0 l .325 kPa mol 0 K 1 atm ( 1 02.5 kPa )

J

Check: Dry air would consist of about 80% N2 and 20% O2. Estimating a molar mass for this mixture gives (0.80 x 28) + (0.20 x 32) 28.8 g/mol, which is close to the calculated value. =

5.9

1

Plan: Calculate the number of moles of each gas present and then the mole fraction of each gas. The partial pressure of each gas equals the mole fraction times the total pressure. Total pressure equals atm since the problem specifies STP. This pressure is an exact number, and wil l not affect the significant figures in the answer. No intermediate values, such as the moles of each gas, will be rounded. This will avoid intermediate rounding. Solution: 1 mol He = 1 .37397 mol He nH e = ( 5.50 g He ) 4.003 g He

( J ( 5.0 )( 1 J 0.7433 1 20. 1 8 n Kr = (35.0 g Kr) ( J = 0.4 1 766 83.80

nN e = I

g Ne

mol Ne g Ne

I mol Kr g Kr

=

mol Ne mol He

Total number of moles of gas = l.37397 + 0.7433 1 + 0.4 1 766 = 2.53494 mol P A = X A X PIOla l 1 .37397 mol He PH e = ( 1 atm ) = 0.5420 1 = 0.542 atm He 2.53494 mol 0.7433 1 mol Ne atm) = PNe = (I 0.29322 = 0.293 atm Ne 2.53494 mol 0.4 1 766 mol PKr = ( I atm ) = 0 . 1 64 76 = 0 . 1 65 atm Kr 2.53494 mol Check: One way to check is that the partial pressures add to the total pressure, 0.542 + 0.293 + 0. 1 65 = 1 .000 atm, which agrees with the total pressure of atm at STP.

( ( (

5. 1 0

) ) Kr)

1

Plan: The gas collected over the water will consist of H2 and H20 gas molecules. The partial pressure of the water can be found from the vapor pressure of water at the given temperature (Table 5.2). Subtracting this partial pressure of water from total pressure gives the partial pressure of hydrogen gas collected over the water. Calculate the moles of hydrogen gas using the ideal gas equation. The mass of hydrogen can then be calculated by converting the moles of hydrogen from the ideal gas equation to grams. Solution : From Table 5.3, the partial pressure of water is 1 3 .6 torr at 1 6°C. P = 752 torr - 1 3 .6 torr = 738.4 = 738 torr H2 The unrounded partial pressure (738.4 torr) will be used to avoid rounding error. ( 738.4 torr ) ( 1 495 m ) 1 atm -Moles of hydrogen = n = PV/RT = 0.082 1 atm 0 L 760 torr Im (( 273. 1 5 + 1 6) K) mol o K = 0.06 1 1 86 mol H2 (unrounded)

(

)

68

L

(

J[10-3LL )

Mass of hydrogen

= ( 0.06 1 1 86 mol H 2 ) ( 2.0I mol1 6 gHH2 2 J

=

0. 1 2335 1

=

0. 1 23 g H 2

Check: Since the pressure and temperature are close to STP, and the volume is near 1 .5 L, the molar volume at STP (22.4 Llmol) may be used to estimate the mass of hydrogen. ( 1 moll22.4 L) ( 1 .5 L) (2 g/mol) 0. 1 3 g, which is close to the calculated value. =

5.1 1

P lan: Write a balanced equation for the reaction. Calculate the moles of HCI(g) from the starting amount of sodium chloride using the stoichiometric ratio from the balanced equation. Find the volume of the HCI(g) from the molar volume at STP. Solution : The balanced equation is H 2S04(aq) + 2 NaCI(aq) -j Na2S04(aq) + 2 HCI(g). 3 I mol NaCI 2 mol HCI ( 0. 1 1 7 kg NaCI ) 11 0k gg 58.44 2.00205 mol H C I g NaCI 2 mol NaC I

[ )(

(

J( J= )( � )

I L 22.4 L 4.4846 X 1 0 4 4.48 X 1 0 4 mL Hel I mol HCI 1 0- L Check: 1 1 7 g NaCI is about 2 moles, which would form 2 moles of HCI(g). Twice the molar volume is 44.8 L, which is the answer as calculated.

A t S T P : ( 2.00205 m o l HCI )

5. 1 2

=

=

P lan: Balance the equation for the reaction. Determine the I imiting reactant by finding the moles of each reactant from the ideal gas equation, and comparing the values. Calculate the moles of remaining excess reactant. This is the only gas left in the flask, so it is used to calculate the pressure inside the flask. There will be no intermediate rounding. Solution: The balanced equation is N H3(g) + HCI(g) -j N H4CI(s). The stoichiometric ratio ofNH3 to HCI is 1 : 1 , so the reactant present in the lower quantity of moles is the limiting reactant. ( 0.452 atm ) ( 1 0.0 L ) PV Moles ammonia 0. 1 8653 mol N H3 0.082 1 atm · L RT 273 . 1 5 + 22 ) K ) (( mol · K ( 7.50 atm ) ( 1 55 mL ) 1 O-3 L 0.052249 mol HCI Moles hydrogen chloride 0.082 I atm · L ( 27 1 K ) I mL mol · K The HCI is limiting so the moles of ammonia gas left after the reaction would be 0. 1 8653 - 0.052249 0. 1 3428 1 mol NH 3 ( 0. 1 3428 1 mOI ) 0.082 1 atm · L ( ( 273 . 1 5 + 22 ) K ) mol · K nRT . Pressure ammOnIa ( 1 0.0 L ) V 0.325387 0.325 atm NH3 Check: Doing a rough calculation of moles gives for N H3 (0.5 x 1 0) 1 ( 0. 1 x 300) 0. 1 7 mol and for HCI (8 x 0. 1 )/(0. 1 x 300) 0.027 mol which means 0. 1 4 mol NH3 is left. Plugging this value into a rough calculation of the pressure gives (0. 1 4 x 0. 1 x 300)/ 1 0 0.4 atmospheres. This is close to the calculated answer.

= =( --

=(

=

=

=

--

=

=

)

=

)

[ )= --

(

)

=

69

=

=

5. 1 3

Plan: Graham' s Law can be used to solve for the effusion rate of the ethane since the rate and molar mass of helium is known, along with the molar mass of ethane. In the same way that running slower increases the time to go from one point to another, so the rate of effusion decreases as the time increases. The rate can be expressed as I 1time. Solution: Rate He M Mc, H ' = Rate C 2 H 6 He

(

�

mol H e 1 .2 5 min

0.0 1 0

)

( 3 0 .07

( 4 . 003

g/mol ) g/mol )

Time for C2H 6 = 3 . 4 2 5 9 7 = 3 .43 min Check: The ethane should move slower than the helium since ethane has a larger molar mass. This is consistent with the calculation that the ethane molecule takes longer to effuse. The second check is an estimate. The square root of 30 is estimated as 5 and the square root of 4 is 2. The time 1 .2 5 min x 5/2 = 3 . 1 min, which validates the calculated answer of 3 .42 min. END-OF-CHAPTER PROBLEMS 5.1

5 .6

a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger container. b) The volume of the container holding the gas sample increases when heated, but the volume of the container holding the liquid sample remains essentially constant when heated. c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced. The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the densities of the two liquids. h H ,o

h Hg

h 5 .8

d Hg

d H ,o

= � xh Hg = d H ,o

(

1 3 .5 g/mL

1 . 00 g / m L

]

( 725 mmH g )

( l( ) 1 0- m 3

I

rum

� 1 O -2 m

= 9 7 8 . 7 5 = 979 c m H2 O

Plan: Use the conversion factors between pressure units: I atm = 760 mmHg = 760 torr = 1 0 1 . 32 5 kPa = 1 .0 1 32 5 bar Solution: 760 mmHg a) ( 0.745 atm ) = 566.2 = 566 mmHg 1 atm b)

( 99 2 torr )

c)

( 3 6 5 kPa )

d) ( 804 5. 1 4

",0

=

( (

(

1 .0 1 3 2 5 760

bar torr

] ]

l atm 1 0 1 . 3 2 5 kPa

mmH ) g

(

1 0 1 .325 760

]

= 1 . 3 2 2 5 6 = 1 .3 2 bar = 3 . 60227 = 3.6 0 a tm

kP a mmHg

]

= 1 07. 1 9 1 = 1 07 kP a

At constant temperature and volume, the pressure of the gas i s directly proportional to the number of moles of the gas. Verify this by examining the ideal gas equation. At constant T and V, the ideal gas equation becomes P = n(RTN) or P = n x constant.

70

5.16

a) A s the pressure on a gas increases, the molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to one third of the original volume at constant temperature (Boyle's Law). b) As the temperature of a gas increases, the gas molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with greater force, which increases the size (volume) of the container. lf the temperature is increased by a factor of 2.5 (at constant pressure) then the volume will i ncrease by a factor of 2.5 (Charles's Law). c) As the number of molecules of gas increase, the force they ex ert on the container increases. This results in an increase in the volume of the container. Adding two moles of gas to one mole increases the number of moles by a factor of three, thus the volu m e increases by a factor of three (Avogadro's Law).

5. 1 8

Plan: This is Charles's Law. Charles's Law states that at constant pressure and with a fixed amount of gas, the volume of a gas is directly proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a gas. Solution:

� =�

at constant n and P

T-?

=

T

I

�

-

VI T I = 1 98°C + 273 = 47 1 K; V2 = 2.50 L T 2 = ? 2.50 L = 230.88 K - 273 = - 42. 1 2 T 2 = 47 1 K = 42°C 5. 1 0 L

T I T2 VI = 5. 1 0 L;

5.20

( )

Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Solution: TI V2

5.22

=

VI

T2 T2

( )(�) P2 T

I

P I = 1 53 .3 kPa; =

( 25.5

q( 298273 KK )( 1 01 53.3 k Pa ) = 35.3437 = l .325 kPa

35.3 L

Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the ideal gas equation, n = PV/RT. The gas constant, R = 0.082 1 Loatm/moloK, gives pressure in atmospheres and temperature in Kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to Kelvin. Solution: I atm = 0.300 atm; V = 5.0 L; T = 27°C + 273 = 300 K P = (228 torr) PV = nRT or n = PV/RT 760 torr (0.300 atm)(5 .0 L) = 0.06090 1 = 0 . 06 1 mol chlorin e n PV = RT (0.082 1 L . atm )(300 K) mol - K =

5.24

-

)

(

Plan: Solve the ideal gas equation for moles and convert to mass using the molar mass of ClF3 · Volume must be converted to L , pressure to atm and temperature to K. Solution: l atm = 0.9 1 974 atm; v = 0.207 L PV = nRT or n = PV/RT P = (699 mmHg) 760 mmHg T = 45°C + 273 = 3 1 8 K

(

)

71

PV (0.9 1 974 atm)(0.207 L) = 0.0072923 mol CIF) R T - (0.082 1 L - atm )(3 1 8 K) mol - K 92.45 g CIF) = 0.674 1 7 = 0.674 g CIF3 mass CIF) = (0.0072923 mol CIF) ) 1 mol CIF) n-

_

(

J

5 .28

The molar mass of Hz is less than the average molar mass of air (mostly Nz, Oz, and Ar), so air is denser. To collect a beaker of Hz(g ), invert the beaker so that the air will be replaced by the l ighter Hz. The molar mass of COz is greater than the average molar mass of air, so COz(g) is more dense. Collect the COz holding the beaker upright, so the l ighter air will be displaced out the top of the beaker.

5.3 1

Plan: Using the ideal gas equation and the molar mass of xenon, 1 3 1 .3 g/mol, we can find the density of xenon gas at STP. Standard temperature is O°C and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and wil l not affect the significant figures. Solution: ( 1 3 1 .3 g/mol ) ( 1 atm ) = 5 . 85 8 1 = 5.86 gIL d = M P I RT = L · atm 0.082 1 ( 273 K ) mol · K

5.33

(

)

Plan: Apply the ideal gas equation to determine the number of moles. Convert moles to mass and divide by the volume to obtain density in giL. Do not forget that the pressure at STP is exact and wil l not affect the significant figures. Solution : ( I atm ) ( 0.0400 L ) = 1 .78465 x 1 0-) = 1 .78 X 1 0-) mol AsH 3 PV n= - = RT L · atm ( 273 K ) 0.082 1 mol · K ( 1 .78465 X 1 0-3 mol )( 77.94 g/mol ) mass d= = 3 .47740 = 3.48 giL volume ( 0.0400 L )

)

(

5.35

Plan : Rearrange the formula PV = (m I M)RT to solve for molar mass: M = mRT I P V . Convert the mass in ng to grams and volume in il L to L. Temperature must be in Kelvin and pressure in torr. Solution : l atm =0.5 1 0526 atm P = (388 torr ) T = 45°C + 273 = 3 1 8 K 760 torr { 1 O -9 g = 2.06 X 1 0-7 g { 1 O -6 L = 2.06 X 1 0-7 L Y = (0.206 ,uL mass = (206 ng \ I� \ 1� L - atm (2.06 X 1 0 -7 g)(0.082 1 )(3 1 8 K) T R = mol e K M= � = 5 1 . 1 390 = 5 1 . 1 g/mol PY (0.5 1 0526 atm)(2.06 x 1 0 -7 L )

(

[

J

)

j

72

J

5.37

Plan: Use the ideal gas equation to determine the number of moles of Ar and O2. The gases are combined (nTOT = n Ar + no) into a 400 mL flask (V) at 27°C (T) . Determine the total pressure from nTOT, V, and T. Pressure must be in units of atm, volume in units of L and temperature in K. Solution: PV PV = nRT n=RT ( 1 .20 atm ) ( 0.600 L ) PV Moles Ar = - = = 0.0 1 7539585 mol Ar (unrounded) L o atm RT ) ( ( 273 + 227 K ) 0.082 1 mol o K ( 50 1 torr ) ( 0.200 L ) PV 1 atm = 0.0040 1 4680 mol O 2 (unrounded) Moles O2 = - = L atm RT (( 273 + 1 27 ) K ) 760 torr 0.082 1 mol o K

(

(

)

)

0

(

]

)

NTOT = n Ar + no = 0.0 1 7539585 mol + 0.0040 1 4680 mol = 0.02 1 554265 mol ( 0.02 1 554265 mol ) 0.082 1 L o atm (( 273 + 27 ) K ) mol K I mL nRT . P mixture = = -------"--------"------400 mL V 1 0-3 L = 1 .32720 = 1 .33 atm

(

--

5.41

( )

Plan: The problem gives the mass, volume, temperature, and pressure of a gas, so we can solve for molar mass using M = mRTIPV. The problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms. We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound. Convert pressure to atm and temperature to K. Solution: ( 0.482 g ) 0.082 1 L atm (( 273 + 1 0 1 ) K ) 760 torr mol K = 7 1 .8869 g/mol (unrounded) M = mRTIPY = ( 767 torr ) ( 0.204 L ) 1 atm The carbon accounts for [5 ( 1 2 g/mol) ] = 60 g/mol, thus, the hydrogen must make up the difference (72 - 60) = 1 2 glmol . A value of 1 2 glmol corresponds to 1 2 H atoms. (Since fractional atoms are not possible, rounding is acceptable.) Therefore, the molecular formula is CsH12 .

(

5 .43

0

)

0

0

(

]

Plan: Since you have the pressure, volume and temperature, use the ideal gas equation to solve for the total moles of gas. Solution: PV = n RT a) ( 850. torr ) ( 2 1 L ) I atm = 0.8996 1 = 0. 9 0 mol gas n = PY I RT = L atm 0.082 1 (( 273 + 45 ) K ) 760 torr mol o K b) The information given i n ppm is a way of expressing the proportion, or fraction, of S0 2 present in the mixture. Since n is directly proportional to Y, the volume fraction can be used in place of the mole fraction used in equation 6 5 . 1 2. There3 are 7.95 x 1 0 3 parts S0 2 in a million parts of mixture, so volume fraction = (7.95 x 1 0 3 I I X 1 0 ) 7.95 X 1 0- . 3 Therefore, Pso , = volume fraction x PTOT = (7.95 x 1 0- ) (850. torr) = 6.7575 = 6.76 torr.

(

0

(

)

]

=

73

occupies 22.that4 Lwilat l L ofesgasof phosphorus me of gases molaor nvoltoudetermi fromththee balstandard es ofcoxygen thecmol Plan: Weand canuse find 5.44 STP) mol e h t e n equati nced a from o rati metri o hi stoi e h t react Sol utiwion:thPthe4(S) oxygen. + 5 02 (g) p40J Q ( s ) Mass P4 = (35. 5 L O 2 )( 22.1 mol4 L OO22 )( 51 molmol OP42 )( 123.I mol8 8 gP4P4 ) = 39.2655 5. 46 reactant. Plan: To findThe thesmalmassler number ofPH3, ofwrimolte thees ofbalproduct anced equati on andthe findl i m ittheingnumber ofSolmolveesforofPH3 produced byngeach i n di c ates reagent. mol e s of Hz usi the standard mol a r vol u me (or use i d eal gas equati o n). Solution:P4(S) + 6 Hz(g) 4 PH3(g) Moles hydrogen = (83. 0 L ) (� 22.4 L ) = 3. 705357 mol Hz (unrounded) PH3 from P4 = (37. 5 g P4 ) ( 123.I mol88 gP4P4 )( 41 moImolPHP4 3 ) = 1. 2 1085 mol PH3 PH 3 from Hz (3. 7 05357 mol Hz ) ( 46molmol PHH 23 ) = 2.470238 mol PH 3 P4 is the li miting reactant. Mass PH3 = (37. 5 g P4 )( 123.1 mol8 8 gP4P4 )( 41 molmol PHP4 3 )( 33.I mol99 gPHPH 3 ) = 41 . 15676 3 Pl an: Fionrst,andwrithente thethebalstoianced equatic oratin. oThefrommolthees balof hydrogen produced cantobedetermi calculantede thefrommolthees ofidealalugasminum 5.48 equati c hi o metri a nced equati o n i s used that reacted. TheTablproble 5.e2mreports specifipressure es "hydrogen gas(25.col2letorr) cted andover28°C water,(28." so3 torr), the partiso atakel pressure of waterof themusttwofirst bevaluessubtracted. at 26°C the average Solutioton:2obtain the+ 6 partial pressure2 of water +at 327°e. Hz(watge)r vapor = Hydrogen pressure = total pressure -pressure of Moles of hydrogen: (751 mmHg) - [(28. 3 + 25.2) torr / 2] = 724.25 torr (unrounded) 10-3 L ) = 0. 00138514 mol Hz PY =nRT n = -PYRT = ( (724.2L5o torratm)()35.( 8 mL+ ) ) ( 760I atmtorr )( -I mL 0. 0821 mol o K 273 27 K Mass of Al = (0. 00138514 mol H 2 ) ( 32 molmol HAl2 )( 26.I mol98 gAl l ) 0. 024914 = (l

�

= 39.3

g P4

�

=

= 4 1 .2

A I (s)

H C I (aq)

�

g PH3

AICI3(aq )

r

j

A

=

0.024 9

g

AI

5. 5 0 toPlafindn: Tomolfindes ofmLSOz,ofSand02, wriusetthee theidbalealagasncedequati equationon,to convert theume.given mass ofP4S3 to moles, use the molar ratio find vol Sol utio n:P4S3( ) + 8 Oz(g) P40I (S) + 3 SOz(g) S 0 P4S, 3 mol SO 2 l mol Moles SOz = ( 0. 800 g P4S3 ) ( 220.09 g P4S3 )( I mol P4S3 ) = 0. 0 10905 mol SOz �

J

74

nRT Volume S02 -P 286.249

=

=

5.5 1

=

=

)

( 0.0 1 0905 mol S02 ) 0.082 1 L o atm (( 273 + 32 ) K )

(

mol K 725 torr 0

286 mL S02

(

760 torr 1 atm

J[ ) I mL 1 0-3 L

Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the number of moles of silicon tetrafluoride gas formed. Then, using the ideal gas equation with the moles of gas, the temperature and the volume, we can calculate the pressure of the silicon tetrafluoride gas. Solution: 2 XeF 6 (S) + Si02(s) -? 2 XeOF4(l) + SiF4(g) 1 mol SiF4 I mol XeF6 0.0040766 mol SiF4 Mole SiF4 n ( 2.00 g XeF6 ) 245 .3 g XeF6 2 mol XeF6

==

(

J

(

(-'--

J=

----'-) (( 273

( 0.0040766 mol SiF4 ) 0.082 1 L atm K . nRT -------- --mol -Pressure SIF4 P = V 1 .00 L 0.099737 0.0997 atm SiF4

= -- = = =

0

0

+

25 ) K )

------

5 .54

At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical. This is because at the same temperature, all gases have the same average kinetic energy, resulting in the same pressure.

5.57

Plan: The molar masses of the three gases are 2.0 1 6 for H2 (Flask A), 4.003 for He (Flask B), and 1 6.04 for CH4 (Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of Hz will contain more gas molecules (about 2 mole' s worth) than 4 g of He or 4 g of CH4• Since helium has a smaller molar mass than methane, 4 g of He will contain more gas molecules (about I mole' s worth) than 4 g of CH4 (about 0.25 mole's worth). Solution: a) PA > PD > Pe The pressure of a gas is proportional to the number of gas molecules. So, the gas sample with more gas molecules will have a greater pressure. b) EA ED Ee Average kinetic energy depends only on temperature. The temperature of each gas sample is 273 K, so they all have the same average kinetic energy. c) rateA > rateD > ratee When comparing the speed of two gas molecules, the one with the lower mass travels faster. d) total EA > total ED > total Ee Since the average kinetic energy for each gas is the same (part b of this problem) then the total kinetic energy would equal the average times the number of molecules. Since the hydrogen flask contains the most molecules, its total kinetic energy will be the greatest. e) dA dD de Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same mass, 4 g. Thus, the density of each is 4 gl5 L 0.8 giL. =

=

5.58

=

=

=

To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses (equation 5 . 1 4, Graham' s Law). 352.0 g/mol Molar Mass UF6 Rate H2 1 3 .2 1 3 7 = 13.21 2.0 1 6 g/mol Rate UF6 Molar Mass H 2

=

5 .60

a) The gases have the same average kinetic energy because they are at the same temperature. The heavier Ar atoms are moving slower than the lighter He atoms to maintain the same average kinetic energy. Therefore, Curve 1 better represents the behavior of Ar. b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice. c) Fluorine gas exists as a diatomic molecule, F2, with .M = 38.00 glmol. Therefore, F2 is much closer in size to Ar (39.95 glmol) than He (4.003 glmol), so Curve 1 more closely represents Fz 's behavior.

75

squareon.roots of the molar masses oftithemeratiforotheof theF2 effusi calcoulnateratesthetoinverse on orates, ratio ofuseeffusi Plan: Toonfind5.14).the Then 5. 62 (equati the find effusi of rati the Sol uti o n: g/mol = 3. 08105 (unrounded) Rate He Mol ar Mass F2 4.38.00030 g/mol Rate F2 Molar Mass He 3. 08105 Time F2 Time F2 = 1 4.0 1 878 = Rate He Ti me F2 1. 00 4. 5 5 min He Rate F2 Time He ofar mass of number, ting oftosome consiandsneon phosphorus of theonelofement ar formof effusi isearelmolatievcule rates phosphorus 5. 64 phosphorus Plan: White atoms. mol the e n determi phosphorus e t whi h t Use the mol ar mass of white phosphorus, determine the number of phosphorus atoms, in From e phosphorous. tmol whi one e cul e of whi t e phosphorus. Sol ution: Rate Px Mol ar Mass e Rate N e = 0.404 = Molar Mass NPx Ne 20.18 g/mol (0.404)2 = MMololaarr Mass Mass Px Molar Mass Px g/mol 0.163216 = Mol20.18ar Mass Px Molar Mass Px = 123. 6398 gimol ( 123.mol6398Px g )( 30.mol97 gPP ) = 3. 992244=4 mol P/mol Px or 4 atoms P/molecule Px Thus, so Px = P 5. 66 The Intermol the realon pressure pressure,ng tosoTabl it causes n. size eofculthear attracti intermolonseculcause ar attracti is relatedtotobethe constantidealAccordi e 5 .4,a a N , 1 . 39devi, atio2.32 and aco, = 3. 5 9. Therefore, CO2 experiences greater negative deviation i n pressure than the other two gases: N2 < Kr< CO2. 5. 6 8 Nfarther itrogenapart.gas behaves more iddefined eal l y atasI atm thanstingatof500gasatmmolbecause at loactweri npressures they ofgasthmol ehculer gases are An i d eal gas is consi e cul e s that dependentl e ot mol ecules.andWhenthe gasvol umolmeeofculthees aremolfareculapart, eallyo, nbecause i ntermolnerevolculuame. r attractions are less important es is atheysmalactleri dfracti of the contai 5. 7 1 4:PlIan:ratiUseo . DithveidIdeal GasmassEquati onbytotfind the number ofesmolto eobtai s ofnO2.molMolar mass, es of O2g/mol.combine with Hb in a e t h e of Hb h e number of mol SolPV=utinoRn:T Moles O2 = PVRT = (743L torratm) (1. 5 3 mL ) ) ( 760I atmtorr J [I1O-3 L J ( 0.0821 mol K ) ( (273 37 K ) = 5. 8 7708 10-5 mol O2 (umounded) Moles Hb = ( 5. 87708 10.5 mol O2 ) ( 41 molmol HbO2 ) = 1.46927 1 0-5 mol Hb (unrounded) Mol ar mass hemoglobin = (1. 00 g Hb) I (1.46927 10-5 mol Hb) = 6.806098 1 04 1 4.0

min

x,

x,

I

4

4.

atoms per molecule,

less than

a

0

0

negative

a.

=

mL

+

X

X

X

X

X

76

=

6.81

X

1 04

g/mol

a K, =

5 . 73

Plan: Convert the mass of CI2 to moles and use the ideal gas law and van der Waals equation to find the pressure of the gas. Solution: 1 03 g I mol CI 2 a) Moles Ch: ( 0.5850 kg CI 2 ) ___ = 8.25 1 0578 mol I kg 70.90 g CI 2 Ideal gas equation: PY = nRT L - atm 8.25 1 0578 mol 0.082 1 (( 273 + 225 ) K ) mol - K IGL = nRTN = --'P ----'� 1 5 .00 L = 22.490 1 = 22.5 atm -

b)

_

[ p :: } +

Y - nb )

( )( (

=

)

-

)

---

nRT

2 nRT n a atm- e From Table 5.4: a = 6.49 2 Y nb y mof n = 8.25 1 0578 mol from Part (a) P YDW -

_

(

)

L b = 0.0562 mol 2

[

2

atm - L ( 8.25 1 0578 mol CI2 ) 0.082 1 L - atm (( 273 + 225 ) K ) ( 8.25 1 0578 mol Cl2 ) 6.49 mol 2 mol - K P yDW = ----------'-------,--:------:--2 ( 1 5.00 L ) 1 5.00 L - ( 8.25 1 0578 mol CI2 ) 0.0562 �

5.75

= 2 1 .24378 = 21.2 atm

(

mol

)

)

Plan: Partial pressures and mole fractions are calculated from Dalton's Law of Partial Pressures: P A = XA(Ptota l) Solution: a) Convert each mole percent to a mole fraction by dividing by 1 00%. PNitrogen = XNitrogen PTotal = (0.786) ( 1 .00 atm) (760 torr / I atm) = 597.36 = 597 torr Nz 1 59 torr Oz P Oxygen = XOxygen PTotal = (0.209) ( 1 .00 atm) (760 torr / I atm) = 1 58.84 P C arOOn Di oxide = XCarOOn Di oxide PTotal = (0.0004) ( 1 .00 atm) (760 torr / I atm) = 0.304 = 0.3 torr COz P Water = XWater PTotal = (0.0046) ( 1 .00 atm) (760 torr / atm) = 3 .496 = 3.5 torr Oz b) Mole fractions can be calculated by rearranging Dalton's Law of Partial Pressures: XA = P AlPtotal and multiply by 1 00 to express mole fraction as percent. PTotal = (569 + 1 04 + 40 + 47) torr = 760 torr [(569 torr) / (760 torr) ] x 1 00% = 74.8684 = 74.9 mol% Nz N2: O2: [( 1 04 torr) / (760 torr) ] x 1 00% = 1 3 .6842 = 1 3.7 mol% Oz [(40 torr) / (760 torr) ] x 1 00% = 5.263 = 5.3 mol% COz CO2: [(47 torr) / (760 torr)] x 1 00% = 6. 1 842 = 6.2 mol% COz H20: c) Number of molecules 0[ 02 can be calculated using the Ideal Gas Equation and Avogadro's number. PY = nRT ( 1 04 torr ) ( 0.50 L ) I atm PY - 0.0026883 mol O2 Moles O2 = - = L - atm RT (( 273 + 37 ) K ) 760 torr 0.082 1 mol - K 6.022 x 1 0 23 molecules 0 2 Molecules O2 = ( 0.0026883 mol O 2 ) I mol O 2 21 Z1 = 1 .6 1 89 X 1 0 = 1.6 X 1 0 molecules Oz =

I

(

)

(

(

77

j_

J

5 . 77

Plan: For part a, since the volume, temperature, and pressure of the gas are changing, use the combined gas law. For part b, use the ideal gas equation to solve for moles of air and then moles ofN2. Solution: 1 atm P V P 2 V2 = 1 .842 1 atm; VI = 208 mL; P I = ( 1 400. mmHg ) a) I I 760 mmHg T2 TI T I = 286 K; P2 = l atm; T2 = 273 K; V2 = ? 2 298 K 1 .842 1 atm = 399.23 mL = 4 x l 0 m L V2 = V I � � = ( 208 mL) 286 K latm TI P2

( ( )(

=

( )(

)

(

)

)

( 1 .842 1 atm) ( 0.208 L ) = 0.0 1 63 1 8 mol air L · atm ( 286 K ) 0.082 1 mol · K 77% N 2 = 0.0 1 25649 = O. 0 13 mol N2 Mole N2 = ( 0.0 1 63 1 8 mol ) 1 00%

PV b ) Mole air = n = - = RT

5 .78

(

J

(

)

)(

)( ) )[ ) ( J(

The balanced equation and reactant amounts are given, so the first step is to identify the limiting reactant. 2 mol N02 8.95 g c u 1 mol Cu = 1 .394256 mol N02 (unrounded) Moles N 02 from Cu = ( 4.95 cm 3 ) 63 .55 g Cu 1 mol Cu cm 3

(

)(

)

2 mol N02 68.0% HN0 3 1 cm 3 1 .42 g 1 mol HN0 3 1 00% 1 mL mL 63 .02 g 4 mol HN0 3 = 1 .7620 mol N02 (unrounded) Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after the reaction goes to completion. Use the calculated number of moles ofN02 and the given temperature and pressure in the ideal gas equation to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the only gas involved in the reaction. L · atm ( ( 273.2 + 28.2) K ) ( 1 .394256 mol N0 2 ) 0.082 1 760 torr mol ·K V = nRT I P = ( 735 torr) 1 atm = 35 .67427 = 3 5.7 L N02 Moles N02 from HN0 3 = ( 230.0 mL)

)

(

5.82

( )

Plan: The empirical formula for aluminum chloride is AICh (Ae+ and Cn The empirical formula mass is 1 33 .33 glmol). Calculate the molar mass of the gaseous species from the ratio of effusion rates. This molar mass, divided by the empirical weight, should give a whole number multiple that will yield the molecular formula. Solution: Molar Mass He Rate Unk --= 0. 1 22 = Molar Mass Unk 0. 1 22 = Molar mass Unknown = 268.9465 glmol The whole number multiple is 268.946511 33.33, which is about 2. Therefore, the molecular formula of the gaseous species is 2 x (AICI3) = A12C16.

5 . 84

Plan: First, write the balanced equation for the reaction: 2 S02 + O2 � 2 S03' The total number of moles of gas will change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From the volume, temperature and pressures given, we can calculate the number of moles of gas before and after the reaction using ideal gas equation. For each mole of S03 formed, the total number of moles of gas decreases by mole. Thus, twice the decrease in moles of gas equals the moles of S03 formed.

78

1 /2

Solution : Moles of gas before and after reaction ( 1 .95 atm ) ( 2 . 00 L ) PV Initial moles 0.05278 1 1 6 mo l (u nrounded ) · RT 0.082 1 L atm ( 900. K ) mol · K ( 1 .65 atm ) ( 2 .00 L ) PV . Fmal moles 0.04466098 mol (unrounded) atm RT ' L 0.082 1 ( 900. K ) mol · K Moles of S03 produced 2 x decrease in the total number of moles 2 x (0.05278 1 1 6 mol - 0.04466098 mol) 0.0 1 624036 1 .62 x 1 0-2 mol Check: Jf the starting amount is 0.0528 total moles of S02 and 02, then x + y 0.0528 mol, where x mol of S02 and y mol of O2. After the reaction : (x - z) + (y - 0.5z) + z = 0.0447 mol Where z mol of S03 formed mol of S02 reacted 2(mol of O2 reacted). Subtracting the two equations gives: x - (x - z) + Y - (y - 0.5z) - z 0.0528 - 0.0447 z 0.0 1 63 mol S03 The approach of setting up two equations and solving them gives the same result as above. =

=

=

-

=

=

=

(

(

)

=

)

=

=

=

=

=

=

=

=

=

=

=

5.88

a) A preliminary equation for this reaction is CxHyNz + n O 2 --j 4 CO2 + 2 N2 + 1 0 H 20. Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To form 4 volumes of CO2 would require 4 volumes of O2 and to form 1 0 volumes of H 20 would require 5 volumes of O 2 , Thus, 9 volumes of O2 was required. b) Since the volume of a gas is proportional to the number of moles of the gas, we can equate volume and moles. From a volume ratio of 4 CO2 :2 N2: 1 0 H20 we deduce a mole ratio of 4 C:4 N :20 H or I C: 1 N : 5 H for an empirical formula of CHsN.

5 .90

Plan: To find the factor by which a diver's lungs would expand, find the factor by which P changes from 1 25 ft to the surface, and apply Boyle 's L aw. To find that factor, calculate Pseawater at 1 25 ft by converting the given depth from ft-seawater to mmHg to atm and adding the surface pressure ( 1 .00 atm). Solution : 2 1 2 in 2.54 cm 1 0- m I � 4 = 3 . 8 1 X 1 0 nunH 20 P (H20) ( 1 25 ft ) 3 1 ft 1 III 1 cm 10 m 3 . 8 1 X 1 0 4 mmH 2 0 1 3 . 5 g/m L 2935 . 1 1 1 1 mmHg h Hg 1 .04 g/m L h Hg

_

( )(

=

( I

)(

)( ) _

)

=

atm 3 .86 1 988 atm (unrounded) 760 mm Hg Ptotal ( 1 .00 atm) + (3 .86 1 988 atm) 4.86 1 988 atm (unrounded) Use Boyle 's Law to find the volume change of the diver's lungs: P

( Hg)

=

( 2935 . 1 1 1 1 1 mmHg )

=

PIVI

=

=

P2V

=

4.86 1 988 atm 4.86 VI P2 VI 1 atm To find the depth to which the diver could ascend safely, use the given safe expansion factor ( 1 .5) and the pressure at 1 25 ft, P 1 2 5 , to find the safest ascended pressure, P safe. P 1 2 5 / Psafe 1 .5 Psa fe P 1 2 5 / 1 . 5 = (4.86 1 988 atm) / 1 .5 3 .24 1 325 atm (unrounded) Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance from the initial depth to find how far the diver could ascend. V2

_

=

=

=

=

79

hH 2 0

(

( 4.86 1 988 - 3 .24 1 325 atm ) 760 mmHg

h (Rg): =

hHg

dHg

dH 2 0

)

1 23 1 .7039 mmHg 1 atm 1 3 . 5 g/mL hH 2 0 1 23 1 .7039 mmHg 1 .04 g/mL

(

J(

_

-3 ( 1 5988.464 mmH 2 0 ) 1 0 m I .094 Yd 1 mm

1m

)( J

Therefore, the diver can safely ascend 52.5

3 ft 1 yd

ft to

=

=

52.474 1

ft

a depth of ( 1 25 - 52.474 1 )

=

72.5259

=

73 ft.

5 .97

Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive forces between the substances. Solution: a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than xenon. The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe's larger size also means that the volume the gas occupies becomes a greater proportion of the container's volume at high pressures. b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at room temperature while neon is a gas at room temperature). c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury are greater because it is a liquid at room temperature while radon is a gas. d) Water is a liquid at room temperature; methane is a gas at room temperature (think about where you have heard of methane gas before - Bunsen burners in lab, cows' digestive system). Therefore, water molecules have stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than methane molecules.

5 . 1 00

Plan: V and T are not given, so the ideal gas equation cannot be used. The total pressure of the mixture is given. Use P A X A X P total to find the mole fraction of each gas and then the mass fraction. The total mass of the two gases is 35.0 g. Solution: Ptotal P krypton + Pc arbo n diox ide 0.708 atm The NaOR absorbed the CO2 leaving the Kr, thus P krypton 0.250 atm P c arbon diox ide P total - P kry pton 0.708 atm - 0.250 atm 0.458 atm Determining mole fractions: P A X A X Ptotal Peo 2 0.458 atm Carbon dioxide: X 0.64689 (unrounde d ) 0.708 atm Ptota) 0.250 atm Krypton : X PKr 0.353 1 07 (unrounded) 0.708 atm Ptotal =

=

=

=

=

=

=

=

=

=

---

Re i ative mass fract;oo

=

=

�

l

=

(

( 0.353 1 07 ) 83 .80 g ( 0.64689 )

=

(

Kr) j )

mo i 44.01 g CO 2 mol

�

1 .039366 (uomuoded)

3 5 .0 g x g CO2 + ( 1 .039366 x) g Kr 3 5 .0 g 2.039366 x Grams CO2 x (35.0 g) / (2.039366) 1 7. 1 62 1 95 8 1 1 7.2 g CO2 Grams Kr 35.0 g - 1 7. 1 62 g CO2 1 7.837804 1 9 1 7.8 g Kr =

=

=

=

=

=

=

80

=

5. 1 03

�

3 RT M Set the given relationships equal to each other.

a) Derive

U

=

rms

..!.

--

mu2

2

�(l!...-) T 2

Multiply each side by 2 and divide by m.

NA

2 (�) T

=

�

=

m

NA

3 RT

�

- --

mNA

Solve for u by taking the square root of each side; substitute molar mass, M, for mNA (mass of one molecule x Avogadro's number of molecules). u

,m.,

-

�3RT

--

M

�

rate2 � ....; M2 rate, At a given T, the average kinetic energy is equal for two substances, with molecular masses ml and m2:

b) Derive Graham's L aw

E

k

=

..!.

/

2

-2 mjUj

..!!!..!....-

mju

I

U 22

_

m2

U

/

--

/

m2u

2 -2 m2u 2

=

�

----..

U2

=

�

Uj

The average molecular speed, u , is directly proportional to the rate of effusion. Therefore, substitute "rate" for each "u." In addition, the molecular mass is directly proportional to the molar mass, so substitute M for each m:

� f.M:

5 . 1 07

=

rate2 rate,

Plan: Find the number of moles of carbon dioxide produced by converting the mass of glucose in grams to moles and using the stoichiometric ratio from the balanced equation. Then use the T and P given to calculate volume from the ideal gas equation. Solution: a) C6H I 206(S) + 6 02(g) -; 6 C O2(g) + 6 H 20(g) 6 mol C O2 mol C 6 H 1 2 0 6 = 0 . 599467 mol CO2 Moles CO2: ( 1 8 . 0 g C6 H I 2 0 6 ) mol C6 H , 2 06 1 80. 1 6 g C6 H , 2 06

(

(I

)

)( I

)

atm (0.599467 mOI) 0.082 1 L · ( ( 27 3 + 3 5 ) K ) nRT = mol-.,-· ---'K ----'-----;V = ( 760 torr ) atm ( 780. torr ) P = 14.76992 = 14.8 Liters CO2 This solution assumes that partial pressure of 02 does not interfere with the reaction conditions. --

-

-

-

-

81

I

b) Plan: From the stoichiometric ratios in the balanced equation, calculate the moles of each gas and then use Dalton's law of partial pressures to determine the pressure of each gas. Solution : 6 mol 1 mol C 6 H 12 0 6 Moles CO2 Moles O2 ( 9.0 g C 6 H 12 0 6 ) 1 80. 1 6 g C 6 H 12 0 6 I mol C 6 H 12 0 6 0.299734 mol CO2 mol O 2 At 35°C, the vapor pressure of water is 42.2 torr. No matter how much water is produced, the partial pressure of H2 0 wi II stil l be 42.2 torr. The remaining pressure, 780 torr - 42.2 torr 73 7.8 torr (unrounded) is the sum of partial pressures for O2 and CO2. Since the mole fractions of 02 and CO2 are equal, their pressures must be equal, and be one-half of sum of the partial pressures just found. =

Pwater

=

42.2

=

torr

(737.8 torr) / 2 368.9

5.1 1 1

=

=

)(

(

=

=

3.7 x 1 0 2 torr Poxygen

=

=

)

Pcarbon dioxide

Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of effusion rates. This ratio is the enrichment factor for each step. Solution : Molar Mass 2 3 8 UF6 352.04 g/mol 1 ��� 3 2 Rate 23 8 UF6 349.03 g/mol Molar Mass 5 UF6 =

__ __ __ __

1 .004302694 enrichment factor (unrounded) Therefore, the 23abundance of 23 5 UF6 after one membrane is 0. 72% x 1 .004302694; Abundance of 5 U F6 after "N" membranes 0.72% * ( 1 .004302694)N Desired abundance of 23 5 UF6 3 .0% 0.72% * ( 1 .004302694)N Solving for N : 4. 1 6667 ( 1 .004302694)N In 4. 1 6667 In ( 1 .004302694)N In 4. 1 6667 N * In ( 1. 004302694) N (In 4. 1 6667) / (In 1 .004302694) 332.392957 332 steps =

=

=

=

=

= =

=

=

5. 1 12

=

Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 45% as the rate for H2, the rate for O2 can be determined from Graham's Law. Solution : Rate 0 2 Molar Mass H 2 2.0 1 6 g/mol Rate O 2 Rate H 2 Molar Mass 0 2 32.00 g/mol 45 = Rate O2 0.250998(45) 1 1 .2949 (unrounded) Amount of H 2 that leaks 45%; 1 00-45 = 55% H2 remains Amount of O2 that leaks 1 1 .2949%; 1 00- 1 1 . 2949 88.705% O2 remains � 88.705 1 .6 1 28 1 .6 55 H2 =

=

=

=

=

=

=

=

=

82

Chapter 6 Thermochemistry: Energy Flow and Chemical Change FOLLOW-UP PROBLEMS

6. 1

Plan: The system is the reactant and products of the reaction. Since heat is absorbed by the surroundings, the system releases heat and q is negative. Because work is done on the system, w is positive. Use equation 6.2 to calculate 11E. Solution : 1 .055 kJ 4. 1 84 kJ I1E = q + w = ( -26.0 kcal ) + ( + 1 5 .0 Btu ) = -92.959 = - 93 kJ 1 kcal I Btu Check: A negative I1E seems reasonable since energy must be removed from the system to condense gaseous reactants into a liquid product.

(

(

)

)

6.2

Plan: Since heat is a "product" in this reaction, the reaction is exothermic (f1H < 0) and the reactants are above the products in an enthalpy diagram. Solution:

6.3

Plan: Heat is transferred away from the ethylene glycol as it cools. Table 6.2 lists the specific heat of ethylene glycol as 2.42 J/gK. This value becomes negative because heat is lost. The heat released is calculated using equation 6.7. Solution: �T = 37.0°C - 25.0°C = 1 2°C = 1 2.0 K 1 J I L � q = c X mass X �T = ( 2 .42 J/gK ) ( 5 .50 L ) � ( l 2.0 K ) m L 1 0- L 10 J = - 1 77.289 = -1 77 kJ

[ l ( )l

[

6.4

[ �1

Plan: To find �T, find the T final by applying equation 6.7. The diamond loses heat (-q d iamon d ) whereas the water gains heat (q w ater) . Although the Celsius degree and Kelvin degree are the same size, be careful when interchanging the units. To be safe, always convert the Celsius temperature to Kelvin temperatures. Use the initial temperatures, masses, and specific heat capacities to solve the expression below. Solution: -q d iamon d = q water

-( m Cd iamo nd � T)

=

m Cwater �T

(

( 1 0.25 carat ) 0.2000 g

)

= 2.050 g 1 carat = 74.2 1 + 273 . 1 5 347.36 K Tinit (diamond) = 27.20 + 273 . 1 5 = 300.35 K Tinit (water) - (2.050 g) (0.5 1 9 J/g-K) (T final - 347.36) = (26.05 g) (4. 1 84 J/g-K) (T final - 300.35) (-1 .06395 J/ K) (T final - 347.36) = ( 1 08.9932 J/ K) (T final - 300.35) - 1 .06395 T final + 369.5737 = 1 08.9932 T final - 32736. 1 08 33 1 05 .68 1 7 = 1 1 0.057 1 5 Tfina l

mass of diamond

=

=

83

T final = 300.80446 K �Td iamon d = 300.80446 - 347.36 = -46.55554 = -46.56 K �Twater = 300.80446 - 300.35 = 0.45446 K = 0.45 K

Check: Rounding errors account for the slight differences in the final answer when compared to the identical calculation using Celsius temperatures. The temperature of the diamond decreases as the temperature of the water increases. The temperature of the water does not increase significantly because its specific heat capacity is large in comparison to the diamond, so it takes more heat to raise I g of the substance by I OC.

6.5

Plan: The bomb calorimeter gains heat from the combustion of graphite, so -qgraphi te = q c alorimeter Convert the mass of graphite from grams to moles and use the given kllmol to find qgrap h ite ' The heat lost by graphite equals the heat gained from the calorimeter, or L'lT multiplied by C calorimeter ' Solution: - (mol graphite x kllmol graphite) = Ccalorimeter �Tcalorimeter I mol C -393.5 kJ = Ccalorimete/2 . 6 1 3 K) - ( 0.8650 g C ) I mol C 1 2.0 1 g C -28.34 1 1 7 = Ccalorimeter(2 .6 1 3 K) Ccalorimeter = 1 0.8462 2 = 10.85 kJ/K

(

6.6

)(

)

Plan: To find the heat required, write a balanced thermochemical equation and use appropriate molar ratios to solve for required heat. Solution: Mf = - 1 37 kJ C2H4(g) + H2(g) -7 C2H 6 (g) I mol C 2 H6 - 1 37 kJ H eat = ( 1 5 .0 kg C 2 H 6 ) 1 0 3 g = -6.83405 X 1 0 4 = --6.83 X 1 0 4 kJ I kg 30.07 g C 2 H6 I mol C 2 H6

[ )(

)

)(

6.7

Plan: Manipulate the two equations so that their sum will result in the overall equation. Reverse the first equation (and change the sign of M!); reverse the second equation and multiply the coefficients (and M!) by two. Solution: Mf = -(223 .7 kl)= -223.7 kJ 2 �lO(g) + 3/2 02(g) -7 N20S(s) 2 N02(g) -7 2 NOEg) + Olg} Mf = -2(-57 . 1 kl) = 1 1 4.2 kJ Total: 2 N02(g) + 1 12 02(g) -7 N20S(s) Mf = -1 09.5 kJ

6.8

Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on the product side. Balance the equation with the following differences from "normal" balancing - only one mole of the desired product can be on the right hand side of the arrow (and nothing else), and fractional coefficients are allowed on the reactant side. The values for the standard heats of formation ( � H ; ) may be found in the appendix. Solution: a) C(graphite) + 2 H2(g) + 1 /2 02(g) -7 C H) O H ( l) Mf; = -238.6 kJ b) Ca(s) + 1 /2 02(g) -7 CaO(s) Mf ; = --635. 1 kJ c) C(graphite) + 1 14 Sg(rhombic) -7 CS 2 ( l) Mf ; = 87.9 kJ

6.9

Plan: Apply equation 6.8 to this reaction, substitute given values, and solve for the Mf; (C H) O H ). Solution: Mf;omb = 'L Mf; (products) - 'L Mf; (reactants) Mf;omb = [ Mf; (C02(g» + 2 Mf; (H20(g» ] - [ Mf; (CH)OH(l) + 3/7 Mf; (02(g» ] -63 8.5 kJ = [ 1 mol(-393 .5 kllmol) + 2 mol(-24 1 .826 kllmol)] - [ Mf ; (CH)OH(l) + 3/2(0)]

-638.5 kJ = (-877. 1 52 kl) - Mf ; (CH)OH( l) Mf; (CH)OH(l) = -23 8.652 = -238.6 kJ Check: You solved for this value in Follow-Up 6.8. Are they the same? 84

END-OF-CHAPTER PROBLEMS

6.5

The change in a system' s energy is M q + w. If the system receives heat, then its qfinal is greater than qinitial so q is positive. Since the system performs work, its Wfinal < Winitial so W is negative. The change in energy is (+425 J) + (- 425 J) 0 J. =

=

6. 7

C(s) + 02(g) � CO2(g) + 3 . 3 x 1 0 1 0 J ( 1 .0 ton) 1 J a) M(kJ) (3 . 3 x 1 0 1 0 3.3 X 107 kJ 10 J =

b) M(kcal) c) M(Btu) 6. 1 0

6. 1 1

J) [ � ) ( J[ � ) ( J =

=

=

(3.3

(3 . 3

x x

10

10

10

10

J)

J)

1

cal

4 . 1 84

Btu 1 05 5 J 1

1

J

10

=

cal cal

3 . 1 2 7 96

=

X

X

1 06

3.1

X

7 . 8 87

10

7

=

=

7.9

X

106 kcal

107 Btu

a) Exothermic, the system (water) i s releasing heat in changing from liquid to solid. b) E ndothermic, the system (water) is absorbing heat in changing from liquid to gas. c) Exothermic, the process of digestion breaks down food and releases energy. d) Exothermic, heat is released as a person runs and muscles perform work. e) Endothermic, heat is absorbed as food calories are converted to body tissue. f) Endothermic, the wood being chopped absorbs heat (and work). g) Exothermic, the furnace releases heat from fuel combustion. Alternatively, if the system is defined as the air in the house, the change is endothermic since the air's temperature is increasing by the input of heat energy from the furnace. An exothermic reaction releases heat, so the reactants have greater H (Hini tial) than the products (Hfinal) . /).}f Hfinal - Hinitial < O. =

Reactants

!

Products 6. 1 3

-

Mf � ( ) , (exothwn;,)

a) Combustion of methane : CH 4 (g) + 2 02(g) � CO2(g) + 2 H20(g) + heat C H 4 + 2 O2 (initial)

!

-

Mf � ( ) , (oxothenn;,)

85

b) Freezing of water: H20(l) � H20(s) + heat H20(l) (initial) M! = (-), (exothermic)

6. 1 5

a) Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon dioxide gas, water vapor, and heat. C2HsOH(l) + 3 02(g) � 2 CO2 (g) + 3 H20(g) + heat C2HsOH + 3 O2 (initial)

2 CO2 + 3 H20 (final)

!

tlli � ( - ), (exothenn;c)

b) Nitrogen dioxide, N02, forms from N2 and O2. 1 /2 N2(g) + 02(g) + heat � N02(g) N02 (final)

f

tlli � (+), (endothenn;c)

1 12 N2 + O2 (initial)

6. 1 8

To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change in temperature.

6.20

Plan: The heat required to raise the temperature of water is found by using the equation q = c x mass x LH . The specific heat capacity, Cwate" is found in Table 6.2. Because the Celsius degree is the same size as the kelvin degree, �T = 1 00.oC - 20.oC = 80.oC = 80. K.

(

q(J) = c x mass x �T = 4. 1 84 6.22

:

g C

}

1 2.0 g) ( ( I OO. - 20.) OC ) = 40 1 6.64 = 4.0 x 1 03 J

q(J) = c x mass x �T Tj = 3 .00°C Tr = ? 3 10 J 4 q = (85.0 kn = 8.50 x 1 0 J I kJ 4 8.50 x 1 0 J = (0.900 J/g0C) (295 g) (Tr - 3 .00)OC ( 8.50 x 1 04 J ) (T r - 3 . OO)OC = ----'----:-----'-.,... J O (295 g

( )

mass = 295 g

{ :�� J

(Tr - 3.00tC = 320. 1 5°C (unrounded) Tr = 323°C

86

c = 0.900 J/gOC

6.24 Since the bolts have the same mass, and one must cool as the other heats, the intuitive answer is [(7; : T, ) 1 [(l OO" C + 55" C )] 2 6.26 The heat l ost by Both the water origiarenal lconverted y at 82°e toi s gaimassnedusibyngthethewater thatty. is originally at 22°e. Therefore volumes densi Mass of85 mL = ( 85 mL ) ( I1.OmLO g ) = 85 g Mass of 165 mL ( 165 mL ) ( \. o� ) 165 g x mass84 Jx/g0C)(82°e water)=82)c x°emass(4.184 x (22°e water) J/g g) (85 -(4.1 0C) ( 165 g) (Tf -22)Oe 165 82) Tr -6970[ 85](Tf[ -22) ] ( Tr -3630 -+ 853630Tr= 165165 Tr+ 6970 85 Tr 10600. 250. T f Tr= (10600. /250. ) 42.4 on has a positive because thi s reaction requires the input of energy to break the oxygen-oxygen 6. 3 1 The bond:reacti02(g) + energy 2 0(g substanceochanges from the gaseous state to the l i q ui d state, energy i s rel eased so woul d be negative for 6. 3 2 Asthe acondensati n of I mol of water. The value of f..H for the vapori z ati o n of2 mol of water woul d be twi c e the value of H20(g) for the condensati on of 1 mol of water an opposite sign (+f..H) . 2 H20(l)vapor+ butEnergywoul d have 2 H20(g) H20(l) + Energy The enthalo npof2 y formol1 moles ofli e ofqwater would be opposite i n sign to and one-half the value for the conversi uid H20condensi to H20ngvapor. 6. 3 3 a)b) ThiBecause s reactif..Honi si sa state functiobecause tlHenergy i s negatirequi ve. red for the reverse reaction, regardless of how the change n, the total occurs, magni tudethe butreactidi foferent sigtten,n ofmeani the forward reacti okJn. iTherefore, speci fi c for n as wri n g that 20. 2 rel easedmorewhenenergy118 wioflal bemolrele eofased.sulfur c)reacts. The Inis thisthe issame case, 3. 2 moles of sul fur react and we therefore expect thatsmuch ( 3. 2 molSg ) ( ( 118-20.) mol2 kJSg ) 517 12 5 2 x 1 d) The mass of Ss requires conversion to mol e s and then a calculation identical to c) can be performed. ( 20. 0 g Sg ) ( 256.I mol5 6 gSgSg J(( 11-28 )0.mol2 kJSg ) 1 2 5974 12 6 90.29 kJ 6. 3 5 a) 112 N2(g) + 112 02(g) b) Mf. ( 1. 5 0 g NO) ( 30.1 mol0 1 gNONO J ( -I 90mol.29NOkJJ 5129957 4 5 1 �

-ql osl

=

�

77.50C

qgai n ed '

=

-q losl

=

=

qgai n ed

tJ. T

C

=

tJ.T

=

( Tf -

=

=

=

=

42°C

f..Hrx n ,

�

)

f..H

f..H

f..Hc ond

=

�

�

f..Hvap

(-)

=

( +) 2 [ f..Hc ond]

exothermic

f..H

tlHrx n

=

f..Hrx n =

-

.

=

=

� NO(g)

tlH

0 2 kJ

.

=

f..Hrx n =

rx n

-

.

-

= -

.

kJ

=

=

87

-4

.

=

-

.

kJ

=

+20.2 kJ.

6.37

6.42

6.44

For the reaction written, 2 moles of H202 release 1 96. l kl of energy upon decomposition. tillrxn = - 1 96. 1 kJ 2 H202(l) � 2 H20(l) + 02(g) 1 mol H202 1 03 g - 1 96. 1 kJ - -2 . 1 097 x 1 0 6 - -2 . 1 1 Heat - q - ( 732 kg H 2 0 2 -1 kg 34.02 g H202 2 mol H 202

) [ )(

J

(

J

x

6 1 0 kJ

To obtain the overall reaction, add the first reaction to the reverse of the second. When the second reaction is reversed, the sign of its enthalpy change is reversed from positive to negative. till = -635. 1 kJ Ca(s) + 1 12 02(g) � GaGW till = - 1 78.3 kl GaGW + COlg) � CaC03W till = -8 1 3.4 kJ Ca(s) + 1 /2 02(g) + CO2(g) � CaC03 (s) till = 1 80.6 kJ N2(g) + 02(g) � 2NO(g) till = - 1 1 4.2 kJ + Oig) � 2N02{g} tillrxn = +66.4 kJ N2(g) + 2 02(g) � 2 N02(g) In Figure P6.44, A represents reaction 1 with a larger amount of energy absorbed, B represents reaction 2 with a smaller amount of energy released, and C represents reaction 3 as the sum of A and B .

2NO(g)

6.47

The standard heat of reaction, t:Jf:n , is the enthalpy change for any reaction where all substances are in their standard states. The standard heat of formation, .:1H ; , is the enthalpy change that accompanies the formation of one mole of a compound in its standard state from elements in their standard states. Standard state is I atm for gases, 1 M for solutes, and pure state for liquids and solids. Standard state does not include a specific temperature, but a temperature must be specified in a table of standard values.

6.48

a) 1 12 Ch(g) + Na(s) � NaCI(s) The element chlorine occurs as C12, not Cl. b) H2(g) + 1 12 02(g) � H20( l) The element hydrogen exists as H2, not H, and the formation of water is written with water as the product in the liquid state. c) No changes

6.49

Formation equations show the formation of one mole of compound from its elements. The elements must be in their most stable states ( till ; =0) a) Ca(s) + CI2(g) � CaCI2(s) b) Na(s) + 1 12 H2(g) + C(graphite) + 3/2 02(g) � NaHC03 (s) c) C(graphite) + 2 CI2(g) � CCI4(l) d) 1 12 H2(g) + 1 12 N2(g) + 3/2 02(g) � HN0 3 (l)

6.5 1

The enthalpy change of a reaction is the sum of the tillf of the products minus the sum of the t:Jff of the reactants. Since the t:Jff values (Appendix B) are reported as energy per one mole, use the appropriate coefficient to reflect the higher number of moles. till :n = Im [ t:Jf; (products) ] - In [ till; (reactants) ] a) till:xn = { 2 t:Jf; [ S02(g)] + 2 till; [ H20(g) ] } - {2 till; [ H2S(g) ] + 3 t:Jf; [ 02(g) ] } = 2 mol(-296.8 kJ/mol) + 2 mol(-24 1 .826 kl/mol) - [2 mol(-20.2 kl/mol) + 3(0.0) ] = -1 036.8 kJ

b) The balanced equation is CH4(g) + 4 CI2(g) � CCI4(l) + 4 HCI(g) till;", = { I .:1H; [ CCI4(l)] + 4 t:Jf; [ HCI(g)] } - { 1 till; [CH4(g) ] + 4 t:Jf; [CI2(g) ] } t:Jf:xn = I mol(- 1 39 kJ/mol) + 4 mol(-92.3 1 kJ/mol) - [ I mol(-74.87 kl/mol) + 4 mol(O) ] = -433 kJ

88

6.53

!:l.H:n = Lm [ Mf; (products)] - Ln [ !:l.H; (reactants)] !:l.H:n - 1 46.0 kJ Cu20(s) + 1 12 02(g) � 2 CuO(s) !:l.H :n = {2 mol ( !:l.H; , CuO(s» } - { I mol ( Mf; , Cu20(s» + 1 12 mol ( Mf; , 02(g» } kJ - 1 46.0 kJ 2 mol ( !:l.H; , CuO(s» - [ 1 mol (- 1 68.6 ) + 1 /2 mol (0) ] mol - 1 46.0 kJ = 2 mol ( !:l.H; , CuO(s» + 1 68.6 kJ 3 1 4.6 kJ A LJ O tin ' C U O(S ) -157.3 kJ/mol I 2 mol =

=

=

6.56

-

=

-

2 PbS04(s) + 2 H20(!) � Pb(s) + Pb02(s) + 2 H2S04(!) a) !:l.H:" { I mol ( Mf ; , Pb(s» + I mol ( !:l.H; , Pb02(s» + 2 mol ( !:l.H; , H2S04(!) } - {2 mol ( !:l.H; , PbS04(s» + 2 mol ( Mf; , B20(!) } kJ kJ kJ = [ 1 mol (0 - ) + 1 mol (-276.6 - ) + 2 mol (-8 1 3 .989 - ) ] mol mol mol kJ kJ - [ 2 mol (-9 1 8.39 - ) + 2 mol (-285.840 - ) ] = 503.9 kJ mol mol b) Use Hess' s Law to rearrange equations ( 1 ) and ( 2) to give the equation wanted. Reverse the first equation (changing the sign of Mf:,, ) and multiply the coefficients (and !:l.H:,, ) of the second reaction by 2 . 2 PbS04(s) � Pb(s) + Pb02(s) + 2-8G;,W tili" = -(-768 kJ) tili" = 2{- 1 32 kJ) 2-8G;,(g) + 2 B20 fD � 2 B2S04(£L Reaction: 2 PbS04(s) + 2 B20(!) � Pb(s) + Pb02(s) + 2 H2S04(!) Mf:xn 504 kJ =

=

6.57

a) C l sH 36 02(S) + 26 02(g) � 1 8 CO2(g) + 1 8 H20(g) b) tili"comb = { 1 8 mol ( Mf; CO2(g)) + 1 8 mol ( !:l.H; H20(g» } - { I mol ( !:l.H; C l sH 36 02(S» + =

26 mol ( Mf; 02(g» } 1 8 mol (-393.5 kJ/mol) + 1 8 mol (-24 1 .826 kJ/mol) - [ 1 mol (-948 kJ/mol) + 26 mol (0 kJ/mol) ]

= -1 0,488 kJ

c) q(kJ) = ( 1 .00 g C l sH 36 02 = -36.9 kJ

mol CI s H 36 02 ( - 1 0, 488 kJ /l 2184.4 7 CI s H 36 02 J I I mol CI sH 36 02 J

q(kcal) = (-36.9 kJ

[

/l 4.11 kcal J 84 kJ

=

-8.81 kcal

]

- 8.8 1 kcal = 96.9 kcal 1 .0 g fat The calculated calorie content is consistent with the package information. d) q(kcal) = ( 1 1 .0 g fat )

6.58

a) A first read of this problem suggests there is insufficient information to solve the problem. Upon more careful reading, you find that the question asks volumes for each mole of helium. T = 273 + 1 5 = 288 K or T = 273 + 30 303 K L atm 0.082 1 ( 288 K ) V R T mol K IS = = 23 .6448 = 23.6 Ll mol P n ( 1 .00 atm) V30 n

=

RT P

(

=

(

0

0

0.082 1

)

)

L o atm (303 K ) mol K ( 1 .00 atm ) 0

=

=

=

24.8763 = 24 . 9 L/mol

89

b) Internal energy is the sum of the potential and kinetic energies of each He atom in the system (the balloon). The energy of one mole of hel ium atoms can be described as a function of temperature, E nRT, where n mole. Therefore, the internal energy at l 5°C and 30°C can be calculated. The inside back cover lists values of R with different units. 1 87 J nRT ( I mol) J/moloK) c) When the bal loon expands as temperature rises, the balloon performs work. However, the problem specifies that pressure remains constant, so work done on the surroundings by the balloon is defined by the equation : When pressure and volume are multiplied together, the unit is Loatm. Since we would like to express work in Joules, we can create a conversion factor between Loatm and 1 . IJ ? 2 Pa L atm m-Is

= 3/2 1 E = 3/2 = (3/2) .00 (8. 3 14 (303 -288)K = 187.065PY= w = -P�Y. 2 s a 101325 kg/m [ ) p ) [ 10-13 m3 )][ kg ( [ 1 ) ) ( ( 763 448 8 6 0 0 atm 1. 23. 24. = w=-P�Y ) 1 ( =-124. 7 8 = d) = M + P�Y = (187. 065 J) + (124. 7 8 J) = 31 1. 8 45 = e) = =

=

_

f)

1 .2

qp

till qp

x

2

10 J

3.1

310 J.

x

2 10 J

0

)

When a process occurs at constant pressure, the change in heat energy o f the system can be described by a state function called enthalpy. The change in enthalpy equals the heat (q) lost at constant pressure: M

6.66

�E - w = + w) -w = (q

qp

Chemical equations can be written that describe the three processes. Assume one mole of each substance of interest so that units are expressed as kJ . till ; till :x,. kJ H2(g) � CH4(g) ( I ) C(graphite) CH4(g) � C(g) R (g) till :,. kJ MfDatoJ11 till :,. kJ MfDatom ( 3 ) H 2(g) � H (g) The third equation is reversed and its coefficients are multiplied by to add the three equations. till :,. kJ C(graphite) + �W � GR4W till :,. kJ �W � C(g) � till :", kJ 4-:Hfgt � 2--#;ltrl­

(2)

+2

2

=

+4

+

C(graphite) � C(g)

6. 7 3

till = + P�Y =

Ml:n

= -74.9 = = 1660 = = 432 = -74.92 = 1660 = -2(432 kJ = -864 = = 72l.l = MfD ato m

72 1

kJ per one mol C(graphite)

a) The heat of reaction is calculated from the heats of formation found in Appendix B . The Ml; 's for all of the species, except SiCI4, are found in Appendix B. Use reaction with its given Ml:xn ' to find Ml; [ SiC I4(g)) . SiCI4(g) 2 H20(g) � Si02(s) 4HC I(g) Ml:,. Ml; [ H20(g) ] } [ SiCI4(g)] [ HCI(g) ] } [ Si02(s) ] kJ) } kJ kJ) kJ) { �H; [ SiCI4(g)] kJ) } kJ - { Ml ; [ SiCI4(g)] kJ) } kJ { M; [ SiCI4(g)] Ml; [ SiCI4(g)] kJ/mol (unrounded) The heats of reaction for the first two steps can now be calculated. ( I ) Si C s) � SiCI4(g) Ml; [ SiCI4(g)] till° rxn l [ Si Cs) ] Ml ; [ C I2(g)] } MfDrxn l 2(0) ] -657.0 kJ (2) Si02(s) 2C(graphite) 2CI2(g) � SiCI4(g) 2CO(g) Ml; [ C(gr) ] �H ; [ C h(g) ] } MfDrxn2 [ CO(g)]} � [ Si02(g) ] [ SiCI4(g)] 32.9 kJ kJ kJ) MfD rxn2 2(0) }

3,

(3) = + + - {till; +2 + 4 �H; { Ml; + 2(-241. 826 - 139. 5 = (-910. 9 + 4(-92. 3 1 + (---483. 652 -139. 5 = -1280.14 + (---483.652 1 140. 64 = = -656. 9 88 = + 2CI2(g) - { Ml; + 2 = -656.+ 9 88 kJ - [0 + + = -656.988 = + +2 = {Ml; +2 + 2 till; {M; = -656. 9 88 + 2(-1 10. 5 - { (-910.9 kJ) + 2(0) + = 32. 9 12 = 90

b) Adding reactions 2 and 3 yields: (2) �ts1 + 2C(graphite) + 2CI2(g) � 8iG4tgj + 2 C O ( g) f1ffO rxn = 3 2 .9 1 2 kJ (3) 8iGl4tgj + 2H20(g) � £iG;,fst + 4 HC I(g) f1ffOrxn = - 1 39.5 kJ 2 C(gr) + 2 Cb(g) + 2 H20(g) -> 2 C O (g) + 4 H CI(g) f1ffO rxn2+3 = -1 06.588 kJ kJ 06.588 1 -106.6 = = 2+3 f1ffO rxn Confirm this result by calculating Mi:. using Appendix B values. 2 C(gr) + 2 C b (g) + 2 H20(g) -> 2 CO(g) + 4 HC I(g) f1ffO rxn2+3 = { 2 Mi; [ CO(g) ] + 4 Mi ; [ HCI(g)] } - { 2 Mi; [ C( gr) ] + 2 Mi ; [ CI2(g) ] + 2 Mi; [ H20(g)] } f1ffO rxn2+3 = 2(- 1 1 0.5 kJ) + 4(-92.3 1 kJ) - [ 2(0) + 2(0) + 2(-24 1 .826 kJ) ] = -1 06.588 = -1 06.6 kJ 6.76

a) LlT = 8 1 9°C - O°C = 8 1 9°C = 8 1 9 K 3 w = PLl V = IlRLlT = ( 1 mol) x (8.3 1 4 J/mol o K) x (8 1 9 K) = --6809. 1 66 = -6.8 1 x 1 0 J b) q = (mass)(c)( LlT) 6.809 1 66 X 1 0 3 J LlT = --q"---­ (mass)(c) 28.02 g N 2 I mO I N 2 ( l . O O J/g o K) 1 mol N 2

[

Jl

(

6.77

Only reaction 3 contains N 204(g) , and only reaction 1 contains N 203(g) , so we can use those reactions as a starting point. N20 S appears in both reactions 2 and 5, but note the physical states present: solid and gas. As a rough start, adding reactions 1 , 3, and 5 yield the desired reactants and products, with some undesired intermediates: N203(g) � NO(g) + N02(g) Reverse ( 1 ) f1ffO rxn 1 = -(-39.8 kJ) = 39.8 kJ 4 N 02(g) � 2 N204 (g) Multiply (3) by 2 f1ffO rxn3 = 2(-57.2 kJ) = - 1 1 4.4 kJ Reverse (5) f1ffO rxns = -(-54. 1 kJ) = 54. 1 kJ lliQs(s) � N2�(g} N203(g) + 4N 02 (g) + N20 S( S) � NO(g) + N02( g) + 2 N204(g) + N20S( g) To cancel out the N20S(g) intermediate, reverse equation 2. This also cancels out some of the undesired N 02(g) but adds NO(g) and 02 (g) . Finally, add equation 4 to remove those intermediates: N203(g) � NG(gj + NQ;o(gj Reverse ( 1 ) f1ffO rxn 1 = -(-39.8 kJ) = 39.8 kJ Multiply (3) by 2 4-NG;o(gt-� 2 N204(g) f1ffO rxn3 = 2(-57.2 kJ) = - 1 1 4.4 kJ Reverse (5) N20S(s) � N;oG�(gj f1ffO rxn s = -(-54. 1 kJ) = 54. 1 kJ Reverse (2) N;oG�(gt-� NG(gj + �(gj + G;o(gj f1ffO rxn2 = --(-1 1 2.5 kJ) = 1 1 2.5 (4) 2 �+O(g) + G;"fgt-� 2-NG;ofrlf1ffOrxn4 = - 1 1 4.2 kJ Total: N 203(g) + N 20s(s) � 2 N204(g) Mi,:" = -22.2 kJ

6.79

a) The balanced chemical equation for this reaction is CH 4(g) + 2 02(g) � C O2(g) + 2 H20(g) Instead of burning one mole of methane, ( 25.0 g

CH 4 )

(

1 mol CH 4 1 6 .04 g CH 4

J

= 1 .5586 mol CH4 (unrounded) of methane

are burned. CH4(g) + 2 02(g) -> C O 2 (g) + 2 H20(g) f1ffOc omb = [ I mol ( Mi; C02(g» + 2 mol ( Mi; H20(g» ] - [ I mol ( Mi; C H4 (g» + 2 mol ( Mi; 02(g» ] f1ffO c omb = 1 mol (-393.5 kJ/mol) + 2 mol (-24 1 .826 kJ/mol) - [ 1 mol (-74.87kJ/mol) + 2mol (0.0 kJ/mol) ] = -802.282 kJ/mol CH4 (unrounded) ) -802.282 kJ 3 = - 1 250.4 = -1.25 x 1 0 kJ ( 1 .5586 mol C H 4 1 mol C H 4

(

)

91

b) The heat released by the reaction is "stored" in the gaseous molecules by virtue of their specific heat capacities, c, using the equation AH = mc�T. The problem specifies heat capacities on a molar basis, so we modify the equation to use moles, instead of mass. The gases that remain at the end of the reaction are CO2 and H20. All of the methane and oxygen molecules were consumed. However, the oxygen was added as a component of air, which is 78% N2 and 2 1 % O2, and there is leftover N2. 1 mol CO 2 = 1 .5586 mol Moles of C02(g) = ( 1 .5586 mol CH4 ) 1 mol CH4

( )(

J J

2 mol H 2 O = 3 . 1 1 72 mol 1 mol CH4 Mole fraction N2 = (79% 1 1 00%) = 0.79 Mole fraction O2 = (2 1 % I 1 00%) = 0.2 1 Moles ofN2(g) = (3 . 1 1 72 mol O2 used) (0.79 mol N2 1 0.2 1 mol O2) = 1 1 .7266 mol N2 (unrounded) Q = (mol)(c)(�T) 1 250.4 kJ ( 1 03 JI kJ) = 1 .5586 mol CO2 (57.2 J I mol°C) (Tr - O.O)OC + 3 . 1 1 72 mol H20 (36.0 J I mol°C) (T r - O.O)OC + 1 1 .7266 mol N2 (30.5 J I mol°C) (Tr - O.O)OC 1 .2504 x 1 06 J = « 89. 1 5 1 9 + 1 1 2.2 1 92 + 357.66 1 3) J 1°C)Tr = (559.0324 J 1°C)Tf Tr = ( 1 .2504 x 1 0 6 J) I (559.0324 J 1°C) = 2236.72 = 2.24 x l030C Moles of H20(g) = ( 1 .5586 mol CH4

92

Chapter 7 Quantum

Theory

and Atomic

Structure The value for the speed of light will be 3 .00 x 1 08 mls except when more significant figures are necessary, in which cases, 2.9979 x 1 08 mls will be used. FOLLOW-UP PROBLEMS

7. 1

Plan: Given the frequency of the light, use the equation c = AV to solve for wavelength. Solution: 3 x 08 = 4 1 4. 938 = 4 15 nm A = c I v = . 00 1 m/s 7.23 x 1 01 4 S - 1 1 0-9 m

(�) ( A)

3 . 00 x 1 08 m/s I = 4 1 49 . 3 8 = 4 1 5 0 A 7.23 x 1 01 4 S - I 1 0 - 1 0 m Check: The purple region of the visible light spectrum occurs between approximately 400 to 450 nm, so 4 1 5 nm is in the correct range for wavelength of purple light.

A c/v =

7.2

=

Plan: To calculate the energy for each wavelength we use the formula E = hc I A Solution: ( 6.626 x 1 0 -34 J s)(3 .00 X 1 08 m/s ) 1 7 = 2 X 1 0- 1 7 J = 1 .9878 X 1 0E = hc I A = 1 x 1 0-8 m •

E

=

hc I A =

( 6.626 x 1 0-34 J s)(3 .00 5 X 1 0- 7 m •

1 08 m/s )

X

=

3 .9756 X

1 0- 1 9 = 4

X

1 0- 1 9 J

( 6.626 x 1 0 -34 J . s)(3 .00 X 1 08 m/s ) 21 1 = 1 .9878 X 1 0= 2 X 1 0-2 J 1 x 1 0-4 m As the wavelength of light increases from ultraviolet (uv) to visible (vis) to infrared (ir) the energy of the light decreases. Check: The decrease in energy with increase in wavelength follows the inverse relationship between energy and wavelength in the equation E = hc I A .

E

7.3

=

hc I A =

P lan: With the equation for the de Broglie wavelength, A = hlmu and the given de Broglie wavelength, calculate the electron speed. Solution: 6.626 X 1 0 -34 J S kg · m 2 /s 2 = 7273.3 7.27 x 1 03 mls u=himA= J ' m ( 9 1 ' x 1 0 -" kg ) ( ' OO nm

[

•

{ �:m )1

[

Check: Perform a rough calculation to check order of magnitude: 1 0 -34 = 1 X 1 04

[

( 1 0 -" ) ( I 00 )

[ l O,-" )1

93

)

=

7.4

Plan: Following the rules for / (integer from 0 to n - I) and m, (integer from -/ to +i) write quantum numbers for n = 4. Solution: / = 0, 1 , 2, 3 For n = 4 For / = 0, m, = ° For / = 1 , m, = -1, 0, 1 For / = 2, m, = -2, -1, 0, 1 , 2 For / = 3 , m, = -3, -2, -1, 0, 1, 2, 3 2 Check: The total number of orbitals with a given n is n . For n = 4, the total number of orbitals would be 1 6. Adding the number of orbitals identified in the solution, gives I (for / = 0) + 3 (for / = 1 ) + 5 (for / = 2) + 7 (for / = 3) = 16 orbitals.

7.5

Plan: Identify n and / from the subshell designation and knowing the value for /, find the m, values. The subshells are given a letter designation, in which s represents / = 0, p represents / = 1 , d represents / = 2,/ represents / = 3 . Solution· m, values / value n value Sublevel name 1 - 1 , 0, 1 2 2p -3 , -2, - 1 , 0, 1 , 2, 3 3 5 Sf Check: The number of orbItals for each sublevel equals 21 + 1 . Sublevel 2p should have 3 orbitals and sublevel Sf should have 7 orbitals. Both of these agree with the number of m, values for the sublevel.

7.6

Plan: Use the rules for designating quantum numbers to fill in the blanks. For a given n, / can be any integer from 0 to n- l . For a given /, m, can be any integer from - / to + /. The subshells are given a letter designation, in which s represents / = 0, p represents 1 = 1 , d represents 1 = 2,/ represents / = 3. Solution: The completed table is: / m, n Name 4 0 a) I 4p 1 b) 2 0 2p 2 c) 3 -2 3d 2 d) 0 0 2s Check: All the given values are correct deSIgnatIons.

END-OF-CHAPTER PROBLEMS

7.2

a) Figure2 7.3 describes the2 electromagnetic spectrum by wavelength and frequency. Wavelength increases from left ( 1 0- nm) to right ( 1 0 1 nm). The trend in increasing wavelength is: x-ray < ultraviolet < visible < infrared < microwave < radio waves . b) Frequency is inversely proportional to wavelength according to equation 7. 1 , so frequency has the opposite trend: radio < microwave < infrared < visible < ultraviolet < x-ray. c) Energy is directly proportional to frequency according to equation 7.2. Therefore, the trend in increasing energy matches the trend in increasing frequency: radio < microwave < infrared < visible < ultraviolet < x-ray. High­ energy electromagnetic radiation disrupts cell function. It makes sense that you want to limit exposure to ultraviolet and x-ray radiation.

7.4

In order to explain the formula he developed for the energy vs. wavelength data of blackbody radiation, Max Planck assumed that only certain quantities of energy, called quanta, could be emitted or absorbed. The magnitude ofthese gains and losses were whole number multiples of the frequency: M = nhv.

94

7. 6 Pl1 Isan: WavelAssume ength thati s reltheatednumber to frequency through thesigequati onfigures. c Recall that a Hz i s a reci procal second, or 9 " 60" has three nificant cSol ution: 3. 00 x l08 m/s 3 Hz I (�H Z I ( 960. kH Z) ( 101 kHz ) ) 3. 00 x 108 m/s [ 1 �m 1 (nm) 3 H Z ] [ � ] 10 9 m ( 960. kH z ) [ 101kHz Hz 3. 00 x 108 m/s I ( I O 3 H Z ][ S - ] IO- m J ( 960. kH Z) [ 101 kHz Hz. . SolPlan:uhvtiFrequency on: i s related to energy through the equation hv. N ote that 1 Hz 1 10-34 Jos) 10 1 0 10-23 Si nce energy is directly proporti onal to frequency hv) and frequencyhcand wavelength are inversel y related (v it fol l ows that energy is inversely rel ated to wavel ength ) . As wavelength decreases, energy increases. In terms of i ncreasing energy the order is Plan: a) The shortest wavell eeastngthenergeti c photon has the longest wavelength nm). b) The most energetic photon has the 00 x 108 m/s [� l 1. 23 9669 10 1 5 a) v nm 10-9 m 1 nm ] hc ( 10-34 J s)( 108 m/s) [ -10-9 m. m/s [ 1 1 1. 3 636 101 5 b) v 00 10-10 m hc ( 1 0-34 J s)( 108 m/s ) [ 1 ] 9 1 10 00 10-1 0 m The number ishigrelheratedenergy to the(hienergy l evelvingofantheabsorpti electron.on spectrum. An electronAn electronenergy toenergy changeas from ldrops owerquantum energy (l o w to g h gi it from a hi g her energy l e vel to a l o wer one gi v i n g an emi s si o n spectrum. a) b) c) d) = AV .

I = S- .

= AV

A ( m) = � =

= 3 1 2.5 = 3 1 2 m

v

A

�

=

v

II = 3 . 1 25 x l O = 3.1 2 x l Ol l nm

=

I A

A (A) = �

12 12 = 3 . 1 25 x I 0 = 3. 1 2 x 1 0 A

v

E=

7.8

E=

E = ( 6 . 626

X

(3 . 6

X

I S- ) = 2 . 3 8 5 x

= 2 .4

X

=

1 0-

23

(E =

7. 1 0

= Ci A ) ,

red

<

yellow

7. 1 3

I S- .

J

(E=

<

A,

blue.

(242

(2200 A ) .

3. = � =

A,

=

242

6 . 62 6

X

E= - =

A,

6 . 62 6

A

X

E= - =

A,

n)

absorption

0

X

10

X

3 . 00

= 1 .4

X

X

1 015

A

A

e m ission

15

S

-I

= 8 . 2 1 40 x l 0-

nm

n

e mission

= 1 .24

X

=

22

7. 1 6

3 . 00

0

242

8 3 . 00 X 1 0 = � = A A, 22

x

S-

19

= 8. 2 1 x 1 0

absorption

95

J

I

= 9.03545 x

= 9.0 x 1 0

absorbs

n)

-1 9

emits

-19

J

7. 1 8

Plan: Calculate wavelength by substituting the given values into equation 7.3, where nl = 2 and n2 = 5 because n 2 > n l . Although more significant figures could be used, five significant figures are adequate for this calculation. Solution: = 1 .096776 X 1 0 7 m- I � A nl n2 n l = 2 n2 = 5 =

�

A

=

R [ � - �) R R [ � - �n 2 ) = ( 1 .096776 1 07 )(�2 - �)5 = 2303229.6 )( � ) = 434. 1 729544 = 434.1 7 om =[ 2303229.6 1 1 0nl

I

A (nm)

7.20

m- I

X

I

m-

m- I (unrounded)

m m

Plan: To find the transition energy, apply equation 7.4 and multiply by Avogadro's number. Solution :

) [ 11£ = ( -2. 1 8 1 0- 1 8 J ) ( 12 - 12 ) = -4.578 1 0- 1 9 2 5 ( x 19 ( x ] = -2.7 5687 x 1 05 = -2.76 x l 05 J/mol b.E = -4.578 1 0- J ] 6.022 0 1 1 11£ = ( -2. 1 8 x 1 0- 1 8 ] ) _2_ _ _2 _ n n final

i n i t i al

X

X

1

2

3

]Iphoton

P hotons

mol photon I Check: The value is negative and so light is emitted.

7.22

Looking at an energy chart will help answer this question . " n=

, (d) (a)

(c)

5 n=4 n=3

, , ... ... ... ... ... -

. '

.

• •

- - - -n

=2

(b) •

•

n=

1

Frequency is proportional to energy so the smallest frequency will be d) n = 4 to n = 3 ; levels 4 and 5 have a smaller 11£ than the levels in the other transitions. The largest frequency is b) n = 2 to n = 1 since levels I and 2 have a larger 11£ than the levels in the other transitions. Transition a) n = 2 to n = 4 will be smaller than transition c) n = 2 to n = 5 since level 5 is a higher energy than level 4. In order of increasing frequency the transitions are d < a < c < b.

96

7.24

Plan : Use the Rydberg equation. A combination o f E = hC/A, and equation 7.4, would also work. Solution : 1 0-9 m 11. = (97.20 nm ) -- = 9.720 X 1 0-8 m ground state: n l = I ; I nm

(

J -1 ) [

I - = ( 1 .096776 x 1 0 7 m A.

9.72 x 1 0-8 m 0.9380 =

( l .096776

I 1 - -2 2 nl n2

X

1 0 7 m- I

[ �-�J I

n2

J )[ �-�J 1

n2

I , = I - 0.9380 = 0.0620 n -2 n � = 1 6. 1 n2 = 4 7.27

Macroscopic obj ects have significant mass. A large m in the denominator of A = hlmu will result in a very small wavelength. Macroscopic obj ects do exhibit a wavelike motion, but the wavelength is too small for humans to see.

7.29

Plan: Use the de Broglie equation. Mass is pounds must be converted to kg and velocity in mi/h must be converted to m/s because a j oule is equivalent to kg.m 2 I S 2 . Solution: � ( 6.626 X 1 0 -34 J s ) kg . m 2 /s 2 2.205 I b 0.62 mi 1 � 3600 s = a) A = Ih I km mu l kg 10 m J ( 220 I b ) 1 9 .6 :1

(

•.

)

[

= 7.562675 x 1 0-37 = 7. 6 X 1 0-3 7 m 7.3 1

7.33

7.37

h 11. = mu

[

62_6_x--1 0_ h ...:.( 6.. ._ ---,- -_34_J_.--,s...:.) ... kg . m 2 /s 2 u = __ = _ J (56.5 g ) ( 5400 A) rnA.

J [ J(

J(

J(

J[ [ J 1 03 _g _ I kg _

1

A

l O- l o m

J

J

= 2. 1 7 1 7 X 1 0-2 6 = 2.2 X 1 0-2 6 m/s

Plan: The de Broglie wavelength equation wil l give the mass equivalent of a photon with known wavelength and velocity. The term "mass-equivalent" is used instead of "mass of photon" because photons are quanta of electromagnetic energy that have no mass. A light photon 's velocity is the speed of light, 3 .00 x 1 08 mls. Solution: h 11. = mu ( 6.626 X 1 O-34 J . s ) I m k g ' m 2 /s 2 h = 3 . 7498 x 1 0-36 = 3 . 75 X 1 0-36 kg/photon : m= - = J 10 m A.U ( 5 89 nm) ( 3 .00 x I 08 m/s

)[

J[

J

a) Principal quantum number, n , relates to the size of the orbital. More specifically, it relates to the distance from the nucleus at which the probability of finding an electron is greatest. This distance is determined by the energy of the electron. b) Angular momentum quantum number, I, relates to the shape of the orbital. It is also called the azimuthal quantum number. c) Magnetic quantum number, ml, relates to the orientation of the orbital in space in three-dimensional space. 97

7.38

a) one b) five c) three d) nine a) There is only a single s orbital in any shell. b) All d-orbitals consists of sets of five (m, = -2, - 1 , 0, + 1, +2). c) All p-orbitals consists of sets of three (m, = -1, 0, + 1 ). d) If n = 3, then there is a 3s ( 1 orbital), a 3p set (3 orbitals), and a 3d set (5 orbitals) giving 1 + 3 + 5 = 9.

7.40

Magnetic quantum numbers (m, ) can have integer values from -I to + l. a) m,: -2, -1 , 0, + 1 , +2 b) m,: 0 (if n = I , then 1 = 0) c) m,: -3 , -2, -1 , 0, + 1 , +2, +3

7.42

Plan : The fol lowing letter designations correlate with the following 1 quantum numbers: 1 = 0 = s orbital; 1 = I = P orbital; 1 = 2 = d orbital; 1 = 3 = f orbital. Remember that allowed m, values are - 1 to + I. Solution: sublevel a) d (I = 2)

al lowable m,

-2, -1 , 0, + 1 , +2 -1 , 0, + 1 -3, -2, - 1 , 0, + 1 , +2, +3

b) p (l = I ) c) f (l = 3)

7.44

7.46

7.48

no of possible orbitals

5 3 7

P lan: The integer in front of the letter represents the n value. The letter designates the 1 value: 1 = 0 = s orbital ; 1 = I = P orbital; 1 = 2 = d orbital; 1 = 3 = f orbital. Solution: a) For the 5s subshell, n = 5 and 1 = O. Since m, = 0, there is one orbital. b) For the 3p subshell, n = 3 and 1 = I . Since m, = -1 , 0, + I , there are three orbitals. c) For the 4fsubshell, n = 4 and 1 = 3. Since m, = -3, -2, - 1 , 0, + 1 , +2, +3, there are seven orbitals. Plan: Allowed values of quantum numbers : n = positive integers; 1 = integers from 0 to n - I ; = integers from - I through 0 to + I. Solution : a) With 1 = 0, the only allowable m, value is O. To correct, either change l or m, value. Correct n = 2, 1 = 1 , m, = - I ; n = 2, 1 = 0, m, = o. b) Combination is allowed. c) Combination is al lowed. d) With 1 = 2, m, = -2, -1 , 0, + I , +2; +3 is not an allowable m, value. To correct, either change 1 or m, value. Correct: n = 5, 1 = 3, m, = +3 ; n = 5, 1 = 2, m, = O.

m,

a) The mass of the electron is 9. 1 094 x 1 0-3 1 kg. h2 I h2 E= 2 8n 2 m e a n 2 m 8n ea � )2 4 6.626 x 1 0 -3 J . s _

_

6

(

6

( )

) 8n2 ( 9. 1 094 x 1 0-3 1 kg ( 52.92

C)

X

10- 1 2 m

f

(

kg . m 2 /s 2 J

C)

= - (2. 1 7963 x 1 0- 1 8 J) I2 = - (2. 1 80 1 0- 1 8 J) I2 X

)( 1 ) n2

This is identical with the result from Bohr' s theory. For the H atom, Z = 1 and Bohr's constant = -2. 1 8 x 1 0- 1 8 J. For the hydrogen atom, derivation using classical principles or quantum-mechanical principles yields the same constant. b) The n = 3 energy level is higher in energy than the n = 2 level. Because the zero point of the atom's energy is defined as an electron's infinite distance from the nucleus, a larger negative number describes a lower energy level. Although this may be confusing, it makes sense that an energy change would be a positive number. I M = - (2. 1 80 x 1 0- 1 8 J). "22 - }2I = -3 .027778 x 1 0- 1 9 = 3.028 x l O- 1 9 J

(

)

98

( 6.626 X 1 0-3 4 J s ) ( 2.9979 x 1 0 8 m/s) hc c) E = � A (m) = - = = 6. 5606 1 x 1 0-7 = 6.561 (3 . 027778 X 1 0-19 J ) E A ( 6.5606 1 x 1 0.7 m) = 656.06 1 = 656. 1 om m This is the wavelength for the observed red line in the hydrogen spectrum. h

0

x

10-7m

C��m )

7.49

a) The lines do not begin at the origin because an electron must absorb a minimum amount of energy before it has enough energy to overcome the attraction of the nucleus and leave the atom. This minimum energy is the energy of photons of light at the threshold frequency. b) The lines for K and Ag do not begin at the same point, because the amount of energy that an electron must absorb to leave the K atom is less than the amount of energy that an electron must absorb to leave the Ag atom, where the attraction between the nucleus and outer electron is stronger than in a K atom. c) Wavelength is inversely proportional to energy. Thus, the metal that requires a larger amount of energy to be absorbed before electrons are emitted will require a shorter wavelength of light. Electrons in Ag atoms require more energy to leave, so Ag requires a shorter wavelength of light than K to eject an electron. d) The slopes of the line show an increase in kinetic energy as the frequency (or energy) of light is increased. Since the slopes are the same, this means that for an increase of one unit of frequency (or energy) of light, the increase in kinetic energy of an electron ejected from K is the same as the increase in the kinetic energy of an electron ejected from Ag. After an electron is ejected, the energy that it absorbs above the threshold energy becomes kinetic energy of the electron. For the same increase in energy above the threshold energy, for either K or Ag, the kinetic energy of the ejected electron will be the same.

7.52

The Bohr model has been successfully applied to predict the spectral lines for one-electron species other than H. Common one-electron species are small cations with 2 all but3 one electron removed. Since the problem specifies a metal ion, assume that the possible choices are Li + or Be + , and solve Bohr's equation to verify if a whole number for Z can be calculated. Recall that the negative sign is a convention based on the zero point of the atom's energy; it is deleted in this calculation to avoid taking the square root of a negative number. The highest-energy line from n = I to n = I 3 corresponds to16the transition 1 E = hv = (6.626 X 1 0- 4 Js) (2.96 1 X 1 0 Hz) (s- /Hz) = - 1 .96 1 9586 x 1 0- 7 J (unrounded) 8 Z = charge of the nucleus E = (-2. 1 8 x Wl J )

00

( �: J

.

1\ - 1 .96 1 9586 x 1 0-8 1 2 ) En 2 18 = 8.99998 Z = 1 -2 . 1 8 x 1 0- J 2. 1 8 X 1 0- J 2 Then Z = 9 and Z = 3 . Therefore, the ion is Li2+. 2

7.55

thus A = hclE E = hc/A a) The energy of visible light is lower than that of U V light. Thus, metal A must be barium since, of the three metals listed, barium has the smallest work function indicating the attraction between barium's nucleus and outer electron is less than the attraction in tantalum or tungsten. The longest wavelength corresponds to the lowest energy (work function). 8 ) )( ( 6.626 X I O- 34 jOS 3 .00X I 0 rn!S I nm hc 19 -- 29 1 .894 - 292 om Ta.. A = - 9 E 1 0- m 6.8 I x I 0- J ( 6.626 x 1 3 4 J o s)(3 .00 x 1 0 8 m/s) OI nm hc 19 462.279 = 462 om Ba: A - = 9 E 1 0- m 4.30 x 1 0- J (6.626 x 1 3 4 J o )(3 .00 x 1 0 8 m/s) OS I nm hc 19 9 = 277.6257 = 278 om W: A = - = E 1 0- m 7 . 1 6 x 1 0- j Metal A must be barium, because barium is the only metal that emits in the visible range (462 nm). _

=

(

l( 1 ( 1 --

---

99

-

=

b) A U V range of 278 run to 292 nm is necessary to distinguish between tantalum and tungsten. 7.57

7.60

( ](

)

The speed of light is necessary; however, the frequency is irrelevant. 7 ) 103 m 1s . = 270 = 2.7 x 102 s a) Time = ( 8. 1 x 1 0 km -1 km 3.00 x 1 08 m

(

]

( ) 3 . 00 X 108 m = 3.6 x 108 m . b) Distance = 1 .2 s s calculation of the amount of heat energy absorbed by a substance from its heat the Plan: Refer to Chapter 6 for capacity and temperature change (q = C x mass x �T). Using this equation, calculate the energy absorbed by the water. This energy equals the energy from the microwave photons. The energy of each photon can be calculated from its wavelength: E = hc/A. Dividing the total energy by the energy of each photon gives the number of photons absorbed by the water. Solution : q = C x mass x � T 4 q = (4. 1 84 J/g°C) (252 g) (98 - 20.)OC = 8.22407 x 1 0 J (unrounded) ( 6.626 X 10-34 J s )(3.00 X 1 08 m/s) hc 3 = 1 .28245 X 10-2 J/photon (unrounded) E = - = 2 A, 1 .55 x W- m 4 1 photon Number of photons = ( 8.22407 x 1 0 J ) 2 = 6.4 1 278 X 1027 = 6.4 X 1027 photons 1 .28245 x 1 0· 3 J 1 Check: The order of magnitude appears to be correct for the calculation of total energy absorbed: 1 0 x 102 X 1 01 = 4 4 34 10 . The order of magnitude of the energy for one photon also appears correct: 10- x 108/10-2 = 10-2 . I n addition, 4 4 2 2 the order 2 of magnitude of the number of photons is estimated as 10 /10- = 10 8, which agrees with the calculated 6x l 0 7 . 0

)

(

7.63

Plan: The energy differences sought may be determined by looking at the energy changes in steps. Solution: a) The difference between levels 3 and 2 (E32) may be found by taking the difference in the energies for the 3 transition (E31) and the 2 � 1 transition (E21). 1 E3 = E31 - E 1 = (4.844 X 10- 7 J) - (4.088 X 1 0- 1 7 J) = 7.56 X 10-18 J 2 2 ( 6.626 x 1 0- 34 J o s )( 3 .00 x 1 0 8 m/s) hc -s A= - = = 2.629365 x 1 0-8 = 2.63 x 10 m ( 7.56 x l 1 8 J ) E O-

� 1

b) The difference between levels 4 and I (E41) may be found by adding the energies for the 4 � 2 transition (Ed and the 2 � 1 transition (E21). 1 1 17 E 41 = E4 + E 1 = ( 1 .022 X 10- 7 J) + (4.088 X 10- 7 J) = 5.110 X 10- J 2 2 ( 6.626 X 10-34 J s )(3.00 X 1 0 8 m/s) hc 1 = 3 .8900 X 10-9 = 3.890 X 10-9 m A= - = E ( 5.110 X 10- 7 J ) 0

c) The difference between levels 5 and 4 (E54) may be found by taking the difference in the energies for the 5 � 1 transition (E51) and the 4 � 1 transition (see part b). 1 1 E54 = E51 - E41 = (5.232 X 10- 7 J) - (5 . 1 1 0 X 10- 7 J) = 1.22 X 10-18 J ( 6.626 X 10-34 J s)(3 .00 x 1 0 8 m/s) hc A= - = = 1 .629344 X 10-7 = 1.63 X 10-7 m ( 1 .22 x 1 0-18 ) E J 0

1 00

7.65

a) Figure 7.3 indicates that the 64 1 nm wavelength of Sr falls in the red region and the 493 nm wavelength of Ba falls in the green region. b) Use the formula E = hc/A to calculate kJ/photon. Convert the 1 .00 g amounts of BaCl2 (M = 208.2 gimol) and SrCh (M = 1 5 8.52 gimol) to moles, then to atoms, and assume each atom undergoes one electron transition (which produces the colored light) to find number of photons. M ultiply kJ/photon by number of photons to find total energy. SrC12 1 mol SrCl 2 6.022 x 1 0 23 Photons N urnb er 0 f p h otons = ( 1 . 00 g S rCl 2 ) 1 mol SrC1 2 1 5 8.52 g SrCl2 21 = 3 . 7988897 X 1 0 photons (unrounded) 8 ( ( 6.626 X 1 0 -34 J s ) 3 .00 X 1 0 m/s ) � 1 kJ h n hc p oo = E t = A, 64 1 nm 1 0 9 m 1 03 J 22 = 3 . 1 0 1 09 X 1 021 kJ/photon (unrounded)2 o Et tal = (3 .7988897 x 1 0 photons) (3 . 1 0 1 09 X 1 0- 2 kJ/photon) = 1 . 1 7807 = 1.18 kJ (Sr) BaCh 1 mol BaCl 2 6.022 x 1 0 23 Photons N urnb er 0 f p h otons = (1 . 00 g B a Cl 2 ) 1 mol BaCl 2 208.2 g BaCl2 21 = 2 . 8924 1 1 x 1 0 photons (unrounded) 8 ( 6.626 X 1 0 34 J s ) ( 3 .00 x 1 0 m/s ) h n hc oo = EP t = A, 493 nm 1 0 9 m 1 03 J 22 = 4.0320487 X2 1 1 0 kJ/photon (unrounded)22 o Et tal = (2.8924 1 1 X 1 0 photons) (4.0320487 x 1 0- kJ/photon) = 1 . 1 6623 = 1.17 kJ (Ba)

(

] J[ ( )( ) -

(

] J[ (�)(�) -

•

-

7.67

•

a) At this wavelength the sensitivity to absorbance of light by Vitamin A is maximized while minimizing interference due to the absorbance of light by other substances in the fish-liver oil. b) The wavelength 3 29 nm lies in the ultraviolet region of the electromagnetic spectrum. c) A known quantity of vitamin A ( 1 .67 x 1 0 3 g) is dissolved in a known volume of solvent (250. mL) to give a standard concentration with a known response ( 1 .018 units). This can be used to find the unknown quantity of V itamin A that gives a response of 0.724 units. An equality can be made between the two concentration-to­ absorbance ratios: � = A2 C1 C2 1 .67 X 1 0 -3 g (0.724 ) 250. mL AC CI = � = -----':-----,----'- = 4.7508 X 1 0-6 gimL (unrounded) ( 1 .0 1 8 ) A2 ( 4.7508 X 1 O 6 g Vitamin A/mL ( 500. mL ) ) --- = 1 .92808 1 0 2 = 1.93 X 10-2 g Vitamin A/g Oil -----x (0. 1 232 g Oil )

-

�

7.7 1

-

(

- ...,---

)

---: -'-

-

(---) )( ) W) ( IJ/S )( W )( - )

The amount of energy is calculated from the wavelength of light: 8 ( h n hc = 6.626 X 1 0 34 J s ) ( 3 .00 x 1 0 m/s ) 1 nm p oo = 3 .6 1 4 1 8 X 1 0 E t =A, 550 nm 10 9 m •

Amount of E from the bulb = ( 75 Number of photons:

(

0.375 J s

-1

9

J/photon (unrounded)

5% = 0.375 J/s 100 % 1 = l .0376 x 1 0 8 = 1.0 x 10 18 photons/s

1 0% 1 00%

1 Photon 1 3.6141 8 x 1 0 9 J

1

01

Chapter 8 Electron Configuration and Chemical Periodicity FOLLOW-UP PROBLEMS 8.1

Plan: The superscripts can be added to indicate the number of electrons in the element, and hence its identity. A horizontal orbital diagram is written for simplicity, although it does not indicate the sublevels have different energies. Based on the orbital diagram, we identify the electron of interest and determine its four quantum numbers. Solution : The number of electrons = 2 + 2 + 4 = 8; element oxygen. Orbital diagram: =

[U] [ill It.lt It I Is

2p

2s

The electron in the middle 2p orbital is the sixth electron (this electron would have entered in this position due to H und 's rule). This electron has the following quantum numbers: n = 2, I I (for p orbital), m, 0, m,. + 112. By convention, + I /2 is assigned to the first electron in an orbital. Also, the first p orbital is assigned a a m, value of \ the middle p orbital a m, value of 0 and the last p orbital a m, value of + I . =

=

=

-

8.2

,

Plan: The atomic number gives the number of electrons. The order of filling may be inferred by the location of the element on the periodic table. The partial orbital diagrams shows only those electrons after the preceding noble gas except those used to fill inner d and! subshells. The number of inner electrons is simply the total electrons minus those electrons in the partial orbital diagram. Solution: a) For Ni (Z = 28), the full electron configuration is li2i2 p63i3p64i3tf. The condensed configuration is [Ar)4i3tf. The partial orbital diagram for the valence electrons is

[ ] []] It + I f + It+ I tit I 1,--,-----,--, Ar

4s

4p

3d

There are 28 l O(valence) = 18 inner electrons. b) For Sr (Z = 38), the full electron configuration is li2i2p63i3p64i3d104p6Si. The condensed configuration is [Kr)Si. The partial orbital diagram is -

[ ] []] I

I I I I 1""-----'--'

Ar

Ss

Sp

4d

There are 38 2(valence) 36 inner electrons. c) For Po (84 electrons), the full configuration is li2i2l3i3p64i3io4p6si4iosp66i4j4sio6p4. The condensed configuration is [Xe)6i4j4sio6p4. The partial orbital diagram represents valence electrons only; the inner electrons are those in the previous noble gas (Xe or li2i2l3i3p64i3d,04lSs24d,oSl) and filled transition (SdlO) and inner transition (414) levels. -

=

[ ] []] It+lt It I Ar

6s

6p

There are 84 - 6( valence) = 78 inner electrons.

102

8.3

Plan: Locate each of the elements on the periodic table. All of these are main-group elements, so their sizes increase down and to the left in the periodic table. Solution : a) CI < Br < Se. CI has a smal ler n than Br and Se. Br experiences a higher Zeff than Se and is smaller. b) Xe < I < Ba. Xe and I have the same n, but Xe experiences a higher Zeff and is smaller. Ba has the highest n and is the largest.

8.4

Plan: These main-group elements must be located on the periodic table. The value of I EI increases toward the top (same column) and the right of the periodic table (same n). Solution : a) Sn < Sb < I. These elements have the same n, so the values increase to the right on the periodic table. Iodine has the highest IE, because its outer electron is most tightly held and hardest to remove. b) Ba < Sr < Ca. These elements are in the same column, so the values decrease towards the bottom of the column. Barium 's outer electron receives the most shielding; therefore, it is easiest to remove and has the lowest I E.

8.S

Plan: We must look for a large "jump" in the IE values. This jump occurs after the valence electrons have been removed. The next step is to determine the element in the designated period with the proper number of valence electrons. Solution : The exceptionally large jump from I E3 to I E4 means that the fourth electron is an inner electron. Thus, Q has three valence electrons. Since Q is in period 3, it must be aluminum, AI: li2i2 p63i3pl.

8.6

Plan: We need to locate each element on the periodic table. Elements in the first two columns on the left or the two columns to the left of the noble gases tend to adopt ions with a noble gas configuration. Elements in the remaining columns may use either their ns and np electrons, or just their np electrons. Solution : i 2 a) Barium loses two electrons to be isoelectronic with Xe: Ba i ([Xe] 6 ) --7 Ba + ( [Xe] ) + 2e4 2 b) Oxygen gains two e- to be isoelectronic with Ne: 0 ([He] 2 2p + 2e- --7 0 - ( [Ne]) c) Lead can lose two to form an "inert1 opair" configuration: I 0 electrons i /4 i4/4 2 2 Pb ([Xe]6 Sd 6p ) --7 Pb + ( [Xe]6 4 Sd ) + 2eor lead canilose four electrons form a "pseudo-noble gas" configuration: 4/4 j0 / 4/4 j 4 Pb ([Xe]6 S 6 ) --7 Pb + ( [Xe] S O) + 4e-

8.7

Plan: Write the condensed electron configuration for each atom, being careful to note those elements, which are irregular. The charge on the cation tells how many electrons are to be removed. The electrons are removed beginning with the ns electrons. If any electrons in the final ion are unpaired, the ion is paramagnetic. If it is not obvious that there are unpaired electrons, a partial orbital diagram might help. Solution: i � 3 a) The Y atom ( [Ar]4 3 3) loses the two 4s electrons and one 3d electron to form y + ([Ar]3d2). There are two + unpaired d electrons, soi yJl is paramagnetic. b) The2 Ni atom ( [Ar]4 3 ) loses the two 4s electrons to form N i2+ ([Ar]3Ji). There are two unpaired d electrons, so Ni + is paramagnetic.

it. lt.it.it It I

3d 3 2 l c) The La atom 3( [Xe]6s Sd ) loses all three valence electrons to form La + ([XeJ). There are no unpaired electrons, so La + is diamagnetic.

1 03

8.8

Plan: Locate each of the elements on the periodic table. Cations are smaller than the parent atoms, and anions are larger than the parent atoms. I f the electrons are equal, anions are larger than cations. The more electrons added or removed; the greater the change in size. Solution: a) Ionic size increases down a group, so F < CI- < Br . b) These species are isoelectronic (all have 1 0 electrons), so size increases with increasing negative charge: -

Mg

2+

<

Na+

<

-

F-.

c) Ionic size increases as charge decreases for different cations of the same element, so C r3+ < CrH END-OF-CHAPTER PROBLEMS

8. 1

Elements are listed in the periodic table in an ordered, systematic way that correlates with a periodicity of their chemical and physical properties. The theoretical basis for the table in terms of atomic number and electron configuration does not allow for an "unknown element" between Sn and Sb.

8.3

Plan: The value should be the average of the elements above and below the one of interest. Solution : a) Predicted atomic mass (K) = Na + Rb 22.99 + 85.47 54. 3 (actual value = 39. 1 0 amu) = 2 amu = 2 2 b) Predicted melting point (Br2) = CI2 + 1 2 - 1 0 1 .0 + 1 1 3.6 3. 0 (actual value = -7.2°C) =6 C = 2 2 c) Predicted boiling point ( H Br) = HCI + H I -84.9 + (-3 5.4) (actual value =-67.0°C) =-60. 1 5 =-60.20C 2 2

8.5

The quantum number m s relates to just the electron; all the others describe the orbital.

8.8

Shielding occurs when inner electrons protect or shield outer electrons from the ful l nuclear attractive force. The effective nuclear charge is the nuclear charge an electron actually experiences. As the number of inner electrons increases, the effective nuclear charge decreases.

8. 1 0

a) The 1= 1 quantum number can only refer to a p orbital. These quantum numbers designate the 2p orbital set, which hold a maximum of 6 electrons, 2 electrons in each of the three 2p orbitals. b) There are five 3d orbitals, therefore a maximum of 10 electrons can have the 3d designation. c) There is one 4s orbital which holds a maximum of2 electrons.

8. 1 2

a) 6 electrons can be found in the three 4p orbitals, 2 in each orbital. b) The 1= 1 quantum number can only refer to a p orbital, and the m, value of + I specifies one particular p orbital, which hold a maximum of2 electrons with the difference between the two electrons being in the ms quantum number. c) 1 4 electrons can be found in the 5forbitals (I = 3 designates f orbitals; there are 7 forbitals in a set).

8. 1 5

H und ' s rule states that electrons will fill empty orbitals in the same sublevel before filling half-filled orbitals. This lowest-energy arrangement has the maximum number of unpaired electrons with parallel spins. The correct electron configuration for nitrogen is shown in (a), which is contrasted to an incorrect configuration shown in (b). The arrows in the 2p orbitals of configuration (a) could alternatively all point down. (b) - incorrect (a) - correct

[lllliJ It It It I

[lllliJ Ittlt I

Is 2p 2p 2s In (a), there is one electron in each of the 2p orbitals; in (b) which is incorrect, 2 electrons were paired in one of the 2p orbitals while leaving one 2p orbital empty. Is

2s

1 04

8. 1 7

For elements in the same group (vertical column in periodic table), the electron configuration of the outer electrons are identical except for the n value. For elements in the same period (horizontal row in periodic table), their configurations vary because each succeeding element has one additional electron. The electron configurations are similar only in the fact that the same level (principal quantum number) is the outer level.

8. 1 8

Plan: Assume that the electron is in the ground state configuration and that electrons fill in a P.-Py-Pz order. By convention, we assign the first electron to fill an orbital an m s = + 1 12. Also by convention, m, = -1 for the Px orbital, m, = 0 for the Py orbital, and m, = + 1 for the pz orbital. Solution: a) The outermost electron in a rubidium atom would be in a 5s orbital (rubidium is in Row 5, Group I). The quantum numbers for this electron are n = 5, 1= 0,im/ , = 0, and m s = + 1 12. b) The S- ion would have the configuration [Ne]3 3 . The electron added would go into the 3pz orbital and is the second electron in that orbital. Quantum numbers 1 I are n = 3, 1= 1, m, = + 1 , and ms = -112. c) Ag atoms have the configuration [Kr] 5s 4d O • The electron lost would be from the 5s orbital with quantum numbers n = 5, 1= 0, m, = 0, and ms = + 1 12. i / d) The F atom has the configuration [He]2 2 . The electron gained would go into the 2pz orbital and is the second electron in that orbital. Quantum numbers are n = 2, 1= 1, m, = + 1 , and ms = -112.

8.20

a) Rb: b) Ge: c) Ar:

8.22

Valence electrons are2those ct electrons beyond the previous noble gas configuration and unfilled d and fsublevels. a) Ti (Z= 22); [Ar]4S 3

l s2 2i2 p63s23p64i3dl04p6 5 s1 1 i2i2 p63i3p64i3d1o4p2 1i2i2 p63i3p6

[ ] [OJ I f I f I

I IL--L--..JL.....-.I

Ar

4s i 5 b) CI (Z= 1 7); [Ne] 3 3p

[ ] [OJ Ne

3s i cf c) V (Z= 23); [Ar]4 3

4p

3d

1ttl ttl t

1

3p

[ ] [OJ 1f 1fit 1 Ar

4s

8.24

3d

a) Element = 0, Group 6A( 1 6), period 2

[ ] [OJ He

1ttl tit

2s 2p b) Element = P, Group 5A( 1 5), period 3

[ ] Ne

1

IT] ""-I t I"--- t 1----' 1 t ""3s

3p

1 05

4p

8.26

a) The orbital diagram shows the element i 10 isl in period 4 ( n = 4 as outer level). The configuration is li2i2p63i3p64s23dl04pl or [Ar]4 3d 4p . One electron in the p level indicates the element is in group 3A(13). The element is Ga. b) The orbital diagram shows the 2s andi2plorbitals filled which would represent the last element in period 2, Ne. The configuration is li2i2p6 or [He]2 2 . Fil led s and p orbitals indicate the group 8A(18).

8.28

Inner electrons are those seen in the previous noble gas and completed transition series (d orbitals). Outer electrons are those in the highest energy level (highest n value). Valence electrons are the outer electrons for main-group elements; for transition metals, valence electrons also include electrons in the unfil led d set of orbitals. It is easiestito/determine the types of electrons by writing a condensed electron configuration. a) 0 (Z= 8); [He]2 2 . There are 2 inner electrons (represented by [He)) and 6 outer electrons. The number of valence electrons (6) iequals the outer electrons in this case. 0 / b) Sn (Z = 50); [Kr] 5 4dI 5 . There are 36 (from [Kr)) + 1 0 (from the fil led d level) = 4 6 inner electrons. There are 4 outer electrons (highest energy level is n = 5) and 4 valence electrons. 2 c) Ca (Z = 20); [Ar]4si . cf There are 2 outer electrons, 2 valence electrons, and 18 inner electrons. d) Fe (Z = 26); [Ar]4 3 . There are 2 outer electrons (from n = 4 level), 8 valence electrons (the d orbital electrons count in thisi case the sublevel is not full), and 18 inner electrons. 'o because / e) Se (Z = 34); [Ar]4 3d 4 . There are 6 outer electrons (2 + 4 in the n = 4 level), 6 valence electrons (filled d sublevels count as inner electrons), and 28 « 1 8 + 1 0) or (34 - 6» inner electrons. i ' a) The electron configuration [He ]2 2p has a total of 5 electrons (3 + 2 from He configuration) which is element boron with symbol B. Boron is in group 3A( 1 3) . Other elements in this group are AI, Ga, In, and Tl. b) The electrons in this element total 1 6, 1 0 from the neon configuration plus 6 from the rest of the configuration. Element 1 6 is sulfur, S, in group 6A( 1 6). Other elements in group 6A( 1 6) are 0, Se, Te, and Po. c) Electrons total 3 + 54 (from xenon) = 57. Element 57 is lanthanum, La, in group 38(3). Other elements in this group are Sc, Y, and Ac.

8.30

8.33

Atomic size increases down a main group and decreases across a period. Ionization energy decreases down a main group and increases across a period. These opposite trends result because as the atom gets larger, the outer electron is further from the attraction of the positive charge of the nucleus, which is what holds the electron in the atom. It thus takes less energy (lower lE) to remove the outer electron in a larger atom than to remove the outer electron in a smal ler atom. As the atomic size decreases across a period due to higher Zeff, it takes more energy (higher IE) to remove the outer electron.

8.35

For a given element, successive ionization energies always increase. As each successive electron is removed, the positive charge on the ion increases, which results in a stronger attraction between the leaving electron and the ion. When a large jump between successive ionization energies is observed, the subsequent electron must come from a lower energy level. Thus, by looking at a series of successive ionization energies, we l can determine the number of valence electrons. For instance, the electron configuration for potassium is [Ar]4s . The first electron lost is the one from the 4s level. The second electron lost must come from the 3p level, and hence breaks into the core electrons. Thus, we see a significant jump in the amount of energy for the second ionization when compared to the first ionization.

8.37

A high, endothermic lE I means it is very difficult to remove the first outer electron. This value would exclude any metal, because metals lose an outer electron easily. A very negative, exothermic EAI suggests that this element easily gains one electron. These val ues indicate that the element belongs to the halogens, Group 7A(17), which form -I ions.

8.39

a) Increasing atomic size: K < Rb < Cs; these three elements are all part of the same group, the alkali metals. Atomic size increases down a main group (larger outer electron orbital), so potassium is the smallest and cesium is the largest. b) Increasing atomic size: 0 < C < Be; these three elements are in the same period and atomic size decreases across a period (increasing effective nuclear charge), so beryllium is the largest and oxygen the smallest. c) Increasing atomic size: CI < S < K; chlorine and sulfur are in the same period so chlorine is smaller since it is further to the right in the period. Potassium is the first element in the next period so it is larger than either CI or S.

1 06

d) I ncreasing atomic size : Mg < Ca < K; calcium is larger than magnesium because Ca is further down the alkaline earth metal group on the periodic table than Mg. Potassium is larger than calcium because K is further to the left than Ca in period 4 of the periodic table. 8.41

a) Ba < Sr < Ca The "group" rule applies in this case. Ionization energy decreases down a main group. Barium's outer electron receives the most shielding; therefore, it is easiest to remove and has the lowest IE. b) B < N < Ne These elements have the same n, so the "period" rule applies. Ionization energy increases across a period. B experiences the lowest Ze fT and has the lowest IE. Ne has the highest IE, because it's very difficult to remove an electron from the stable noble gas configuration. c) Rb < Se < Br IE decreases with increasing atomic size, so Rb (largest atom) has the smallest IE. Se has a lower IE than Br because IE increases across a period. d) So < Sb < As IE decreases down a group, so Sn and Sb will have smaller I E's than As. The "period" rule applies for ranking Sn and Sb.

8.43

The successive ionization energies show a significant jump between the first and second I E's and between the third and fourth IE's. This indicates that 1) the first electron removed occupies a different orbital than the second electron removed, 2) the second and third electrons occupy the same orbital, 3) the third and fourth electrons occupy different orbitals, and 4) the fourth and fifth electrons occupy the same orbital. The electron configurations for period 2 elements range from li2s1 for lithium to li2i2l for neon. The three different orbitals are Is, 2s, and 2p. The first electron is removed from the 2p orbital and the second from the 2s orbital in order for there to be a significant jump in the ionization energy between the two. The third electron is removed from the 2s orbital while the fourth is removed from the Is orbital since I E4 is much greater than I E 3• The configuration is li2i2pl which represents the five electrons in boron, B.

8.45

a) Na would have the highest I E 2 because ionization of a second electron would require breaking the stable [Ne] configuration: F irst ionization: Na ([Ne]3sl) � Na+ ([NeD + e- (low I E) Second ionization: Na+ ([NeD � Na+2 ([He]2i2i ) + e- (high IE) b) Na would have the highest I E 2 because it has one valence electron and is smaller than K. c) You might think that Sc would have the highest I E2 , because removing a second electron would require breaking the stable, fi lled 4s shell. However, Be has the highest I E 2 because Be's small size makes it difficult to remove a second electron.

8.47

Three of the ways that metals and nonmetals differ are : 1 ) metals conduct electricity, nonmetals do not; 2) when they form stable ions, metal ions tend to have a positive charge, nonmetal ions tend to have a negative charge; and 3) metal oxides are ionic and act as bases, nonmetal oxides are covalent and act as acids. How many other differences are there?

8.48

Metallic character increases down a group and decreases toward the right across a period. These trends are the same as those for atomic size and opposite those for ionization energy.

8.52

Metallic behavior increases down a group and decreases across a period. a) Rb is more metallic because it is to the left and below Ca. b) Ra is more metallic because it lies below Mg in Group 2A(2). c) I is more metallic because it lies below Br in Group 7 A( 1 7).

8.54

The reaction of a nonmetal oxide in water produces an acidic solution. An example of a Group 6A(I6) nonmetal oxide is SOz(g) : S0 2 (g) + H 2 0(g) � H 2S0 3(aq).

8.56

For main group elements, the most stable ions have electron configurations identical to noble gas atoms. a) C I : li2i2l3i3p5; cr, li2i2p63i3l, chlorine atoms are one electron short of the noble gas configuration, so a -1 ion will form by addiny an electron to have the same electron configuration as an argon atom. b) Na: Is22s22p63s1; Na+ , Is 2i2p6, sodium atoms contain one more electron than the noble gas configuration of neon. Thus, a sodium atom loses one electron to form a + 1 ion. 2+ c) Ca: li2s22l3i3p64s2; Ca , li2i2p63i3p6, calcium atoms contain two more electrons than the noble gas configuration of argon. Thus, a calcium atom loses two electrons to form a +2 ion.

1 07

8.58

To find the number of unpaired electrons look at the electron configuration expanded to include the different orientations of the orbitals, such as Px and py and pz. Remember that one electron will occupy every orbital in a set (p, d, orf) before electrons wil l pair in an orbital in that set. In the noble gas configurations, all electrons are paired because all orbitals are filled. a) Configuration of 2A(2) group elements: [noble gas]ni , no unpaired electrons. b) Configuration of 5A(l5) group elements: [noble gas]n inpx1np/ np,l . Three unpaired electrons, one each in p" Py , and pz. c) Configuration of 8A( 18) group elements: noble gas configuration with no half-filled orbitals, no unpaired electrons. d) Configuration of 3A( l 3) group elements: [noble gas]ninpl. There is one unpaired electron in one of the p orbitals.

(a)

(c) 8.60

ns

(b )

[TI I t�1 t� It� I ns

np

(d)

[TIlt I t It I ns

np

ns

np

Substances are paramagnetic if they have unpaired electrons. This problem is more challenging if you have difficulty picturing the orbital diagram from the electron configuration. Obviously an odd number of electrons will necessitate that at least one electron is unpaired, so odd electron species are paramagnetic. However, a substance with an even number of electrons can also be paramagnetic, because even numbers of electrons do not guarantee all electrons are paired. Remember that all orbitals in a p, d, orfset will each have one electron before electrons pair in an orbital. a) Y: [Ar]4i3tf ; y3+: [Ar]3J1 Transition metals first lose the s electrons in forming ions, so to form the +3 ion a vanadium atom loses two 4s electrons and one 3d electron. Paramagnetic

4s 3d b) Cd: [Kr]5i4d1o ; Cd2+: [Kr]4d1o Cadmium atoms lose two electrons from the 4s orbital to form the +2 ion. Diamagnetic

[ ] D If�lf�lt�lt�lt� I Kr

5s c) Co: [Ar]4i3d7 ; C03+: Paramagnetic

4d [Ar]3cf Cobalt atoms lose two 4s electrons and one 3d electron to form the +3 ion.

[ ] D If � I tit I tit Ar

4s

3d

108

I

d) Ag: [Kr]Ssi4diO ; Ag+ : [Kr]4diO Silver atoms lose the one electron in the Ss orbital to form the + I ion.

Diamagnetic

Ss 8.62

4d

For palladium to be diamagnetic, i Jlall of its electrons must be paired. You might first write the condensed electron configuration for Pd as [Kr]S 4 . However, the partial orbital diagram is not consistent with diamagnetism.

[ Kr] [] I t .If.ltl. tIt I IL--

.L...-....I..

Sp Ss 4d Promoting an s electron into the d sublevel (as in (c)) still leaves two electrons unpaired.

[ ] [] I t .If.lt.lt .lt I I�� Kr

Ss 4d The only configuration that supports diamagnetism is (b)

Ss

[Kr]4dlO.

Sp

Sp

4d

8.64

The size of ions increases down a group. For ions that are isoelectronic (have the same electron configuration) size decreases with increasing atomic number. a) Increasing size: Lt < Na+ < K+,2 size increases down group I A. b) I ncreasing size: Rb+ < Br- < Se -, these three ions are isoelectronic with the same electron configuration as krypton. S ize decreases with increasing atomic number in an isoelectronic series. 3 2 c) I ncreasing size: y - < 0 - < N -, the three ions are isoelectronic with an electron configuration identical to neon. Size decreases with increasing atomic number in an isoelectronic series.

8.67

Plan: Write the formula of the oxoacid. Remember that in naming oxoacids (H + polyatomic ion), the suffix of the polyatomic changes: -ate becomes -ic acid and -ite becomes -ous acid. Determine the oxidation state of the nonmetal in the oxoacid - hydrogen has an O.N. of + I and oxygen has an O.N. of -2. Based on the oxidation state of the nonmetal, and the oxidation state of the oxide ion (-2), the formula of the nonmetal oxide may be determined. The name of the nonmetal oxide comes from the formula; remember that nonmetal compounds use prefixes to indicate the number of each type of atom in the formula. Solution : a) hypochlorous acid HCIO has3 CI+ so the oxide is CI20 dichlorine oxide or dichlorine monoxide b) chlorous acid HCl0 2 has C1 5 + so the oxide is CIz03 dichlorine trioxide c) chloric acid HCl0 3 has Cl + so the oxide is CI20s dichlorine pentaoxide d) perchloric acid HCI04 has6 CI7 + so the oxide is CI20? dichlorine heptaoxide e) sulfuric acid H 2 S04 has S +4 so the oxide is S03 sulfur trioxide f) sulfurous acid H 2 S0 3 has 5 S + so the oxide is S02 sulfur dioxide 3 g) nitric acid HN0 has N +3 so the oxide is N20S dinitrogen pentaoxide h) nitrous acid HN0 2 has N +4so the oxide is N203 dinitrogen trioxide i) carbonic acid H 2 C0 3 has C + 5so the oxide is CO2 carbon dioxide j) phosphoric acid H 3 P04 has p + so the oxide is P20S diphosphorus pentaoxide =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

109

8.69

Remember that i soelectronic species have the same electron configuration. a) A chemically unreactive Period 4 element would be Kr in Group 8A(l8). Both the Sr2+ ion and Br- ion are isoelectronic with Kr. Their combination results in SrBr2, strontium bromide. 2 2+ b) Ar is the Period 3 noble gas. Ca and S - are isoelectronic with Ar. The resulting compound is CaS, calcium sulfide.

c) The smallest filled d subshell is the 3d shell, so the element must be in Period 4. Zn forms the Zn2+ ion by losing its two s subshell electrons to achieve a pseudo-noble gas configuration ([Ar] 3d1o). The smallest halogen is fluorine, whose anion is F-. The resulting compound is ZnF2' zinc fluoride. d) Ne is the smallest element in Period 2, but it is not ionizable. Li is the largest atom whereas F is the smallest atom in Period 2. The resulting compound is LiF, lithium fluoride. 8.70

Plan: Determine the electron configuration for iron, and then begin removing one electron at a time. Filled subshells give diamagnetic contributions. Any partially filled subshells g ive a paramagnetic contribution. The more unpaired electrons, the greater the attraction to a magnetic field. Solution: partially fi lled 3d= paramagnetic number of unpaired electrons 4 [Ar]4i3 Ji Fe Fe+ partially filled 3d= paramagnetic number of unpaired electrons = 5 [Ar]4s1 3 d' 2+ partially filled 3d paramagnetic number of unpaired electrons 4 Fe [Ar] 3 Ji partially filled 3d paramagnetic number of unpaired electrons 5 [Ar] 3 d' Fe 3+ 4+ partially filled 3d paramagnetic number of unpaired electrons 4 Fe [Ar] 3 � partially filled 3d paramagnetic number of unpaired electrons 3 [Ar] 3 Jl Fes+ partially filled 3d paramagnetic number of unpaired electrons 2 Fe6+ [Ar] 3 tf partially fi lled 3d = paramagnetic number of unpaired electrons I Fe7+ [Ar] 3 i filled orbitals = diamagnetic number of unpaired electrons 0 [Ar] Fe8+ / partially filled 3p = paramagnetic number of unpaired electrons 1 Fe9+ [Ne]3 i 3 Fe 1o+ partially filled 3p paramagnetic number of unpaired electrons 2 [Ne] 3 i 3p4 / ll+ partially filled 3p paramagnetic number of unpaired electrons 3 Fe [Ne] 3 i3 2 partially filled 3p paramagnetic number of unpaired electrons 2 [Ne] 3s 3p2 Fe 12+ J + i 3 partially filled 3p paramagnetic number of unpaired electrons 1 Fe [Ne] 3 3 pl Fe14+ filled orbitals diamagnetic number of unpaired electrons 0 [Ne] 3 i Fe + and Fe 3+ would both be most attracted to a magnetic field. They each have 5 unpaired electrons. =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

110

Chapter 9 Models of Chemical Bonding FOLLOW-UP PROBLEMS

9.1

Plan: First, write out the condensed electron configuration and electron-dot structure for magnesium atoms and chlorine atoms. In the formation of the ions, each magnesium atom wi l l lose two electrons to form the +2 ion and each chlorine atom will gain one electron to form the - I ion. Write the condensed electron configuration and electron-dot structure for each of the ions. The formula of the compound is found by combining the ions in a ratio that gives a neutral compound. Solution : Condensedielectron configuration: i/ 2 2 / Mg ( [Ne]3 ) + CI ([Ne] 3 3 ) -7 Mg+ ( [ Ne)) + c r ( [ Ne]3s 3 ) In order to ibalance the charge (or the number of electrons lost and two chlorine atoms are needed: 6 i gained) i/ 2 Mg ( [Ne]3 ) + 2CI ([Ne]3 3 ) -7 Mg+ ([Ne)) + 2C r ( [Ne]3 3p ) Lewis electron-dot symbols: ......----... . CI :

/

Mg ·

..

+

\.. CI : � ..

•

• •

-

•

The formula of the compound would contain two chloride ions for each magnesium ion, MgCI2. 9.2

Plan: a) All bonds are single bonds from silicon to a second row element. The trend in bond length can be predicted from the fact that atomic radii decrease across a row, so fluorine will be smaller than oxygen, which is smaller than carbon. Bond length will therefore decrease across the row while bond energy increases. b) All the bonds are between two nitrogen atoms, but differ in the number of electrons shared between the atoms. The more electrons shared the shorter the bond length and the greater the bond energy. Solution : Bond strength: Si-C < Si-O < Si-F a) Bond length: Si-F < Si-O < Si-C b) Bond length: N=N < N=N < N-N Bond energy: N-N < N=N < N=N Check: (Use the tables in the chapter.) Bond lengths from Table: Si-F, 156 pm < Si-O, 161 pm < Si-C, 1 86 pm Bond energies from Table: Si-C, 301 kJ < Si-O, 368 kJ < Si-F, 565 kJ Bond lengths from Table: N=N, 1 1 0 pm < N N , 1 22 pm < N-N, 1 46 pm Bond energies from Table: N-N, 160 kJ < N N 4 1 8 kJ < N=N , 945 kJ The values from the tables agree with the order predicted. =

=

9.3

,

Plan: Assume that all reactant bonds break and all product bonds form. Use Table 9.2 to find the values for the different bonds. Sum the values for the reactants, and sum the values for the products (these are all negative values). Add the sum of product to the values from the sum of the reactant values. Solution : Calculating !1H0 for the bonds broken : 1 x N=N ( 1 mol) (945 kJ/mol) 945 kJ 3 x H-H (3 mol) (432 kJ/mol) 1 296 kJ Sum broken 224 1 kJ Calculating !1H0 for the bonds formed: 1 x NH3 (3 mol) (-391 kJ/mol) -\ 173 kJ I x NH3 (3 mol) (-39\ kJ/mol) -1173 kJ -2346 kJ Sum formed !1H:n 224 1 kJ + (-2346 kJ) 1 05 kJ =

=

=

=

=

=

=

=

=

=

=

= -

III

9.4

Plan: Bond polarity can be determined from the difference in electronegativity of the two atoms. The more electronegative atom holds the 0- charge and the other atom holds the 0+ charge. Solution: a) From F igure 9.20, EN of CI 3 .0, EN of F 4.0, EN of Br = 2.8. CI-CI will be the least polar since there is no difference between the electronegativity of the two chlorine atoms. CI-F will be more polar than C I-Br since the electronegativity difference between CI and F (4.0 - 3 .0 1 .0) is greater than the electronegativity difference between CI and Br (3 .0 - 2.8 =0.2) 0+ 0- 0+ 0CI-CI < Br-CI < CI-F b) From Figure 9.20, EN of CI = 3 .0, EN of Si 1 .8, EN of P 2. 1 , EN of S 2.5. Si-Si bond is nonpolar, therefore the least polar of the four bonds. The most polar is Si-CI with an EN difference of l .2, next is P-Cl with an EN difference of 0.9 and next is S-CI with an EN of 0.5 . 0+ 0- 0+ 0- 0+ 0Si-Si < S-CI < P-CI < Si-CI Check: For a) the order follows the increase in electronegativity going up the group of halogens. for b) the order follows the increase in electronegativity across a row. The closer to the end of row 3 (chlorine being at the end of row 3 next to the noble gas argon) that the element that is bonded to chlorine is found, the smaller the electronegativity difference of the bond. Therefore, the order of bond polarity fol lows the order of elements in row 3 : Si, P, S. =

=

=

=

=

=

EN D-Of-CHAPTER PROBLEMS 9. 1

a) Larger ionization energy decreases metallic character. b) Larger atomic radius increases metallic character. c) Larger number of outer electrons decreases metallic character. d) Larger effective nuclear charge decreases metallic character.

9.4

Metallic behavior increases to the left and down on the periodic table. a) Cs is more metallic since it is further down the alkali metal group than Na. b) Rb is more metallic since it is both to the left and down from Mg. c) As is more metallic since it is further down Group SA than N .

9.6

Ionic bonding occurs between metals and nonmetals and covalent bonding between nonmetals. a) Bond in CsF is ionic because Cs is a metal and F is a nonmetal. b) Bonding in N2 is covalent because N is a nonmetal. c) Bonding in H2S(g) is covalent because both H and S are nonmetals.

9.8

Lewis electron-dot symbols show valence electrons as dots. Place one dot at a time on the four sides (this method explains the structure in b) and then pair up dots until all valence electrons are used. a)

Rb·

b)

•

Si

•

c)

: I

•

9. 1 0

a ) Assuming X i s a n A-group element, the number o f dots (valence electrons) equals the group number. Therefore, X is a 6A(16) element with 6 valence electrons. Its general electron configuration is [noble gas]ninl , where n is the energy level. b) X has three valence electrons and is a 3A( 1 3) element with general e- configuration [noble gas]ninp' .

9. 1 3

Because the lattice energy i s the result of electrostatic attractions among the oppositely charged ions, its magnitude depends on several factors, including ionic size, ionic charge, and the arrangement of ions in the solid. for a particular arrangement of ions, the lattice energy increases as the charges on the ions increase and as their radii decrease.

1 12

9. 16

a) Barium is a metal2 and loses 2 2 electrons to achieve a noble gas configuration: Ba ( [Xe]6s ) � Ba + ( [XeD + 2 e2+ Ba Ba .. + 2 e-

[ ]

•

•

Chlorine is a nonmetal a noble gas configuration: 6 i 5and gains 1 electron toiachieve CI ( [Ne]3 3p ) + 1 e- � Cl- ( [Ne]3 3p )

[ ] : �� : ] [

]-

Two CI atoms gain the two electrons lost by Ba. The ionic compound formed is BaCI 2 .

�

: �'CI ..

/

' Ba .

+

----I....

CI . : ••

•

[

: �� : • •

2+

Ba

.

••

i� 2 b) Sr ( [Kr] 5 ) � Sr + ([KrD + 2 e­ The ionic compound formed is SrO. .

•

Sr·

+

•

. 0:

�

i l 3 c) Al ( [Ne]3 3p ) � AI +� ( [NeD + 3 e­

.

•

: F

.. � Al •

•

� .. •

••

The ionic compound formed is AIF3. l d) Rb ([Kr]5s ) � Rb + ( [KrD + I e••

0:

9.18

F : .. o

: :. : [ Al ] .

.

]

-

3+

[:y.:r

[ : : : ]••

.

i 4 i 6 2 ([He]2 2p ) + 2 e- � 0 - ( [He]2 2p )

�.

Rb . ---./ . The ionic compound formed is Rb2 0. •

F :

•

•

Rb · �

[

.

[

Rb

r [:?> r

[ r Rb

a) X in X 2 03 is a cation with +3 charge. The oxygen in this compound has a -2 charge. To produce an electrically neutral compound, 2 cations with +3 charge bond with 3 anions with -2 charge: 2(+3) + 3 (-2) O. Elements in Group 3A(l3 ) form +3 ions. 2 b) The carbonate ion, C03 -, has a -2 charge, so X has a +2 charge. Group 2A(2) elements form +2 ions. c) X in Na2 X has a -2 charge, balanced with the +2 overall charge from the two Na + ions. Group 6A( \6) elements gain 2 electrons to form -2 ions with a noble gas configuration. 2 2 a) BaS has the lower lattice energy because the ionic radius of Ba + is larger than Ca + . A larger ionic radius results in a greater distance between ions. The lattice energy decreases with increasing distance between 2 ions. 2 b) NaF has the lower lattice energy since the charge on each ion (+ I, - 1) is half the charge on the Mg + and 0 ions. Lattice energy increases with increasing ion charge. =

9.20

113

9.23

When two chlorine atoms are far apart, there is no interaction between them. Once the two atoms move closer together, the nucleus of each atom attracts the electrons on the other atom. As the atoms move closer this attraction increases, but the repulsion of the two nuclei also increases. When the atoms are very close together the repulsion between nuclei is much stronger than the attraction between nuclei and electrons. The final internuclear distance for the chlorine molecule is the distance at which maximum attraction is achieved in spite of the repulsion. At this distance, the energy of the molecule is at its lowest value.

9.24

The bond energy is the energy required to overcome the attraction between H atoms and CI atoms in one mole of HCI molecules in the gaseous state. Energy input is needed to break bonds, so bond energy is always endothermic and !1Ho bond brea ki ng is positive. The same amount of energy needed to break the bond is released upon its formation, so f1HD bond form i n g has the same magnitude as !!.H\ond brea ki ng, but opposite in sign (always exothermic and negative).

9.28

a) I-I < Br-Br < CI-CI. Bond strength increases as the atomic radii of atoms in the bond decreases. Atomic radii decrease up a group in the periodic table, so I is the largest and CI is the smallest of the three. b) S-Br < S-CI < S-H . Radius of H is the smal lest and Br is the largest, so the bond strength for S-H is the greatest and that for S-Br is the weakest. c) C-N < C=N < C=N. Bond strength increases as the number of electrons in the bond increases. The triple bond is the strongest and the single bond is the weakest.

9.30

a) For given pair of atoms, in this case carbon and oxygen, bond strength increases with increasing bond order. The C=O bond (bond order = 2) is stronger than the C-O bond (bond order = I ). b) 0 is smaller than C so the O-H bond is shorter and stronger than the C-H bond.

9.33

Reaction between molecules requires the breaking of existing bonds and the formation of new bonds. Substances with weak bonds are more reactive than are those with strong bonds because less energy is required to break weak bonds.

9.35

For methane: C H4(g) + 2 02(g) � CO2(g) + 2 H20(l) which requires that 4 C-H bonds and 2 0=0 bonds be broken and 2 C=O bonds and 4 O-H bonds be formed. For formaldehyde: CH20(g) + 02(g) � CO2(g) + H20(l) which requires that 2 C-H bonds, 1 C=O bond and 1 0=0 bond be broken and 2 C=O bonds and 2 O-H bonds be formed. Methane contains more C-H bonds and fewer C=O bonds than formaldehyde. Since C-H bonds take less energy to break than C=O bonds, more energy is released in the combustion of methane than of formaldehyde.

9.37

To find the heat of reaction, add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. Reactant bonds broken : I (C=C) + 4(C-H) + I (CI-CI) = ( I mol) (614 kJ/mol) + (4 mol) (413 kJ/mol) + (l mol) (243 kJ/mol) = 2509 kJ Product bonds formed: 1(C-C) + 4(C-H) + 2(C-CI) = ( \ mol) (-347 kJ/mol) + (4 mol) (--413 kJ/mol) + (2 mol) (-339 kJ/mol) = -2677 kJ !!.H° rxn = !!.H° bonds bro ken + !!.H° bonds formed = 2509 kJ + (-2677 kJ) = -168 kJ Note: It is correct to report the answer in kJ or kJ/mol as long as the value refers to a reactant or product with a molar coefficient of 1.

114

9.39

The reaction: H

I I H

H -- C -- O -- H + : C = I C-O 3 C-H 1 0-H 1 C= O

DJiO bonds bro ken

t:J!° rxn = DJiO bonds bro ken

9.40

_

O:

I �

----1 � ..

:

:

H -- C -- C -- O -- H

I

H

= 1 mol (358 kl/mol) = 3 mol (4 1 3 kllmo!) = 1 mol (467 kJ/mo!) = 1 mol ( 1 070 kl/mo!) = 3 1 34 kJ

DJiO bond s formed

= 3 C-H I C-C 1 C=O I C-O 1 0-H

+ DJiO bond s formed = 3 1 34 kl + (-3 1 56 kJ) = -22 kJ

= 3 mol (--4 1 3 kJ/mo!) = I mol (-347 kllmo!) = 1 mol (-745 kl/mol) = 1 mol (-35 8 kJ/mo!) = 1 mol (--467 kl/mo!) = -3 1 56 kl

Plan: Examine the structures for all substances and assume that all reactant bonds are broken and all product bonds are formed. Add the energy required to break all the bonds in the reactants to the energy released to form all bonds in the product. Remember to use a negative sign for the energy of the bonds formed since bond formation is exothermic. Solution :

\C / H

H

=

/ + \H

C

H

I I H

H

H

I I H

H -- Br : ----... H -- C -- C -- Br :

..

..

Bonds broken: I (C=C) + 4(C-H) + 1 (H-Br) = 6 1 4 + 4(4 1 3) + 363 = +2629 kJ Bonds formed: I (C-C) + 5(C-H) + 1 (C-Br) = (-347) + 5(--4 1 3) + (-276) = -2688 kJ DJiOrxn = DJiO bonds bro ken + DJiO bonds formed = 2629 + (-2688) = -59 kJ

9.4 1

Electronegativity increases from left to right across a period (except for the noble gases) and increases from bottom to top within a group. F luorine (F) and oxygen (0) are the two most electronegative elements. Cesium (Cs) and francium (Fr) are the two least electronegative elements.

9.43

The H-O bond in water is polar covalent, because the two atoms differ in electronegativity which results in an unequal sharing of the bonding electrons. A nonpolar covalent bond occurs between two atoms with identical electronegativities where the sharing of bonding electrons is equal. Although electron sharing occurs to a very small extent i n some ionic bonds, the primary force in ionic bonds is attraction of opposite charges resulting from electron transfer between the atoms.

9.46

a) Si < S < 0, sulfur is more electronegative than silicon since it is located further to the right on the table. Oxygen is more electronegative than sulfur since it is located nearer the top of the table. b) Mg < As < P, magnesium is the least electronegative because it lies on the left side of the periodic table and phosphorus and arsenic on the right side. Phosphorus is more electronegative than arsenic because it is higher on the table.

9.48

The arrow points toward the more electronegative atom. none .. .. c) C -- S b) N - O a) N -- B d)

I

..

S -- O

e)

..

I

N -- H

f)

I

..

Cl -- O

1 15

9.50

The more polar bond will have a greater difference in electronegativity, .-1EN. a) N: EN 3 .0; B : EN 2.0; .-1ENa 1 .0 b) N : EN 3 .0; 0: EN 3 . 5 ; .-1EN b 0.5 c) C: EN 2 . 5 ; S : EN 2.5; .-1EN c 0 d) S : EN 2.5; 0: EN 3 . 5 ; .-1EN d 1 .0 e) N : EN 3 .0; H : EN 2. 1 ; .-1ENe 0.9 f) CI: EN = 3 .0; 0: EN = 3 . 5 ; .-1ENr = 0.5 (a), (d), and (e) have greater bond polarity. =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

9.52

a) B onds in Ss are nonpolar covalent. All the atoms are nonmetals so the substance is covalent and bonds are nonpolar because all the atoms are of the same element and thus have the same electronegativity value. b) B onds in RbCI are ionic because Rb is a metal and CI is a nonmetal. c) B onds in PF 3 are polar covalent. All the atoms are nonmetals so the substance is covalent. The bonds between P and F are polar because their electronegativity differs. (By 1 .9 units for P-F) d) B onds in SCI2 are polar covalent. S and CI are nonmetals and differ in electronegativity. (By 0.5 unit for S-CI) e) Bonds in F2 are nonpolar covalent. F is a nonmetal. Bonds between two atoms of the same element are nonpolar since .-1EN = O. f) Bonds in S F2 are polar covalent. S and F are nonmetals that differ in electronegativity. (By 1 .5 units for S-F) I ncreasing bond polarity: SCI2 < SF2 < PF 3

9.54

I ncreasing ionic character occurs with increasing .-1EN. a) .-1ENH B r 0.7, .-1EN HC1 0.9, .-1EN H I = 0.4 b) .-1EN HO 1 .4, .-1EN cH = 0.4, .-1EN HF 1.9 c) .-1ENscl 0.5 , .-1EN pCI 0.9, .-1EN s i CI 1 .2 =

=

=

=

a) b) c) 9.58

=

=

=

I

..

I

..

I

..

H -- I

H-C S -- C I

< < <

I

..

I

..

I

..

<

H -- Br H -O P -- CI

I

..

I

..

H -- C I

<

H -- F

<

S i -- C I

t1H - 1 259 kllmol a) C2H2 + 5/2 O2 � 2 CO2 + H20 5 H-C == C-H + - 0=0 � H-O-H + 2 O C O 2 5 t1Hrxn [2 B Ec-H + BEcEc + "2 B E o, ] + [4 (-B Ec=o) + 2 (-B Eo-H)] =

=

=

=

- 1 259 kl [2(4 1 3) + B EcEc + 5/2 (498)] + [4(-799) + 2(--467)] - 1 259 kl [826 + B EcEc + 1 245] + [--4 1 30)] kl B EcEc = 800. kllmol Table 9.2 lists the value as 839 kllmol I mol C 2 H 2 - 1 259 kl 4 4 = -2.4 1 74347 X 1 0 = -2.4 1 7 X 1 0 kl b) heat (kJ) (500.0 g C 2 H 2 ) 26.04 g C 2 H 2 1 mol C 2 H 2 1 mol C 2 H 2 2 mol CO 2 44.01 g CO 2 c) CO2 produced (g) (500.0 g C 2 H 2 ) I mol CO 2 26.04 g C 2 H 2 1 mol C 2 H 2 1 690.092 1 690. g CO2 (5/2) mol O 2 1 mol C 2 H 2 d) mol O2 ( 500.0 g C 2 H 2 ) 48.00307 mol O2 (unrounded) 26.04 g C 2 H 2 1 mol C 2 H 2 L atm ( 48.00307 mol O 2 ) 0.08206 ( 298 K ) mol ' K = 65 .2 1 45 = 65 .2 L 02 V = nRT / P = 1 8.0 atm =

=

(

=

=

=

=

=

(

(

J(

)(

)( (

•

1 16

J

J )

)(

=

)

9.59

I ) Mg(s) � Mgtg1 2) 1 12 CI2(g) � GIW 3 ) Mgtg1 � Mg+(g) + e4) GIW + e- � Gl-(g) 5) Mg+(g) + Gl-fgt � MgCI(s)

a)

MIlO 1 48 kJ MI; 1 /2 (243 kJ) MI; 738 kJ MI; = -349 kJ MI; -783 .5 kJ ( MI,:";,, (MgCI)) =

=

=

=

=

6) Mg(s) + 1 12 C I2(g) � MgCI(s) MI; (MgCI) ? MI; (MgCI) MI lo + MI; + MI; + MI; + MI; 1 48 kJ + 1 12 (243 kJ) + 738 kJ - 349 kJ - 783.5 kJ -1 25 kJ b) Yes, since f1Ho f for MgCI is negative, MgCI(s) is stable relative to its elements. c) 2 MgCI(s) � MgCI2(s) + Mg(s) Mlo [ I mol ( MI; , MgCI2(s)) + I mol ( MI; , Mg(s))] - [2 mol ( MI; , MgCI(s))] [ I mol (-64 1 .6 kJ/mol) + I mol (0)] - [2 mol (- 1 25 kJ/mol)] -39 1 .6 -392 kJ d) No, MI; for MgCl2 is much more negative than that for MgCI. This makes the Mlo value for the above reaction very negative, and the formation of MgCl2 would be favored. =

=

=

=

=

=

=

9.60

=

a) Find the bond energy for an H-I bond from Table 9.2. Calculate wavelength from this energy using the relationship from Chapter 7: E hc I A. Bond energy for H-I is 295 kJ/mol (Table 9.2). ( 6.626 x 1 0-3 4 J o s ) ( 3 .00 x 1 08 m/s ) I nm -A_ - hc I E -_ 9 - -_ 405 .7807 -_ 406 nm 1 0- m mol kJ 1 0 3 J 295 - -mol 1 kJ 6.022 x 1 0 2 3 =

) [ )[

(

)

(

)

b) Calculate the energy for a wavelength of254 nm and then subtract the energy from part a) to get the excess energy. 19 1 03 J mol X 1 0E (HI) 295 � 4. 8987 J (unrounded) mol 1 kJ 6.022 x 1 0 2 3 ( 6.626 X 1 0-3 4 J s )( 3.00 x 1 08 m/s ) 1 n m 19 -7 . 82598 X 1 0- J (unrounded) E (254 nm) hc I A. 9 254 nm 1 0- m 9 19 19 19 1 X 1 0X 1 0Excess energy 7 . 82598 x 1 0- J - 4.8987 X 1 0- J 2.92728 2.93 J 2 c) Speed can be calculated from the excess energy since Ek 1 /2 mu . 1 �g mol 1 .008 g 2 2E 2 1 .67386 X 1 0- 7 kg (unrounded) m Ek 1 I2mu thus, u 2 3 mol m 6.022 x 1 0 10 g 19 2(2.92728 X 1 0-7 J) kg m 2 /s 2 4 4 1 .870 1 965 x 1 0 1 .87 x 1 0 m/s u 2 J 1 .67386 x 1 0- kg =

) [ )[

(

)

[

=

0

=

=

=

=

=

)

=

[

=

9.63

( 0

)

H)[ =

)

=

=

=

=

)[ )

=

=

Find the appropriate bond energies in Table 9.2. Calculate the wavelengths using E hc I A. C-Cl bond energy 3 3 9 kJ/mol ( 6.626 x 1 0-34 J s ) ( 3 .00 X 1 08 m/s ) = 3.53 I I 3 x I O-7 3 53 x I 0-7 m A= 3 � � mol 339 2 mol 1 kJ 6.022 x 1 0 3 =

=

(

) [ )[ 0

)

=

1 17

.

O2 bond energy = 498 kllmol ( 6.626 x 1 0-34 ] s )( 3 .00 x 1 08 m/s ) = 2.40372 x I 0-7 = 2.40 x I 0-7 m A= 3 � 10 1 mol 498 mol 1 kl 6.022 x 1 0 2 3

)

( . ) [ )[ 0

9.64

Write balanced chemical equations for the formation of each of the compounds. Obtain the bond energy of fluorine from Table 9.2 ( 1 59 kl/mol). Determine the average bond energy from !1H = bonds broken + bonds formed. Remember that the bonds formed (Xe-F) have negative values since bond formation is exothermic. XeF2 Xe(g) + F2 (g) --7 X e F2 ( g) !1H = [ 1 mol F 2 ( 1 59 kllmol)] + [2 (-Xe-F)] = - 1 05 kllmol Xe-F = 1 32 kllmol XeF4 X e(g) + 2 F2 (g) --7 XeF4(g) !1H = [2 mol F 2 ( 1 59 kl/mol)] + [4 (-Xe-F)] = -284 kllmol Xe-F = 1 50.5 = 1 50. kllmol Xe(g) + 3 F 2 (g) --7 XeF6(g) XeF6 !1H = [3 mol F 2 ( 1 59 kllmol)] + [6 (-Xe-F)] = -402 kllmol Xe-F = 1 46.5 = 1 46 kllmol

9.66

a) The presence of the very electronegative fluorine atoms bonded to one of the carbons makes the C-C bond polar. This polar bond will tend to undergo heterolytic rather than homolytic cleavage. More energy is required to force homolytic cleavage. b) Since one atom gets both of the bonding electrons in heterolytic bond breakage, this results in the formation of ions. In heterolytic cleavage a cation is formed, involving ionization energy; an anion is also formed, involving electron affinity. !1H = (homolytic cleavage + electron affinity + first 3 ionization energy) !1H = (249.2 kllmol + (- 1 4 1 kl/mol) + 1 .3 1 x 1 0 kllmol) = 1 420 kJ/mol It would require 1 420 kJ to heterolytically cleave 1 mol of O2.

9.69

Use the equations E = hv, and E = hc I A. mol kl 1 0 3 l 347 mol 1 kJ 6.022 x 1 023 E = 8.6963556 X 1 0 1 4 = 8.70 X 1 0 1 4 S- 1 V - - 4 h 6.626 X 1 0-3 l o s ( 6.626 x 1 0-34 l s )( 3 .00 x 1 08 m/s ) _ A = hc I E = = 3 .44972 x I 0_7 = 3 .45 x I 0 7 m 3 � 10 1 mol 347 mol 1 kJ 6.022 X 1 0 2 3 This is in the ultraviolet region of the electromagnetic spectrum. _

_

(

)[ )[ (

9.7 1

)

) [ )[

)

0

a) 2 CH4(g) + 0 2 (g) --7 CH 3 0CH 3 (g) + H 20(g) !1Hrxn = LB Ereactants + LB E prod ucts !1Hrxn =[ 2 x 4 B Ec-H + BEo�o)] + [6 (-BEc-H) + 2 (-BEc-a) + 2 (-B EO-H)] !1Hrxn = [8 (4 1 3 kJ) + 498 kl] + [6 (-4 1 3 kJ) + 2 (-3 5 8 kJ) + 2 (-467)] !1Hrxn = -326 kl 2 CH4(g) + 0 2 (g) --7 CH 3 CH 2 0H(g) + H 2 0(g) !1Hrxn = LB Ereactants + LB E prod ucts !1Hrxn = [2 x 4 B Ec_H + BEo=a )] + [5(- B Ec_H) + (-B Ec-e) + (-BEc-a) + 3(- B E o-H)] !1Hrxn = [8 (4 1 3 kl) + 498 k]+ [5 (-4 1 3 kl) + (-347 kl) + (-35 8 kl) + 3 (-467)] !1Hrxn = -369 kl b) The formation of gaseous ethanol is more exothermic. c) !1Hrxn = -326 kJ - (-369 kJ) = 43 kJ

1 18

Chapter 1 0 The Shap es of Molecules F O L LOW-U P PRO B L E M S

1 0. 1

Plan: Follow the steps outlined in the sample problem. Solution : a) The sulfur is the central atom, as the hydrogen is never central. Each of the hydrogens is placed around the sulfur. The actual positions of the hydrogens are not important. The total number of valence electrons available is [2 x H( l e-)] + [ I x S(6 e-)] 8 e-. Connect each hydrogen atom to the sulfur with a single bond. These bonds use 4 of the electrons leaving 4 electrons. The last 4 e- go to the sulfur because the hydrogens can take no more electrons. =

H-S :

I

H Check: Count the electrons. The sulfur has an octet (4 from two lone pairs and 4 from two bonding pairs). Each hydrogen has its 2 electrons (from the bonding pair). Solution : b) The oxygen has the lower group number so it is the central atom. Each of the fluorine atoms will be attached to the central oxygen. The total number of valence electrons available is [2 x F(7 e-)] + [ 1 x 0(6 e-)] 20 e-. Connecting the two fluorine atoms to the oxygen with single bonds uses 2 x 2 = 4 e-, leaving 20 - 4 = 1 6 e-. The more electronegative fluorine atoms each need 6 electrons to complete their octets. This requires 2 x 6 1 2 e-. The 4 remaining electrons go to the oxygen. =

=

: F-O :

I

: F: Check: Count the electrons. Each of the three atoms has an octet. Solution : c) Both S and ° have a lower group number than CI, thus, one of these two elements must be central. Between S and 0, S has the higher period number so it is the central atom. The total number of valence electrons available is [2 x CI(7 e-)] + [1 x S(6 e-)] + [ 1 x 0(6e-)] 26 e-. Begin by distributing the two chlorine atoms and the oxygen atom around the central sulfur atom. Connect each of the three outlying atoms to the central sulfur with single bonds. This uses 3 x 2 = 6 e-, leaving 26 - 6 20 e-. Each of the outlying atoms stil l needs 6 electrons to complete their octets. Completing these octets uses 3 x 6 1 8 electrons. The remaining 2 electrons are al l the sulfur needs to complete its octet. =

=

=

: CJ :

I .. l /"�. . 0. .: :� S

Check: Count the electrons. Each of the four atoms has an octet. Note: Later we will see that the presence of sulfur may lead to some complications.

1 19

1 0.2

Plan: Follow the steps outlined in the sample problem. Solution : a) As in C H40, the N and 0 both serve as "central" atoms. The N is placed next to the 0 and the H 's are distributed around them. The N needs more electrons so it gets two of the three hydrogens. You can try placing the N in the center with all the other atoms around it, but you will quickly see that you will have trouble with the oxygen. The number of valence electrons is: [3 x H( l e-)] + [ 1 x N(5 e-)] + [ 1 x 0(6 e-)] = 1 4 e-. Four single bonds are needed (4 x 2 = 8 e-). This leaves 6 electrons. The oxygen needs 4 electrons, and the nitrogen needs 2. These last 6 electrons serve as three lone pairs. H

I I

.

H - N - O- H

H-N-O '

H correct

incorrect

I

H

Solution: b) The hydrogens cannot be the central atoms. The problem states that there are no O-H bonds, so the oxygen must be connected to the carbon atoms. Place the 0 atom between the two C ' s, and then distribute the H ' s equally around each of the C 's. The total number of valence electrons is [6 x H( I e-)] + [2 x C(4 e-)] + [ I x 0(6 e-)] 20 e- Draw single bonds between each of the atoms. This creates six C-H bonds, and two C-O bonds, and uses 8 x 2 1 6 electrons. The four remaining electrons will become two lone pairs on the 0 atom. H H =

=

I I H

I I H

H - C -- O -- C - H Check: Count the electrons. Both the C ' s and the 0 have octets. Each H has its pair. 1 0.3

P lan: Follow the steps in the example, and pay attention to the hint for CO. Solution: a) In CO there are a total of [ I x C(4 e-)] + [ I x 0(6 e-)] = 1 0 e-. The hint states that carbon has three bonds. Since oxygen is the only other atom present, these bonds must be between the C and the O. This uses 6 of the 1 0 electrons. The remaining 4 electrons become two lone pairs, one pair for each of the atoms. : C == O : Check: Count the electrons. Both the C and the 0 have octets. Solution: b) In HCN there are a total of [ 1 x H( 1 e-)] + [ 1 x C(4 e-)] + [ 1 x N(5 e-)] = 1 0 electrons. Carbon has a lower group number, so it is the central atom. Place the C between the other two atoms and connect each of the atoms to the central C with a single bond. This uses 4 of the 1 0 electrons, leaving 6 electrons. Distribute these 6 to nitrogen to complete its octet. However, the carbon atom is 4 electrons short of an octet. Change two lone pairs on the nitrogen atom to bonding pairs to form two more bonds between carbon and nitrogen for a total of three bonds. H-C=N :

Check: Count the electrons. Both the C and the N have octets. The H has its pair. Solution: c) I n CO2 there are a total of [ 1 x C(4 e-)] + [2 x 0(6 e-)] 1 6 electrons. Carbon has a lower group number, so it is the central atom. P lacing the C between the two O's and adding single bonds uses 4 electrons, leaving 1 6 - 4 12 e-. Distributing these 1 2 electrons to the oxygen atoms completes those octets, but the carbon atom does not have an octet. Change one lone pair on each oxygen atom to a bonding pair to form two double bonds to the carbon atom, completing its octet. =

=

O === C ===g Check: Count the electrons. Both the C and the O's have octets.

1 20

l OA

Plan: Modify the structure shown in the text to produce additional resonance structures. Solution: The structure drawn in the text is shown below on the right. Just shift the double bond and electron pairs. 222: 0 :0: : 0:

I C

"

0/ .

""0 .

..

II C

..

' / "".0" . .

: .0. .

"

. .

•

.

1 0.5

..

•

I

..

C

.' / � " "0

_ . 0. .

Plan: The presence of available d orbitals makes checking formal charges more important. Solution: a) In POCI 3 , the P is the most l ikely central atom because all the other elements have higher group numbers. The molecule contains: [ I x P(5 e-)] + [ 1 x 0(6 e-)] + [3 x CI(7 e-)] = 32 electrons. Placing the P in the center with single bonds to all the surrounding atoms uses 8 electrons and gives P an octet. The remaining 24 can be split into 1 2 pairs with each of the surrounding atoms receiving three pairs. At this point, structure I below, all the atoms have an octet. The central atom is P (smallest group number, highest period number) and can have more than an octet. To see how reasonable this structure is, calculate the formal (FC) for each atom. The + I and -1 formal charges are not too unreasonable, however, 0 charges are better. If one of the lone pairs is moved from the ° (the atom with the negative FC) to form a double bond to the P (the atom with the positive FC), structure II results. The calculated formal changes in structure II are all 0, thus, this is a better structure even though P has 1 0 electrons. : 0: : 0: : CI

I .. . CII :

--

P -- C I :

: CI

•

I

--

IIP ..CI: I --

II

FCp = 5 - (0 + 1 12( 1 0» = 0 FCp = 5 - (0 + 1 12(8» = + 1 FCo = 6 - (6 + 1 /2(2» = - 1 FCo = 6 - (4 + 1 12(4» = 0 FCcl = 7 - (6 + 1 12(2» = 0 FCCI = 7 - (6 + 1 12(2» = 0 Check: Count the electrons. All the atoms in structure I I , except P, have an octet. The P can have more than an octet, because it has readily available d orbitals. Solution: b) In C102, the C I is probably the central atom because the O's have a lower period. The molecule contains: [ 1 x CI(7 e-)] + [2 x 0(6 e-)] = 19 electrons. The presence of an odd number of electrons means that there will be an exception to the octet rule. Placing the O's about the CI and using 4 electrons to form single bonds leaves 1 5 electrons, 1 4 of which may be separated into 7 pairs. If 3 of these pairs are given to each 0, and the remaining pair plus the lone electron are given to the CI, we have the following structure: : O

-

CI -- O :

FCo = 6 - (6 + 1 /2(2» = - 1 FCCI = 7 - (3 + 1 12(4» = +2 Calculating formal charges: The +2 charge on the C I is a little high, so other structures should be tried. Moving a lone pair from one of the ° atoms (negative FC) to form a double bond between the CI and one of the oxygens gives either structure I or II below. If both O ' s donate a pair of electrons to form a double bond, then structure I I I results. The next step is to calculate the formal charges.

121

CI

· · O� · ..

..

:-.�.

•

___ CI � . . -':::: 0 ·

. o. . .

•

_ CI __

•

· O� · �O . .

.

III

II

I

•

.

FCc l 7 - (3 + 1 /2(6)) + I FCc l 7 - (3 + 1 12(8)) 0 FCCI 7 - (3 + 1 /2(6)) + 1 FCo 6 - (4 + 1 12(8)) 0 Left FCo 6 - (4 + 1 /2(4)) 0 FCo 6 - (6 + 1 /2(4)) -1 FCo 6 - (4 + 1 /2(8)) 0 Right FCo 6 - (6 + 1 12(2)) -1 FCo 6 - (4 + 1 12(2)) 0 Check: Pick the structure with the best distribution of formal charges (structure I I I ). Solution : c) XeF4 is a noble gas compound, thus, there will be an exception to the octet rule. The molecule contains: [ 1 x Xe(8 e-)] + [4 x F(7 e-)] 36 electrons. The Xe is in a higher period than F so Xe is the central atom. lf it is placed in the center with a single bond to each of the four fluorine atoms, 8 electrons are used, and 28 electrons remain. The remaining electrons can be divided into 1 4 pairs with 3 pairs given to each F and the last 2 pairs being given to the Xe. This gives the structure: .. : . F. ,....... . . / .F. : /Xe . . ", . . .. F: :F =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

=

• •

Check: Determine the formal charges for each atom: FCF 7 - (6 + 1 12(2)) 0 FCxe 8 - (4 + \ /2(8)) 0 =

=

1 0.6

=

=

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution : a) The Lewis structure for CS2 is shown below. The central atom, C, has two pairs (double bonds only count once). The two pair arrangement is li near with the designation, AX2• The absence of lone pairs on the C means there is no deviation in the bond angle (1 80°). S ==== C ==== S Check: Compare the results in the figures in the chapter. Solution : b) Even though this is a combination of a metal with a nonmetal, it may be treated as a molecular species. The Lewis structure for PbCI 2 is shown below. The molecule is of the AX2E type; the central atom has three pairs of electrons ( 1 lone pair and two bonding pairs. This means the electron-group arrangement is trigonal planar ( 1 2 0°), with a lone pair giving a bent or V-shaped molecule. The lone pair causes the ideal bond angle to decrease. : CI -- Pb

I : CI :

coPb

.. / �.. : CI

CI :

Check: Compare the results in the figures in the chapter.

1 22

Solution : c) The Lewis structure for the CBr4 molecule is shown above. [t has the AX4 type formula which is a perfect tetrahedron (with 1 09.5° bond angles) because all bonds are identical, and there are no lone pairs. " "" " : Br " : Br"

I "I

"

: Br -- C -- Br"

I

\C B" "r " ' \ \ \\ � " " Br: . .: Br.

"

:

I

. .

" Sr"" "" Check: Compare the results in the figures in the chapter. Solution: d) The S F 2 molecule has the Lewis structure shown below. This is a AX2 E 2 molecule; the central atom is surrounded by four electron pairs, two of which are lone pairs and two of which are bonding pairs. The electron group arrangement is tetrahedral . The two lone pairs give a V-shaped or bent arrangement. The ideal tetrahedral bond angle is decreased from the ideal 1 09.5° value.

(j B S . / � F F

" 0

: F -- S :

"I

o

. F:

. ..

',

"/

.

0

. .

. ..

Check: Compare the results in the figures in the chapter. 1 0. 7

Plan: Draw a Lewis structure. Determine the electron arrangement by counting the electron pairs around the central atom. Solution: a) The Lewis structure for the ICI2- is shown below. This is a AX2 E3 type structure. The five pairs give a trigonal b ipyramidal arrangement of electron groups. The presence of 3 lone pairs leads to a linear shape ( 1 80°) . The usual distortions caused by lone pairs cancel in this case. In the trigonal bipyramidal geometry, lone pairs always occupy equatorial positions. " "

[ : CI ..

.. . -CI:

" 1°_ 0 .

__

0

" Cl o

�I� 0f �

]-

: .C:1:

Check: Compare the results in the figures in the chapter. Solution: b) The Lewis structure for the CIF3 molecule is given below. Like ICI 2- there are 5 pairs around the central atom; however, there are only 2 lone pairs. This gives a molecule that is T-shaped. The presence of the lone pairs decreases the ideal bond angles to less than 90°.

:

F -- C I.-- F : .

:

. .

.

I �

:

: F:

� CII F : \9: FI : .. --

Check: Compare the results in the figures in the chapter.

1 23

Solution: c) The SOF4 molecule has several possible Lewis structures, two of which are shown below. In both cases, the central atom has 5 atoms attached with no lone pairs. The formal charges work out the same in both structures. The structure on the right has an equatorial double bond. Double bonds require more room than single bonds, and equatorial positions have this extra room. : 0: : F: :F: : F .. 'F : , .. "p" \\ \ .. \ \ ' \ \ F -- S\\ .. \\ + .O==S . S == O . .F . . F� . ' /'" : : F : F: : .F. : .

'

•

.

.

I I

..

.

•

II I �F.:

..

I "" I

The molecule is trigonal bipyramidal, and the double bond causes deviation from ideal bond angles. All of the F atoms move away from the O. Thus, all angles involving the 0 are increased, and all other angles are decreased. Check: Compare the results in the figures in the chapter. 1 0.8

Plan: Draw the Lewis structure for each of the substances, and determine the molecular geometry of each. Solution : a) The Lewis structure for H2S04 is shown below. The double bonds ease the problem of a high formal charge on the sulfur. Sulfur is al lowed to exceed an octet. The S has 4 groups around it, making it tetrahedral. The ideal angles around the S are 1 09.5°. The double bonds move away from each other, and force the single bonds away. This opens the angle between the double bonded oxygens, and results in an angle between the single bonded oxygens that is less than ideal. : 0: :0:

. O -- ISI == O ·· H/ I .. 0: H/

.

, \ \\\\�==o

H�· ' .0 I H -O · .

.

.

.

..

. .

Solution: b) The hydrogens cannot be central so the carbons must be attached to each other. The problem states that there is a carbon-carbon triple bond. This leaves only a single bond to connect the third carbon to a triple-bonded carbon, and give that carbon an octet. The other triple-bonded carbon needs one hydrogen to complete its octet. The remaining three hydrogens are attached to the single-bonded carbon, which allows it to complete its octet. This structure is shown below. The single-bonded carbon has four groups tetrahedrally around it leading to bond angles -1 09.5° (little or no deviation). The triple-bonded carbons each have two groups (the triple bond counts as one electron group) so they should be linear ( 1 80°) . H H H . ' 1/ , '" -11 / C --- C === C -- H H --- C --- C ==== C -- H

1 I H

-

/

H

1 24

Solution: c) Fluorine, like hydrogen, is never a central atom. Thus, the sulfurs must be bonded to each other. Each F ends with 3 lone pairs, and the sulfurs end with 2 lone pairs. All 4 atoms now have an octet. This structure is shown below. Each sulfur has four groups around it, so the electron arrangements is tetrahedral. The presence of the lone pairs on the sulfurs results in a geometry that is V-shaped or bent and in a bond angle that is < 1 09.5°. : F: F:

I

: S -- S :

I : F:

1 0.9

-- S / . . (Q S() () Q) : .F. / •

Plan: Draw the Lewis structures, predict the shapes, and then examine the positions of the bond dipoles. Solution: a) Dichloromethane, C H2CI2, has the Lewis structure shown below. It is tetrahedral, and if the outlying atoms were identical, it would be nonpolar. However, the chlorines are more electronegative than hydrogen so there is a general shift in their direction resulting in the arrows shown. H

I ICI: ..

H - C - Cl: :

: CI :

iC H �t � . : CI H

/

.

Solution: b) Iodine oxide pentafluoride, rOF s , has the Lewis structure shown below. The overall geometry is octahedral. All six bonds are polar, with the more electronegative 0 and F atoms shifting electron density away from the L The 4 equatorial fluorines counterbalance each other. The axial F is not equivalent to the axial O. The more electronegative F results in an overall polarity in the direction of the axial F .

« ��):

:·F/ I��: : �

t

F ",

i

..- F

F/ � F F

The lone electron pairs are left out for simplicity. Solution: c) Nitrogen tribromide, NBr3 , has the trigonal pyramidal Lewis structure shown below. The nitrogen is more electronegative than the B r, so the shift in electron density is towards the N, resulting in the dipole indicated below.

o

: Br -- N -..Br :

..

.. ;1

I : Br:

Br

N��: : 1 Br :

.

•

END-OF-CHAPTER PROBLEMS

1 0. 1

To be the central atom in a compound, the atom must be able to simultaneously bond to at least two other atoms. He, F, and H cannot serve as central atoms in a Lewis structure. H elium ( I i) is a noble gas, and as such, it does not need to bond to any other atoms. Hydrogen ( l S i ) and fluorine ( l i2s22p s ) only need one electron to complete their valence shells. Thus, they can only bond to one other atom, and they do not have d orbitals available to expand their valence shells. 1 25

1 0. 3

F o r an element t o obey the octet rule i t must b e surrounded b y 8 electrons. To determine the number of electrons present ( 1 ) count the individual electrons actually shown adj acent to a particular atom, and (2) add two times the number of bonds to that atom. Using this method the structures shown give : (a) 0 + 2(4) = 8 ; (b) 2 + 2(3) = 8; (c ) 0 + 2(5) = 1 0; ( d ) 2 + 2(3) = 8 ; ( e) 0 + 2(4) = 8 ; (f) 2 + 2(3) = 8 ; (g ) 0 + 2(3 ) = 6; (h) 8 + 2(0) = 8 . All the structures obey the octet rule except: c and g.

1 0. 5

Plan: Count the valence electrons and draw Lewis structures. Solution: Total valence electrons: SiF4: [1 x Si(4e-] + [4 x F(7e-)] = 32; SeC\z: [1 x Se(6e-)] + [2 x CI(7 e-)] = 20; COF2 : [1 x C(4e-)] + [1 x 0(6e-)] + [2 x F(7e-)] = 24. The Si, Se, and the C are the central atoms, because these are the elements in their respective compounds with the lower group number (in addition, we are told C is central). Place the other atoms around the central atoms and connect each to the central atom with a single bond. SiF4: At this point, 8 electrons (2e- in 4 Si-F bonds) have been used with 32 - 8 = 24 remaining ; the remainin g electrons are placed around the fluorines (3 pairs each). All atoms have an octet. SeC\z : The 2 bonds use 4 e- (2e- in 2 Se-CI bonds) leaving 20 - 4 = 1 6 e-. These 1 6 e- are used to complete the octets on Se and the CI atoms. COF 2 : The 3 bonds to the C use 6 e- (2e- in 3 bonds) leaving 24 - 6 = 1 8 e-. These 1 8 e- are distributed to the surrounding atoms first to complete their octets. After the 1 8 e- are used, the central C is 2 electrons short of an octet. Forming a double bond to the 0 (change a lone pair on 0 to a bonding pair on C) completes the C octet. (a) SiF4 : (b) SeCI 2 : F: : F

--

I I

Si

--

F: : C I -- Se -- C I :

:F: (c) COF 2

: F

--

C -- F :

:F

I

: 0:

1 0. 7

--

C

II

--

F:

: 0:

Plan: Count the valence electrons and draw Lewis structures. Solution: a) PF3 : [ 1 x P(5 e-)] + [3 x F(7e-)] = 26 valence electrons. P is the central atom. Draw single bonds from P to the 3 F atoms, using 2e- x 3 bonds = 6 e-. Remaining e-: 26 - 6 = 20 e-. Distribute the 20 e- around the P and F atoms to complete their octets. b) H 2 C03 : [2 x H(2e-)] + [ 1 x C(4e-)] + 3 x 0(6e-)] =24 valence electrons. C is the central atom with the H atoms attached to the 0 atoms. Place appropriate single bonds between all atoms using 2e- x 5 bonds = 1 0 e- so that 24 - 1 0 = 1 4 e- remain. Use these 1 4 e- to complete the octets of the 0 atoms (the H atoms already have their two electrons). After the 1 4 e- are used, the central C is 2 electrons short of an octet. Forming a double bond to the 0 that does not have a H bonded to it (change a lone pair on 0 to a bonding pair on C) completes the C octet. c) CS 2 : [ I x C(4 e-)] + [2 x S(6 e-)] = 1 6 valence electrons. C is the central atom. Draw single bonds from C to the 2 S atoms, using 2e- x 2 bonds = 4 e-. Remaining e-: 1 6 - 4 = 1 2 e-. Use these 1 2 e- to complete the octets of the surrounding S atoms; this leaves C 4 electrons short of an octet. Form a double bond from each S to the C by changing a lone pair on each S to a bonding pair on C.

1 26

: F-P

I

--

··o·

F :

H/

: F:

_-

C

I

· · ·· O

-

""H

: 0:

c) CS2 ( 1 6 valence e-) S === C === S

1 0.9

Plan: The problem asks for resonance structures, so there must be more than one answer for each part. Solution : a) NO / has [ 1 x N(5 e-)] + [2 x 0(6 e-)] - I e- (+ charge) 1 6 valence electrons. Draw a single bond from N to each 0, using 2e- x 2 bonds 4 e-; 1 6 - 4 1 2 e- remain. Distribute these 1 2 e- to the 0 atoms to complete their octets. This leaves N 4 e- short of an octet. Form a double bond from each 0 to the N by changing a lone pair on each 0 to a bonding pair on N. No resonance is required as all atoms can achieve an octet with double bonds. =

=

[: �- N - �:r

=

[O�N�ol+ . . .

�

b ) N02F has [ I x N ( 5 e-)] + [ 2 x 0(6 e-)] + [ 1 x F(7e-)] 2 4 valence electrons. Draw a single bond from N to each surrounding atom, using 2e- x 3 bonds = 6 e-; 24 - 6 = 1 8 e- remain. Distribute these 1 8 e- to the 0 and F atoms to complete their octets. This leaves N 2 e- short of an octet. Form a double bond from either 0 to the N by changing a lone pair on 0 to a bonding pair on N. There are two resonance structures since a lone pair from either of the two 0 atoms can be moved to a bonding pair with N : =

:F:

I

.

N

. ./'" � . .

. 0. .

1 0. 1 1

: F:

: F: . . ./'"

. 0.

O.

.

.

.

NI

..

� ..

.

. .o�

.

0

.

NI

� .. . 0 .

.

.

Plan: Count the valence electrons and draw Lewis structures. Additional structures are needed to show resonance. Solution: a) N 3- has [3 x N(5e-)] + [ 1 e-(from charge)] = 1 6 valence electrons. Place a single bond between the nitrogen atoms. This uses 2e- x 2 bonds 4 electrons, leaving 1 6 - 4 1 2 electrons (6 pairs). Giving 3 pairs on each end nitrogen gives them an octet, but leaves the central with only 4 electrons as shown below:

N

=

[ : � -H ·

.

-

N

=

.

The central N needs 4 electrons. There are three options to do this: ( I ) each of the end N ' s could form a double bond to the central N by sharing one of its pairs; (2) one of the end N ' s could form a triple bond by sharing two of its lone pairs; (3) the other end N could form the triple bond instead.

[.�= N =� r

.. ·

·

1 27

b) N02- has [ I x N(5e-)] + [2 x 0(6 e-)] + [ I e- (from charge)] 1 8 valence electrons. The nitrogen should be the central atom with each of the oxygens attached to it by a single bond (2e- x 2 bonds = 4 electrons). This leaves 1 8 - 4 1 4 electrons (7 pairs). If 3 pairs are given to each 0 and I pair is given to the N, then both O ' s have an octet, but the N only has 6. To complete an octet the N needs to gain a pair of electrons from one 0 or the other (from a double bond). The resonance structures are: =

=

1 0. 1 3

H-N-�r

Plan: Initially, the method used in the preceding problems may be used to establish a Lewis structure. The total of the formal charges must equal the charge on an ion or be equal to 0 for a compound. The FC only needs to be calculated once for a set of identical atoms. Solution: a) I Fs has [ I x [(7 e-)] + [5 x F(7 e-)] 42 valence electrons. The presence of 5 F ' s around the central I means that the I will have a minimum of 1 0 electrons; thus, this is an exception to the octet rule. The 5 I-F bonds use 2e- x 5 bonds 1 0 electrons leaving 42 - 1 0 32 electrons ( 1 6 pairs). Each F needs 3 pairs to complete an octet. The 5 F ' s use 1 5 of the 1 6 pairs, so there is I pair left for the central I. This gives: .. : F: : . F. � / F. . : .. I : F -- �· · .. F: =

=

..

.

=

..

I

.

Calculating formal charges: FC valence electrons - (lone electrons + 1 /2 (bonded electrons» For iodine: FC, 7 - (2 + 1 /2( 1 0» 0 For each fluorine: FCF 7 - (6 + 1 12(2» Total formal charge 0 charge on the compound. b) AIH4- has [ I x AI(3 e-)] + [4 x H( I e-)] + [ I e- (from charge)] 8 valence electrons. The 4 AI-H bonds use all the electrons and leave the AI with an octet. H =

=

=

=

=

=

0

=

=

H

-

I I H

AI

--

H

For aluminum : FCA, 3 - (0 + 1 12(8» For each hydrogen FCH I - (0 + 1 /2(2» Total formal charge - I = charge on the ion. =

=

= -

=

0

1

=

1 0. 1 5

Plan: The general procedure is simi lar to the preceding problems, plus the oxidation number determination. Solution : a) Br03- has [ I x Br(7 e-)] + 3 x 0(6 e-)] + [ I e-- (from charge)] 26 valence electrons. Placing the O ' s around the central Br and forming 3 Br-O bonds uses 2e- x 3 bonds 6 electrons and leaves 26 - 6 20 electrons ( 1 0 pairs). Placing 3 pairs on each 0 (3 x 3 9 total pairs) leaves 1 pair for the Br and yields structure 1 below. In structure I, all the atoms have a complete octet. Calculating formal charges: FCB, = 7 - (2 + 1 /2(6» = +2 FCo 6 - (6 + 1 12(2» = -1 The FCo is acceptable, but FCB, is larger than is usually acceptable. Forming a double bond between any one of the O ' s gives Structure I I . Calculating formal charges: FCB, 7 - (2 + 1 /2(8» + 1 FCo 6 - (6 + 1 /2(2» -I FCo 6 - (4 + 1 /2(4» 0 Double bonded 0 =

=

=

=

=

=

=

=

=

1 28

=

=

The FCBr can be improved further by forming a second double bond to one of the other O's (structure I II ) . FCBr = 7 - ( 2 + 1 /2( 1 0)) = 0 FCo = 6 - (6 + 1 /2(2)) = -1 FCo = 6 - (4 + 1 /2(4)) = 0 Double bonded O's Structure I I I has the more reasonable distribution of formal charges .

. ..

. -- 8r=== 0 I

. : O -- Br - O :

..

.0

I : 0:

I: 0 : ..

o === Br === 0

: 0:

II

III

-6 The oxidation numbers (O.N. ) are: O.N . Br = +5 and O.N.o = -2 . +5 -2 Check: The total formal charge equals the charge on the ion (-1 ). Br03b) SO /- has [ 1 x S(6 e-)] + 3 x 0(6 e-)] + [2e- (from charge)] = 26 valence electrons. Placing the O ' s around the central S and forming 3 S-O bonds uses 2e- x 3 bonds = 6 electrons and leaves 26 - 6 = 20 electrons ( 1 0 pairs). Placing 3 pairs on each 0 (3 x 3 = 9 total pairs) leaves 1 pair for the S and yields structure I below. In structure I all the atoms have a complete octet. Calculating formal charges: FCs = 6 - (2 + 1 12(6)) = + 1 ; FCo = 6 - (6 + 1 12(2)) = - 1 The FCo is acceptable, but FCs is larger than is usually acceptable. Forming a double bond between any one of the O's (Structure II) gives: FCo = 6 - (4 + 1 12(4)) = 0 FCs = 6 - (2 + 1 12(8)) = 0 FCo = 6 - (6 + 1 12(2)) = -I Double bonded 0 22O

-- S -- O : I o

�

O

- S -- O : " : 0:

II -6 I Structure I I has the more reasonable distribution of formal charges. +4 -2 2 S03 The oxidation numbers (O.N. ) are: O.N.s = +4 and O.N.o = -2. Check: The total formal charge equals the charge on the ion ( 2 ) -

1 0. 1 7

.

Plan: We know that each of these compounds does not obey the octet rule. Solution: a) B H3 has [ 1 x B(3e-)] + [3 x H( 1 e-)] = 6 valence electrons. These are used in 3 B-H bonds. The B has 6 electrons instead of an octet - electron deficient. b) AsF 4- has [ I x As(5e-)] +[4 x F(7e-)] + [ 1 e- (from charge)] = 34 valence electrons. Four As-F bonds use 8 electrons leaving 34 - 8 = 26 electrons ( 1 3 pairs). Each F needs 3 pairs to complete its octet and the remaining pair goes to the As. The As has an expanded octet with 10 electrons. The F cannot expand its octet.

1 29

c) SeCI4 has [ 1 x Se(6e-)] + 4 x CI(7e-)] = 34 valence electrons. The SeC14 is isoelectronic (has the same electron structure) as AsF 4-, and so its Lewis structure looks the same. Se has an expanded octet of 1 0 electrons.

1

H

1 As 1 : F:

/ 8""

:F

0

--

H

(a) 1 0. 1 9

: CI :

:F:

H

0 0

: CI

0

--

F:

1 I : CI : 0

__ o

Se -- Cl :

(c)

(b)

P lan: We know that each of these compounds does not obey the octet rule. Solution: a) BrF3 has [ 1 x Br(7e-)] + [3 x F(7 e-)] = 28 valence electrons. P lacing a single bond between Br and each F uses 2e- x 3 bonds = 6 e-, leaving 28 - 6 = 22 electrons ( 1 1 pairs). After the F ' s complete their octets with 3 pairs each, the Br gets the last 2 lone pairs. The Br expands its octet to 1 0 electrons. b) ICI2- has [ 1 x I(7e-)] + [2 x Cl(7e-)] + [ l e- (from charge)] = 22 valence electrons. P lacing a single bond between I and each Cl uses 2e- x 2 bond = 4e-, leaving 22 - 4 = 1 8 electrons (9 pairs). After the Cl's complete their octets with 3 pairs each, the iodine finishes with the last 3 lone pairs. The iodine has an expanded octet of 1 0 electrons. c) BeF2, [ 1 x Be(2e-)] + [2 x F(7e-)] = has 1 6 valence electrons. Placing a single bond between Be and each of the F atoms uses 2e- x 2 bonds = 4e-, leaving 16 - 4 = 1 2 electrons (6 pairs).The F's complete their octets with 3 pairs each, and there are no electrons left for the Be. Formal charges work against the formation of double bonds. Be, with only 4 electrons, is electron deficient. oo o Br -- F :

: F -- Be -- F :

0

:F

__

1

:F: (a) 1 0.2 1

(b)

(c)

Plan: Draw Lewis structures for the reactants and products. Solution: Beryllium chloride has the formula BeCI2• BeCI2 has [ I x Be(2e-)] + [2 x CI(7e-)] = 1 6 valence electrons. 4 of these electrons are used to place a single bond between Be and each of the Cl atoms, leaving 1 6 - 4 = 1 2 electrons (6 pairs). These 6 pairs are used to complete the octets of the Cl atoms, but Be does not have an octet - it is electron deficient. Chloride ion has the formula cr with an octet of electrons. BeCI /- has [ I x Be(2e-)] + [4 x CI(7e-)] + [ 2e- (from charge)] = 32 valence electrons. 8 of these electrons are used to place a single bond between Be and each CI atom, leaving 32 - 8 = 24 electrons ( 1 2 pairs). These 1 2 pairs complete the octet of the C I atoms (Be already has an octet). 2: Cl : : CI

--

Be

--

CI:

+

[: �I :r [: �I r

: CI

:

1 30

1 1 : CI :

--

Be

--

Cl o 0

1 0.24

Plan: Use the structures in the text to determine the formal charges. Solution: Structure A : FCc 4 - (0 + 1 12(8» 0; FCo 6 - (4 + 1 12(4» 0; FCCI 7 - (6 + 1 12(2» Total FC 0 Structure B : FCc 4 - (0 + 1 12(8» 0; FC o 6 - (6 + 1 12(2» - 1 ; FCCI 7 - (4 + 1 /2(4» + I (double bonded); FCCI 7 - (6 + 1 12(2» 0 (single bonded) Total FC 0 Structure C : FCc 4 - (0 + 1 12(8» 0; FCo 6 - (6 + 1 /2(2» - I ; FCCI 7 - (4 + 1 12(4» + 1 (double bonded); FCcl 7 - (6 + 1 12(2» 0 (single bonded) Total FC 0 Structure A has the most reasonable set of formal charges. =

=

=

=

=

=

=

=

=

=

=

=

0

=

=

=

=

=

=

=

=

=

=

=

=

=

1 0.26

The molecular shape and the electron-group arrangement are the same when there are no lone pairs on the central atom.

1 0.28

Plan: Examine a list of all possible structures, and choose the ones with four electron groups. Solution: AX4 Tetrahedral AX3 E Trigonal pyramidal AX 2 E 2 Bent or V -shaped

1 0.30

Plan: Begin with the basic structures and redraw them. Solution: a) A molecule that is V -shaped has two bonds and generally has either one (AX2 E) or two (AX2 E 2 ) lone electron pairs. b) A trigonal planar molecule follows the formula AX3 with three bonds and no lone electron pairs. c) A trigonal bipyramidal molecule contains five bonding pairs (single bonds) and no lone pairs (AXs). d) A T-shaped molecule has three bonding groups and two lone pairs (AX 3 E 2 )' e) A trigonal pyramidal molecule follows the formula AX3 E with three bonding pairs and one lone pair. t) A square pyramidal molecule shape follows the formula AXsE with 5 bonding pairs and one lone pair. X " X '\'\' X - A' .,;: A X

/ "'-

X

X

/ "'-

(a) X

Ie 1 '0 X

X -- A

(d)

I

I 'x

X

X

o (e)

131

x.

(c) X

I .X X�G ' X '1

III \\" " A

/

(t)

1 0.32

Plan: First, draw a Lewis structure, and then apply YSEPR. Solution : a) 0 3 : The molecule has [3 x 0(6e-)] = 1 8 valance electrons. 4 electrons are used to place a single bond between the oxygen atoms, leaving 1 8 - 4 = 1 4e- (7 pairs). 6 pairs are required to give the end oxygen atoms an octet; the last pair is distributed to the central oxygen, leaving this atom 2 electrons short of an octet. Form a double bond from one of the end 0 atoms to the central 0 by changing a lone pair on the end 0 to a bonding pair on the central O. This gives the following Lewis structure: : 0 -- 0==== 0

or

..

H, [ H /O��H r

There are three groups around the central 0, one of which is a lone pair. This gives a trigonal planar electron­ group arrangement, a bent molecular shape, and an ideal bond angle of 1 20°. b) H 3 0 + : This ion has [3 x H( l e-)] + [ I x 0(6e-)] - [ I e- (due to + charge] = 8 valence electrons. 6 electrons are used to place a single bond between 0 and each leaving 8 6 = 2e- ( l pair). Distribute this pair to the 0 atom, giving it an octet (the H atoms only get 2 electrons). This gives the following Lewis structure:

H[ -!-t

-

There are four groups around the 0, one of which is a lone pair. This gives a tetrahedral electron-group arrangement, a trigonal pyramidal molecular shape, and an ideal bond angle of 1 09.5°. c) N F 3 : The molecule has [ I x N(5e-)] + [3 x F(7e-)] = 26 valence electrons. 6 electrons are used to place a single bond between N and each F, leaving 26 - 6 = 20 e- ( l O pairs). These 1 0 pairs are distributed to all of the F atoms and the N atoms to give each atom an octet. This gives the following Lewis structure: : F -- N -- F :

. ..F

I.

•

: F·

• •

// N �I': � ': � .. F:

'. .

There are four groups around the N, one of which is a lone pair. This gives a tetrahedral electron-group arrangement, a trigonal pyramidal molecular shape, and an ideal bond angle of 1 09.5°. 1 0.34

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution : 2 a) C0 3 -: This ion has [ 1 x C(4e-)] + [3 x 0(6e-)] + [2e- (from charge)] 24 valence electrons. 6 electrons are used to place single bonds between C and each 0 atom, leaving 24 - 6 = 1 8 e- (9 pairs). These 9 pairs are used to complete the octets of the three 0 atoms, leaving C 2 electrons short of an octet. Form a double bond from one of the 0 atoms to C by changing a lone pair on an 0 to a bonding pair on C. This gives the following Lewis structure: =

[

.

.

.. : � : ..

O

-

C-O :

]

2-

.' '. ' . 2'. 0"" /� . C

II

: 0: +

2 additional resonance forms. There are three groups of electrons around the C, none of which are lone pairs. This gives a trigonal planar electron-group arrangement, a trigonal planar molecular shape, and an ideal bond angle of 1 20°.

1 32

(b) S0 2 : This molecule has [ 1 x S(6 e-)] + [ 2 x S(6 e-)] = 18 valence electrons. 4 electrons are used to place a single bond between S and each 0 atom, leaving 18 - 4 14 e- (7 pairs). 6 pairs are needed to complete the octets of the 0 atoms, leaving a pair of electrons for S. S needs one more pair to complete its octet. Form a double bond from one of the end 0 atoms to the S by changing a lone pair on the 0 to a bonding pair on the S. This gives the fol lowing Lewis structure: · · S .O === S -- O. : =

.

.�"" . . ..

.

. 0

0 . .. .

There are three groups of electrons around the C, one of which is a lone pair. This gives a trigonal planar electron-group arrangement, a bent (V-shaped) molecular shape, and an ideal bond angle of 120°. (c) CF4 : This molecule has [I x C(4 e-)] + [ 4 x F(7 e-)] = 32 valence electrons. 8 electrons are used to place a single bond between C and each F, leaving 32 - 8 24 e- (12 pairs). Use these 12 pairs to complete the octets of the F atoms (C already has an octet). This gives the fol lowing Lewis structure: =

:F:

I I :F:

.

: F -- C--F·

There are four groups of electrons around the C, none of which is a lone pair. This gives a tetrahedral electron-group arrangement, a tetrahedral molecular shape, and an ideal bond angle of 1 09.5°. 10.36

Plan: Examine the structure shown, and then apply YSEPR. Solution : a) This structure shows three electron groups with three bonds around the central atom. There appears to be no distortion of the bond angles so the shape is trigonal planar, the classification is AX3, with an ideal bond angle of 1 20°. b) This structure shows three electron groups with three bonds around the central atom. The bonds are distorted down indicating the presence of a lone pair. The shape of the molecule is trigonal pyramidal and the classification is AX3E, with an ideal bond angle of 1 09.5°. c) This structure shows five electron groups with five bonds around the central atom. There appears to be no distortion of the bond angles so the shape is trigonal bipyramidal and the classification is AXs, with ideal bond angles of 90° and 1 20°.

10.38

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution : a) The CI0 2- ion has [ 1 x CI(7 e-)] + [ 2 x 0(6 e-)] + [ I e- (from charge)] = 20 valence electrons. 4 electrons are used to place a single bond between the CI and each 0, leaving 20 - 4 = 16 electrons (8 pairs). All 8 pairs are used to complete the octets of the CI and 0 atoms. There are two bonds (to the O's) and two lone pairs on the Cl for a total of 4 electron groups (AX2 E 2 ). The structure is based on a tetrahedral electron-group arrangement with an ideal bond angle of 1 09.5°. The shape is bent (or V-shaped). The presence of the lone pairs wil l cause the remaining angles to be less than 1 09.5°. b) The PFs molecule has [1 x P(5 e-)] + [5 x F(7 e-)] = 40 valence electrons. 10 electrons are used to place single bonds between P and each F atom, leaving 40 - 10 = 30 e- ( 1 5 pairs). The 1 5 electrons are used to complete the octets of the F atoms. There are 5 bonds to the P and no lone pairs (AXs). The electron-group arrangement and the shape is trigonal bipyramidal . The ideal bond angles are 90° and 1 20°. The absence of lone pairs means the angles are ideal.

133

c)single Thebonds SeF4 molecule hasandxeach Se(6Fe-)]atom,+ leaving F (7 e-)] valence electrons. electrons are tousedcomplete to placethe between Se 8 6 epairs). pairs are used octets ofto thethe FSeatoms which leaves pairpairof(AX4E). electrons.TheThisstructure pair isisplaced onon thea trigonal centralbipyramidal Se atom. There are with bonds which also has a lone based structure ideal angles of The shape is The presence of the lone pairs means the angles are KrF2 molecule hasK[1 x Kr(8 e-)] + x F(7 e-)] valence electrons. electrons are used to place a d)Single The bond between the The r atomremaining and each Fpairsatom,of electrons leaving are placed e-on thepairs). 6 pairsTheareKrused to central complete Kr atom. the octets of the F atoms. is the atom. Therewithareidealbonds toofthe Krandand loneThepairsshape is The The structure is basedof theon aF'strigonal bipyramidal structure angles placement makes their idealthusbond angle to be x The placement of the lone pairs is such that they cancel each other's repulsion, the actual . F: : [: �-Cl-� :]K ",,-/ .. .. .. .. I.. : : F-- r--F: : :F--Se--F :..F--P--F 1 1 : F: (a) (c) (d) (b) : F: : F: F: : 0: : F :' F: . 1 1 1 1 : F--P\\ .. . : "'7Cl" '. F· .. I" 1 ,: :-Y�f' '. ..0 : F: F: : (a) (b) (c) (d) The Lewis structures must be drawn, and VSEPR applied to the structures. Plan: Solution: a) CH30Hboth : Thiscarbon molecule has xservee-)]as central + x H(1 e-)] + x 0(6 e-)] valence electrons. Into theplaceCHa30H molecule, and oxygen atoms. ( can never be central.) Use electrons H single bondthe between thelastC andatom. the 0This atomleaves and of the atoms andpairs). another electrons to placeto complete a single bond between 0 and the eUse these two pairs the octet of(AX4), the 0soatom. C already has an octet and each H only gets electrons. The carbon has bonds and no lone pairs with so it is (no lone pairs)withfromthetheangles ideal angle of Theangleoxygen has bonds andit islone Hpairs (AX2E2) or of ' I . H--C--O · I ""-H [1

[4

x

34

1

90° and 120°.

=

-

34 =

2

8 12

(13

4

see-saw.

less than

ideal.

=

[2

3 3 120°.

2

90°

2 90° = 180°. bond angle is ideal.

.

: F·

•

22 22 - 4

4

=

18

(9

(AX2E3).

linear.

F:

: F:

..

\\ ' "

.

\\\\\'", ..

\\\\\\\"' . ' -- Kr

..

10.40

F:

[1

H

tetrahedral

2

.

C(4

[4

[1

3 H 14 - 10 = 4

no deviation V-shaped

bent

H

134

=

14

8

2

(2

2

4 109.5°. less than the ideal

2 109.5°.

b) N 204: This molecule has [2 x N(5 e-)] + [4 x 0(6 e-)] = 34 valence electrons. Use 1 0 electrons to place a single bond between the two N atoms and between each N and two of the 0 atoms. This leaves 34 - 10 = 24 e­ ( 1 2 pairs). Use the 12 pairs to complete the octets of the oxygen atoms. Neither N atom has an octet however. Form a double bond from one 0 atom to one N atom by changing a lone pair on the 0 to a bonding pair on the N. Do this for the other N atom as well. In the N 2 04 molecule, both nitrogens serve as central atoms. This is the arrangement given in the problem. Both nitrogens are equivalent with 3 groups and no lone pairs (AX3 ), so the arrangement is trigonal planar with no deviation (no lone pairs) from the ideal angle of 1 20°. The same results arise from the other resonance structures.

�

O

. .

.' /

O === N --N===O

.. 10.42

I : 0:

I .. : 0:

N

;

-- N

-. q.

0..

"" -.

Q

.

Plan: The Lewis structures must be drawn, and VSEPR applied to the structures. Solution: a) CH 3 COOH has [2 x C(4 e-)] + [4 x H(l e-)] + [2 x 0(6 e-)] = 24 valence electrons. Use 14 electrons to place a single bond between all of the atoms. This leaves 24 - 1 4 = 10 e- (5 pairs). Use these 5 pairs to complete the octets of the 0 atoms; the C atom bonded to the H atoms has an octet but the other C atom does not have a complete octet. Form a double bond from the 0 atom (not bonded to H) to the C by changing a lone pair on the 0 to a bonding pair on the C . In the CH 3 COOH molecule, the carbons and the 0 with H attached serve as central atoms. The carbon bonded to the H atoms has 4 groups and no lone pairs (AX4), so it is tetrahedral with no deviation from the ideal angle of 1 09.5°. The carbon bonded to the 0 atoms has 3 groups and no lone pairs (AX3 ), so it is trigonal planar with no deviation from the ideal angle of 1 20°. The H bearing 0 has 2 bonds and 2 lone pairs (AX 2 E 2 ) so the arrangement is V-shaped or bent with an angle less than the ideal values of 109.5°. ' : 0·· H :0 : H

-- CI --IIC -- O : I I H H

H H"

""

jc

"

/ C � /H

\,\

.

.

O. .

H b) H 2 0 2 has [2 x H( I e-)] + [2 x 0(6 e-)] = 1 4 valence electrons. Use 6 electrons to place single bonds between the 0 atoms and between each 0 atom and a H atom. This leaves 14 - 6 = 8 e- (4 pairs). Use these 4 pairs to complete the octets of the 0 atoms. In the H 2 02 molecule, both oxygens serve as central atoms. Both O ' s have 2 bonds and 2 lone pairs (AX2 E 2 ) ' so they are V-shaped or bent with angles less than the ideal values of 109.5°. : 0--0: " 0 -- 0"

1

10.44

\

/

1

H

H

H

H

Plan: First, draw a Lewis structure, and then apply VSEPR. Solution:

.. .. 1 : F -- C -- F : 1 F: F'

: F

-- B -- F : F -- Be-F : I F:

Bond angles:

1 09.5°

OF2

<

NF3

<

CF4

<

BF3

<

BeF2

135

: F-- N·--F: : F--O :

I.

.. F1 ·.

<

109.5°

: F· «

109S

BeF 2 is an AX2 type molecule, so the angle is the ideal 180°. BF 3 is an AX3 molecule, so the angle is the ideal 120°. CF4, NF 3 , and OF 2 all have tetrahedral electron-group arrangements of the fol lowing types: AX4, AX3E, and AX2 E 2 , respectively. The ideal tetrahedral bond angle is 109.5°, which is present in CF4• The one lone pair in NFJ decreases the angle a little. The two lone pairs in OF 2 decrease the angle even more. 10.46

Plan: The ideal bond angles depend on the electron-group arrangement. Deviations depend on lone pairs. Solution: a) The C and N have 3 groups, so they are ideally 1 20°, and the 0 has 4 groups, so ideally the angle is 1 0 9.5°. The Nand 0 have lone pairs, so the angles are less than ideal. b) All central atoms have 4 pairs, so ideally all the angles are 1 09.5°. The lone pairs on the 0 reduce this value. c) The B has 3 groups (no lone pairs) leading to an ideal bond angle of 1 20°. All the O ' s have 4 pairs (ideally 109.5°),2 of which are lone, and reduce the angle.

10.49

Plan: The Lewis structures are needed to predict the ideal bond angles. Solution: The P atoms have no lone pairs in any case so the angles are ideal. PCls: PCI6-: pcV: : CI :

, I CI'... . : .CI--P/ -- I: .. I � : �I : '

: .CI: .. PI : CI I : CI:

+

.

--

--

: CI :

:�I�PI/�!: :�)/: �I1 � 9.I: :

Cl :

"

.. .. I \\\\\\,.CI : .·CI--P .. : CII "'c. (: :

+

: CI:

•.

The original PCls is AXs, so the shape is trigonal bipyramidal, and the angles are 120° and 90°. The pcV is AX4, so the shape is tetrahedral, and the angles are 109.5°. The PCI6- is AX6, so the shape is octahedral, and the angles are 90°. Half the PCls (trigonal bipyramidal, 120° and 90°) become tetrahedral pcV (tetrahedral, 109.5°), and the other half become octahedral PCI6- (octahedral, 90°). 10.5 1

Plan: To determine if a bond is polar, determine the electronegativity difference of the atoms participating in the bond. To determine if a molecule is polar, it must have polar bonds, and a certain shape determined by YSEPR. Solution: a) The greater the difference in electronegativity the more polar the bond: Molecule Bond Electronegativities Electronegativity difference SCI 2 S-CI S = 2.5 CI = 3.0 3.0 - 2.5 = 0.5 4.0 - 4.0 = 0.0 F = 4.0 F = 4.0 F-F F2 2.5 - 2.5 = 0.0 CS 2 C = 2.5 S = 2.5 C-S CF4 C = 2.5 F = 4.0 C-F 4.0 - 2.5 = 1.5 3.0 - 2.8 = 0.2 Br-CI Br = 2.8 CI = 3.0 BrCI The polarities of the bonds increases in the order: F-F = C-S < Br-CI < S-CI < C-F. Thus, CF4 has the most polar bonds.

136

b) The F2 and CS2 cannot be polar since they do not have polar bonds. CF4 is a AX4 molecule, so it is tetrahedral with the 4 polar C-F bonds arranged to cancel each other giving an overall nonpolar molecule. BrCI has a dipole moment since there are no other bonds to cancel the polar Br-Cl bond. SCI2 has a dipole moment (is polar) because it is a bent molecule, AX2E2, and the S-Cl bonds both pull to one side .

.- -. . /\

: CI 1 0.53

: CI :

Plan: If only 2 atoms are involved, only an electronegativity difference is needed. If there are more than 2 atoms, the structure must be determined. Solution: a) All the bonds are polar covalent. The SO) molecule is trigonal planar, AX), so the bond dipoles cancel leading to a nonpolar molecule (no dipole moment). The S02 molecule is bent, AX2E, so the polar bonds both pull to one side. S02 has a greater dipole moment because it is the only one of the pair that is polar.

.'

0'·

# SI I � : of/ .. �.o"

b) ICI and I F are polar, as are all diatomic molecules composed of atoms with differing electronegativities. The electronegativity difference for rCI (3.0 - 2.5 = 0.5) is less than that for IF (4.0 - 2.5 = 1 .5). The greater difference means that IF has a greater dipole moment. c) All the bonds are polar covalent. The SiF4 molecule is nonpolar (has no dipole moment) because the bonds are arranged tetrahedrally, AX4. SF4 is AX4E, so it has a see-saw shape, where the bond dipoles do not cancel. SF4 .

has the greater dipole moment.

I

: F:

: F -- Si -- F:

: F -- S - F :

:F:

: F:

1

\ . , : F - S - ,I: ' .. :F" G: p',

I, ,

:F:

..

1

..

.

'l

.•

.

1

d) H20 and H2S have the same basic structure. They are both bent molecules, AX2E2, and as such, they are polar. The electronegativity difference in H20 (3.5 - 2. 1 = 1 .4) is greater than the electronegativity difference in H2S (2.5 - 2. I = 0.4) so H20 has a greater dipole moment. 1 0.55

Plan: Draw Lewis structures, and then apply YSEPR. Solution : There are 3 possible structures for the compound C2H2C12:

I

II

III

137

The presence of the double bond prevents rotation about the C=C bond, so the structures are "fixed." The C-CI bonds are more polar than the C-H bonds, so the key to predicting the polarity is the positioning of the C-CI bonds. Compound 1 has the C-CI bonds arranged so that they cancel leaving I as a nonpolar molecule. Both I I and !I I have C-CI bonds on the same side so the bonds work together making both molecules polar. Both I and I I will react with H 2 to give a compound with a CI attached to each C (same product). Compound III will react with H2 to give a compound with 2 CI's on one C and none on the other (different product). Compo u nd I must be X as it is the only one that is nonpolar (has no dipole moment). Compound I I must be Z because it is polar and gives the same product as compound X. This means that compound I I I must be the remaining compound - Y. Compound Y

1 0.56

( I II) has a dipole moment.

Plan: The Lewis structures are needed to do this problem. For part (b), obtain bond energy values from Table 9.2. Solution : a) The H atoms cannot be central, and they are evenly distributed on the N's. N2H4 has [2 x N(5 e-)] + [4 x H(l e- )] = 14 valence electrons, 10 of which are used in the bonds between the atoms. The remaining two pairs are used to complete the octets of the N atoms. N2H2 has [2 x N(5 e-)] + (2 x H ( I e- )] = 1 2 valence electrons, 6 of which are used in the bonds between the atoms. The remaining three pairs of electrons are not enough to complete the octets of both N atoms, so one lone pair is moved to a bonding pair between the N atoms. N2 has [2 x N(5 e-)] = 10 valence electrons, 2 of which are used to place a single bond between the two N atoms. Since only 4 pairs of electrons remain and 6 pairs are required to complete the octets, two lone pairs become bonding pairs to form a triple bond. H-N-N-H

I

H

I

H -- N === N - H

H

Nitrogen Diazene Hydrazine The single (bond order = I) N-N bond is weaker and longer than any of the others are. The triple bond (bond order 3) is stronger and shorter than any of the others. The double bond (bond order = 2) has an =

b) N4H4 has [4 x N(5 e-)] + [4 x H(1 e-)] = 24 valence electrons, 1 4 of which are used for single bonds between the atoms. When the remaining 5 pairs are distributed to complete the octets, one N atom lacks two electrons. A lone pair is moved to a bonding pair for a double bond. intermediate strength and length.

H -- N - N - H

H - N - N === N - N - H

I

I

I

I

H H H Mrbonds broken = 4 N-H = 4 mol (39 1 kJ/mol) Mrbonds formed = 4 N-H = 4 mol (39 1 klima I) 1 N-N = I mol ( 1 60 kJ/mot) 2 N-N = 2 mol ( 1 60 kJ/mol) 1 N=N = 1 mol (4 1 8 kl/mol) I N=N = 1 mol (945 kJ/mol) = 2302 kJ = 2669 kJ -367 kJ kJ = kJ) 2302 = = (2669 onds n formed onds r roken WOb b WOb WO x Note: It is correct to report the answer in kJ or kJ/mol as long as the value refers to a reactant or product with a molar coefficient of I. H

1 0.57

Plan: Use the Lewis structures shown in the text. FC = valence electrons - [lone electrons + 1 12 (bonded electrons)] Solution: a) Formal charges for A12C16: FCA1 3 - (0 + 1 12(8)) =-1 FCCI, ends = 7 - (6 + 1/2(2) ) = 0 FCCI, bridging = 7 - (4 + 1 12 ( 4)) = + I (Check: Formal charges add to zero, the charge on the compound.) =

138

Formal charges for 12Ck FCI = 7 - (4 + 112(8» = -1 FCcl, ends = 7 - (6 + 112(2» = 0 FCCI, bridging = 7 - (4 + 112(4)) = + I (Check: Formal charges add to zero, the charge on the compound.) b) The aluminum atoms have no lone pairs and are AX4, so they are tetrahedral. The 2 tetrahedral AI's cannot give a planar structure. The iodine atoms have two lone pairs each and are AX4E2 so they are square planar. Placing the square planar I's adjacent can give a planar molecule. 10.65

Plan: Assume all reactants and products are assumed to be gaseous. Ethanol burns (combusts) with O2 to produce CO2 and H20. Solution: a) To save time the balanced equation includes the Lewis structures: H H

I I

I .. I I

+

H-C--C--O: H

H

3

0=0

----l.. �

H

..

2 ..O=C=O

+

3 :O--H

I

H 4(799 kllmol) 6(467 kllmol)

1(347 kllmol) Bonds formed: 2 x 2 C=O 1 C-C 2 x 3 O-H 5(413 kllmol) 5 C-H 1(358 kllmol) I C-O 1(467 kllmol) 1 00H 3(498 kllmol) 3 0=0 Totals: 5998 kl 4731 kJ tJi = Bonds broken - Bonds formed = 4731 kJ - 5998 kJ = - 1267 kJ for each mole of ethanol burned. b) If it takes 40.5 kllmol to vaporize the ethanol, part of the heat of combustion must be used to convert liquid ethanol to gaseous ethanol. The new value becomes: tJicomb(liquid)= -1267 kJ + (I mol) (40.5 kllmol) = -1226.5 = - 1226 kJ per mol of liquid ethanol burned c) tJicomb(liquid) = L ntJi; (products) - L mtJi; (reactants) tJicOmb(liquid) = [2(Mf't
\_/ C-C "H /

I I H

+

I I H

H--C--C--O

H

The energy to break the reactant bonds is: BE(C=C) + 4 BE(C-H) + 2 BE(O-H) = (I mol) (614 kJ/mo!) + (4 mol) (413 kllmol) + (2 mol) (467 kllmo!) = 3200. kl The energy from forming the product bonds is: BE(C-C) + 5 BE(C-H) + BE(C-O) + BE(O-H) = (I mol) (347 kllmo!) + (5 mol) (413 kllmol) + (1 mol) (358 kllmol) + (1 mol) (467 kllmol) = 3237 kl Mf'rxn = Mf'bonds broken - Mf'bonds Cormed = 3200. kl - (3237 kl) = -37 kJ

139

.

•

"H

10.6 7

Plan: Determine the empirical formula from the percent composition (assuming 1 00 grams of compound). Use the titration data to determine the mole ratio of acid to the NaOH. This ratio relates the number of acidic H ' s to the formula of the acid. Finally, combine this information to construct the Lewis structure. Solution : 1 mOl 2.222 mol H ( 2.24 g H 1 .00 l l.008 g H 2.222 mol H 2.222 mol C o

) / 1 mol ) ( 26.7 g C /l 12.01 gC mol (7 1 . 1 g O/ l 1 6.00 g 0 ) 1

=

=

2.223 mol 2.222 mol

2.223 mol C

=

4.444 mol 2.222 mol

4.444 mol 0

=

=

l.00

=

2.00

The empirical formula is H C02. The titration required: I mmol 0.040 mol NaOH 50.0 mL 1L 2.0 mmole NaOH 1 000 mL 0.00 1 mol L Thus, the ratio is 2.0 mmole base I 1 .0 mmole acid, or each acid molecule has two hydrogens to react (diprotic). The empirical formula indicates a monoprotic acid, so the formula must be doubled to: H2C204. There are 34 valence electrons to be used in the Lewis structure. 1 4 of these electrons are used to bond the atoms with single bonds, leaving 1 0 pairs of electrons. When these 1 0 pairs of electrons are distributed to the atoms to complete octets, neither C atom has an octet; a lone pair from the oxygen without hydrogen is changed to a bonding pair on C.

(

H

-

)(

'"

/ . � �O· :O

O:

f r

·. o 1 0.68

)(

)(

-

)

=

H

C -

.

Plan: Draw the Lewis structure of the OH species. Find the additional bond energy information in Table 9.2. Solution: a) The OH molecule has 7 valence electrons. Thus, no atom can have an octet, and one electron is left unpaired. The Lewis structure is: ·O

H

b) The formation reaction is: 1 12 02(g) + 1 12 H2 (g) � OH(g). The heat of reaction is: !1Hreaction !1Hbroken - !1Hfonned 39.0 kl [ B Eo� + B EH-H] - [BEo_H] 39.0 kJ [( 112 mol) (498 kJ/mol) + ( 1 12 mol) (432 kJ/mol)] - [BEQ-H] 39.0 kl B Eo_H 426 kJ c) The average bond energy (from the bond energy table) is 467 kJ/mol . There are 2 O-H bonds in water for a total of 2 x 467 kl/mol 934 kJ . The answer to part b accounts for 426 kl of this, leaving: 934 kJ - 426 kJ 508 kJ --

=

=

=

=

=

=

=

1 0.70

Plan: The basic Lewis structure will be the same for all species. The CI atoms are larger than the F atoms. Solution: a) The F atoms will occupy the smaller axial positions first. The molecules containing only F or only CI are nonpolar, as all the polar bonds would cancel. The molecules with one F or one CI would be polar since there would be no corresponding bond to cancel the polarity. The presence of 2 axial F ' s means that their polarities will cancel (as would the 3 CI' s) giving a nonpolar molecule. The molecule with 3 F ' s is also·polar.

1 40

b)

:.. F:

\. CI : I ' :CI -- P ' 1: CI:"'�I: \\\'

..

..

1 0.73

Plan: Draw the Lewis structures and look for potential problems. Calculate the heat of reaction from the bond energies. Solution: S0 3 (g) H2S04(l) -7 H2S207(l)

+

o. · '·O. =S= II ' ·.0.

'

I1H

=

:0:

+

.. II .. II \ HI

."

: 0:

II .. .. II

:0 :

II .... II

'O--S--O---S--O I ""H H

:O--S--O: : 0:

5(552 kJ/mol) 5 S=O 2(265 kJ/mol) 2 S-O 2(467 kJ/mol) 2 0-H Totals: 4224 kJ Bonds broken - Bonds formed = 4224 kJ - 4202 kJ = 22 kJ

Bonds broken:

1 0.75

Non-polar

Polar

Polar

Non-polar

Polar

:0:

Bonds formed:

4 S=O 4 S-0 2 0-H

: 0:

4(552 kJ/mol) 4(265 kJ/mol) 2(467 kJ/mol) 4202 kJ

Plan: Pick the VSEPR structures for A Y3 substances. Then determine which are polar. Solution: Y Y

0

1 A

Y

/ '-...,.

Y

Y

/

A

ID I� Y

�,.y

Y-A

Y

(c) (a) (b) 5 groups 4 groups 3 groups (AX3 ) (AX3 E) (AX 3 E2) T-shaped trigonal pyramidal Trigonal planar The trigonal planar molecules, such as (a), are nonpolar, so it cannot be A Y3 . Trigonal pyramidal molecules (b) and T-shaped molecules (c) are polar, so either could represent AY 3 .

141

Chapter 11 Theories of Covalent Bonding FOLLOW-UP PROBLEMS

1 1.1

P lan: Draw a Lewis structure. Determine the number and arrangement of the electron pairs about the central atom. This process leads to the hybridization. Solution : a) In BeF2 , Be is surrounded by two electron groups (two sing le bonds) so hybridization around Be is sp.

: F

Be-F :

-

2p

2s

sp

2p

Hybridized Be atom I so lated Be atom b) In SiC I4 , Si is surrounded by four electron groups (four single bonds) so its hybridization is

Sp3.

: Cl : :Cl

-

I I

Si-Cl:

: Cl:

[ill It It I 3s

3p

------.�

it it it if I

Hybridized Si atom Isolated Si atom c) In XeF4, xenon is surrounded by 6 electron groups (4 bonds and 2 lone pairs) so the hybridization i s sicf- .

:F :

I .. .. I

: F-Xe-F : : F :

5s

5p

5d

it tit tit It it I t li L.....-- 5d

L...--.....L..-

Hybridized Xe atom

Isolated Xe atom

142

I 1 .2

Plan: First, determine the Lewis structure for the molecule. Then count the number of electron groups around each atom. Hybridization is sp if there are two groups, Sp 2 if there are three groups and sp3 if there are four groups. No hybridization occurs with only one group of electrons. The bonds are then designated as sigma or pi. A single bond is a sigma bond. A double bond consists of one sigma and one pi bond. A triple bond includes one sigma bond and two pi bonds. Hybridized orbitals overlap head on to form sigma bonds whereas pi bonds form through the sideways overlap of p or d orbitals. Solution: a) Hydrogen cyanide has H-C=N: as its Lewis structure. The single bond between carbon and hydrogen is a sigma bond formed by the overlap of a hybridized sp orbital on carbon with the 1 s orbital from hydrogen. Between carbon and nitrogen are three bonds. One is a sigma bond formed by the overlap of a hybridized sp orbital on carbon with a hybridized sp orbital on nitrogen. The other two bonds between carbon and nitrogen are pi bonds formed by the overlap of p orbitals from carbon and nitrogen. One sp orbital on nitrogen is filled with a lone pair of electrons. b) The Lewis structure of carbon dioxide, CO 2 , is O === C === O Both oxygen atoms are s/ hybridized (the 0 atoms have three electron groups - one double bond and two lone pairs) and form a sigma bond and a pi bond with carbon. The sigma bonds are formed by the overlap of a hybridized sp orbital on carbon with a hybridized s/ orbital on oxygen. The pi bonds are formed by the overlap of an oxygen p orbital with a carbon p orbital. Two s/ orbitals on each oxygen are filled with a lone pair of electrons.

1 1 .3

Plan: Draw the molecular orbital diagram. Determine the bond order from calculation: BO = 1 /2(# e- in bonding orbitals - #e- in antibonding orbitals). A bond order of zero indicates the molecule will not exist. A bond order greater than zero indicates that the molecule is at least somewhat stable and is likely to exist. H 2 2- has 4 electrons ( 1 from each hydrogen and two from the -2 charge). Solution:

Jill. G* Is

,

[IT]

M O Hl

-

I s H- ,

lill I sR

Iill

,

Gis Configuration for H /- is (G Is) 2 (G* Is ) 2 . Bond order of H 2 2- is 1 12(2 - 2) = O. Thus, it is not l ikely that two W ions would combine to form the ion H/-. I 1 .4

Plan: To find bond order it is necessary to determine the molecular orbital electron configuration from the total number of electrons. Bond order is calculated from the configuration as 1 12(# e- in bonding orbitals - #e- in antibonding orbitals). Solution : F/-: total electrons = 9 9 2 = 20 Configuratio n: (G Is ) 2 (G* Li( G2 s )2 (G*2s ) 2 (G2p )2 ( 1t2p )4 ( 1t*2p )4 (G*2p ) 2 Bond order = I 12( I 0 - 1 0) = 0

++

1 43

F2-: total electrons = 9

9 I 19 Configuration : ((JIs)2((J*ls)2((J2,)\ (J*2s)2((J2p)2(n:2Pt(n:*2p)4((J*2p)I Bond order = 1 12( 1 0 - 9) 0.5 F2: total electrons 9 9 = 1 8 Configuration: ((JIs)2((J*IJ2((J2J2((J*2s)\ (J2p)2(n:2Pt(n:*2p)4 Bond order 1 12(10 - 8) = 1 F/: total electrons 9 9 - I 1 7 Configuration: ((JIs)2((J*ls)2((J2s)2((J*2s)2((J2p)\ n:2p)4(n:*2p)3 Bond order 1 12( I 0 - 7) 1.5 F22+: total electrons = 9 9 - 2 = 1 6 Configuration: ((J 1,)2((J*ls)2( (J2s)2((J*2s)2((J2p)\ n:2p)4(n:*2P)2 Bond order = 1 12( 1 0 - 6) = 2 Bond energy increases as bond order increases: F/- < F2- < F2 < F/ < F/+ Bond length decreases as bond energy increases, so the order of increasing bond length will be opposite that of increasing bond energy. Increasing bond length : F/+ < F/ < F2 < F2-. F/- will not form a bond, so it has no bond length and is not included in the list. Check: Since the highest energy orbitals are anti bonding orbitals, it makes sense that the bond order increases as electrons are removed from the anti bonding orbitals. =

++ + + +

=

=

=

=

=

=

=

END-OF-CHAPTE R PROBLE M S

1 1 .1

1 1 .3

1 1 .5

1 1 .7

Table 1 1 . 1 describes the types of shapes that form from a given set of hybrid orbitals. a) trigonal planar: three electron groups - three hybrid orbitals : Sp2 b) octahedral: six electron groups - six hybrid orbitals: sid2 c) linear: two electron groups - two hybrid orbitals: sp d) tetrahedral : four electron groups - four hybrid orbitals: Sp3 e) trigonal bipyramidal : five electron groups - five hybrid orbitals: Sp3d Carbon and silicon have the same number of valence electrons, but the outer level of electrons is n = 2 for carbon and n = 3 for silicon. Thus, silicon has 3d orbitals in addition to 3s and 3p orbitals available for bonding in its outer level, to form up to 6 hybrid orbitals, whereas carbon has only 2s and 2p orbitals available in its outer level to form up to 4 hybrid orbitals. The number of orbitals remains the same as the number of orbitals before hybridization. The type depends on the orbitals mixed. a) There are six unhybridized orbitals, and therefore six hybrid orbitals result. The type is Sp3d2. b) Fou r Sp3 hybrid orbitals form from three p and one s atomic orbitals. To determine hybridization, draw the Lewis structure and count the number of electron groups. Hybridize that number of orbitals. a) The three electron groups (one double bond, one lone pair and one unpaired electron) around nitrogen require three hybrid orbitals. The hybridization is si .

. . N==:O

b) The nitrogen has three electron groups (one single bond, one double bond and one unpaired electron), requiring three hybrid orbitals so hybridization is Sp2. N

"0/ . ., � . .0" .

1 44

c) The nitrogen h a s three e lectron groups (one single bond, one double bond and one lone pair so hybridization is sp . 2

11.9

a) Sp The C I ha s four e lectron groups (I lone pair , I lone e lectron, and 2 double bonds) and therefore four hybrid orbita ls are required ; the hybridization is s . Note that in CI02 , the 1t bond is formed by the overlap of d-orbita ls from c hlorine with p-orbita ls from oxygen . ' ' I

/

3

q.

C

.�

.' � �"

b) Sp The CI ha s four e lectron groups (I lone pair and 3 bonds) and therefore four hybrid orbita ls are required ; the hybr idization is s . 3

/

[:�-�1=§] : 0:

/

3 Sp

c) The CI h a s four e lectron gro ups (4 bonds) and therefore four hybrid orbitals are required; the hybridization is s . :0:

II .. "

:O

..

CI =O

/ hybrid orbitals are made

: 0:

11.11

a) Silicon h a s four e lectron groups (four bonds) requiring four hybrid orbitals; four s from one s and three p atomic orbitals. H

-- I .. HI

:C I

Si--H

b) Carbon has two e lectron groups (two double bonds) requiring two hybrid orbitals; two hybrid orbita l s are m ade from one s and one p orbital. S ====C ==== S

1l.13

I

sp

a) Sulfur is surrounded by 5 e lectron groups (4 bonding pair s, lone pair) requiring five hybrid orbitals; five sid hybrid orbita ls are formed from one s orbital, three p orbitals, and one d orbital.

145

b) Nitrogen is surrounded by 4 electron groups (3 bonding pairs, 1 lone pair) requiring four hybrid orbitals; four / hybrid orbitals are formed from one s orbital and three p orbitals.

s

11. 15

a) Germanium is the central atom in GeCl 4 . Ge has four electron groups (four bonds), requiring four hybrid orbitals. Hybridization is Sp 3 around Ge. : CI :

I I : Cl :

: CI -- Ge - CI :

I

[ill It It s

p

--..�

If If It It I

Isolated Ge atom Hybridized Ge atom c) Boron is the central atom in BCI 3 . B has three electron groups (three bonds), requiring three hybrid orbitals. Hybridization is Sp 2 around B.

It It It I D

-----.�

s

p p Isolated B atom Hybridized B atom d) Carbon is the central atom in CH 3 +' C has three electron groups (three bonds), requiring three hybrid orbitals. Hybridization is s/ around C. +

H

I

/C", H

[ill It It I s

p Isolated C atom

H

-----.�

It It It I

Sp2 p Hybridized C atom

146

1 1.17

a) Tn SeCI 2, Se is the central atom and has four electron groups (two single bonds and two lone pairs), requiring four hybrid orbitals. Se is si hybridized. Two si orbitals are filled with lone electron pairs and two si orbitals bond with the chlorine atoms. : CI -Se-CI :

[ill IHlt It I s

�.

p

If.lf.lt I t I �

b) In H 3 0+, 0 is the central atom and has four electron groups (three single bonds and one lone pair, requiring four hybrid orbitals. 0 is Sp3 hybridized. One si orbital is filled with a lone electron pair and three si orbitals bond with the hydrogen atoms. +

H

H

--

I

O--H

If�lf It It I

[ill Itt It It I s

p

�

c) I is the central atom in IF4- with 6 electron groups (four single bonds and two lone pairs) surrounding it. Six hybrid orbitals are required and I has si J2 hybrid orbitals. The si J2 hybrid orbitals are composed of one s orbital, three p orbitals, and two d orbitals. Two si J2 orbitals are filled with a lone pair and four si J2 orbitals bond with the fluorine atoms. :F: :F

--

I·.

I--F:

.I

:F

s

p

d

If � I f � Itit Itit 11,-----,----,---, cf

11.20

a) False, a double bond is one sigma and one pi bond. b) False, a triple bond consists of one sigma (cr) and two pi (1t) bonds. c) True d) True e) False, a 1t bond consists of one pair of electrons; it occurs after a cr bond has been previously formed. t) False, end-to-end overlap results in a bond with electron density along the bond axis.

1 47

1 1 .2 1

a) Nitrogen is the central atom in N0 3- . Nitrogen has 3 surrounding electron groups (two single bonds and one double bond), so it is Sp2 hybridized. Nitrogen forms three cr bonds (one each for the N-O bonds) and one 1t bond (part of the N=O double bond).

9

.

[:

-

I

===

: 0:

-

Q

]

b) Carbon is the central atom in CS 2. Carbon has 2 surrounding electron groups (two double bonds), so it is sp hybridized. Carbon forms two cr bonds (one each for the C-S bonds) and two 1t bonds (part of the two C=S double bonds).

�� C��

c) Carbon is the central atom in CH 2 0. Carbon has 3 surrounding electron groups (two single bonds and one double bond), so it is sl hybridized. Carbon forms three cr bonds (one each for the two C-H bonds and one C-O bond) and one 1t bond (part of the C=O double bond). H "" C === .. O H 1 1 .23

/

a) Examine the Lewis structure. Three electron groups (one lone pair, one single bond and one double bond) surround the central N atom. Hybridization is sl around nitrogen. One sigma bond exists between F and N, and one sigma and one pi bond exists between N and O. Nitrogen participates in a total of2 cr and I 1t bonds. F--N ===O : .. .. b) Examine the Lewis structure. Each carbon has three electron groups (two single bonds and one double bond) with Sp2 hybridization. The bonds between C and F are sigma bonds. The C-C bond consists of one sigma and one pi bond. Each carbon participates in a total of3 cr and I 1t bonds. : ..F � - / .. F: C-C "" ../ F: :F . .

c) Examine the Lewis structure. Each carbon has two electron groups ( I single bond and one triple bond) and is sp hybridized with a sigma bond between the two carbons and a sigma and two pi bonds comprising each C-N triple bond. Each carbon participates in a total of2 cr and 2 1t bonds. :N - C -- C - N: 1 1 .25

Four molecular orbitals form from the four p atomic orbitals. In forming molecular orbitals, the total number of molecular orbitals must equal the number of atomic orbitals. Two of the four molecular orbitals formed are bonding orbitals and two are antibonding.

1 1 .27

a) Bonding MO's have lower energy than antibonding MO's. The bonding MO's lower energy, even lower than its constituent atomic orbitals, accounts for the stabil ity of a molecule in relation to its individual atoms. However, the sum of energy of the MO's must equal the sum of energy of the AO 's. b) The node is the region of an orbital where the probability of finding the electron is zero, so the nodal plane is the plane that bisects the node perpendicular to the bond axis. According to Figure 1 1 . 1 3A, there is no node along the bond axis (probability is positive between the two nuclei) for the bonding MO. The anti bonding MO does have a nodal plane, shown as "waves cancel" on Figure 1 1 . 1 38. c) The bonding MO has higher electron density between nuclei than the antibonding MO.

1 48

1 1 .29

a) Two electrons are required to fill a cr-bonding molecular orbital. Each molecular orbital requires two electrons. b) Two electrons are required to fill a 1t -antibonding molecular orbital. There are two 1t -antibonding orbitals, each holding a maximum of two electrons. c) Four electrons are required to fill the two cr molecular orbitals (two electrons to fill the cr-bonding and two to fill the cr-antibonding) formed from two Is atomic orbitals.

1 1 .3 1

The horizontal line in all cases represents the bond axis. Bonding s p a)

+

�

Antibonding s

p

+ -(J--cx3+ •

b)

-

-

�

Bonding p p

Antibonding p - p

�

c::::> --t:>oc::::>

+ � •

-

I I

c::::> � c:=>

I

� I

1 1 .33

a) The molecular orbital configuration for Be/ with a total of 7 electrons is (crls) 2 (cr*ls) 2 (cr2s)\cr* 2s)1 and bond order = 1 /2(4 - 3) = 1 12. With a bond order of 1 12 the Be/ ion will be stable. (Bond order = Yz(number of electrons in bonding MO - number of electrons in antibonding MO)) b) No, the ion has one unpaired electron in the cr*2s MO, so it is paramagnetic, not diamagnetic. c) Valence electrons would be those in the molecular orbitals at the n = 2 level, so the valence electron configuration is (cr2s)2((cr*2s)l.

1 1 .35

The sequence of MO ' s for C 2 is shown in Figure 1 1 . 1 9 (the (cr l s) 2 and (cr l :) 2 MO ' s are not shown for convenience). For each species, determine the total number of electrons, the valence molecular orbital electron configuration and bond order. Bond order = Y2(number of electrons in bonding M O - number of electrons in antibonding MO) Total valence electrons = 4 4 1 = 9 C 24 2 Valence configuration: (cr2s)\cr*2s ( 1t2p ) ( cr2p)1 Bond order = 1 /2(7 - 2) = 2.5 Total valence electrons = 4 4 = 8 C2 4 2 2 Valence configuration: (cr2s) ((cr*2s) ( 1t2p ) Bond order = 1 /2(6 - 2) = 2 Total valence electrons = 4 4 - 1 = 7 C/ Valence configuration: (cr2s) 2 ((cr*2s ) 2 ( 1t2Pi Bond order = 1 /2(5 - 2) = 1 .5 a) Bond energy increases as bond order increases: C2+ < C2 < C2b) Bond length decreases as bond energy increases, so the order of increasing bond length will be opposite that of increasing bond energy. I ncreasing bond length: C2- < C2 < C/

++ + +

1 1 .39

a) Each of the six C atoms in the ring has three electron groups (two single bonds and a double bond) and has Sp2 hybridization; all of the other C atoms have four electron groups (four single bond) and have si hybridization; all of the 0 atoms have four electron groups (two single bonds and two lone pairs and have Sp3 hybridization; the N atom has four electron groups (three single bonds and a lone pair) and has Sp3 hybridization. b) Each of the single bonds is a sigma bond; each of the double bonds has one sigma bond for a total of 26 sigma bonds. 1 49

c) The ring has three double bonds each of which is composed of one sigma bond and one pi bond; so there are three pi bonds each with 2 electrons for a total of 6 pi electrons. 1 1 .4 1

1 1 .43

a) Every single bond is a sigma bond. There is one sigma bond in each double bond as well. There are 17 cr bonds in isoniazid. Every atom-to-atom connection contains a cr bond. b) All carbons have three surrounding electron groups (two single and one double bond), so their hybridization is Sp2 . The ring N also has three surrounding electron groups (one single bond, one double bond and one lone pair), so its hybridization is also Sp2 . The other two N ' s have four surrounding electron groups (three single bonds and one lone pair) and are Sp3 hybridized. a) B changes from si -7 sf3. Boron in BF 3 has three electron groups with sp2 hybridization. In BF4-, 4 electron groups surround B with sp hybridization. :F

:F

: F-B-F

: F-B -F

I I

I

:F

b) P changes from Sp3 -7 sid. Phosphorus in PCh is surrounded by 4 electron groups (3 bonds to CI and I lone pair) for Sp 3 hybridization. In PCls, phosphorus is surrounded by 5 electron groups for sid hybridization. :CI :

I P

:CI-

:Cl:

Cl· . / .. .. ·CI-P

I . .. I ""'Cl: : CI: ..

Cl:

-

H

H / � /C = C H/ "'H

c) C changes from sp -7 si. Two electron groups surround C in C2H2 and 3 electron groups surround C in C2H4• H- C==C -H

�

d) Si2 changes from Sp3 -7 sp3d2 . Four electron groups surround Si in SiF4 and 6 electron groups surround Si in SiF6 -. :F :

I Si I

: F- -F

:F :

.. I �""'-...Si /".. ·· ··/. 1"' F ·

: .

: F

F

.

:F . •

F:

.

.

• •

e) No change, S in S02 is surrounded by three electron groups (one single bond, one double bond and one lone pair) and in S0 3 is surrounded by 3 electron groups (two single bonds and one 2 double bond); both have Sp hybridization. :O-S _O:

:0 : �

I

:O - S

1 50

0:

I 1 .44

1 1 .48

Count the electron groups, both bonds and lone pairs, about each atom. (Do not forget to count double bonds only once.) 3 P (3 single bonds and one double bond) Sp AX4 tetrahedral 3 N (3 single bonds and one lone pair) Sp AX3 E trigonal pyramid 3 Sp AX4 tetrahedral C1 and C 2 (4 single bonds) 2 trigonal planar AX3 Sp C 3 (2 single bonds and one double bond) H

"

"

/

C === C === C·

\\ H

,\\

,

H H The central C is sp hybridized, and the other two C ' s are sl . 1 1 .53

a) The 6 carbons in the ring each have three surrounding electron groups (two single2 bonds and one double bond) with sl hybrid orbitals. The two carbons participating in the C=O bond are also Sp hybridized. The single carbon in the -CH 3 group has 4 electron groups (four single bonds)and is Sp3 hybridized. There are only 2 central oxygen atoms, one in a C-O-H configuration and the other in a C-O-C configuration. Both of these atoms have 4 surrounding electron groups (two single bonds and two lone pairs) and are Sp 3 hybridized. Summary: C in -CH 3 : Sp3, all other C atoms (8 total): si, 0 (2 total): Sp3 b) The two C=O bonds are localized; the double bonds on the ring are delocalized as in benzene. c) Each carbon with three surrounding groups has sl hybridization and trigonal planar shape; therefore, 8 carbons have this shape. Only 1 carbon in the C H 3 group has four surrounding groups with Sp 3 hybridization and tetrahedral shape.

lSI

Chapter 12 Intermolecular Forces: Liquids, Solids, and Phase Changes FOLLOW-UP PROBLEMS 1 2. 1

Plan: The variables !:J.Hvap, P" T" and T2 are given, so substitute them into Equation both temperatures from °c to K. Convert !:J.Hvap to J so that the units cancel with R. Solution: T2 = 8 5 . 5 + 273 . 1 5 = 3 5 8 .6 K T, = 3 4 . 1 + 273 . 1 5 = 307.2 K In .!1.. PI In

-AHvap (_I - �J T T

=

R

P 2 torr

=

40. 1 P

_-=-2 40. 1

_

torr

=

I kJ -40.7 mol 8 . 3 1 4 J/mol- K 2

9 . 8 1 68 9

(I

358.6

K

-

I 3 07 . 2

K

(unrounded)

J

[ ) 1 03 J

--

1 kJ

=

2 . 2 84 1 0

12. 1

and solve for P2 . Convert

(unrounded)

P2 = (9 . 8 1 689) (40 . 1 torr) = 3 9 3 . 657 = 394 torr Note: Watch significant figures when subtracting numbers: 1 /3 5 8 . 6 = 0.002789 and 1 /307.2 = 0.003 2 5 5 0 . 002789 - 0 . 0 0 3 2 5 5

= 0 .000466

In the subtraction, the number of significant figures decreased from 4 to 3 . Check: The temperature increased, so the vapor pressure should be higher. 1 2.2

a) The

A:

-

H

· · ··"

H

B

\C -

-

sequence is present in both structures below:

..", -- /

··O�-----H-O

-- �

H

C � \"H 'H HI �·O-H-----� I H . O.� IH H . jQ. . \ / C, C-C /C H I '" '··OT----H-O../ H / H

C

•

__

b) The

B

/

H

••

H sequence can only be achieved in one arrangement.

H H H"", I I/H \C --\C-- H C--C, / .. ,H / H HI "'··O-i----H--O H/ The hydrogens attached to the carbons cannot form H bonds because they do not satisfy the sequence. c) Hydrogen bonding is not possible because there are no O-H, N-H, or F-H bonds. A:

-

H

H

· · · ··

-

-

H

.•

1 52

A:

-

"

"H-B­

1 2.3

1 2.4

a) Both CH 3 Br and CH 3 F are polar molecules that experience dipole-dipole and dispersion intermolecular interactions. Because C H 3 Br ( M = 94.9 glmol) is �3 times larger than CH 3 F ( M = 34.0 glmol), dispersion forces result in a higher boiling point (BP) for CH3Br. b) CH 3 CH 2 CH 2 0H, n-propanol, forms hydrogen bonds and has dispersion forces whereas C H 3 CH 2 0CH3, ethyl methyl ether, has only dipole-dipole and dispersion attractions. C H3CH2C H20 H has the higher BP. c) Both C 2 H6 and C 3 HS are nonpolar and experience dispersion forces only. C3Hg, with the higher molar mass, experiences greater dispersion forces and has the higher boiling point. Plan: The density of Fe is 7 . 874 glcm 3 , but Fe atoms occupy only 68% of the volume in a body-centered cubic structure. Calculate the volume/mole Fe ratio, and multiply by 0.68 to determine the volume/mol Fe atoms ratio. Dividing by the volume of a single Fe will yield the units of atoms/mol, which is Avogadro ' s number. Solution: Molar volume of Fe = ( 5 5 . 8 5 glmol) / (7.874 glcm3 ) = 7.09296 cm3 /mol Fe (unrounded) The volume of just the atoms (not including the empty spaces between atoms) is: Volume/mole Fe atoms = (7.09296 cm3 /mol Fe) (68% / 1 00%) = 4.8 2 3 2 1 cm3 /mol Fe atoms (unrounded) The number of atoms in one mole of Fe is obtained by dividing by the volume of one Fe atom: Atoms/mol = (4.823 2 1 cm3 /mol Fe 23atoms) ( I Fe atom / 8.38 x 1 0-24 cm3 ) 23 = 5.8 X 1 0 atoms/mol = 5 .7556 x 1 0 Check: The calculated value is within 4.4% error of the true value, 6.02 x 1 023 atoms/mol (% error = (true - measured) -;- true).

END-OF-CHAPTER PROBLEMS

1 2. 1

The energy of attraction is a potential energy and denoted Ep. The energy of motion is kinetic energy and denoted The relative strength of Ep vs. Ek determines the phase of the substance. In the gas phase, Ep « Ek because the gas particles experience little attraction for one another and the particles are moving very fast. In the solid phase, Ep » Ek because the particles are very close together and are only vibrating in place. Ek.

Two properties that differ between a gas and a solid are the volume and density. The volume of a gas expands to fill the container it is in while the volume of a solid is constant no matter what container holds the solid. Density of a gas is much less than the density of a solid. The density of a gas also varies significantly with temperature and pressure changes. The density of a solid is only slightly altered by changes in temperature and pressure. Compressibility and ability to flow are other properties that differ between gases and solids. 1 2.4

a) Heat of fusion refers to the change between the solid and the liquid states and heat of vaporization refers to the change between liquid and gas states. In the change from solid to liquid, the kinetic energy of the molecules must increase only enough partially to offset the intermolecular attractions between molecules. In the change from liquid to gas, the kinetic energy of the molecules must increase enough to overcome the intermolecular forces. The energy to overcome the intermolecular forces for the molecules to move freely in the gaseous state is much greater than the amount of energy needed to allow the molecules to move more easily past each other but still stay very close together. b) The net force holding molecules together in the solid state is greater than that in the liquid state. Thus, to change solid molecules to gaseous molecules in sublimation requires more energy than to change liquid molecules to gaseous molecules in vaporization. c) At a given temperature and pressure, the magnitude of I:!. Hvap is the same as the magnitude of I:!.Hc ond ' The only difference is in the sign: I:!.Hvap = -I:!. Hc ond'

1 2.5

a) Condensation b) Fusion (melting) c) Evaporation

1 2.7

The water vapor in the air condenses to liquid when the temperature drops during the night. Solid ice melts to liquid water. Liquid water on clothes evaporates to water vapor.

The propane gas molecules slow down as the gas is compressed. Therefore, much of the kinetic energy lost by the propane molecules is released to the surroundings upon liquefaction.

1 53

1 2. 1 1

I n closed containers, two processes, evaporation and condensation occur simultaneously. Initially there are few molecules in the vapor phase, so more liquid molecules evaporate than gas molecules condense. Thus, the numbers of molecules in the gas phase increases causing the vapor pressure of hexane to increase. Eventually, the number of molecules in the gas phase reaches a maximum where the number of liquid molecules evaporating equals the number of gas molecules condensing. In other words, the evaporation rate equals the condensation rate. At this point, there is no further change in the vapor pressure.

12. 1 4

The total heat required i s the sum of three processes: 1 ) Warming the ice from -5 .00DC to O.OODC q l = CrrlLH = (2.09 J/gDC) ( 1 2.00 g) [0.0 - (-5 .00)]DC = 1 25.4 J (unrounded) 2) Phase change of ice at O.OODC to water at O.OODC 3 6.02 kJ 1 0 J 1 mol = 4008. 879 J (unrounded) q 2 = �Hfus = ( 1 2.0 g ) 1 8 .02 g mol 1 kJ 3) Warming the liquid from O.OODC to 0.500DC q 3 C�T = (4.2 1 J/gDC) ( 1 2.00 g) [0.500 - (O.OWC = 25 .26 J (unrounded) The three heats are positive because each process takes heat from the surroundings (endothermic). The phase change requires much more energy than the two temperature change processes. The total heat is q I + q2 + q3 = ( 1 25 .4 J 4008.879 J + 25 .26 J) = 4 1 60 J = 4 1 59.539 = 4. 1 6 x 1 03 J .

(

J(

)( ]

=

+

1 2. 1 6

The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature. Boiling point is defined as the temperature when vapor pressure of liquid equals atmospheric pressure, usually assumed to be exactly 1 atm. So, the number of significant figures in the pressure of 1 atm given is not one, but is unlimited since it is an exact number. In the calculation below, 1 .00 atm is used to emphasize the additional significant figures. PI

Ln

P2 1 .00 atm

_

=

1 .00 atm;

T2 = 1 09°C

-3 5 . 5-'� 1 1 m"'o"-'I_ 8.3 1 4 J/mol · K 382 K 395 K

__ _

(

__ _ __

1 54

273 = 382 K;

+)[ 1

P --=2_ = 0.692203 (unrounded) 1 .00 atm P 2 = (0.692203) ( 1 .00 atm) = 0.692203 = 0.692 atm

T I = 1 22°C

+

273 = 395 K;

P2 = ?

1 03 J = -0.3678758 (unrounded) 1 kJ

__

1 2. 1 8

,, , ,,

50.5

.-..... s

�

�

,,

L

S

- 20 0

"

G

1

1 0-3

,,

- 1 00 T

a

(0C)

The pressure scale is distorted to represent the large range in pressures given in the problem, so the liquid-solid curve looks different from the one shown in Figure 1 2 . 8 . The important features of the graph include the distinct ion between the gas, liquid and solid states, and the melting point T, which is located d irectly above the critical T. Solid ethylene is denser than liquid ethylene since the solid-liquid line slopes to the right with increasing pressure. 1 2 .20

This problem is also an app licat ion of the C lausius-Clapeyron equation. Convert the temperatures from °C to K and Mivap from kl/mol to J/mol to allow cancellation with the units in R. In

In

H = -� vap

P2

PI

R

atm

_P-=2_

=

8.3 1 4

PI

TI

�

J / mol - K

(

= 2 . 3 atm; 1 5 0°C + 273

T2 = 1 423

1

_ _ _

_

K

298

_

K

)[ ) 1 03 J I kJ

TI

= 2 5 . 0°C + 273 = 298

= 423 K ;

K;

P2 = ?

= 2 . 89834

1 8 . 1 4404

= ( 1 8 . 1 4404) (2 . 3 atm) = 4 1 .73 1 = 42 atm

2.3

P2

_

T2

-24.3

P2 2.3

(...!... ...!...)

atm

1 2 .24

To form hydrogen bonds, the atom bonded to hydrogen must have two characteristics: small size and high electronegativity (so that the atom has a very high electron density) . With this high electron density, the attraction for a hydrogen on another molecule is very strong. Selenium is much larger than oxygen (atomic radius of 1 1 9 pm vs. 73 p m) and less electronegat ive than oxygen (2.4 for Se and 3 . 5 for 0) resulting in an electron density on Se in H2 Se that is too small to form hydrogen bonds.

1 2 .25

All particles (atoms and molecules) exhibit dispersion forces, but these are the weakest of intermolecular forces. The dip ole-dipole forces in p olar molecules dominate the dispersion forces.

1 2.28

a) Hydrogen bonding will be the strongest force between methanol molecules since they contain O-H bonds. Dipole-dip ole and disp ersion forces also exist . b) Dispersion forces are t h e only forces between nonpolar carbon tetrachloride molecules and, thus, are the strongest forces. c) Dispersion forces are the only forces between nonpolar chlorine molecules and, thus, are the strongest forces.

1 55

1 2.30

a) has a dipoleinteractions will be the strongest forces between methyl bromide molecules because the C-Br bond moment. dominate because is a symmetrical molecule. c)F)b) that participateforcesin hydrogen dominatesbonding. becauseCH3CH3 hydrogen(ethane) is bonded to nitrogen, nonpolar which is one of the 3 atoms 0, or Hydrogen bonds ofareanformed whenleads a hydrogen atom is bonded to 0,bonds or F.in a)hydrogen The presence OH group to the formation of hydrogen There are no bonds in CH3SCH3. H H H H .... .. I H""" I H / / C H C H ......C--C I I/ H \C--C \ H H / I ?-O�---- H -/ """ H H H/ The presence ofH attached to F in HF leads to the formation of hydrogen bonds. There are no hydrogen bonds inb) HBr. Dipole-dipole

Dispersion Hydrogen bonding

1 2.32

(N,

N,

CH3CH(OH)CH3.

H

__

• •

1 2.34

Polarizability increases. increases down a group and decreases from left to right because as atomic size increases, polarizability a)polarized overhasa larger greatervolume polarizability in a largerthanatomthe orbromide ion. ion because the iodide ion is larger. The electrons can be b) has greater polarizability than ethane (CH3CH3) because orbitals are more easily c)polarizedhasthangreaterorbitals. polarizability than water because the selenium atom is larger than the oxygen atom. Weakerto attractive forces result ininto a higher vaporphase.pressure because the molecules have a smaller energy barrier in order escape the liquid and go the gas a) C2H6 isCH3CH2F a smallerhasmolecule exhibiting weaker dispersion forces than C4H 1 0exhibits • dipole-dipole forces, b)which no H-F bonds (F is bonded to C, not to H), so it only than theintermolecular hydrogen bonding CH3CH20H. than NH3 (hydrogen bonding). c) arePH3weaker has weaker forcesin(dipole-dipole) The stronger thehaveintermolecular forces,pointthethan higherClthebecause boilingthepoint.ions in lithium chloride are held together by ionic a)forces, would a higher boiling Hintermolecular forces between hydrogen chloride molecules in which are stronger than the dipole-dipole theb) liquid phase. would have a higher boiling point than phosphine (PH3) because the intermolecular forces in ammonia are strongerforces than those inphosphine phosphinemolecules are. Hydrogen bonding exists between ammonia molecules but weaker dipole-dipole hold together. c)between iodine wouldmolecules have a higher boiling point than xenon. Bothxenonare nonpolar with dispersion forces, but theareforces would be stronger than between atoms are since the iodine molecules more polarizable because of their larger size. Iodide ion

1t

Ethene (CH2= CH2)

H2Se

1 2.36

cr

C2H6 CH3CH2F

PH3

1 2.38

LiCI

Ammonia, NH3,

Iodine

1 56

1 2.40

The molecule in the pair with the weaker intermolecular attractive forces will have the lower boiling point. a) C4Hs, the cyclic molecule, cyclobutane, has less surface area exposed, so its dispersion forces are less than the straight chain molecule, C 4 H 1 0 • b) PBr3, the dipole-dipole forces of phosphorous tribromide are weaker than the ionic forces of sodium bromide. c) HBr, the dipole-dipole forces of hydrogen bromide are weaker than the hydrogen bonding forces of water .

1 2.45

The shape of the drop depends upon the competing cohesive forces (attraction of molecules within the drop itself) and adhesive forces (attraction between molecules in the drop and the molecules of the waxed floor). If the cohesive forces are strong and outweigh the adhesive forces, the drop wi ll be as spherical as gravity will allow. If, on the other hand, the adhesive forces are significant, the drop wil l spread out. Both water (hydrogen bonding) and mercury (metallic bonds) have strong cohesive forces, whereas cohesive forces in oil (dispersion) are relatively weak. F igure 1 2. 1 9 explains the difference between these forces when glass, not wax, is used. Neither water nor mercury will have significant adhesive forces to the nonpolar wax molecules, so these drops wil l remain nearly spherical. The adhesive forces between the oil and wax can compete with the weak, cohesive forces of the oil (dispersion) and so the oil drop spreads out.

1 2.47

The stronger the intermolecular force, the greater the surface tension. All three molecules exhibit hydrogen bonding, but the extent of hydrogen bonding increases with the number of O-H bonds present in each molecule. Molecule b) with 3 O-H groups can form more hydrogen bonds than molecule c) with 2 O-H groups, which in tum can form more hydrogen bonds than molecule a) with only I O-H group. CH3CH2CH20H

<

HOCH2CH20H

<

HOCH2CH(OH)CH20H

The greater the number of hydrogen bonds, the greater the energy requirements. 1 2.5 1

Water is a good solvent for polar and ionic substances and a poor solvent for nonpolar substances. Water is a polar molecule and dissolves polar substances because their intermolecular forces are of similar strength. Water is also able to dissolve ionic compounds and keep ions separated in solution through ion-dipole interactions. Nonpolar substances will not be very soluble in water since their dispersion forces are much weaker than the hydrogen bonds in water. A solute whose intermolecular attraction to a solvent molecule is less than the attraction between two solvent molecules will not dissolve because its attraction cannot replace the attraction between solvent molecules.

1 2.52

A

1 2.54

Water exhibits strong capillary action, which allows it to be easily absorbed by the plant's roots and transported to the leaves.

1 2.60

A

1 2.63

The energy gap is the energy difference between the highest filled energy level (valence band) and the lowest unfilled energy level (conduction band). In conductors and superconductors, the energy gap is zero because the valence band overlaps the conduction band. In semiconductors, the energy gap is small but greater than zero. In insulators, the energy gap is large and thus insulators do not conduct electricity.

1 2.64

The simple cubic structure unit cel l contains 1 atom since the atoms at the 8 comers are shared by eight cells for a total of 8 atoms x 1 /8 atom per cell = 1 atom.; the body-centered cell also has an atom in the center, for a total of2 atoms; the face-centered cell has 6 atoms in the faces which are shared by two cells: 6 atoms x Y2 atom per cell = 3 atoms plus another atom from the 8 corners for a total of 4 atoms. a) Ni is face-centered cubic since there are 4 atoms/unit cell. b) Cr is body-centered cubic since there are 2 atoms/unit cell. c) Ca is face-centered cubic since there are 4 atoms/unit cell .

single water molecule can form 4 hydrogen bonds. The two hydrogen atoms form a hydrogen bond each to oxygen atoms on neighboring water molecules. The two lone pairs on the oxygen atom form hydrogen bonds with hydrogen atoms on neighboring molecules.

solid metal is a shiny solid that conducts heat, is malleable, and melts at high temperatures. (Other answers include relatively high boiling point and good conductor of electricity.)

1 57

1 2.66

a) Tin is a metal (malleable, excellent electrical conductor) that forms metallic bonds. b) Silicon is in the same group as carbon, so it exhibits similar bonding properties. Since diamond and graphite are both network covalent solids, it makes sense that Si forms the same type of bonds. c) Xenon, Xe, is a noble gas and is monatomic. Xe is an atomic solid.

1 2.68

Figure P 1 2.68 shows the face-centered cubic array of zinc blende, ZnS. Both ZnS and ZnO have a I : I ion ratio, so the ZnO unit cell will also contain four Zn2+ ions.

1 2.70

a) To determine the number of Zn2+ ions and Se2- ions in each unit cell count the number of ions at the comers, faces, and center of unit cell. Looking at se1enide ions, there is one ion at each comer and one ion on each face. The total number of selenide ions is 1/8 (8 corner ions) + 112 (6 face ions) = 4 Se2- ions. There are also 4 Zn2+ ions due to the 1: I ratio of Se ions to Zn ions. b) Mass of unit cell = (4 x mass of Zn atom) + (4 x mass of Se atom) = (4 x 65.41 amu) + (4 x 78.96 amu) = 577.48 amu c) Given the mass of one unit cell and the ratio of mass to volume (density) divide the mass, converted to grams (conversion factor is 1 amu = 1.66054 x 1 0-24 g), by the density to find the volume of the unit cell. 4 Volume = � 1.66054 x 10- 2 g ( 577.48 amu) I amu 5.42 g 22 = 1.76924 x 10- = 1.77 X 1 0-22 cm3 d) The volume of a cube equals (length of sidel Side = �1.76924 x I 0- 2 2 cm3 = 5.6139 x 10-8 = 5.61 X 10-8 cm

[ )[

)

1 2.72

To classify a substance according to its electrical conductivity, first locate it on the periodic table as a metal, metalloid, or nonmetal. In general, metals are conductors, metalloids are semiconductors, and nonmetals are insulators. a) Phosphorous is a nonmetal and an insulator. b) Mercury is a metal and a conductor. c) Germanium is a metalloid in Group 4A(14) and is beneath carbon and silicon in the periodic table. Pure germanium crystals are semiconductors and are used to detect gamma rays emitted by radioactive materials. Germanium can also be doped with phosphorous (similar to the doping of silicon) to form an n-type semiconductor or be doped with lithium to form a p-type semiconductor.

1 2.73

First, classify the substance as an insulator, conductor, or semiconductor. The electrical conductivity of conductors decreases with increasing temperature, whereas that of semiconductors increases with temperature. Temperature increases have little impact on the electrical conductivity of insulators. a) Antimony, Sb, is a metalloid, so it is a semiconductor. Its electrical conductivity increases as the temperature increases. b) Tellurium, Te, is a metalloid, so it is a semiconductor. Its electrical conductivity increases as temperature increases. c) Bismuth, Bi, is a metal, so it is a conductor. Its electrical conductivity decreases as temperature increases.

1 2.77

The Clausius-Clapeyron equation can be solved for the temperature that will give a vapor pressure of 5.0x10-5 torr. In .!2. PI

=

-LlH yap R

(...!..- J...) T2

5.0 x 1 0-5 torr In 1 .20 x 1 0-3 torr

-

PI = 1 .20 X 10-3 torr;

TI

kJ -;;;t 8.314 J/mol -59.1

•

K

(

P2 = 5.0 X 10-5 torr; 1 T

2

I 293 K

)[ )

1 58

1 03 J 1 kJ

T I = 20.0°C + 273 = 293 K ;

[ ( T; K ( 29�) K) (unrounded) (-3. 1 7805) I (-7108.49) 4.47078 x 10-4 [ ( T� K ( 29�) K) (unrounded) 7078 x 10-4 + I 1 293 1 I T T 12.79 a)ofThe cell contents one Ca, x (1/8) a I Ti, and 12 x unit (114)cell.30, or one CaTi03 formula unit. The presence one formula unit perare unit cell 8indicates 135.96 g CaTi03 ) 1 CaTi03 J[ 1 mol CaTi03 3 )( ( Unit Cell 6. 022 x 102 3CaTi03 3I mol CaTi03 3. 98728 b) Density A) 1 cm ( [ Unit Cell l [ 1 A ) m Both furfuryl 12. 8 1 a)directly bonded toalcohol oxygen.and 2-furoic acid can for hydrogen bonds since these two molecules have hydrogen H 2-furoic acid 0 "" C �C - Ci -- \\ H :CI ",C \ I I '\/C -- H !I" �C - C� )0: I H/ C · I furfuryl alcohol -3 . 1 780 5

=

-7 1 08.49

=

4.4

=

2 5 9 . 064

=

=

=

259 K

=

=

simple cubic

=

3 . 84

=

0 -8

H - - �- O.

'

'

'

. .O . . --

. O -:- - - H

H

H

1 59

""

= 3.99 g/cm

3

b) Both furfuryl alcohol and 2-furoic acid can form internal hydrogen bonds by forming a hydrogen bond between the O-H and the 0 in the ring. 2-furoic acid H

'

H

\C / C '-- / C �

· C?'

"-

�O:

c --

II

H

H

-

I

�0..

C

..

furfuryl alcohol

1 2.85

This problem involves carefully examining the figures showing the different cells pictured in the chapter. a) The atoms touch along the body diagonal. The two corner atoms each contribute one radius (r), and the center atom contributes a diameter (2r). The total for the body diagonal = 4r. b) The face diagonal is the hypotenuse of a right triangle with the other two sides being the unit cell edge (a). Using the Pythagorean Theorem (a2 + b2 = c2 ) with a = b = the unit cell edge, and c = the face diagonal : a2 + b 2 = c 2 a2 + a2 = 2 i = c 2 C = face diagonal = J2 a c) The body-diagonal is the hypotenuse ( c) of a triangle with one of the other sides being a face-diagonal (b) and the remaining side being a unit cell edge (a). Again, the Pythagorean Theorem is applied. a2 + b 2 = c 2 From part a: a2 + b2 = (4r) 2 From part b: or a2 + 2 a2 = 1 6r2 a2 + ( J2 a) 2 = (4r) 2 Rearranging: 2 4r 1 6r a2 = a= 3 J3 d) A body-centered cubic unit cell contains 2 atoms (eight atoms in the corners are each 1 /8 in the cell (i.e., shared by eight cells), so there is a net of one atom - 1 /8 x 8 and one atom in the center of the cell . e ) Fraction filled = (volume of atoms present) / (volume o f unit cell) = [2 atoms x volume of one atom] / (answer from part c) 4 3 Find the volume of one atom using the equation for the volume of a sphere: - 7rr 3 Find the volume of the unit cell by cubing the value of the edge length from part c: Volume of a cube = length x width x height --

Fraction filled =

2[�m3] [�r

8.37758r 3 ----:-3 = 0.680 1 7 1 2.3 1 68r

1 60

Chapter 1 3 The Properties of Mixtures : Solutions and C olloids FOLLOW-UP PROBLEMS

13. 1

Plan : Compare the intermolecular forces in the solutes with the intermolecular forces in the solvent. The intermolecular forces for the more soluble solute will be more similar to the intermolecular forces in the solvent than the forces in the less soluble solute. Solution: a) 1 ,4-Butanediol is more soluble in water than butanol. Intermolecular forces in water are primarily hydrogen bonding. The intermolecular forces in both solutes also involve hydrogen bonding with the hydroxyl groups. Compared to butanol, each 1 ,4-butanediol molecule will form more hydrogen bonds with water because 1 ,4butanediol contains two hydroxyl groups in each molecule, whereas butanol contains only one -OH group. Since 1 ,4-butanediol has more hydrogen bonds to water than butanol, it will be more soluble than butanol . b) Chloroform is more soluble in water than carbon tetrachloride because chloroform is a polar molecule and carbon tetrachloride is nonpolar. Polar molecules are more soluble in water, a polar solvent.

1 3 .2

Plan: Solubility of a gas can be found from Henry's law: Sgas = kH X P gas. The problem gives kH for N2 but not its partial pressure. To calculate the partial pressure, use the relationship from Chapter 5 : Pgas = Xgas X Plolal where X represents the mole fraction of the gas. Solution : To find partial pressure use the 78% N2 given for the mole fraction: P N, = 0.78 x 1 atm = 0.78 atm Use Henry ' s law to find solubil ity at this partial pressure: S N, = 7 X 1 0-4 mollL-atm x 0.78 atm = 5 .46 x 1 0-4 mollL = 5 x 1 0-4 mollL Check: The solubil ity should be close to the value of kH since the partial pressure is close to I atm.

1 3 .3

Plan : Molality ( m ) is defined a s amount (mol) o f solute per k g o f solvent. U s e the molality and the mass of solvent given to calculate the amount of glucose. Then convert amount (mol) of glucose to mass (g) of glucose. Solution : Mass of solvent must be converted to kg: 563 g x ( 1 kg/ 1 000 g) = 0.563 kg 2 2.40 x 1 0 - mol C 6 H I 2 0 6 1 80. 1 6 C 6H I2 0 6 = 2.4343 = 2.43 g C6H I 206 ( 0.563 kg ) I mol C 6 H I 2 0 6 1 kg Check: To check these results estimate the molality from the calculated mass of glucose and given mass of solvent. The number of moles of glucose is approximately (2.4 g) ( l mol / 1 80 g) = 0.0 1 3 mol glucose. Using the calculated value for amount of glucose and the given mass of solvent, calculate the molality: (0.0 1 3) / (0.56 kg) = 0.023 m, which is close to the given value of 0.0240 m.

[

1 3 .4

)(

J

Plan : Mass percent is the mass (g) o f solute per 1 00 g o f solution. For each alcohol , divide the mass o f the alcohol by the total mass. M ultiply this number by 1 00 to obtain mass percent. To find mole fraction, first find the amount (mol) of each alcohol , then divide by the total moles. Solution: Mass percent: 3 5 .0 g Mass% propanol = x 1 00% = 1 8 .9 1 89 = 1 8 .9% propanol ( 3 5 . 0 + 1 50. ) g 1 50. g Mass% ethanol = x 1 00% = 8 1 .08 1 08 = 8 1 . 1 % ethanol ( 3 5 . 0 + 1 50.) g

161

Mole fraction: Moles propanol = ( 35.0 g C 3 H70H )

(

(I

1 mol C 3 H 7 OH 60.09 g C 3 H 7 0H

)

)

=

0.5824596 mol propanol (unrounded)

mol C 2 H sOH 3 .2559 1 49 mol ethanol (unrounded) 46.07 g C 2 HsOH 0.5824596 mol propanol = 0. 1 5 1 746 = 0.1 52 X propanol 0.5824596 mol propanol + 3.2559 1 49 mol ethanol 3 .2559 1 49 mol ethanol = 0.84825 0.848 Xethano l 0.5824596 mol propanol + 3.2559 1 49 mol ethanol Check: S ince the mass of ethanol is 4 times that of propanol, the mass percent of ethanol should be about 4 times greater. 8 1 % is about 4 times 1 9%. The mole fraction should be similar, but the fraction that is ethanol should be slightly greater than the mass fraction of 0.8 1 - and it is at 0.85. Moles ethanol = ( 1 50. g C 2 HsOH )

=

=

=

1 3 .5

ILL

Plan: To find the mass percent, molality and mole fraction of HCI, the following is needed. 1 ) Moles of HCI in solution (from molarity) 2) Mass of HCI in 1 solution (from molarity times molar mass of HCI) 3) Mass of solution (from volume times density) 4) Mass of solvent in 1 solution (by subtracting mass of solute from mass of solution) 5) Moles of solvent (by multiplying mass of solvent by molar mass of water) Mass percent is calculated by dividing mass of HCI by mass of solution. Molality is calculated by dividing moles of HCI by mass of solvent in kg. Mole fraction is calculated by dividing mol of HCI by the sum of mol of HCI plus mol of solvent. Solution: Assume the volume is exactly I L. ' = ( 1 .0 ) l l .8 mo I HCI = 1 1 .8 mol HCI ' I L so I utlOn ) Mo I e H CI In 1 .0 36.46 g HCI 2) Mass HCI in 1 solution = ( 1 1 .8 mol HCI ) I mol HCI = 430.228 g HCI 1 l . l 90 g 3) Mass of 1 solution ( 1 .0 10 I mL 1 1 90. g solution

IL

I

L

L

L(

L

=

( 1 1 90 g ) 1 �g

( ) L )( I � ) L

L )( �LL J (

(

J

L

J

)

=

1 . 1 90 kg solution 10 g 4) Mass of solvent in 1 solution = 1 1 90. g - 430.228 g = 759.772 g solvent (H 2 0) g 0.759772 kg solvent ( 759.772 g 10 g 5) Mole of solvent in I

=

=

solution

=

( 759.772 g H 2 0 )

(I

mol H 2 0 1 8.02 g H 2 0

)

=

42. 1 627 mol solvent

Mass HCI ( 1 00 ) = 430.228 g HCl ( 1 00 ) = 36. 1 536 = 36.2 % 1 1 90 g solution Mass solution Mole HCl 1 1 .8 mol HCI . = = 1 5 .530975 = 1 5.5 m MolalIty of HCI = Kg solvent 0.759772 kg M oI H C I 1 1 . 8 mol Mole fraction of HCI = __ _ ___ 0.2 1 866956 = 0.2 1 9 Mol HCI + Mol H 2 0 1 1 .8 mol + 42. 1 627 mol Check: The mass of HCI is about 1 13 of the mass of the solution, so 36% is in the ballpark for the mass% HC\. Since the moles of HCI is about 1 14 the moles of water, the mole fraction of HCI should come out around 0.2. Mass percent of HCI =

_ _

__

------

1 62

=

13. 6 Plan: Raoult's law states that the vapor pressure ofa solvent is proportional to the mole fraction of the solvent: To calcforulathat te theofdrop invvapor pressure, a similar relationship is used with the mole fraction of the solute substituted the sol ent. Sol Moleution: fraction of aspirin in methanol: mol Aspirin ) ( 2.00 g ) (I180.15 g Aspirin 1 mol Methanol ) 1 mol Aspirin ( 2.00 g ) ( 180.15 g Aspirin ) ( 50. 0 g ) ( 32.04 g Methanol 7.06381 10-3= (unrounded) = P x 10-3)to be(101insignificant torr) = 0.71344 (7.06381 � The drop Check: in vapor pressure is expected since the amount of aspirin in solution is so small - only 0.01 mol aspirin in 1.5 mol solvent. Plan: TheO.OO°F 13.7 convert question the concentration of ethylofewater ne glywill col that freezingsubtracted at O.OO°F.fromFirst,the to DC.asksTheforchange in freezing point thenwould be thisprevent temperature freezialnity.g point of pure water, 0.00°c. Use this value for �T in 1.86°C/m molality of solution and solve for mol Sol ution: conversion Temperature T (°C) = (T ( OF ) - 320F) ( �:� ) = ( 0.000F - 320F) ( �:� ) = -17.7778°C (unrounded) °C = �Tf = -17.7778 1.86° C/m = 9.557956 = The minimum concentration of ethylene glycol would have to be 9.56 in order to prevent the water from freezi n g at O.OO°F. Check: In Samplbye Probl em 13.7,thea concentration 3.6 solutionmust lowered the freezing point bym.about 7ulated degrees.valueThus,of9.56 to loweris the temperature 18 degrees, be 18/7 times 3.6 or 9.3 Cal c close to the estimated value. Plan:ution: The osmotic pressure is calculated as the product of molarity, temperature, and the gas constant. 13.8 Sol L · atm ) ( ( 273 37 ) ) 7.6353 = = MR.T = ( 0.30LmOI )( 0.0821 mol K · Check: solutionTheof 0.3answer, 7.6 atm, is a significant osmotic pressure, which is expected from a relatively concentrated ride o chl magnesium the of pressure osmotic the te a ul c cal then solution, the of molarity 13. 9 solPlan:utioCaln. cDoulatenottheforget unit. a formul per ions three d yiel to ionizes and te, y ctrol e el ectrolyte (glucose) solution of equal concentration. This will is a strong thattimesMgCh nonel a as great as three pressure a in t resul Solution: 10.3 L I 0.1003499 L (unrounded) ( ) a) Liters of solution = (100. + 0.952 g SOlution ) (I1. 0mL 06 g 1 mL ) 1 mol MgCI2 = 9.9989 x 10-3 mol MgCh (unrounded) Moles MgCI2 ( 0.952 g MgCI2 ) ( 95.21 g MgCI2 J . 2 = Mol MgCI2 0. 0099989 mol = 0.099640 MMgCI2 (unrounded) Molanty. MgCI Volume solution 0.1003499 L Psolvent = Xsolvent X P Osol vent.

+

X Xaspi r in = Xaspi r i n POmethanol

=

0.713 torr

�Tf =

�Tf = m Kf m

9.56

Kf

x

m

m

m

m

n

+

K

=

7.6 atm

M.

=

=

163

(

J(

J

0.099640 m O I L atm 2 . 2 + 2 0.0) 0 .0 8 2 1 ' K ) = 7 . 1 955 = 7 . 2 0 atm ( ( 73 mol ' K L b) The magnesium chloride solution has a higher osmotic pressure, thus, l iquid wil l diffuse from the glucose solution into the magnesium chloride solution. The glucose side wil l lower and the magnesium chloride side wil l rise. Scene C represents this situation. Check: Repeat the calculations using approximate values (M = 0. 1 , R = 0.08, and T = 300). n

=

iMRT = 3

END-OF-CH APTER PROB L E M S

1 3 .2

When a salt such as NaCI dissolves, ion-dipole forces cause the ions to separate, and many water molecules cluster around each of them in hydration shells. Ion-dipole forces hold the first shell . Additional shells are held by hydrogen bonding to inner shells.

1 3 .4

a)

1 3 .6

To identify the strongest type of intermolecular force, check the formula of the solute and identify the forces that could occur. Then look at the formula for the solvent and determine if the forces i dentified for the solute would occur with the solvent. The strongest force is ion-dipole followed by dipole-dipole (including H bonds). Next in strength is ion-induced dipole force and then dipole-induced dipole force. The weakest intermolecular interactions are dispersion forces. a) Ion-dipole forces are the strongest intermolecular forces in the solution of the ionic substance cesium chloride in polar water. b) Hydrogen bonding (type of dipole-dipole force) is the strongest intermolecular force in the solution of polar propanone (or acetone) in polar water. c) Dipole-induced dipole forces are the strongest forces between the polar methanol and nonpolar carbon tetrachloride.

1 3.8

a ) Hydrogen bonding occurs between the H atom o n water and the lone electron pair o n the 0 atom in dimethyl ether (CH30CH3). However, none of the hydrogen atoms on dimethyl ether participates in hydrogen bonding because the C-H bond does not have sufficient polarity. b) The dipole in water induces a dipole on the Ne(g) atom, so dipole-induced dipole interactions are the strongest intermolecular forces in this solution. c) Nitrogen gas and butane are both nonpolar substances, so dispersion forces are the principal attractive forces.

13. 1 0

CH3CH 2 0CH 2 CH3 is polar with dipole-dipole interactions as the dominant intermolecular forces. Examine the solutes to determine which has intermolecular forces more similar to those for the diethyl ether. This solute is the one that would be more soluble. a) HCI would be more soluble since it is a covalent compound with dipole-dipole forces, whereas NaCI is an ionic solid. Dipole-dipole forces between HCI and diethyl ether are more similar to the dipole forces in diethyl ether than the ion-dipole forces between NaCI and diethyl ether. b) CH3CHO (acetaldehyde) would be more soluble. The dominant interactions in H 2 0 are hydrogen bonding, a stronger type of dipole-dipole force. The dominant interactions in CH3CHO are dipole-dipole. The solute-solvent interactions between CH3CHO and diethyl ether are more similar to the solvent intermolecular forces than the forces between H 2 0 and diethyl ether. c) C H3CH2MgBr would be more soluble. CH3CH 2 MgBr has a polar end (-MgBr) and a nonpolar end (CH3CHr), whereas MgBr2 is an ionic compound. The nonpolar end of CH3CH 2 MgBr and diethyl ether would interact with dispersion forces, while the polar end of CH3CH 2 MgBr and the dipole in diethyl ether would interact with dipole-dipole forces. Recall , that if the polarity continues to increase, the bond will eventually become ionic. There is a continuous sequence from nonpolar covalent to ionic.

A more concentrated solution will have more solute dissolved in the solvent. Potassium nitrate, KN03, is an ionic compound and therefore soluble in a polar solvent like water. Potassium nitrate is not soluble in the nonpolar solvent CCI4. Because potassium nitrate dissolves to a greater extent in water, KN03 in H 20 wil l result in the more concentrated solution.

1 64

13.12

Gluconic acid i s a very polar molecule because it has -OR groups attached to every carbon. The abundance of -OR bonds allows gluconic acid to participate in extensive H-bonding with water, hence its great sol ubility in water. On the other hand, caproic acid has a 5-carbon, nonpolar, hydrophobic ("water hating") tail that does not easily dissolve in water. The dispersion forces in the nonpolar tai l are more similar to the dispersion forces in hexane, hence its greater solubility in hexane.

13. 1 7

This compound would be very soluble in water. A large exothermic value in tJls o luti on (enthalpy of solution) means that the solution has a much lower energy state than the isolated solute and solvent particles, so the system tends to the formation of the solution. Entropy that accompanies dissolution always favors solution formation. Entropy becomes important when explaining why solids with an endothermic tJlsol ut i on (and higher energy state) are sti ll soluble in water.

13. 1 8

tJlhydralion

tJlsolution

KCI(s) --�----��----------------------------- �

Lattice energy values are always positive as energy is required to separate the ions from each other. Hydration energy values are always negative as energy is released when intermolecular forces between ions and water form. Since the heat of solution for KCI is endothermic, the lattice energy must be greater than the hydration energy for an overall input of energy.

1 3 .20

1 3 .22

1 3 .24

Charge density is the ratio of an ion ' s charge (regardless of sign) to its volume. An ion ' s volume is related to its radius. F or ions whose charges have the same sign (+ or -), ion size decreases as a group in the periodic table is ascended and as you proceed from left to right in the periodic table. a) Both ions have a + I charge, but the volume of Na + is smal ler, so it has the greater charge density. b) SrH has a greater ionic charge and a smaller size (because it has a greater Zefr), so it has the greater charge density. c) Na + has a smaller ion volume than cr, so it has the greater charge density. 2 d) 0 - has a greater ionic charge and simi lar ion volume, so it has the greater charge density. e) OH- has a smaller ion volume than SH-, so it has the greater charge density. The ion with the greater charge density will have the larger tJlhydrati on . a) Na + would have a larger tJlhydration than Cs + since its charge density is greater than that of Cs + . 2 b) Sr + would have a larger tJlhydrati on than Rb+ . c) Na+ would have a larger tJlhydration than cr. 2 d) 0 - would have a larger tJlhydration than F-. e) OH- would have a larger tJlhydrati on than SH-.

a) The two ions in potassium bromate are K+ and 8r03-' The heat of solution for ionic compounds is tJlsoln = tJllattice + tJlhydr or the i ons' Therefore, the combined heats of hydration for the ions is (tJlsoln - tJllatti ce) or 4 1 . 1 kJ/mol - 745 kJ/mol -703 .9 -704 kJ/mol. b) K+ ion contributes more to the heat of hydration because it has a smaller size and, therefore, a greater charge density. =

=

1 65

1 3 .26

Entropy increases as the possible states for a system increases. a) Entropy increases as the gasoline is burned. Gaseous products at a higher temperature form. b) Entropy decreases as the gold is separated from the ore. Pure gold has only the arrangement of gold atoms next to gold atoms, while the ore mixture has a greater number of possible arrangements among the components of the mixture. c) Entropy increases as a solute dissolves in the solvent.

1 3 .29

Add a pinch of the solid solute to each solution. A saturated solution contains the maximum amount of dissolved solute at a particular temperature. When additional solute is added to this solution, it will remain undissolved. An unsaturated solution contains less than the maximum amount of dissolved solute and so will dissolve added solute. A supersaturated solution is unstable and addition of a "seed" crystal of solute causes the excess solute to crystallize immediately, leaving behind a saturated solution.

1 3 .3 1

a) Increasing pressure for a gas increases the solubility of the gas according to Henry ' s law. b) Increasing the volume of a gas causes a decrease in its pressure (Boyle ' s Law), which decreases the solubility of the gas.

1 3 .33

a) Solubility for a gas is calculated from Henry ' s law: Sgas kH X Pgas. Sgas is expressed as mollL, so convert moles of O 2 to mass of O2 using the molar mass. mol ( S gas = 1 .28 x 1 0 -3 1 . 00 atm ) = 1 .28 X 1 0-3 mollL L atm 1 .28 x 1 0.3 mol 0 2 32.0 g 0 2 ( 2.00 L ) 0.08 1 92 0.08 1 9 g O 2 1 mol 02 L b) The amount of gas that will dissolve in a given volume decreases proportionately with the partial pressure of the gas, so mol ( S gas = 1 .28 x 1 0 -3 0.209 atm ) 2.6752 x 1 0-4 moUL L atm =

(

(

(

1 3 .36

(

)

)( 0

0

2.6752 X 1 0-4 mol 0 2 L

)(

)

)

=

=

=

32.0 g 0 2 1 mol 02

)(

2.00 L ) 0.0 1 7 1 2 =

=

0.0 1 7 1

g

O2

Solubility for a gas is calculated from Henry ' s law: Sgas kH X P gas. S gas = (3.7 X 1 0-2 mol/Loatm) (5.5 atm) = 0.2035 0.20 mol/L =

=

1 3 .39

Converting between molarity and molality involves conversion between volume of solution and mass of solution. Both of these quantities are given so interconversion is possible. To convert to mole fraction requires that the mass of solvent be converted to moles of solvent. Since the identity of the solvent is not given, conversion to mole fraction is not possible if the molar mass is not known. Why is the identity of the solute not necessary for conversion?

1 3 .4 1

Convert the masses to moles and the volumes to liters and use the definition of molarity: M = a) Molarity

=

b) Molarity =

( (

)[ � )( )( � )( I

42.3 g C 1 2 H 22 0 I l 1 00. m L 5 . 50 g LiN0 3 505 mL

1 L 1 0- L

1 1 0- L

1 mol C 1 2 H 22 0 1 1 342.30 g C I 2 H 22 0 1 1

mol Li�0 3 68.95 g LIN0 3

1 66

)

=

)

=

mol of solute VeL) of solution

------

1 .235758 = 1 .24 M C 1 2 H 22 0 ))

0. 1 5 7956 = 0. 1 58 M LiN 0 3

1 3 .43

Dilution calculations can be done using Meone Veone = MdilVdil a) Meone = 0.250 M N aOH Veone = 75.0 mL Mdil = ? Vdil = 0.250 L ( 0.250 M ) ( 75.0 mL) 1 O -3 L = 0.0750 M Mdil = Me one Veone / Vdil = 1 mL ( 0.250 L. ) b) Meone = 1 .3 M HN0 3 V eone = 3 5 . 5 mL Mdil = ? Vdil = 0. 1 50 L ( 1 .3 M ) ( 35.5 mL ) 1 0 -3 L = 0.3076667 = 0.3 1 M Mdil = Me one V eone / Vdil = 1 mL ( 0. 1 50 L )

[ ) [ ) --

--

1 3 .45

1 3 .47

a) Find the number of moles KH 2 P04 needed to make 355 mL of this solution. Convert moles to mass using the molar mass of KH 2 P04 (Molar mass = 1 36.09 gimol) 2 1 0 -3 L 8.74 x 1 0 - mol KH 2 P0 4 1 36.09 g KH 2 P0 4 Mass KH 2 P04 = ( 355 mL ) L I mol KH 2 P0 4 I mL = 4.22246 = 4.22 g KH 2 P04 Add 4.22 g KH2P04 to enough water to make 355 mL of aqueous solution. b) Use the relationship Meone V eone = MdilVdil to find the volume of 1 .25 M NaOH needed. Me one = 1 .25 M NaOH Veone = ? Mdil = 0.3 1 5 M NaOH Vdil = 425 mL V eone = Mdil Vdil / Meone = (0.3 1 5 M) (425 mL) / ( 1 .25 M) = 1 07. 1 = 1 07 mL Add 1 0 7 mL of 1 .25 M N aOH to enough water to make 425 mL of solution. Molality, m =

moles of solute kg of solvent

88.4 g Glycine (

-----

1 mol GlYCine l 75.07 g Glycine ( 1 .250 kg )

a) m glycine = b) m glycerol =

1 3 .49

M o I a I tty, · m=

m

1 3 .5 1

)(

[ )[

benzene =

8.89 g GIYCero /

)

I mol Glycerol

1 92.09 g Glycerol

( 75.0 g )

moles o f solute kg of solvent

= 0.942054 = 0.942

)[ )

glycine

1 03 g = 1 .287 1 466 = I kg

1 .29

m

glycerol

. U se d enstty to convert vo I ume to mass.

( )( J[ ) ( ) ( J( )( )

1 mol C6 H6 78. 1 1 g C6H6 0.660 g ( 1 87 mL C 6 H 1 4 ) mL

( 34.0 mL C6H 6 )

m

J

0.877 g mL

1 03 g I kg

--

=

3 .0930 = 3.09 m C6H6

)

2 2 a) The total weight of the solution is 3 .00 x 1 0 g, so mass so lute + maSS sol vent = 3 .00 x 1 0 g 0. 1 1 5 mol C 2 H 6 ° 2 62.07 g C 2 H60 2 1 kg Mass of C 2 H602 in 1 000 g ( 1 kg) of H 2 0 = I mol C 2 H60 2 1 kg H 2 0 1 039 = 7. 1 3 805 g C 2 H60 2 in 1 000 g H 2 0 (unrounded) Grams of this solution = 1 000 g H 2 0 + 7. 1 3805 g C 2 H60 2 = 1 007. 1 3 805 (unrounded) 2 7 1 3 805 g 2 3 .00 X 1 0 g solution = 2. 1 262378 = 2 . 1 3 g C 2 H60 2 H60 2 Mass C 2 H60 2 = ( . I I 007. 1 3 805 gC solution 2 2 2 Masssol vent = 3 .00 x 1 0 g - masssolute = 3 .00 x 1 0 g - 2. 1 262378 g C 2 H602 = 297. 87376 = 2.98 x 1 0 g H 2 0 Therefore, add 2 . 1 3 g C2H602 to 298 g of H20 to make a 0. 1 1 5 m solution.

1 67

J[

b) This is a disguised dilution problem. First, determine the amount of solute in your target solution: 2.00% ( 1 00% ( l .00 kg ) 0.0200 kg RN03 (solute) Then determine the amount of the concentrated acid solution needed to get 0.0200 kg solute: 62.0% ( mass needed ) 0.0200 kg 1 00% Mass solute needed 0.032258 0.0323 kg Mass solvent Mass solution - Mass solute 1 .00 kg - 0.032258 kg 0.96774 0.968 kg Add 0.0323 kg of the 62.0 % (w/w) UN03 to 0.968 kg H20 to make 1 .00 kg of2.00% (w/w) H N 03

=

) ( )

=

=

1 3 .53

=

=

=

=

=

=

=

a) Mole fraction is moles of isopropanol per total moles. 0.30 mol Isopropanol . . . 0.2727272 0.27 (NotIce that mole fractIons have no UnIts.) X isopropan o l ( 0.30 + 0.80 ) mol b) Mass percent is the mass of isopropanol per 1 00 g of solution. From the mole amounts, find the masses of isopropanol and water: 60.09 g C3 H 7 0H . . Mass Isopropanol ( 0.30 mol C3 H 70H ) 1 8.027 g Isopropanol (unrounded) 1 mol C3 H 7 0H Mass water

=

=

( 0.80 mol

H20

=

)

(

=

(

1 8.02 g H 2 0 1 mol H 2 0

)=

)=

1 4.4 1 6 g water (unrounded)

=

( 1 8 .027 g Isopropanol ) x 1 00% 55 .565 1 ( 1 8.027 + 1 4 . 4 1 6 ) g c) Molality of isopropanol is moles of isopropanol per kg of solvent. . . 0.30 mol Isopropanol 1 0 3 g 20.8 1 02 Molaltty Isopropanol 1 4.4 1 6 g Water 1 kg Percent isopropanol

[

=

1 3 .55

--

J=

=

=

56%

21

m

isopropanol . Isopropanol

The information given is 8.00 mass % N H 3 solution with a density of 0.965 1 g/mL. For convenience, choose exactly 1 00.00 grams of solution. Determine some fundamental quantities: 8.00% N H 3 Mass 0 f N H 3 ( 1 00 g solutIOn ) 8.00 g N H 3 1 00% solutIOn Mass H20 mass of sol ution - mass NH3 ( 1 00.00 - 8.00) g 92.00 g H20 I mol N H 3 Moles NH3 ( 8.00 g N H 3 ) I 0.469759 mol N H 3 (unrounded) 1 7.03 g N H 3 ) Moles

= = =

H20

=

(

.

(

( 92.00 g

H20

)

.

=

(

)=

=

=

I mol H 2 0 1 8 .02 g H 2 0

)=

5 . 1 054 mol

H20

)(

(unrounded)

I mL solutio 1 0-3 L I 0. 1 036 1 6 L (unrounded) � 0.965 1 g solutIOn 1 mL ) Using the above fundamental quantities and the definitions of the various units: Moles solute 0.469759 mol NH 3 I 03 g Molality 5 . 1 06076 5. 1 1 m NH3 kg solvent 92.00 g H 2 0 I kg Volume solution

=

( 1 00.00 g solution )

( =(

(

)[ = J )= =

_

-

=

Molarity Mole fraction

=

= = X

=

--

=

Moles solute 0.469759 mol NH 3 4.53365 4.53 M NH3 0. 1 036 1 6 L L solution Moles substance 0.469759 mol N H 3 0.084259 0.0843 total moles ( 0.469759 + 5 . 1 054 ) mol

----

------

1 68

=

=

1 3 .57

ppm

=

(

)

mass solu te . mass solutIOn

x

1 06

The mass of 1 00.0 L of waste water solution is ( 1 00.0 ppm Ca

2+

=

2 ppm Mg + =

( (

2 0.22 g Ca + 1 .00 I x 1 0 5 9 solution 2 0.066 g Mg + s . 1 .00 1 x 1 0 9 solutIOn

J )

10

X

x

6

1 06

L

=

=

solution )

2 . 1 978

=

0.65934

( � )( I

1 0-

1 .00 1 g

L

2.2 ppm Ca

=

I mL

2+

0.66 ppm M g

)

=

1 .00 I

x

1 05 g.

2+

1 3 .60

The boiling point temperature is higher and the freezing point temperature is lower for the solution compared to the solvent because the addition of a solute lowers the freezing point and raises the boiling point of a l iquid.

1 3 .64

Strong electrolytes are substances that produce a large number of ions when dissolved in water; strong acids and bases and soluble salts are strong electrolytes. Weak electrolytes produce few ions when dissolved in water; weak acids and bases are weak electrolytes. Nonelectrolytes produce no ions when dissolved in water. Molecular compounds other than acids and bases are nonelectrolytes. a) Strong electrolyte When hydrogen chloride is bubbled through water, it dissolves and dissociates completely into H+ (or H30+) ions and cr ions. b) Strong electrolyte Potassium nitrate is a soluble salt. c) Nonelectrolyte Glucose solid dissolves in water to form individual C6H I 206 molecules, but these units are not ionic and therefore do not conduct electricity. d) Weak electrolyte Ammonia gas dissolves in water, but is a weak base that forms few NH/ and OH- ions.

1 3 .66

To count solute particles in a solution of an ionic compound, count the number of ions per mole and multiply by the number of moles in solution. For a covalent compound, the number of particles equals the number of molecules. 0.2 mol Kl 2 mol particles ( 1 L ) 0.4 mo I 0 f partlc · I es a) I mol Kl L KI consists of K+ ions and r ions, 2 particles for each Kl . 0.070 mol HN0 3 2 mo l particles ( 0. 1 4 mol of particles I L) b) 1 mol HN0 3 L HN03 is a strong acid that forms hydronium ions and nitrate ions in aqueous solution. 4 1 0 - mol K2S0 4 3 mol particles (1 L ) 3 x 1 0-4 mol of particles c) 1 mol K2S0 4 L Each K2S04 forms 2 potassium ions and I sulfate ion in aqueous solution. 0.07 mol C2 H sOH 1 mo l particles ( L ) 0.07 mol of particles d) I L 1 mol C2 HsOH Ethanol is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is the same as the number of moles of molecules, 0.07 mol in I L.

( (

)(

J( )( )(

(

(

1 3 .68

)

=

)

)

=

=

)

=

(

)[ )

The magnitude of freezing point depression is directly proportional to molality. ( 1 0. O g CH 3 0H ) I mol CH 3 0H 1 03 g = 3 . 1 2 1 0986 = 3 . 1 2 m CH30H a) Molality of CH30H = ( 1 00. g H 2 0) 32.04 g C H 3 0H 1 kg ( 20.0 g CH 3 C H 2 0H ) 1 mol C H 3 CH 2 0H 1 03 g . MolalIty of CH3CH20H ( 200. g H 2 0 ) 46.07 g CH 3 CH 2 0H I kg 2 . 1 706 2 . 1 7 m CH3CH20H =

=

(

)[ ) --

=

1 69

The molality of methanol, CH 3 0H, in water is 3 . 1 2 m whereas the molality of ethanol, CH 3 CH 2 0H, in water is 2 . 1 7 m . Thus, CH 3 0H/H 2 0 solution has the lower freezing point. ( 1 0.0 g H 2 0 1 mol H 2 ° = 0.5549 = 0.55 m H 2 0 b) Molality of H 2 0 = ( l .00 kg CH 3 0H 1 8.02 g H 2 0 ( 1 0.0 g CH 3 CH 2 OH 1 mol CH 3 CH 2 OH Molality of CH 3 CH 2 0H = = 0.2 1 706 = 0.2 1 7 m CH3CH 2 0H ( l .00 kg CH 3 0H 46.07 g CH 3 CH 2 0H

)[ )

)

)[ J

J

The molality of H 2 0 in CH 3 0H is 0.555 m , whereas CH 3 CH 2 0H in CH 3 0H is 0.2 1 7 m . Therefore, HzO/CH30H solution has the lower freezing point. 1 3 . 70

To rank the solutions in order of increasing osmotic pressure, boiling point, freezing point, and vapor pressure, convert the molality of each solute to molality of particles in the solution. The higher the molality of particles the higher the osmotic pressure, the higher the boiling point, the lower the freezing point, and the lower the vapor pressure at a given temperature. 2 mol particles (I) ( 0. 1 00 m NaN0 3 = 0.200 m IOns 1 mol NaN0 3 NaN0 3 consists ofNa+ ions and N0 3- ions, 2 particles for each NaN0 3 . mol particles = 0.200 m molecules (II) ( 0.200 m glucose 1 mol glucose Ethanol is not an ionic compound so each molecule dissolves as one particle. The number of moles of particles is the same as the number of moles of molecules. 3 mol particles = 0.300 m ions (III) ( 0. 1 00 m CaCl 2 1 mol CaCI 2 CaCI 2 consists of Ca+2 ions and cr ions, 3 particles for each CaCI 2 • a) Osmotic pressure: III = IIn < II I I I b) Boiling point: bpI = bP n < b Pm c) Freezing point: fp l I I < fpl = fp n d) Vapor pressure at 50°C: VP l n < VPI = vPn

)[

J

) [1

)[

1 3 .72

J

J

)[

The mol fraction of solvent affects the vapor pressure according to the equation: P solvent = XsolventPO solvent 1 mol C 3 Hg0 3 Moles C 3 Hs0 3 = ( 44.0 g C 3 Hg0 3 = 0.47779 mol C 3 Hs0 3 (unrounded) 92.09 g C 3 Hg0 3 1 mol H 2 0 = 27.7469 mol H 2 0 (unrounded) Moles H 2 0 = ( 500.0 g H 20 ) 1 8.02 g H 2 0 mol H 2 0 27.7469 mol H 2 0 = 0.98307 (unrounded) Xsolvent mol H 2 0 + mol glycerol 27.7469 mol H z O + 0.47779 mol glycerol Psolvent = Xsol ventPOsolvent = (0.98307) (23 .76 torr) = 23.35774 = 23.36 torr

[

_

1 3 .74

.

J

J

The change in freezing point is calculated from i1Tr = iKttn, where Kr is 1 .86°Clm for aqueous solutions, i is the van ' t Hoff factor, and m is the molality of particles in solution. Since urea is a covalent compound and does not ionize in water, i = I . Once i1T r is calculated, the freezing point is determined by subtracting it from the freezing point of pure water (O.OO°C). i1Tr = iKr m = ( 1 ) ( 1 .86°Clm) (0. 1 1 1 m ) = 0.20646°C (unrounded) The freezing point is O.OO°C - 0.20646°C = -0.20646 = -O.206°C.

1 70

1 3 . 76

The boiling point of a solution is increased relative to the pure solvent by the relationship � T b = iKwn . Vanillin is

)[

a nonelectrolyte so i = l . The molality must be calculated, and Kb is given ( l .22°C ( 3 .4 g vanillin )

(

I mol Vanillin

1 5 2 . 1 4 g Vanillin Molality of Vanillin = -------;------'------'-:---.:.... ( 5 0 . 0 g Ethanol )

The boiling point is 7 8 . 5 °C + 0 . 54528 7°C = 79.045287 = 1 3 .78

1 kg

)

m Vanillin (unrounded) m = ( l ) ( l .22°C I m) (0.44695675 m) = 0 . 545287°C (unrounded)

= 0.44695675 �Tb = i Kb

3

10 g

1 m).

79.0°C .

The molality of the solution can be determined from the relationship �Tf = iKrm with the value l . 86°Clm inserted

for Kr and i = 1 for the nonelectrolyte ethylene glycol (ethylene glycol is a covalent compound that will form one particle per molecule when dissolved) . Convert the freezing point of the solution to °C and find �Tf: °C = (5/9) CF - 3 2 . 0) = (5/9) ((- 1 0. 0)OF - 3 2 . 0) = -23 . 3 3 3 3°C (unrounded)

�Tr = (0.00 - (-23 . 3 3 3 3 » OC = 2 3 . 3 3 3 3 °C �T m = _f Kf

=

23 . 3 3 3 3° C

1.86 °Clm

=

12.54478 m (unrounded)

Ethylene glycol will be abbreviated as E O

(

)

(

)

Multiply the molality by the given mass of solvent to find the mass of ethylene glycol that must be in solution. M ass eth y I ene g I yco I = =

12.54478 mol

EG

I kg H 2 0

( 1 4 . 5 kg H 2 0 )

1.129049 x 104 = 1 . 1 3 X 1 0 4 g ethylene glycol

62.07 g EO 1 mol E G

To prevent the solution from freezing, dissolve a minimum of 1 . 1 3 x 1 3 .80

104 g ethylene glycol in

1 4 . 5 kg water.

Convert the mass percent t o molality and use � T = iKbm t o find the van ' t Hoff factor. a)

Assume exactly

Mass NaCI =

(100 g solutIon )

(

1 00% solution

)

)

= 1 . 00g NaCI

1 mol NaC I = 0 . 0 1 7 1 1 1 6 mol NaCI 58.44 g NaCI H20 = 100.00 g solution - 1.00g NaC I = 99.00 g H20

Moles NaCI = ( l . 00 g NaC l ) Mass

(

100 grams of solution. 1.00% NaCI .

Molal ity NaCI =

mo le NaC I kg H z O

=

0 . 0 1 7 1 1 1 6 mol NaCI

( ) 10

3 g

1 kg

99.00 g H 2 0

Calculate �T = (0. 000 - (-0 . 5 9 3 ))DC = 0 . 5 9 3 °C �Tf = i f

= 0. 1 72 844 =

0.1 73

m

NaC I

K m 0 . 5 93 ° C �Tf = = 1 . 8445 3 7 = 1 .84 i = K f m (1.86°Clm ) (0.172844 m) The value of i should b e close t o 2 because NaCI dissociates into 2 particles when dissolving i n water. b) For acetic acid, CH3COOH : Assume exactly

100 grams

Mass CH3COOH =

.

(100 g solutIOn )

Moles CH)COOH = ( 0 . 5 00 g Mass H20 =

(

of solution. 0 . 5 00% CH COOH 3 . 1 00% solutI On

CH3COOH)

(

)

= 0 . 5 00 g C H)COOH

I mol CH 3 COOH

60.05 g

CH3COOH

J

=

100.00 g solution - 0 . 500 g CH)COOH = 9 9 . 5 00 g H20

171

0.0083264 mol CH)COOH

( )

0.0083264 mol CH 3 COOH 1 03 g 1 kg 99.500 g H 2 0

= 0.083682 = 0.0837 m

CH3COOH

Calculate �T = (0.000 - (-0. I 59))OC = 0. 1 59°C �Tf = i Kf m �Tf 0. 1 59 ° C = 1 .02 1 53 = 1 .02 i= = Kfm ( 1 . 86°Clm ) ( 0.083682 m ) Acetic acid is a weak acid and dissociates to a small extent in solution, hence a van ' t Hoff factor that is close to 1 . 1 3 .83

The pressure of each compound i s proportional to its mole fraction according to Raoult ' s law: P A = XA po A moles C H 2CI2 1 .50 mol = 0.600 X CH,Ci, = moles CH2CI2 + mol CCI 4 1 .50 + 1 .00 mol 1 .00 mol moles CCI 4 = 00400 1 .50 + 1 .00 mol PA = XA p eA = (0.600) (352 torr) = 2 1 1 .2 = 211 torr CH2CI2 = (00400) ( 1 1 8 torr) = 47.2 torr CCI4

1 3 .85

To find the volume of seawater needed, substitute the given information into the equation that describes the ppb concentration, account for extraction efficiency, and convert mass to volume using the density of seawater. mass Gold x 1 09 1 . 1 x 1 0-2 ppb = mass seawater 3 1 . 1 g Au 9 x 10 1 . 1 x 1 0-2 ppb = mass seawater mass seawater =

[

]

31.1 g 9 2 2 x 1 0 = 2.827273 x 1 0 1 g (with 1 00% efficiency) 1 . 1 x 1 0-

(

1 00% 2 Mass of seawater = ( 2 .827273 x 1 01 g ) = 3.5563 X 1 0 1 2 g seawater (unrounded) (79.5% efficiency) 79.5% 2 1 mL 1 0.3 L Volume seawater = ( 3 . 5 563 x 1 01 g ) = 3 046956 X 1 09 = 3.5 X 109 L 1 .025 g 1 mL

(

1 3 .90

J )(---)

--

---

a) First, find the molality from the freezing point depression and then use the molality, given mass of solute and volume of water to calculate the molar mass of the solute compound. Assume the solute is a nonelectrolyte (i = 1 ). �T f = iK f m = ( 0 . 0 0 0 - ( - 0 . 2 0 1 ) ) = 0 . 2 0 1 ° C �T 0.20 1 ° C m = _f = = 0. 1 080645 m (unrounded) Kfi ( 1 . 86°Clm ) ( 1 )

( J( )

1 .00 g 1 g moles solute � = 0.0250 kg water ( 25.0 mL ) kg solvent 1 mL 1 0 g moles solute = (m)(kg solvent) = (0. 1 080656 m)(0.0250 kg) = 0.00270 1 6 mol 0.243 g = 89.946698 = 89.9 g/mol molar mass = 0.00270 1 6 mol

m=

1 72

b) Assume that 1 00.00 g of the compound gives 53.3 1 g carbon, 1 1 . 1 8 g hydrogen and 1 00.00 - 53.3 1 - 1 1 . 1 8 = 3 5 . 5 1 g oxygen. I mol C = 4.43 880 mol C (unrounded) Moles C = ( 53.3 1 g C ) 1 2.0 1 g C

Moles H = ( 1 1 . 1 8 g H

Moles 0 = ( 3 5 . 5 1 g 0

( )( )(

) ) )

1 mol H = 1 1 .09 1 27 mol H (unrounded) 1 .008 g H

1 mol 0 = 2 .2 1 9375 mol 0 (unrounded) 1 6.00 g 0 Dividing the values by the lowest amount of moles (2.2 1 9375) will give 2 mol C, 5 mol H and 1 mol 0 for an empirical formula C2HsO with molar mass 45.06 g/mol. Since the molar mass of the compound , 89.9 g/mol from part (a), is twice the molar mass of the empirical formula, the molecular formula is 2(C2HsO) or C4HI O02 . c) There is more than one example in each case. Possible Lewis structures: Does not form hydrogen bonds Forms hydrogen bonds H H H H H H H H

I I H

I I H

I I H

I I H

HO- C -- C -- C -- C -- OH

1 3 .92

H

-- I -- -- I C

I

O

I I H

I -- H I H

C -- O -- C -- C

I

H

H

a) Use the boiling point elevation of 0.45°C to calculate the molality of the solution. Then, with molality, the mass of solute, and volume of water calculate the molar mass. �T = iKbm i = 1 (nonelectrolyte) �T = ( 1 00.45 - 1 00.00)OC = 0.45°C ° �Tb 0.45 C m = _. = = 0.878906 m = 0.878906 mol/kg (unrounded) Kbl ( 0.5 1 2 ° C/m ) ( 1 )

moles solute 0.997 g I g � = 0.0249250 kg water ( 25.0 mL ) kg solvent 1 mL 10 9 moles solute = (m)(kg solvent) = (0.878906 m)(0.0249250 kg) = 0.02 1 9067 mol 1 .50 g = 68.4722 = 68 g/mol molar mass = 0.02 1 9067 mol b) The molality calculated would be the moles of ions per kg of solvent. If the compound consists of three ions the molality of the compound would be 1 132 of 0.878906 m and the calculated molar mass would be three times greater: 3 x 68 .4722 = 205 .4 1 7 = 2.1 x 1 0 g/mol . c) The molar mass of CaN 2 06 is 1 64. 1 0 g/mol. This molar mass is less than the 2 . 1 x 1 0 2 g/mol calculated when the compound is assumed to be a strong electrolyte and is greater than the 68 g/mol calculated when the compound is assumed to be a nonelectrolyte. Thus, the compound is an electrolyte, since it dissociates into ions in solution. However, the ions do not dissociate completely in solution. d) Use the molar mass of CaN 2 06 to calculate the molality of the compound. Then calculate i in the boiling point elevation formula. 1 .50 g caN 2 06 mL 1 03 g I mol = 0.3667309 m (unrounded) m= 25.0 mL 0.997 g 1 kg 1 64. 1 0 g CaN 2 06 �T = iKbm ( 0.45°C ) -- 2.396597 = 2. 4 . SIb I = -- = ,----Kbm ( 0.5 1 2°/m ) ( 0.3667309 m )

(

)( )

(

m=

)(

) [ )(

-,--- -'--:-- ---'-- --:-

J

=

1 73

13.95

a) From the osmotic pressure, the molarity of the solution can be found. The ratio of mass per volume to moles per volume gives the molar mass of the compound. nV = nRT n = MRT

M=

n

-

RT

(

=

0.0821

L · atm

K

mol ·

)

+

I atm

760 torr ((273 25 ) K)

)

( J

= (1.828546 x 10-5 M (30.0 mL ) 10-3 L I mL

(M)(V) = moles Moles

(

(0.340 torr )

(10.0 mg )

( J 10-

3

g

=

)

= 1.828546 x 10-5 M (unrounded)

5.48564 x 1 0-7 mol

1 mg

= 1.82294 x 104 = 1 .82 X 1 0 4 g/moi mol b) To find the freezing point depression, the molarity of the solution must be converted to molality. Then use �Tf = iKtffl · (i = 1) Mass solvent = mass of solution - mass of solute Molar mass =

Mass solv.n!

�

Moles solute =

[(

[

5.48564 x 10-7

30.0 mL )

(

: ) - (10.0

09 ;

g

1.828546 X 10-5 mOI L

mg

)[ )

l( \0::)1 U;g ) g

10-3 L (30.0 mL ) 1 mL

=

�

0.0299 kg

5.485638 x 10-7 mol (unrounded)

5.485638 X 10-7 mol = 1 . 83466 x 10-5 m ( unroun de d) 0.0299 kg �Tf = iKtffl = (1) (1.86°C/m) (1.83466 x 10 5 m) = 3.412 x 10-5 = 3.41 X to-SOC (So the solution would freeze at 0 - 3.41 X 1O 5 o c = -3.41 x 1 O-5 0C.) M o I a I Ity · =

-

-

13.101

a) Assume a 100 g sample of urea. This leads to the mass of each element being equal to the percent of that element.

( )( )(

Moles

C = (20.1 g C)

Moles

H = (6.7 g H

Moles N =

( 46.5 g N

) ) )

1 mol C 12.01 g C

= 1.6736 mol

C (unrounded)

I mol H = 6.6468 mol H (unrounded) 1.008 g H

I mol N = 3.31906 mol N (unrounded) 14.01 g N

(

I mol 0

)

= 1.66875 mol 0 (unrounded) 16.00 g 0 Dividing all by the smallest value (1.66875 mol 0) gives: C = I , H = 4, N = 2, 0 = I . Thus, the empirical formula is CH4N20. The empirical formula weight is 60.06 glmo!.

Moles

0 = ( C I OO - 20.1 - 6.7 - 46.5) g 0 )

b) Use n = MRT to solve for the molarity of the urea solution. The solution molarity is related to the concentration expressed in % w/v by using the molar mass.

M = n / RT =

(

( 2.04 atm )

0.082 1

L · atm mol ' K

)

= 0.083 3 8 1 7 M (unrounded)

+

((273 25) K )

174

( e ) 1 ·

Mol", mass =

Og

L

0.083 3 8 1 7

mol

= 5 9 . 965 = 60. g1mol

L

Because the molecular weight equals the empirical weight, the molecular formula is also CH4NzO.

1 75

Chapter 1 4 Periodic Patterns in the Main-Group Elements FOLLOW-UP PROBLEMS

There are no fol low-up problems for this chapter. END-OF-CHAPTER PROBLEMS

1 4. 1

Ionization energy i s defined as the energy required to remove the outermost electron from an atom. The further the outermost electron is from the nucleus, the less energy is required to remove it from the attractive force of the nucleus. I n hydrogen, the outermost electron is in the n = 1 level and in lithium the outermost electron is in the n = 2 level. Therefore, the outermost electron in lithium requires less energy to remove, resulting in a lower ionization energy.

1 4.2

a) 2 AI(s) + 6 HCI(aq) � 2AICI 3 (aq) + 3 H 2 (g) Active metals displace hydrogen from HCI by reducing the H+ to H 2 . b) LiH(s) + H 2 0(l) � LiOH(aq) + H 2 (g) In water, H- (here in LiH) reacts as a strong base to form H 2 and OH-.

1 4.4

a)

Na = + l B = +3 B = +3 Al = +3 Al = +3 Li = + 1 b) The polyatomic ion in NaBH4 is [BH4r. therefore, its shape is tetrahedral. H H

-

I I H

B

-

H = -l in NaBH4 H = -1 in AI(BH4)3 H = - I in LiAIH4 Boron is the central atom and has four surrounding electron groups;

H

1 4.7

a) Alkali metals generally lose electrons (act as reducing agents) in their reactions. b) Alkali metals have relatively low ionization energies, meaning they easily lose the outermost electron. The electron configurations of alkali metals have one more electron than a noble gas configuration, so losing an electron gives a stable electron configuration. c) 2 Na(s) + 2 H 2 0(l) � 2 Na+(aq) + 2 Otr(aq) + H 2 (g) 2 Na(s) + CI 2 (g) � 2 NaCI(s)

1 4.9

a) Density increases down a group. The increasing atomic size (volume) is not offset by the increasing size of the nucleus (mass), so ruN increases. b) Ionic size increases down a group. Electron shells are added down a group, so both atomic and ionic size mcrease. c) E-E bond energy decreases down a group. Shielding of the outer electron increases as the atom gets larger, so the attraction responsible for the E-E bond decreases. d) l E I decreases down a group. Increased shielding of the outer electron is the cause of the decreasing lEI . e) bJih ydr decreases down a group. bJihydr i s the heat released when the metal salt dissolves in, or i s hydrated by, water. Hydration energy decreases as ionic size increases. Increasing down Decreasing down

a and b c, d, and e

1 76

1 4. 1 1

Peroxides are oxides in which oxygen has a -1 oxidation state. Sodium peroxide has the formula Na2 02 and is formed from the elements Na and O2 , 2 Na(s) + 02 (g) � Na2 02 (S)

1 4. 1 3

The problem specifies that an alkali halide is the desired product. The alkali metal i s K (comes from potassium carbonate, K 2 C0 3 (S)) and the halide is I (comes from hydroiodic acid, HI(aq)). Treat the reaction as a double displacement reaction. K2 C0 3 (S) + 2 H I (aq) � 2 KJ(aq) + H 2 C03(aq) However, H 2 C0 3 (aq) is unstable and decomposes to H 2 0(!) and CO 2 (g), so the final reaction is: K2 C0 3 (S) + 2 HI(aq) � 2 K J (aq) + H P (!) + CO 2 (g)

1 4. 1 7

Metal atoms are held together by metallic bonding, a sharing of valence electrons. Alkaline earth metal atoms have one more valence electron than alkali metal atoms, so the number of electrons shared is greater. Thus, metallic bonds in alkaline earth metals are stronger than in alkali metals. Melting requires overcoming the metallic bonds. To overcome the stronger alkaline earth metal bonds requires more energy (higher temperature) than to overcome the alkali earth metal bonds. First ionization energy, density, and boiling points will be larger for alkaline earth metals than for alkali metals.

1 4. 1 8

a) A base forms when a basic oxide, such as CaO, i s added to water. CaO(s) + H 2 0(!) � Ca(OHMs) b) Alkaline earth metals reduce O 2 to form the oxide. 2 Ca(s) + 0 2 (g) � 2 CaO(s)

1 4.20

Beryllium is generally the exception to properties exhibited by other alkaline earth elements. a) Here, Be does not behave like other alkaline earth metals: BeO(s) + H 2 0(!) �NR. The oxides of alkaline earth metals are strongly basic, but BeO is amphoteric. BeO will react with both acids and bases to form salts, but an amphoteric substance does not react with water. b) Here, Be does behave like other alkaline earth metals: BeCI 2 (!) + 2 Cr(solvated) � BeCI /-(solvated) Each chloride ion donates a lone pair of electrons to form a covalent bond with the Be in BeCh. Metal ions form similar covalent bonds with ions or molecules containing a lone pair of electrons. The difference in beryllium is that the orbital involved in the bonding is a p orbital, whereas in metal ions it is usually the d orbitals that are involved.

1 4.22

For period 2 elements in the first four groups, the number of covalent bonds equals the number of electrons in the outer level, so it increases from 1 covalent bond for lithium in group 1 A( l ) to 4 covalent bonds for carbon in group 4A( l 4). For the rest of period 2 elements, the number of covalent bonds equals the difference between 8 and the number of electrons in the outer level. So for nitrogen, 8 - 5 = 3 covalent bonds; for oxygen, 8 - 6 = 2 covalent bonds; for fluorine, 8 - 7 = 1 covalent bond; and for neon, 8 - 8 = 0, no bonds. For elements in higher periods, the same pattern exists but with exceptions for groups 3A( 1 3) to 7 A( 1 7) when an expanded octet allows for more covalent bonds.

1 4.25

The electron removed in Group 2A(2) atoms is from the outer level s orbital, whereas in Group 3A( l 3) atoms the electron is from the outer level p orbital. For example, the electron configuration for Be is Is 2 2i and for B is l i2i2p l . It is easier to remove the p electron of B than the s electron of Be, because the energy of a p orbital is slightly higher than that of the s orbital from the same level. Even though the atomic size decreases from increasing ZeIT, the IE decreases from 2A(2) to 3A( l 3).

1 4.26

a) Compounds of Group 3 A( l 3) elements, like boron, have only six electrons in their valence shell when combined with halogens to form three bonds. Having six electrons, rather than an octet, results in an "electron deficiency." b) As an electron deficient central atom, B is trigonal planar. Upon accepting an electron pair to form a bond, the shape changes to tetrahedral. BF 3 (g) + NH 3 (g) � F 3 B-N H 3 (g) B(OH) 3 (aq) + O W (aq) � B(OH)4-(aq)

1 77

1 4.30

Halogens typically have a -1 oxidation state in metal-halide combinations, so the apparent oxidation state of TI +3 . However, the anion 1 3- combines with TI in the + 1 oxidation state. The anion 13- has [3 x (I)7e-] + [ l e- from the charge] 22 valence electrons; 4 of these electrons are used to form the two single bonds between iodine atoms and 1 6 electrons are used to give every atom an octet. The remaining two electrons belong to the central I atom; therefore the central iodine has five electron groups (2 single bonds and 3 lone pairs) and has a general formula of AX2E 3 • The electrons are arranged in a trigonal bipyramid with the three lone pairs in the trigonal plane. It is a linear molecule with bond angles 1 80°. (TI 3+) (n 3 does not exist because of the low strength of the Tl-I bond. =

=

=

O.N. = +3 (apparent) ; = +1 (actual)

[: ·i: J._ . 'i:] .. . ..

.

1 4.35

Network covalent solids differ from molecular solids in that they are harder and have higher melting points than molecular solids. A network solid is held together by covalent bonds while intermolecular forces hold molecules in molecular solids together. Covalent bonds are much stronger than intermolecular forces and require more energy to break. To melt a network solid, more energy (higher temperature) is required to break the covalent bonds than is required to break the intermolecular forces in the molecular solid, so the network solid has a higher melting point. The covalent bonds also hold the atoms in a much more rigid structure than the intermolecular forces hold the molecules. Thus, a network solid is much harder than a molecular solid.

1 4.36

Oxide basicity is greater for the oxide of a metal atom. Tin(IV) oxide is more basic than carbon dioxide since tin has more metallic character than carbon.

1 4.38

a) I E I values generally decreases down a group. b) The increase in ZeIT from Si to Ge is larger than the increase from C to Si because more protons have been added. Between C and Si an additional 8 protons have been added, whereas between Si and Ge an additional 1 8 (includes the protons for the d-block) protons have been added. The same type o f change takes place when going from Sn to Pb, when the 1 4 fblock protons are added. c) Group 3A( 1 3) would show greater deviations because the single p electron receives no shielding effect offered by other p electrons.

1 4.4 1

Atomic size increases moving down a group. As atomic size increases, ionization energy decreases so that it is easier to form a positive ion. An atom that is easier to ionize exhibits greater metallic character.

1 4.43 (a) : O

-

. ". 0/ :

(b)

8-

:0:

I ' : 0: Si - O. 1 '. "- Si

'

--

O:

I I " : 0: I ""O· Si/ .O. : : 0: . I

: O--Si

•

--

H

H

-

H

-

H

1 1 1 1 C-C 1 1 H C-C

-

H

-

H

H

--

o

There is another answer possible for C 4 Hg.

1 78

14.46

a) Diamond, C, a network covalent solid of carbon b) Calcium carbonate, CaC03 (brands that use this compound as an antacid also advertise them as an important source of calcium) c) Carbon dioxide, CO2, is the most widely known greenhouse gas; CH4 is also implicated. d) CO binds t o Fe(II) in blood, p reventing t h e normal binding of O2, e) Lead, Pb ( in old plumbing as lead solder, and in paint as a pigment)

1 4.50

a) In Group 5A(15), all elements except bismuth have a range of oxidation states from -3 to +5. b) For nonmetals, the range of oxidation states is from the lowest at group number (A) 8, which is S 8 = -3 for Group SA, to the highest equal to the group number (A), which is +S for Group SA. -

-

1 4.S 1

a) Element oxides are either ionic or covalent molecules. Elements with low electronegativity (metals) form ionic oxides, whereas elements with relatively high electronegativity (nonmetals) form covalent oxides. The greater the electronegativity of the element, the more covalent the bonding is in its oxide. b) Oxide acidity increases to the right across a period and decreases down a group, so increasing acidity directly correlates with increasing electronegativity.

1 4.S4

a) With excess oxygen arsenic will form the oxide with arsenic in its highest possible oxidation state, +S . 4 As(s) + S 0 2 (g) � 2 AS2 0S(S) b) Trihalides are formed by direct combination of the elements (except N). 2 B i(s) + 3 F 2 (g) � 2 B i F 3 (s) c) Metal phosphides, arsenides, and antimonides react with water to form Group SA hydrides. Ca3 As 2 (S) + 6 H 2 0(l) � 3 Ca(OH)z(s) + 2 AsH 3 (g)

1 4.60

a) Both groups have elements that range from gas to metalloid to metal. Thus, their boiling points and conductivity vary in similar ways down a group. b) The degree of metallic character and methods of bonding vary in similar ways down a group. c) Both P and S have allotropes and both bond covalently with almost every other nonmetal. d) Both N and 0 are diatomic gases at normal temperatures and pressures. Both N and 0 have very low melting and boiling points. e) Oxygen, O 2 , is a reactive gas whereas nitrogen, N 2 , is not. Nitrogen can exist in multiple oxidation states, whereas oxygen has 2 oxidation states.

1 4.62

a) To decide what type of reaction will occur, examine the reactants. Notice that sodium hydroxide is a strong base. Is the other reactant an acid? If we separate the salt, sodium hydrogen sulfate, into the two ions, Na+ and HS04-, then it is easier to see the hydrogen sulfate ion as the acid. The sodium ions could be left out for the net ionic reaction. NaHS04(aq) + NaOH(aq) � Na2 S04(aq) + H 2 0(l) b) As mentioned in the book, hexafluorides are known to exist for sulfur. These will form when excess fluorine is present. S 8 (S) + 24 F 2 (g) � 8 SF6(g) c) Group 6A( 1 6) elements, except oxygen, form hydrides in the following reaction. FeS(s) + 2 HCI(aq) � H 2 S(g) + FeCI 2 (aq)

1 4.64

a) Se is a nonmetal; its oxide is acidic. b) N is a nonmetal; its oxide is acidic. c) K is a metal; its oxide is basic. d) Be is an alkaline earth metal, but all of its bonds are covalent; its oxide is a mphoteric. e) Ba is a metal ; its oxide is basic.

1 4.66

a) 03, ozone b) S03, sulfu r trioxide (+6 oxidation state) c) S02, sulfu r dioxide

1 79

1 4.67

S2F I O(g) -7 SF4(g) + SF6(g) O.N. of S in S2F I O: - ( 1 0 x - 1 for F) I 2 +5 O.N. of S in SF 4 : - (4 x -I for F) = +4 O.N. of S in SF6: - (6 x -1 for F) = +6 =

1 4.68

a) Polarity is the molecular property that is responsible for the difference in boiling points between iodine monochloride and bromine. The stronger the intermolecular forces holding molecules together, the higher the boiling point. More polar molecules have stronger intermolecular forces and higher boiling points. Molecular polarity is a result of the difference in electronegativity, an atomic property, of the atoms in the molecule. b) ICI is a polar molecule while Br2 is nonpolar. The dipole-dipole forces between polar molecules are stronger than the dispersion forces between nonpolar molecules. The boiling point of the polar ICI is higher than the boiling point of Br2.

1 4.70

a) Bonding with very electronegative elements: + 1 , +3, + 5 , +7. Bonding with other elements: - 1 b) The electron configuration for CI is [Ne]3i3i. By adding on electron, CI achieves an octet similar to the noble gas Ar. By forming covalent bonds, CI completes or expands its octet by maintaining its electrons paired in bonds or lone pairs. c) Fluorine only forms the -1 oxidation state because its small size and no access to d orbitals prevent it from forming multiple covalent bonds. Fluorine's high electronegativity also prevents it from sharing its electrons.

1 4.7 1

a) The CI-CI bond is stronger than the Br-Br bond since the chlorine atoms are smaller than the bromine atoms, so the shared electrons are held more tightly by the two nuclei . b) The Br-Br bond is stronger than the I-I bond since the bromine atoms are smaller than the iodine atoms. c) The CI-CI bond is stronger than the F-F bond. The fluorine atoms are smaller than the chlorine but they are so small that electron-electron repulsion of the lone pairs decreases the strength of the bond.

1 4. 72

a) A substance that disproportionates serves as both an oxidizing and reducing agent. Assume that OH- serves as the base. Write the reactants and products of the reaction, and balance like a redox reaction. 3 Br2(l) + 6 O W(aq) -7 5 Br-(aq) + Br0 3-(aq) + 3 H20(l)

1 4.73

a) Iz(s) + H20(l) -7 H I (aq) + H I O(aq) b) Br2(l) + 2 I -(aq) -7 12(s) + 2 Br-(aq) c) CaF2(s) + H2S04 (l) -7 CaS0 4 (s) + 2 HF(g)

1 4.76

Heliu m is the second most abundant element in the universe. Argon is the most abundant noble gas in Earth 's atmosphere, the third most abundant constituent after N2 and O2,

1 4.77

Whether a boiling point is high or low is a result of the strength of the forces between particles. Dispersion forces, the weakest of all the intermolecular forces, hold atoms of noble gases together. Only a relatively low temperature is required for the atoms to have enough kinetic energy to break away from the attractive force of other atoms and go into the gas phase. The boiling points are so low that all the noble gases are gases at room temperature.

1 4.79

a) Alkali metals have an outer electron configuration of ns I . The first electron lost by the metal is the ns electron, giving the metal a noble gas configuration. Second ionization energies for alkali metals are high because the electron being removed is from the next lower energy level and electrons in a lower level are more tightly held by the nucleus. The metal would also lose its noble gas configuration. b) The reaction is 2 CSF2(S) -7 2 CsF(s) + F2(g) You know the t!Jf1 for the formation of CsF: Cs(s) + 1 I2F2(g) -7 CsF(s) t!Jf1 = -530 kJ/mol You also know the t!Jfl for the formation of CsF2: Cs(s) + F2(g) -7 CSF2(S) t!Jf1 = - 1 25 kJ/mol

1 80

To obtain the heat of reaction for the breakdown of CsF 2 to CsF, combine the formation reaction of CsF with the reverse of the formation reaction of CsF 2 , both multiplied by x kJ) f.?W � CsF(s) CsFls) � -¥#H ;?,-fJ{g} = x kJ) (Note sign change) CSF 2 (S) � CsF(s) F 2 (g) = kJ kJ of energy are released when moles of CsF 2 convert to moles of CsF, so heat of reaction for one mole of CsF is or -405 kJ/mol.

2:MfMf 22 ((-+513205 Mf2 -8 1 0

�

2 + 2 + 2 2 8 1 0 - 8 1 012 2 + 14.8 1 31 .02 3 1 .02,

=

a) Empirical formula HNO has a molar mass of so hyponitrous acid with a molar mass of glmol which is twice would have a molecular formula twice the empirical formula, H 2N 2 02 . In addition, the molecular formula of nitroxyl would be the same as the empirical formula, HNO, since the molar mass of nitroxyl is the same as the empirical formula. b) H

-- -- == -- -O

N

N

O

6 2 . 04

H

H

- == N

O

c) In both hyponitrous acid and nitroxyl, the nitro gens are surrounded by electron groups (one single bond, one double bond and one unshared pair), so the electron arrangement is trigonal planar and the molecular shape is bent.

d)

[ : (:

..

..

N == N

3

]2>� :

.'N

"/ o

2-

=/

: 0:

N

In a disproportionation reaction, a substance acts as both a reducing agent and oxidizing agent because an atom within the substance reacts to form atoms with higher and lower oxidation states.

14.84 -1 -l/3 + 2+4 + +5 -1 + +3+1 +4 +3 +4 +5. + + + 2 0 -1 0 +1 . -3 + 3 0 + 2 3 0 , + 3 +63 + 2 2+7 ++4 + 4 +6 +7 +6 +4. 3 +1 +3 + 2 0 + +3 +1 181 o

a) I 2 (s) KI(aq) � KI 3 (aq) 1 in 1 2 reduces to 1 in KI 3 . 1 in K I oxidizes to 1 in KI 3 . This is not a disproportionation reaction since different substances have atoms that reduce or oxidize. The reverse direction would be a disproportionation reaction because a single substance (I in KI) both oxidizes and reduces. b) CI0 2 (g) H 2 0(l) � H C10 3 (aq) HCI02 (aq) Yes, CI02 disproportionates, as the chlorine reduces from o

c ) CI 2 (g) NaOH(aq) � NaCI(aq) NaCIO(aq) H 2 0(l) Yes, CI 2 disproportionates, as the chlorine reduces from to

to

and oxidizes from

to

and oxidizes from to

d) NH 4N0 2 (s) � N 2 (g) H 2 0(g) Yes, NH 4N0 2 disproportionates, the ammonium (NH/) nitrogen oxidizes from nitrogen reduces from to O.

to

and the nitrite (N0 2-)

O H-(aq) H 2 0(l) � Mn04-(aq) Mn0 2(s) e) MnO /-(aq) Yes, MnO/- disproportionates, the manganese oxidizes from to and reduces from f) AuCI(s) � AuCIJCs) Au(s) Yes, AuCI disproportionates, the gold oxidizes from I to

and reduces from

to O.

to

1 4.85

a) Group 5A(1 5) elements have 5 valence electrons and typically form three bonds with a lone pair to complete the octet. An example is N H 3 . b) Group 7 A(1 7) elements readily gain an electron causing the other reactant to be oxidized. They form monatomic ions of formula X- and oxoanions. Examples would be cr and ClO-. c) Group 6A(1 6) elements have six valence electrons and gain a complete octet by forming two covalent bonds. An example is H20. d) Group l A( l ) elements are the strongest reducing agents because they most easily lose an electron. As the least electronegative and most metallic of the elements, they are not likely to form covalent bonds. Group 2A(2) have similar characteristics. Thus, either Na or Ca could be an example. e) Group 3A(13) elements have only three valence electrons to share in covalent bonds, but with an empty orbital they can accept an electron pair from another atom. Boron would be an example of an element of this type. t) Group SA(l S), the noble gases, are the least reactive of all the elements. Xenon is an example that forms compounds, while helium does not form compounds.

1 4.86

Find I1Hrx n for the reaction 2 B rF(g) � Br2(g) + F2(g). Apply Hess ' s Law to the equations given: 3 BrF(g) � Br2(g) + BrF3(l) I1Hrx n - 1 25.3 kJ I) 2) 5 BrF(g) � 2 Br2(g) + BrFs(l) I1Hrx n - 1 66. 1 kJ BrF3( l) + F2(g) � BrFs(l) I1Hrx n - 1 58.0 kJ 3) Reverse equations I and 3 , and add to equation 2: B r2(g) + �f/1 � 3 BrF(g) I) I1Hrx n + 1 25.3 kJ (note sign change) 2) 5 BrF(g) � 2 Br2(g) + BW"f/1 I1Hrx n - 1 66. 1 kJ 3) BW"ffi � BW"ffi.(gl Mirx n + 1 5 8.0 kJ (note sign change) ..±b Total: 2 BrF(g) � Br2(g) + F2(g) !illr xn = + 1 1 7.2 kJ =

=

=

=

=

=

1 4.87

To answer these questions, draw an initial structure for AI 2 Ch-:

a) Aluminum uses its 3s and 3p valence orbitals to form s/ hybrid orbitals for bonding. b) With formula AX4, the shape is tetrahedral around each aluminum atom. c) It is sp, since the ion is linear, the central atom must be sp hybridized. d) The sp hybridization suggests that there are no lone pairs on the central chlorine. Instead, the extra four electrons participate in bonding with the empty d-orbitals on the aluminum to form double bonds between the chlorine and each aluminum.

1 82

1 4.88

The Lewis structures are

O==N

-

[ - .. .. J .. ) O

•

O:

:

O

+

- ..Q N ==

N-O

The nitronium ion (N02 + has a linear shape because the central N atom has 2 surrounding electron groups, which achieve maximum repulsion at 1 80°. Both the nitrite ion (N02-) and nitrogen dioxide (N02) have a central N surrounded by three electron groups. The electron group arrangement would be trigonal planar with an ideal bond angle of 1 20°. The bond angle in N02- is more compressed than that in N02 since the lone pair of electrons in N02- takes up more space than the lone electron in N02. 1 4.90

a) Determine the mass of As in a formula mass of the compound. 74.92 g As x 1 00% = 3 9.96 1 60 = 39.96% As % As in CuHAs03 = 1 87.48 g CuHAs0 3 01 /0

. A S I n (CH 3 ) 3 A S

b) Mass As =

(

0.50

�g AS

- ----==---)[ \ :: } U )( (

_

74.92 g As x 1 00% 1 20.02 g (CH 3 ) 3As

---

0

--

=

62.4229 =

2.3 5 m )( 7.52 m )( 2.98 m ))

Mass CuHAs03 = 0. 1 3 838 g As

1 00 g CuHAsO 3 39.96 1 60 gAs

1 83

J

=

=

62.42%

As

0. 1 3 8 3 8 g As

0.346282 =

0.35

g CuHAs03

Chapter 15 Organic Compounds and the Atomic Properties of Carbon FOLLOW-UP PROBLEMS 15.1

Plan: a) For a five-carbon compound start with 5 C atoms in a chain and place the triple bond in as many unique places as possible. Solution: H

H

--

C

_

I I

I I

C--C--C

(1)

H

H

H

--

H

I I

C

-

H H

--

H

I I

C

H

H

H

--

C

_

C

--

I I

C H

H

--

I I

C

-

H

H

(2) 4 C atoms in chain: There are two unique placements for the triple bond: one between the first and second carbons and one between the second and third carbons in the chain. C::=:C--C--C

C

--

c

_

c

--

C

The fifth carbon is added as a methyl group branched off the chain. With the triple bond between the first and second carbons, the methyl group is attached to the third carbon to give structure (3). H

I

H-C

-

H H

H

--

I I

C===C--C--C--H (3)

I

H

H

With the triple bond between the second and third carbons, the methyl group cannot be attached to either the second or the third carbons because that would create 5 bonds to a carbon and carbon has only 4 bonds . Thus, a unique structure cannot be formed with a 4 C chain where the triple bond is between the second and third carbons. There are only 3 structures with 5 carbon atoms, one triple bond, and no rings.

1 84

1 5.2

Plan: Analyze the name for chain length and side groups, then draw the structure. Solution : a) 3-ethyl-3-methyloctane. The end of the name, octane, indicates an 8 C chain (oct- represents 8 C) with only single bonds between the carbons (-ane represents alkanes, only single bonds). 3--ethyl means an ethyl group (-CH2CH3) attached to carbon #3 and 3-methyl means a methyl group (-CH3) attached to carbon #3 . CH3

,I --

H3C-CH2-

CH2-CH2-CH2-CH2-CH3

CH2

I

CH3

b) I--ethyl-3-propylcyclohexane. The hexane indicates 6 C chain (hex-) and only single bonds (-ane). The cyclo­ indicates that the 6 carbons are in a ring. The I-ethyl indicates that an ethyl group (-CH2CH3) is attached to carbon # 1 . Select any carbon atom in the ring as carbon # 1 since all the carbon atoms in the ring are equivalent. The 3-propyl indicates that a propyl group (-CH2CH2CH3) is attached to carbon #3 . CH2CH3

c)

3,3--diethyl- l -hexyne. The I-hexyne indicates a 6 C chain with a triple bond (-yne) between C #1 and C #2 . The 3 ,3-diethyl means two (di-) ethyl groups (--CH2CH3) both attached to C #3 . CH3

I - -- I I I

CH2

H

-- C===C

C--CH2-CH2-CH3

CH2 CH3

d) trans-3-methyl-3-heptene. The 3-heptene indicates a 7 C chain (hept-) with one double bond (-ene) between the 3rd and 4th carbons. The 3-methyl indicates a methyl group (-CH3) attached to carbon #3 . The trans indicates that the arrangement around the two carbons in the double bond gives the two smaller groups on opposite sides of the double bond. Therefore, the smaller group on the third carbon (which would be the methyl group) is above the double bond while the smaller group, H, on the fourth carbon is below the double bond.

H3C

/CH2-CH2- CH3

""C===C"

/ H3C--CH2

H

1 85

1 5.3

Plan: a) An addition reaction involves breaking a multiple bond, in this case the double bond in 2-butene, and adding the other reactant to the carbons in the double bond. The reactant Cl2 will add -Cl to one of the carbons and -Cl to the other carbon. b) A substitution reaction involves removing one atom or group from a carbon chain and replacing it with another atom or group. For I-bromopropane the bromine will be replaced by hydroxide. c) An elimination reaction involves removing two atoms or groups, one from each of two adjacent carbon atoms, and forming a double bond between the two carbon atoms. For 2-methyl-2-propanol, the -OH group from the center carbon and a hydrogen from one of the terminal carbons will be removed and a double bond formed between the two carbon atoms. Solution: (a) Cl Cl

I

+

I

(b) CH3-CH 2-CH 2

I

Br

(c) CH3

I

CH 3 -C

I O

15.4

--

CH 3 +

-

H

Plan: Reaction a) is an elimination reaction where H and Cl are eliminated from the organic reactant and a double bond is formed. Determine which carbons were involved in formation of the double bond and those two carbons would have been bonded to the H and Cl. Reaction b) is an oxidation reaction because Cr20 / - and H2S04 are oxidizing agents. An alcohol group (-OH) is oxidized to an acid group (--COOH), so determine which carbon in the product is in the acid group and attach an alcohol group to it for the reactant. Solution: a) The organic reactant must contain a single bond between the second and third carbons and the second and third carbons each have an additional group: an H atom on one carbon and a Cl atom on the other carbon. H CH3 CH3

I I I HI Cl

H3C--C--C--CH3

or

H3C

--

I I Cl

CH2-C

--

CH3

b) The third carbon from the left in the product contains the acid group. Therefore, in the reactant this carbon should have an alcohol group. H

I I H

H 3C--CH 2-C

--

OH

Check: When the reactants are combined, the products are identical to those given.

1 86

15.5

Plan: a) The product is an ester because it contains the unit: o

II

--C--O R Reacting an alcohol with an acid forms an ester. Since the reactant shown is an alcohol the other reactant must be the acid. Take the -OR group off the carbon in the ester product and replace with a hydroxyl group to get the structure of the acid. b) The product is an amide because it contains the unit: -

o

II

--C

--

N--

I

An amide forms from the reaction between an ester and an amine. This is the indicated reaction because the other product is an alcohol that results when the -OR group on the ester breaks off To draw the reactants first take the -NH-R group off the amide and add another hydrogen to the nitrogen to form the amine. Attach the -OR group to the remaining carbonyl (-C=O) group of the amide molecule so that an ester linkage is formed. Solution: a) The -OR group attached to the oxygen in the product is -OCH3 . Removing this and adding the -OH gives the acid reactant: O

� CH2-C�

OH b) Take the -NHCH2CH3 group off the amide product, attach a hydrogen to the nitrogen to make the amine reactant, and attach the -OCH3 from the alcohol product in its place to make the ester reactant. o

II

CH2-CH2-C H2N CH2-CH3 O CH3 + ester amine Check: When the reactants are combined, the products are identical to those given. H3C

15.6

--

--

--

--

Plan: Examine the structures for known functional groups: alkenes (C=C), alkynes (C=C), haloalkanes (C-X, where X is a halogen), alcohols (C-OH), esters (-COOR), ethers (C-O-C), amines (NR3), carboxylic acids (-COOH), amides (-CONR2), aldehydes (O=CHR), and ketones (O=CR2 where R cannot be a H). Solution: a) The structure contains an aromatic ring, alkene bond, and an aldehyde group. o

IIC

-

H

aldehyde aromatic ring

187

b) The structure contains an amide and a haloalkane. o

NH2

--

IIC- CH2

amide

CH

I

Br haloalkane

END-OF-CHAPTER PROBLEMS

1 5. 1

a) Carbon ' s electronegativity i s midway between the most metallic and nonmetallic elements of period 2 . To attain a filled outer shell, carbon forms covalent bonds to other atoms in molecules (e.g., methane, CH 4 ), network 2 covalent solids (e.g., diamond), and polyatomic ions (e.g., carbonate, C03 -). b) Since carbon has 4 valence shell electrons, it forms four covalent bonds to attain an octet. c) Two noble gas configurations, He and Ne, are equally near carbon's configuration. To reach the He 4 configuration, the carbon atom must lose 4 electrons, requiring too much energy to form the C + cation. This is confirmed by the fact that the value of the ionization energy for carbon is very high. To reach the Ne 4 configuration, the carbon atom must gain 4 electrons, also requiring too much energy to form the C - anion. The fact that a carbon anion is unlikely to form is supported by carbon' s electron affinity. The other possible ions would not have a stable noble gas configuration. d) Carbon is able to bond to itself extensively because carbon' s small size allows for closer approach and greater orbital overlap. The greater orbital overlap results in a strong, stable bond. e) The C-C bond is short enough to allow the sideways overlap of unhybridized p orbitals on neighboring C atoms. The sideways overlap of p orbitals results in double and triple bonds.

1 5.2

a) The elements that most frequently bond to carbon are other carbon atoms, hydrogen, oxygen, nitrogen, phosphorus, sulfur, and the halogens, F, CI, Br, and I . b ) I n organic compounds, heteroatoms are defined as atoms of any element other than carbon and hydrogen. The elements 0, N, P, S, F, Ct, B r, and I listed in part a) are heteroatoms. c) Elements more electronegative than carbon are N, 0, F, CI, and Br. Elements less electronegative than carbon are H and P. Sulfur and iodine have the same electronegativity as carbon. d) The more types of atoms that can bond to carbon, the greater the variety of organic compounds that are possible.

1 5 .4

Chemical reactivity occurs when unequal sharing of electrons in a covalent bond results i n regions of high and low electron density. The C-H, C-C and C-I bonds are unreactive because electron density is shared equally between the two atoms. The C=O bond is reactive because oxygen is more electronegative than carbon and the electron rich pi bond is above and below the C-O bond axis, making it very attractive to electron-poor atoms. The C-Li bond is also reactive because the bond polarity results in an electron-rich region around carbon and an electron-poor region around Li.

1 5.5

a) An alkane is an organic compound consisting of carbon and hydrogen in which there are no multiple bonds between carbons, only single bonds. A cycloalkane is an alkane in which the carbon chain is arranged in a ring. An alkene is a hydrocarbon with at least one double bond between two carbons. An alkyne is a hydrocarbon with at least one triple bond between two carbons.

1 88

b) The general formula for an alkane is CnH2 n+2 . The general formula for a cycIoalkane is CnH2 n. Elimination of two hydrogen atoms is required to form the additional bond between carbons in the ring. For an alkene, assuming only one double bond, the general formula is CnH2 n. When a double bond is formed in an alkane, two hydrogen atoms are removed. For an alkyne, assuming only one triple bond, the general formula is CnH2 n-2 . Forming a triple bond from a double bond causes the loss of two hydrogen atoms. c) For hydrocarbons, "saturated" is defined as a compound that cannot add more hydrogen. An unsaturated hydrocarbon contains multiple bonds that react with H2 to form single bonds. The alkanes and cycloalkanes are saturated hydrocarbons since they contain only single C-C bonds. 1 5.8

An asymmetric molecule has no plane of symmetry. a) A circular clock face numbered 1 to 1 2 o'clock is asymmetric. Imagine that the clock is cut in half, from 1 2 to 6 or from 9 to 3 . The one-half of the clock could never be superimposed on the other half, so the halves are not identical. Another way to visualize symmetry is to imagine cutting an object in half and holding the half up to a mirror. If the original object is "re-created" in the mirror, then the object has a plane of symmetry. b) A football is symmetric and has two planes of symmetry - one axis along the length and one axis along the fattest part of the football. c) A dime is asymmetric. Either cutting it in half or slicing it into two thin diameters results in two pieces that cannot be superimposed on one another. d) A brick, assuming that it is perfectly shaped, is symmetric and has three planes of symmetry at right angles to each other. e) A hammer is symmetric and has one plane of symmetry, slicing through the metal head and down through the handle. t) A spring is asymmetric. Every coil of the spring is identical to the one before it, so a spring can be cut in half and the two pieces can be superimposed on one another by sliding (not flipping) the second half over the first. However, if the cut spring is held up to a mirror, the resulting image is not the same as the uncut spring. Disassemble a ballpoint pen and cut the spring inside to verify this explanation.

15.9

To draw the possible skeletons it is useful to have a systematic approach to make sure no structures are missed. a) Since there are 7 C atoms but only a 6 C chain, there is one C branch off of the chain. First, draw the skeleton with the double bond between the first and second carbons and place the branched carbon in all possible positions starting with C #2. Then move the double bond to between the second and third carbon and place the branched carbon in all possible positions. Then move the double bond to between the third or and fourth carbons and place the branched carbon in all possible positions. The double bond does not need to be moved further in the chain since the placement between the second and third carbon is equivalent to placement between the fourth and fifth carbons and placement between the first and second carbons is equivalent to placement between the fifth and sixth carbons. The only position to consider for the double bond is between the branched carbon and the 6 C chain. Double bond between first and second carbons: C= C

I

-

C

-

C--C-C

C=C

-

C C=C

-

C

I

-

C

-

C-C

C C

-

C

I

--

C

-

C

C=C

C

-

C

-

C

-

C

I

C

1 89

-

C

Double bond between second and third carbons: C

-

C=C

I

--

C-C

-

C

C

-

C= C

I

-

C

-

C

-

C

-

C

-

C

-

C

C

C C

-

C=C

-

C

I

-

C

-

C

C

-

C= C

Double bond between third and fourth carbons: C

-

C

I

-

C=C

-

C

-

C

C

-

C

C=C

I

-

C ---

C

-

C

-

C

C

Double bond between branched carbon and chain: C

I

C

C

---

C

II

---

C

---

C---C

C

The total number of unique skeletons is II. To determine if structures are the same, build a model of one skeleton and see if you can match the structure of the other skeleton by rotating the model and without breaking any bonds. If bonds must be broken to make the other skeleton, the structures are not the same. b) The same approach can be used here with placement of the double bond first between C # 1 and C #2, then between C #2 and C #3. Since there are 7 C atoms but only 5 C atoms in the chain, there are two C branches. Double bond between first and second carbons: C= C

I

-

C

C=C

C-C--C

I

C C

-

I I

C

C =C

I

-

-

-

C

-

C-C

C

I

-

C

C

C=C-C

I

-

C

C

C=C

C

C-C

I

C

C

I I

C-C

-

C

C=C

-

C

190

C

-

C

C

---

C

I

-

C

Double bond between second and third carbons: C

--

C===C

I

I

C

--

C

--

C

C

--

C===C

I

--

C

C

C--C

I

C C

C

--

C===C

I

--

C

C

-

C===C

I

C

I

--

I I

C C

-

C===C

-

C

C

--

-

C

C

C-C

C--C

c) Five of the carbons are in the ring and two are branched off the ring. Remember that all the carbons in the ring are equivalent and there are two groups bonded to each carbon in the ring.

C

C\/C

C

/C"'"

/C"'"

c

\C--C/

15.11

C

I

c

I

/c

c

\C--C/

c

/C"'"

c

\C--C/

""

C

Add hydrogen atoms to make a total of 4 bonds to each carbon. a) CH2= C

I

--

CH2-CH2-CH2-CH3

CH2=CH-CH-CH2-CH2-CH3

I

CH3

CH3

CH2=CH-CH2-CH-CH2-CH3

i

I

CH2=CH-CH2-CH2-CH-CH3

CH3

CH3- ===CH-CH2-CH2-CH3

CH3-CH==

--

CH2-CH2-CH3

CH3

CH3 CH3-CH==CH-CH-CH2-CH3

I

CH3-CH ==CH-CH2-CH-CH3

,

CH3 CH3-CH-CH==CH-CH2-CH3

I

,

I

CH3

I

CH3

CH3-CH2- ===CH-CH2-CH3 CH3

CH3

191

II

CH3-CH2-C

-- CH2-CH2-CH3

CH2 b) CH2=

r -- r CH3

-- CH2-CH -CH 3

I

C H2=C

H -CH2-CH3

I

CH3

CH3

CH3

CH3

I -I

CH2=CH-C

CH2-C H3

CH2=CH-CH-CH-CH3

I

I

CH3

CH3

CH3

CH3

I -I

CH2=CH-CH2-C

CH3

CH2=CH-CH-CH2-CH3

I

CH2-CH3

CH3

C H 3-C=C

I

I

CH3

-- CH2-C H 3

CH 3-C

I

CH3

= CH-CH-CH3

I

CH3

CH3

I -I

CH3 CH 3 -CH == C

I

-- CH-CH3

I

CH3

CH3-CH ==CH-C

CH3

CH3

CH3

I

CH3-CH ==C--CH2-CH3 CH2-CH3 c) CH3

\/ C

/ CH2

\

CH3

I

CH3

",,-

/

CH2

C H2-CH2

CH3

CH CH3 .... / " CH2 CH

\

/

CH2-CH2

I

CH

......

\

CH2

""

/

CH2

CH2-CH

"CH3

1 92

/ CH2

\

CH

�

/

CH

CH2-CH2

/

CH2-CH3

1 5. 1 3

a) The second carbon from the left in the chain is bonded to five groups . Removing one of the groups gives a correct structure. CH3

I I

CH3-C--CH2-CH3 CH3

b) The first carbon in the chain has five bonds, so remove one of the hydrogens on this carbon. H2C === CH

--

CH2

-

CH3

c) The second carbon in the chain has five bonds, so move the ethyl group from the second carbon to the third. To do this, a hydrogen atom must be removed from the third carbon atom. HC==C--CH--CH3

I I

CH2 CH3

d) Structure is correct. 15 . 1 5

a) Octane denotes an eight carbon alkane chain. A methyl group (-CH3) is located at the second and third carbon position from the left. CH3

I

f

CH3-CH- H-CH2-CH2-CH2-CH2-CH3 CH3

b) Cyclohexane denotes a six-carbon ring containing only single bonds. Numbering of the carbons on the ring could start at any point, but typically numbering starts at the top carbon atom of the ring for convenience. The ethyl group (-CH2CH3) is located at position I and the methyl group is located at position 3 . CH2-CH3

6

5 4 c) The longest continuous chain contains seven carbon atoms, so the root name is "hept". The molecule contains only single bonds, so the suffix is "ane". Numbering the carbon chain from the left results in side groups (methyl groups) at positions 3 and 4. Numbering the carbon chain from the other will result in side groups at positions 4 and 5 . Since the goal is to obtain the lowest numbering position for a side group, the correct name is 3 ,4dimethylheptane. Note that the prefix "di" is used to denote that two methyl side groups are present in this molecule. d) At first glance, this molecule looks like a 4 carbon ring, but the two -CH3 groups mean that they cannot be bonded to each other. Instead, this molecule is a 4-carbon chain, with two methyl groups (dimethyl) located at the position 2 carbon. The correct name is 2,2-dimethylbutane .

1 93

1 5. 1 7

a) 4-methylhexane means a 6C chain with a methyl group on the 4th carbon: CH3

I

2 5 4 6 CH3-CH2-CH2-CH-CH2-CH3 I 3 Numbering from the end carbon to give the lowest value for the methyl group gives correct name of 3-methylhexane. b) 2-ethylpentane means a 5 C chain with an ethyl group on the 2nd carbon: CH3

I

CH2

I

CH3--CH--CH2CH2CH3 1 2 3 4 5 Numbering the longest chain gives the correct name, 1 CH3

CH3

I

CH2

-- I -- -- -2

3-methylhexane.

CH

CH2

CH2

CH3

3 4 6 5 c) 2-methylcyclohexane means a 6 C ring with a methyl group on carbon #2: CH3

In a ring structure, whichever carbon is bonded to the methyl group is automatically assigned as carbon # 1 . Since this is automatic, it is not necessary to specify I -methyl in the name. Correct name is methylcyclohexane. d) 3 ,3-methyl-4-ethyloctane means an 8 C chain with 2 methyl groups attached to the 3 rd carbon and one ethyl group to the 4th carbon. CH3

,I

CH3-CH2- --CH-CH2-CH2-CH2-CH3 CH3 CH2-CH3 Numbering is good for this structure, but the fact that there are two methyl groups must be indicated by the prefix di- in addition to listing 3 ,3 . The branch names appear in alphabetical order. Correct name is 4-ethyl-3,3dimethyloctane.

1 94

1 5. 1 9

A carbon atom i s chiral i f it i s attached to four different groups. The circled atoms below are chiral. �

H

""'/

�

H

C

H chiral carbon

Cl H Both can exhibit optical activity. 1 5.2 1

An optically active compound will contain at least one chiral center, a carbon with four distinct groups bonded to it. a) This compound is a 6 carbon chain with a Br on the third carbon. 3-bromohexane is optically active because carbon #3 has four distinct groups bonded to it: 1 ) -Br, 2) -H, 3) -CH2CH3, 4) -CH2C H2CH3. I carbon C H3-CH2

� y

C H2

- CH2-CH3

Br b) This compound is a 5 carbon chain with a CI and a methyl (CH3) group on the third carbon. 3--<:hloro-3methylpentane is not optically active because no carbon has four distinct groups. The third carbon has three distinct groups: I) -CI, 2) -CH3 3) two -CH2CH3 groups. C H3 C H3

- C H2 - I -- CH2 - C H3

r

CI c) This compound is a 4 carbon chain with Br atoms on the first and second carbon atoms and a methyl group on the second carbon. 1 ,2-dibromo-2-methylbutane is optically active because the 2nd carbon is chiral, bonded to the four groups: I) -CH2Br, 2) -CH3, 3) -Br, 4) -CH2C H3. CH3 Chiral carbon C H2

I

-f= C

Br 1 5.23

CH2

- C H3

Br

Geometric isomers are defined as compounds with the same atom sequence but different arrangements of the atoms in space. The cis-trans geometric isomers occur when rotation is restricted around a bond, as in a double bond or a ring structure, and when two different groups are bonded to each atom in the restricted bond. a) Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur. The double bond occurs at position 2 in a five-carbon chain. H H CH3 H

"", / C - C "" /

/ "", C -C "", /

_

C H3CH2

CH3

_

CH3CH2 trans-2-pentene

cis-2-pentene

1 95

H

b) Cis-trans geometric isomerism occurs about the double bond. The ring is named as a side group (cyclohexyl) occurring at position I on the propene main chain. H H CH3 H

,,_/ C C -

,,

C

'"CH3

-

/ C"

H

trans- I-cyclohexylpropene cis- l -cyclohexylpropene c) No geometric isomers occur because the left carbon participating in the double bond is attached to two identical methyl (-CH3) groups. 1 5 .25

a) The structure of propene is CH2=CH-CH3. The first carbon that is involved in the double bond is bonded to two of the same type of group, hydrogen. Geometric isomers will not occur in this case. b) The structure of 3-hexene is CH3CH2CH=CHCH2CH3 . Both carbons in the double bond are bonded to two distinct groups, so geometric isomers will occur. H CH3-CH2 CH2-CH3 CH3-CH2

/ "'-/C C "

"'_/

/C

-

H

H

H

-

C

", CH2-CH3

cis-3-hexene trans-3-hexene c) The structure of 1 , I-dichloroethene is CCI2=CH2• Both carbons in the double bond are bonded to two identical groups, so no geometric isomers occur. d) The structure of 1 ,2-dichloroethene is CHCI=CHCI. Each carbon in the double bond is bonded to two distinct groups, so geometric isomers do exist. CI Cl

", / / C_C " -

H 1 5 .27

H

cis- l ,2-dichloroethene trans- l ,2-dichlorethene Benzene is a planar, aromatic hydrocarbon. It is commonly depicted as a hexagon with a circle in the middle to indicate that the 1t bonds are delocalized around the ring and that all ring bonds are identical. The artha, meta, para naming system is used to denote the location of attached groups in benzene compounds only, not other ring structures like the cycloalkanes. CI CI CI CI

CI

1 ,2-dichlorobenzene (a-dichlorobenzene)

1 ,3 -dichlorobenzene (m-dichlorobenzene)

1 96

CI 1 ,4-dichlorobenzene (p-di chI orobenzene)

1 5 . 29

Analyzing the name gives benzene as the base structure with the following groups bonded to it: I) on carbon # 1 a hydroxy group, -OH; 2) on carbons #2 and #6 a tert-butyl group, -CCCH3)3; and 3 ) on carbon #4 a methyl group, -CH3' HO

\5 .30 The compound 2-methyl-3-hexene has cis-trans isomers.

trans-2-methyl-3-hexene cis-2-methyl-3-hexene The compound 2-methyl-2-hexene does not have cis-trans isomers because the #2 carbon atom is attached to two identical methyl (-CH3) groups: H3C H

� / /C====C�

C H 2CH 2C H3

H3C

2-methyl-2-hexene 1 5 . 32

a) HBr is removed in this elimination reaction, and an unsaturated product is formed. b) This is an addition reaction in which hydrogen is added to the double bond, resulting in a saturated product.

1 5 . 34

a) Water (H20 or H and OH) is added to the double bond: ----J.� CH3CH 2C H 2CHCH 2 CH3

I

OH b) H and Br are eliminated from the molecule, resulting in a double bond: CH3CHBrCH3 + CH3CH20K � CH3CH=CH2 + CH3CH20H + KBr c) Two chlorine atoms are substituted for two hydrogen atoms in ethane: CH3CH3 + C I2

hv

) CHCI2CH3

+

2 HCI

1 97

15.37

a) The structures for chloroethane and methylethylamine are given below. The compound methylethylamine is more soluble due to its ability to form H-bonds with water. Recall that N-H, O-H , or F-H bonds are required for H-bonding. H H H H H H

-

I I H

C

--

I I I H-C--C-N-C-H

I I H

C--CI

I

I

I

I

H H H H b) The compound I-butanol is able to H-bond with itself (shown below) because it contains covalent oxygen­ hydrogen bonds in the molecule. Diethylether molecules contain no O-H covalent bonds and experience dipole­ dipole interactions instead of H-bonding as intermolecular forces. Therefore, I-butanol has a higher melting point because H-bonds are stronger intermolecular forces than dipole-dipole attractions. CH3-CH2-CH2-CH2-O:

\H

11111111111

1

H

.O·-CH2-CH2-CH2-CH3

"""

•

Hydrogen Bond

H

H

I

I

H3C-C-O-C-CH3

HI

I

H

diethyl ether c) Propylamine has a higher boiling point because it contains N-H bonds necessary for hydrogen bonding. Trimethylamine is a tertiary amine with no N-H bonds, and so its intermolecular forces are weaker.

2

CH3-CH2-CH -N

I

-

H

CH3-N-CH

I

CH3

H

Propylamine 15.39

3

Trimethylamine

The C=C bond is nonpolar while the C=O bond is polar, since oxygen is more electronegative than carbon. Both bonds react by addition. In the case of addition to a C=O bond, an electron-rich group will bond to the carbon and an electron-poor group will bond to the oxygen, resulting in one product. In the case of addition to an alkene, the carbons are identical, or nearly so, so there will be no preference for which carbon bonds to the electron-poor group and which bonds to the electron-rich group. This may lead to two isomeric products, depending on the structure of the alkene. When water is added to a double bond, the hydrogen is the electron-poor group and hydroxyl is the electron-rich group. For a compound with a carbonyl group only one product results: CH 3 -CH 2 -C--CH

II

3

OH I CH 3 -CH 2 -C

+

--

I

o

OH

198

CH 3

But when water adds to a C=C two products result: CH3-CH2 -CH-CH3 +

1

OH +

In this reaction very little of the second product forms. 1 5.4 1

Esters and acid anhydrides form through dehydration-condensation reactions. Dehydration indicates that the other product is water. In the case of ester formation, condensation refers to the combination of the carboxylic acid and the alcohol. The ester forms and water is the other product. o

II

1 5.43

o

II

a) Halogens, except iodine, differ from carbon in electronegativity and form a single bond with carbon. The organic compound is an alkyl halide. b) Carbon forms triple bonds with itself and nitrogen. For the bond to be polar, it must be between carbon and nitrogen. The compound is a n itrile. c) Carboxylic acids contain a double bond to oxygen and a single bond to oxygen. Carboxylic acids dissolve in water to give acidic solutions. d) Oxygen is commonly double-bonded to carbon. A carbonyl group (C=O) that is at the end of a chain is found in an aldehyde.

1 5.45 a) CH3

-E§

�

CH2

Alkene

� �

b)

Alcohol Carboxylic Acid Haloalkane

1 99

d)

�)

Ketone

Nitrile

Alkene e)

1 5 .47

To draw all structural isomers, begin with the longest chain possible. For C5HI20 the longest chain is 5 carbons: fH2-CH2-CH2-CH2-CH3

CH3-fH-CH2-CH2-CH3

OH

CH3-CH2-fH-CH2-CH3

OH

OH

The three structures represent all the unique positions for the alcohol group on a 5 C chain. Next use a 4 C chain and attach to side groups, -OH and -CH3. CH2-CH-CH2-CH3

CH2-CH2-CH-CH3

OH

OH

I

I

CH3

I

I

CH3 CH3

CH3-CH-CH-CH3

I

OH

I

I I

CH3-C--CH2-CH3

CH3

OH

Next use a 3 C chain with three side groups: CH3

I I

CH3-C--CH2-OH

CH3

1 5.49

The total number of different structures is eight.

First, draw all primary amines (formula R - N H2 ). Next, draw all secondary amines (formula R- N H R ). There is only one possible tertiary amine structure (formula R3 N). Eight amines with the formula C4H11N exist. -

-

CH3-CH2-CH2-CH2-NH2

CH3-CH2-CH-NH2

I

CH3 CH3-CH-CH2-NH2

I

CH3

200

'

CH 3-CH 2 -N--CH 3

I

CH 3-CH-NH-CH 3

I

CH3

CH3 CH3

I I C

CH 3-C

--

NH2

H3

1 5.5 1

a) This reaction is a dehydration-condensation reaction. o

"

o

"

N--CH3

CH 3-C--N

I

H

I

-

H +

H20 eliminated

b) An alcohol and a carboxylic acid undergo dehydration-condensation to form an ester. o C H3

"

CH3-C H2-CH2-C

O

--

I I

CH

o

"

CH 3-CH 2-CH 2-C

CH3 --

c) This reaction is also ester formation through dehydration-condensation. CH3 o H

--

II

C

I

O--CH2-CH-CH3

20 1

O

--

I I CH3

CH

+

H20

CH3

1 5.53

/" Add -H

To break an ester apart, break the -C-O- single bond and add water (-H and -OH) as shown. o

R

o

II -- OH

•

I l t O - R' --t '" Bond Broken

+

R--C

H

- O - R'

Add -OH

a) o

I I -- OH

CH3-(CH 2)4--C b)

+

HO-CH 2-CH3

+

HO-CH 2-CH 2-CH3

o

II - OH

C c)

o

C H3-CH2 -OH

1 5.55

+

II -- CH2-CH 2

HO-C

a ) Substitution of Br- occurs b y the stronger base, OIr. Then a substitution reaction between the alcohol and carboxylic acid produces an ester: o

CH3-CH 2-Br

OH-

I I - OH

CH3-CH 2-C •

C H3-CH 2-OH

•

H+

o

II -- -_

CH3-CH 2-C 0 C H 2-CH3 b) The strong base, CN", substitutes for Br. The nitrile is then hydrolyzed to a carboxylic acid. Br C N CN• CH3-CH 2-CH-CH3 CH3-CH 2-CH-CH3

I

I

OH

H30+, H 20 •

I I CH3-CH2-CH-CH3 C=O

202

15.59

Addition reactions and condensation reactions are the two reactions that lead to the two types of synthetic polymers that are named for the reactions that form them.

15.61

Polyethylene comes in a range of strengths and tlexibilities. The intermolecular dispersion forces (also called London forces) that attract the long, unbranched chains of high-density polyethylene (HDPE) are strong due to the large size of the polyethylene chains. Low-density polyethylene (LDPE) has increased branching that prevents packing and weakens intermolecular dispersion forces.

15.63

Nylon is formed by the condensation reaction between an amine and a carboxylic acid resulting in an amide bond. Polyester is formed by the condensation reaction between a carboxylic acid and an alcohol to form an ester bond.

15.64

Both PVC and polypropylene are addition polymers. The general formula for creating repeating units from a monomer is given in the chapter. (b) a) H

I I

C

H

H

H

C

C

C

-

CI

H 15.66

I I

I I

-

H

n

I I

CH n

A carboxylic acid and an alcohol react to form an ester bond. 0 o

II

II

C

n HO-C

--

OH

+

n

HO-CH2-CH2-OH

o

HO

II

H

C

n

15.68

a) Amino acids form condensation polymers called proteins. b) Alkenes form addition polymers, the simplest of which is polyethylene. c) Simple sugars form condensation polymers called polysaccharides. d) Mononucleotides form condensation polymers called nucleic acids.

15.70

The amino acid sequence in a protein determines its shape and structure, which determine its function.

203

1 5 .72

Locate the specific amino acids in F igure 1 5.28. a) alanine b) histidine

c) methionine CH3

I I CH 2 I CH 2

r

s

r 1 5 .74

I

a) A tripeptide contains three amino acids and two peptide (amide) bonds.

II

o

II

H 2N - CH-C -- OH

I I C I OH

I

I

C H2

C H2 O

�

N

L

HN

NH

tryptophan

histidine aspartic acid Join the above three acids to give the tripeptide: o H H 0

II

I

II

o

I

II

H 2N - CH-C - N - C H-C - N - CH-C--OH

I I C I OH

I .

C H2 ====

I

CH 2 .

CH 2 O N

�

L

NH

HN

204

II

H2N - CH-C -- OH

H 2N-CH-C - OH

C H2 ====

0

0

b) Repeat the above, with charges on the terminal groups. tyrosine glycine cysteine o H 0 H 0 +

H3 N

-

II

CH-C

I

H

--

I

N

--

"I

CH-C

I I

-

,,

N-CH-C

I.

C�

--

_

O

C�

SH

OH IS.76

Base A always pairs withBase T; Base C always pairs withBase G. a) Complementary DNA strand is AATCGG. b) Complementary DNA strand is TCTGTA.

IS.78

a)Both side chains are part of the amino acid cysteine. Two cysteine R groups can form a disulfide bond (covalent bond). b) The R group (-{CH2)cNH/) is found in the amino acid lysine and the R group (-CH2COO-) is found in the amino acid aspartic acid. The positive charge on the amine group in lysine is attracted to the negative charge on the acid group in aspartic acid to form a salt link. c) The R group in the amino acid asparagine and the R group in the amino acid serine. The -NH- and -OH groups will hydrogen bond. d)Both the R group -CH(CH3)-CH3 from valine and the R group C6Hs-CH2- from phenylalanine are nonpolar so their interaction is through dispersion forces.

IS.80

The desired reaction is the displacement ofBr- by OIr: CH3- H-CH2-CH3 OH CH3..

f

Br

�

r

-CH2-CH3

OH CH3-CH==CH-CH3

However, the elimination of HX also occurs in the presence of a strong base to give the second product, 2-butene.

20S

15.8 1

a) Functional groups injasmolin II alkene

ester alkene ester ketone

3

b) Carbon atoms surrounded by four electron regions (4 single bonds) are S p hybridized. Carbon atoms 2 surrounded by three electron regions (2 single bonds and I double bond) are S p hybridized. 3 2 Carbon 2 is Sp hybridized. Carbon 1 is Sp hybridized. 3 2 Carbon 3 is Sp hybridized. Carbon 4 is Sp hybridized. 2 3 Carbon 6 and 7 are Sp hybridized. Carbon 5 is Sp hybridized. c) Carbons 2, 3, and 5 are chiral centers as they are each bonded to four different groups. 15.83

A vanillin molecule includes three functional groups containing oxygen: ether

aromatic ring The shortest carbon-oxygen bond is the double bond in the aldehyde group. Bond length decreases as the number of bonds increases.

206

Chapter 16 Kinetics: Rates and Mechanisms of Chemical Reactions FOLLOW-UP PROBLEMS

1 6. 1

a) Plan: The balanced equation is 4 NO(g) + 02(g) -7 2 N203 . Choose O2 as the reference because its coefficient is 1 . Four molecules of NO (nitrogen monoxide) are consumed for every one O2 molecule, so the rate of O2 disappearance is 1 /4 the rate of NO decrease. By similar reasoning, the rate of O2 disappearance is 1 /2 the rate of N203 ( dinitrogen trioxide) increase. A negative sign is associated with the rate in terms of the reactant concentrations since these concentrations are decreasing. Solution : Li [ 0 2 ] 1 Li [ NO ] I Li [ N2 0 3 J Rate = = - --- = + --'='----'---=. --=. 4 Lit Lit 2 Lit b) Plan: Because NO is decreasing; its rate of concentration change is negative. Substitute the negative value into the expression and solve for Li[02]/Lit. Solution : Li z l Rate = - - 1 .60 x 1 0-4moI l L o s =

-

±(

Li [O 2 ]

=

)

_

�

-4.00 X 1 0-5 mol/Vs

Lit Check: a) Verify that the general formula shown in Equation 1 6.2 produces the same result when a = 4, b = I, and c = 2. b) The rate should be negative because O2 is disappearing. The rate can also be expressed as -Li[02]/Lit = 4.00 x 1 0-5 mol/Losec, where the negative sign on the left side of the equation denote reactant disappearance as well . It makes sense that the rate of NO disappearance is 4 times greater than the rate of O2 disappearance, because 4 molecules of NO disappear for every one O2 molecule. 1 6.2

The exponent of [Brl is 1 , so the reaction is first order with respect to Br- . Similarly, the reaction is first order with respect to Br03- , and second order with respect to H+ since the exponent of H + is 2. The overall reaction order is ( 1 + I + 2) = 4, or fou rth order overall.

1 6.3

Solution : Assume that the rate law takes the general form rate = k[H2r[I2]". To find how the rate varies with respect to [H2] , find two experiments in which [H2] changes but [h] remains constant. Take the ratio of rate laws for Experiments 1 and 3 to find m. rate 3 [H2]�n -----_

rate 1

[H2r

2 1 0- 3 2 1 .9 x 1 0- 3

9.3

X

[ 0.0550 ]: [O.Ol13r

4. 8947 = (4. 8672566)m Therefore, m = 1 If the reaction order was more complex, an alternate method of solving for m is: log (4.8947) = m log (4. 8672566); m = log (4. 8947)/log (4. 8672566) = I

207

Take the ratio of rate laws for Experiments 2 and 4 to find n.

[12]: [I2]�

rate 4 = rate 2

1 .9 l.l

[ 0.0056]: 1 0 -22 = 22 x 1 0[ 0.0033]� X

1 . 72727 = ( 1 .69697)" Therefore, n = I The rate law is rate = k[H2] [12] and is second order overall. Check: The rate is first order with respect to each reactant. If one reactant doubles, the rate doubles; if one reactant triples, the rate triples. I f both reactants double (each reactant x 2), the rate quadruples (2 x 2). Comparing reactions 1 and 4, the [H2] doubles (x 2) and the [12] increases by 5 times (x 5). Accordingly, the rate increases by 1 0 (2 x 5). This exercise verifies that the determined rate law exponents are correct. 1 6.4

Plan: The rate expression indicates that the reaction order is two (exponent of [HI] = 2), so use the integrated second-order law. Substitute the given concentrations and the rate constant into the expression and solve for time. Solution: 1 I _ = kt _ _

_

_

[A ] t

[A]o I

[ 0.00900 mol/ q

I

[ 0.0 1 00 mol/ L]o

-

= (2.4 X 1 0

21

Llmol-s) t

0.00900 moll L :------:- 0.0 1 00 moll L t = --------: 2 1 2.4 X 1 0 - Ll mol - s 2

t = 4.6296 X 1 0 1 s = 4.6 X 10 21 s The Check: problem indicates this decomposition reaction i s very slow, so it should take a long time to decrease the HI concentration. The incredibly small rate constant also indicates that the reaction rate is very slow. A hundred thousand billion years is indeed a long time. 16. 5

Plan: Rearrange the first-order half-life equation to solve for k. Solution: k = In 2 I t 1 /2 = In 2 I 1 3 . 1 h = 0.0529 1 1 998 = 0.0529 h-I Check: Verify that the units on k reflect the data given.

1 6.6

Plan: Activation energy, rate constant at TJ, and a second temperature, T2, are given. Substitute these values into Equation 16. 9 and solve for k2' the rate constant at T2. Solution : Rearranging Equation 1 6.9 to solve for k2:

( -�J

1 In � = _ � _ k, R T2 kl = 0.286 Llmol-s k2 = ? Llmol-s In In

k2

0.286 Llmol - s

TI

= -

T I = 500. K T2 = 490. K 1 .00 xI0 2 kJ / mol 8.3 1 4 Ilmol - K

k2 = -0.49093 0.286 L l mol - s

(

Ea

2

= 1 .00 X 1 0 kllmol

1 I 490. K 500. K

J

[IkJ) 1 0 31

Raise each side to eX to eliminate the In term

k=2

= 0.6 1 2054 0.286 Llmol - s k2 = (0.6 1 2054) (0.286 Llmol-s) = 0. 1 75047423 = 0. 1 75 Llmol-s ___

__

208

Check: The temperature only changes by 1 0 K degrees, so it is likely that the rate constant at T z would not change by orders of magnitude; 0. 1 75 S-I is in the same "ballpark" as 0.286 S-I. Furthermore, the rate should decrease at the lower temperature, so the lower rate of 0. 1 75 S-I at 490 K is expected. 1 6.7

Plan: Begin by using Figure 1 6. 1 6 as a guide for labeling the diagram. Solution: The reaction energy diagram indicates that O(g) + HzO(g) � 2 OH(g) is an endothermic process, because the energy of the product is higher than the energy of the reactants. The highest point on the curve indicates the transition state. In the transition state, an oxygen atom forms a bond with one of the hydrogen atoms on the HzO molecule (hashed line) and the O-H bond (dashed line) in HzO weakens. Ea(fwd) is the sum of I'lHrxn and Ea(rev). Transition state:

.- o·

..

O : ------H� "".,

Ea(fwd)=

' "

H

+78 kJ

Reaction progress 1 6.8

Plan: Sum the elementary steps to obtain the overall equation. Think of the individual steps as reactions along the "reaction progress" axis, i.e., sum each equation as written (this is different from the application of Hess's law in which reactions are reversed as necessary to reach a final reaction). Solution: b) Molecularity c) Rate law z 2 rate 1 = kl [NO] ( I) 2 NO(g) � N;!�� I (2) 2[Hz(g) � � ratez = kz[Hz] (3) ��� + H(g1 � N;!G(g) + HG(g) rate3 = k3 [NzOz][H] 2 2 rate 4 = k4 [HO][H] (4) 2[HG(g1 + H(g1 � HzO(g)] (5) � + N;!G(g) � HG(g1 + Nz(g) 2 rates = ks[H] [NzO] a) 2 NO(g) + 2 Hz(g) � 2 HzO(g) + Nz(g) Check: The overall equation is balanced.

END-OF-CHAPTER PROBLEMS

1 6.2

Rate is proportional to concentration. An increase in pressure will increase in number of gas molecules per unit volume. In other words, the gas concentration increases due to increased pressure, so the reaction rate increases. Increased pressure also causes more collisions between gas molecules.

1 6.3

The addition of more water will dilute the concentrations of all solutes dissolved in the reaction vessel. If any of these solutes are reactants, the rate of the reaction will decrease.

1 6.5

An increase in temperature affects the rate of a reaction by increasing the number of collisions, but more importantly the energy of collisions increases. As the energy of collisions increase, more collisions result in reaction (i.e., reactants � products), so the rate of reaction increases.

209

1 6.8

a) For most reactions, the rate of the reaction changes as a reaction progresses. The instantaneous rate is the rate at one point, or instant, during the reaction. The average rate is the average of the instantaneous rates over a period of time. On a graph of reactant concentration versus time of reaction, the instantaneous rate is the slope of the tangent to the curve at any one point. The average rate is the slope of the line connecting two points on the curve. The closer together the two points (shorter the time interval), the more closely the average rate agrees with the instantaneous rate. b) The initial rate is the instantaneous rate at the point on the graph where time = 0, that is when reactants are mixed.

1 6. 1 0

At time t = 0 , no product has formed, so the B(g) curve must start at the origin. Reactant concentration (A(g» decreases with time; product concentration (B(g» increases with time.Many correct graphs can be drawn. Two examples are shown below. The graph on the left shows a reaction that proceeds nearly to completion, i.e., [products] » [reactants] at the end of the reaction. The graph on the right shows a reaction that does not proceed to completion, i.e., [reactants] > [products] at reaction end. B (g) c o

c o

.� c Q) u c o

.�

c Q) u ::: o

A(g)

B (g)

U

U

Time

T ime 1 6. 1 2

A(g)

a) The average rate from t = 0 to t = 20.0 s i s proportional to the slope of the l ine connecting these two points: ( 0.0088 M-0.0500M ) 1 .1[AX 2 ] Rate = = _ .!. = 0.00 1 03 = 0.00 10 Mis 2 .1t 2 ( 20.Os-Os )

The negative of the slope i s used because rate is defined as the change in product concentration with time. If a reactant is used, the rate is the negative of the change in reactant concentration. The 1 12 factor is included to account for the stoichiometric coefficient of 2 for AX2 in the reaction.

b)

[AX21

vs

time

0.06 0.05 0.04 ......

X 0.03 «

......

0.02 0.01 0 0

5

10

15

time,s

210

20

25

=

The slope of the tangent to the curve (dashed line) at t 0 is approximately -0.004 Mis. This initial rate is greater than the average rate as calculated in part (a). The initial rate is greater than the average rate because rate decreases as reactant concentration decreases. 1 6. 1 4

=

=

=

� �[B] �[A] �[C] �t 2 �t �t Rate is defined as the change in product concentration with time. If a reactant is used, the rate is the negative of the change in reactant concentration. The 1 12 factor is included for reactant B to account for the stoichiometric coefficient of 2 for B in the reaction. 1 ( 0.50 mollL) �[A] � �[B] --0 . 25 mo IlL os ( unroun d e d ) �t 2 �t 2 s [B] decreases twice as fast as [A] , so [A] i s decreasing at a rate of 1 12 (0.5 mol/Los) or 0.2 mollLos. Rate

_

_

=

_

= -'----

_

.:.- =

-

1 6. 1 6

A term with a negative sign i s a reactant; a term with a positive sign is a product. The inverse of the fraction becomes the coefficient of the molecule: 2 N 20S(g) � 4 N 0 2 (g) + 0 2 (g)

1 6. 1 9

a) Rate

_

=

� �[0 3 ]

)(

3 mol O / Los

)=

1. 4 5

X

10-5 mollLos

a) k is the rate constant, the proportionality constant in the rate law. k represents the fraction of successful collisions which includes the fraction of collisions with sufficient energy and the fraction of collisions with correct orientation. k is a constant that varies with temperature. b) rn represents the order of the reaction with respect to [A] and n represents the order of the reaction with respect to [B] . The order is the exponent in the relationship between rate and reactant concentration and defines how reactant concentration influences rate. The order of a reactant does not necessari ly equal its stoichiometric coefficient in the balanced equation. If a reaction is an elementary reaction, meaning the reaction occurs in only one-step, and then the orders and stoichiometric coefficients are equal. However, if a reaction occurs in a series of elementary reactions, called a mechanism, then the rate law i s based on the slowest elementary reaction in the mechanism. The orders of the reactants will equal the stoichiometric coefficients of the reactants in the slowest elementary reaction but may not equal the stoichiometric coefficients in the overall reaction. 2 c) For the rate law rate k[A] [B] substitute in the units: 2 Rate (moIlLomin) = k[A] I [B] mollLomin mollLomin Rate k 2 [A] I [B] m m

=

k-

k

1 6.2 1

.!.. �[0 2 ]

3 �t 2 �t b) Use the mole ratio in the balanced equation: 2. 1 7 x 1 0 -5 mol 0 2 IL O S 2 mol 0 3 IL O S

(

1 6.20

=

=

=

[t [t J J --- (--) r} mol L o min mol 3

2 L /mo e · min

=

=

a) The rate doubles. If rate k[ A] I and [ A] is doubled, then the rate law becomes rate k[2 x A] I . The rate increases by 2 I or 2 . 2 2 b ) The rate decreases b y a factor of four. If rate = k[ 1 I2 x B] , then rate decreases t o ( 1 12) or 1 /4 o f its original value. 2 2 c) The rate increases by a factor of nine. If rate = k[3 x C] , then rate increases to 3 or 9 times its original value.

21 1

1 6.22

The order for each reactant is the exponent on the reactant concentration in the rate law. The orders with respect to [Br03-] and to [Brl are both 1 . The order with respect to [ H+ ] is 2. The overall reaction order is the sum of each reactant order: 1 + 1 + 2 = 4. + first order with respect t o Br03- , first order with respect to Br- , second order with respect to H , fou rth order overall

1 6.24

a) The rate is first order with respect to [Br03-] . If [Br03-] is doubled, rate = k[2 x Br03-] , then rate increases to i or 2 times its original value. The rate doubles. b) The rate is first order with respect to [Brl . If [Brl is halved, rate = k[ 1 12 x Br-] , then rate decreases by a factor of 1 12 1 or 1 12 times its original value. The rate is halved. 2 2 c) The rate is second order with respect to [H+] . If [H +] is quadrupled, rate = k[4 x H + ] , then rate increases to 4 or 1 6 times its original value.

1 6.26

a) To find the order for reactant A, first identify the reaction experiments in which [A] changes but [B] is rate exp l [A] exP l m · · . · · clor rates an d concentratIOns an d constant. S et up a proportIOna I Ity: = . F I II In t he va I ues given · rate exp 2 [A ] exp 2 solve for m, the order with respect to [A] . Repeat the process to find the order for reactant B . Use experiments 1 and 2 (or 3 and 4 would work) to find the order with respect to [A] . m [A ] exp I rate exp I = [A] exp 2 rate exp 2 45 .0 moUL.min 0.300 mOUL m = 5 .00 moUL·min 0. 1 00 moUL 9.00 = (3 .00)m log (9.00) = m log (3 .00) m=2 Using experiments 3 and 4 also gives 2nd order with respect to [A) . Use experiments 1 and 3 with [A] = 0. 1 00 M or 2 and 4 with [ A] = 0.300 M to find order with respect to [B] .

[ l ---

[ l (

rate exp I rate exp 2

_

[ l"

)

[B] exp I

[B] exp 2

(

)

1 0.0 moUL·min 0.200 mOUL " = 5 . 00 mollL·min 0. 1 00 mollL 2 .00 = (2.00)" log (2.00) = n log (2.00) n= 1 The reaction is first order with respect to [B) . 2 b) The rate law, without a value for k, is rate k[A) [B) . c) Using experiment 1 to calculate k: rate 5 .00 moUL·min 2 2 k= = 5 .00 x 1 03 L Imol -min 2 [Af [B] [0. 1 00 mollL] [0. 1 00 mollL] =

1 6.29

The integrated rate law can be used to plot a graph. I f the plot of [reactant] versus time is l inear, the order is zero. I f the plot of ln [reactant] versus time is linear, the order is first. I f the plot of inverse concentration ( \ /[reactant]) versus time is linear, the order is second. a) The reaction is first order since In [reactant] versus time is linear. b) The reaction is second order since \ I [reactant] versus time is linear. c) The reaction is zero order since [reactant] versus time is linear.

212

1 6.3 1

The rate expression indicates that this reaction is second order overall (the order of [ABJ is 2), so use the second order integrated rate law to find time. ( [ABJ = 1 /3 [ABJo = 1 13 ( 1 .50 M) = 0.500 M) 1 1_ _ = /ct [ AB ] t [ AB ] o _

t�

t�

_

_

(rfsr -rfsr) k

(osdo �Ml M

0.2 L 1 mol o s t = 6 .6667 = 7 s 1 6.33

a) The given information is the amount that has reacted in a specified amount of time. With this information, the integrated rate law must be used to find a value for the rate constant. For a first order reaction, the rate law is In [AJ( = In [AJo - /ct. Using the fact that 50% has decomposed, let [AJo = 1 M and then [AJ( =1 M/2 : = 0.5 M : I n [AJ( = I n [AJo - /ct In [0.5J = In [ I J - k( l 0.5 min) -0.693 1 47 = 0 - k( l 0. 5 min) 0.693 1 47 = k( l O. 5 min) k = 0.0660 min- I Alternatively, 50.0% decomposition means that one half-life has passed. Thus, the first order half-life equation may be used: In 2 t l/2 = k In 2 In 2 = 0.0660 1 4 = 0.0660 min- I k= = t l /2 1 0. 5 min b) Use the value for k calculated in part a. Let [AJo = 1 M; since 75% of A has decomposed, 25% of A remains and [AJt = 25% of [AJo or 0.25 {AJo = 0.25 [ I J = 0.25 M In [AJ( = In [AJo - /ct In[AJ( - In[AJo =t -k I n [ 0 .2 5 J - ln [ 1:.. J =t _-,,-_-...:'-_....:.,.::. -0.0660 min-I t = 2 1 .0045 = 2 1 .0 min If you recognize that 75 .0% decomposition means that two half-lives have passed, then t = 2 ( 1 0.5 min) = 2 1 .0 min

1 6.36

The Arrhenius equation, k = Ae -E, / RT , can be used directly to solve for activation energy at a specified temperature if the rate constant, k, and the frequency factor, A, are known. However, the frequency factor is usually not known. To find Ea without knowing A, rearrange the Arrhenius equation to put it in the form of a linear plot: In k = In A - Ea I RT where the y value is In k and the x value is 1 I T. Measure the rate constant at a series of temperatures and plot In k versus 1 I T. The slope equals -Ea I R.

213

1 6. 3 8

Substitute the given values into Equation 1 6.9 and solve for k2• 3 T , = 25°C = (273 + 25) = 298 K k, = 4.7 X 1 0- s-' T2 = 75°C = (273 + 75) = 348 K k2 = ? Ea = 3 3 . 6 kJ/mol = 33600 J/mol

( -�J

k 1 In 2 = _ ..S... _ k, R T2 T, In In

(

I 1 33600 J / mol k2 = 3 K K K 298 348 mol / J 4 1 3 . 8 4.7 x l O- s '

k2

4.7 x 1 0-3 s-'

= 1 .9485 1 5 (unrounded)

J

R aise each side to eX

k2 ---"--:- - 7.0 1 82577 4.7 x 1 0 -3 s- ' 3 k2 = (4. 7 X 1 0- s-' ) (7.0 1 82577) = 0.0329858 = 0.033 S-I _

1 6.42

No, coll ision frequency is not the only factor affecting reaction rate. The collision frequency is a count of the total

number of collisions between reactant molecules. Only a small number of these collisions lead to a reaction. Other factors that influence the fraction of collisions that lead to reaction are the energy and orientation of the collision. A collision must occur with a minimum energy (activation energy) to be successful. In a coll ision, the orientation - which ends of the reactant molecules collide - must bring the reacting atoms in the molecules together in order for the collision to lead to a reaction. 1 6.45

Collision frequency is proportional to the velocity of the reactant molecules. At the same temperature, both 2 reaction mixtures have the same average kinetic energy, but not the same velocity. Kinetic energy equals 1 12 mv , where m i s mass and v velocity. The methylamine (N(CH3)3) molecule has a greater mass than the ammonia molecule, so methylamine molecules will collide less often than ammonia molecules, because of their slower velocities. Collision energy thus is less for the N(CH3)3(g) + HCl(g) reaction than for the NH3(g) + HCl(g) reaction. Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between methylamine and hydrogen chloride. The fraction of successful coll isions also differs between the two reactions. In both reactions the hydrogen from H C I is bonding to the nitrogen in NH3 or N(CH3h The difference between the reactions is in how easily the H can collide with the N, the correct orientation for a successful reaction. The groups (H) bonded to nitrogen in ammonia are less bulky than the groups bonded to nitrogen in trimethylamine (CH3). So, collisions with correct orientation between HCl and NH3 occur more frequently than between HCI and N(CH3)3 and NH3(g) + HCl(g) � NH 4 C I(s) occurs at a higher rate than N(CH3)3(g) + HCI(g) � (CH3)3NHCI(s). Therefore, the rate of the reaction between ammonia and hydrogen chloride is greater than the rate of the reaction between methylamine and hydrogen chloride.

1 6.46

Each A particle can collide with three B particles, so (4 x 3) =

1 6.48

12

unique collisions are possible.

a The fraction of collisions with a specified energy is equal to the e-£ l RT term in the Arrhenius equation. 1 03 1 00 kJ/ moi a f = e-£ lRT (25 0 C = 273 + 25 = 298 K) - Ea I R T = _ ( 8 . 3 1 4 J / mol - K ) ( 298 K ) I kJ - Ea I RT = - 40.362096 32 Fraction = e-£alR T = e-40 6 096 = 2.9577689 x 1 0-' 8 = 2.96 X t o-18

214

[ J)

1 6.50

a) The reaction is exothermic, so the energy of the reactants is higher than the energy of the products.

Ea (fwd) Ea (rev)

I�,," . . . . . . . . . . . .

ABC + D

b) E.(rev) c)

=

Ea(fwd) - t-Jlrxn

=

Reaction coordinate 2 1 5 kJ/ mol - (-5 5 kJ/mol) bond forming

I

=

AB + CD

.===----====-

2.70

x

1 02

kJ/mol

B / II II ! IC I I I I I I I D A bond weakening 1 6.52

a) The reaction is endothermic since the enthalpy change is positive. Ea (rev)

-

] ��

_

_ _ _

_ N O_ C_ 1 +_C _l_ "-:.II,.--H

NO + CI2 -------

=

+83 kJ

Ea(fwd) = +86 kJ

----------- -----' Reaction coordinate

=

=

b) Activation energy for the reverse reaction: Ea(rev) Ea(fwd) - Mi" 86 kJ - 83 kJ c) To draw the transition state, look at structures of reactants and products: ..

: Cl - Cl :

+

.. .. . N = O:

.. : Cl '

+

..

./

: Cl /

=

3

kJ.

N

� . ". 0' .

The collision must occur between one of the chlorines and the nitrogen. The transition state would have weak bonds between the nitrogen and chlorine and between the two chlorines . : Cl - - - - 1 6.53

.. /.. N �.O. · C1 '

The rate of an overall reaction depends on the slowest step. Each individual step' s reaction rate can vary widely, so the rate of the slowest step, and hence the overall reaction, will be slower than the average of the individual rates because the average contains faster rates as well as the rate-determining step.

215

16.57

A bimolecular step (a collision between two particles) is more reasonable physically than a termolecular step (a collision involving three particles) because the likelihood that two reactant molecules will collide with the proper energy and orientation is much greater than the likelihood that three reactant molecules will collide simultaneously with the proper energy and orientation.

16.58

No, the overall rate law must contain reactants only (no intermediates) and is determined by the slow step. If the first step in a reaction mechanism is slow, the rate law for that step is the overall rate law.

16.59

a) The overall reaction can be obtained by adding the two steps together: CO2 (aq) + O Jr(aq) ---7 HC03-(aq) 2 HC03-Caq) + OJrCaq) ---7 C03 -Caq) + H2Qffi 2 Total : CO2 (aq) + OJr(aq) + HGG:;-� + O Jr(aq) ---7 HGG:;-� + C03 -(aq) + H2 0(!) 2 Overall reaction: CO2 (aq) + 20J r(aq) ---7 C03 -(aq) + H2 0(!) b) Intermediates appear in the mechanism first as products, then as reactants. HC03 is an intermediate. c) Molecularity Rate law Step: rate I = kl [C02 ] [Of1] bimolecular CO2 ( aq)+O W( aq)---7 HC03(aq) bimolecular HC03-(aq) + OJr(aq) ---7 CO /-(aq) + H2 0(!) d) Yes, the mechanism is consistent with the actual rate law. The slow step in the mechanism is the first step with the rate law: rate = kl [C02 ] [Of1] . This rate law is the same as the actual rate law. -

16.60

a) The overall reaction can be obtained by adding the two steps together: CI2 (g) + N02 (g) ---7 CI(g) + N02 CI(g) CI(g) + N02(g) ---7 N02Q(g} Total : C I2 (g) + N02 (g) + Q(gj + N02 (g) ---7 Q(g1 + N02 CI(g) + N02 CI(g) Overall reaction: CI2 (g) + 2N02 (g) ---7 2N02 CI(g) b) Intermediates appear in the mechanism first as products, then as reactants. CI is an intermediate. c) Step: Molecularity Rate law bimolecular rate l = kl [CI2 ] [N02 ] C I2 (g) + N02 (g) ---7 CI(g) + N02 CI(g) rate2 = k2 [CI][N02 ] CI(g) + N02 (g) ---7 N02 CI(g) bimolecular d) Yes, the mechanism is consistent with the actual rate law. The slow step in the mechanism is the first step with the rate law: rate = kl [CI2 ] [N02 ]. This rate law is the same as the actual rate law.

16.6 1

a) The overall reaction can be obtained by adding the three steps together: ( 1 ) A(g) + B(g) !:; X (g) fast (2) X(g) + C(g) ---7 Y(g) slow (3) Y(g) ---7 D(g) fast ru� : � + � + � + � + � ---7 � + � + � A(g) + B(g) + C(g) ---7 D (g) b) Intermediates appear in the mechanism first as products, then as reactants. Both X and Y are intermediates in the given mechanism. c) Step: Molecularity Rate law A(g) + B (g) !:; X(g) bimolecular rate I = kl [A] [B] bimolecular X(g) + C(g) ---7 Y(g) rate2 = k2 [X][C] unimolecular rate3 = k3 [Y] Y(g) ---7 D(g)

216

d) Ye s, the mechanism is consistent with the actual rate law. The slow step in the mechanism is the second step with rate law: rate = k 2 [X] [C] . Since X is an intermediate, it must be replaced by using the first step. For an equilibrium, rateforwardrxn = ratereverserxn. For step then, k,[A] [B] = k_,[X] . Rearranging to solve for [X] gives [X] = (k/k_,)[A] [B] . Substituting this value for [X] into the rate law for the second step gives the overall rate law as rate = (k2k,lk_,)[A] [B] [C] which is identical to the actual rate law with k = k2k/k+ e) Yes, The one step mechanism A(g) + B(g) + C(g) � D(g) would have a rate law of rate = k[A][B] [C], which is the actual rate law.

I

1 6.63

Nitrosyl bromide is NOBr(g). The reactions sum to the equation 2 NO(g) + Br2(g) � 2 NOBr(g), so criterion I (elementary steps must add to overall equation) is satisfied. Both elementary steps are bimolecular and chemically reasonable, so criterion 2 (steps are physically reasonable) is met. The reaction rate is determined by the slow step; however, rate expressions do not include reaction intermediates (NOBr2). Derive the rate law. The slow step in the mechanism is the second step with rate law: rate = k2 [NOB r2] [NO] . Since NOBr2 is an intermediate, it must be replaced by using the first step. For an equilibrium like Step I , rateforwardrxn = ratereverse rxn. Rate, (forward) = k, [NO] [Br2 ] (2) Rate, (reverse) = k_,[NOBr2] (3) Rate2 (forward) = k2[NOBr2] [NO] Solve for [NOBr2 ] in Step Rate, (forward) = Rate, (reverse) k,[NO] [Br2 ] = k_,[NOBr2 ] [NOBr2 ] = (k, I k_,)[NO] [Br2] Substitute the expression for [NOBr2] into equation (3), the slow step: Rate2 (forward) = k2 (k, / k_,)[NO] [Br2 ] [NO] Combine the separate constants into one constant: k = k2 (k, I k_,) Rate2 = k[NO]2 [Br2 ] The derived rate law equals the known rate law, so criterion 3 is satisfied. The proposed mechanism is valid.

(I)

1 6.66 1 6.69

I:

a catalyst changes the mechanism of a reaction to one with lower activation energy. Lower activation energy means a faster reaction. An increase in temperature does not influence the activation energy, but instead increases the fraction of collisions with sufficient energy to react.

No,

The activation energy can be calculated using E quation 1 6.9. Although the rate constants, k, and k2' are not expressly stated, the relative times give an idea of the rate. The reaction rate is proportional to the rate constant. At T, = 20°C = (273 + 20) = 293 K, the rate of reaction is apple/4 days while at T2 = O°C = (273 + 0) = 273 K, the rate is apple/l 6 days. Therefore, rate, = apple/4 days and rate2 = 1 apple/l 6 days are substituted for k, and k2' respectively. T, = 293 K k, = 114 T2 = 273 K k2 = 11 1 6 Ea = ? In k 2 = k, R T2 T, /1 6 In ( 1 ) 8.3 1 4 1 R ln k 2 mol· K ( 1/4 ) k, - ':;-- = --- Ea = -....,-"'27 K 29 K 2 Ea = 4.6096266 10 4 llmol = 4.61 104 J/mol The significant figures are based on the Kelvin temperatures.

I

I

I

_�(_I _�) ( ) ( ( � ;J ( X

)[

�

X

�

J

]

217

1 6.72

Use the given rate law, and enter the given values: rate = k [H+] [sucrose] [W]; = 0.0 1 M [sucrose]; = 1 .0 M The glucose and fructose are not in the rate law, so they may be ignored. a) The rate is first order with respect to [sucrose]. The [sucrose] is changed from 1 .0 M to 2.5 M, or is increased by a factor of 2.51 1 .0 or 2.5. Then the rate = k [H+] [2.5 x sucrose]; the rate i ncrea se s bya fact or of2.5. b) The [sucrose] is changed from 1 .0 M to 0.5 M, or is decreased by a factor of 0.5/ 1 .0 or 0.5. Then the rate = k [W][ O .5 x sucrose]; the rate decreases by a factor of Y2 or hal fthe origi nal rate. c) The rate is first order with respect to [W]. The [H+] is changed from 0.01 M to 0.0001 M, or is decreased by a factor 0[0.0001/0.01 or 0.0 1 . Then the rate = k[ O.O I xW][sucrose] ; the rate d ecr ea sesby a factor of 0.01. Thus, the reaction will decrease to 1 1 1 00 th eorigi nal. d) [sucrose] decreases from 1.0 M to 0. 1 M, or by a factor of (O. 1 M / 1 .0 M) = 0. 1 [H+] increases from 0.0 1 M to 0. 1 M, or by a factor of (0.1 M / 0.0 I M) = 1 0. Then the rate wil l increase by k [ l O x W][ O . I x sucrose]= 1.0 times as fast. Thus, there will be no cha nge .

1 6.75

First, find the rate constant, k , for the reaction by solving the first order half-life equation for k. Then use the first order integrated rate law expression to find t, the time for decay. In 2 In 2 to k = -Rearrange t 1/2 = -t /2 k k = � = 5.7762 1 0-2 yr-I (unrounded) 1 2 yr Use the first-order integrated rate law: In [A]( = In [A]o - kt In[A]( - In[A]o =t -k In[ l O. ppbm] - In[275 ppbm] = t �����-�-�-� -5 .7762 x 10-2 yr-I t = 57.3765798 = 57 yr X

1 6.77

1

a) The rate constant can be determined from the slope of the integrated rate law plot. To find the correct order, the data should be plotted as I ) [sucrose] versus time - linear for zero order, 2) In[ sucrose] versus time - linear for first order, and 3) 1 /[ sucrose] versus time - linear for second order. All three graphs are linear,2 so picking the correct order is difficult. One way to select the order is to compare correlation coefficients (R ) - you may or may not have experience with this. The best correlation coefficient is the one closest to a value of 1 .00. Based on this selection criterion, the plot of ln[ sucrose] vs. time for the first order reaction is the best. Another method when linearity is not obvious from the graphs is to examine the reaction and decide which order fits the reaction. For the reaction of one molecule of sucrose with one molecule of liquid water, the rate law would most likely include sucrose with an order of one and would not include water. The plot for a first order reaction is described by the equation In[A]( = kt + In[A]o. The slope of the plot of In [sucrose ] versus t equals -k. The equation for the straight line in the first order plot is y = -0.2 1 x - 0.6936. So, k = -(-0.2 1 h-I ) = 0.2 1 h-I. Half-life for a first order reaction equals In 2/k, so tl/2 = In 2 / 0.2 1 h-I = 3.3007 = 3.3 h . -

218

Integrated rate law plots

4 --

3

:0 c CD CI

2

.!!!.

-

0.5897x + 1.93 R2=0.993

�

�.-.A

---

�

-

y=-O.077x + 0.4896 R2=0.9875

-

-

.!!

CD CD

y

--

0 0.5 -1

1

1.5

2

2.5

--

y = -0.21x - 0.6936 R2 = 0.9998

-2

3

3 5

-

time, h

Legend:

y-axis is [sucrose] - axis is In[sucrose] & - y-axis is 1I[sucrose] •

•

b) 1f75% of the sucrose has been reacted, 25% of the sucrose remains. The time to hydrolyze 75% of the sucrose can be calculated by substituting l.0 M for [A]o and 0.25 M for [A]t in the integrated rate law equation: In[A]t (-0.21 h-I)t + In[A]o. In[A]t -In[A]o (-O.21h-l)t In[0.25] -In[l.O] (-0.21 h-I)t t 6.6014 6.6 h c) The reaction might be second order overall with first order in sucrose and first order in water. If the concentration of sucrose is relatively low, the concentration of water remains constant even with small changes in the amount of water. This gives an apparent zero-order reaction with respect to water. Thus, the reaction appears to be first order overall because the rate does not change with changes in the amount of water. =

=

=

=

16.81

kl k2 E.

=

= = =

=

=

1 egg I 4.8 min 1 egg I 4.5 min

TI (273.2 + 90.0) K 363.2 K T2 (273.2 + 100.0) K 373.2 K =

?

=

The number of eggs (1) is exact, and has no bearing on the significant figures. 1 ln� _�(_ _�) R T2 TI kl =

_

Ea -

-

R ( ln�)

kl ....,.---'---...:;..,-

( ;2

Ea

=

-

;J

7.2730 X 103 J/mol

(8.314

J )[In (1 egg/4.5min)) mol- K (1 egg/4.8min) (373�2 K

=

7.3

X

103 J/mol

-

363�2 K

219

J

16.82

a) Starting with the fact that rate of formation of 0(rate of step I ) equals the rate of consumption of 0(rate of step 2), set up an equation to solve for [0] using the given values of kl' k2' [N02], and [02]. rate 1 rate2 kl[N02] k2[0][02] ( 6.0 x 1O-3s-1 ) [ 4.0 X 10-9 MJ kl [ N02 ] [0] 2.4 X 10-15 M k2 [ 0 2 ] (1.0 x 106 L l mol · s )[1.0 x 10-2 MJ b) Since the rate of the two steps is equal, either can be used to determine rate 2 of formation of ozone. rate2 k2[0][02] (1.0 x 106 Llmol·s)(2.4 x 10-15 M) (1.0 X 10- M) 2.4 X 10-11 mollL·s =

=

=

=

=

=

=

=

220

Chapter 17 Equilibrium: The Extent of Chemical Reactions FOLLOW-UP PROBLEMS

1 7. 1

Plan: First, balance the equations and then use the coefficients to write the reaction quotient as outlined in E quation 1 7.4. Solution: a) Balanced equation: 4 NH3(g) + 5 02(g) !:; 4 NO(g) + 6 H20(g)

[NOt [H2ot [NH31 [02] b) Balanced equation: 3 NO(g) !:; N20(g) + N02(g) [N20][N0 ] Reaction quotient: Qc = -"---.::...:...-.,-2-....:. [Nof 5

Reaction quotient: Qc =

Check: Are products in numerator and reactants in denominator? Are all reactants and products expressed as concentrations by using brackets? Are all exponents equal to the coefficient of the reactant or product in the balanced equation? 1 7.2

1 7.3

Plan: Kp and Kc for a reaction are related through the ideal gas equation as shown in Kp = Kc(RT )"'" . Find Llngas, the change in the number of moles of gas between reactants and products (calculated as products minus reactants). Then, use the given Kc to solve for Kp . Solution: The total number of product moles of gas is I (PCIs) and the total number of reactant moles of gas is 2 (PCh + CI2). Lln = I - 2 = -1. Kp = Kc(RT)fuJ Kp = 1 .67[(0.08206) (500.W' 1 .67 Kp (0.08206)(500) Kp = 0.04070 1 925 = 4.07 1 0-2 x

Plan: To decide whether CH3CI or CH4 are forming while the reaction system moves toward equilibrium, calculate Qp and compare it to Kp . If Qp > Kp , the reaction is proceeding to the left and reactants are forming. If Q p Kp , the reaction is proceeding to the right and products are forming. Solution: ( 0.24 atm )( 0.47 atm ) = 24.79 1 2 = 25 Qp = PCH3CIPHCI = ( 0. 1 3 atm )( 0.035 atm ) PCH4 PC1z Kp for this reaction is given as 1 .6x 1 04 . Qp is smaller than Kp (Qp Kp ) so more products will form. CH3CI is one of the products forming. Check: Since number of moles of gas is proportional to pressure, a new calculation of Qp can be done with increased pressures for products and decreased pressures for reactants. If the conclusion that formation of products moves the reaction system closer to equilibrium then the new Qp should be closer to Kp than the 25 found for the original Qp . I ncreasing and decreasing by 0.0 1 atm: ( 0.25 atm )( 0.48 atm ) - 40 Qp ( 0. 1 2 atm )( 0.025 atm ) 40 is closer to 1 6,000 than 25, so the answer that CH3CI is forming is correct. <

<

_

_

22 1

17.4

Plan: The information given includes the equilibrium constant, initial pressures of both reactants and equilibrium pressure for one reactant. First, set up a reaction table showing initial partial pressures for reactants and 0 for product. The change to get to equilibrium is to react some of reactants to form some product. The mole ratio between NO:02:N02 is 2:1:2. Use the equilibrium quantity for O2 and the expression for O2 at equilibrium to solve for the change. From the change find the equilibrium partial pressure for NO and N02. Calculate Kp using the equilibrium values. Solution: 2 NO(g) + Pressures (atm) 1.000 Initial +2 x (2:1:2 mole ratio) -2 x -x Change 1.000 - 2 x 1.000 - x 2x Equilibrium At equilibrium Po, 0.506 atm 1.000 - x; so x 1.000 - 0.506 0.494 atm 1.000 - 2 x 1.000 - 2(0.494) 0.012 atm P P 2 x 2 (0.494) 0.988 atm Use the equilibrium pressures to calculate Kp. � (0.988)2 Kp P 02 1.339679 104 1.3 104 P�OP02 (0.012/ (0.506) Check: One way to check is to plug the equilibrium expressions for NO and N02 into the equilibrium expression along with 0.506 atm for the equilibrium pressure of O2. Then solve for x and make sure the value of x by this calculation is the same as the 0.494 atm calculated above. 2 1.3 x 104 (2x) (0.506)2 (1.000-2x) (2x/ (1.3 104) / 0.506 2.569 x 104 (1.000-2x) 2 (2x) �2.569 x 104 (1.000 - 2x) x 0.497 The two values for x agree to give Kp 1.3x104. Plan: Convert Kc to Kp for the reaction. Write the equilibrium expression for Kp and insert the atmospheric pressures for P and Po, as their equilibrium values. Solve for PNO• Solution: The conversion of Kc to Kp: Kp Kc(RT)fu1. for this reaction /).n 2-2 0 (2 mol of product gas, N2 and O2 and 2 mol of reactant gas, NO) Kp Kc(RT)o so Kp Kc. Kp PN22P02 Kc 2.3 x 1030 (0.781)(0.209) 2 PNO 2.6640 10-16 2.7 10-16 atm The equilibrium partial pressure of NO in the atmosphere is 2.7xlO-16 atm. Check: Insert the pressure of NO and calculate the equilibrium constant to check the result. (0.781)(0.209) --'---'-'--' - ----'--=- 2 2.300 1030 2.3 1030 (2.6640 10-16) NO

NO

=

=

=

,

=

=

=

=

=

=

=

=

=

=

17.5

N

,

=

=

=

=

=

X =

X

=

=

=

X

=

=

=

X

=

=

=

X

=

X

X

=

X

222

X

=

X

1 7.6

Plan: Set up a reaction table and use the variables to find equilibrium concentrations in the equilibrium expression. ' . . I [ H I ] 2.50 mol 0.242248 M S o I utlOn · : T h e IllItla 1 0.32 L Concentration (M) 2 HI(g) + Initial 0.242248 .=C:!.! h"""an'-"g ""e'--2 x"---+-'-'x"---+�x (2 : I I mole ratio) Equilibrium 0.242248 - 2x x x (U11founded values) Set up equilibrium expression: [x][ x] 3 [ H 2 ] [12 ] Take the square root of each side Kc 1 .26 X 1 02 [HI] [0.242248-2x] 2 [x] 0.035496 [0.242248 - 2x] 0.0085988 - 0.070992x x x 8.0288 X 1 0-3 8.033 X 1 0-3 =

=

+­ -+

""

_______

=

_______ _

_____

:

=

=

=

=

[H2]

=

=

[12]

=

8.03

x

1 0- M

into the equilibrium expression and calculate Kc. Check: Plug equilibrium concentrations 3 [HI] 0.242 - 2(8.03 x 1 0- ) 0.226 M (8.03 x 1 0-3 ) 2 1 (0.226) 2 1 .2624 X 1 0-3 1 .26 xLO-3 =

=

=

1 7.7

=

Plan: First set up the reaction table, then set up the equilibrium expression. To solve for variable, x, first assume that x is negligible with respect to initial concentration ofl2 . Check the assumption by calculating the % error. If the error is greater than 5%, calculate x using the quadratic equation. The next step is to use x to determine the equilibrium concentrations Ofl2 and I. Solution: [I2]init 0.50 mol 1212.5 L 0.20 M a) Equilibrium at 600 K 2 1(g) Concentration (M) Initial o +2 x Change -x 2x 0.20 - x Equilibrium =

=

+­ -+

Equilibrium expression:

Kc

=

f�:2]

=

2.94 x 1 0-10

[2X ] 2 2.94 X 1 0-1 0 [0.20 - x ] Assume x is negligible so 0.20 - x 0.20. 4x 2 2.94 1 0-1 0 0.20 x 3 . 834 X 1 0-6 3.8 X 1 0-6 M Check the assumption by calculating the % error: (3 .8x 1 0-61 0.20) x 1 00% 1 .9x I 0-3 % which is smaller than 5%, so the assumption is valid. At equilibrium [I] eq 2x 7.668 X 1 0-6 7.7 1 0-6 M and [h]eq 0.20 - 3 . 834 x 1 0-6 0. 1 99996 0.20 M b) Equilibrium at 2000 K -::-''--�-:;-

=

X

=

=

'"

=

=

=

=

=

=

=

=

Equilibrium expression: [2x ] 2 [0.20-x]

-;---=.----=---:;-

=

0.209

223

X

A ssume x i s negl igible so 0.20 - x i s approximately 0.20 .

4x 2 = 0.209 0.20 x = 0. 102225 = 0. 102 M

Check the a ssumption by calculating the % error: 0. 1027- 0 .20 = a ssumption i s not valid. Solve u sing quadratic equation.

-x [0 .20 2

4x

a=

-b

5 1 % which i s l arger than 5 % so the

] = 0.209

+ 0.209 x - 0.0 4 1 8 = 0 4; b = 0.209; c = -0.0 4 1 8 ±

Jb2

-

4ac

=x

2a -0 .209 ± (0.209) 2 - 4 ( 4) (-0.0 4 1 8) x= = 0.0 793 857 or -0. 1 3 1 6 2 (4) Choo se the po sitive value , x = 0.0 79 At equil ibrium [I]eq = 2 x = 0. 1 5877 = 0. 1 6 M and [I2] eq = 0.20 - x = 0 . 1 206 1 = 0. 1 2 M Check: Calculate equil ibrium con stant s u sing calculated equi l i brium concentration s: 2 a) K= ( 7 . 7x 1 0--6 ) /(0.20) = 2.9645 x 10-1 0 = 3.0 10-1 0 2 b) K= (0. 1 6) /(0. 1 2) = 0.2 1 333 = 0.2 1 3

�

X

1 7.8

Plan : Calculate the i nitial concentration s o f each sub stance . For part (a) , calculate Qc and compare t o given Kc . I f Qc> Kc then the reaction proceed s t o the le ft to make reactant s from product s. I f Qc < Kc then the rea ction proceed s to right to make product s from reactant s. For part (b) , u se the re sult of part (a) and the given equi li bri u m concentration of PCI s to find the equi l i brium concentration s of P C I 3 and C1 2 . So lutio n: [ P C I s] = 0. 1050 mol /0. 5 000 L = 0.2 100 M I n itial concentration s: [ PC I 3 ] = [CI2 ] = 0.0450 mol/0.5000 L = 0.0 900 M a) Qc

-

_

[ P C I 3 ][ C1 2 ] - [0 .0 900 ] [0 .0 900] - 0.03857 1 - 0.0 386 [ ] _

_

_

[0 .2 100 ] Qc, 0 .0386, i s le ss than Kc, 0.0 42, so the reaction wi l l proceed to the right to ma ke more pro du cts. PCI s

b) To reach eq u i l i brium , concentration s wil l i ncrea se for the product s, PCI 3 and C 12 , and de crease for the reactant , P C I s. PCI s !:; PCI 3 + CI2 0.2 100 M 0.0900 M 0 .0 900 M I n itial Chan ge -x +x +x Equil ibrium 0.2 100 - x 0.0 900 + x 0 .0900 + x [ P C l s] = 0.2065 M = 0.2 100 - x ; x = 0.0035 M [ PC I 3 ] = [CI2 ] = 0 .0 900 + x = 0.0900 + 0.0035 = 0.0935 M Chec k : Calculate equi l i brium con stant s u sing the calculated equi l i brium concentration s:

Kc =

[ P C I 3 ][ C I 2 ] [ PC l s ]

[0.0935][0 .0 935] = 0.042335 = 0.0423 [0 .2065 ]

224

1 7.9

P lan: Exam ine each change for its i mpact on Qc' Then decide how the sy stem would re spond to re-e sta bl i sh eq ui li bri u m . 2 = [ S i F4 H 2 0 S o l ution : Q c HF

][

[ t

]

a) Decrea sing [ H20] lead s to Qc < Kc, so the reaction would shi ft to ma ke more product s from reactant s. There fore, the S i F 4 concentration, a s a product, would inc reas e. b) Add i ng l iq u i d water to thi s sy stem at a temperature a bove the boi l i n g point o f water would re sult i n an increa se in the concentration o f water vapor. The increase in [ H 20] increa se s Qc to ma ke it greater than Kc To re -e sta bl i sh equi li brium product s w i l l be converted to reactant s and the [S i F4 ] w i l l dec reas e. c ) Removing the reactant H F i ncrea se s Qc, which cau se s the product s to react to form more reactant s. Thu s, [ S i F4 ] dec reas es . d) Removal o f a so l i d product ha s no impact on the equi l i bri u m ; [ S i F4 ] do es not ch ang e. Chec k: Loo k at each change and decide which direction the equi l i bri um would shi ft u sing Le Ch atel i er 's princ iple to chec k the change s predi cted a bove. a ) Remove product, equi l i bri u m shi fts to right. b) Add product, eq u i l i brium sh i fts to l e ft. c) Remove reactant, eq u i l i bri um shi fts to le ft. d) Remove so l i d reactant, equ i l i brium doe s not sh i ft. 1 7. 1 0

P lan: Change s i n pre ssure ( and volume) a ffect the concentration of ga seou s reactant s and product s. A decrea se in pre ssure, i . e . , increa se i n volume, favor s the production of more ga s molecule s wherea s an i ncrea se i n pre ssure favor s the production o f fewer ga s molecule s. Examine each reaction to decide whether more or fewer ga s mo lec u l e s w i l l re sult from producing more product s. I f more gas molec u l e s re sult, then the pre ssure should be increa sed ( vol ume decrea sed) to reduce product formation. I f fewer ga s molecule s re su l t, then pre ssure should be decrea sed to produce more reactant s. S o l ution : a) In 2 S02( g) + 02( g) !:; 2 S0 3( g) three mo lec u l e s o f ga s form two molec u l e s o f ga s, so there are fewer ga s molecule s in the product. Dec reas in g p ressu re ( increa sing vol ume ) wi l l favor the reaction direction that produce s more mole s o f ga s: toward s the reactant s and away from product s w i l l decrea se the product yield . b) I n 4 N H 3( g) + 5 02( g) !:; 4 NO( g) + 6 H P( g) 9 molecu le s o f reactant ga s convert t o 1 0 molecu l e s o f product ga s. Inc reasing p ress u re (decrea sing volume) w i l l favor the reaction direction that produce s fewer mole s o f ga s: toward s the reactant s and away from product s. c) In CaC204 ( s) !:; CaC0 3( s) + CO( g) there are no reactant ga s mo l ecul e s and one product ga s molec u l e s. Inc reas ing p ressu re (decrea sing vo l ume) w i l l favor the reaction direction that produce s fewer mole s o f ga s: toward s the reactant s and away from product s.

1 7. 1 1

Plan: A decrea se i n temperature favor s the exothermic direction o f an equ i l i brium reaction, the direction that produce s heat. F i r st, identi fy whether the forward or rever se reaction i s exothermi c fro m the given entha lpy change . t1H < 0 mean s the forward reaction i s exothermic, and t1H> 0 mean s the rever se reaction i s exothermic. I f the forward reaction i s exotherm ic then a decrea se in temperature w i l l sh i ft the equi l i brium to the right to ma ke more product s from reactant s and increa se Kp . I f the rever se reaction i s exothermic then a decrea se i n temperature wi l l sh i ft the equ i l i brium to the le ft to ma ke more reactant s from product s and decrea se Kp S o l ution : a) t1H < 0 so the forward reaction i s exotherm ic. A decrea se i n temperature increa se s the partial pre ssure o f product s and decrea se s the parti al pre ssu re s o f reactant s, so P " dec reas es . With i ncrea se s i n product pre ssure s , and decrea se s in reactant pre ssure s, Kp inc reas es. b) t1H> 0 so the rever se reaction i s exothermi c . A decrea se i n temperature decrea se s the partial pre ssure o f product s and increa se s the partial pre ssure s o f reactant s, so P N, in c reas es . Kp dec reas es with decrea se i n product pre ssure s and i ncrea se in reactant pre ssure s. c) !1H < 0 so the forward reaction i s exothermic. Decrea sing temperature inc reas es P PC!, and inc reas es

225

Kp-

Chec k: Solve the problem for increa sing i n stead o f decrea sing temperature and ma ke sure the an swer s are oppo site . a) I ncrea sed temperature favor s the rever se endothermi c reaction leading to an i ncrea se in reactant pre ssure s, so P,." increa se s and Kp decrea se s. b) I ncrea sed temperature favor s the forward endothermi c reaction leading to a decrea se in reactant pre ssure s. P N, decrea se s and Kp increa se s. c) B oth Prel and ,

17.12

Kp decrea se with increa sed temperature .

Plan: Given the balanced equi librium equation, it i s po ssible to set up the ap propr iate equ il ibr ium ex pre ssion CQc). For the equation given f..n = 0 meaning that Kp = Kc. The value of K may be found for scene 1 , and value s for Q may be determined for the other two scene s. The reaction w i l l sh i ft toward s the reactant side if Q > K, and the reaction w i l l shi ft toward s the product side if Q < K. The reaction i s exothermic, thu s, heat may be con si dered a product . I ncrea si ng the temperature add s a product and decrea sing the temperature remove s a product . Solution : a) Kp require s the equ i l i brium val ue o f P for each ga s. The pre ssure may be found from P = n RT / V.

(nc�RT)2 (nc�RT)( � )

2

Kp =�

n o RT

P C2 P 02

Thi s equation may be simpl i fi ed becau se for the sample Kp b) Scene 2

�

Qp =

� =4

2n co _)2 1

n o n c2 n O2

=

=

n C2 n 02

R,

T, and

V are con stant .

U sing scene

I:

( 2 )( 2 ) =

(6

(1)( )

= 36

Q> K, thu s, the reacti on wi l l shift to th e l eft (toward s the reactant s) . Scene

3 Qp =

�

n o n C2 n 02

=

� =1 (3)(3)

ha s Q < K, thu s, it w i l l shift to th e right (toward s the product s) . c) I ncrea sing the te mperature i s equivalent to adding a product (heat) to the equi l i bri u m . The reaction will shi ft to con sume the added heat . The reaction w i l l shi ft to the le ft (toward s the reactant s) . However, since there are 2 mo le s o f ga s on each side of the equation, the shi ft ha s no effect on total mole s of ga s. Chec k: a) Remember to u se the coe fficient s a s exponent s, and to multiply, not add the value s for the different reactant s. b) The relation ship between Q and K determine s the direction the reaction w i l l precede . c) Rever se the sign on !1H (ma ke the reaction endothermic), and see if you get the oppo si te re sult . END-OF-CHAPTER PROBLEMS

17.1

If the rate of the forward reaction exceed s the rate of rever se reaction, product s are formed fa ster than they are con sumed . The change i n reaction condition s re sult s in more product s and le ss reactant s. A change in reaction condition s can re sult from a change in concentration or a change i n temperature . If concentration change s, product concentration i ncrea se s whi l e reactant concentration decrea se s, but the Kc re mai n s unchanged becau se the ratio of product s and reactant s remain s the same . If the increa se in the forward rate i s due to a cha nge in temperature, the rate of the rever se reaction al so i ncrea se s. The equil ibrium ratio o f product concentration to reactant concentration i s no lo nger the same . S i nce the rate of the forward react ion i ncrea se s more than the rate of the rever se reaction, Kc increa se s (numerator, [product s] , i s larger and denominator, [reacta nt s] , i s sma l ler) .

-

[ product s] [ reactant s]

Kc = -=-

--=-

226

17.6

The equilibrium constant expression is K [02]. If the temperature remains constant, K remains constant. If the initial amount of Li2 0 2 present was sufficient to reach equilibrium, the amount of O 2 obtained will be constant, regardless of how much Li 2 0 2(s) is present.

17.7

On the graph the concentration of HI increases at twice the rate that H 2 decreases because the stoichiometric ratio in the balanced equation is I H 2 : 2HL Q for a reaction is the ratio of concentrations of products to concentrations of reactants. As the reaction progresses the concentration of reactants decrease and the concentration of products increase, which means that Q increases as a function oftime. [ HI f H 2 (g) + I 2 (g) !:; 2HI(g) Q [ H 2 ] [1 2 ]

=

=

c

.g �c

(1) u C o

U

Time The value of Q increases as a function of time until it reaches the value of K. b) No, Q would still increase with time because the [1 2 ] would decrease in exactly the same way as [H 2 ] decreases. 17.10

Yes, the Q 's for the two reactions do differ. The balanced equation for the first reaction is 3/2 H 2 (g) + 1 /2 N2(g) !:; NH3(g) (1) The coefficient in front ofNH3 is fixed at 1 mole according to the description. In the second reaction, the coefficient in front ofN 2 is fixed at one mole. 3 H2 (g) + N 2(g ) !:; 2 NH3(g) (2) The reaction quotients for the two equations and their relationship are: [NH 3 1 Q2 [H2 f [ N2 ] _

17. ) )

--'=-----::-"-='--

Check that correct coefficients from balanced equation are included as exponents in the mass action expression. a) 4 NO(g) + 0 2 (g) !:; 2 N 2 03(g) [ N 20 3 1 Qc [ NO t [ 0 2 ] b) SF6 (g) + 2 S03(g) !:; 3 S02F 2 (g) [ S0 2F2 f Qc ':" ----::[SF6 IS0 3 1 c) 2 SCI Fs(g) + H 2 (g) !:; S 2FIO(g) + 2 HCl(g) [S2 FIO ][ HCl f Qc [SClFs ] [ H 2 ] _

=----=--,----=

---=-----=

_

-' ==O---'...:C !=.--'-'-

2 27

1 7. 1 3

. . . . Th e Q "'lor t h e on g ina Ireact Ion IS Qref=

[ H 2 ] 2 [ S2 ] [ H2S] 2

a) The given reaction 1 12 S2(g) + H2(g)!:; H2S(g) is the reverse reaction o f the original reaction multipl ied by a factor o f 1 12 . The eq ui l i brium constant for the reverse reaction is the inverse o f the original constant . When a reaction is multipl ied by a factor, K, o f the new equation is equal to the K o f the original equi l i brium raised to a 2 power equal to the factor . For the reaction given in part a ta ke ( 1 IK) 1 1 . 2 Qa = (1/ Qref)1/ =

[�2S] [S2 ]X [ H 2 ]

2 2 K= ( l / 1 .6 X 10- )1/ = 7 .90569 = 7.9 b) The given reaction 5 H2S(g)!:; 5 H2(g) + 5/2 S2(g) is the original reaction multiplied by 5/2 . Ta ke the original K to the 5/2 power to find K o f given reaction. Qa

=

(Qref)5/ K

1 7. 1 5

2

=

=

[ H 2 ]5 [ S2 ]7i [ H 2 S]5

2 2 ( 1 . 6 x 10- )51

=

3 .238 1 7

X

1 0-5

=

3 .2

X

1 0-5

The concentration o f solids and pure liquids do not change, so their concentration terms are not written i n the reaction q uotient expression . a) 2 Na202(S) + 2 CO2(g)!:; 2 Na2C03(S) + 02(g) Qc=

[ 02 ] [ C02 ] 2

---=------=----

b) H20(l)!:; H20(g) Only the gaseous water is used . The "(g)" is for emphasis . Qc = [H20(g)] c ) NH4CI(s)!:; NH3(g) + HCI(g) Qc=[NH3][HCI] 1 7. 1 8

a) The balanced equations and corresponding reaction quotients are given below . Note the second equation must occur twice to get the appropriate overal l equation .

[ C LF f [ C I2 H F2 ] [C IF3 ] Q2 - [ C IF H F2 ] [ C IF3 ] Q2 - [ C IF ] [ F2 ] [C IF3 J Qoverall [ C 12 ] [ F2 ] 3 QI

=

_

_

_

b) The second equation occurs twice, thus it could simply be multipl ied by two and its reaction quotient squared . The reaction quotient for the overa l l reaction, Qoverall/, determined from the reaction i s : _

Qovcrall -

[C IFJ [ C I2 ] [ F2 ] 3

228

17.20

Kc and Kp are related by the equation Kp Kc(RTy"n where /').n represents the change in the number of moles of gas in the reaction (moles gaseous products - moles gaseous reactants). When /').n is zero (no change in number of moles of gas) the term ( RT)Lln equals 1 and Kc Kp . When /').n is not zero, meaning that there is a change in the number of moles of gas in the reaction, then Kc of. Kp . =

=

17.21

a) Kp Kc(RTy"n. Since /').n number of moles gaseous products - number of moles gaseous reactants, /').n is a positive integer. If /').n is a positive integer, then (RT)LlI1 is greater than I . Thus, Kc is multiplied by a number that is greater than 1 to give Kp . Kc is s mall er thanKpb) Assuming that RT > I (which occurs when T > 12.2 K, because 0.0821 (R) x 12.2 I ) , Kp Kc if the number of moles of gaseous products exceeds the number of moles of gaseous reactants. Kp Kc when the number of moles of gaseous reactants exceeds the number of moles of gaseous product. =

=

>

=

<

17.22

17.24

a) Number of moles of gaseous reactants 0; Number of moles of gaseous products 3; /').n 3 - 0 b) Number of moles of gaseous reactants I; Number of moles of gaseous products 0; /').n 0 - 1 c) Number of moles of gaseous reactants 0; Number of moles of gaseous products 3; /').n 3 - 0 =

=

=

=

=

=

=

= -

=

=

=

=

3 1

3

First, determine /').n for the reaction and then calculate Kc using Kp Kc(RT)LlI1. a) /').n Number of product gas moles - Number of reactant gas moles 1 - 2=-1 3.9x 10.2 Kc � =3.2019=3.2 ( RT)Lln [(0.0821)(1000.)r1 b) /').n Number of product gas moles - Number of reactant gas moles I - 1 0 =

=

=

=

=

=

Kc

=

=

� ( RT)Lln

=

28.5 [(0.0821)(500.) t

=

28

.

=

5

17.26

When Q K, the reaction proceeds to the rig ht to form more products. The reaction quotient and equilibrium constant are determined by [products] / [reactants] . For Q to increase and reach the value of K, the concentration of products (numerator) must increase in relation to the concentration of reactants (denominator).

17.27

To decide if the reaction is at equilibrium, calculate Qp and compare it to Kp . I f Qp Kp then the reaction is at equilibrium. If Qp > Kp then the reaction proceeds to left to produce more reactants. I f Q p Kp then the reaction proceeds to right to produce more products.

<

=

Qp

=

PH2 PBr2 p2

HBr

=

(0.010)(0.010) (0.20)2

=

<

2.5 X 10 3 > K p 4.18 X 10-9 -

=

Qp > Kp, thus, the reaction is not at equilibrium and will proceed to the l eft (towards the reactants). Thus, the numerator will decrease in size as products are consumed and the denominator will increase in size as more reactant is produced. Qp will decrease until Qp Kp . =

17.31

a) The approximation applies when the change in concentration from initial to equilibrium is so small that it is insignificant. This occurs when K is small and initial concentration is large. b) This approximation will not work when the change in concentration is greater than 5%. This can occur when [reactant] initial is very small, or when [reactant]change is relatively large due to a large K.

17.32

Since all equilibrium concentrations are given in molarities and the reaction is balanced, construct an equilibrium expression and substitute the equilibrium concentrations to find Kc. 2 [1.87 x 10-3 J -;=---=------,�---='-----=c -

[ 6.50 x 10-5 J [1.06

X

10-3 J

22 9

=

50.753

=

50.8

17.34

The reaction table requires that the initial [PCIs] be calculated: [PC Is] =0.15 mo 1/2.0 L 0.075 M Since there is a 1:1:1 mole ratio in this reaction: x = [PCIs] reacting (-x), and the amount of PCI 3 and of CI 2 forming (+x). Concentration (M) PCIs(g) =t PCI3 (g) + CI 2 (g) 0 0 0.075 Initial +x +x -x Change x x 0.075 - x Equilibrium

17.36

Two of the three equilibrium pressures are known. Construct an equilibrium expression and solve for PNOC1 . p2 Kp =6.5 104 = 2NOC1 PNO PCI 2 p2

=

X

6.5 X 104=

NOCI

(0.35)2 (0.10)

J

PNOC1 ( 6.5 x 104 )(0.35)2(0.1O)=28.2179=28atm A high pressure for NOCI is expected because the large value of Kp indicates that the reaction proceeds largely to the right, i.e., to the formation of products. =

17.38

The ammonium hydrogen sulfide will decompose to produce hydrogen sulfide and ammonia gas until Kp =0.11: NH4HS(s) =t H2 S(g) + NH 3 (g) x = [NH4HS] reacting (-x), and the amount of H 2 S and ofNH 3 forming (+x) since there is a 1:1:1 mole ratio between the reactant and products. (It is not necessary to calculate the molarity ofNH4HS since, as a solid, it is not included in the equilibrium expression). Concentration (M) NH4 HS(s) =t H2 S(g) + N H 3 (g) Initial Molarity 0 0 �

Q��

Equilibrium

Molarity - x

�

�

x

x

Kp =0.1 I = (PH 2S )(PNH 3 ) (The solid N H4HS is not included ) 0.11 = (x)(x) x = PNH 3 = = 0.33166=0.33 atm 17.40

The initial concentrations ofN2 and O2 are (0.20 mol/l .O L) 0.20 M and (0.15 mol/ I .O L) 0.15 M, respectively. N 2(g) + 02 (g) =t 2 NO(g) There is a I : 1:2 mole ratio between reactants and products. Concentration (M) N 2 (g) 2 NO(g) 02 (g) =t 0 0.15 0.20 Initial Change -x -x + 2x ( I : I :2 mole ratio) 2x 0.15 - x 0.20 - x Equilibrium [ NO]2 [ 2x]2 =

-4

Kc=4.10xlO =

[N ][0 ] [0.20-x][0.15-x] 2 2 Assume 0.20 M - x"" 0.20 M and 0.15 M - x"" 0.15 M 4x2 4.10 x 10-4 = [0.20][0.15] x =1.753568 X 10-3 M (unrounded) [NO] =2 x =2 (1.753568 X 10-3 M) =3.507136 X 10-3=3.5 X 10-3 M (Since (1.8 x 10-3 ) / (0.15) < 0.05, the assumption is OK.)

230

=

17.42

Construct a reaction table, using [TCI]init=(0.500 mol l 5.00 L)=0.100 M, and substitute the equilibrium concentrations into the equi librium expression. There is a 2: I: I mole ratio between reactant and products: Concentration (M) 2 TCl(g) I2(g) + CI2(g) !:; 0.100 Initial 0 0 Change -2 x +x +x (2:1: 1 mole ratio) Equilibrium 0.100 - 2 x x x

[x][x]

[I2][CI2] [ICI]2 2 [x] 0.110= [ 0.100 - 2 x]2

[ 0.100 - 2 x]2

Kc=O.IIO=

Take the square root of each side:

0.331662 =

[x ]

[0.100 - 2 x]

x=0.0331662 - 0.663324 x 1.663324 x=0.0331662 x=0.0199397 =

=

[12]eq [CI 2]eq 0.0200 M [ICI]eq=0.100 - 2 x=0.601206= 0.060 M ICI

17.44

4 NH (g) + 3 02(g) !:; 2 N2(g) + 6 H20(g) 3 the equilibrium constant, determine the equilibrium concentrations of each reactant and product and insert To find into the equilibrium expression. Since [N2] increases from 0 to 1.96 x 10-3 M, the concentration of H20 gas will increase by 3 times as much (stoichiometric ratio is 6 H20 : 2 N2) and the concentration of reactants will decrease by a factor equivalent to the stoichiometric ratio (2 for NH3 and 3/2 for O2). Since the volume is 1.00 L, the

concentrations are equal to the number of moles present. + Concentration (M) 4 NH (g ) + 3 02(g) 0.0150 0.01503 Initial +2x Change -4x -3x Equilibrium +2x 0.0150 - 4 x 0.0150 - 3 x All intermediate concentration values are ul1founded. [N2] eq=2x=1.96 x 10-3 M [H20]eq (6 mol H20/2 mol N2) (1.96 x 10-3) =5.8800 x 10-3 M [NH3]eq=(0.0150 mol NH3/1.OO L) - (4 mol NHi2 mol N2) (1.96 x 10-3) =1.1080 2x 10-2 M [02]eq=(0.0150 mol 02/1.00 L) - (3 mol 02/2 mol N2) (1.96 x 10-3)=1.2060 x 10- M +­ .....

+6x +6x

=

_

Kc -

[N2]2[H20]6 [NH31 [O2 f

_

-

[1.96 1O-3J2 [ 5.8800 1O-3J6 =6.005859 x 10-{i=6.0 1 [1.1080 10-2 r [1.2060 x 10-2 J X

X

X

x

1 0-{i

If values for concentrations were rounded to calculate Kc, the answer is 5.90 x 10-{i. 17.46

Equilibrium position refers to the specific concentrations or pressures of reactants and products that exist at equilibrium, whereas equilibrium constant refers to the overall ratio of equilibrium concentrations and not to specific concentrations. Reactant concentration changes cause changes in the specific equilibrium concentrations of reactants and products (equilibrium position), but not in the equilibrium constant.

17.47

A positive Wrxn indicates that the reaction is endothermic, and that heat is consumed in the reaction: NH4Cl(s) + heat!:; NH3(g) + HCI(g) a) The addition of heat (high temperature) causes the reaction to proceed to the right to counterbalance the effect of the added heat. Therefore, more products form at a higher temperature and container (B) with the largest number of product molecules best represents the mixture.

231

b) When heat is removed (low temperature), the reaction shifts to the left to offset that disturbance. Therefore, NH 3 and HCI molecules combine to form more reactant and container (A) with the smallest number of product gas molecules best represents the mixture. 17.50

An endothermic reaction can be written as: reactants + heat!:; products. A rise in temperature (increase in heat) favors the forward direction of the reaction, i . e., the formation of products and consumption of reactants. Since K [products]/[reactants], the addition of heat increases the numerator and decreases the denominator, making K2 larger than Ki. =

17.51

a) Equi librium position shifts tow ard products. Adding a reactant (CO) causes production of more products as the system will act to reduce the increase in reactant by proceeding toward the product side, thereby consuming additional CO. b) Equilibrium position shifts tow ard products. Removing a product (C02 ) causes production of more products as the system acts to replace the removed product. c) Equilibrium position do es not sh ift. The amount ofa solid reactant or product does not impact the equilibrium as long as there is some solid present. d) Equilibrium position shifts tow ard react ants. When product is added, the system will act to reduce the increase in product by proceeding toward the reactant side, thereby consuming additional CO2; dry ice is solid carbon dioxide that sublimes to carbon dioxide gas. At very low temperatures, CO2 solid will not sublime, but since the reaction lists carbon dioxide as a gas, the assumption that sublimation takes place is reasonable.

17.53

An increase in container volume results in a decrease in pressure ( Boyle's Law). Le Chatelier's principle states that the equilibrium will shift in the direction that forms more moles of gas to offset the decrease in pressure. a) Mo reF forms (2 moles of gas) and l ess Fz ( I mole of gas) is present as the reaction shifts towards the right. b) Mo re CzHz and Hz form (4 moles of gas) and l ess CH4 (2 moles of gas) is present as the reaction shifts towards the right.

17.55

The purpose of adjusting the volume is to cause a shift in equilibrium. a) Because the number of reactant gaseous moles (4H 2 ) equals the product gaseous moles (4H 2 0), a change in volume will have no effect on the yield. b) The moles of gaseous product (2 CO) exceed the moles of gaseous reactant ( I O 2 ). A decrease in pressure favors the reaction direction that forms more moles of gas, so inc reas ethe reaction vessel volume.

17.57

An increase in temperatur e (addition of heat) causes a shift in the equilibrium away from the side of the reaction with heat. a) CO(g) + 2H 2 (g) !:; CH 3 0H(g) + heat Mfrxn -90.7 kJ A negative f<,.Hrxn indicates an exothermic reaction. The equilibrium shifts to the left, away from heat, towards the reactants, so amount of product dec reas es. 131 kJ Mfrxn b) C(s) + H 2 0(g) + heat !:; CO(g) + H 2 (g) A positive Mfrxn indicates an endothermic reaction. The equilibrium shifts to the right, away from heat, towards the products, so amounts of products inc reas e. c) 2N02 (g) + heat !:; 2NO(g) + 0ig) The reaction is endothermic. The equilibrium shifts to the right, away from heat, towards the product, so amounts of products inc reas e. d) 2C(s) + 02(g) !:; 2CO(g) + heat The reaction is exothermic. The equilibrium shifts to the left, away from heat, towards the reactants; amount of product dec reas es . =

=

17.60

a) S0 2 (g) + 12 0 2 (g) !:; S03 (g) + heat The forward reaction is exothermic (f<,.Hrxn is negative), so it is favored by low ert emp eratu res. Lower temperatur es will cause a shift to the right, the heat side of the reaction. There are fewer moles of gas as products ( I S03) than as reactants ( l S0 2 (g) + 12 O2 ), so products are favored by h ig h er p ressu re. H igh pressure will cause a shift in equilibrium to the side with the fewer moles of gas. [SO ] ) b) Addition of O 2 would dec reas e Q ( Q ) and have no impact o nK. [S02][02]i 2 =

/

232

c) To enhance yield of S03 , a low temperature is used. Reaction rates are slower at lower temperatures, so a catalyst is used to sp eed u p th e react io n. 17.65

a) You are given a value of Kc but the amounts of reactant and product is given in units of pressure. Convert Kc to Kp. Kp=Kc(RT)t." tin =1 - 2=-I (I mol of product, C2HsOH and 2 mol of reactants (C2H 4 + H20) Kp=Kc(RTrl =(9 x 103)[(0.0821 L-atmlmol-K) (600. K) t =1.8270 x 102 (unrounded) Substitute the given values into the equilibrium expression and solve for Pc,,,, . Kp=

PC HsOH 2 PC H4 P H 0 2 2

Pc,,,, =2.7367 X 10-3 =3 X

10-3 at m

b) The forward direction, towards the production of ethanol, produces the least number of moles of gas and is favored by high p r essu re. A low t emp eratu re favors an exothermic reaction. c) No, condensing the C2HsOH would not increase the yield. Ethanol has a lower boiling point (78.5°C) than water ( I OO°C). Decreasing the temperature to condense the ethanol would also condense the water, so moles of gas from each side of the reaction are removed. The direction of equilibrium (yield) is unaffected when there is no net change in the number of moles of gas. 17.68

a) You are given a value of Kc but the amounts of reactant and product is given in units of pressure. Convert Kc to Kp. Kp=Kc(RT)"''' tin =2 - 3=-I (2 mol of product, S0 3 and 3 mol of reactants, 2 S02 + O2) Kp=Kc(RT) t.n=Kc(RTrl =(1.7 x 108)[(0.0821 L-atm / mol-K) (600. K) rl =3.451 x 106 (unrounded) 2 p20 (300 ' ) S Kp= 3 2 (100.) =3.451 106 P P

i

p 02 02

Pso,

X

S 02

=0.016149= 0.0 1 6

at m

b) Create a reaction table that describes the reaction conditions. Since the volume is 1.0 L, the moles equals the molarity. Note the 2: 1:2 mole ratio between S02: 02: S03 ' Concentration (M) + 02(g) 2 S02(g) Initial 0.0040 0.0028 ---"'2=--x'-'--"'C"-'h -"' a n'-"g"' e'-(2: 1:2 mo Ie ratio) --'-'x'------'-+.::.2�x 2 x=0.0020 (given) 0.0028 - x Equilibrium 0.0040 - 2 x x= O.OO l O, therefore: [S02] =0.0040 - 2(0.0010)=0.0020 M [02] =0.0028 -0.0010=0.0018 M [S03 ] =2(0.00I0)=0.0020 M Substitute equilibrium concentrations into the equil ibrium expression and solve for Kc. [0.0020]2 [SO 1 Kc= =555.5556= 5.6 1 02 2 2 [ S0 2 ] [ 0 2 ] [0.0020] [0.0018] The pressure of S02 is estimated using the concentration of S02 and the ideal gas law (although the ideal gas law is not wel l behaved at high pressures and temperatures). L - atm (1000. K) (0.0020 mOI) 0.0821 nRT mol -K =0.1642= 0. 1 6 at m = Pso, = (1.0 L ) V +­ -+

___

_____

________

3

x

(

17.69

J

The equi librium constant for the reaction is Kp= Peo,=0.220 atm The amount of calcium carbonate solid in the container at the first equilibrium equals the original amount, 0.100 mol, minus the amount reacted to form 0.220 atm of carbon dioxide. The moles of CaC0 3 reacted is equal to the number of moles of carbon dioxide produced. Use the pressure of CO2 and the Ideal Gas Equation to calculate the moles of CaC03 reacted:

233

Moles CaC03 = moles CO2 = n = PY I RT (0.220 atm) ( I O . O L) n= 0.06960 mol CaC0 3 lost (unrounded) 0.082 1 L · atm (385 K) mol · K 0. 1 00 mol CaC0 3 - 0.06960 mol CaC0 3 = 0.0304 mol CaC03 at first equilibrium As more carbon dioxide gas is added, the system returns to equilibrium by shifting to the left to convert the added carbon dioxide to calcium carbonate to maintain the partial pressure of carbon dioxide at 0.220 atm (Kp). Convert the added 0.300 atm of CO2 to moles using the Ideal Gas Equation. The moles of CO2 reacted equals the moles of CaC03 formed. Moles CaC0 3 = moles CO2 = n = PY I RT ( 0.300 atm ) ( 1 0.0 L) 0.0949 1 mol CaC03 formed (unrounded) n= L · atm (385 K) 0.082 1 mol · K Add the moles of CaC03 formed in the second equilibrium to the moles of CaC03 at the first equilibrium position and convert to grams. 0.0304 mol + 0.0949 1 mol CaC0 3 1 00.09 g CaCO J = 1 2.542 = 1 2.5 g CaC03 Mass CaC03 = 1 mol CaC0 3

(

)

(

)

(

1 7.73

)(

J

a) Calculate the partial pressures of oxygen and carbon dioxide because volumes are proportional to moles of gas, so volume fraction equals mole fraction. Assume that the amount of carbon monoxide gas is small relative to the other gases, so the total volume of gases equals Yeo, + Yo, + Y N , = 1 0.0 + 1 .00 + 50.0 = 6 1 .0. 1 0.0 mol CO 2 (4.0 atm ) = 0.6557377 atm (unrounded) pc o .. = 6 1 .0 mol gas 1 .00 mol 0 2 ( 4.0 atm) = 0.06557377 atm (unrounded) Po , = 6 1 .0 mol gas b) Use the partial pressures and given Kp to find Peo . 2 CO2(g) !:; 2 CO(g) + 02 (g) P�o Po2 P�o ( 0.06557377) Kp = = = 1 .4 x 1 0 2 Pe0 ( 0.6557377t 2 Peo = 3 .0299 x 1 0-1 4 = 3 . 0 1 0- 1 4 atm c) Convert partial pressure to moles per liter using the ideal gas equation, and then convert moles of CO to grams. ( 3.0299 x 1 0-1 4 atm ) nco I Y = P I RT = = 4.6 1 3 1 1 0-1 6 moUL (unrounded) · atm L 0.082 1 (800 K) mol · K 4.6 1 3 1 x 1 0 - 1 6 mol CO 28.0 I g CO l �g = 0.0 1 292 = 0.0 1 3 pg COIL I mol CO 1 0 _ g L

( (

J

J

-28

?

X

(

1 7.76

(

J(

.

)

)( J

X

a) Write a reaction table given that PC H , (init) = Pco , (init) = 1 0.0 atm, substitute equilibrium values into the equilibrium expression, and solve for P H , . Pressure (atm) CH 4(g) + 2 CO(g) + 1 0.0 Initial o Change -x -x +2x +2x 1 0.0 - x 1 0.0 - x 2x 2x 2 2 (2x) (2x) ( 2x t---'---'-- = 3 .548 1 06 (take square root of each side) ( 1 0.0 - x ) ( l O . O - x ) ( 1 0.0 - x ) 2 X

234

(2x)2 = 1.8836135 x 103 (10.0 - x) A �uadratic is necessa�r 4 x + (1.8836135 x 10 x) - 1.8836135 X 104 = 0 (unrounded) a=4 b = 1.8836135 X 103 c = - 1.8836135 X 104 b 2-4ac x = -b ± J2a -1.8836135 x 103 ± �(1.8836135 x 103 / - 4(4) ( -1.8836135 X 104 ) x= 2(4) x = 9.796209 (unrounded) P (hydrogen) = 2 x = 2 (9.796209) = 19.592418 atm (unrounded) If the reaction proceeded entirely to completion, the partial pressure of Hz would be 20.0 atm (pressure is proportional to moles, and twice as many moles of Hz form for each mole of CH4 or CO2 that reacts). atm (100%) = 97.96209 = 98.0%. The percent Yle. Id I· S 19.592418 20.0 atm b) Repeat the calculations for part (a) with the new Kp value. The reaction table is the same. (2X)2(2X)2 (2X)4 = 2.626 X 107 P�o P� z KpPCH4 PC02 (10.0-x)(I O.0-x) (1O.0-x)2 (2x)2 = 5.124451 X 103 (10.0 - x) A quadratic is needed: 4 x2 + (5.124451 x 103 - 5.124451 X3 104 = 0 (unrounded) a=4 b = 5.124451 x 10 c = - 5.124451 X 104 -5.124451 X 103 ± �( 5.124451 x 103 ) 2 - 4(4) ( -5.124451 X 104 ) x = ------------�------��------------------2(4) x = 9.923138 (unrounded) P (hydrogen) = 2 x = 2 (9.923138) = 19.846276 atm (unrounded) If the reaction proceeded entirely to completion, the partial pressure of H 2 would be 20.0 atm (pressure is proportional to moles, and twice as many moles of H 2 form for each mole of CH4 or CO2 that reacts). atm (100%) = 99.23138 = 99.0%. The percent yield is 19.846276 20.0 atm _

17.77 a) Multiply the second equation by 2 to cancel the moles of CO produced in the first reaction. Kp for the second reaction is then (Kp l 2 CH4(g) + 02(g) � �) + 4 H 2(g) Kp = 9.34 x 102 28 H?Q(g) + 2 H20(g) � 2 CO2(g) + 2 H2(g) Kp =(1.374) = 1.888 2 CH 4(g) + 02(g) + 2 H 20(g) � 2 CO2(g) + 6 H 2(g) b) Kp = (9.34x I 028) (1.888) = 1.76339 x 1029 = 1.76 1029 c) !Y.n = 8 - 5 = 3 (8 moles of product gas - 5 moles of reactant gas) K Kc = [RT ]"" X

p--

K = c

1.76339 X 1029 = 3.18654 X 1023 = 3. 19 [(0.0821 atmoLimoloK)(l000)]3

235

X

102 3

d) The initial total pressure is given as 30. atm. To find the final pressure use relationship between pressure and number of moles of gas: nillitial P initial n al fP na Total mol of gas initial = 2.0 mol CH 4 + 1 .0 mol O 2 + 2.0 mol H 2 0 = 5.0 mol Total mol of gas final = 2.0 mol CO2 + 6.0 mol H 2 = 8.0 mol 30. atm reactants 8 mol products - 48 a t m P fill a l 5 mol reactants

(

_

1 7.78

=

fill

fi

l

j

a) Careful reading of the problem indicates that 3the given Kp occurs at 1 000. K and that the initial pressure ofN2 Kp = 1 0-4 1 0 = 7.94328 1 0-44 is 200. atm. Log Kp = -43. 1 0; Pressure (atm) N 2 (g) =+ 2N(g) 0 200. Initial +2x -x Change 2x 200-x Equilibrium 2 ) Kp = ( PN ) = 7 .94328 x 1 0-44 X

PN 2

( 2x ) 2 = 7.94328 x 1 0-44 ( 200. - x)

b)

Assume 200. - x

==

200.

= 2x = �( 200.) (7.94328 1 0-44 ) = 3 .985795 10-2 1 = 4.0 Log Kp = -1 7.30; Kp = 1 0-1 7 3 0 = 5.0 1 1 87 x 1 0-1 8 Pressure (atm) H 2 (g) =+ 2H(g) 0 600. Initial -x Change +2x 2x 600-x Equilibrium X

PN

K, �

(::,;;

�

X

X

2 1 0- 1

5.0 1 1 87 x 1 0 ' "

( 2x) 2 = 5 .0 1 1 87 x 1 0-1 8 (600. - x)

Assume 600. - x

==

600.

= 2x = �( 600.) (5.0 1 1 87 x 1 0-18 ) = 5 .48372 x 1 0-8 = 5.5 1 0� c) Convert pressures to moles using the ideal gas law. PY = nRT. Convert moles to atoms using Avogadro's number. ( 3 .985795 1 0-2 1 atm ) 6.022 x 1 023 atoms Moles N I Y = P I RT = -,--'-- ----,----'-mol 0.082 1 L o atm ( I OOO . K) mol o K = 29.2356 = 29 N atoms/L ( 5.48372 x 1 0-S atm 1 0 23 atoms Moles H I Y = P I RT = ...,.-'.- - L atm. ---c- --'-) 6.022 xmol ( 1 000. K) 0.082 1 o mol K = 4.022 1 0 1 4 = 4.0 1 0 1 4 H atoms/L d) The more reasonable step is N2(g) + H(g) � NH(g) + N (g) . With only 29 N atoms in 1 .0 L, the first reaction would produce virtually no NH(g) molecules. There are orders of magnitude more N 2 molecules than N atoms, so the second reaction is the more reasonable step. x

PH

( (

X

--0

X

X

j

j

-

236

[

)

[

)

-

1 7.79

The mole ratio H 2 :N2 = 3 : 1 ; ifN 2 = H 2 = 3x

a) 3 H 2(g) + N 2 (g) !:; 2 NH3(g) Kp =

X,

( PN H 3 f = l . OO x I O-4 ( PN2 ) ( PH 2 t

(50 .) 2 = 1 . 00 x 1 0-4 3 (x)(3x) = 3 1 .020 1 6 = 3 1 atm N2 3 x = 3 (3 1 .020 1 6) = 93 .06048 = 93 atm H2 Ptota l = P nitrogen + Phydrogen + Pammonia = (3 1 .020 1 6 atm) + (93 .06048 atm) + (50. atm) = 1 74.08064 = 1 74 atm total

K = X

P

(soy

= 1 .00 x 1 0-4 ( x) ( 6 x) x = 1 8.44 = 1 8 atm N2 6 x = 6 ( 1 8.44) = 1 1 0.64 = 1 1 1 atm H2 Ptotal = Pnitrogen + P hydrogen + P ammonia = ( 1 8.44 atm) + ( 1 1 0.64 atm) + (50. atm) = 1 79.09 = 1 79 atm total This is not a valid argument. The total pressure in (b) is greater than in (a) to produce the same amount ofN H3 .

b) Kp =

1 7.80

3

a) Equilibrium partial pressures for the reactants, nitrogen and oxygen, can be assumed to equal their initial partial pressures because the equilibrium constant is so small that very l ittle nitrogen and oxygen will react to form nitrogen monoxide. After calculating the equilibrium partial pressure of nitrogen monoxide, test this assumption by comparing the partial pressure of nitrogen monoxide with that of nitrogen and oxygen. N 2 (g) + 02 (g) !:; 2NO(g) Pressure (atm) Initial 0.780 0.2 1 0 0 +2x -x -x Change 0.780 - x 0.2 1 0 - x 2x Equilibrium K = ( PNO )2 = 4.35 1 0-3 1 P

X

( PN2 ) ( P02 )

( 2 x )2 -:-----'--:--;-'---------,- = 4.35 x 1 0-3 1 Assume x is small because K is small. ( 0.780 - x) (0.2 1 0 - x) ( 2 x) 2 = 4.35 x l O-3 1 ( 0.780) (0.2 1 0) x = l .33466 1 0-16 (unrounded) Based on the small amount of nitrogen monoxide formed, the assumption that the partial pressures of nitrogen and oxygen change to an insignificant degree holds. Pnitrogen (equilibrium) = (0.780 - 1 .33466 1 0-1 6) atm = 0.780 atm N2 P oxygen (equilibrium) = (0.2 1 0 - l .33466 x 1 0-16) atm = 0.2 1 0 atm O2 PNO (equilibrium) = 2 ( 1 .33466 x 1 0-1 6) atm = 2.66932 x 1 0-16 = 2.67 1 0-1 6 atm NO b) The total pressure is the sum of the three partial pressures: 0.780 atm + 0.2 1 0 atm + 2.67 x 1 0-1 6 atm = 0.990 atm c) Kp = KcCRT)"'n f m = 2 mol NO product - 2 mol reactant (I N 2 + 1 O2) = 0 Kp = Kc(RT)o Kc = Kp = 4.35 1 0-3 1 because there is no net increase or decrease in the number of moles of gas in the course of the reaction. X

x

X

X

237

1 7 . 84

The reaction i s :

CO(g)

+ H20(g) � CO2(g) + H2(g) CO and initial H2 = 0. 1 00 mol I 20.00 L

a) Set up a table with the initial Initial

0 . 00500 M

C�nR

CO2

H20

CO

0.00500 M

�

�

0 . 00500

[CO] equi l ibrium

=

[C02 ] [H2 ]

[ 0.00276 ] [ 0.00276 ]

=

=

1 . 5 1 8 1 76

= 1 . 52

= [CO] + [H20] + [C02] + [ H2] = (0. 00224 M) + (0. 00224 M) + (0.00276 M) + (0.00276 M) =

ntotal

O �

0 . 00500

c = [CO] [H20] = [ 0. 00224 ] [ 0. 00224 ]

K

Motal

O �

0 . 00 5 00 M

-x 0 . 00500 - x x x 3 M (given in problem) x [H20] = 24 0. 1 2 = x x = 0 . 00276 M = [C02] = [H2]

Equilibrium

b)

=

H2

O . O I OOO M

Motal V

(

) ((

= (0 . 0 1 000 mol/L) (20.00 L) = 0.2000 mol total

P total = ntot alRT I V =

( 0 . 2000 mol )

0. 08206

L · atm mol · K

( 20.00 L)

c) Initially, an equal number of moles must be added

27

3 + 900.) K )

= 0.9625638

=

0.9626 atm

= 0.2000 mol CO

d) Set up a table with the initial concentrations equal to the final concentrations from part (a), and then add 0.2000 mol

CO I 20.00 L CO

Initial Added

= 0 . 0 1 000 M to compensate for the added CO.

CO

Change

0 . 0 1 000 M

-x

CO2

0.00276 M

H2

0. 00224 M

-x

+x

+x

H20

0 . 00224 M

0.00276 M

-x 0.00276 + x 0.00276 + x [ CO 2 ] [ H 2 ] [ 0. 00276 + x ] [ 0 . 00276 + x ] = = 0 . 962 5 6 3 8 Kc = [CO] [H20] [ 0 . 0 1 224 - x ] [ 0 . 00224 - x ] [7.61 76 x 1 0-6 + 5 . 5 2 X 1 0-3 x + x 2 ] _::''_ = 0 . 962563 8 2 5 2 . 74 1 76 x 1 0- - 1 .448 X 1 0- X + X2 2 2 2 ..{i 3 5 7 . 6 1 76 x 1 0 + 5 . 5 2 X 1 0- x + x = (0.9625638) (2 . 74 1 76 x 1 0- - 1.1448 X 1 O- x + x ) 2 2 3 5 ..{i 7.6176 X 1 O 2 + 5 . 5 2 X 1 0- x + x2 = 2 . 639 1 1 89 X 1 0- 5- 1 .3937923 x 1 0- x + 0.9625638 x2 0.0374362 x + 1.9457923 X 1 0- x - 21 . 8773 5 8 9 1 0- = 0 5 c = -1 .8773589 1 0b = 1 . 945 7923 X 1 0a = 0.03 74362 2 -b ± b 0 . 0 1 224

Equilibrium

-x

-;:--

---::-

-'=---

0 . 00224

----

---

-

X

X

� --'--2 x 1 /-4(0.0374362)(-1.8773589 1 0-5 ) x ± �( -------------� ----------� �--------------------x= 2(0.0374362) x=

- 4ac

2a

-(1 . 9457923

1 0-2 )

1 . 9457923

0-

x

x = 9 . 6304567 X 1 0-4 (unrounded) [CO] = 0 . 0 1 224 - x = 0 . 0 1 224 - (9.6304567 x 1 0-4) = 0.01 1276954 = 0.0 1 1 28 M

238

Chapter 1 8 Acid-Base Equilibria FOLLOW-UP PROBLEMS

1 8. 1

a) Chloric acid, HCI03, is the stronger acid because acid strength increases as the number of 0 atoms in the acid increases. b) Hydrochloric acid, HCI, is one of the strong hydrohalic acids whereas acetic acid, CH 3 COOH, is a weak carboxylic acid. c) Sodium hydroxide, NaOH, is a strong base because Na is a Group I A( I ) metal. Methylamine, CH 3N H 2 , is an organic amine and, therefore, a weak base.

1 8.2

Plan: The product of [H 3 0+] and [OIr] remains constant at 25°C because the value of Kv is constant at a given temperature. Use Kw [H 3 0+] [OH-] 1 .0 x 1 0-1 4 to solve for [H 3 0+] . Solution: Calculating [H 30+] : . 14 [H 3 0+] � 1 .0 x 1 0 1 4925 x 1 0- 1 3 1 .5 1 0- 1 3 M 2 [OR ] 6.7 x 1 0Since [OIr] > [H 3 0+], the solution is basic . Check: The solution is neutral when either species concentration is 1 x 1 0-7 M. Since the problem specifies an [OH-] much greater than this, the solution is basic. =

=

=

1 8.3

=

=

=

.

X

P lan: NaOH is a strong base that dissociates completely in water. Subtract pH from 1 4.00 to find the pOH, and calculate inverse logs of pH and pOH to find [H 3 0+] and [OH-], respectively. Solution: pH + pOH 1 4.00 pOH 1 4.00 9.52 4.48 pH -log [H 3 0+] [ H30+] l O-p H 1 0-9 5 2 3 . 0 1 995 1 0- 1 0 3.0 1 0- 1 0 M pOH -log po[OIr] [011] 1 0- H 1 0-4 4 8 3 .3 1 1 3 x 1 0-5 3 .3 1 0-5 M Check: At 25°C, [H 3 0+] [OH-] should equal 1 .0 x 1 0- 1 4 . (3.0 x 1 0-1 0) (3.3 x 1 0-5) 9.9 x 1 0- 1 5 ", 1 .0 1 0-1 4 . The significant figures in the concentrations reflect the number of significant figures after the decimal point in the pH and pOH values. =

=

=

=

=

=

=

=

-

=

=

X

=

=

X

=

X

=

X

1 8.4

Plan: Identify the conj ugate pairs by first identifying the species that donates H+ (the acid) in either reaction direction. The other reactant accepts the H+ and is the base. The acid has one more H and + I greater charge than its conj ugate base. Solution: a) CH 3 COOH has one more H+ than CH 3 COO-. H 3 0+ has one more H+ than H 2 0. Therefore, CH 3 COOH and H 3 0+ are the acids, and CH3COO- and H 2 0 are the bases. The conjugate acid/base pairs are CH3COOH/CH3COO- and H30 +1H20 . b) H 2 0 donates a H+ and acts as the acid. F- accepts the H+ and acts as the base. In the reverse direction, HF acts as the acid and OIr acts as the base. The conjugate acid/base pairs are H20IOH- and HF/F -.

1 8.5

P lan: Identify the conjugate acid-base pairs. To predict the direction, consult Figure 1 8.9 to see which acid and base are stronger. The reaction will proceed in the direction which forms the weaker acid and base. Solution: H 2 0(!) + H S-(aq) !:; OIr(aq) + H 2 S (aq) The conj ugate pairs are H 2 0IOH- and H 2 S/HS-. H 2 S is higher than H 2 0 on the l ist of acids and OIr is lower than HS- on the list of bases. Thus, we have OIr(aq) + H 2 S (aq) H 2 0(!) + HS-(aq) !:; weaker acid + weaker base f- stronger base + stronger acid The net direction is to the left to form the weaker acid and base. Since [reactants] wil l be greater than [products], Kc < 1 .

239

1 8.6

Plan: Write a balanced equation for the dissociation ofN H/ in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and make assumptions where possible to simplify the calculations. Solution: + NH lg) H 30+(aq ) H 20(l) NH/(aq) + � 0.2 M O O Initial +x +x -x Change Equilibrium 0.2 - x x x The initial concentration ofN H/ = 0.2 M because each mole ofNH4CI completely dissociates to form one mole ofNH/. x = [H 3 0+] = [NH 3 ] = 1 0-p H = I O-s = 1 .0 x 1 0-s M [NH 3 J[ H 3 0+ ] ( 1 .0 x 1 0'5 ) ( 1 .0 x 1 0'5 ) x2 Ka = (0.2 - x ) ( 0.2 - 1 .0 x I O,5 ) [N H; ] = 5 x 1 0-1 0 Check: The small value of Ka is consistent with the value expected for a weak acid. .

1 8.7

o

Plan: Write a balanced equation for the dissociation of HOCN in water. Using the given information, construct a table that describes the initial and equilibrium concentrations. Construct an equilibrium expression and solve the quadratic expression for x, the concentration of H 3 0+. Solution: HOCN( aq ) + H 20(l) OCN'(aq ) + Initial O. I O M o Change -x +x Equilibrium 0. 1 0 - x x x [ OCN- ][ H 3 0+ ] x2 Ka = = = 3.5 1 0-4 (0. 1 0 - x ) [ HOCN ] In this example, the dissociation of HOCN is not negligible in comparison to the initial concentration: Therefore,2 [HOCN] eq = 0. 1 0 - x and the equilibrium expression is solved using the quadratic formula. x2 = 3 .5 x 1 0-4 (0. 1 0 - x) x2 = 3.5 1 0-5 - 3.5 1 0-4 xs x + 3.5 1 0-4 x - 3 .5 I O- = 0 (a x2 + b x + c = 0) 4 s a = 1 b = 3.5 x 1 0= -3 .5 I O-b± �b 2 - 4ac x = --'----2a -3 .5 x 1 0--4 ± ( 3.5 x 1 0--4 ) 2 - 4 ( 1 ) ( -3 .5 x I O-S ) x = -------'-----:-:--------2(I) x 5.7436675 1 0-3 = 5.7 1 0-3 M H3 0+ 3 pH = -log [ H30+] = -log [5 .7436675 x 1 0- ] = 2.2408 = 2.24 +­ -+

X

X

X

X

X

C

X

)

=

1 8.8

X

X

Plan: Pyridine contains a nitrogen atom that accepts H+ from water to form OH- ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, convert pKb to Kb, make simplifying assumptions (if valid) and solve for [OW] . Calculate [H 30+] using [H 30+] [OHl = 1 .0 x 1 0- 1 4 and convert to pH. Solution: CsHsN(aq) + H 2 0(l) � CsHsNH\aq) + OW(aq) Kb = = 1 0-pKb = 1 0-8 77 = 1 .69824 x 1 0-9 Kb =

[ CsNsNH+ J [ OH- ] [C sNsN ]

= 1 .69824 1 0-9 X

240

Assumptions: 1 ) Since Kb » Kw, disregard the [OH-] that results from the autoionization of water. 2) Since Kb is small, [CsHsN] = 0. 1 0 x 0. 1 0. [ Cs NsNH+ J [ OH- ] x 2 Kb = = = 1 .69824 1 0-9 (unrounded) . ( 0. 1 0 ) [ C sNs N ] x = 1 .303 1 65 I O-s = 1 .3 I O-s M = [Ofr] = [CsHsNH+] Os Since [OH- ] ( 1 00 ) = 1 .303265 I - ( 1 00 ) = 0.0 1 3 1 3 which 5%, the assumption that the dissociation of [C s H s N s ] 0. 1 0 CsHsNs is small is valid. K", - 1 .0 1 0 - 14 [H 0+] = 7.67362 1 0-10 M (unrounded) [OH- ] 1 .303 1 65 I O -s pH = -log (7.67362 x 1 0-1 °) = 9. 1 1 49995 = 9. 1 1 Check: Since pyridine is a weak base, a pH > 7 is expected. -

'"

X

--

X

3

1 8.9

-

X

X

X

-

<

X

x

Plan: The hypochlorite ion, CIO-, acts as a weak base in water. Write a balanced equation a n d equil ibrium expression for this reaction. The Kb of CIO- is calculated from the Ka of its conj ugate acid, hypochlorous acid, HCIO (from Appendix C, Ka = 2.9 1 0-8). Make simplifying assumptions (if valid), solve for [ O f ] , convert to [H 3 0+] and calculate pH. Solution: + CIO-(aq) + HCIO(aq) H 2 O(l) Ofr(aq) Initial 0.20 M 0 0 +x +x Change 0.20 x Equilibrium x [ HCIO ] [ OH- ] Kb = [ ClO- ] K", 1 .0 1 0- 14 Kb = 3 .448276 1 0-7 (unrounded) --;z: 2.9 1 0-8 Assumptions: I ) Since Kb Kw, the dissociation of water provides a negligible amount of [Ofr]. 2) Since Kb is very small, [CIO-]eq = 0.20 x 0.2. [ HC IO ] [ OW ] 2 = _x_ = 3.448276 1 0-7 Kb ( 0.20 ) CIO - ] 4 x = 2.626 1 1 0Therefore, [HCIO] = [OW] = 2.6 1 0-4 M. 4 Since [OH - ] ( 1 00 ) = 2.626 1 x 1 0- ( 1 00 ) = 0. 1 3 1 3 which 5%, the assumption that the dissociation of CIO- is 0.20 [CIO- ] smal l is valid. 1 .0 1 0 - 14 = 3 . 8079 1 0- 1 1 M and pH = -log (3.8079 x 1 0- 1 1 ) = 1 0.4 1 93 = 10.42 [H 0+] = 2.626 1 x 1 0-4 Check: Since hypochlorite ion is a weak base, a pH > 7 is expected. x

r

+-+

-x

X

_

-

X

-

-

X

X

»

-

_

-

X

3

1 8. 1 0

X

'"

X

---;=---'=-=,----=-

[

X

<

X

a) The ions are K+ and CI0 2-; the K+ is from the strong base KOH, and does not react with water. The C102- is from the weak acid HC102 , so it reacts with water to produce OH- ions. Since the base i s strong and the acid is weak, the salt derived from this combination will produce a ba sic solution. K+ does not react with water CI02-(aq) + H 2 0(l) � HCI02 ( aq ) + OW(aq)

24 1

1 8. 1 1

b) The ions are CH 3NH/ and N03-; CH3NH/ is derived from the weak base methylamine, CH 3 NH 2 . Nitrate ion, N03-, is derived from the strong acid RN03 (nitric acid). A salt derived from a weak base and strong acid produces an acidic sol ution. N03- does not react with water CH3NH3\aq) + H 2 0(1) !.:; CH3NH 2 (aq) + H30+(aq) c) The ions are Cs+ and r. Cesium ion is derived from cesium hydroxide, CsOH, which is a strong base because Cs is a Group I A( l ) metal. Iodide ion is derived from hydroiodic acid, HI, a strong hydrohalic acid. Since both the base and acid are strong, the salt derived from this combination wil l produce a neutral solution. Neither Cs+ nor r react with water. a) The two ions that comprise this salt are cupric ion, Cu2 +, and acetate ion, CH 3 COO-. Metal ions are acidic in water. Assume that2 the hydrated cation is Cu(H 2 0)i+. The Ka is found in Appendix C. Cu(H 2 0)6 \aq) + H 2 0(1) !.:; Cu(H 20) 5 0H\aq) + H30\aq) Ka = 3 1 0-8 Acetate ion acts likes a base in water. The Kb is calculated from the Ka of acetic acid ( 1 .8 x 1 0-5 ): KIV 1 .0 1 0- 14 = 5.6 1 0-1 0• = Kb = K" 1 .8 1 0-5 CH3COO-(aq) + H 2 0(1) !.:; CH3COOH(aq) + OW( aq) Kb = 5 .6 1 0- 1 0 Cu( H 2 0)62+ is a better proton donor than CH3COO- is a proton acceptor (i .e., Ka > Kb), so a solution of Cu(CH 3 COO) 2 is acidic . b) The two ions that comprise this salt are ammonium ion, N H/, and fluoride ion, F-. Ammonium ion is the acid K w _ 1 .0 x l O - 14 o f NH 3 with Ka - 5 7 1 0-10 Kb 1 .76 x 1 0-5 Ka = 5.7 1 0-1 0 N H/(aq) + H 2 0(1) !.:; NH 3 (aq) + H 3 0+( aq ) 14 K, . 1 .0 1 0' . t h e base Wit. h Kb = = 1 .5 x 1 0- . ' = Fl uon' d e Ion Ka 6.8 1 0-4 Kb = 1 .5 x 1 0-1 1 F-(aq) + H 2 0(1) !.:; H F(aq) + O W(aq ) Since Ka > Kb, a solution of (N H 4 F) is acidic . X

X

X

X

X

.

'

X

IS

1 8. 1 2

X

X

I I

X

Plan: Begin by drawing Lewis dot structures for the reactants to see which molecule donates the electron pair ( Lewis base) and which molecule accepts the electron pair ( Lewis acid). Solution : a) OH

� OH -H 8: l '\ AlI ++

[ j • •

HO

/ ""

/

I

AI " 'IIII

OH

�OH

HO OH trigonal planar tetrahedral Hydroxide ion, O H-, donates an electron pair and is the Lewis base; AI(OH) 3 accepts the electron pair and is the Lewis acid. Note the change in geometry caused by the formation of the adduct. b) ·0'

. � \\ .

r:-:--.:.

�O

\\\ \ \ \

\ '

S

J

�

1/�C3

.. ..

0

Sulfur trioxide accepts the electron pair and is the Lewis acid. Water donates an electron pair and is the Lewis base. 242

c) NH3

[� ] 3+ + I Co

6 H

N

/ � H

I � I '

H 3N /111 ,

H

H 3N

'Co'

\ \ \\\ N H 3

3+

NH3

NH 3 3 C0 + accepts six electron pairs and is the Lewis acid. Ammonia donates an electron pair and is the Lewis base. END-OF-CHAPTER PROBLEMS

18.2

All Arrhenius acids contain hydrogen and produce hydronium ion (H 3 0+) in aqueous solution. All Arrhenius bases contain an OH group and produce hydroxide ion (OH-) in aqueous solution. Neutralization occurs when each H 3 0+ molecule combines with an OH- molecule to form 2 molecules of H 2 0. Chemists found that the /).firxn was independent of the combination of strong acid with strong base. In other words, the reaction of any strong base with any strong acid always produced 56 kllmol (/).fi -56 kJ/mol). This was consistent with Arrhenius's hypothesis describing neutralization, because all other counter ions (those present from the dissociation of the strong acid and base) were spectators and did not participate in the overall reaction. =

18.4

Strong acids and bases dissociate completely into their ions when dissolved in water. Weak acids only partially dissociate. The characteristic property of all weak acids is that a significant number of the acid molecules are not dissociated. For a strong acid, the concentration of hydronium ions produced by dissolving the acid is equal to the initial concentration of the undissociated acid. For a weak acid, the concentration of hydronium ions produced when the acid dissolves is less than the initial concentration of the acid.

18.5

a) Water, H 20, is an Arrhenius acid because it produces H 3 0+ ion in aqueous solution. Water is also an Arrhenius base because it produces the OH- ion as well. b) Calcium hydroxide, Ca(OH) 2 is a base, not an acid. c) Phosphorous acid, H 3 P03 , is a weak Arrhenius acid . It is weak because the number of 0 atoms equals the number of ionizable H atoms. d) Hydroiodic acid, HI, is a strong Arrhenius acid.

18.7

a) HN02 (aq) H 2 0(1) !:; H3 0+(aq) N02-(aq) N0 2 H 3 0+ ] Ka [ HN0 2 ] b) CH 3 COOH( aq) H 2 0(1) !:; H 3 0\ aq) CH 3 COO-( aq) CH 3 COO- H 3 0+ ] K [C H 3 COOH ] c) HBrOiaq) + H 2 0(1) !:; Hp\aq) Br02-( aq) 8r0 2 - H 3 0+ ] Ka [ H Br0 2 ] =

a =

=

+[

[

[

+

J[

+

+

J[

J[

+

18.9

Appendix C lists the Ka values. The larger the Ka value, the stronger the acid. Hydroiodic acid, H I , is not shown in Appendix C because Ka approaches infinity for strong acids and is not meaningful. Therefore, H I is the strongest acid and acetic acid, CH 3COOH, is the weakest: CH3 COOH < HF < HI03 < HI.

18.11

a) Arsenic acid, H 3 As04 , is a weak acid. The number of 0 atoms is 4, which exceeds the number of ionizable H atoms, by one. This identifies H 3 As04 as a weak acid. b) Strontium hydroxide, Sr(OHh, is a strong base. Soluble compounds containing OH- ions are strong bases. c) HIO is a weak acid . The number of 0 atoms is I , which is equal to the number of ionizable H atoms identifying H I O as a weak acid.

3,

24

3

d) Perchloric acid, HCI04 , is a strong acid . HCI04 is one example of the type of strong acid in which the number of 0 atoms exceeds the number of ionizable H atoms by more than 2. 1 8. 1 5

The lower the concentration of hydronium (HP+) ions, the higher the pH: a) At equal concentrations, the acid with the larger Ka will ionize to produce more hydronium ions than the acid with the smaller Ka. The solution of an acid with the smaller K. 4x 1 0-5 has a lower [H 3 0+] and higher pH. b) pKa is equal to -log Ka. The smaller the Ka, the larger the pKa is. So the acid with the larger pK., 3 .5, has a lower [H 3 0+] and higher pH. c) Lower concentration of the same acid means lower concentration of hydronium ions produced. The 0.0 1 M solution has a lower [H 3 0+] and higher pH. d) At the same concentration, strong acids dissociate to produce more hydronium ions than weak acids. The 0. 1 M solution of a weak acid has a lower [H 3 0+] and higher pH. e) Bases produce OH- ions in solution, so the concentration of hydronium ion for a solution of a base solution is lower than that for a solution of an acid. The 0. 1 M base solution has the higher pH. t) pOH equals -log [OH-] . At 25°C, the equilibrium constant for water ionization, Kw, equals I xI 0-1 4 so 14 = pH pOH. As pOH decreases, pH increases. The solution of pOH = 6.0 has the higher pH. =

+ 1 8. 1 6

a) This problem can be approached two ways. Because NaOH is a strong base, the [OH-]eq = [NaOH]init. One method involves calculating [H 3 0+] using from Kw [H 3 0+] [OH-], then calculating pH from the relationship pH -log [H 3 0+] . The other method involves calculating pOH and then using pH + pOH = 1 4.00 to calculate pH. First method: 14 K 9.0090 x 1 0- 1 3 M (unrounded) [H 3 0+] __IV_ 1 .0 1 00.0 1 1 1 [OH - ] pH -log [H 30+] = -log (9.0090 x 1 0-1 3 ) = 1 2.04532 1 2.05 Second method: pOH -log [OW] = -log (0.0 I I I ) 1 .954677 (unrounded) pH = 1 4.00 - pOH = 1 4.00 - 1 .954677 = 1 2.04532 = 1 2.05 With a pH > 7, the solution is basic. b) There are again two acceptable methods analogous to those in part a; only one will be used here. For a strong acid: [H 30+] [HCI] 1 .233 x 1 0-3 M pH -log ( 1 .23 x 1 0- ) = 2.9 1 009 (unrounded) pOH = 1 4.00 - 2.9 1 009 = 1 1 .0899 1 = 1 1 .09. With a pH 7, the solution is acidic. =

=

X

=

=

=

=

=

=

=

=

=

=

1 8. 1 8

<

a) [H 3 0+] 1 0-p H 1 0-9 7 8 1 .659587 x 1 0- 1 0 1 .7 X t o - I O M H30+ pOH 1 4.00 - pH 1 224.00 - 9.78 = 4.22 [OW] l O-pOH = 1 0-4 6.025596 1 0-5 = 6.0 X 1 0-5 M OU­ b) pH 1 4.00 - pOH = 1 4.00 3- 1 0.43 = 3.57 [H 3 0+] l O-p H = 1 0- 5 7 3 2.69 1 53 x 1 0-4 = 2.7 X 1 0-4 M H30+ [OW] l O-po H 1 0-1 0 4 = 3.75 1 535 1 0- 1 1 3.7 X t o- I I M O H=

=

=

=

=

=

=

X

=

=

=

=

=

1 8.20

X

=

=

The pH is increasing so the solution is becoming more basic. Therefore, OH- ion is added to increase the pH. Since 1 mole of H 3 0+ reacts with2 1 mole of OH-, the difference in [H 3 0+] would be equal to the [OW] added. [ H 3 0+] = l O-p H 1 0-4 228 1 .5 1 356 1 0-5 M H 3 0+ (unrounded) 6.02560 1 0-6 M H 3 0+ (unrounded) [H3 0+] l O-p H 1 0-5 1 .5 1 356 x 1 0-5 M - 6.02560 x 1 0-6 M = 9. 1 1 x 1 0-6 M OH- must be added. 9. 1 1 x 1 0-ii mol ----- ( 6.5 L ) 5.92 1 5 1 0-5 = 6 X t o-5 m o l o( OH­ L =

=

=

=

=

=

X

X

X

244

1 8.23

a) Heat is absorbed in an endothermic process: 2 H 2 0(l) + heat � H 30+(aq) + OJ1(aq). As the temperature increases, the reaction shifts to the formation of products. Since the products are in the numerator of the Kw expression, rising temperature increases the value of Kw. b) Given that the pH is 6.80, the [H+] can be calculated. The problem specifies that the solution is neutral, meaning [H+] [OW]p. A new Kw can then be calculated. [H 3 0+] 1 0- H 1 0-<5 80 1 .58489 x 1 0-7 M H 3 0+ 1 .6 X 1 0-7 M [ OH-I Kw [H30+] [OW] ( 1 .58489 x 1 0-7 ) ( 1 .58489 x 1 0-7 ) 2.5 1 1 876 x 1 0 1 4 2.5 X 1 0- 1 4 For a neutral sol ution: pH pO H 6.80 =

=

=

=

=

=

=

=

=

=

=

-

=

1 8.24

The Bf0nsted-Lowry theory defines acids as proton donors and bases as proton acceptors, while the Arrhenius definition looks at acids as containing ionizable H atoms and at bases as containing hydroxide ions. In both definitions, an acid produces hydronium ions and a base produces hydroxide ions when added to water. Ammonia, N H 3 , and carbonate ion, CO/-, are two Bf0nsted-Lowry bases that are not Arrhenius bases because they do not contain hydroxide ions. Bnmsted-Lowry acids must contain an ionizable H atom in order to be proton donors, so a Bf0nsted-Lowry acid that is not an Arrhenius acid cannot be identified. (Other examples are also acceptable. )

1 8.27

An amphoteric substance can act as either an acid or a base. In the presence of a strong base (OH-), the dihydrogen phosphate ion acts like an acid by donating hydrogen: H 2 P04-(aq) + OJ 1(aq) � H 20(aq) + HPO/-(aq) In the presence of a strong acid (HCI), the dihydrogen phosphate ion acts like a base by accepting hydrogen: H 2 P04-(aq) + HCI(aq) � H3P04 (aq) + C r( aq)

1 8.28

To derive the conj ugate base, remove one H and decrease the charge by 1 . Since each formula is neutral, the conj ugate base wil l have a charge of -I . a) CI- b) HC03- c) O H-

1 8.30

To derive the conj ugate acid, add an H and increase the charge by I . a) NH/ b) NH3 c) CI OH I 4N 2H+

1 8.32

The acid donates the proton to form its conj ugate base; the base accepts a proton to form its conj ugate acid: a) NH 3 + H 3P04 !:; H 2 P04NH/ + base acid conj ugate acid conjugate base Conj ugate acid-base pairs: H 3 POJH 2 P04-; NH /IN H 3 b) CH 3 0- + N H3 CH 3 0H + N H 2!:; base acid conj ugate acid conjugate base Conjugate acid-base pairs: N H 3 IN H 2-; CH 3 0H/CH 3 0c) H PO/- + H S0 4- !:; H 2 P04- + SO/base acid conj ugate acid conj ugate base Conj ugate acid-base pairs: HS04-/SO/-; H 2 P04-/HPO/-

1 8.34

Write total ionic equations and then remove the spectator ions to write the net ionic equations. The (aq) subscript denotes that each species is soluble and dissociates in water. H 2 0( l) + � � + HPO/-(aq) a) Na"'� + OI1(aq) + Na"'� + H 2 P04-(aq) !:; H PO/-(aq) Net: OJ1(aq) + H 2 P04-(aq) !:; H 2 0(l) + base acid conj ugate acid conjugate base Conj ugate acid/base pairs: H 2 P04-/ HPO/- and H 2 0/OHb) K"� + HS04-(aq) + � � + CO/-(aq) !:; � � + S042-( aq) + K"� + HC03-(aq) HC03-(aq) + SO/-(aq) Net: HS04-(aq) + CO/-(aq) !:; acid base conj ugate base conjugate acid Conj ugate acid/base pairs: HS04-/SO/- and HC0 3-/C03 2...

..

..

245

1 8.36

The conj ugate pairs are H 2 S (acid)/HS- (base) and HCI (acid)/Cr (base). The reactions involve reacting one acid from one conj ugate pair with the base from the other conjugate pair. Two reactions are possible: ( I ) HS- + HCI !:; H 2 S + CI- and (2) H 2 S + CI- !:; HS- + HCI The first reaction is the reverse of the second. To decide which will have an equilibrium constant greater than I , look for the stronger acid producing a weaker acid. HCI is a strong acid and H 2 S a weak acid. The reaction that favors the products (Ke > I ) is the first one where the strong acid produces the weak acid. Reaction (2) with a weaker acid forming a stronger acid favors the reactants and Ke < I .

1 8.38

a)

b)

1 8.40

HCI NH/ + cr + NH3 !:; weaker base weak acid stronger base strong acid HCI is ranked above NH/ in Figure 1 8.9 and is the stronger acid. NH3 is ranked above cr and is the stronger base. NH3 is shown as a "stronger" base because it is stronger than cr, but is not considered a "strong" base. The reaction proceeds towards the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and Kc 1 . The stronger acid is more likely to donate a proton than the weaker acid. H 2 S03 + NH3 !:; HS03+ NH/ weaker acid weaker base stronger base stronger acid H 2 S03 is ranked above NH/ and is the stronger acid. NH3 is a stronger base than HS03-. The reaction proceeds towards the production of the weaker acid and base, i.e., the reaction as written proceeds to the right and Kc 1 . >

>

a) The concentration of a strong acid is very different before and after dissociation since a strong acid exhibits 1 00% dissociation . . After dissociation, the concentration of the strong acid approaches 0, or [HA] O. b) A weak acid dissociates to a very small extent « < 1 00%), so the acid concentration after dissociation is nearly the same as before dissociation. c) Same as (b), but the percent, or extent, of dissociation is greater than in (b). d) Same as (a) '"

1 8.43

Butanoic acid dissociates according to the following equation: CH3CH 2 CH2 COOH(aq) + H 2 0(l) !:; H30\aq) + CH3CH 2 CH 2COO-(aq) 0. 1 5 M O O Initial: Change: -x +x +x Equilibrium: 0. 1 5 - x +x +x According to the information given in the problem, [H30+]eq 1 .5 1 x 1 0-3 M x 3 Thus, [H30+] [CH3CH 2 CH 2 COO-] 1 .5 1 x 1 0- M 3 [CH3CH 2 C H2 COOH] (0. I 5 - x) (0. 1 5 - 1 .5 1 1 0- ) M 0. 1 4849 M [ H 3 0+ J[ C H 3 C H 2 CH 2 COO- ] Ka [CH 3 CH 2 CH 2 COOH ] ( 1 .5 1 x 1 0-3 )( 1 .5 1 x 1 0-3 ) l .53552 1 0-5 1 .5 X 1 0-5 Ka ( 0. 1 4849 ) =

=

=

=

=

=

=

x

=

=

1 8.45

=

X

=

For a solution of a weak acid, the acid dissociation equilibrium determines the concentrations ofthe weak acid, its conj ugate base and H30+. The acid dissociation reaction for HN02 is: Concentration HN0 2 (aq ) + H 20(l) !:; H30+(aq) + N02-(aq ) lnitial 0.50 0 0 Change -x +x +x x Equilibrium 0.50 - x x (The H30+ contribution from water has been neglected.)

246

Ka 7. I x I 0-4 =

Ka 7. 1 =

X

=

[ H O+ ][ NO 2 - ] 3

[ HN0 2 ] ( )( ) 1 0-4 ( 0.50 - x ) X

=

X

Assume x is small compared to 0.50: 0.50 - x 0.50. =

( )( ) 0-4 (0.50 ) 0.0 1 884 1 (unrounded) Check assumption : (0.0 1 884 1 / 0.50) 1 00% 4% error, so the assumption is valid. [H 30+] [N02-] 1.9 1 0-2 M The concentration of hydroxide ion is related to concentration of hydronium ion through the equilibrium for water: 2 H 2 0(1) !:; H 3 0+(aq) + OIr(aq) with3 Kw 1 .0 1 0- 1 4 [OH-] 1 .0 1 0-1 4 / 0.0 1 884 1 5 .30757 1 0-1 5.3 X 1 0- 1 3 M OH-

K

a =

7. 1

X

=

I

X

X

X =

=

[O W] 1 8.4 7

=

=

=

X

x

X

=

X

=

=

X

1 .0 1 0- 1 4 / 0.0 1 884 1 5.30757 x 1 0-1 3

=

=

X

1 5.3 X 1 0- 3 M O H-

Write a balanced chemical equation and equilibrium expression for the dissociation of chloroacetic acid and convert pKa to Ka. Ka 1 0-pK 1 0-2 8 7 1 .34896 x 1 0-3 (unrounded) CICHzCOOH(aq) + H 2 0(1) !:; H 3 0+(aq) + CICHzCOO-(aq) x x 1 .05 - x ( H30+ )( CICH 2 COO- ) Ka - 1 .34896 x 1 0-3 ( CICH 2 COOH ) (x)(x) Ka 1 .34896 1 0-3 Assume x is small compared to 1 .05 . ( 1 .05 - x ) (x)(x) Ka 1 .34896 1 0-3 ( 1 .05 ) x 0.03 7635 (unrounded) Check assumption : (0.037635 / 1 .05) x 1 00% 4%. The assumption is good. [H 3 0+] [CICH 2 COO-] 0.038 M [CICH 2 COOH] 1 .05 - 0.037635 1 .0 1 2365 1 .0 1 M pH -log [H 30+] -log (0.037635 ) 1 .4244 1 .42 =

=

=

_

_

=

X

=

x

=

=

=

=

1 8.49

=

=

=

=

=

=

=

=

=

=

Percent dissociation refers to the amount of the initial concentration of the acid that dissociates into ions. Use the percent dissociation to find the concentration of acid dissociated. HA will be used as the formula of the acid. a) The concentration of acid dissociated is equal to the equilibrium concentrations of A- and H 3 0+. Then pH and [OIr] are determined from [ H 3 0+] . P ercent H A - Dissociated Acid x 1 00°1 Initial Acid 10

3.0 % _x_ ( 1 00) 0.25 [Dissociated Acid] x 7.5 1 0-3 M Concentration HA(aq ) + H 2 0(1) !:; H 3 0+(aq ) + A-(aq) 0 0 0.25 Initial : +x +x -x Change: Equilibrium : 0.25 - x x x X 1 0-3 M [Dissociated Acid] x [H 3 0+] 7.5 pH -log [H 30+] -log (7.5 x 1 0-3 ) 2. 1 249 2. 1 2 =

=

=

=

=

=

=

=

X

=

=

247

14 = 1 .0 1 0- 3 = 1 .3333 1 0-1 2 = 1 .3 X 1 0- 1 2 M [ H 3 0 + ] 7.5 1 0pOH = -log [OW] = -log ( 1 .3333 x 1 0-1 2 ) = 1 1 .87506 = 1 1 .88 b) ]n the equilibrium expression, substitute the concentrations above and calculate KaH 3 0 + A- ] 7.5 x 1 0-3 7.5 x 1 0-3 - 2.3 1 9588 x 1 0-4 - 2.3 1 0 Ka [H A] 0.25 - 7.5 x 1 0-3 _

1 8_5 1

X

Kw

[OW] =

J[

[

X

X

(

_

)(

(

)

)

_

_

x

-4

a) Begin with a reaction table, then use the Ka expression as in earlier problems. Concentration HZ(aq) + H 2 0(l) !::; H30+(aq) + T(aq) 0 0 0.075 Initial +x +x C�n� -x x Equilibrium 0.075 - x (The H30+ contribution from water has been neglected.) H 3 O + z-' K = I . 5 5 x I 0-4 = =---_-;---=7----=-

[

a

Ka = 1 .55 x I 0-4 =

x

J[ J

[ HZ] (x)(x)

Assume x is small compared to 0.075.

( 0.075 - x ) (x) ( ) Ka = 1 .55 x 1 0-4 = ( 0.075 ) [H30+] = x = 3 .4095 1 0-3 (unrounded) X

X

Check assumption: (3 .4095 x 1 0-3 / 0.075) x 1 00% = 5% error, so the assumption is valid. p H = -log [H30+] = -log (3.4095 x 1 0-3 ) = 2.4673 = 2.47 b) Begin this part like part a_ Concentration HZ(aq) + + 0.045 Initial -x Change +x +x 0.045 - x Equilibrium x x (The H30+ contribution from water has been neglected.) H 3 0 + zKa = I . 5 5 x I 0-4 = =--_-=---''-7----=-' +­ -+

[

Ka = 1 .55 x 1 0-4 =

J[ J

[ HZ] (x)(x)

Assume x is small compared to 0.045 .

( 0.045 - x ) (x) (x) Ka = 1 .55 x 1 0-4 = ( 0.045 ) [H30+] = x = 2.64 1 0 1 0-3 (unrounded) X

Check assumption: (2.64 1 0 x 1 0-3 / 0.045) x 1 00% = 6% error, so the assumption is not valid_ Since the error is greater than 5%, it is not acceptable to assume x is small compared to 0.045, and it is necessary to use the quadratic equation_ x2 = ( 1 .55 1 0-4) (0.045 - x) = 6.975 x 1 0--6 - 1 .55 x 1 0-4 x2 + 1 .55 x 1 0-4 - 6.975 x 1 0--6 = 0 a = 1 b = 1 .55 x I 0-4 c = -6.975 x I 0--6 -b ± b 2 - 4ac x = -----'---2a - 1 .55 x 1 O-4 ± 1 .55 x 1 0-4 - 4 ( 1 ) -6.975 x 1 0-6 � � x= 2 (I ) X

�

X

X

--------

�(

----

f

(

)

�--------�

-----

248

x 2.564659 x 1 0-3 M HP+ =

14

1 .0 1 0 [ H 3 0+ ] 2.564659 1 0 3 pOH -log [Otr] -log (3.899 1 53 1 0- 1 2 ) _

=

[OH ] [O t ] r

=

=

1 8.54

K

=

>V

X

'

'

=

-----

=

X

X

=

1

3 . 899 1 5 3

1 .4090

=

X

2

1 0- 1 M

1 1 .4 1

First, find the concentration of formate ion at equilibrium. Then use the initial concentration of formic acid and equilibrium concentration of formate to find % dissociation. Concentration HCOOH(aq) + H20(l) !:; H 3 0\aq) + HCOO-(aq) 0 0 0.50 Initial Change +x -x +x 0.50 Equilibrium x x [ H 3 0+ J[ HCOO - ] Ka - 1 .8 x 1 0 [ HCOOH ] ( )( ) Assume x is small compared to 0.50. Ka 1 .8 x 1 0-4 ( 0.50 - x ) (x) (x) Ka 1 .8 x 1 0-4 ( 0.50 ) 3 9.4868 1 0- (unrounded) Check assumption: (9.4868 x 1 0-3 / 0.50) x 1 00% 2% error, so the assumption is valid. Percent HCOOH Dissociated Dissociated Acid x 1 00% Initial Acid 9.4868 1 0-3 x 1 00% 1 .89736 1 .9 % Percent HCOOH D issociated _x_ x 1 00% 0.50 0.50 - X

-4 _

_

=

=

=

=

X =

X

X

X

=

=

=

=

X

=

=

1 8.55

All Bnmsted-Lowry bases contain at least one lone pair of electrons. This lone pair binds with an H+ and allows the base to act as a proton-acceptor.

1 8.58

a) A base accepts a proton from water in the base dissociation reaction: CsHsN(aq) + H20(l)!:; Otr(aq) + CsHsNH+(aq) [ CsH s NH + J [ OH- ] Kb - =---,=---�=------=[ CsH sN] b) The primary reaction is involved in base dissociation of carbonate ion is: C0 32 -(aq) + H20(l)!:; O tr(aq) + HC03-(aq) [ HC0 3- J[ OH- ] Kb CO/- ] The bicarbonate can then also dissociate as a base, but this occurs to an insignificant amount in a solution of carbonate ions. _

_

1 8.60

-'=----=-..::.-:.::. ----=-

[

The formula of dimethylamine has two methyl (CHr) groups attached to a nitrogen: CH - N - H 3

I

CH3 The nitrogen has a lone pair of electrons that will accept the proton from water in the base dissociation reaction: The value for the dissociation constant is from Appendix C.

249

(CH3) 2NH(aq) + H 2 0(l) !:; OW(aq) + (CH3) 2NH /(aq) 0.050 0 0 +x +x -x x x 0.050 - x + ( C H 3 ) 2 N H 2 .= O_H_--== 5.9 x I 0-4 Kb = __-;:-_=-__-,=.,-:; ( CH 3 ) 2 NH ]

Concentration Initial Change Equilibrium

.=.[

X

[

J[ J

X

[ ][ ]

Assume 0.050 - x =0.050. = 5.9 1 0-4 [0.050 - x ] [x ] [ x ] = 5.9 1 0-4 [0.050 ] x = 5.43 1 4 x 1 0-3 M Check assumption: (5 .43 1 4 x 1 0-3 / 0.050) x 1 00% = I I % error, so the assumption is invalid. The problem will need to be solved as a quadratic. [ ] [ x ] = 5.9 x I 0-4 [0.050 - ] x22 = (5.9 x 1 0-4) (0.050 - x) = 2.95 1 0-5 - 5.9 x 1 0-4 x + 5.9 1 0-4 x - 2.95 1 0-5 = 0 a = J b = 5 .9 x I 0-4 c = -2.95 x l O-5 -b ± )b 2 - 4ac x = --'----2a -5 .9 x 1 0-4 ± 5.9 x J o-4 - 4 ( 1 ) -2.95 x 1 0-5 '-x= 2 (1) 3 x = 5 . 1 443956 x 1 0- M Of1 (unrounded) � = 1 .0 1 0- 1 4 = 1 .94386 1 0-1 2 M H 3 0+ (unrounded) [H 3 ot = [ OH - ] 5 . 1 443956 x 1 0-3 pH = -log [H 3 0+] = -log ( 1 .943 86 x 1 0-12 ) = 1 1 .7 1 1 22 = 1 1.71 Kb

=

X

X

x

x

X

X

X

�(

-

X

f

)

(

��---------

-

X

1 8.62

X

a) The Ka of chlorous acid, HCI02 , is reported in Appendix C. HCI02 is the conjugate acid of chlorite ion, CI02-. The Kb for chlorite ion is related to the K. for chlorous acid through the equation Kw = Ka x Kb, and pKb = -log Kb . 14 Kw = 1 .0 1 0- = 9.0909 x 1 0 3 (unrounded) Kb of CI0 2- = Ka l. 1 x 1 03 pKb = -log (9.0909 x 1 0-1 ) = 1 2.04 1 39 = 1 2.04 b) The Kb of dimethylamine, (CH 3)2NH, is reported in Appendix C. (CH 3)2NH is the conjugate base of (CH3) 2NH/. The K. for (CH3) 2 NH/ is related to the Kb for (CH 3)2NH through the equation Kw = K. Kb, and pKa = -log Ka. K 1 .0 1 0- 14 = = 1 .6949 1 5 x 1 0 (unrounded) Ka of (CH 3)2N H 2 = Kb 5.9 x 1 0-4 pKa = -log ( 1 .6949 1 5 x 1 0-" ) = 1 0.77085 = 10.77 X

_

_I

2

X

+

1 8.64

_ IV

X

_I I

a) Potassium cyanide, when placed in water, dissociates into potassium ions, K+, and cyanide ions, CN". Potassium ion is the conjugate acid of a strong base, KOH, so K+ does not react with water. Cyanide ion is the conj ugate base of a weak acid, HCN, so it does react with the base dissociation reaction: CN"(aq) + H 2 0(l) � HCN(aq) + OW (aq) To find the pH, first set up a reaction table and use Kb for CN" to calculate [O W ] .

250

Concentration (M) CN-(aq) + H 20(l) !:; HCN(aq) + OW(aq) Initial 0.050 0 o Change -x +x +x x Equilibrium 0.050-x x 1 .0 10-14 Kw 1 .6 1 2903 10-5 (unrounded) Kb of CN=

=

Ka

Kb

Kb

IGb

X

[HCN] [ OH- ] [ CN - ]

_

-

=

[

x x [ ][ ] 0.050 - x

=

X

1 .6 1 2903 X 10-5

1.612903 X 1 0-5

=

]

1 .6 1 2903 X 10-5

=

=

6.2 X 10-10

=

Assume x is small compared to 0.050: 0.050-x

=

0.050.

( x )(x ) ( 0.050 )

x 8.9803 X 10-4 MOW (unrounded) Check assumption: ( 8.9803 x 1 0-4 / 0.050) x 1 00% 2% error, so the assumption is valid. 1.0 X 10-14 [H3 0t Kw 1 . 1 1 35857 X 1 0-1 1 MH3 0+ (unrounded) OR 8.9803 x 10-4 pH -log [H3 0+] -log (1.1135857 x 1 0-1 1 ) 10.953276 10.95 =

=

=

=

=

=

=

=

=

b) The salt triethylammonium chloride in water dissociates into two ions: (CH3 CH2)3NH+ and cr. Chloride ion is the conjugate base of a strong acid so it will not influence the pH of the solution. Triethylammonium ion is the conjugate acid of a weak base, so the acid dissociation reaction below determines the pH of the solution. Concentration (M) (CH3 CH2)3NH\aq) + H 20(l) !:; (CH3 CH2)3N(aq) + H3 0+(aq) 0 0 Initial 0.30 Change -x +x +x x x Equilibrium 0.30-x 1 .0 10-14 Kw 1 .9230769 1 0-1 1 (unrounded) K. of (CH 3 CH 2) 3NH+ 5.2 x I0-4 Kb =

X

=

X

=

[ H 3 0+ ][(CH 3 CH2 )3 NJ [ (CH 3 CH2 h NH+ ] (x )(x ) 1.9230769 10- 1 1 Assume x is small compared to 0.30: 0.30 -x 0.30. ( 0.30 - x ) ( x )( x ) 1.9230769 10-1 1 ( 0.30 ) [H30+] x 2.4019223 x 10-6 (unrounded) Check assumption: (2.4019223 x 1 0-6 / 0.30) x 1 00% 0.0008% error, so the assumption is valid. pH -log [H30+] -log (2.40 1 9223 x 10-6) 5.6 1 944 1 5.62

K.

K.

K•

=

1.9230769 x 10- 1 1 X

=

=

=

18.66

[

"'---,=---=-----.,,;-­

=

=

=

X

=

=

=

=

)(

J'(

=

=

J(

=

J[

)

First, calculate the initial molarity of ClO-. Then, set up reaction table with base dissociation of OCr: Solution 1.0 g Solut�on 5.0% NaO�1 I mol NaOCI I mol OCl[CIOl I mL 3 10- L Solution 1 mL SolutIOn 1 00% SolutIOn 74.44 g NaOCI 1 mol NaOCI 0.67168 MOCr (unrounded) The sodium ion is from a strong base; therefore, it will not affect the pH, and can be ignored. Concentration (M) OCnaq) + H20(l) !:; HOCI(aq) + OW(aq) 0 0 0.67168 Initial +x +x -x Change x 0.67168-x Equilibrium =

=

x

25 1

Kb of ocr = Kb = Kb =

K

"

Ka

=

X

[HOCI] [ OH- ] [ CIO- ] [x][x] [0.67 1 68 x] -

1 .0 1 0- 14 2.9 x 1 0-8 =

=

=

3.448275862 1 0-7 X

3.448275862 1 0-7 X

3.448275862 1 0-7

Assume is small compared to 0.67 1 68. X

X

(x)(x) 3.448275862 1 0-7 ( 0.67 1 68 ) 4 4.8 1 2627 1 0- 4.8 X 1 0-4 M OHCheck assumption: (4. 8 1 2627 x 1 0-4 ; 0.67 1 68) x 1 00% 0.074% error, so the assumption is valid. K 1 0-1 4 2.077867 1 0-1 1 M H30+ (unrounded) [H3 O r - w - 1 .0 [OR ] 4.8 1 2627 x 1 0-4 pH -log [H30+] -log (2.077867 x 1 0-1 1 ) 1 0.68238 1 0.68 =

X

Kb = X =

X

=

X

=

=

=

X

=

=

=

1 8.68

Electronegativity increases left to right across a period. As the nonmetal becomes more electronegative, the acidity of the binary hydride increases. The electronegative nonmetal attracts the electrons more strongly in the polar bond, shifting the electron density away from H+ and making the H+ more easily transferred to a surrounding water molecule to make H 3 0+.

1 8.7 1

The two factors that explain the greater acid strength of HCI04 are: I ) Chlorine is more electronegative than iodine, so chlorine more strongly attracts the electrons in the bond with oxygen. This makes the H in HCI04 less tightly held by the oxygen than the H in HIO. 2) Perchloric acid has more oxygen atoms than HIO, which leads to a greater shift in electron density from the hydrogen atom to the oxygens making the H in HCI04 more susceptible to transfer to a base.

1 8.72

a) H2Se, hydrogen selenide, is a stronger acid than H 3 As, arsenic hydride, because Se is more electronegative than As. b) B (OH)3, boric acid also written as H3B03, is a stronger acid than AI(OH)3 , aluminum hydroxide, because boron is more electronegative than aluminum. c) HBr02, bromous acid, is a stronger acid than HBrO, hypobromous acid, because there are more oxygen atoms in H Br02 than in HBrO.

1 8.74

Acidity increases as the value of Ka increases. Determine the ion formed from each salt and compare the corresponding Ka values from Appendix C. a) Copper(II) sulfate, CUS04 , contains Cu2+ ion with Ka 33 1 0-8. Aluminum sulfate, AI 2 (S04)3, contains A13 + ion with Ka 1 1 0-5. The concentrations of Cu2+ and A1 + are equal, but the Ka of AI2 (S04)3 is almost three orders of magnitude greater. Therefore, 0.05 M Ah(S04)3 is the stronger acid. b) Zinc chloride, ZnCI2, contains the Zn2 + ion with Ka I x 1 0-9. Lead chloride, PbCI 2 , contains the Pb2+ ion with Ka 3 x 1 0-8. Since both solutions have the same concentration, and Ka (Pb 2+) > Ka (Zn 2 +), 0 . 1 M PbCl2 is the stronger acid. =

=

1 8.77

=

X

X

=

Sodium fluoride, NaF, contains the cation of a strong base, NaOH, and anion of a weak acid, HF. This combination yields a salt that is basic in aqueous solution as the F- ion acts as a base: F-(aq)+ H 2 0(l) !:; HF(aq) + OW(aq) Sodium chloride, NaCI, is the salt of a strong base, NaOH, and strong acid, HCI. This combination yields a salt that is neutral in aqueous solution as neither Na+ or cr react in water to change the [H+] .

252

18.79

For each salt, first break into the ions present in solution and then determine if either ion acts as a weak acid or weak base to change the pH of the solution. Cations are neutral if they are from a strong base; other cations will be weakly acidic. Anions are neutral if they are from a strong acid; other anions are weakly basic. a) KBr(s) + H 2 0( l) -7 K+(aq) + Br-(aq) K+ is the conjugate acid of a strong base, so it does not influence pH. Br- is the conj ugate base of a strong acid, so it does not influence pH. Since neither ion influences the pH of the solution, it will remain at the pH of pure water with a neutral pH. b) NH4 I(s) + H 2 0( l) -7 N H/(aq) + qaq) N H/ is the conjugate acid of a weak base, so it will act as a weak acid in solution and produce H 3 0+ as represented by the acid dissociation reaction: N H/(aq) + H 2 0( l) !:; NH 3 (aq) + H 3 0+(aq) r is the conjugate base of a strong acid, so it will not influence the pH. The production of H 3 0+ from the ammonium ion makes the solution of NH 4 1 acidic. c) KCN(s) + H 2 0( l) -7 K+(aq) + C�(aq) K+ is the conjugate acid of a strong base, so it does not influence pH. C� is the conjugate base of a weak acid, so it will act as a weak base in solution and impact pH by the base dissociation reaction: C�(aq) + H 2 0( l) !:; HCN(aq) + OW(aq) Hydroxide ions are produced in this equilibrium so solution will be basic.

18.81

a) Order of increasing pH : KN03 < K2S03 < K2S (assuming concentrations equivalent) Potassium nitrate, KN03 , is a neutral solution because potassium ion is the conjugate acid of a strong base and nitrate ion is the conjugate base of a strong acid, so neither influences solution pH. Potassium sulfite, K2 S03 , and potassium sulfide, K2 S, are similar in that the potassium ion does not influence solution pH but the anions do because they are conjugate bases of weak acids. Ka for HS0 3- is 6 . 5 1 0-8 from Table 18.5, so Kb for S03- is 1.5 1 0-7 , which indicates3 that sulfite ion is a weak base. Ka for HS- is 1 1 0- 1 7 from Table 18.5, so sulfide ion has a Kb equal to I x 1 0 . Sulfide ion is thus, a strong base. The solution of a strong base wil l have a greater concentration of hydroxide ions (and higher pH) than a solution of a weak base of equivalent concentrations. b) In order of increasing pH: NaHS04 < NH4N03 < NaHC03 < Na2C03 In solutions of ammonium nitrate, only the ammonium will influence pH by dissociating as a weak acid: N H/(aq) + H 2 0( l) !:; NH3 (aq) + H 3 0+(aq) with Ka 1 .0 1 0- 1 4 ! 1 . 8 1 0-5 = 5 . 6 1 0- 1 0 Therefore, the solution of ammonium nitrate is acidic. In solutions of sodium hydrogen sulfate, only H S04- will influence pH. The hydrogen sulfate ion is amphoteric so both the acid and base dissociations must be evaluated for influence on pH. As a base, HS04- is the conjugate base of a strong acid, so it wil l not influence pH. As an acid, H S04- is the conjugate acid of a weak base, so the acid dissociation applies HS04-(aq) + H 2 0( l) !:; S O/-(aq) + H 3 0\aq) Ka2 1 .2 1 0-2 In solutions of sodium hydrogen carbonate, only the HC0 3- will influence pH and it, like H S04-, is amphoteric: HC03-(aq) + H 2 0( l) !:; C03 2-(aq) + H 3 0+(aq) As an acid: Ka 4 . 7 x 1 0- 1 1 , the second Ka for carbonic acid from Table 1 8.5 As a base: H C03-(aq) + H 2 0( l) !:; H 2C0 3 ( aq) + OW(aq) Kb 1 . 0 1 0- 1 4 ! 4 . 5 1 0-7 2.2 1 0-8 Since Kb > K., a solution of sodium hydrogen carbonate is basic. In a solution of sodium carbonate, only C0 32- will influence pH by acting as a weak base: C0 3 2-(aq) + H 2 0( l) !:; HC03-(aq) + OW(aq ) Kb 1 . 0 1 0- 1 4 ! 4 . 7 1 0- 1 1 2. 1 1 0-4 Therefore, the solution of sodium carbonate is basic. X

X

=

X

X

X

=

=

=

X

X

=

X

X

=

=

253

X

X

X

X

Two of the solutions are acidic. Since the Ka of HS04- is greater than that ofNH/, the solution of sodium hydrogen sulfate has a lower pH than the solution of ammonium nitrate, assuming the concentrations are relatively close. Two of the solutions are basic. Since the Kb of CO/- is greater than that of HC03-, the solution of sodium carbonate has a higher pH than the solution of sodium hydrogen carbonate, assuming concentrations are not extremely different. 1 8 .84

1 8.85

1 8.88

1 8 .90

A Lewis acid is defined as an electron pair acceptor, while a Bmnsted-Lowry acid is a proton donor. If only the proton in a Bmnsted-Lowry acid is considered, then every Bnmsted-Lowry acid fits the definition of a Lewis acid since the proton is accepting an electron pair when it bonds with a base. There are Lewis acids that do not include a proton, so all Lewis acids are not Bnmsted-Lowry acids. A Lewis base is defined as an electron pair donor and a Bmnsted-Lowry base is a proton acceptor. In this case, the two definitions are essentially the same except that for a Bnmsted-Lowry base the acceptor is a proton. a) No, a weak Bmnsted-Lowry base is not necessarily a weak Lewis base. For example, water molecules solvate metal ions very well : Zn2+(aq) + 4 H 2 0( l) !:; Zn(H2 0)/+(aq) Water is a very weak Bmnsted-Lowry base, but forms the Zn complex fairly well and is a reasonably strong Lewis base. b) The cyanide ion has a lone pair to donate from either the C or the N, and donates an electron pair to the Cu(H2 0)/+ complex. It is the Lewis base for the forward direction of this reaction. In the reverse direction, water donates one of the electron pairs on the oxygen to the Cu(CN)/- and is the Lewis base. c) Because Kc> I , the reaction proceeds in the direction written (left to right) and is driven by the stronger Lewis base, the cyanide ion. a) Cu2+ is a Lewis acid because it accepts electron pairs from molecules such as water. b) cr is a Lewis base because it has lone pairs of electrons it can donate to a Lewis acid. c) Tin(II) chloride, SnCh, is a compound with a structure similar to carbon dioxide so it will act as a Lewis acid to form an additional bond to the tin. d) Oxygen difluoride, OF2 , is a Lewis base with a structure similar to water, where the oxygen has lone pairs of electrons that it can donate to a Lewis acid. a) Sodium ion is the Lewis acid because it is accepting electron pairs from water, the Lewis base. + 6 H 20 Na+ !:; Na(Hp)/ Lewis acid Lewis base adduct b) The oxygen from water donates a lone pair to the carbon in carbon dioxide. Water is the Lewis base and carbon dioxide the Lewis acid. CO 2 + Lewis acid c) Fluoride ion donates an electron pair to form a bond with boron in BF4-. The fluoride ion is the Lewis base and the boron trifluoride is the Lewis acid. F+ BF3 Lewis base Lewis acid +­ -+

+­ -+

1 8 . 92

a) Since neither H+ nor OH- is involved, this is not an Arrhenius acid-base reaction. Since there is no exchange of protons, this is not a Bmnsted-Lowry reaction. This reaction is only classified as Lewis acid-base reaction, where Ag+ is the acid and NH3 is the base. b) Again, no OH- is involved, so this is not an Arrhenius acid-base reaction. This is an exchange of proton, from H 2 S04 to NH3 , so it is a Bronsted-Lowry acid-base reaction. Since the Lewis definition is most inclusive, anything that is classified as a Bnmsted-Lowry (or Arrhenius) reaction is automatically classified as a Lewis acid­ a

base reaction.

c) This is not an acid-base reaction. d) For the same reasons listed in (a), this reaction is only classified as Lewis acid-base reaction, where AICI 3 is the acid and cr is the base.

254

1 8.94

Calculate the [H30+] using the pH values given. Determine the value of Kw from the pKw given. The [ H 3 0+] is combined with the Kw value at 37°C to find [OH-] using Kw = [H30+][OW]. 3 3 Kw = l O-pKIV = 10-1 6 = 2.3442 x 1 0- 1 4 (unrounded) Kw = [H30+] [OW] = 2.3442 X 1 0- 1 4 at 3 7°C [ H 3 0+] range 3 H igh value ( l o w p H ) = l O-p H = 1 0-7. 5 = 4.4668 X 1 0-8 = 4.5 X 10-8 M H30+ H Low value (h igh pH) = 1 0-p = 1 0-745 = 3 .548 1 X 1 0-8 = 3 .5 X 1 0-8 M H30+ Range : 3 . 5 x 10-8 to 4. 5 X 1 0--1l M H30 + [OW] range Kw = [ H 3 0+] [0 11] = 2.3442 x 1 0- 1 4 at 3 7°C

[011] -

Kw [ H3 0r

H igh value (high pH) =

1 8.97

1 8 .98

x

2.3442 x

1 0- 1 4 = 4.4668 x 1 0-8 1 0 -7 to 6.6 X 1 0-7 M O H-

Low value ( l o w pH) = Range : 5.2

2 . 3 442 x 10 - 1 4 = 6.6069 3 . 548 1 x 1 0-8

X

5.24805 x

X

1 0-7 M OH-

5.2 X

1 0-7 M Ol1

1 0-7 = 6 . 6 1 0-7 =

a) SnC14 i s t h e L e w i s aci d accepting an electron p a i r from (CH3)3N, the L e w i s base. b) Tin is the element in the Lewi s acid accepting the electron pair. The e lectron configuration of tin is 3 2 [ K r] 5s 4d o5/. The four bonds to tin are formed by sp hybri d orbitals, which completely fi l l the 5s and orbitals. The Sd orbitals are empty and available for the bond with tri methylamine.

5p

Hydrochloric acid is a strong aci d that almost completely di ssoc iates i n water. Therefore, the concentration of H 3 0+ is the same as the starting acid concentration: [H30+] = [HCI]. The original solution p H : p H = -log ( 1 .0 x 1 0-5 ) = 5.00 p H . A 1 : 1 0 d i l ution means that the chemist takes I m L o f the 1 .0 x 1 0-5 M solution and d i lutes i t t o 1 0 m L (or d i l ute 1 0 mL to 1 00 m L ) . The chemist then d i l utes the d i luted sol ution in a I : 1 0 ratio, and repeats thi s process for the next two successive d i l utions. M I V = M2 V2 can be used to find the molarity after each d i l ution : 1 Di lution 1 : M V = M2V2 1 1 ( 1 . 0 X 1 0-5 M) ( 1 .0 m L) = (x)( I O . m L) [ H30+] H CI = 1. 0 x 1 0--{j M H30+ pH = -log ( 1 .0 x 1 0--{j) = 6.00. D i l ution 2 : ( 1 .0 x 1 0--{j M) ( 1 .0 m L ) = (x)( I O . mL) [ H 3 0+] H CI = 1 .0 x 1 0-7 M H 30+ Once the concentration of strong acid is c lose to the concentration of H30+ from water autoionization, the [ H 3 0+] in the solution does not equal the initial concentration of the strong acid. The calculation of [H30+] must be based on the water ionization equi l ibriu m : 2 HP(l) => H30\aq) + Ol1(aq) with Kw = 1 .0 x 1 0- 1 4 a t 2 5 ° C . T h e d i l ution gives an initial [H 3 0+] of l . O x 1 0-7 M. Assuming that t h e i nitial concentration of hydroxide i o n s i s zero, a reaction table i s s e t up. Concentration (M) 2 H20(l) => H30+(aq) + OI1(aq) 0 l x l O-7 I nitial +x +x Change Equilibrium I x 10 7 + X x Kw = [ H 3 0+] [011] = ( I x 1 0-7 + x) (x) = 1 . 0 x 1 0- 1 4 Set up as a quadratic equation: x2 + 1 .0 x 1 0-7 X - 1 .0 x 1 0- 1 4 = 0 a= 1 b = 1 .0 x l O-7 c = - 1.0 x I 0- 1 4 =

x=

- 1 .0

X

1 0-7 ±

/ - 1 - x 1 0- 1 4 ) )( �1 . 0 x 1 0-7 4 ( ) ( 1 .0

------

--------------

2 (I )

x = 6 . 1 803 x 1 0-8 ( unrounded)

2 55

[ H 3 0+] = ( 1 . 0 X 1 0-7 + X) M = ( l .0 X 1 0-7 + 6 . 1 803 X 1 0-8) M = l . 6 1 803 X 1 0-7 M H 30+ (unrounded) pH = -log [ H 30+] = -log ( 1 . 6 1 803 X 1 0-7) = 6 . 79 1 0 1 = 6. 79 D i l ution 3 : ( l .0 X 1 0-7 M) ( 1. 0 mL) = (x)( l O. mL) [ H 3 0+] H CI = 1 . 0 X 1 0-8 M H 3 0+ The d i l ution gives an initial [ H 3 0+] of 1 .0 x 1 0-8 M. Assuming that the initial concentration of hydroxide ions i s zero, a reaction table i s set u p . Concentration (M) 2 H 2 0(l) !:; H 30\aq) + OW(aq) l x l O-8 o Initial +x +x Change Equi l i brium I X 1 0-8 + X X Kw = [ H 3 0+] [OH-] = ( I X 1 0-8 + X ) ( X) = 1 . 0 X 1 0- 1 4 Set up as a quadratic equation: x2 + 1 .0 x 1 0-8 X - 1 .0 X 1 0- 1 4 = 0 a= 1 b = L O x 1 0-8 c = - l . O x l O- 1 4

�(

/

)

(

4 - 1 . 0 X I O-S ± 1 .0 X l O-s - 4 ( 1 ) - 1 .0 X 1 0- 1 X = -----'--------;-----;-------2(1)

X = 9 . 5 1 249 X I O-s ( unrounded) [ H 3 0+] = ( 1 . 0 X 1 0-8 + X) M = ( l .0 X 1 0-8 + 9 . 5 1 249 X 1 0-8) M = 1 . 05 1 249 X 1 0-7 M H 3 0+ ( unrounded) pH = -log [ H 30+] = -log ( 1 .05 1 249 x 1 0-7) = 6 . 97829 = 6.98 D ilution 4 : ( 1 . 0 X 10-8 M) ( 1 . 0 m L ) = (x)( 1 0. mL) [ H 3 0+] H CI = l . 0 X 1 0-9 M H 3 0+ The d i l ution gives an i n itial [ H 3 0+] of 1 . 0 x 1 0-9 M. Assuming that the initial concentration of hydroxide ions i s zero, a reaction table i s set up. Concentration (M) 2 H 2 0(l) !:; H 30+(aq) + lni tial 1 x 1 0-9 C hange +x +x 1 x 1 0-9 + X x Equil i brium Kw = [ H 3 0+] [O W] = ( I X 1 0-9 + X) (X) = 1 . 0 x 1 0- 1 4 Set up as a quadratic equation: x2 + 1 .0 x 1 0-9 X - 1 .0 X 1 0- 1 4 = 0 4 a= I b = 1 .0 X 1 0-9 C = - 1 . 0 X 1 0- 1 2 4 - 1 .0 X 1 O- 9 ± 1 .0 X 1 0- 9 - 4 ( 1 ) - 1 . 0 X 1 0- 1 -----'--------;-----;-------X= 2 (1)

�(

)

(

)

X = 9 . 9 5 0 1 2 X 1 0-8 ( unrounded) [ H 3 0+] = ( 1 . 0 X 1 0-9 + X ) M = ( 1 . 0 X 1 0-9 + 9.950 1 2 x 1 0-8 ) M = 1 . 005 1 2 X 1 0-7 M H 3 0+ ( unrounded) pH = -log [ H 30+] = -log ( 1 . 005 1 2 x 1 0-7) = 6.99778 = 7.00 As the H C I s o l ution is d i l uted, the pH of the solution becomes closer to 7 . 0 . Continued d i l utions w i l l not significantly c hange the p H from 7 . 0 . Thus, a solution with a basic pH cannot be made b y adding acid to water. 1 8 .99

Compare the contrib ution of each acid by calculating the concentration of H 3 0+ produced by each. For 3 % hydrogen peroxide, first find initial molarity of H 2 0 2 , assuming the density i s 1 .00 glmL (the density of water).

( )( 1 . 00 g

3 % H 2 02

)(

(J � )

1 mol H 2 0 2 1 = 0 . 8 8 1 834 M H 2 02 (unro u nded) 1 00% mL 34.02 g H 2 0 2 1 0- L Find Ka from pKa: Ka = l O-pKa = 1 0- 1 1 7 5 = 1 . 778279 X 1 0- 1 2 (unrounded) H 2 02 + H 20 !:; t := 2 ':) f := ':) 2 <:::: nitial 0 0 . 8 8 1 834 0 Change -x +x +x Equi l i brium 0 . 8 8 1 834 - x x x M H 2 02 =

256

Ka = 1 . 7 78279

X

1 0- 1 2 =

Ka = 1 . 7 7 82 7 9

X

1 0- 1 2 =

Ka = 1 . 7 7 82 79

X

1 0- 1 2 =

[

H 3 O+

J[

HO 2 -

J

[ H 202 ]

(X)(X)

( 0 . 8 8 1 834 - x )

Assume x is small compared to 0 . 8 8 1 834.

(X)(X)

( 0 . 8 8 1 834 )

[ H 3 0+] = X = 1 . 2522567 X 1 0-6 ( unrounded) Check assumption: ( 1 . 2 5 2 2 5 6 7 x 1 0-6 I 0 . 8 8 1 834) x 1 00% = 0 . 000 1 % error, so the assumption is val id. M H 3 P04 =

( )( 1 . 00 g

o . OO I % H 3 P0 4

mL

1 00%

)(

1 mol H 3 P0 4 97.99 g H 3 P0 4

)[ ) 1 mL 3 1 0- L

= 1 . 0205 X 1 0-4 M H 3 P0 4 ( unrounded) 3 From Appendix C , Ka for phosphoric acid is 7 . 2 x 1 0- . The subsequent Ka values may be ignored . In thi s calcu lation x i s not negligible since the i n itial concentration of acid i s less than the Ka. f ::: 2 P '=> 4 T 0 0 4 nitial 1 .0205 x 1 0-4 Change -x +x +x X x Equi l ibrium 1 . 0205 x 1 0-4 _ 3 Ka - 7 . 2 x 1 0- -

_

Ka = 7 . 2

X

3 1 0- =

[

[

H 3 0+

[

J[

H 2 P0 4 -

H 3 P0 4

]

J

[x] [x] 1 . 0205

X

4 1 0- - x

J

The problem w i l l need to be solved as a quadratic. 3 3 x2 = ( 7 . 2 x 1 0- ) ( 1 . 0205 x 1 0-4 - x) = 7 . 3476 x 1 0-7 - 7 . 2 x 1 03 2 X + 7 . 2 X 1 0- x - 7 . 3476 X 1 0-37 = 0 a= I b = 7 . 2 X 1 0c = -7 .3476 X 1 0-7

�(

f

(

X

)

3 3 - 4 ( 1 ) -7.3476 x 1 0-7 -7 . 2 x 1 O- ± 7 . 2 x 1 O------'-----------x= -;-:2(1) 4 x = 1 . 00643 x 1 0- M H 3 0+ (unrounded) The concentration of hydronium ion produced by the phosphoric acid, 1 x 1 0-4 M, is greater than the concentration produced by the hydrogen peroxide, 1 x 1 0-6 M. Therefore, the phosphoric acid contributes more H 3 0+ to the so l ution. 1 8 . 1 04

=

The freezing point depression equation is required to determine the molality of the solution. �T = [ 0 . 00 - (- 1 . 9 3 °C)] 1 . 93°C = = i Krm Temporari l y assume i = 1 . �T 1 . 93° C = 1 . 037634 m = 1 . 03 7634 M ( unrounded) m= - = iKf ( I )( 1 . 86°C/m) This molal ity i s the total molal ity of al l species i n the solution, and is equal to their molarity. From the equil ibri u m : C I C H 2 C OO H (aq) + H 2 0(1) !:; H 3 0+(aq ) + C I C H 2 COO-(aq ) x x I nitial 1 . 000 M +x +x -x Change x x Equi l i brium 1 . 000 - x The total concentration of a l l species i s : [ C I C H 2 COO H ] + [ H 3 0+] + [CICH 2 COO-] = 1 . 03 7634 M [ 1 . 000 - x] + [x] + [x] = 1 . 000 + x = 1 . 037634 M x = 0 . 0 3 7634 M ( unrounded)

257

[ Ka =

J

[

C H 3 COO-

[C H 3 COOH ]

]

( 0 .03 7634 ) ( 0. 03 7634 )

Ka = 1 8 . 1 05

H 3 0+

( 1 .000 - 0.037634)

= 0.00 1 47 1 7 = 0.00 1 47

a) The two ions that comprise thi s salt are Ca2+ (derived from the strong base Ca( O H ) 2 ) and C H 3 C H 2 COO­ (derived from the weak acid, propionic acid, C H 3 C H 2 COOH ) . A salt derived from a strong base and weak acid produces a basic solution. Ca2 + does not react with water. C H 3 C H 2 COO-(aq) + H 2 0(l) !:; C H 3 C H 2 COOH(aq) + O W(aq ) b) Calcium propionate is a soluble salt and dissolves in water to yield two propionate ions: Ca( C H 3 C H 2 COO)z(S) + H 2 0(/) � Ca2\aq) + 2 C H 3 C H 2 COO-(aq) The molarity of the solution i s . M o l anty =

(

7 . 05 g C a ( C H 3 C H ? COOh 0. 500 L

)(

1 mol Ca( C H 3 C H 2 COOh 1 86.22 g Ca(CH 3 C H 2 COOh

= 0 . 1 5 1 43 3 7 8 8 M C H 3 C H 2 COO- (unrounded) C H 3 C H 2 COO+ H20 !:; 0 0. 1 5 1 43 3 7 8 8 M I n itial +x -x C hange Equi l i brium 0 . 1 5 1 43 3 7 8 8 - x x Kb = Kw / Ka = ( 1 . 0 x 1 0- 1 4 ) / ( 1 . 3 x 1 0-5 ) = 7 . 6923 x 1 0- 1 0 ( unrounded) _

Kb - 7 .6923 x 1 0-

Kb = 7 . 6923

X

_ [C H[3 C H 2 COOH J 10 -

1 0- 1 0 =

Kb = 7 . 6923 x l O- 1 0 =

[

OH-

C H 3 C H 2 COO-

(x)(x) ( 0. 1 5 1 43 3 7 8 8

-

x)

)[

2 mol C H 3 C H 2 COO1 mol C a ( C H 3 C H 2 COO) 2 +

1

mr

+x x

]

] Assume x is small compared to 0 . 1 5 1 43 3 7 8 8 .

(x) (x)

( 0. 1 5 1 43 3 7 8 8 )

x = 1 . 079293 x 1 0-5 = [O W] (unrounded) Check assumption : [ 1 .079294 x 1 0-5 / 0. 1 5 1 43 3 7 88] x 1 00% = 0 . 007%, therefore, the assumption is good. K lV 1 . 0 X 1 0. 1 4 = 9.26532 X 1 0- 1 0 M H 3 0+ ( unrounded) [H 3 ot 1 . 079294 x 1 0 .5 [OR ] pH = -log [ H 30+] = -log (9.26532 x 1 0- 1 0 ) = 9.03 3 1 = 9.03 1 8. 1 1 1

a) The concentration of oxygen i s h igher i n the l ungs so the equ i l i brium shifts to the right. b) In an oxygen deficient environment, the equ i l i brium would shift to the left to rel ease oxygen. c) A decrease i n the [H 3 0+] concentration would shift the equi l i brium to the right. M ore oxygen i s absorbed, but it wi l l be more difficult to remove the O2 . d) An increase in the [HP+] concentration would shift the equ i l i bri um to the left. Less oxygen is absorbed, but it w i l l be easier to remove the O 2 .

1 8. 1 1 3

Note that both pKb values only have one sign i ficant figure. This w i l l l i mit the final answers. Kb (te rtiary amin e N) = l O-pKb = 1 0-5 . 1 = 7 . 94328 X 1 0-6 ( unrounded) Kb (aromatic ring N) = 1 0-pKb = 1 0-9 . 7 = 1 . 995262 x 1 0- 1 0 (unrounded) a) I gnoring the smal ler Kb : C 2 0 H 24N 2 0 2 (aq) + H 2 0(l) !:; OW(aq) + H C 2 0 H 24N 2 0/(aq) 3 Initial 1 .6 x 1 0- M O O Change -x +x +x 3 X X Equi l i brium 1 . 6 x 1 0 - X

258

_ Kb -

[

HC20 H 24 N 202 +

J[ OH - ]

[ H 2 C 20 H 24 N 2 0 2

_ - 7.94328 x 1 0--{i

]

[ x ] [ x ] ] = 7.94328 x I O--{i 1 .6 1 0-3 The problem will need to be solved as a quadratic. x22 = (7.94328 x 1 0--{i) ( 1 .6 x 1 0-3 - x) = 1 .27092 1 0-8 - 7.94328 I O--{i x x + 7.94328 x 1 0--{i - 1 .27092 1 0-8 = 0 a = l b = 7.94328 x I 0--{i c = -1 .27092 x 1 0-8 -b ± �b 2 - 4ac x = -----'--2a 2 -7.94328 x 1 O-6 ± 7.94328 x 1 0-6 - 4 ( 1 ) - 1 .27092 x 1 0-8 x = --------'---------;--;----------2(1) x = 1 .08833 1 0-4 M OW (unrounded) K w - 1 .0 1 0- 1 4 = 9. 1 8838955 1 0-1 1 M H P+ (unrounded) [ H 3 ot 1 .08833 x 1 0-4 [OR ] pH = -log [ H30+] = -log (9. 1 8838955 x 1 0-1 1 ) = 1 0.03676 = 10.0 b) (Assume the aromatic N is unaffected by the tertiary amine N . ) Use the Kb value for the aromatic nitrogen. C20H24N202(aq) + H20(l) !:; OW(aq) + HC20H24N202\aq) 1 .6 x 1 0-3 ] Kb =

[

X

X

X

X

X

)

J(

X

Kb = Kb =

X

X

= 1 .995262 x 1 0-1 0

[ C 2 o H 24 N2 0 2 ]

X

-

X

]

)

X

[ HC 20H 24N 202 + J[ OH [x ] [x ] [ 1 .6 1 0-3

(

X

- X

X

= 1 .995262 1 0-1 0 Assume x is small compared to 1 .6 x 1 0-3 . X

[ x ] [ x ] = 1 .995262 1 0 0 1 .6 1 0-3 ] = 5.650 1 1 0-7 M OW (unrounded) The hydroxide ion from the smaller Kb is much smaller than the hydroxide ion from the larger Kb (compare the powers of ten in the concentration). c) HC20H24N20t(aq) + H20(l) !:; H 30+(aq) + C20H24N202(aq) 1 . 0 1 0- 14 = 1 .2589 1 0-9 (unrounded) K Ka = _w = Kb 7.94328 x 1 0Kb = X

[

X

X

-

1

X

X

_ _ Ka - 1 .2589 x 1 0-9 -

6

X

[ H 3 0+ ][C20 H 24N202 ] [ HC20 H 24N 2 0 2+ ]

(x)(x) Assume x is small compared to 0.53. ( 0.53 - x ) (x) (x) Ka = 1 .2589 1 0-9 = ( 0.53 ) [ H30+] = x = 2.58305 1 0-5 (unrounded) Check assumption: (2.58305 x 1 0-5 / 0.53) x 1 00% = 0.005%. The assumption is good. pH = -log [ H30+] = -log (2.58305 x 1 0-5) = 4.58779 = 4.6

Ka = 1 .2589 x 1 0-9 = X

X

259

( ) ( ) [ )( -[ +

d) Quinine hydrochloride wil l be indicated as QHCI. 1 mol QHCI = 0.04 1 566 M M = 1 .5% 1 .0 g 1 mL (unrounded) 1 00% mL 1 0-3 L 360.87 QHCI H 3 0 ] [ C 2 0 H 24 N 2 0 2 ] K. - 1 .2589 x l 0-9 _ [ HC 20 H 24 N 2 0 2 + ] _

J

(x)(x) Assume x is small compared to 0.04 1 566. (0.04 1 566 - x ) (x) (x) K = 1 .2589 1 0-9 = ( 0.04 1 566 ) [H 3 0+] = x = 7.23377 1 0-6 (unrounded) Check assumption: (7.23377 x 1 0-6 / 0.04 1 566) x 1 00% = 0.02%. The assumption is good. pH = -log [H30+] = -log (7.23377 x 1 0-6) = 5 . 1 406 = 5.1 K = 1 .2589 x l O-9 = •

•

X

X

260

Chapter 1 9 Ionic Equilibria in Aqueous Systems FOLLOW-UP PROBLEMS

1 9. 1

Plan: The problems are both equilibria with the initial concentration of reactant and product given. For part (a), set up a reaction table for the dissociation of HF. Set up an equilibrium expression and solve for [H30+], assuming that the change in [HF] and [F-] is negligible. Check this assumption after finding [H30+] . Convert [H30+] to pH. For part (b), first find the concentration of OH- added. Then, use the neutralization reaction to find the change in initial [HF] and [F-] . Repeat the solution in part (a) to find pH. a) Solution: + Concentration (M) HF(aq) + 0.50 Initial +x +x Change -x x 0.45 + x 0.50 - x Equilibrium Assumptions: 1 ) initial [H30+], from water, at 1 .0 1 0-7 M, can be assumed to be zero and 2) x is negligible with respect to 0.50 M and 0.45 M. [ H 3 0+ K = [ H F] ] [ HF ] ( -4 ) [[ 0.50 [H30+] = Ka = 6.8 x 1 0 -] = 7.5556 x 1 0-4 = 7.6 1 0 4 0.45 FCheck assumptions: I ) 1 .0 1 0-7 « 7.6 x 1 0-4 Thus, the assumption to set the initial concentration of H30+ to zero is valid. 2) Percent error in assuming x is negligible: (7.5556 x 1 0-4/0.45) x 1 00 = 0. 1 7%. The error is less than 5%, so the assumption is valid. Solve for pH: pH = -log (7.5556 x 1 0-4) = 3 . 1 2 1 73 = 3.1 2 The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation: [base]pH = pKa + log -. [acid] ] pKa = -log Ka = -log(6.8 1 0-4 ) = 3 . 1 6749 pH = pKa + log W [HF] [0.45] P H = 3 . 1 6749 + 10 g [0.50] pH = 3 . 1 2 1 73 = 3.1 2 Check: Since [HF] and [F-] are similar, the pH should be close to pKa, which equals log (6.8 x 1 0-4) = 3. 1 7. The pH should be slightly less (more acidic) than pKa because [HF] > [F-] . The calculated pH of 3 . 1 2 is slightly less than pKa of 3 . 1 7. b) Solution: What is the initial molarity of the mr ion? I mol O W 0.40 g NaOH I mol NaOH = O.O I O M OW L 40.00 g NaOH 1 mol NaOH Set up reaction table for neutralization of 0.0 1 0 MOH- (note the quantity of water is irrelevant). Concentration (M) HF( aq ) + OW( aq ) F-( aq ) + H 2 0(l) 0.45 0.50 Before addition 0.0 1 0 Addition + 0.0 1 0 - 0.0 1 0 - 0.0 1 0 Change 0.46 After addition 0.49 0 a

x

J[F ]

[ ]

X

X

X

(

)(

)

)(

-7

26 1

Following the same solution path with the same assumptions as part (a): 0.49 ] = 7.2434782 x 1 0-4 = 7.2 x 1 0-4 M ]] = ( 6.8 x 1 0 -4 ) [[ 0.46 [[ HF [H30+] = Ka F] Check assumptions: I ) 7.2 x 1 0-4 » 1 .0 x 1 0-7, the assumption is valid. 2) (7.2 x 1 0-4/0.46) 1 00 = 0. 1 6%, which is less than the 5% maximum � assumption acceptable. Solve for pH: pH = -log (7.2434782 x 1 0-4) = 3 . 1 4005 = 3 . 1 4 or using the Henderson-Hasselbalch equation: [F- ] [0.46] pH = pKa + I og -- = 3. 1 6749 + log -[0.49] [HF] pH = 3 . 1 4005 = 3 . 1 4 Check: With addition of base, the pH should increase and it does, from 3 . 1 2 to 3 . 1 4. However, the pH should still be slightly less than pKa: 3 . 1 4 is still less than 3. 1 7. --

1 9.2

Plan: Sodium benzoate is a salt so it dissolves in water to form Na+ ions and C 6HsCOO- ions. Only the benzoate ion is involved in the buffer system represented by the equilibrium: C 6HsCOOH(aq) + H 2 0(l) !:+ C 6HsCOO-(aq) + H30+(aq) Given in the problem are the volume and pH of the buffer and the concentration of the base, benzoate ion. The question asks for the mass of benzoic acid to add to the sodium benzoate solution. First, find the concentration of C 6 HsCOOH needed to make a buffer with a pH of 4.25 . Multiply the volume by the concentration to find moles of C 6 HsCOOH and use the molar mass to find grams of benzoic acid. Solution: From the given2 s pH, calculate [Hs3 0+] : [H 3 0+] = 1 O-4. = 5.6234 1 l O- M (unrounded) The concentration of benzoic acid is calculated from the Henderson-Hasselbalch equation: [C HsCOO- ] p H = p K a + 10 g 6 [C 6 HsCOOH] pKa = -log Ka = -log 6.3 x 1 0-s = 4.20066 4.25 = 4.20066 + log [0.050] [x] 0.04934 = log [0.050] Raise each side to l OX . [x] 1 . 1 203 1 46 = [0.050] [x] x = 4.46303 x 1 0-2 M The number of2moles of benzoic acid is determined from the concentration and volume: (4.46303 x 1 0- M C6 HsCOOH) x (5.0 L) = 0.223 1 5 mol C 6 HsCOOH (unrounded) Mass of C 6HsCOOH: ( 0.223 1 5 mol C 6 H sCOOH ) 1 mol C 6 HsCOOH = 27.25 1 1 = 27 g C 6HsCOOH 1 22. 1 2 g C 6 HsCOOH Prepare a benzoic acid/benzoate buffer by dissolving 27 g of C 6HsCOOH into 5.0 L of 0.050 M C6 HsCOONa. Using a pH meter, adjust the pH to 4.25 with strong acid or base. Check: pKa for benzoic acid is 4.20. The buffer pH is slightly greater than the pKa, so the concentration of the acid should be slightly less than the concentration of the base. The calculation confirms this with [C6HsCOOH] at 0.045 M, which is slightly less than [C 6HsCOO-] at 0.050 M. X

(

)

262

1 9.3

P lan: The titration i s of a weak acid, H BrO, with a strong base, NaO H . The reactions involved are : 1 ) Neutralization of weak aci d with strong base : H B rO(aq) + OW(aq) � H20(l) + B rO-(aq) N ote that the reaction goes to completion and produces the conj ugate base, B rO-. 2) The weak acid and its conj ugate base are in equil i brium based on the ac id d issociation reaction: a) H B rO(aq) + H20(l) !::; B rO-(aq) + H 3 0+(aq) or b) the base dissociation reaction: H B rO(aq ) + O H-(aq) BrO-(aq) + H20(l) !::; The p H of the solution is contro l l ed by the rel ative concentrations of H B rO and BrO-. For each step in the titration, first think about what is present initial l y in the solution. Then use the two reactions to determine solution p H . I t i s usefu l i n a titration problem t o first determine at what volume of titrant the equivalence point occurs. e) Use the p H values from (a) - (d) to plot the titration curve. Solution : a) Before any base i s added, the solution contains only H BrO and water. E q u i l ibrium reaction 2a appl ies and pH can be found i n the same way as for a weak acid sol ution : Concentration (M) H BrO(aq) + H20(aq) !::; B rO-(aq) + H30+(aq) 0 0 0 .2000 Initial Change -x +x +x x x Equil i brium 0.2000 - x K = a

[ H 3 0 + J [ BrO- J [ H B rO ]

= 2 . 3 x 1 0-9 =

Assume 0 .2000 - x = 0 .2000

[x][ ] X

2 . 3 x 1 0-9 =

[x] [x]

[ 0.2000 - x ]

M

[ 0 .2000 ]

x = [ H 3 0+] = 2 . 1 4476 1 X 1 0-5 M (unrounded) p H = -log [ H 30+] = -log ( 2 . 1 4476 1 x 1 0-5) = 4 . 66862 = 4.67 Check the assumption by calculating the percent error in [ H B rO]eq. Percent error = ( 2 . 1 4476 1 x 1 0-5 /0 .2000) I 00 = 0.0 1 %, thi s i s well below the 5 % maximum. b) When [ H BrO] = [BrO-] , the solution contains significant concentrations of both the weak acid and its conj ugate base. For the concentrations to be equal, half of the equivalence point vol ume has to have been added. Calculate the concentration of the acid and base and use the equi libri um expression for reaction 2a to find pH. Since [ H B rO] = [BrO-] their ratio equals I .

[ H 3 0 + J [ BrO- J [ H BrO ] [1] = 2.3 x 1 0-9 M [ H B rO]] = ( 2 . 3 x 1 0-9 ) [ij [ H 3 0 +] = [ BrO-

Ka =

K

a

p H = -log [ H 3 0+] = -log ( 2 . 3 x 1 0-9 ) = 8 . 63 827 = 8.64 Note that when [ H B rO] = [ B rO-] , the titration is at the midpoint (half the volume to the equivalence point) and pH = pKa• c ) At the equivalence point, the total number of moles of H B rO present initiall y in solution equal s the number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of B rO-. The solution consists of B rO- and water. Calculate the concentration o f B rO-, and then find the pH using the base dissociation equil i brium, reaction 2b. First, find equivalence point volume ofNaO H . 3 I mL IL 0.2000 m o l H BrO 1 0- L l m o l NaOH = 40 . 00 mL NaOH added ( 2 0 . 00 mL ) 3 .. 1 mL L I m o l H BrO 0 . 1 000 m o l NaOH 1 0- L

(

)[ )

(

)(

263

)( )

A l l of the H B rO present at the beginning of the titration is neutral ized and converted to BrO- at the equivalence point. Calculate the concentration of BrO-. I n itial moles of H BrO : ( 0 .2000 M)(0.02000 L) = 0.004000 mol Moles of added NaO H : ( 0 . 1 000 M)(0.04000 L) = 0.004000 mo l � H 2 0(!) + BrO-(aq ) H BrO(aq) + OW(aq) Amount ( m o l ) 0 Before addition 0. 004000 mol Addition 0. 004000 mol +0. 004000 mol - 0.004000 mol - 0 .004000 mol Change After add ition 0 0 0 .004000 mol At the equival ence point, 40.00 m L of NaOH sol ution has been added ( see calculation above) to make the total volume of the sol ution (20.00 + 40.00) mL = 60.00 mL. [ B rO-] =

[

0. 004000 mol Bro60.00 mL

)( ) 1 mL 3 1 0- L

= 0.06666667 M ( unrounded)

Set up reaction table with reaction 2b, since only BrO- and water are present i n itial ly: OW(aq) BrO-(aq) + H 20( !) Concentration (M) ::; H BrO(aq) + 0 0 I nitial 0 . 06666667 +x +x -x Change x X Equ i l i brium 0 .06666667 - x Kb = K,jKa = ( 1 .0 x 1 0- 1 4 / 2 . 3 x 1 0-9 ) = 4.347826 x 1 0-6 ( unrounded) _

Kb -

[ OH - ] [ H BrO] [ BrO- ]

[ x] [ x]

_

-

[ 0.06666667 - x ]

Assume that x is negligible, si nce [BrO-] 4.347826 x 1 0-6 =

[ x ][x ]

»

= 4.347826 x 1 0-6

Kb.

[ 0 .06666667 ]

x = [OW] = 5 . 3 8 3 8 1 9 1 X 1 0-4 = 5 .4 X 1 0-4 M Check the assumption by calculating the percent error in [BrO-]eq . percent error = ( 5 . 3 8 3 8 1 9 1 x 1 0-4/0 .06666667) I 0 0 = 0 . 8 % , wh ich is wel l below t h e 5 % maxi mum. pOH = -log ( 5 . 3 8 3 8 1 9 1 x 1 0-4 ) = 3 .2689 1 pH = 1 4 - pOH = 1 4 - 3 .2689 1 = 1 0 .73 1 09 = 1 0.73 d) After the equivalence poi nt, the concentration of excess strong base determi nes the p H . F ind the concentration of excess base and use it to calculate the p H . I nitial moles of H B rO : ( 0 .2000 M)(0.02000 L) = 0.004000 mo l M o l es of added NaO H : 2 x 0 .004000 mol = 0.008000 mol NaOH OW(aq) � H 2 0( !) + BrO-(aq) H B rO(aq) + Amount ( mol ) Before addition 0 . 004000 mol 0 Addition 0.008000 mol Change - 0. 004000 mol - 0. 004000 mol +0.004000 mol After addition o 0.004000 mol 0 .004000 mol Excess NaOH : 0 . 004000 mol Volume of added NaO H :

( 0 .0080000 mol NaOH )

(

]

L

0. 1 000 mol NaOH

Total volume : 2 0 . 00 mL + 80.00 m L = 1 00.0 mL [NaO H ] = 0. 004000 mol NaOH/O . 1 000 L = 0.0400 M pOH = -l og ( 0 . 0400) = 1 . 3979 pH = 14 - pOH = 14 - 1 . 3979 = 1 2.60

264

)( I ) mL 3 1 0- L

= 80.00 m L

e) Plot the pH values calculated i n the preceding parts of thi s problem as a function of the volume of titrant.

Titration of H B rO with N a O H 14 13 12 11 10 pH

9 8 7 6 5 4

-+-

-+

10

20

3 0

30

40

50

60

70

80

90 1 00

Vo l u m e of 0 . 1 000 M N a O H , m L

Check: The plot and p H values follow the pattern for a weak aci d vs. strong base titration. The p H at the midpoint of the titration does equal pKa. The equivalence point should be, and is, greater than 7 . 1 9 .4

Plan: Write the formula of the salt and the reaction showing the equi l ibrium o f a saturated solution. The ion­ product expression can be written from the stoichiometry of the solution reaction. Solution : a) The formula of calcium s u l fate i s CaS04 . The equ i l i brium reaction i s CaS0 4 (s) =+ Ca2+(aq) + SO/-(aq) I on-product expression: Ksp = [Ca2+] [SO/-] b) Chromium(JIl) carbonate is Cr2 (C0 3 ) 3 3 Cr2 (C0 3Ms) =+ 2 C r +(aq) + 3 C03 2-(aq) 3 3 lon-product expression: Ksp = [Cr +f [CO/-] c) M agnesium hydroxide i s Mg(OH ) 2 Mg(OH Ms) =+ M g2 +(aq) + 2 0W(aq) Ion-product expression: Ksp = [ M g2+] [O H-] 2 d) Arsenic( I I I ) sulfide i s AS 2 S3 . 3 AS2 S3(S) =+ 2 As +(aq ) + 3 S 2- ( aq ) 2 3 { S -(aq) + H 2 0( l) =+ H S-(aq) + OW(aq) } 3 AS 2 S 3 (S) + 3 H 2 0( l) =+ 2 As +(aq) + 3 H S-(aq) + 3 OW(aq) The second equi l i brium must be considered in thi s case because its equil i bri um constant is large, so essentia l l y all the s u l fi de ion is converted to H S- and O W . 3 3 I on-product expression: Ksp [As +] 2 [ H S-] [OH-f Check: A l l equilibria agree with the formulas of salts. =

1 9.5

Plan : Calculate the solubi l ity of CaF 2 as molarity and use molar ratios t o fi n d the molarity of ci+ and F­ d issolved in solution. Calculate Ksp from [Ca2+] and [ F-] using the ion-product expression. Solution : Convert the solubil ity to molar sol ub i l i ty: 4 1 . 5 x 1 0 - g CaF2 I mol CaF2 I mL = 1 . 92 1 1 x 1 0 -4 M CaF (unroun ded) 2 1 0. 0 mL 7 8 . 0 8 g CaF2 1 0-3 L

[

) ( )(

J

265

[Ca2+] = [CaF 2 ] = 1 . 92 1 1 x 1 0-4 M because there is I mol of calcium ions in each mol of CaF 2 . [ F-] 2 [CaF 2 ] 3 . 8422 x 1 0-4 M because there are 2 mol of fl uoride ions i n each mol of CaF 2 . The solub i l ity equi l i brium i s : CaF 2 (s) !:; Ca2+(aq) + 2 F-(aq) Ksp = [Ca2+] [ F-] 2 Calculate Ksp using the solubil ity product expression from above and the saturated concentrations of calcium and fluoride ions. Ksp = [Ca2+] [ F-] 2 = ( 1 . 92 1 1 x 1 0-4 ) ( 3 . 8422 X 1 0-4 ) 2 = 2 . 8 3 6024 X 1 0- 1 1 = 2 . 8 X 1 0- 1 1 The Ksp for CaF 2 i s 2 . 8 x 1 0- 1 1 at 1 8°C. Check : Appendix C reports a Ksp value at 25°C for CaF 2 o f 3 . 2 x 1 0- 1 1 . T h e calculated value at 1 8°C i s very c lose to thi s l iterature val ue . =

=

1 9.6

Plan: Write the solub i l i ty reaction for Mg( O H ) 2 and set up a reaction table, where S i s the unknown molar solub i l i ty of the Mg2 + ion. U se the ion-product expression to solve for the concentration of Mg(OH) 2 in a saturated solution (also cal led the solub i l ity of Mg( O H ) 2 ) . S o l ution : Concentration ( M) M g(O H Ms) !:; Mg2 \aq) + 2 0W(aq) 0 0 I n itial +2S +S Change 2 S Equil ibrium S 3 Ksp = [ M g2 +] [OH -] 2 = ( S ) ( 2 S) 2 = 4 S = 6.3 x 1 0- 1 0 S = 5 . 4004 1 1 4 X 1 0-4 = 5 .4 X 1 0-4 M Mg(O H ) 2 The solub i l ity of M g( O H ) 2 i s equal to S, the concentration of magnesium ions at equil ibrium, so the molar solubil ity of magnesium hydroxide in pure water is 5.4 x 1 0-4 M. Check: To check the calculation, work it backwards (also a good technique when using problems as study tools). From the calculated sol ubil ity, find Ksp[ M g2+] = 5 .4 X 1 0-4 M and [OW] = 2(5 .4 X 1 0-4 M) = 1 . 08 X 1 0-4 M Ksp = ( 5 .4 x 1 0-4 ) ( 1 .08 x 1 0-4 ) 2 = 6 . 5 x 1 0- 1 0 The backwards calculation gives back given Ksp.

1 9.7

P lan: Write the solub i l ity reaction of BaS04 . For part (a) set up a reaction table i n which [ B a2+] = [SO/-] = S, which also equals the solub i lity of BaS04 . Then, solve for S using the ion-product expression. For part (b), there is an i n itial concentration of sulfate, so set up the reaction tabl e including this i n itial [SO/-] . Solve for the solub i l ity, S, which equals [Ba 2+] at equ i l i bri um. Soluti o n : a) S e t u p reaction tab le. Concentration ( M) BaS0 4 (s) !:; Ba2+(aq) + S0 42 -(aq ) 0 I nitial 0 Change +S +S Equi l i brium S S Ksp = [ B a2+] [SO/-] = S 2 = 1 . 1 X 1 0- 1 0 S 1 . 0488 x 1 0-5 = 1 . 0 X 1 0-5 M The molar solubi l ity of BaS0 4 in pure water is 1 .0 x 1 0-5 M b) Set up another reaction table with i n itial [S04 2-] = 0. 1 0 M Concentration ( M) BaS04 (s) !:; Ba2+(aq) + 0 I nitial Change +S +S Equi l i brium S 0. 1 0 + S Ksp = [ B a2+] [ SO/-] = S(O. 1 0 + S) = 1 . 1 X 1 0- 1 0 =

Assume that 0 . 1 0 + S is approxi matel y equal to 0. 1 0, which appears to be a good assumption based on the fact that 0 . 1 0 > 1 x 1 0- 1 0 , KspKsp = S(O. I 0 ) = 1 . 1 X 1 0- 1 0 S = 1 . 1 X 1 0-9 M Molar solubil ity of BaS0 4 in 0. 1 0 M Na2 S0 4 is 1 . 1 x 1 0-9 M

266

Check: The solubility of BaS04 decreases when sulfate ions are already present in the solution. The calculated decrease is from 10-5 M to 10-9 M, for a I O,OOO-foid decrease. This decrease is expected to be large because of the high concentration of sulfate ions. 19.8

Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if they will react with acid. Three cases are possible: I) If OH- is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions. 2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and neutralization reactions. 3) If the anion from the salt is the conjugate base ofa strong acid, it does not react with a strong acid. The solubility of the salt is unchanged by the addition of acid. Write the solubility reaction. Solution: a) Calcium fluoride,CaF2 Solubility reaction: CaF2 (s)!:; Ca2+(aq) + 2 F-(aq) Fluoride ion is the conjugate base of HF, a weak acid. Thus, it will react with H30+ from the strong acid, HN03· Neutralization reaction: F-(aq) + H30\aq) � HF(aq) + H20(l) The neutralization reaction decreases the concentration of fluoride ions, which causes the solubility equilibrium to shift to the right and moreCaF2 dissolves. The solubility ofCaF2 increases with the addition of HN01,

b) Zinc sulfide, ZnS Solubility reaction: ZnS(s) + H2 0(l) !:; Zn2+(aq) + HS-(aq) + OH-(aq) Two anions are formed because the sulfide ion from ZnS reacts almost completely with water to form HS­ and OW. The hydroxide ion reacts with the added acid: Neutralization reaction: OH-( aq) + H 3 0\aq) � 2 H20(l) In addition, the hydrogen sulfide ion, the conjugate base of the weak acid H2 S, reacts with the added acid: Neutralization reaction: HS-(aq) + H30+(aq) � H2 S(aq) + H20(l) Both neutralization reactions decrease the concentration of products in the solubility equilibrium, which causes a shift to the right, and more ZnS dissolves. The addition ofHN03 will increase the solubility of ZnS. c) Silver iodide, AgI. Solubility reaction: A gI(s) !:; Ag+(aq) + r(aq) The iodide ion is the conjugate base of a strong acid, HI. So, C will not react with added acid. The solubility of Agi will not change with added HN03, 19.9

Plan: First, write the solubility equilibrium equation and ion-product expression. Use the given concentrations of calcium and phosphate ions to calculate Qsp. Compare Qsp to Ksp. If Ksp < Qsp, precipitation occurs. If Ksp;::: Qsp then, precipitation will not occur. Solution: Write the solubility equation: Ca3(P04Ms) !:; 3Ca2+(aq) + 2PO/-(aq) and ion-product expression: Qsp [Ca2+] 3 [P 043-f [Ca2+] [ PO/-] 1.0 x 10-9 M 3 3 Qsp [Ca2+] [PO/-] 2 (1.0 X 1O-9) (1.0 X 10-9)2 1.0 X 10-45 4 Compare Ksp and Qsp. Ksp 1.2 x 10-29 > 1.0 x 10- 5 Qsp Precipitation will not occur because concentrations are below the level of a saturated solution as shown by the value of Qsp. Check: The concentrations of calcium and phosphate ions are quite low, but Ksp is also small so it is difficult to predict whether precipitation will occur by just looking at the numbers. One way to check is with an estimate of the concentrations of ions in a saturated solution. 3 3 Ksp = [Ca2 +] [ PO/-f (3 S) (2 S) 2 108 S5 I X 10-2 9 7 S = 6.213 X 10- M (unrounded) =

=

=

=

=

=

=

=

=

=

=

267

In a saturated solution [Ca2+] = 3(6.2 1 3 x 1 0-7 M) = 1 . 86 x 1 0-{j M and [PO/ -] = 2(6.2 1 3 x 10-7) = 1 .24 x 10-{j M Both of these concentrations are much greater than the actual concentrations of 1 x 10-9 M for both ions, so the solution is unsaturated and no Ca 3 (P04)2 will precipitate. 1 9. 1 0

Plan: Write equations for the solubility equilibrium and formation of the silver-ammonia complex ion. Add the two equations to get the overall reaction. Set up a reaction table for the overal l reaction with the given value for initial [NH3J. Write equilibrium expressions from the overal l balanced reaction and calculate Koverall from Kr and Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and calculate solubility. Equilibria: Solution: Ksp = 5.0 X 1 0-13 AgBr(s ) !:; Ag"'faqi + Br-( aq ) "' ----"-" K.f.= 1.7 x 10 7 q) L... Ag faq1 + 2 NH3(aq) !:; Ag(NH3h.,-l+c."'a>;l. AgBr(s) + 2 NH3 ( aq ) !:; Ag(NH3 )2+( aq) + Br-( aq) Set up reaction table: AgBr(s) Concentration (M) + 2 NH3(aq ) !:; Ag(NH3)2+( aq ) + Br-( aq ) 0 0 1.0 Initial +S -2 S +S Change S S 1 .0 - 2 S Equilibrium [ Ag(NH 3 ) ; J[ Br=KspKr=(5.0x 1 0-13 ) ( 1 .7 x 1 0 7 ) = 8.5x 10-{j Koverall= . . [ NH 3 r Calculate the solubility of AgBr: ___

]

Koverall =

[S][S]

[ 1 .0 - 2S ] 2

= 8.5 1 0-6 X

Assume that 1 .0 - 2S is approximately equal to 1 .0, which appears to be a good assumption based on the fact that 1 .0 » 8.5 x I O-{j, Koverall. [S][ S] 2 = 8.5x 1 0-{j

[1.0]

3 3 S = 2.9 1 54759 X 10- = 2.9 X 1 0- M The solubility of AgBr in ammonia is less than its solubility in hypo (sodium thiosulfate). Check: Since the formation constant for Ag(NH3) / is less than the formation constant of Ag(S203)/-, the addition of ammonia will increase the solubility of AgBr less than the addition of thiosulfate ion increases its solubility.

END-OF-CHAPTER PROBLEMS

1 9.2

The weak acid component of a buffer neutralizes added base and the weak base component neutralizes added acid so that the pH of the buffer solution remains relatively constant. The components of a buffer do not neutral ize one another when they are a conjugate acid/base pair.

1 9.4

A buffer is a mixture of a weak acid and its conj ugate base (or weak base and its conjugate acid). The pH of a buffer changes only slightly with added H30+ because the added H3 0+ reacts with the base of the buffer. The net result is that the concentration of H30+ does not change much from the original concentration, keeping the pH nearly constant.

1 9. 7

The buffer component ratio refers t o the ratio o f concentrations o f the acid and base that make u p the bu ffer . When this ratio is equal to I , the buffer resists changes in pH with added acid to the same extent that it resists changes in pH with added base. The buffer range extends equally in both the acidic and basic direction. When the ratio shifts with higher [base] than [acid] , the buffer is more effective at neutralizing added acid than base so the range extends further in the acidic than basic direction. The opposite is true for a buffer where [acid] > [base]. Buffers with a ratio equal to I have the greatest buffer range. The more the buffer component ratio deviates from I , the smaller the buffer range.

268

19.9

The buffer components are propanoic acid and propanoate ion. The sodium ions are spectator ions and are ignored because they are not involved in the buffer. The reaction table that describes this buffer is: Concentration (M) HC 3 H60 2 (aq) + H20(!)!:; C 3H5 02-(aq) + H 3 0+(aq) Initial O. I S 0.2S 0 Change -x +x +x x Equilibrium O. I S - x 0.2S + x Assume that x is negligible with respect to both O. I S and 0.2S since both concentrations are much larger than Ka. [H 0 + J[ C H s0 2 - ] [ x ] [ 0.2S x ] [ x ] [ 0.2S] + 3 Ka = 3 = = = 1 . 3 X 1 0 -5 ] [ O. l S - x ] [ O. I S] [HC 3 Hs0 2 ...{) [HC 2 H s0 2 ] ( = 1 .3 x 1 O '5 ) 0. IS = 7.8xlO M [H3 0 +] =Ka [C 2 H S0 2 - ] 0.2S Check assumption: percent error = (7.8 x 1 0...{)/ O. I S) 1 00% = 0.00S2%. The assumption is valid. pH = -log [H 3 0+] = -log (7.8 x 1 0...{) = S. I0790S = 5. 1 1. Another solution path to find pH is using the Henderson-Hasselbalch equation: pH = pKa + 10g [ba�e] pKa = -log( I .3 x 1 0-s) = 4.89 [acid] pH = 4.89 + 10g [C3Hs02'] = 4.89 + 1 0g [0.2S] [O. I S] [HC3Hs02]

( )

( ) ( J

pH= 5. 1 1

1 9. 1 1

( )

The buffer components are phenol, C6H5 0H, and phenolate ion, C6H5 0- . The sodium ions are ignored because they are not involved in the buffer. Calculate Ka from pKa and set up the problem with a reaction table. Ka = lO-pKa = 1 0-10.00 = 1 .0 X 1 0-10 Concentration (M) C6H5 0-(aq) + H3 0+(aq) C6H 5 0H(aq) + H 2 0(!) !:; 0 1 .0 Initial 1.2 +x +x Change -x x 1 .0 + x Equilibrium 1 .2 - x Assume that x is negligible with respect to both 1 .0 and 1 .2 because both concentrations are much larger than Ka. [H 0 + J[ C 6H sO[ x ] [ 1 .0 + x ] [ x ] [ 1 .0 ] = 1 .0 x 1 0-10 = = Ka = 3 [C 6H sOH ] [1 .2 ] [ 1 .2 - x ] � [C 6HsOH] ( = 1 .2 X 1 0-10 M = LOX I O 'IO) [H3 0+] = Ka 1 .0 [ C 6HsO- ] Check assumption: percent error = ( 1 .2 x 1 0-10/ 1 .0) 1 00% = 1 .2 x 1 0-8%. The assumption is valid. pH = -log ( 1 .2 x 1 0-1°) = 9.9208 = 9.92 . Verify the pH using the Henderson-Hasselbalch equation: [ba�e] pH = pKa + 10g [acid] [ 1 .0] [C Hs pH = 1 0.00 + 1 0g 6 O'] = 1 0.00 + IOg [ 1 .2] [C 6HsOH]

]

( )

p H= 9.92

1 9. 1 3

( ) [ J

( )

The buffer components are phenol, NH3 , and ammonium ion, NH/. The chloride ions are ignored because they are not involved in the buffer. Calculate Kb from pKb and set up the problem with a reaction table. Kb = lO-pKb = 1 0-4 7 5 = 1 .7782794 X 1 0-5 (unrounded) NH 4+(aq) + OW(aq) Concentration (M) NH 3 (aq) + H 2 0(!) o 0. 1 0 Initial 0.20 +x +x -x Change x 0. 1 0 + x Equilibrium 0.20 - x +­ .....

26 9

Assume that x is negligible with respect to both 0.20 and 0.10 because both concentrations are much larger than Kb· Kb = lO-pKb = 10-4·75 = 1.7782794 X 10-5 (unrounded) [NH/][OH-] [0.10 + x][OH-] [O.IO][OH-] = 1.7782794 X 10-5 (unrounded) = = Kb = [NH3J [0.20] [0.20 - x]

tHJl]

( )

(1.7782794 x 10" 1 0.20 0.10

= 3.5565588 x 10-5 M(unrounded) NH/ Check assumption: percent error = (3.5565588 x 10-5/0.10)100% = 0.036%. The assumption is valid. pOH = -log [OHl = -log (3.5565588 x 10-5) = 4.448970 (unrounded) 14.00 = pH + pOH pH = 14.00 - pOH = 14.00 - 4.448970=9.551030 = 9.55 Verify the pH using the Henderson-Hasselbalch equation. To do this, you must find the pK. of the acid NH/: 14 pK. + pKb pK. = 14 - pKb = 14 - 4.75 = 9.25 [ba�e] pH = pK. + 10g [aCId]

[OH1- K.

-

=

pH = 9.25 + 10g

pH = 9.55 19.15

[NH3] [NH/]

=

9.25 + 10g

( ) [0.20] [0.10]

Given the pH and K. of an acid, the buffer-component ratio can be calculated from the Henderson-Hasselbalch equation. pKa = -log Ka = -log (1.3 x 10-5) = 4.8860566 (unrounded) [ba�e] pH=pK. +IOg [aCId]

( ) ( ) ( )

5.11 = 4.8860566 + 10g 0.2239434 = 10g [Pr·] [HPr] 19.17

( ) l )

=

[Pr·] [HPr]

[Pr·] [HPr]

1.674724 =

Raise each side to l OX

1.7

Determine the pK. of the acid from the concentrations of the conjugate acid and base, and the pH of the solution. This requires the Henderson-Hasselbalch equation. [ba�e] pH = pKa + 10g [aCId]

( J ( )

)

(

[k] [0.1500] = pKa + 10g [0.2000] [HA] 3.35 = pKa - 0.1249387 pKa = 3.474939 = 3.47 Determine the moles of conjugate acid (HA) and conjugate base (A-) using (M)(V) = moles Moles HA = (0.5000 L) (0.2000 mol HA/L) 0.1000 mol HA Moles A- = (0.5000 L) (0.1500 mol A-/L) = 0.07500 mol A3.35 = pKa + 10g

=

270

The reaction is: HA(aq) + NaOH(aq) � Na+(aq) + A-(aq) + H 2 0(l) Initial: 0.1000 mol 0.0015 mol 0.07500 mol Change: - 0.0015 mol + 0.0015 mol -0.0015 mol Final: 0.0985 mol o mol 0.0765 mol NaOH is the limiting reagent. The addition of 0.0015 mol NaOH produces an additional 0.0015 mol A- and consumes 0.0015 mol ofHA. Then 0.0765molk [A-] = =0.153MA0.5000 L 0.0985 mol HA = 0.197MHA [HA] = 0.5000 L

pH = pK

a

+ 10g

[k] ) ( [HA]

pH = 3A74938737 + IOg([ 0.153])

=3.365163942 =3.37

[ 0.197]

19.19

a) The hydrochloric acid will react with the sodium acetate, NaC2 H 302 , to form acetic acid, HC2 H 302 : HC I +NaC 2 H 30 2 � HC2 H 302 + NaCI Calculate the number of moles of HCI and NaC 2 H 302 . All of the HCI wil l be consumed to form HC 2 H 302 , and the number of moles of C 2 H 302- wil l decrease. OA42 mol HCI 10-3 L Initial moles HCI = ( (184 mL ) = 0.081328 mol HCI (unrounded) L I mL OAOO mol NaC2H.02 ( 0.500 L ) = 0.200 mol NaC 2 H 30 2 Initial moles NaC2 H 30 2 = L HCI + NaC 2 H 30 2 � HC 2 H 302 + NaCI Initial: 0.081328 mol 0.200 mol 0 mol + 0.081328 mol Change: -0.081328 mol -0.081328 mol Final: 0 mol 0.118672 mol 0.081328 mol

)( )

(

'

)

Total volume = 0.500 L + (184 mL) (1 0 3 L / I mL) = 0.684 L [HC 2 H 30 2 ] = 0.081328 mol = 0.118900584 M(unrounded) 0.684 L 0.118672 mol =0.173497076 M (unrounded) [C2 H 302 -] = 0.684 L pKa �Iog Ka = -log (1.8 10-5) =4.744727495 [C2H)02·] pH = pK + 10g [HC2H302] -

a

P

(

x

)

0.173497076 ]) =4.9088 = 4.91 [ 0.118900584]

H = 4.744727495 + 1 0g ([

b) The addition of base would increase the pH, so the new pH is (4.91 + 0.15) =5.06. The new [C 2 H 302-] / [ HC 2H302 ] ratio is calculated using the Henderson-Hasselbalch equation. [C2H302-] pH =pK + 10g [HC2H P2] [C2H)02· ] 5.06 =4.744727495 + 10g [HC2H302] a

(

(

)

)

271

[C 2 H 3°2- ] [ HC2H 302]

=

2.06676519 (unrounded)

From part (a), we know that [ H CzH 3 02] + [C2H 3 02-] (0.118900584 M + 0.173497076 M) 0.29239766 M. Although the ratio of [CZH 3 02 -] to [HC2H 3 02] can change when acid or base is added, the absolute amount does not change unless acetic acid or an acetate salt is added. Given that [CZH 3 02-] / [ HC2H 3 0Z] 2.06676519 and [ HCzH 3 0Z] + [C2H 3 02-] 0.29239766 M, solve for [CZH 3 0Z-] and substitute into the second equation. [CZH 3 0Z-] 2.06676519 [ HCzH 3 02] and [HCzH 3 0Z] + 2.06676519 [ H CzH 3 0Z] 0.29239766 M [HCzH3 0Z] = 0.095343999 M and [C2H 3 0Z-] 0.19705366 M. Moles of CzH 3 0z- needed = (0.19705366 mol CZH 3 02- / L) (0.500 L) 0.09852683 mol (unrounded) Moles of CzH 3 0z- initially = (0.173497076 mol C2H3 0Z- / L) (0.500 L) 0.086748538 mol (unrounded) This would require the addition of (0.09852683 mol 0.086748538 mol) 0.011778292 mol C2H 3 0Z- (unrounded) The KOH added reacts with HCzH 3 0Z to produce additional CZH 3 0Z-: HC2H 3 02 + KOH � C2H 3 0Z- + K+ + H20(!) To produce 0.011778292 mol CZH 3 0Z- would require the addition of 0.011778292 mol KOH. =

=

=

=

=

=

=

-

=

Mass KOH 19.21

=

( 0.011778292 mol KOH )

(

56.11 g KOH I mol KOH

)

=

=

= 0.660879964 = 0.66

g

KOH

Select conj ugate pairs with Ka values close to the desired [H3 0 +] . a) For pH "" 4.0, [ H 3 0 +] 10-40 1.0 X 10-4 M. The best selection is the HCOOH / H COO- conj ugate pair with Ka equal to 1.8 x 10-4. From the base list, the C 6HSNH2 / C 6 H SNH 3 + conj ugate pair comes close with Ka = 1.0 X 10-14/4.0 X 10-10 2.5 X IO-s. b) For pH"" 7.0, [HP+ ] 10-70 = 1.0 X 10-7 M. Two choices are the HZP04- / HPO/ - conj ugate pair with Ka of 6.3 x 10-8 and the HzAs04- / HAsO/- conjugate pair with Ka of l .l x 10-7. =

=

=

=

19.24

The value of the Ka from the appendix: Ka = 6.3 X 10-8 (We are using Kaz since we are dealing with the equi librium in which the second hydrogen ion is being lost). Determine the pKa using pKa = -log 6.3 x 10-8 7.200659451 (unrounded) U se the H enderson-Hasselbalch equation: 2 [ HPO4 -] pH pKa + log [ H 2P04-]

(

=

7.40 = 7.200659451 + 10g 0.19934055 = IOg

2 [HPO4 - ] -=------,'--...::. [ H2P04- ] 19.26

=

(

] (

=

[ H PO/ ]

]

[ H2P04- ] 2 [ H P04 -]

[ H2P04-]

1.582486

=

]

1.6

To see a distinct color in a mixture of two colors, you need one color to be about 10 times the intensity of the other. For thi s to take p lace, the concentration ratio [ HIn] / [In-] needs to be greater than 10: I or less than I: 10. This wil l occur when pH pKa - I or pH pKa + 1, respectively, giving a transition range of about two units. =

=

272

19.28

The equivalence point in a titration is the point at which the number of moles of O� equals the number of moles of H 30+ (be sure to account for stoichiometric ratios, e.g., I mol of Ca(OH)2 produces 2 moles of OH-). The endpoint is the point at which the added indicator changes color. If an appropriate indicator is selected, the endpoint is close to the equivalence point, but not normally the same. Using an indicator that changes color at a pH after the equivalence point means the equivalence point is reached first. However, if an indicator is selected that changes color at a pH before the equivalence point, then the endpoint is reached first.

19 . 3 0

a) The initial pH is lowest for flask solution of the strong acid, followed by the weak acid and then the weak base. In other words, strong acid - strong base < weak acid - strong base < strong acid -weak base. b) At the equivalence point, the moles of H 3 0+ equal the moles of OH-, regardless of the type of titration. However, the strong acid - strong base equivalence point occurs at pH 7 . 00 because the resulting cation-anion combination does not react with water. An example is the reaction NaOH + HCI � H20 + NaCI. Neither Na+ nor cr ions dissociate in water. The weak acid - strong base equivalence point occurs at pH> 7, because the anion of the weak acid is weakly basic, whereas the cation of the strong base does not react with water. An example is the reaction HCOOH + NaOH � HCOO- + H20 + Na+. The conj ugate base, HCOO-, reacts with water according to this reaction: HCOO- + H20 � HCOOH + OW. The strong acid - weak base equivalence point occurs at pH < 7, because the anion of the strong acid does not react with water, whereas the cation of the weak base is weakly acidic. An example is the reaction HCI + NH3 � N H/ + cr. The conj ugate acid, N H/, dissociates slightly in water: N H/ + H20 � NH3 + H 3 0+. In rank order of pH of equivalence point, strong acid weak base < strong acid - strong base < weak acid - strong base. =

-

19.32

At the very center of the buffer region of a weak acid-strong base titration, the concentration of the weak acid and its conj ugate base are equal, which means that at this point the pH of the solution equals the pKa of the weak acid.

19 . 3 3

Indicators have a pH range that is approximated by pKa ± I . The pKa of cresol red is -log ( 5 . 0 x indicator changes color over an approximate range of 7.3 to 9.3.

19. 3 5

Choose an indicator that changes color at a pH close to the pH of the equivalence point. a) The equivalence point for a strong acid-strong base titration occurs at pH 7 . 0 . Bromthymol blue is an indicator that changes color around pH 7 . b) The equivalence point for a weak acid-strong base is above pH 7 . Estimate the pH at equivalence point from equilibrium calculations. At the equivalence point, all of the HCOOH and NaOH have been consumed; the solution is 0.050 M HCOO-. (The volume doubles because equal volumes of base and acid are required to reach the equivalence point. When the volume doubles, the concentration is halved.) The weak base HCOO- undergoes a base reaction: Concentration, M COOW( aq ) + H20(l) !:; :=COOH(aq) + OH-( aq) Initial : 0.050 M O O +x +x -x Change: x x Equilibrium: 0.050 - x The Ka for HCOOH is 1 .8 x 1 0- 4 , so Kb 1 . 0 X 10-1 4 / 1 . 8 x 10- 4 5 . 5 5 5 6 X 1 0-11 (unrounded) HCOOH ] [OH- ] [x ][x ] 5 . 5 5 5 6 X 10-11 [x ][x ] = Kb [ [H C OO - ] [0.050 - x ] [0.050 ] [OW] x 1 .666673 x 1 0- 6 M pOH -log ( 1 . 666673 x 1 0 6) 5 . 7 7 8 1 496 (unrounded) pH 1 4 .00 - pOH 1 4 .00 - 5 . 7 7 8 1 496 8.22 1 8504 8 . 22 Choose thymol blue or phenolp hthalein . =

=

=

=

=

=

=

=

=

=

-

=

=

=

273

=

1 0-9 ) = 8.3, so the

1 9.37

The reaction occurring in the titration is the neutralization of H 30+ (from HCI) by OH- (from NaOH ): HCI(aq) + NaOH(aq) � H 2 0(l) + NaCI(aq) or, omitting spectator ions: H 30+(aq) + Of1(aq) � 2 H 20(1) For the titration of a strong acid with a strong base, the pH before the equivalence point depends on the excess concentration of acid and the pH after the equivalence point depends on the excess concentration of base. At the equivalence point, there is not an excess of either acid or base so the pH is 7.0. The equivalence point occurs when 50.00 mL of base has been added. Use (M)(V) to determine the number of moles. The initial number of moles of HCI (O.LOOO mol HCI / L) ( 1 0-3 LI J mL) (50.00 mL) 5.000 x 1 0- 3 mol HCI a) At 0 mL of base added, the concentration of hydronium ion equals the original concentration ofHCI. pH -log (0.1000 M) = 1.0000 b) Determine the moles ofNaOH added: Moles ofNaOH (0. 1 000 mol NaOH / L) ( 1 0-3 L / I mL) (25 .00 mL) 2.500 x 1 0-3 mol NaOH HCI(aq) + NaOH(aq) � H 2 0(1) + NaCI(aq) Initial: 5 .000 x 1 0-3 mol 2.500 x 1 0-3 mol 0 + 2.500 x 1 0-3 mol Change: - 2.500 x 1 0-3 mol -2.500 x 1 0-3 mol 2.500 x LO 3 mol 0 Final: 2.500 x 1 0- 3 mol 3 The volume of the solution at this point is [(50.00 + 25 .00) mL] ( 1 0- L / J mL) 0.07500 L The molarity of the excess HCI is (2.500 x 1 0-3 mol HCI) / (0.07500 L) 0.03333 M(unrounded) pH -log (0.03333) 1 .4772 (Note that theNaCI product is a neutral salt that does not affect the pH ). c) Determine the moles ofNaOH added: Moles ofNaOH (0. 1 000 mol NaOH / L) ( 1 0-3 L / I mL) (49.00 mL) 4.900 x 1 0-3 mol NaOH HCI(aq) + NaOH(aq) � H 2 0(l) + NaCI(aq) Initial: 5 .000 x 1 0-3 mol 4.900 x 1 0-3 mol 0 + 4.900 x 1 0-3 mol Change: - 4.900 x 1 0-3 mol -4.900 x 1 0-3 mol J .000 x 1 0- 4 mol Final: 0 4.900 x 1 0-3 mol The volume of the solution at this point is [(50.00 + 49.00) mL] ( 1 0-3 L / I mL) 0.09900 L The molarity of the excess HCI is ( 1 .00 x 1 0-4 mol HCI) / (0.09900 L) 0.00 1 0 1 M (unrounded) pH -log (0.00 1 0 1 ) = 2.996 d) Determine the moles ofNaOH added: Moles ofNaOH (0. 1 000 mol NaOH / L) ( 1 0- 3 L / I mL) (49.90 mL) 4.990 x 1 0-3 mol NaOH HCI(aq) + NaOH(aq) � H 2 0(l) + NaCI(aq) 0 5 .000 x 1 0- 3 mol 4.990 x 1 0-3 mol Initial: Change: - 4.990 10- 3 mol -4.990 10-3 mol + 4.990 10-3 mol Final: 1.000 x 1 0-5 mol 0 4.900 x 10-3 mol =

=

=

=

=

=

=

=

=

=

=

=

=

=

=

x

=

x

x

The volume of the solution at this point is [(50.00 + 49.90) mL] ( 1 0-3 L / J mL) 0.09990 L The molarity of the excess HCI is ( l .0 x 1 0-5 mol HCI) / (0.09990 L) 0.000 1 00L M(unrounded) pH -log (0.000 1 00 1 ) 4.0000 e) Determine the moles ofNaOH added: Moles ofNaOH (0. 1 000 mol NaOH / L) ( 1 0-3 L / 1 mL) (50.00 mL) 5.000 x 1 0-3 mol NaOH . HCI(aq) + NaOH(aq) � H 2 0(1) + NaCI(aq) Initial: 5 .000 J 0-3 mol 5.000 x 1 0-3 mol 0 Change: - 5.000 1 0-3 mol -5.000 x 1 0-3 mol + 5.000 x 1 0-3 mol Final: o 0 5.000 x 1 0-3 mol The NaOH will react with an equal amount of the acid and 0.0 mol HCI wil l remain. This is the equivalence point of a strong acid-strong base titration, thus, the pH is 7.00. Only the neutral saltNaCI is in solution at the equivalence point. t) The NaOH is now in excess. It will be necessary to calculate the excess base after reacting with the HCI. The excess strong base wil l give the pOH, which can be converted to the pH. Determine the moles ofNaOH added: Moles ofNaOH (0. 1 000 mol NaOH / L) ( 1 0-3 L / J mL) (50. 1 0 mL) 5 .0 1 0 x 1 0-3 mol NaOH The HCI will react with an equal amount of the base, and 1 .0 x 1 0-5 mol NaOH will remain. =

=

=

=

=

=

=

=

x

X

274

HCI(aq) 3 + NaOH(aq)3 � H 2 0(l) + NaCI(aq) Initial: 5 .000 x 1 0- mol 5 .0 1 0 x 1 0- mol 0 3 3 Change: - 5 . 000 x 1 0-3 mol -5 .000 x 1 0- mol + 5 . 000 x 1 0- mol 5 3 Final: 0 1 . 000 x 1 0 mol 5 . 000 x 1 0 mol The volume of the solution at this point is [(50.00 + 50. 1 0) mL] ( 1 0-3 L / 1 mL) = 0 . 1 0 1 0 L The molarity of the excessNaOH is ( 1 .0 x 1 0-5 mol NaOH ) / (0 . 1 0 1 0 L) = 0.0000990 1 M (unrounded) pOH = -log (0. 0000990 1 ) = 4.0043 (unrounded) pH = 14 .00 - pOH = 1 4.00 - 4.0043 = 9.9957 = 1 0.00 g) Determine the moles ofNaOH added: Moles ofNaOH = (0 . 1 000 mol NaOH / L) ( 1 0-3 L / 1 mL) (60.00 mL)3 = 6 . 000 1 0-3 mol NaOH The HCl will react with an equal amount of the base, and 1 .000 x 1 0- molNaOH will remain. HCI(aq) 3 + NaOH(aq) 3 � H 2 0(l) + NaCI(aq) Initial: 5 .000 x 1 0- mol 6.000 x 1 0- mol 0 3 3 Change: - 5 . 000 x 10-3 mol -5 .000 x 1 0- mol +5 .000 x 1 0- mol 3 3 Final: 0 1 .000 x 1 0- mol 5 . 000 x 1 0- mol The volume of the solution at this point is [(50.003 + 60.00) mL] ( 1 0-3 L / 1 mL) = 0 . 1 1 00 L The molarity of the excess NaOH is ( 1 .000 x 1 0- mol NaOH) / (0. 1 1 00 L) = 0. 0090909 M (unrounded) pOH = -log (0 .0090909) = 2.04 1393 (unrounded) pH = 1 4 .00 - pOH = 1 4 .00 - 2.04 1 393 = 1 1 .958607 = 1 1 .96 X

1 9. 3 9

This is a titration between a weak acid and a strong base. The pH before addition of the base is dependent on the of the acid (labeled HBut). Prior to reaching the equivalence point, the added base reacts with the acid to form butanoate ion (labeled Bun. The equivalence point occurs when 20.00 mL of base is added to the acid because at this point, moles acid = moles base. Addition of base beyond the equivalence point is simply the addition of excess OH - . The initial number of moles of HBut = (M)(V)3 = 3 (0 . 1 000 mol HBut i L) ( 1 0- L / 1 mL) (20.00 mL) = 2.000 1 0- mol HBut a) At 0 mL of base added, the concentration of [H 3 0+] is dependent on the dissociation of butanoic acid: Hbut + H 20 BuC !:; H 3 0+ + 0 Initial: 0. 1 00 M +x Change: -x x Equilibrium: 0. 1 00 - x [ H 3 0 + [BuC ____ x2 x2 = __ = 1 .54 1 0-5 Ka = 0. 1 000 - x 0. 1 000 [ H But ] x = [H 3 0+] = 1 .2409673 1 0-3 M (unrounded) pH = -log [H 3 0+] = -log ( 1 . 2409673 x 1 0- 3 ) = 2.9062 = 2.91 b) Determine the moles ofNaOH added: Moles ofNaOH = (0. 1 000 mol NaOH / L) ( 1 0-3 L / 1 mL) ( 1 0.00 mL) = 1 .000 1 0-3 molNaOH The NaOH will react with an equal amount of the acid, and 1 .000 x 1 0-3 mol HBut will remain. An equal number of moles ofBuC will form. HBut(aq)3 + NaOH(aq) � H 2 0(l) + BuC(aq) + Na+(aq) Initial: 2 . 000 x 10- mol 1 .000 x 1 0-3 mol 0 3 + 1 . 000 x 1 0-3 mol - 1 .000 x 1 0- mol Change: - 1 . 000 x 1 0-3 mol Final: 1 .000 x 1 0-3 mol 0 1 . 000 x 1 0-3 mol 3 The volume of the solution at this point is [(20.00 + 1 0.00) mL] ( 1 0- L / 1 mL) = 0.03 000 L The molarity of the excess HBut is ( 1 . 000 x 10-33 mol HBut) / (0.03 000 L) = 0.03 3 3 3 M (unrounded) The molarity of the BuC formed is ( 1 .000 x 1 0- mol Bun / (0.03000 L) = 0.03 3 3 3 M (unrounded) Using a reaction table for the equilibrium reaction of HBut: BuC HBut + H20 !:; H 3 0+ + 0.03333 M Initial: 0.03 3 3 3 M +x Change: -x 0.03 3 3 3 + x Equilibrium: 0.03 3 3 3 - x Ka

X

J

]

X

X

X

275

K

a

=

[H30+ J[BuC ] [HBut]

=

x ( 0.0333 + x) 0.03333 - x

=

x ( 0.03333) 0.03333

=

l.54 x 10-5

x [H 0+] l.54 X 10-5M(unrounded) 3 pH -log [H 0+] -log (1.54 x 10-5) 4.812479 4.8 1 3 c) Determine the moles ofNaOH added: Moles ofNaOH = (0.1000 molNaOH /L) (10-3 L /1 mL) (15.00 mL) l.500 x 10-3 molNaOH TheNaOH will react with an equal amount of the acid, and 5.00 x 10-4 molHBut will remain, and l.500 x 10-3 moles of BuC will form. + BuC (aq) + Na+(aq) + HBut(aq) -7 H20(!) NaOH(aq) l.500 x 10-3 mol Initial: 2.000 x 10-3 mol 0 +1.500 x 10-3 mol -l.500 x 10-3 mol Change: - l.500 x 10-3 mol Final: 5.000 x 10-4 mol 0 l.500 x 10-3 mol The volume of the solution at this p oint is [ (20.00 + 15.00) mL] (10-3 L /1 mL) 0.03500 L The molarity of the excessHBut is (5.00 x 10-4 mol HBut) / (0.03500 L) 0.0142857M(unrounded) The molarity of the BuC formed is (1.500 X 10-3 mol Bun / (0.03500 L) 0.0428571M (unrounded) Using a reaction table for the equilibrium reaction ofHBut: BuC HBut + H20 !:; H 0+ + 3 Initial: 0.0142857 M 0.0428571 M Change: -x +x Equilibrium: 0.0142857 - x 0.0428571 + x =

=

=

=

=

=

=

=

=

=

K

a

[H 0+ J [ But- ]

x ( 0.0428571 + x)

3

=

x ( 0.0428571)

0.0142857 - x

[HBut]

0.0142857

=

1.54

X

10-5

x = [H 0+] 5.1333 X 10-6M(unrounded) 3 pH -log [H 0+] = -log (5.1333 x 10-6) 5.2896 5.29 3 d) Determine the moles ofNaOH added: Moles ofNaOH = (0.1000 mol NaOH /L) (10-3 L /1 mL) (19.00 mL) 1.900 x 10-3 molNaOH TheNaOH will react with an equal amount of the acid, and l.00 x 10-4 molHBut will remain, and 1.900 x 10-3 moles of BuC will form. HBut(aq) + NaOH(aq) + BuC(aq) + Na+(aq) -7 H20(!) Initial: 2.000 x 10-3 mol l.900 x 10-3 mol 0 Change: -1.900 x 10-3 mol - 1.900 x 10-3 mol +l.900 x 10-3 mol Final: l.000 x 10-4 mol 0 l.900 x 10 3 mol The volume of the solution at this point is [ (20.00 + 19.00) mL] (10-3 L /1 mL) 0.03900 L The molarity of the excessHBut is (l.00 x 10-4 molHBut) / (0.03900 L) 0.0025641 M(unrounded) The molarity of the BuC formed is (l.900 X 10-3 mol Bun / (0.03900 L) 0.0487179M (unrounded) Using a reaction table for the equilibrium reaction ofHBut: HBut + H20 !:; H 0+ + BuC 3 0.0025641 M Initial: 0.0487179 M Change: -x +x Equilibrium: 0.0025641 - x 0.0487179 + x =

=

=

=

=

=

=

=

Ka

[H30+ J[BuC ]

x ( 0.0487179 + x)

x ( 0.0487179)

l.54 x 10-5 0.0025641 - x 0.0025641 x [H 0+] 8.1052631 x 10-7M (unrounded) 3 pH -log [H 0+] -log (8.1052631 x 10-7) 6.09123 6.09 3 e) Determine the moles ofNaOH added: Moles ofNaOH (0.1000 molNaOH /L) (10-3 Lll mL) (19.95 mL) 1.995 x 10-3 molNaOH TheNaOH will react with an equal amount of the acid, and 5 x 10-6 mol HBut will remain, and 1.995 x 10-3 moles of BuC will form. =

[HBut]

=

=

=

=

=

=

=

=

=

=

=

276

H But(aq) + NaOH(aq) � H20(l) + B uC(aq) + N a+(aq) 3 3 Initial : 2 . 000 x 1 0- mol 1 .995 x 1 0- mol 0 3 3 3 Change: - 1 . 995 x 1 0- mol - 1 .995 x 1 0- mol + 1 . 995 x 1 0- mol 6 3 F inal : 5 .000 x 1 0- mol 0 1 .995 x 1 0- mol 3 The volume of the solution at this point is [(20.00 + 1 9.95) mL] ( 1 0- L / 1 mL) = 0.03 995 L 6 The molarity of the excess H B ut i s (5 x 1 0- mol HBut) / (0.03995 L) = 0.000 1 25 1 56 M(unrounded) 3 The molarity of the B ue formed is ( 1 .995 x 1 0- mol Bue) / (0.03995 L) = 0.0499374 M(unrounded) Using a reaction table for the equ i li brium reaction of H But: H B ut + H20 =+ H 3 0+ + Bue Initial : 0.000 1 25 1 5 6 M 0.0499374 M Change: -x +x 0.04993 74 + x Equil i brium: 0.000 1 25 15 6 - x + BuC H 30 x ( 0.04993 74 + x ) x ( 0.04993 74 ) -5 Ka = = = = 1 .54 xlO 0.000 1 25 1 5 6 0.000 1 25 1 5 6 - x [ H B ut ] 8 + x = [ H 3 0 ] = 3 .85963 7 X 1 0- M(unrounded) pH = -log [H 3 0 + ] = -log (3 .859637 x 1 0-8 ) = 7.4 1 345 = 7.41 t) Determine the moles ofNaOH added : 3 3 Moles of NaOH = (0. 1 000 mol NaOH / L) ( 1 0- L / 1 mL) (20.00 mL) = 2 . 000 x 1 0- mol NaOH The NaOH will react with an equal amount of the acid, and 0 mol H B ut w i l l remain, and 3 2 . 000 x 1 0- moles of Bue w i l l form. This is the equivalence point. HBut(aq) + N aOH(aq) � H20(l) + B uC(aq) + Na+(aq) 3 3 Initial : 2.000 x 1 0- mol 2.000 x 1 0- mol o 3 3 3 Change: - 2 . 000 x 1 0- mol -2 .000 x 1 0- mol +2 . 000 x 1 0- mol 3 Final : o 0 2 . 000 X 1 0- mol The Kb of B ue is now i mportant. 3 The volume of the solution at this point is [(20.00 + 20.00) mL] ( 1 0- L / 1 mL) = 0.04000 L 3 The molarity of the B ue formed i s (2 .000 x 1 0- mol Bue) / (0.04000 L) = 0.05000 M(unrounded) 0 Kb = Kw / Ka = ( 1 . 0 x 1 0-14) / ( 1 .54 x 1 0-5 ) = 6.493 5 x 1 0-1 (unrounded) U s ing a reaction table for the equilibrium reaction of But-: B ue + H20 =+ H B ut + OW Initial : 0.05000 M O O +x -x +x Change: x x Equil ibrium : 0.05 000 - x

[

_

Kb -

][

[ HBut ][ OH- ]

[ B uC ]

]

_

[ x ][ x ]

- -;:----=--=-=-----=--:;--

[ 0.05000 - x ]

[ x ][ x ]

[ 0.05 000 ]

= 6.493 5

X

1 0-1

0

6 [OW] = x = 5 .698025 9 X 1 0- M 6 pOH = -log (5 .698025 9 x 1 0- ) = 5 .244275575 (unrounded) pH = 1 4.00 - pOH = 1 4 .00 - 5 . 2442 7 5 5 7 5 = 8.755724425 = 8 .76 g) After the equivalence point, the excess strong base is the primary factor i nfluencing the pH. Determine the moles o f N aOH added : 3 3 Moles ofNaOH = (0. 1 000 mol NaOH / L) ( 1 0- L / 1 mL) (20.05 mL) = 2 .005 X 1 0- mol NaOH 6 The NaOH will react with an equal amount of the acid, 0 mol HBut w i l l remain, and 5 x 1 0- moles of 3 NaOH w i l l be in excess. There will be 2 . 000 x 1 0- mol of B ue produced, but thi s weak base wil l not affect the pH compared to the excess strong base, NaOH . H But(aq) + NaOH(aq) � H20(l) + B uC(aq) + Na+(aq) 3 3 Initial : 2 . 000 x 1 0- mol 2 .005 x 1 0- mol 0 3 3 3 +2 . 000 x 1 0- mol -2.000 x 1 0- mol - 2 . 000 x 1 0- mol Change: 3 6 2 .000 x 1 0 mol 5 . 000 x 1 0- mol 0 F inal : 3 The volume of the solution at this point is [(20.00 + 20.05) mL] ( 10- L / 1 mL) = 0.04005 L 6 The molarity of the excess OH- i s (5 x 1 0- mol OW) / (0.04005 L) = 1 .2484 X 1 0-4 M(unrounded) pOH = -log ( 1 .2484 x 1 0-4) = 3 .9036 (unrounded) pH = 1 4.00 - pOH = 1 4.00 - 3 .9036 = 1 0.0964 = 1 0. 1 0

277

h) Determine the moles ofNaOH added: Moles ofNaOH = (0. 1 000 molNaOH / L) ( 1 0-3 L / I mL) (25.00 mL) = 2.500 x 1 0-3 molNaOH The NaOH will react with an equal amount of the acid, 0 mol HBut will remain, and 5.00 x 1 0-4 moles ofNaOH will be in excess. H 20(l) + BuC(aq) + Na+(aq) � NaOH(aq) + HBut(aq) 0 2.500 x 10-3 mol 2.000 x 1 0-3 mol Initial: +2.000 x 1 0-3 mol -2.000 x 1 0-3 mol - 2.000 x 1 0- 3 mol Change: Final: 0 5.000 x 1 0-4 mol 2.000 x 1 0-3 mol 3 The volume of the solution at this point is [(20.00 + 25.00) mL] ( 1 0- L / 1 mL) 0.04500 L The molarity of the excess O H- is (5.00 x 1 0-4 mol OH-) / (0.04500 L) = 1 .1 1 1 1 1 0-2 M(unrounded) pOH = -log ( 1 . 1 1 1 1 x 1 0-2) = 1 .9542 (unrounded) pH = 1 4.00 - pOH = 1 4.00 - 1 .9542 = 1 2.0458 = 1 2.05 =

X

1 9.4 1

a) The balanced chemjcal equation is: NaOH(aq) + CH 3 COOH(aq) �Na\aq) + CH 3 COO-(aq) + H20(l) The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume ofNaOH needed: Volume = l mL L 0.0520 mOICH 3 COOH I O - 3 L ( 42.2 ) I moINaOH mL 1 mol CH 3 COOH 0.0372 mol NaOH 1 0- 3 L L 1 mL = 5 8.989247 = 59.0 mL NaOH Determine the moles of CH3 COOH present: 0.0520 mOl CH 3 COOH 0 42.2 mL ) Moles = L = 0.0021 944 mol CH 3 COOH (unrounded) At the equivalence point, 0.0021 944 mol NaOH will be added so the moles acid = moles base. The NaOH will react with an equal amount of the acid, 0 mol CH 3 COOH will remain, and 0.0021 944 moles of CH 3 COO- will be formed. CH 3 COOH(aq) + + CH 3 COO-(aq) + Na+(aq) NaOH(aq) 0.0021 944 mol Initial: 0.002 1 944 mol o Change: -0.0021 944 mol - 0.0021 944 mol +0.0021 944 mol Final: o o 0.0021944 mol Determine the liters of solution present at the equivalence point: Volume = [(42.0 + 5 8.989247) mL] ( 1 0-3 L / 1 mL) = 0.1 00989 L (unrounded) Concentration of CH 3 COO- at equivalence point: Molarity (0.0021 944 mol CH 3 COO-) / (0. 1 00989 L) = 0.02 1 729 M (unrounded) Calculate Kb for CH 3 COO-: Ka CH 3 COOH = 1 .8 x 1 0-5 Kb = Kw / Ka = ( 1 .0 x 1 0-1 4 ) / ( 1 .8 x 1 0-5) 5.556 x 1 0-10 (unrounded) Using a reaction table for the equilibrium reaction of CH 3 COO-: CH 3 COO+ H 20 !::; CH 3 COOH + mr O O 0.021 729 M Initial: Change: -x +x +x Equilibrium: 0.021 729 - x x x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH. [CH 3 COOH J [ OH - ] [ x ][ x ] [ x ][ x ] Kb = 5.556 x lO-10 729 x ] [ 0.021 729 ] [ 0.021 [ CH 3 COO- ] [OW] = x = 3.4745693 x 1 0-
(

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278

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b) The balanced chemical equation is: Na(aq) + HN02 (aq) � Naaq) +N02-(aq) + H 20(1) The sodium ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume ofNaOH needed: Volume 0.0390 mol HN0 2 10- 3 L L I mL ( 23.4 mL) 1 mol NaOH 0.0372 mol NaOH 10- 3 L I mol HN0 2 1 mL L 24.532 = 24.5 mL NaOH Determine the moles of HN0 2 present: 0.0390 mOI HN0 2 -10- 3 L Mo I es = ( 23.4 mL ) L 1 mL 0.0009126 mol HN02 At the equivalence point, 0.0009126 mol NaOH will be added so the moles acid moles base. The NaOH will react with an equal amount of the acid, 0 mol HN02 wil l remain, and 0.0009126 moles ofN02 wil l be formed. + HN0 2 (aq) � NaOH(aq) H 2 0(1) + N02-(aq) + Na+(aq) Initial: 0.0009126 mol 0.0009126 mol o Change: - 0.0009126 mol +0.0009 1 26 mol -0.0009126 mol Final: o 0.0009 1 26mol o Determine the liters of sol ution present at the equivalence point: Volume [(23.4 + 24.532) mL] (10- 3 L / 1 mL) 0.047932 L ( unrounded) Concentration ofN02- at equivalence point: Molarity = (0.0009126 mol N02 -) / (0.047932 L) 0.0190395 M (unrowlded) Calculate Kb forN02-: Ka HN0 2 7.1 x 10-4 Kb Kw / Ka (1.0 x 10-1 4 ) / (7.1 x 10-4 ) 1.40845 x 10-11 (unrounded) Using a reaction table for the equilibrium reaction ofN02-: N02+ H 20 ow !:; HN02 + 0.0190395 M Initial: o 0 Change: +x -x +x x x Equilibrium: 0.0190395 - x Determine the hydroxide ion concentration from the Kb, and then determine the pH from the pOH . [ HN0 2 ] [ OH - ] [x] [x] [x] [ x] = I .40845 x 10-11 Kb [ 0.0190395 - x ] [ 0.0190395 ] [N0 2 - ] [011] x 5 .178431 10- 7 M(unrounded) pOH -log (5.178431 x j 0- 7 ) 6.285802 (unrounded) pH 14.00 - pOH 14.00 - 6.285802 7.714198 7.71

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19.42

---c:-

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a) The balanced chemical equation is: HCI(aq) +NH 3(aq) � NH/ (aq) + Cr(aq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCI needed: Volume 0.234 mol N H 3 1 10- 3 L L I mL mol HCI (55.5 mL) I 3 I mol NH ) 1 mL L 3 0.135 mol HCI 10- L

(

=

=

96 2 mLHCI

[ )

(

.

( :01 )[ \O:t } 55.5 mL)

Determine the moles ofNH 3 present: 0.234 NH 3 Moles = =

0.01298711101 NH 3 (unrounded)

279

J

(

[ ) J

At the equivalence point, 0.0 1 2987 mol HCI will be added so the moles acid = moles base. The H CI will react with an equal amount of the base, 0 mol NH 3 will remain, and 0.0 1 2987 moles ofNH/ will be formed. NH/ (aq) + Creaq) -----7 + NH 3 (aq) HCI(aq) o 0.0 1 2987 mol 0.0 1 2987 mol Initial : -0.0 1 2987 mol +0.0 1 2987 mol - 0.0 1 2987 mol Change: 0.0 1 2987 mol o o Final: Determine the liters of solution present at the equivalence point: Volume = [(55.5 + 96.2) mL] ( 1 0- 3 L / I mL) = 0. 1 5 1 7 L Concentration ofNH/ at equivalence point: Molarity = (0.0 1 2987 mol NH/) / (0. 1 5 1 7 L) = 0.0856098 M (unrounded) Calculate Ka forNH/: KbNH 3 = 1 .76 1 0-5 1. x 1 0-5) = 5.68 1 8 x 1 0-10 (unrounded) Ka = Kw/ Kb = ( 1 .0 x 1 0- 14) / ( 76 Using a reaction table for the equilibrium reaction ofNH/ : NH/ + H2 0 � NH 3 + 0 0.0856098 M Initial : +x +x -x Change: x x Equilibrium: 0.0856098 - x Determine the hydrogen ion concentration from the K., and then determine the pH. H 3 0+ J[NH 3 ] [ Hx] [x] [x] = 5.68 1 8 x 1 0-10 = -=---� --"-----] -=Ka = [ 0.0856098 ] [ 0.0856098 - x NH/ J X

[

[

X ..::.::.

x = [H 3 0+] = 6.974366 1 0-<; M (unrounded) pH = -log [H 3 0+] = -log (6.974366 x 1 0-<;) = 5 . 1 565 = 5. 1 6 b ) The balanced chemical equation is: HCI (aq) + CH 3N H 2 (aq) -----7 CH 3N H/ (aq) + Creaq) The chloride ions on the product side are written as separate species because they have no effect on the pH of the solution. Calculate the volume of HCI needed: Volume = 1 . l l moI CH 3NH 2 1 0- 3 L ( 1 7.8 mL) l mol HCI L l mL I mol CH 3NH 2 0. 1 35 mol HCI 1 0 - 3 L L I mL = 1 46.3556 = 1 46 mL RCt Determine the moles of CH 3N H 2 present: ·l l mOl CH 3NH 2 1 7.8 mL) Moles = L = 0.0 1 9758 mol CH 3NH 2 (unrounded) At the equivalence point, 0.0 1 9758 mol HCI will be added so the moles acid = moles base. The HCI will react with an equal amount of the base, 0 mol CH3N H 2 wil l remain, and 0.0 1 9758 moles of CH 3NH 3 + will be formed. + CH 3NH 2 (aq) HCI(aq) CH 3N H/ (aq) + Creaq) Initial: 0.0 1 9758 mol 0.01 9758 mol o Change: - 0.0 1 9758 mol -0.0 1 9758 mol +0.0 1 9758 mol Final : o o 0.0 1 9758 mol Determine the liters of solution present at the equivalence point: Volume = [( 1 7.8 + 1 46.3556) mL] ( 1 0-3 L l l mL) = 0. 1 64 1 556 L (unrounded) Concentration of CH 3N H/ at equivalence point: Molarity = (0.0 1 9758 mol CH 3NH/) / (0. 1 6 1 556 L) = 0. 1 22298 M (unrounded) Calculate Ka for CH 3NH 3 +: Kb CH 3NH 2 = 4.4 x 1 0-4 Ka = Kw/ Kb = ( 1 .0 x 1 0-1 4) / (4.4 x 1 0-4) = 2.2727 10-1 1 (unrounded) Using a reaction table for the equilibrium reaction of CH 3N H 3 +: X

(

)[ ]

C

l

J

)[ \O�� }

X

280

(

J

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CH3NH3 + + H 20 !::; CH 3NH 2 + H 30+ 0. 1 22298 M O O Initial: Change: +x -x +x x x Equilibrium: 0. 1 22298 - x Determine the hydrogen ion concentration from the Ka, and then determine the pH. [ H30 ] [CH3NH2J [ x] [ x] [ x] [ x] = Ka = 2.2727 x 1 0-1 1 [0. 1 22298 - x] [0. 1 22298] [CH3NH/] x = [H30+] = 1 .667 1 7 x 1 0-6 M (unrounded) pH = -log [H30+] = -log ( 1 .667 1 7 x 1 0-6) = 5.7780 1 9 = 5.78

+

_

19.44

Fluoride ion in CaF2 is the conjugate base of the weak acid H F. The base hydrolysis reaction of fluoride ion F-(aq) + H 2 0(l) !::; HF(aq) + OI1(aq) therefore is influenced by the pH of the solution. As the pH increases, [OH-] increases and the equilibrium shifts to the left to decrease [OH-] and increase the [F-]. As the pH decreases, [OH-] decreases and the equilibrium shifts to the right to increase [OH-] and decrease [F-]. The changes in [F-] influence the solubility of CaF2. Chloride ion is the conjugate base of a strong acid so it does not react with water. Thus, its concentration is not influenced by pH, and solubility of CaCI2 does not change with pH.

1 9.45

Consider the reaction AB(s)!::; A+(aq) + 8-(aq), where Qsp = [A+] [8-]. If Qsp> Ksp, then there are more ions dissolved than expected at equilibrium, and the equilibrium shifts to the left and the compound AB precipitates. The excess ions precipitate as solid from the solution.

19.46

a) Ag2 C03(S)!::; 2 Ag+(aq) + CO/-(aq) lon-product expression: Ksp [Ag+]2[CO/-] b) BaF2 (s) !::; 8a2+(aq) + 2 F-(aq) Ion-product expression: Ksp [Ba2+][V"]2 c) CuS(s) + H 2 0(l)!::; Cu2+(aq) + HS-(aq) + OI1(aq) Ion-product expression: Ksp [Cu2+][HS-][OH-] =

=

19.48

=

Write a reaction table, where S is the molar solubility of Ag2C03: Concentration (M) Ag2 C03(S)!::; 2 Ag\aq) + C032-(aq) 0 0 Initial +S +2 S Change S 2S Equilibrium S = [Ag2 C03] = 0.032 M so [Ag+] = 2 S = 0.064 M and [C032-] = S = 0.032 M -4 2 2 Ksp = [Ag+] [CO/-] = (0.064) (0.032) = 1 .3 1 072 x 1 0-4 = 1.3 10 X

1 9.50

The equation and ion-product expression for silver dichromate, Ag2 Cr2 0 7 , is: AgzCrZ0 7 (s) !:; 2 Ag+(aq) + CrzO/-(aq) Ksp = [Ag+f[CrzOl-] The solubility of Ag2 CrZ07 , converted from gil 00 mL to M is: 3 1 mol Ag 2Cr207 = 0.000 1 922 1 86 1 M (unrounded) 8.3 1 0- g Ag 2Cr207 1 mL Molar solubility = S = 3 1 00 mL 1 0 - L 431 .8 g Ag 2Cr20 7 Since I mol of Ag2 CrZ07 dissociates to form 2 moles of Ag+, the concentration of Ag+ is 2 S = 2(0.000 1 9221861 M) = 0.00038443723 M (unrounded). The concentration of CrzOl- is S = 0.000 1 922 1 86 1 M because I mol of Ag2 CrZ0 7 dissociates to form 1 mol of Cr2 0 72-. z 2 2 Ksp = [Ag+] [CrzOl-] = (2 S) (S) = (0.00038443723) (0.000 1 922 1 86 1 ) = 2.8408 1 0-1 1 = 2.8 to-II.

(

X

) ( J(

)

X

28 1

X

1 9.52

The equation and ion-product expression for SrC0 3 is: Ksp = [Sr2+] [C0 3 2-] SrC0 3 (s) !::; Sr2 +(aq) + CO/-(aq) a) The solubility, S, in pure water equals [S�+] and [CO/l Write a reaction table, where S is the molar solubility ofSrC03 : Concentration (M) SrC03 (s) !::; Sr2+(aq) + C0 3 2-(aq) 0 0 Initial +S +S Change Equilibrium S S Ksp 5 .4 1 0-10 = [Sr2+] [CO/-] = [S][S] = S 2 S = 2.32379 1 0-5 2.3 1 0-5 M b) In 0. 1 3 M Sr( N03 ) 2 , the initial concentration of Sr2+ is 0. 1 3 M. Equilibrium [Sr2+] 0. 1 3 + S and equilibrium [C03 2-] = S where S is the solubility of SrC0 3 . Concent,ation (M) SrC03 (s) !::; Sr2+(aq) + CO/-(aq) 0 0. 1 3 Initial +S +S Change Equilibrium 0. 1 3 + S S Ksp 5 .4 1 0-10 [Sr2+] [C03 2-] = (0. 1 3 + S)S This calculation may be simplified by assuming S is small and setting (0. 1 3 + S) = (0. 1 3). Ksp = 5 .4 1 0-10 = (0. 1 3)S S = 4. 1 538 1 0-9 = 4.2 1 0-9 M The equilibrium is: Ca(I03 )z(s) !::; Ca2+(aq) + 2 103-(aq). From the Appendix, Ksp(Ca(l03 ) 2 ) = 7. 1 x 1 0-7 . a) Write a reaction table that reflects an initial concentration of Ca2+ = 0.060 M. In this case, Ca2+ is the common ion. Ca(I03 )z(s) !::; Ca2\aq) + Concentration (M) 0.060 Initial +2 S +S Change 2S 0.060 + S Equilibrium Assume that 0.060 + S '" 0.060 because the amount of compound that dissolves will be negligible in comparison to 0.060 M. Ksp = [Ca2+] [I0 3-f = (0.060) (2 S)2 = 7. 1 1 0-7 S = 1 .7 1 998 1 0-3 1 .7 1 0-3 M Check assumption: ( 1 .7 1 998 x 1 0-3 M) / (0.060 M) x 1 00% = 2.9% 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(l03 ) 2 , so the molar solubility of Ca(I03 ) 2 is 1 .7 1 0-3 M. b) In this case, Concentration (M) Ca(l0 3 )z(s) !::; Ca2+(aq) + 2 I03-(aq) Initial 0 0.060 Change +S +2 S S 0.060 + 2 S Equilibrium The equilibrium concentration of Ca2+ is S, and the 10 3- concentration is 0.060 + 2 S. The iodate ion is the common ion in this problem. Assume that 0.060 + 2 S '" 0.060 Ksp = [Ca2+] [I0 3-] 2 = (S) (0.060)2 7 . 1 1 0-7 S = 1 .97222 1 0-4 = 2.0 1 0-4 M Check assumption: ( 1 .97222 x 1 0-4 M) / (0.060 M) x 1 00% = 0.3% 5%, so the assumption is good. S represents both the molar solubility of Ca2+ and Ca(l03 ) 2 , so the molar solubility of Ca(l03 ) 2 is 2.0 1 0-4 M. X

=

X

=

X

=

X

=

=

X

X

X

1 9.54

X

X

=

X

<

x

=

X

X

X

<

x

1 9.56

The larger the Ksp, the larger the molar solubility if the number of ions are equal. a) Mg(0 J:I)2 with Ksp 6__ � 1 0- 0: has higher mola.r .solubility than .Ni(OH)2 with Ksg = 6 x 1 0-16• b) PbS with Ksp 3 x 1 0 2 has higher molar solubIlity than CuS with Ksp = 8 1 0-3 . C)Ag2S04 with Ksp 1 .5 1 0-5 has higher molar solubility than MgF2 with Ksp = 7.4 1 0-9• X

=

X

=

=

1 9.58

X

X

a) AgCI(s) !::; Ag\aq) + Cr(aq) The chloride ion is the anion of a strong acid, so it does not react with H 3 0+. The solubility is not affected by pH .

282

b) SrC03(s) !:+ Sr2+(aq) + CO/-(aq) The strontium ion is the cation of a strong base, so pH will not affect its solubility. The carbonate ion is the conjugate base of a weak acid and will act as a base: CO/-(aq) + HzO (l) !:+ HC03-(aq) + OW(aq) and HC03- (aq) + H 2 0(l) !:+ H2 C03(aq) + OW(aq) The H 2C0 3 will decompose to CO2 (g) and H2 0(l). The gas will escape and further shift the equilibrium. Changes in pH will change the [ C0 3 2-], so the solubility of SrC03 will increase with decreasing pH. Solubility increases + with addition of H30 (decreasing pH) .

19.60

The ion-product expression for CU(OH)2 is Ksp = [ Cu2+] [OH-] 2 and, from the Appendix, Ksp equals 2.2 x 10-2°. To decide if a precipitate will form, calculate Qsp with the given quantities and compare it to Ksp3 2 1.0 X 10- mol CU(N03 h I mol Cu + 3 2 = 1.0 X 10- M Cu2 + [ Cu +] = I mol Cu ( N03 h L

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0.075 g KOH I mol OHI mol KOH 3 = 1.33666 X 10- M O W ( unrounded) 56.11 g KOH I mol KOH 1.0 L 3 3 Qsp = [ Cu2 +] [OH-] 2 = (1.0 x 10- ) (1.33666 x 10- )2 = 1.7866599 X 10-9 ( unrounded) Qsp is greater than Ksp (1.8 x 10-9 > 2.2 x 10-2 °), so Cu(OH)2 will precipitate. [OH-] =

19.65

In the context of this equilibrium only, the increased solubility with added OH- appears to be a violation of Le Chatelier's Principle. Adding. OH- should cause the equilibrium to shift towards the left, decreasing the solubility of PbS. Before accepting this conclusion, other possible equilibria must be considered. Lead is a metal ion and hydroxide ion is a ligand, so it is possible that a complex ion forms between the lead ion and hydroxide ion: Pb2\aq) + n OW(aq) !:+ Pb(OH)/-n (aq) This decreases the concentration of Pb 2+, shifting the solubility equilibrium to the right to dissolve more PbS.

19.66

In many cases, a hydrated metal complex (e.g., Hg(H 2 0)/+) will exchange ligands when placed in a solution of another ligand (e.g., C�), Hg(H2 0)/ +(aq) + 4 C�(aq) !:+ Hg(CN)/-(aq) + 4 H 2 0(l) N ote that both sides of the equation have the same "overall" charge of -2. The mercury complex changes from +2 to -2 because water is a neutral molecular ligand, whereas cyanide is an ionic ligand.

19.68

The two water ligands are replaced by two thiosulfate ion ligands. The +1 charge from the silver ion plus -4 charge from the two thiosulfate ions gives a net charge on the complex ion of -3. Ag(H 20)2+(aq) + 2 S 20 /-(aq) !:+ Ag( S 2 03)/ - (aq) + 2 H2 0(l)

19.70

Write the ion-product equilibrium reaction and the complex-ion equilibrium reaction. Sum the two reactions to yield an overall reaction; multiply the two constants to obtain Koverall. Write a reaction table where S = [AgI] d i ssolved = [ A g(NH3)t] . Solubility-product: AgI(s) !:+ Ag"faq1 + r-(aq) " Complex-ion: Ag faq1 + 2 NHiaq) !:+ Ag(NH3h.� Overall: AgI (s) + 2 NH3(aq) !:+ A g(NH3)2+(aq) + r-(aq) Koverall = Ksp x Kf = (8.3 x 10-1 7) (1.7 x 107) = 1. 411 X 10-9 (unrounded) Reaction table: Concentration (M) 2 NH3(aq) !:+ Ag(NH3)t(aq) + r - (aq) A gI (s) + 0 0 2.5 Initial S -2 S +S + Change 2.5 2 S S S Equilibrium Assume that 2.5 - 2S '" 2.5 because Koverall is so small. Ag ( NH3 ) 2 + [ I- ] [S][S] [S][S] Koverall 2 - --2 - 1.411 x 10 9 [NH3 J [2.5 - S] [2.5] S = 9.3908 X 10-5 = 9.4 x 1 0-5 M _

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19.74

The minimum urate ion concentration necessary to cause a deposit of sodium urate is determined by the Ksp for the salt. Convert solubility in gi l mL to molar solubility and calculate Ksp . Substituting [Na+] and Ksp into the ion-product expression allows one to find [Ur-] . Molar solubility ofNaUr: mol NaUr mL g Naur M NaUr (unrounded) [NaUr] = mL mol NaUr L x MNaUr = [Na+] = [Ur-] Ksp = [Na+] [Ur-] = x x = x M (unrounded) When [Na+] = M: M = [0. 1 5][Ur-] Ksp = x (unrounded) [Ur-] = The minimum urate ion concentration that will cause precipitation of sodium urate is 1.3 10-4 M.

00. ( 0.0100.85 J[ 10-31 ( 190.101 4.47133 10-3 J J 4.47133 10-3 (4.47133 10-3) (4.47133 10-3) 1.999279 10-5 0.1510-5 1.91.99279 3 3285 10-4 =

X

X

19. 7 6

x

a) The solubility equilibrium for KCI is: KCI (s) !:; K\aq) + Cqaq) = = 14 Ksp = [K+] [Cr] = b ) Determine the total concentration of chloride ion in each beaker after the H C l has been added. This requires the moles originally present and the moles added. Original moles from the KCI : mol Cl - ion mo l KCl mL mol cr = M oles K+ = Moles cr = L mL mol KCI Original moles from the M HCl: mol CI mol HCI L Moles Cl- = mol ClmL) I = mol HCI L mL This results in + mol = mol c r Original moles from the M HCl: -3 L OO. ) mol Clmol HCI Moles c r = = mol cr (I m l mol HCI I L This results in + mol = mol c r (unrounded) Volume of mixed solutions = mL + mL) L mL) = L After the mixing: [K+] = mol K+) L) = M K+ (unrounded) From M HCl: L) = M CI- (unrounded) [CI -] = mol C r) From M HCI : [Cr] = mol Cr) = M CI- (unrounded) Determine a Qsp value to see if Ksp is exceeded. If Q Ksp, nothing will precipitate. From M HCI Qsp = [K+] [C r ] = = = so no KCI will precipitate. From M HCl Qsp = [K+] [CI -] = = so KCI will precipitate. The mass of KCI that will precipitate when M HCI is added: Equal amounts of K and CI will precipitate. Let x be the molarity change. Ksp = [K+] [CI -] = - x) - x) = x= = This is the change in the molarity of each of the ions. mol + g KCl = Mass KCI K . L) l mol KC I = 1 g KCl I mol KCl mol K+

(3. 7 ) (3. 7 ) 13. 69

J 0. 3 7 ( 3. 7 1 )[ 101 -3 L J ( 100. )[ 1 1 6.0 (6. 0 1 [ 10-31 J ( 100. [ 1 J 0.60 (0.123 7 0.60)J 0. 9 7 (12 )[ 10mL J L [ 11 J 1.2 1. 5 7 / 1 0.200 (100.(0.37 1.100.2) (10-3 6.(0.0 37 (0.97/ (0.200 / (0.1.28005 4. 8 5 12 (1. 5 7 / (0.200 L) 7. 8 5 6. 0 (1. 8 5) (4.85) 8.9725 9. 0 14, 12 (1. 8 5) (7.1285) 14.5225 15 > 14, 0. 0 88697657 0. 09 (1. 8 5 (7. 8 5 13.69 [ 0.088697657L J ( O 200 [ 1 J( 74 55 ) 1.3 2248 <

<

=

.

284

1 9.78

a) Use the Henderson-Hasselbalch equation. Kal = 4.5 X pKa = -lo g Ka = -Iog( 4.5 x 1 0- 7 ) = 6.34679 (unrounded) pH

=

7 .40

pK. +

=

IO g

(

[HCO) ] [H2 CO) ]

6. 34679 + IOg

1 .0532 1

(

[HCO)· ] [H2 CO) ]

[H c ) ] oIOg [H2 CO) ]

=

[HCO) ] [ H2C03 ] [ H2CO) ] [ HCO)- ]

(

1

l

1 0-7 (From Appendix C)

J

1 1 .30342 3 5 (unrounded)

=

=

0.0884688

(

=

0.088

b) Use the Henderson-Hasselbalch equation. pH = pK + •

IOg

[HCO) · ] [H2 CO) ]

(

J

[HCO) · ] 7.20 = 6.34679 + IOg [H2 CO) ] 0 . 8532 1

1 9.80

=

IOg

(

[HCO)- ] [H2 COJ

1

1

An indicator changes color when the buffer component ratio of the two forms of the indicator changes from a value greater than 1 to a value less than I . The pH at which the ratio equals 1 is equal to pKa The midpoint in the pH range of the indicator is a good estimate of the pKa of the indicator. Ka = 1 0-4 1 = 7.943 X 10-5 = 8 X 1 0-5 pK. = (3.4 + 4.8)/2 = 4. 1 .

1 9 .8 1

a) A spreadsheet will help you to quickly calculate �pH/� V and average volume for each data point. At the equivalence point, the pH changes drastically when only a small amount of base is added, therefore, �pH/� V is at a maximum at the equivalence point. V(mL) pH

0.00 1 0.00 20.00 30.00 3 5 . 00 3 9. 00 39.50 39.75 39.90 39.95 3 9.99 40.00 40.0 1

1 .00 1 .22 1 .48 1 .85 2. 1 8 2.89 3 .20 3.50 3 .90 4.20 4.90 7.00 9.40

�pH �V

Va verage( m L )

0.022

5 .00 1 5 .00

0 .0 3 7 0 . 066 0. 1 8 0 . 62 1 .2 2.67 6 18 200

2 5 .00 3 2 .5 0

0.026

200

3 7.00 39.25 39 .63 39.83 39.93

39.97 40.00 40.0 1

285

40.05 40. 1 0 40.25 40. 5 0 4 1 . 00 4 5 . 00 50.00 60.00 70.00 80.00

40.03 40.08 40. 1 8 40. 3 8 40.75 43.00 47.50 5 5 .00 65 .00 7 5.00

10 10 0.67 1 .2 0.60 0. 1 7 0.058 0.025 0.0 1 3 0.009

9 . 80 1 0040 1 0. 5 0 1 0.79 1 1 . 09 1 1 . 76 1 2 .05 1 2 .30 1 2 043 1 2 .5 2

E

b)

::l

0 210

> 1 90 .';:; 1 70 § 1 50 t-. 1 30 � 1 10 :r:: 90 0... 70

50 ..:: - 30 � 10

.:: - 1 0 1l 30

35

40 45 50 Volume. mL Maximum slope (equivalence point) is at V 40.00 mL U

=

1 9. 84

The equation that describes the behavior of a weak base in water is: B(aq) + H20(l)!:; BH \aq ) + O W(aq )

[BH+ B] [OH- ] [ ]

Kb =

-log Kb

=

-log

-log Kb = -log pKb

=

pOH 1 9. 8 9

-log

=

[BW B] ][OH - ] [ [BW -[ B]-] -log [ O H ] -=-------.=-=,.---=

_

[BW -[ B]-]

pKb + log

+ pOH

[BW ] -[ B ]-

K values from the Appendix 2 Ka l = 5 .6 X 1 0H2C204(aq)!:; H +(aq) + HG.!G4-(aq) 5 Ka2 = 504 X 1 0�G4-fatt)!:; H+Caq) + C20/-Caq) K = Ka l Ka2 = 3 . 024 x I O- Ksp = Precipitate forms c) The higher pH would favor precipitation.

286

19. 92

3 To find the volume of rain, first convert the inches to yards and find the volume in yd 3 . Then convert units to cm and on to L. 2 x yd 2 in cm 3 mL 1 O-3 L acres ( ) ( ) x 1 0 6 L (unrounded) . acre 1 111 1 yd mL 1 cm 3 a) At pH X M (unrounded) [ H30+] 5 6 Mol H30+ X 1 0X 1 65 mol 2 -3 L x 1 03 yd 2 in in 2 cm 3 mL b) Volume ( 1 acres ) ft ) (1 1 acre I ft 111 mL yd 1 cm 3 x L (unrounded) Total volume of lake after rain x L + 1 .02790 1 5 x 1 0 6 L x 1 0 8 L (unrounded) [H30+] mol H30+I 1 .243 759 x x 1 O-7 M 7) pH x 6.28 c) Each mol of H30+ requires one mole of H C03- for neutralization. mol HC03 H C03 M ass mol H30+) 4.0 X 1 0 3 g HCO JX mol H 3 0+ I mol H C0 3

36 J 1. 00 m. ( 2. 54 J [-1 [--10. 0 [ 4. 840I 103 J (I J = 1. 0279015 J 10-5 0 L) = 64. 8 562 = = 4. 2 0, = (6. 3=095734 10-4 20 = 6. 3095734 M) (1. 0279015 = 0. 0 [ 4. 840 J ( 36I ) 0.0 ( 12 J ( . 514 ) [ I J [ 101 J = 1.2 3348 108 = 1.233481081 L=08 5. 2 14531 = 1.243759 = 64. 8 562 = -log(5.2 1453I 10- = 6. 2 827847 = [I 61. 02 g ) = 3. 97575 103 = )[ = (64. 8 562 1 19.93 = ( 317 g ) ( 58.1 44 g ) = 5.42436687

The molarity of a saturated NaCI solution must be found. NaCI mol NaCI M NaCI (unrounded) M NaCI NaCI L Determine the Ksp from the molarity j ust calculated. NaCI(s) !:; Na+( aq) + C r( aq ). 2 Ksp [Na+] [Cr] S 1 mol Clmol NaCI mol cr (unrounded) Moles of Cr initially ( mol NaCI L This is the same as the moles ofNa+ in the solution. mol Cl mol HCI 1 0-3 L mol Cl- (unrounded) ( 5 mL ) M oles of CI- added mol HCI mL L L of saturated solution contains mol each Na+ and Cl-, to which you are adding 0.2 mol of additional cr from HCI. Volume of mixed solutions 0. 1 00 L + mL) L l i mL) L (unrounded) Molarity of Cr in mixture + 0.200 1 75) mol Cr] / (0. 1 25 5 L) M Cr (unrounded) Molarity ofNa+ in mixture mol Na+) / (0. 1 255 L) M Na+ (unrounded) Determine a Q value and compare this value to the Ksp to determine if precipitation will occur. Qsp [Na+] [C n S ince Qsp < Ksp, no NaC I will precipitate.

=

0.100

1 9.95

= = (5.42436687)2 = 29.42375594 = 29.4 ) 0.100 L)[ I J = 0. 54236687 = ( 5.4236687 = ( 7. 8 5 ) [ 1 J 25 . [ II J = 0.200175 0. 542 ===[(0.(0. 554236687 4236687(25. 5 (10-3 = =4.0.1255 3 21648 = 5.9 16668 = = (4. 321648) (5. 9 16668) = 25. 569756 = 25. 6

a) For the solution to be a buffer, both HA and A- must be present in the solution. This situation occurs in A and D . b) Box A The amounts of HA and A- are equal. [A - ] [A - ] when the amounts of HA and A- are equal pH pK + [ HA] [ HA] pH pKa + log p H pKa x 4.35

( )=I = 10g ( ) == = -log I(4.5 10-5) = 4. 346787 a

=

287

Box B

Kb Kb = Kw / Ka = 1. 0 X 10-1 4 /4. 5 0.1010-5 = 2. 222 10- 1 0 -xxO -xxO -x0.10O .I O M-x Kb = [ [] [ - ] ] = 2.222 x 10-1 0 x 0.10. Kb = [ 0.10[ x] [-x]x ] = 2.222 x 10-1 0 Kb = 2.222 X 10- 1 0 = ((x0.10) ( x)) x=4. 7 138095 x 10-
Only A- is present at a concentration of The for A- is needed . A-(aq) + H20(l)

I n itial : Change : Equil ibrium:

HA

!:;

X

(unrounded)

X

O W (aq ) + H A(aq)

OH -

A

Assume that

is small compared to

M O W (unrounded)

Check assumption : val i d . [ H3 0r pH

Box C

-log [ H 3

This is a

0 +]

error, so the assumption is

M H30+ ( unrounded)

-log

8 .67

M H A solution. The hydrogen ion, and hence the pH, can be determined from the

Concentration HA(aq) + H20(l) !:; H 3 0+ (aq) + A-(aq) I nitial Change + +x Equil ibrium (The H 3 0+ contribution from water has been neglected . ) H O+ A ::--'=-7----='-

HA

Assume that

- X

is small compared to

a

Box

0

[ H 3 0+] Check assumption : val id. p H -log [ H 3 0+] -log

3

( unrounded)

error, so the assumption is

2.67

This is a buffer with a ratio of [ A-] [ HA] U se the H enderson-H asselbalch equation for this buffer. [A . ] p H pK + lo H A] a

pH

-log

+ log

4.57

c) The initial stage in the titration would only have HA present. The amount of H A will decrease, and the amount of A- will increase until only A- remains. The sequence will be : C, A, D , and B . d) A t the equivalence point, all the H A wi ll have reacted with the added base. T h i s occurs in scene B .

288

C h apter 2 0 Thermodynamics : E ntropy, Free E nergy, and the Direction of C hemical Reactions FOL LOW-U P P RO B L E M S

20. 1

a) PC1s(g) . For substances with the same type of atoms and in the same physical state, entropy increases with increasing number of atoms per mo lecule because more types of molecular motion are available. 2 2 b) BaCI2(s) . Entropy increases with increasing atom ic size. The Ba + ion and c r ion are larger than the Ca + ion and F- ion, respectively. c ) B r2(g) . Entropy increases from solid � l iquid � gas .

20.2

Plan: Predict the sign of �S:" by compari ng the randomness of the products with the randomness of the reactants. Calcu late �S:" using Appendix B values. Solution : a) 2 NaO H (s) + CO2( g ) � Na2C0 3 (s) + H20(l) The �S�, is predicted to decrease ( �S,",., < 0) because the more random, gaseous reactant is transformed into a more ordered, liquid prod uct.

� S ;yS = [( I mol N a2C0 3 ) (SONa2C0 (s)) + I mol H20) ( SO H 2 0( l) )] - [(2 mol NaO H ) (SONaOH(s)) + 1 mol CO2) (SOc02 (g))] 3 �S� , = [( I mol Na2C0 3 ) ( 1 3 9 J/mo l - K ) + ( I mol H20) (69.940 J/mol-K)]

��, <

- [(2 mol NaO H ) ( 64.454 J/mo l - K ) + ( I mol CO2) (2 1 3 . 7 J/mo l - K ) ] = - 1 3 3 .668 = - 1 34 J/K 0 as predicted.

b) 2 Fe(s) + 3 H 20(g ) � Fe20 3 ( S ) + 3 H2( g ) The change in gaseous moles i s zero, so the sign of �S;ys is difficult to predict. r ron( I I I ) oxide has greater entropy than Fe because it is more complex, but this is offset by the greater molecular comp lexity of H20 versus H2. �S�, = [( I mol Fe20 3 ) (SOFe2 0 ) + ( 3 mol H 2 ) (SOH ) ] [(2 mol Fe) (SOFe) + (3 mol H20) (SOH 2 0)]

3

-

�S;." = [( I mol Fe20 3 ) ( 8 7 .400 J/mol - K ) + (3 mol H 2 ) ( 1 30 . 6 J/mo l - K ) ] - [(2 mol Fe) ( 2 7 . 3 J/mo l - K ) + (3 m o l H 2 0 ) (1 8 8 . 72 J/mol - K)] = - 1 4 1 . 5 6 = - 1 4 1 .6 J/K The negative �S�" for equation (b) reflects that greater entropy of H20 versus H 2 does outweigh the greater entropy of Fe20 3 versus Fe. 20.3

Plan: Write the balanced equation for the reaction and calculate the �S�, using Appendix B. Determine the

�S :'m by fi rst fi nding 1'1H,�"" Add �S:'", to �S:." to veri fy that �Suniv is positive. Solution : 2 FeO(s) + 1 /2 02(g ) � Fe20 3 (s) �S :", [( I mol Fe20 3 ) (S'Fe20 3 )] - [(2 mo l FeO) (S'FeO) + ( 1 12 mo l 02)(S'02)] =

�S:."

=

[( 1 mol Fe20 3 ) ( 8 7 .400 J/mol - K)] - [(2 mol FeO) (60 . 7 5 J/mo l - K ) + ( 1 /2 mol O 2 ) (205 .0 J/mo l - K ) ]

�S� , = - 1 3 6.6 J/K (entropy change is expected t o b e negative because gaseous reactant is converted to solid product)

289

Mf;" Mf;"

[( I mol Fe20 3 ) ( Mfj Fe20 3 )] - [(2 mol FeO) ( Mfj FeO) + ( 1 /2 mol O2)( Mfj O2)]

=

[( 1 mol Fe20 3 ) (-825 , 5 kJ/mol)] - [(2 mol FeO) (-272,0 kllmol) + ( 1 /2 mol O2) (0)]

=

Mfosys -28 1 . 5 kl =

° thl '''''

MfO

-� T

=

=

-

-28 1 . 5 k] 298

K

=

3 0.94463 kJ/K( 1 0 11 1 kJ)

=

M;:"i" thl;" + thl:,,,, (- 1 3 6.6 1 1K) + (944,63 J/K) 808.03 Because thl'�"i" is positive, the reaction is spontaneous at 298 K. =

=

=

944 ,63 11K (unrounded)

=

808 J/K.

Check: This process is also known as rusting. Common sense tells u s that rusting occurs spontaneously. Although the entropy change of the system is negative, the increase in entropy of the surroundings is large enough to offset thl;" . 20A

Plan: Calculate the f:..H :" using Mf ; values from Appendix B . Calculate f:..S :" from tabulated SO values and then use the relationship f:.. G:" Mf:n - T f:..S:" . So lution : Mf:',,, [(2 mol N02) ( Mf; (N02))] - [(2 mol NO) ( Mf; (NO)) + ( I mol O2) ( Mf ; (02))] =

=

Mf:'",

=

[(2 mol N02) ( 3 3 . 2 kJ/mol)] - [(2 mol NO) (90.29 kl/mol) + ( I mol O2) (0)]

Mf:'", - 1 1 4 . 1 8 =

thl:"

=

f:.. S:n

20.5

- 1 1 4 .2 kJ

[(2 mol N02) ( S; (N02))] - [(2 mo l NO) ( S; (NO)) + ( I mol O2) ( s; (02))]

f:.. S�'n f:.. G:"

=

=

=

=

[(2 mol N02) (239.9 J/moloK)] - [(2 mol NO) (2 1 0.65 J/moloK) + ( 1 mol O2) (205 .0 J/mol oK)] - 1 46.5 11K

Mf ;x" - T thl'�m

=

3 - 1 1 4.2 kJ - [(298 K) (- 1 46.5 11K) ( I kJI I 0 J)]

Plan: Use f:.. G; values from Appendix B to calculate f:.. G:n • f:.. G�,s

=

=

-70.543

=

-70.5 kJ

1: (m f:.. Gj ( products ) ) - 1:(n f:.. Gj (reac tan ts ) )

Solution : a) f:.. G:'",

=

[(2 mol N02) ( f:.. G; (N02))] - [(2 mol NO) ( f:.. G; (NO)) + ( 1 mol O2) ( f:.. G; (02))]

f:.. G:'",

=

[(2 mol N02) (5 1 kl/mol)] - [(2 mol NO) ( 8 6 . 60 kllmol) + ( 1 mol O2) (0))]

f:.. G:'", -7 1 .2 =

b) f:.. G:'",

=

=

-7 1 kJ (result agrees with answer in 20A)

[(2 mol CO) ( f:.. G; (CO)] - [(2 mo l C) ( f:.. G; (C)) + ( 1 mol O2) ( f:.. G; (02))]

f:.. G:'",

=

[(2 mol CO) (- 1 3 7 . 2 kJ/mol)] - [(2 mol C) (0) + ( 1 mol O2) (0)]

f:.. G:" -274.4 kJ =

20.6

Plan: Examine the equation f:.. Go = f:..HD T/:;.SO and determi ne which combination of enthalpy and entropy will describe the given reaction. Solution : Two choices can already be eliminated : 1 ) When Mf > 0 (endothermic reaction) and thl < 0 (entropy decreases), the reaction is always nonspontaneous, regardless of temperature, so this combination does not describe the reaction. 2) When Mf < 0 (exothermic reaction) and f:..S > 0 (entropy increases), the reaction is always spontaneous, regardless of temperature, so this combination does not describe the reaction. Two combinations remain: 3) f:..HD > 0 and /:;.SO > 0, or 4) MfD < 0 and /:;.SO < O. If the reaction becomes spontaneous at -40°C, thi s means that f:.. Go becomes negative at lower temperatures. Case 3 ) becomes spontaneous at h igher temperatures, when the -T /:;.SO term is larger than the positive enthalpy term. By process of elimination, Case 4) describes the reaction. Check: At a lower temperature, the negative f:..HD becomes larger than the positive (-T /:;.SO) value, so f:.. Go becomes negative. -

290

20.7

Plan : Write a balanced equation for the dissociation of hypobromous acid in water. The free energy of the reaction at standard state (part (a» is calculated using I'1Go = -RT In K. The free energy of the reaction under non­ standard state conditions is calculated using I'1G = I'1Go + RT In Q. Solution: H BrO(aq) + H20(l) !:; BrO-(aq) + H 3 0\ aq) 4 a) I'1Go = -RT In K = -(8 . 3 1 4 llmol oK) (298) In (2.3 x 1 0-9 ) = 4 . 927979 x 1 0 = 4.9 X 1 04 J/mol H 3 0 + BrO 6 . 0 x I 0 -4 [ 0. 1 0 b) Q = [ H B rO ] [ 0.20

J[

[

I'1G

=

I'1Go

J

[

]

J ]

+ RT In Q = (4.927979

X

4 1 0 llmol) + ( 8 . 3 1 4 J/moloK) (298 K) In

[6.0 =--

X

J[O.IO] [ 0.20 ] 1 0-4

--;:-----='-­

4 = 2 . 9 1 82399 x 1 0 = 2.9 X 1 04 J/mol Check: The value of Ka is very small , so it makes sense that I'1Go is a positive number. The natural log of a 4 negative exponent gives a negative number (In 3 .0 x 1 0- ), so the value of I'1G decreases with concentrations lower than the standard state I M values. E N D-OF-C HAPTER PRO BLEMS

20.2

A spontaneous process occurs by itself (possibly requiring an initial input of energy) whereas a nonspontaneous process requires a continuous supply of energy to make it happen. It is possible to cause a nonspontaneous process to occur, but the process stops once the energy source is removed. A reaction that i s found to be nonspontaneous under one set of conditions may be spontaneous under a different set of conditions (different temperature, different concentrations).

20.5

Vaporization i s the change of a liquid substance to a gas so Mvaporization = Sgas - Sliquid. Fusion is the change of a solid substance into a liquid so Mfusion = Sliquid - Ssoli d . Vaporization involves a greater change in volume than fusion. Thus, the transition from l iquid to gas involves a greater entropy change than the transition from solid to liquid.

20.6

In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases (Msurr > 0). In an endothermic process, the system absorbs heat fro m the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases (Msurr < 0). A chemical cold pack for inj uries is an example of a spontaneous, endothermic chemical reaction as is the melting of ice cream at room temperature.

20.8

a) Spontaneou s . E vaporation occurs because a few of the liquid molecules have enough energy to break away from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase. b) Spontaneo u s . A lion spontaneously chases an antelope without added force. This assumes that the lion has not j ust eaten. c) Spontaneo u s . An unstable substance decays spontaneously to a more stable substance.

20. 1 0

a) Msys positive, melting is the change in state from solid to liquid. The solid state of a particular substance always has l ower entropy than the same substance in the liquid state. Entropy increases during melting. b) Msys negative, the entropy of most salt solutions is greater than the entropy of the solvent and sol ute separately, so entropy decreases as a salt precipitates. c) Msys negative, dew forms by the condensation of water vapor to liquid. Entropy of a substance in the gaseous state is greater than its entropy in the liquid state. Entropy decreases during condensation.

20. 1 2

a) Msys negative, reaction involves a gaseous reactant and no gaseous products, so entropy decreases. The number of particles also decreases, indicating a decrease in entropy. b) Msys negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) Msys positive, when salts dissolve in water, entropy generally increases.

29 1

20. 1 4

a) Msys positive, the reaction produces gaseous CO2 molecules that have greater entropy than the physical states of the reactants. b) Msys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system 's entropy decreases. c) Msys positive, the reaction produces a gas from a solid.

20. 1 6

a) Msys positive, decreasing the pressure increases the volume avai lable to the gas molecules, so entropy of the system increases. b) Msys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. c) Msys positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules.

20. 1 8

a) Butane has the greater molar entropy because it has two additional C-H bonds that can vibrate and has greater rotational freedom around its bond. The presence of the double bond in 2-butene restricts rotation. b) Xe(g) has the greater molar entropy because entropy increases with atomic size. c) C H 4(g) has the greater molar entropy because gases in general have greater entropy than liquids.

20.20

a) Diamond < graphite < charcoal. Diamond has an ordered, 3-dimensional crystalline shape, fol lowed by graphite with an ordered 2-dimensional structure, followed by the amorphous (disordered) structure of charcoal. b) Ice < liquid water < water vapor. Entropy increases as a substance changes from solid to liquid to gas. c) 0 atoms < O2 < 03, Entropy increases with molecular complexity because there are more modes of movement (e.g., bond vibration) available to the complex molecules.

20.22

a) Cl04-(aq) > Cl03-(aq) > Cl02-(aq). The decreasing order of molar entropy fol lows the order of decreasing molecular complexity. b) N02(g) > NO(g) > N2(g) . N2 has lower molar entropy than NO because N2 consists of two of the same atoms while NO consists of two different atoms. N02 has greater molar entropy than NO because N02 consists of three atoms while NO consists of only two. c) Fe304(S) > Fe203(S) > Ah03(S), Fe304 has greater molar entropy than Fe203 because Fe304 is more complex and more massive. Fe20 3 and AI203 contain the same number of atoms but Fe203 has greater molar entropy because iron atoms are more massive than aluminum atoms.

20.25

A system at equilibrium does not spontaneously produce more products or more reactants. For either reaction direction, the entropy change of the system is exactly offset by the entropy change of the surroundings. Therefore, for system at equilibrium, L'lSuniv L'lSsy s + L'lSs u rr O. However, for a system moving to equilibrium, Muniv > 0, because the Second Law states that for any spontaneous process, the entropy of the universe increases. =

20.26

=

Since entropy is a state function, the entropy changes can be found by summing the entropies of the products and subtracting the sum of the entropies of the reactants. For the given reaction L'lS : = 2 SOHClO(g) - (SOH 20(g) + S OCI 20(g))' Rearranging this expression to solve for SOCI 2 0(g) gives "

20.27

a) Prediction: � negative because number of moles of (L'ln) gas decreases. L'lSo = [( I mol N20(g)) (SO of N20) + ( I mol N02(g)) (SO of N02] - [ ( 3 mol NO(g)) (SO of NO)] L'lSo [( I mol N20(g)) (2 1 9.7 J/mol oK) + ( 1 mol N02(g)) (239.9 J/mol oK)] - [(3 mol NO(g)) (2 1 0.65 J/mol oK)] L'lSo = - 1 72.35 = - 1 72.4 J/K =

292

b) Prediction: S ign difficult to predict because I'1n = 0, but possibly !lS" positive because water vapor has greater complexity than H2 gas . !).SO [(2 mol Fe(s)) (SO of Fe) + (3 mol H20(g)) (SO of H20)] - [(3 mol H2(g)) (SO of H2) + ( 1 mol Fe20 3 (S)) (SO of Fe20 3 )] !).So [(2 mol Fe(s)) (27.3 J/mol-K) + (3 mol H20(g)) ( 1 8 8 . 72 J/mo l - K)] - [(3 mol H 2 (g)) ( 1 30.6 ]Imo l - K) + ( J mol Fe20 3 (S)) ( 8 7 .400 J/mo l - K )] !).SO 1 4 1 . 5 6 1 4 1 .6 J/K c) Prediction : I'1S:y, negative because a gaseous reactant forms a solid product and also because the number of =

=

=

=

moles of gas (I'1n) decreases. !).SO [( I mol P 4 0 I 0 (S)) (SO of P 4 0 I o)] - [( 1 mol P 4 (S)) (SO of P 4 ) + (5 mol 02(g) ) (SO of O2)] =

M?

=

[( 1 mol P 4 0 I 0 (S)) (229 J/mol -K)] - [( 1 mol P 4 (S)) (4 1 . 1 J/mol-K) + ( 5 mol 02(g)) (205 .0 J/mol-K)] !).SO -83 7 . 1 -83 7 J/K =

20.29

=

The balanced combustion reaction is 2 C2 H6 (g) + 7 02(g) � 4 CO2(g) + 6 H 20(g) !).S:" = [4 mol (SO(C02)) + 6 mol (SO ( H20))] - [2 mol (SO(C2H 6) + 7 mol (SO (C02)]

[4 mol (2 1 3 . 7 J/mo l - K) + 6 mol ( 1 8 8 . 72 J/mo l - K)] - [2 mol (229 . 5 J/mo l - K) + 7 mol (205 .0 J/mol-K)] 93 . 1 2 = 93. 1 J/K The entropy value is not per mole of C2H 6 but per 2 moles. Divide the calculated value by 2 to obtain entropy per mole of C2H 6 . Yes, the positive sign of I'1S is expected because there is a net increase i n the number of gas molecules from 9 moles as reactants to 1 0 moles as products. =

=

20.3 1

The reaction for forming CU20 from copper metal and oxygen gas is 2 Cu(s) + 1 12 02(g) � Cu20(s) I'1S:, = [ I mol (SO(Cu20))] - [2 mol (SO(Cu)) + 1 12 mol (SO(02))] =

[ I mol (93 . 1 J/mol - K)] - [2 mol (3 3 . 1 J/mo l - K) + 1 12 mol (205 .0 J/mo l -K)] 75 6 J/K

= -

20.34

.

Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the products. C2H2(g) + 5/2 02(g ) � 2 CO2(g) + H20(g) I'1S:n [2 mol (SO(C02)) + I mol (SO( H 20))] - [ 1 mol (SO(C2H2)) + 5/2 mol (SO(02))] =

=

20.36

20 . 3 8

[2 mol (2 1 3 . 7 J/mo l - K) + 1 mol ( 1 8 8 . 72 J/mol -K)] - [ 1 mol (200 . 8 5 J/mo l - K) + 5/2 mol (205 .0 J/mo l - K)]

=

-97.23

=

-97.2 J/K

A spontaneous process has !).Suniv > O. Since the Kelvin temperature is always positive, I'1Gsys must be negative (I'1Gsys < 0) for a spontaneous process.

!:Jl :xn positive a n d L\S�., positive. The reaction is endothermic ( M/:" > 0) and requires a lot of heat from its surroundings to be spontaneous. The removal of heat from the surroundings results in I'1S:urr <

O.

The only way

an endothermic reaction can proceed spontaneously is if I'1S:'., » 0, effectively offsetting the decrease in surroundings entropy. In summary, the values of I'1H:" and I'1S �., are both positive for this reaction. Melting is an example. 20.39

The I'1G:" can be calculated from the i ndividual I'1G; ' s of the reactants and products found in Appendix B. I'1G:n

=

I:[m I'1G; (products)] - I:[n I'1G; (reactants)]

a) I'1G:" = [(2 mol M gO) ( I'1G; M gO)] - [(2 mol M g) ( I'1G; M g) + ( I mol O2) ( I'1G; O2)] I'1G:" = [(2 mol M gO) (-569.0 kJ/mol)] - [(2 mol M g) (0) + ( I mol O2) (0)]

293

=

-1 1 38.0 kJ

b) t:J.G:" = [(2 mol CO2) ( t:J.G; CO2) + (4 moI H20)( t:J.G; H20)] - [(2 mol CH30H) ( t:J.G; CH30H) + (3 mol O2) ( t:J.G; O2)] t:J.G:n = [(2 mol CO2) (-3 94.4 kJ/mol) + (4 mol H20) (-228 .60 kJ/mol)] - [(2 mol C H30H) (- 1 6 1 .9 kJ/mol) + (3 mol O2) (0)] t:J.G:" = - 1 3 79.4 kJ c) t:J.G:n = [ ( 1 mol BaC03) ( t:J.G; BaC03)] - [( 1 mol BaO) ( t:J.G; BaO) + ( l mol CO2) ( t:J.G; CO2)] t:J.G:n = [ ( 1 mol BaC03) (- 1 1 3 9 kJ/mol)] - [( 1 mol BaO) (-520.4 kJ/mol ) + ( 1 mol CO2) (-394.4 kJ/mol)] t:J.G:" = -224.2 = -224 kJ 20.4 1

The MI�xn can be calculated from the individual Mil ' s of the reactants and products found in Appendix B .

MI�xn L [ m Mil (products)] - L[n Mil (reactants)] =

The S�xn can be calculated from the individual So ' s of the reactants and products found in Appendix B .

MI�xn = L[m Mil (products)] - L[n Mil (reactants)] t:J.G:" can be calculated using t:J.G:"

=

MI:" - T t:J.S:"

a) MI:" = [2 mol ( MI; (MgO))] - [2 mol ( t:J.H; ( M g)) + I mol ( MI; (02))] = [2 mol (-60 1 .2 kJ/mol)] -[ {2 mol (0) + 1 mol (0)] = - 1 202.4 kJ t:J.S:" = [2 mol (S" ( M gO))] - [2 mol ( S" ( M g)) + 1 mol (S" (02))] = [2 mol (26.9 J/mol o K)] - [2 mol (32.69 J/moloK) + 1 mol (20 5 . 0 J/moloK)] = -2 1 6. 5 8 JIK (unrounded) 3 t:J.G:" = MI:" - T t:J.S:" = - 1 202.4 kJ - [(298 K) (-2 1 6. 5 8 J/K) ( I kJ I 1 0 J)] = - 1 1 3 7 . 8 5 9 = -1 1 3 8 kJ

b) MI::" = [2 mol ( MI; (C02)) + 4 mol ( MI; ( H 20(g))] - [2 mol ( MI; (CH30H)) + 3 mol ( MI; (02))] [2 mol (-3 93 . 5 kJ/mol) + 4 mol (-24 1 .826 kJ/mol)] - [2 mol (-20 1 .2 kJ/mol) + 3 mol (0)] = - 1 3 5 1 . 904 kJ (unrounded) t:J.S:" = [2 mol (S" (C02)) + 4 mol (SO(H20(g))] - [2 mol (SO(CH30H)) + 3 mol (S" (02))]

=

=

=

t:J.G:n

=

[2 mol (2 1 3 . 7 J/mol o K) + 4 mol ( 1 8 8 . 72 J/mol oK)] - [2 mol (23 8 J/moloK) + 3 mol (205 .0 J/moloK)] 9 1 .28 JIK (unrounded) 3 MI:n - T t:J.S:" = - 1 3 5 1 .904 kJ - [(298 K) (9 1 .28 JIK) ( 1 kJ I 1 0 J)] = - 1 3 79. 1 05 = - 1 379 kJ

c) MI:" = [ 1 mol ( MI; (BaC03(s))] - [ I mol ( MI; (BaO)) + 1 mol ( MI; (C02))] = [ 1 mol (- 1 2 1 9 kJ/mol)] - [ 1 mol (-548 . 1 kJ/mol) + I mol (-3 9 3 . 5 kJ/mol)] = -277.4 kJ (unrounded) t:J.S:" = [ 1 mol (S" ( B aC03(s))] - [ I mol (S" (BaO)) + 1 mol (S" (C02))] = [ I mol ( 1 1 2 J/moloK)] - [ 1 mol (72.07 J/mol o K) + 1 mol (2 1 3 . 7 J/moloK)] - 1 7 3 . 7 7 J/K (unrounded) 3 t:J.G:" = t:J.H:n - T t:J.S:n = -277.4 kJ - [(298 K) (- 1 73 . 77 J/K) ( 1 kJ 1 1 0 J)] = -225 .6265 =

20.43

=

-226 kJ

a) Entropy decreases (t:J.S' negative) because the number of moles of gas decreases from reactants ( I Y2 mol) to products ( I mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then releases energy (exothermic, !!JIO negative) which is typical of all combustion reactions. b) M ethod 1 : Calculate t:J.G:,., from t:J.G; ' s of products and reactants. t:J.G:n = L[m t:J.G; (products)] - L[n t:J.G; (reactants)] t:J.G:n = [( I mol CO2) ( t:J.G; (C02)] - [( 1 mol CO) ( t:J.G; (CO)) + ( 112 mol) ( t:J.G; (02))] t:J.G:n = [( 1 mol CO2) (-3 94.4 kJ/mol)] - [( I mol CO) (- 1 3 7.2 kl/mo l) + ( 1 12 mol O2) (0)] = -257.2 kJ

294

M ethod 2 : Calculate /";.G�" from tlH° and /";.so at 298 K (the degree superscript indicates a reaction at standard state, given in the Appendix at 25°C). tlH�xn = L[m tlH'f (products)] - L[n tlH'f (reactants)] tlH�" = [( I mol CO2) ( tlH ; (C02)] - [( I mol CO) ( tlH; (CO)) + ( 1 12 mol) ( tlH; (02))]

tlH:" = [( 1 mol CO2) (-393.5 kJ/mol)] - [( I mol CO) (-1 1 0.5 kJ/mol) + ( 1 12 mol O2) (0)] = -283.0 kl S�rn = L[m So (products)] - L[n So (reactants)] /";.S�" = [( 1 mol CO2) (2 1 3 .7 J/mol·K)] - [( I mol CO) ( 1 97.5 J/mol·K) + ( 1 12 mol O2) (205 .0 J/mol·K)] = -86.3 J/K /";.G:" = tlH�" - T /";.S:" = (-283.0 kJ) - [(298 K) (-86.3 11K) ( I kl I 1 0 3 J)] = -25 7.2826 = -257.3 kJ

20.45

a) tlH�" = [( I mol CO) ( tlH ; (CO)) + (2 mol H2) ( tlH; ( H2))] - [( I mol CH 3 0H) ( tlH; (CH 3 0H))] tlH:", = [( I mol CO) (- 1 1 0.5 kllmol) + (2 mol H2) (0)] - [( I mol CH 3 0H) (-20 1 .2 kllmol)] tlH�" = 90.7 kJ /";.S:" = [(I mol CO) ( S� (CO)) + (2 mol H2) ( S; (H2))] - [( 1 mol CH 3 0H) ( S; (CH 3 0H))] /";.S:n = [( I mol CO) ( 1 97.5 J/mol·K) + (2 mol H2) ( 1 30.6 J/mol·K] - [( 1 mol CH 3 0H) (23 8 Ilmol·K)] /";.S,:" = 220.7 = 22 1 J/K b) TJ = 38 + 273 = 3 1 1 K /";.Go = 90.7 kJ - [(3 1 1 K) (220.7 J/K) ( 1 kJ I 1 0 3 J)] = 22.0623 = 2 2 . 1 kJ T2 = 1 3 8 + 273 = 4 1 1 K /";.Go = 90.7 kJ - [(4 1 1 K) (220.7 JIK) ( I kl 1 1 0 3 J)]= -0.0077 = 0.0 kJ T 3 = 238 + 273 = 5 1 1 K /";.Go = 90.7 kJ - [(5 1 1 K) (220.7 J/K) ( l kJ I 1 0 3 J)] = -22.0777 = -22. 1 kJ c) For the substances in their standard states, the reaction is nonspontaneous at 3 8°C, near equilibrium at 1 3 8°C and spontaneous at 23 8°C. Reactions with positive values of tlH,:" and /";.S�" become spontaneous at high temperatures.

20.47

At the normal boiling point, defined as the temperature at which the vapor pressure of the liquid equals I atm, the phase change from liquid to gas is at equilibrium. For a system at equilibrium, the change in Gibbs free energy is zero. Since the gas is at I atm and the liquid assumed to be pure, the system is at standard state and /";.Go = O. The temperature at which this occurs can be found from /";.G:" = 0 = tlH:" - T /";.S,:" .

T bpt = tlFfO I � tlFfO = [ 1 mol ( tlH ; (Br2(g)))] - [ I mol ( tlH ; (Br2( 1)))] tlFfO = [ I mol(30.9 1 kllmol] - [ I mol ( tlH; (0)] = 30.9 1 kJ 309 1 0 J � = [ I mol (SO( Br2(g)))] - [ I mol (SO(Br2( 1)))] � = [ I mol (SO(245 .38 lIK·mol] - [ I mol (SO( 1 52.23 l/K·mol] = 93. 1 5 J/K 309 1 0 J o = 3 3 1 . 830 = 33 1 .8 K T bpt = tlH = 93 . 1 5 J/K /";.S o =

20.49

a) The reaction for this process is H2(g) + 1 12 02(g) � H20(g). The coefficients are written this way (instead of 2 H2(g) + 02(g) � 2 H20 (g)) because the problem specifies thermodynamic values "per ( 1 ) mol H2," not per 2 mol H2. tlH:" = [( 1 mol H20) ( tlH; (H20))] - [( I mol H2) ( tlH; (H2)) + ( 1 12 mol O2) ( tlH ; (02))] tlH:" = [( 1 mol H20) (-24 1 .826 kllmol)] - [( 1 mol H2) (0) + ( 1 12 mol O2) (0)] tlH :", = -24 1 . 8 26 kJ

295

/:;.s:,. =

[( I

mol H 2 0 )

/:;.S:,. =

( S; ( H 2 0 ) ) ] - [( I mol H 2 ) ( S; ( H 2 ) ) + ( 1 12 mo l O2 ) ( S; ( 0 2 ) )]

( 1 88.72 J/mo l o K ) ] - [( I mol H2 ) ( I 30.6 l/mo l o K ) + ( 1 /2 mol O2) (205.0 l/mo l o K ) ] /:;.S :,. = -44.38 = -44.4 J/K /:;. c,�", = [( I mo l H 2 0 ) ( /:;. C; ( H 2 0 ) )] - [( 1 mol H 2 ) ( /:;. C; ( H 2 ) ) + ( 1 12 mo l O2 ) ( /:;. c; ( 0 2 ) )] /:;. C:,. = [( I mol H2 0 ) (-228.60 kl/mol ) ] - [( I mol H 2 ) (0) + (1 /2 m o l O2 ) ( 0 )] [( I mol H 2 0 )

/:;.G:,. = -22 8.60 kJ Because /:;.H < 0 and /:;.S < 0, the reaction

b) wi l l become non spontaneous at h i gher temperatures because the positive (-T /:;.S) term becomes l arger than the negati ve /:;.H term . c ) The reaction becomes spontaneous below the temperature where

/:;'GO = 0 = /:;.H 0

rXII

"X/1

T=

/:;. C:'" = 0

- T /:;.SO

/:;.H0 /:;.SO

rxn

=

[

-24 1 .826 kJ 1 03 J ) -44.38 J/K I kJ

=

5448.986 = 5.45

x

1 03 K

20.5 1

a ) An equi l i brium constant that is much less than 1 indicates that very l ittl e product i s made to reach equ i l i brium. The reaction, th us, i s not spontaneous i n the forward d i rection and /:;.Go i s a re l ati vely l arge positive value. b) A l arge negative /:;.Go i nd i c ates that the reaction i s quite spontaneous and goes a l most to comp letion. At equi l i brium, much more product i s present than reactant so K > 1 . Q depends on initial conditions, not equi l i bri um conditions, so its val ue cannot be pred i cted from /:;. C o.

20.53

T h e standard free energy change, not confuse th i s with

t,G; , the

t,Go,

occurs w h e n a l l components o f t h e system are i n t h e i r standard states ( do

standard free energy of fonnati o n ) . Standard state is defined as I atm for gases,

I M for s o l utes, and pure so l i ds and l i quids. Standard state does not spec i fy a temperature because standard state can occur at any temperature . t,Co = /:;.G when a l l concentrations equal I M and a l l partial pressures equal 1 atm . T h i s occurs because the val ue of Q = 1 and In Q = 0 in the eq uation t,C = t,C o + RT In Q.

20.54

For e a c h reaction, fi rst fi n d

t,Go, t h e n calculate K fr o m t,Go = - RT

fro m Appen d i x B . a)

t,Go =

[1

mol

( t,C; ( N 0 2 (g) )] - [ I

[

mol

I n K. Calculate

t,G:,. using t h e t,G;

( t,C; (N O(g)) ) + 12 m o l ( t,G; ( 02(g)) ]

(86.60 kJ ) + 12 mol (0 kJ ) ] = -35 .6 kJ ( unrounded) t,Go -35.6 kJ / mo l . ) -1 03 J ) = 1 In K = -- = 4.3689 ( unrounded ) -RT - ( 8.3 1 4 J / mol K ) ( 298 K ) I kl K = e l 4 3 68 9 = 1 .739 1 377 x 1 0 6 = 1 . 7 1 0 6 b ) t,Co = [ 1 mo l ( t,G; ( H 2 (g) ) ) + 1 mol ( t,C; ( C I2(g) ) ) ] - [2 mol ( t,G; ( H C I (g) ) ) ] = [ I mol (0) + I mol ( O )kJ ] - [2 mol (-95.30 kJ/mol ) ] = 1 90.60 k J t,Go 1 90.60 kJ / mo l 1 03 J ) = In K = = -76.930 ( unrounded ) W - RT - ( 8.3 1 4 J / mo l K ) (2 9 8 K ) 34 K = e-7 6 9 3 0 = 3 .88799 x L O-34 = 3.89 1 0c ) t,Co = [2 mol ( t,G; ( C O (g) ) ) ] - [2 mol ( t,G; (C( graph ite ) ) ) + I mol ( t,G; ( 0 2 (g) ) ) ] = [2 mo l (- 1 37.2 kl/mol ) ) - [2 mol (0) + 1 mol (0) kJ ] = -274.4 kJ t,Go -274.4 kJ /mol 1 03 J ) = = In K = 1 1 0.75359 ( unrounded ) - RT - ( 8.3 1 4 J / mol o K ) (298 K ) W 11 . 5 K = e 0 7 3 59 = 1 .2579778 x 1 04 8 = 1 .26 1 0 48 = [ I mol

(5 1

k J ) ] - [ I mol

[

0

X

[

)[

0

X

[

)[

X

N ote : You may get a d i fferent answer depending on how you rounded in earl ier calculations.

296

val ues

20.56

The solubility reaction for Ag2 S is Ag2 S(s) + H 2 0(l) =+ 2 Ag+(aq) + HS-(aq) + O W(aq) /lGo = [2 mol ( /lG; Ag\aq)) + 1 mol ( /lG; H S-(aq ) ) + 1 mol ( /lG; O W (aq)] - [ 1 mol ( /lG; Ag2 S(s )) + I mol ( /lG; H 2 0(l))] = [(2 mol) (77. 1 1 1 kJ/mol) + ( l mol) ( 1 2.6 kJ/mol) + ( I mol) (- 1 57.30 kJ/mol)] - [( 1 mol) (-40.3 kJ/mol) + ( 1 mol) (-237. 1 92 kJ/mol)] = 287.0 1 4 kJ (umounded) /lGo 287.0 1 4 kJ / mol 1 03 J. In K = -= = - 1 1 5.8448 675 (umounded) -RT - ( 8.3 1 4 J /mol K ) ( 298 K ) 1 kJ K = e-1 1 5. 844867 5 = 4.888924 1 x 1 0-5 1 = 4. 89 1 0-5 1

[

][ ]

•

X

20.58

Calculate /lG:.. , recognizing that 1 2 (s), not 1 2 (g), is the standard state for iodine. Solve for Kp using the equation /lGo = -RT In K. /lG:n = [(2 mol) ( /lG; (ICI))] - [( 1 mol) ( /lG; ( 1 2 )) + ( 1 mol) ( /lG; (Cl 2 ))] /lG:.. = [(2 mol) (-6.075 kJ/mol)] - [( 1 mol) ( 1 9.38 kJ/mol) + ( I mol ) (0)] /lG:. = -3 1 . 5 3 kJ I 03 J -3 1 .53 kJ / mol = 1 2.726 1 69 (umounded) In Kp = /lGo I -RT = - (8.3 1 4 J / mol · K ) (298 K ) I kJ Kp = e 1 2 726 1 6 9 = 3.3643794 x l Os = 3.36 1 0 5

][ ]

[

X

In K = -(8.3 1 4 J/mol ·K) (298 K) In ( 1 .7 1 0-5) = 2.72094 x 1 04 Ilmol = 2.7 1 04 J/mol The large positive /lGo indicates that it would not be possible to prepare a solution with the concentrations of lead and chloride ions at the standard state concentration of 1 M A Q calculation using I M solutions will confirm this: PbCI 2 (s) =+ Pb 2+(aq) + CI - (aq) 2 Q = [Pb +] [Cr] 2 = ( I M) ( I M) 2 =1 Since Q > Ksp, it is impossible to prepare a standard state solution of PbCI 2 .

20.60

/lGo = -RT

20.62

a) The equilibrium constant, K, is related to /lGo through the equation /lGo = -RT In K. /lGo = -RT In K = -(8.3 1 4 J/mol·K) (298 K) In (9. 1 x 1 0-6 ) = 2.875776 x 1 0 4 = 2.9 X 1 04 J/mol b) Since /lG:n is positive, the reaction direction as written is nonspontaneous. The reverse direction,

x

X

formation

of reactants, is spontaneous, so the reaction proceeds to the left. c) Calculate the value for Q and then use to find /lG. 2 2 2 2 Fe 2 + Hg 2 + [ 0.0 1 0 ] [ 0.025 ] = 1 .5625 x 1 0-4 ( umounded) Q 2 ] ] [ [ 0.20 0.0 1 0 Fe3 + Hg� + =

[ J[ J [ r[ ]

=

RT In Q = 2.875776 1 04 J/mol + (8.3 1 4 Ilmol·K) (298) In ( 1 .5625 x 1 0-4 ) = 7 .044 1 87 1 03 = 7.0 1 03 J/mol Because /lG29 8 > 0 and Q > K, the reaction proceeds to the left to reach equilibrium.

/lG = /lGo +

X

X

20.64

X

Formation of 0 3 from O 2 : 3 0 2 (g) =+ 2 0 3 (g) or per mole of ozone: 3/2 0 2 (g) =+ 0 3 (g) a) To decide when production of ozone is favored, both the signs of /lH; and /lS; for ozone are needed. From Appendix B , the values of llH; and 5" can be used: llH ; = [ I mol 0 3 ( llH; 0 3 )] - [3/2 mol ( llH; O2 )] llH; = [ I mol( 1 43 kJ/mol] - [3/2 mol(O)] = 1 43 kllmol /lS ; = [ 1 mol 0 3 ( 5" 0 3 )] - [3/2 mol O 2 ( 5"0 2 )] 297

""S; = [ 1 mol (23 8.82 J/mol-K] - [3/2 mol (205 .0 llmol-K)] = -68.68 J/mol-K (unrounded). The positive sign for !'1f1; and the negative sign for M; indicates the formation of ozone is favored at no temperature. The reaction is nonspontaneous at all temperatures. b) At 298 K, ""Go can be most easily calculated from ""G; values for the reaction 3/2 02(g) =+ 03(g). From ""G; : ""Go = [ 1 mol 03 ( ""Gj 03)] - [3/2 mol ( ""G,} O2)] ""Go = [ 1 mol ( 1 63 kllmol 03)] - [3/2 mol (0)] = 1 63 kJ for the formation of one mole of 03. c) Calculate the value for Q and then use to find ""G. 0 3 ] [ 5 X 1 0 -7 atm [ = 5 . 1 95664 x 1 0-6 (unfounded) Q= 3 ti ti 3- = [ [ 02 ] 0.2 1 atm ] 3 ""G = ""Go + RT In Q = 1 63 kllmol + (8.3 1 4 J/mol-K) (298) ( l kJ I 1 0 J) In ( 5 . 1 95664 x 1 0-6) = 1 32.85368 = 1 x 1 0 2 kJ/mol

]

_ _

20.67 (a) (b) (c) (d) (e) ( f)

!'1f1rxn

""Srxn

+

0 + ( -) 0 +

( +) -

0 ( ) -

+

""Grxn

Comment

-

Spontaneous

( +)

-

Spontaneous Not spontaneous Spontaneous

+

Not spontaneous

(-)

T""S > !'1f1

a) The reaction is always spontaneous when ""Grxn < 0, so there is no need to look at the other values other than to check the answer. b) Because ""Grxn = !'1f1 - T ""S = -T ""S, M must be positive for ""Grxn to be negative. c) The reaction is always nonspontaneous when ""Grxn > 0, so there is no need to look at the other values other than to check the answer. d) Because ""Grxn = !'1f1 T M = !'1f1, !'1f1 must be negative for ""Grxn to be negative. e) Because ""Grxn = !'1f1 TM = T""S, M must be negative for ""Grxn to be positive. f) Because T ""S > !'1f1 the subtraction of a larger positive term causes ""Grxn to be negative. -

-

-

,

20.69

.

a) For the reactIOn, K = K=

[ Hb - CO] [ Hb - 02 ]

[ Hb - CO][02 ] [ since the problem states that [ 0 2] = [ CO ] ; the K expression simplifies to: Hb - 02 ][ CO ]

][ )

The equilibrium constant, K, is related to ""Go through the equation ""Go = -RT In K. ""Go - 1 4 kJ /mol 1 03 J In K = -- = -- = 5 .43 1 956979 ( unrounded) RT - ( 8.3 1 4 J /mol - K ) ( (273 + 37)K ) l kJ

[

3 K = e 5 4 1 9569 79 8 = 228 .596 = 2.3

x

1 02 =

[ Hb - CO] [ Hb - 02 ]

�---=-

b) By increasing the concentration of oxygen, the equilibrium can be shifted in the direction of Hb-02• Administer oxygen-rich air to counteract the CO poisoning.

298

20.72 a) 2 N20S(g) + 6 F2(g) 4 NF3(g) + 5 02(g) b) Use the values from Appendix B to determine the value of t:J.Go. t:J.G:n =[(4 mol) (t:J.G; (NF3» + (5 mol) (t:J.G; (02»] - [( 2 mol) (t:J.G; (N20S» + (6 mol) (t:J.G; (F2»] t:J.G:n =[(4 mol) (-83.3 kl/mol» + (5 mol) (0 kl/mol) ] - [(2 mol) ( 1 18 kl/mol) + (6 mol) (0 kl/mol)] -569.2 t:J.G:n = =-569 kJ c) Calculate the value for Q and then use to find t:J.G. _ [ NF3 1 [ 02 f - [ O. 25atm t[ O.50atm f = 47.6837 (unrounded) Q2 [ N20S f [ F2 ]6 [ O.20atm ] [ O.20atm ]6 3 + (I kJ I 10 J) (8.3 14 J/mol-K) ( 298) In ( 47.6837) t:J.G =t:J.Go + RT In Q =-569.2 kl/mol 2 =-559.625 =-5.60 10 kJ/mol 20.75 a) The chemical equation for this process is 3 C(s) + 2 Fe203(S) 3 CO2(g) + 4 Fe(s) tili:n =[(3 mol) (tili; (C02» + (4 mol) (tili; (F e»] - [(3 mol) (tili; (C» + (2 mol) (tili; (F e203»] tili:n =[(3 mol) (-393.5 kl/mol) + (4 mol) (0)] - [(3 mol) (0) + (2 mol) (-825.5 kl/mol)] 470.5 kJ t:J.S:n =[(3 mol) (S;(C02» + (4 mol) (S;(Fe»] - [(3 mol) (S;(C» + (2 mol) (S;(Fe203»] t:J.S:n =[(3 mol) (213.7 J/mol-K) + (4 mol) (27.3 J/mol-K) ] - [(3 mol) (5.686 J/mol-K) + (2 mol) (87.400 J/mol-K)] M:n =558. 4 42 =558.4 JIK b) The reaction will be spontaneous at higher temperatures, where the -Tt:J.S term will be larger in magnitude than tili. c) t:J.G;98 =tili:n - T t:J.S:n =470.5 kl - [(298 K) (558.442 JIK) (1 kl I 103 J)] = 304.084 = 304.1 kJ Because t:J.G is positive, the reaction is not spontaneous. d) The temperature at which the reaction becomes spontaneous is found by calculating t:J.G:n =0 =tili:n - T t:J.S:n tili:n =T t:J.S:n ° 470.5 kJ [ 103 J 1 = 842.5 225896 =842.5 T = tili = 558.442 J/K 1 kJ MO �

_

x

�

=

K

20.77 a) The balanced chemical equation is N20S(s) + H20(/) 2 HN Oi/) Calculate t:J.G:n for the reaction and see if the value is positive or negative. t:J.G:n =[(2 mol) (t:J.G; (HN03»] - [(1 mol) (t:J.G; (N20S» + (1 mol) (t:J.G; (H20»] t:J.G:n =[(2 mol) (-79.914 kJ/mol)] - [(1 mol) (1 14 kl/mol) + (1 mol) (-237.192 kl/mo!)] t:J.G:n =-36.636 =-37 kl Yes, the reaction is spontaneous because the value of t:J.G:n is negative. b) The balanced chemical equation is 2 N20s(s) 4 N02(g) + 02(g) The value of t:J.G:n indicates the spontaneity of the reaction, and the individual tili:n and t:J.S:n values are necessary to determine the temperature. t:J.G:n =[(4 mol) (t:J.G; (N02» + (1 mol) (t:J.G; (02 )) ] - [(2 mol) (t:J.G; (N20s )) ] t:J.G:n =[(4 mol) (51 kl/mol» + (I mol) (0)] - [(2 mol) (1 14 kl/mol) ] t:J.G:n =-24 kl Yes, the reaction is spontaneous because the value of t:J.G:n is negative. �

�

299

MI:" = [ (4 mol) (MI; (N02» + ( 1 mol) (MI; (02»] - [( 2 mol) (MI; (N20 S))] MI:" = [(4 mol) (33 .2 kl/mol) + (I mol) (0)] - [(2 mol) (-43 . lkJ/mol)] MI:,, = 2 1 9.0 kJ fhl:" = [(4 mol) (S"(N0 2» + ( 1 mol) (S"(02»] - [(2 mol) (S"(N20 S))] i1S:n = [(4 mol) (239.9 J/mol oK) + (I mol) (205 J/mol oK)] - [(2 mol) ( 1 78 J/moloK)] fhl :" = 808.6 JIK

[

MIo - 2 1 9.0 kJ 1 03 J = 270.83 8 = 2 70. 8K 808.6 J/K 1 kJ i1So c) The balanced che mical e quati on is 2 N 20 S( g) 44 N02(g) + 02( g) MI:" = [(4 mol) ( MI; (N02» + (I mol) (MI; (02)]- [ ( 2 mol) (MI; (N20 S))] MI:" = [(4 mol) (3 3 .2 kJ/mot) + (I mol) (0)] - [(2 mol) ( llkJ/mol)] MI:,, = 1 1 0.8 = III kJ fhl :" = [( 4 mol) (S"(N02» + (I mol) (S"(02»]- [(2 mol) (S°(N20 S))] fhl:" = [(4 mo t) (239.9 J/moloK) + (I mol) (205 J/mol oK)] - [(2 mol) (346 J/moloK)] fhl :n = 472.6 = 473 JIK T-

]

MIo l l O.8 k J l O 3 J = 234.4477 = 234 K = 472.6 J/K 1 kJ fhl o The te mpe rat ure is d iffe rent because the val ues fo r N 20S vary w ith physical state .

[

T=

20.8 1

]

a) Kp = 1 .00 when i1G = 0; co mb ine th is with i1G = MI- Tfhl. F irst, calculate MIand fhl, using values in the Append ix. MfD = [(2 mol NH ) (MI; N H ) ] - [( 1 mol N 2) (MI; N 2) + (3 mol H2) (MI; H 2)] 3 3 = [ ( 2 mol NH 3 ) (-45.9 kJ/mol)] - [( I mol N 2) (0) + ( 3 mol H 2) (0)] = -9 1 .8 kJ t1S" = [(2 mol NH ) ( SO N H ) ] - [( 1 mol N 2 ) ( SO N 2) + (3 mol H 2) ( SO H 2)] 3 3 = [(2 mol NH 3 ) ( 1 93 J/moloK)] - [( 1 mol N 2 ) ( l 9 l .50 J/mol oK) + (3 mol H 2) ( 1 30.6 J/moloK)] = -1 97.3 J/ K ( unro unded) i1G = 0 ( at e quil ib rium) i1G = 0 = MI- Tfhl MI= Tfhl -9 1 .8 kJ 1 0 3 J MIo = 465.28 1 = 465 K T - 1 97.3 J/K 1 kJ i1So b) U se the relationsh ip s: i1G = MI- Tfhl and i1G = -RT I n K w ith T = (273 + 400.) K 673 K i1G = MI- Ti1S= (-9 1 . 8 kJ) ( 1 0 3 J 1 1 kl) - (673 K) (- 1 97.3 JIK) i1G = 4.09829 x 1 04 J ( un rounded) i1Go = -RT I n K 4.09829 X 1 0 4 ll mol In K = i1Go I -RT = = -7.3 2449 (unrounded) - ( 8.3 1 4 J/ mol o K)( 673 K ) 44 = e -7 .3 2 9 = 6.59 1 934 x l O -4 = 6.59 X 10-4 c) The reaction rate is h igher at the h igher te mperature. The time re quired (kinetics) overshadows the lower yield (the rmodynamic s). _

_ _

[

]

=

K

[

]

300

Chapter 21 Electrochemistry: Chemical Change and Electrical Work FOLLOW-UP PROBLEMS 21.1

Plan: Follow the steps fo r balancin g a redox reacti on in acidic s ol uti on : 1 . Divide into half- reactions 2. Fo r each half- reaction balance a) Ato ms othe r than 0 and H, b) 0 ato ms with H2 0 , c) H a to ms with H+ and d) Charge with e -. 3. Multiply each half- reac ti on by an integer that will make the n umbe r of electrons l os t e qual to the nu mber o f electrons gained . 4. Add the half- reactions and cancel s ub stance s appearing as both reactants and p roducts. Then, add anothe r step fo r basic solution: 5. Add hyd roxide ion s to neutralize H+ . Cancel wate r. Solution: 1 . Divide into half- reactions: group the reac tan ts and p roduc ts wi th si milar ato ms . Mn0 4-(aq) � MnO /-(aq) naq) � lO ) -(aq) 2. For each half- reaction balance a) A to ms othe r than 0 and H Mn and r are balanced so no chan ge s needed b) 0 ato ms with H 2 0 , Mn0 4-(aq) � MnO/-(aq) 0 already balanced naq) + 3 H20( l) � rO) -(aq) Add 3 H20 to balance oxygen c) H ato ms with H+ Mn0 4-(aq) � MnO/-(aq) H al ready balanced naq) + 3 H2 0( l) � rO )-(aq) + 6 H +(aq) Add 6 H+ to balance hydro gen d) Cha rge with eMn0 4-(aq) + e- � Mn0 42-(aq) Total charge of reactants is -I and of p roducts is -2, so add I e- to reactants to balance charge: naq) + 3 H2 0( l) � rO) -(aq) + 6 H +(aq) + 6 e - Total charge is -I fo r reactants and +5 for p roducts, so add 6 e- as p roduct: 3. Multiply each half- reac tion by an inte ger that will make the n umbe r of electrons lost e qual to the n umbe r of elec tron s gained . One electron is gained and 6 are lost so reductio n must be multiplied by 6 fo r the n umbe r of elec tron s to be e qual. 6 {M n0 4-(aq) + e - � MnO /-(aq) } naq) + 3 H2 0( l) � rO)-(aq) + 6 H \aq ) + 6 e4. Add half- reactions and cancel s ubstance s appearing as both reactants and p roducts . 6 Mn0 4-(aq) + �- � 6 Mn042-(aq) nag) + 3 H20 CD � rO)-Cag) + 6 H \ag) + �Ove rall: 6 Mn0 4-(aq) + naq ) + 3 H2 0( l) � 6 MnO /-(aq) + rO)-(aq) + 6 H\aq) 5 . Add hyd roxide ions to neutralize H+ . Cancel wate r. The 6 H+ are neutralized by adding 6 O Ir. The same nu mbe r of hydroxide ions mus t be added to the reactants to keep the balance of 0 and H ato ms on both side s of the reaction. 6 Mn0 4-(aq) + naq) + 3 H2 0( l) + 6 O Ir(aq) � 6 MnO /-(aq) + lO)-(aq) + 6 H +(aq) + 6 OI1(aq) The neutralization reaction p roduce s wa ter: 6 { H+ + O Ir � H2 0 } . 6 Mn0 4-(aq) + naq) + 3 H2 0( l) + 6 O Ir(aq) � 6 MnO/-(aq) + 10) -(aq) + 6 H2 0( l)

301

Cancel water: 6 Mn0 4-(aq) + naq) + ��GtI1 + 6 OW(aq) -76 MnO /-(aq) + I0 3-(aq) + 6-H2 0( l) Balanced reaction i s 2 6 Mn0 4-(aq) + naq) + 6 OW(aq) -76 Mn0 4 -(aq) + I0 3-(aq) + 3 H2 0( l) Balanced reaction includin g spectato r ion s i s 6 KMn0 4(aq) + KI(aq) + 6 KO H(aq ) -7 6 K2 Mn0 4(aq) + KI0 3 (aq) + 3 H 2 0( l) Check: Check balance of atom s and charge: Reactants: P roducts: 6 Mn atoms 6 Mn atoms 1 I ato m I I ato m 30 0 ato ms 30 0 atoms 6 atoms 6 H atoms - 1 3 charge - 1 3 charge 2 1 .2

Plan: G iven the solution and electrode co mpo sition s, the two half cells involve the transfer of electrons I) between chromium in C r2 0 - and C r3 + and 2) between Sn and Sn2+ . The negative electrode is the anode so the tin half-cell i s where oxidation occurs. The graphite e lectrode with the chromium io n/chromate solution is where reduction occurs. In the cell dia gram, show the electrode s and the solutes involved in the half-reactions. I nclude the salt bridge and wi re connection between electrode s. Set up the two half-reactio ns and balance. ( Note that the Cr 3 +/Cr20 - half-cel l i s i n acidic solution.) W rite the cell notation placing the anode half-ce ll first, then the salt bridge , then the cathode half-cell. Solution: Cell dia gram: Voltmete r

/

/

ft

\\

Salt bridge

Sn ( )

C (+)

-

Cr 3 + , W C r20 7 2-

Sn2+

Balanced e quation s: Anode i s Sn/Sn2+ half-cell. Oxidation of Sn produces Sn2+ : Sn(s) -7Sn2 aq) All that needs to be balanced is charge: Sn (s) -7Sn2 aq) + 2 eCathode i s the C r3 +/C r2 0 7 2- half-cell. Check the oxidation number of chromium i n each substance to determine which i s reduced. C r3 + oxidation number i s +3 and chromium in Cr2 0 7 2- ha s oxidation number +6. Going from +6 to +3 involve s gain of electron s so C r2 0 - i s reduced. C rp -(aq) -7C r3 +( aq) Balance C r: C r2 0 -(aq) -72 C r3 +(aq) Balance 0: Cr2 0 -(aq) -72 C r3+(aq) + 7 H 2 0( l) Balance H : Cr2 0 -(aq) + 1 4 H +(aq) -72 C r3 aq) + 7 H2 0( l) Balance charge: C r2 0 -(aq) + 1 4 H + (aq) + 6 e- -72 C r3 aq) + 7 H2 0( l)

\ \

/ / l / l

l

\

\

302

Add two half-reactions2multiplying the tin half-reaction by 3 to equalize the number of electrons transferred. 3 {Sn(s) Sn \aq) + 2 e-} CrzO/-(aq) + 14 H\aq) + 6 e- 2 Cr3\aq) + 7 HzO(!) 3 Sn(s) + CrzO/-(aq) + 14 H +(aq) 3 Sn2+(aq) + Cr3\aq) + 7 HzO(!) Cell notation: Sn(s) I SnZ\aq) II H +(aq), CrzO/-(aq), Cr3+(aq) I C(graphite) 21. 3 Plan: Divide the reaction into half-reactions showing that Br2 is reduced and y3+ is oxidized. Use the equation E;ell = E;o''''''I< - E:', Oe to solve for E:,lOde . O Solution: Half-reactions: Reduction (cathode): Brz(aq) + 2 e- 2 Br-(aq) E:o,,,
�

�

�

2

�

�

_

21. 4 Plan: To determine if the reaction is spontaneous, divide into half-reactions and calculate E:ell. If E;ell is negative, the reaction is not spontaneous, so reverse the reaction to obtain the spontaneous reaction. Reducing strength increases with decreasing EO. Solution: Divide into half-reactions and balance: E:"," Oe = -0.44 Y Reduction: FeZ\aq) + 2 e- Fe(s) O 3 E:nlNle = 0. 7 7 Y Oxidation: 2 {FeZ\aq) Fe \aq) + e-} The first half-reaction is reduction, so it is the cathode half-cell. The second half-reaction is oxidation, so it is the anode half-cell. Find the half-reactions in Appendix D. E;ell = E;aI"OOe - E:"O
�

2

�

Fe

>

Fe

2

>

Fe

3

�

I'lGo,

I'lGo

I'lGo

[

I'lGO

][

2

X

1.17

X

1025

)[

_[

303

0.741 V

K.

Check: Reverse the calculation of K: = -RT In K = - [(8.3 1 4 J/mol·K) (298 K) In ( 1 . 1 655237 x 1 0 2 5)] ( 1 kJ 1 1 0 3 J) = 1 43 The potential can easily be checked against the value calculated from Appendix D. E:ell = E;orhode - E:node = (0.34 V) - (-0.40 V) = 0.74 V The potential value agrees with this literature value.

�Go

21 .6

P lan: The problem is asking for the concentration of iron ions when . RT Use the Nernst equatIOn, Ecel! = E:'II - � In Q to find [Fe2+] .

Ecell

=

E:ell +

kJ

0.25 V.

Solution: Determining the cell reaction and E;'II : Fe(s) � Fe2 \aq ) + 2eEO = -0.44 V Oxidation: Reduction: Cu2+( aq ) + 2e- � Cu(s) EO = 0.34 V Overall Fe(s) + Cu2+(aq) � Fe 2+( aq) + Cu(s) E;ell = E:arhode - E:node = 0.34 V - (-0.44 V) = 0.78 V For the reaction Q = [Fe2+]/[Cu2+], so Nemst equation is 0.0592 [Fe 2+ ] O I og --2 - - or Eeel! = E cell n [Cu + ] Ecel! = E:
_

_

21 .7

Plan: Half-cell B contains a higher concentration of gold ions, so the ions will be reduced to decrease the concentration while in half-cell A, with a lower [Au 3 +], gold metal will be oxidized to increase the concentration of gold ions. In the overall cell reaction, the lower [Au3 +] appears as a product and the higher [Au 3 + ] appears as a reactant. This means that Q = [Au 3+] lowe,![Au 3 +] higher. Use the Nemst equation with E;ell = 0 to find Ecell' Oxidation of gold metal occurs at the anode, half-cell A, which is negative. Solution: RT 0.0592 [Au 3+ lower ] O Ecell = Ecell In Q = Ecaell log 3+ n = 3 mol e- for the reduction of Au3 + to Au nF n [Au higher ] 0.0592 [7.0 x 1 0 -4] Ecell = 0 log = 0.0306427 = 0.031 V 3 [2.5 x 1 0-2 ] _

_

304

21.8

Check: The potential of the cell can be estimated by realizing that each tenfold difference in concentration between the two half cells gives a 20 mY change in potential. The difference is more than tenfold but less than 100 times, so potential should fall between 20 and 40 mY. Plan: In aqueous AuBr3 , the species present are Au3\3aq), Br-(aq), H 20, and very small amounts of H+ and OH-. The possible half-reactions are reduction of either Au + or H20 and oxidation of either Br- or H Whichever reduction and oxidation half-reactions are more spontaneous will take place, with consideration2of0. the overvoltage. Solution: The3 two possible reductions are 1.50 Y Au \aq) + 3 e- Au(s) -1 ( with overvoltage) 2 H 2 0(Z) + 2 eH 2(g) + 2 OH-(aq) The reduction of gold ions occurs because it has a higher reduction potential (more spontaneous) than reduction of water. The two possible oxidations are 1.07 Y 2 Br-(aq) Br (Z) + 2 eE 1.4 Y (wi t h overvol t age) 2 H 2 0(Z) 02(g)2 + 4 H +(aq) + 4 eThe oxidation with the less positive reduction potential is the more spontaneous oxidation so bromide ions are oxidized at the anode. The two half-reactions that are3 predicted are Cathode: Au +(aq) + 3 e- Au(s) 1.50 Y 1.07 Y 2 Br-(aq) Br2(Z) + 2 eAnode: Plan: Current is charge per time, so to find the time, divide the charge by the current. To find the charge transferred, first write the balanced half-reaction, then calculate the charge from the grams of copper and moles of electrons transferred per molar mass of copper. Solution: 2 Cu \aq) + 2 e- Cu(s), so 2 mol e- per mole of Cu (1.50 g cu ) ( 1 mol CU J [ 2 mol e- ) [ 96485 C ) ( � J( _I- J ( I min ] 15.9816 63.55 g Cu I mol Cu 1 mol e- Cis 4.75 A 60 s �

EO = E=

�

£D

�

= =

�

�

21.9

Y

�

£D

= EO =

�

=

= 16.0 min

END-OF-CHAPTER PROBLEM S

21.1

21.2

21.3

21.5

Oxidation is the loss of electrons (resulting in a higher oxidation number), while reduction is the gain of electrons (resulting in a lower oxidation number). In an oxidation-reduction reaction, electrons transfer from the oxidized substance to the reduced substance. The oxidation number of the reactant being oxidized increases while the oxidation number of the reactant being reduced decreases. one half-reaction cannot take place independently of the other because there is always a transfer of electrons from one substance to another. If one substance loses electrons (oxidation half-reaction), another substance must gain those electrons (reduction half-reaction). Spontaneous reactions, 0, take pl a ce in vol t aic cells, which are also called galvanic cells. Nonspontaneous reactions take place in electrolytic cells and result in an increase in the free energy of the cell 0). a) To decide which reactant is oxidized, look at oxidation numbers. is oxidized because its oxidation number increases from - I in to in C12 . b) is reduced because the oxidation number of Mn decreases from +7 in Mn04- to +2 in Mn2+. c) The oxidizing agent is the substance that causes the oxidation by accepting electrons. The oxidizing agent is the substance reduced in the reaction, so is the oxidizing agent. d) is the reducing agent because it loses the electrons that are gained in the reduction. e) which is losing electrons, which is gaining electrons. f) 8 H2S04(aq) + 2 KMn04(aq) + 10 KCI(aq) 2 MnS04(aq) + 5 C!z(g) + 8 H 20(Z) + 6 K2 S04(aq) No,

t1.Gsys

cr

<

(t1.Gsys >

C I-

a

Mn04-

Mn04-

CI-

From C I-,

to M n04-, �

305

21.7

a) Divide into half-reactions: CI03-(aq) -7 Cnaq) qaq) -712(s) 0 H I C 03-(aq) -7 Cnaq) 2 qaq) -7I2(s) H20 0 IO}-(aq) -7 Cnaq) + H20(l) C 2 qaq) -7I2(s) H+ H CIO}-(aq) + 6 H\aq) -7 Cnaq) + H20(l) 2 qaq) -7 12(s)

Balance elements other than and

chlorine is balanced iodine now balanced Balance by adding add 3 waters to add 3 0's to product 3 no change Balance by adding 3 add 6H+ to reactants no change Balance charge by adding eCIO}-(aq) + 6 H+(aq) + 6 e- -7 Cnaq) + 3 H20(l) add 6 e- to reactants for a -I charge on each side 2 qaq) -7I2(s) + 2 ee- to products for a -2 charge on each side Multiply each half-reaction by an integer to equalize the numberaddof2electrons CIO}-(aq) + 6 H+(aq) + 6e- -7 CI-(aq) + 3 H20(l) multiply by 1 to give 6 emultiply by 3 to give 6 e3 {2 qaq) -7 12(S) + 2 e-} Add half-reactions to give balanced equatiol] in acidic solution. 3 31 I Check balancing: C 03-(aq) + 6H+(aq) + 6 qaq) -7 Cnaq) + HP(l) + 2(s) Reactants: I CI Products: CI 3I 0 30 6 6H 6T

H 61

-I

-1

charge Oxidizing agent is CI03- and reducing agent is 1-. b) Divide into half-reactions: Mn04-(aq) -7 Mn02(s) SO/-(aq) -7 SO/-(aq) 0 H Mn04-(aq) -7 Mn02(S) SO/-(aq) -7 SO/-(aq) 0 H20 Mn04-(aq) -7 Mn02(S) + 2 H20(l) SO/-(aq) + H20(l) -7 SO/-(aq) H+ H Mn04-(aq) + 4 H+(aq) -7 Mn02(s) + 2 H20(l) SO/-(aq) + H20(l) -7 SO/-(aq) + 2 H+(aq)

Balance elements other than and

charge

is balanced is balanced Balance by adding add 2 H20 to products add H20 to reactants Balance by adding add 4 H+ to reactants add 2 H+ to products Balance charge by adding eMn04-(aq) + 4 H+(aq) + 3 e- -7 Mn02(S) + 2 H20(l) add 3 e- to reactants for a 0 charge on each side 2 SO/-(aq) + H20(l) -7 S04 -(aq) + 2 H\aq) + 2 e- add 2 e- to products for a -2 charge on each side Multiply each half-reaction by an integer to equalize the number of electrons multiply by 2 to give 6e2 {Mn04-(aq) + 4 H+(aq) + 3 e- -7Mn02(s) + 2 H20(l)} 3 {SO/-(aq) + H20(l) -7 SO/-(aq) + 2 H+(aq) + 2 e-} multiply by 3 to give 6eAdd half-reactions and cancel substances 2that appear as both reactants and products 2 Mn04-(aq) + &-H+(aq) + 3 S03 -(aq) + -7 2 Mn02(s) + 4=H20(l) + 3 SO/-(aq) + M4"� The balanced equation in acidic solution is: 2 Mn04-(aq) + 2 H+(aq) + 3 SO/-(ag) -7 2 Mn02(S) + H20(l) + 3 SO/-(aq) To change to basic solution, add OH- to both sides of equation to neutralize H+. 2 Mn04-(aq) + 2 H+(aq) + 2 OW(aq) + 3 S032-(aq) -7 2 Mn02(s) + H20(l) + 3 SO/-(aq) + 2 OW(aq) Mn04-(aq) + H20( l) + 3 SO/-(aq) -7 2 Mn02(S) + mGfl1 + 3 SO/-(aq) + 2 OW(aq) Balanced2equation in basic solution: 2 Mn04-(aq) + H20(l) + 3 S032-(aq) -7 2 Mn02(s) + 3 SO/-(aq) + 2 OW(aq) J-H;oGfA

�

306

Mn S

I

Check balancing: Reactants:

Products: 2Mn 2Mn 18 0 18 0 2H 2H 3S 3S -8 charge -8 charge Oxidizing agent is Mn04- and reducing agent is SO/-. c) Divide into half-reactions:

Mn04-(aq) � Mn2\aq) HZ02(aq) � 02(g) Balance elements other than 0 and H Mn04-(aq) � Mn2+(aq) Mn is balanced H20Z(aq) � 02(g) N o other elements to balance Balance 0 by adding H20 add 4 H20 to products Mn04-(aq) � Mn2+(aq) + 4 HzO(l) HZ02(aq) � 02(g) o is balanced Balance H by adding H+ add 8 H+ to reactants Mn04-(aq) + 8 H\aq) � Mn2\aq) + 4 H20(l) add 2 H + to products H202(aq) � 02(g) + 2 H+(aq) Balance charge by adding e add 5 e- to reactants for +2 on each side Mn04-(aq) + 8 W(aq) + 5 e- � Mn2+(aq) + 4 H20(l) add 2 e- to products for 0 charge on each side H202(aq) � 02(g) + 2 H+(aq) + 2 eMultiply each half-reaction by an integer to equalize the number of electrons multiply by 2 to give 1 0 e2 {Mn04-(aq) + 8 H+(aq) + 5 e- � Mn2+(aq) + 4 H2 0( l)} multiply by 5 to give 1 0 e5 {H202(aq) � 02 ( g) + 2 H+(aq) + 2 e -} Add half-reactions and cancel substances that appear as both reactants and products 2 Mn04-(aq) + M-H+(aq) + 5 H20Z(aq) � 2 Mn2\aq) + 8 H20(l) + 5 02(g) + -W-H"'taqj The balanced equation in acidic solution 2 2 Mn04-(aq) + 6 H+(aq) + 5 H20Z(aq) � 2 Mn +(aq) + 8 H20(l) + 5 02 ( g) Check balancing: Reactants: Products: 2 Mn 2Mn 18 0 18 0 16 16 H +4 charge +4 charge Oxidizing agent is Mn04- and reducing agent is H202.

H

21 . 1 0

a) Balance the reduction half-reaction: balance 0 N03-(aq) � NO(g) + 2 H 2 0(l) balance H N03-(aq) + 4 H+(aq) � NO(g) + 2 H20(l) balance charge N03-(aq) + 4 H+(aq) + 3 e- � NO(g) + 2 H20(l) Balance oxidation half-reaction: balance Sb 4 Sb(s) � Sb406 (s) balance 0 4 Sb(s) + 6 H20(l) � Sb406(s) balance H 4 Sb(s) + 6 H20(l) � Sb406(s) + 1 2 H+(aq) balance charge 4 Sb(s) + 6 H 2 0( l) � Sb406 (s) + 1 2 H\aq) + 1 2 eMultiply each half-reaction by an integer to equalize the number of electrons Multiply by 4 to give 1 2 e4{ N03-(aq) + 4 H+(aq) + 3 e- � N O( g) + 2 H 2 0( l)} Multiply by 1 to give 1 2 eI {4 Sb(s) + 6 H 20(l) � Sb406 (s) + 1 2 H+(aq) + 1 2 e- } Add half-reactions. Cancel common reactants and products. 4 N03-(aq) + M-H+(aq) + 4 Sb(s) + �G( l) � 4 NO(g) + 8-HzO(l) + Sb406(s) + �"'taqj Balanced equation in acidic solution: 4 N03-(aq) + 4 H\aq) + 4 Sb(s) � 4 NO(g) + 2 H20(l) + Sb406(s) Oxidizing agent is N 03- and reducing agent is Sb.

307

b) Balance reduction half-reaction: balance 0 Bi03-(aq) Bi 3+\aq) + 3 H320(l) balance H Bi03-(aq) + 6 H+(aq) Bi \aq) 3++ 3 H 20(l) balance charge to give +3 on each side Bi 0(l) e6 (aq) + 3 H (aq) + 2 Bi0 -(aq) + H 2 3 half-reaction: Balance oxidation balance 0 Mn22+(aq) + 4 H 20(l) Mn04-(aq) balance H Mn2\+ aq) + 4 H 20(l) Mn04-(aq) + 8 H +\aq) balance charge to give +2 on each side Mn + 4 H 2 0(l) Mn04-(aq) + 8 H (aq) + 5 e(aq) Multiply each half-reaction by an integer to equalize of electrons 3+(aq) +the3 Hnumber Multiply by 5 to give 1 0 eBi 0(l)} 5 {Bi0 3-(aq) + 6 H \ aq) + 2 e2 Mul tiply by 2 to give 1 0 e2 {Mn2+(aq) + 4 H 2 0(l) Mn04-(aq) + 8 H +(aq) + 5 e-} Add half-reactions. Cancel H 20 and H + in2 reactants and products. 3 Bi03-(aq)in acidic + W-H +(aq) + 2 Mn \ aq) + �G(l) 5 Bi +(aq) + B.-H 2 0(l) + 2 Mn0 4-(aq) + � Balanced5reaction solution: 2 5 Bi 3 +(aq) + 7 H 2 0( l) + 2 Mn0 4-(aq) 5 Bi0 3 -(aq) + 1 4 H+(aq) + 2 Mn \ aq) 2 Bi0 - is the oxidizing agent and Mn + is the reducing agent. c) Balance the3 reduction half-reaction: balance 0 Pb(OH) -(aq) Pb(s) + 3 H 20(l) balance H Pb(OH)33-(aq) + 3 H+(aq) Pb(s) + 3 H 2 0(l) Pb(OH)3-(aq) + 3 H \aq) + 2 e- Pb(s) + 3 H 20(l) balance charge to give 0 on each side Balance the oxidation half-reaction Fe(OH hCs) + H 20(l) Fe(O HMs) balance 0 Fe(OHhCs) + H 20(l) Fe(O HMs) + H \+ aq) balance H Fe(O H hCs) + H 20(l) Fe(O H)3(S) + H (aq) + ebalance charge to give 0 on each side Multiply each half-reaction by an integer to equalize the number of electrons I {Pb(O H)3-(aq) + 3 H \aq) + 2 e- Pb(s) + 3 H 20(l)} Multiply by I to give 2 eMultiply be 2 to give 2 e2 {Fe(OHh(s) + H 2 0(l) Fe(O HMs) + H +(aq) + e-} Add the two half-reactions. Cancel 0 and H Pb(OH)3-(aq) + �H +(aq) + 22 Fe(OHhCs) + ;!...H;;G(l) Pb(s) + �H20(l) + 2 Fe(O HHs) + ;!...H"� Pb(s) + H 20(l) + 2 Fe(O H)3(S) + H \ aq) + 2 Fe(O Hh(s) 3-(aq)sides Add one Pb(OH) OH- to both to neutralize H +. Pb(OH)3-(aq) + H\eq) OH-(aq) + 2 Fe(OHhCs) Pb(s) + H 20(l) + 2 Fe(OHMs) + OW(aq) GfI) + 2 Fe(O HhCs) Pb(s) + H;;GfI) + 2 Fe(O H Ms) + OW(aq) BalancedPb(OH) reaction3-(aq) in basic+ H;;solution: Pb(OH)3-(aq) + 2 Fe(OHhCs) Pb(s) + 2 Fe(OHMs) + OW(aq) Pb(OH)3- is the oxidizing agent and Fe(OH)2 is the reducing agent. a) Balance reduction half-reaction: balance 0 N03-(aq) N0+2(g) + H 20(l) balance H N03-(aq) + 2 H (aq) N02(g) + H 20(l) balance charge to give 0 on each side N0 -(aq) + 2 H \ aq) + e- N0 2 (g) + H 2 0(l) 3 Balance oxidation half-reaction: Au(s) + 4 Cr(aq) AuCI4-(aq) balance CI Au(s) + 4 Cr(aq) AuCI4-(aq) + 3 ebalance Multiply each half-reaction by an integer to equalize the number of electrons charge to on each side Multiply by 3 to give e3 {N0 3-(aq) + 2 H +(aq) + e- N0 2 (g) + H 2 0(l)} Multiply be 1 to give 3 e1 {Au(s) + 4 Cr(aq) AuCI 4-(aq) + 3 e-} Add half-reactions. Au(s)agent + 3 N0 3-(aq) + 4 Cr(aq) + 6 H +(aq) AuCI 4-(aq) + 3 N02 (g) + 3 H 2 0( l) b) Oxidizing is and reducing agent is c) The HCI provides chloride ions that combine with the unstable gold ion to form the stable ion, AuCLt-. -?

-?

-?

-?

-?

-?

-?

-?

-?

M-H"

-?

-?

-?

-?

-?

-? -?

-?

-?

W.

-?

-?

=i=

-?

-?

-?

2l . 1 2

-?

-?

-?

-?

-4

-?

-?

N03-

3

-?

-?

Au.

308

21.13 a) is the anode because by convention the anode is shown on the left. b) is the cathode because by convention the cathode is shown on the right. c) is the salt bridge providing electrical connection between the two solutions. d) is the anode, so oxidation takes place there. Oxidation is the loss of electrons, meaning that electrons are the anode.a positive charge because it is the cathode. e)leavingis assigned gains mass because the reduction of the metal ion produced the metal. 21.16 An active electrode is a reactant or product in the cell reaction, whereas an inactive electrode is neither a reactant nor a product. An inactive electrode is present only to conduct electricity when the half-cell reaction does not include a metal. Platinum and graphite are commonly used as inactive electrodes. 21.17 a) The metal is being oxidized to form the metal cation. To form positive ions an atom must always lose electrons, so this half-reaction is always an oxidation. b) The metal ion is gaining electrons to form the metal B, so it is displaced. c) The anode is the electrode at which oxidation takes place, so metal is used as the anode. d) Acid oxidizes metal B and metal B oxidizes metal A, so acid will oxidize metal A and when metal A is placed in acid. The same answer results if strength of reducing agents is considered. The fact that metal A is a better reducing agent than metal B indicates that if metal B reduces acid then metal A will also reduce acid. 21 .l8 a)Oxidation: Zn(s) Zn2+(aq) + 2e- (oxidation takes place at the negative electrode) 2 Reduction: Sn \aq) + 2e- Sn(s) Overall reaction: Zn(s) + Sn2+(aq) Zn2+(aq) + Sn(s) b) Voltmeter e e A

E

C

A

E f) E

A

B

A

b ubbles will form

�

�

�

\\

Zn () 1M Zn2+

�

Salt bridge

Sn (+)

-

It Amon flow

ICatIon t. Sn1 M2+ flow

21 .20 a) Electrons flow from the anode to the cathode, so left to right in the figure. By convention, the anode appears on the left and the cathode on the right. b) Electrons Oxidationenter occurstheatreduction the anode,half-cell, which isthethe electrode in the half-cell. c) half-cell in this example. d) Electrons are consumed in the reduction half-reaction. Reduction takes place at the cathode, electrode. e) The anode is assigned a negative charge, so the electrode is negatively charged. Metal is oxidized in the oxidation half-cell, so the electrode will decrease in mass. g) The solution must contain nickel ions, so any nickel salt can+ be added. is one choice. h) KN03 is commonly used in salt bridges, the ions being K Other salts are also acceptable answers. because an inactive electrode could not replace either electrode since both the oxidation and the i) reduction half-reactions include the metal as either a reactant or a product. j ) Anions will move toward the half-cell in which positive ions are bei n g produced. The oxidation half-cell produces Fe2+, so salt bridge anions move (nickel half-cell) (iron half-cell). Fe(s)2+ Fe2+(aq) + 2 ek) Oxidation half-reaction: Reduction half-reaction: Ni (aq) +22 e- Ni(s)2 Overall cell reaction: Fe(s) + Ni \aq) Fe \aq) + NiCs) from the iron half-cell to the nickel half-cell, iron

nickel

nickel

iron iron

f)

1 M NiS04

and N03-.

Neither

from right �

to left

�

�

309

21.22 fn cell notation, the oxidation components of the anode compartment are written on the left of the salt bridge and the reduction components of the cathode compartment are written to the right of the salt bridge. A double vertical line separates the anode from the cathode and represents the salt bridge. A single vertical line separates species of different phases. Anode I Cathode a) Al is oxidized,3+so it is the3+ anode and appears first in the cell notation: AI(s)IAI (aq)IICr (aq)ICr(s) reduced, so Cu is the cathode and appears last in the cell notation. The oxidation of S02 does not b)include Cu2+ ais metal , so an inactive electrode must be present. H ydrogen ion must be included in the oxidation half­ cell. 21.25 A negative E:ell indicates that the cell reaction is not spontaneous, > O. The reverse reaction is spontaneous with E;'II > O. 21.26 Similar to other state functions, the sign of E O changes when a reaction is reversed. Unlike and EO is an intensive property, the ratio of energy to charge. When the coefficients in a reaction are multiplied by a factor, the values of and are multiplied by the same factor. However, E O does not change because both the energy and charge are mUltiplied by the factor and their ratio remains unchanged. 21.27 a) Divide the balanced equation into reduction and oxidation half-reactions and add electrons. Add water and hydroxide ion to the half-reaction that includes oxygen. Oxidation: Se2-(aq) -7Se(s) + 2 eReduction: 2 SO/-(aq) + 3 H20(l) + 4 e--7S2032-(aq) + 6 OI1(aq) b) E ;'II = E �""ode - E:node E'�node = E�,,'ux'e - E :ell = -0.57 V - 0.35 V = 21.29 The greater (more positive) the reduction potential, the greater the strength as an oxidizing agent. a) 3From Appendix D:2 Fe +(aq) + e--7 Fe +(aq) EO = 0.77 V EO = 1.07 V Br22(l) + 2 e- -7 2Br-(aq) EO = 0.34 V Cu +(aq) + e- -7 Cu(s) + When placed in order of decreasing strength as oxidizing agents: b) 2From Appendix D: E O = -2.87 V Ca +(aq) + 2e- -7 Ca(s) Cr20l-(aq) + 14H\aq) 6e- -7 2Cr3\aq) + 7H20(l) EO = 1.33 V EO = 0.80 V Ag\aq) + e- -7 Ag(s) When placed in order of increasing strength as oxidizing agents: + 21.31 EOcell = Ecafhod E aOnode O e EO val u es are found in Appendix D . Spontaneous reactions have E:ell> O. a) Oxidation: Co(s)+ -7C02\aq) + 2 eEO = -0.28 V Reduction: 2 H (aq) + 2 e- -72 H2(g) E O = 0.00 V Overall reaction: Co(s) + 2 H+(aq) -7C0 +(aq) + H 2(g) E :ell = 0.00 - (-0.28 V) = Reaction is under standard state conditions because E :ell is positive. b) Oxidatio n: 2 {Mn2+(aq) + 4 H 20(l) -7Mn04-(aq) + 8 H\aq) + 5 e-} EO = +1.51 V 5 {Br2(l) + 2 e- -72 Br-(aq)} Reduction: 2 E O = +1.07 V Overall: 2 Mn \aq) + 5 Br2(l) + 8 H20(l) -72 Mn04-(aq) + 10 Br-(aq) + 16 H+(aq) E ;ell = 1.07 - 1.51 V = Reaction is under standard state conditions with E ,�ell O. tlGo

tlGo,!lff'

tlGo, !lff'

S",

S"

-0.92 V

3

Br2 > Fe + > C u 2 .

Ca2+

<

Ag

<

Cr20/-.

_

0.28 V

spontaneous

-0.44 V

<

not spontaneous

310

Oxidation: Hg/\aq) � 2 Hg2+(aq) + 2 e­ EO=+0.92V Reduction: Hg/+(aq) + 2 e- � 2 Hg(l) EO=+0.85V Overall: 2 Hg22+(aq) � 2 Hg2 2+(aq) + 2 H g(l) or H g/\aq) � Hg \aq) + H g(l) E:fI = 0.85 - 0.92V= Negative E:efl indicates reaction is under standard state conditions. EO values are found i n Appendix D. Spontaneous reactions have E:efl> O . E:efl = E;""OOe - E:"OOe Ag\aq) + e-} a) Oxidation: 2 {Ag(s) EO=0.80V Reduction: Cu2\aq) + 2 eCu(s) EO=+0.34V Overall: 2 Ag(s) + Cu2\aq) 2 Ag\aq) + Cu(s) E:fI = +0.34 - 0.80V= The reaction is b) Oxidation: 3 {Cd(s) Cd2\aq) + 2 e-} 3 EO= -0.40V Reduction: Cr20/-(aq) + 14 H+(aq) + 6 e-� 2 Cr 3+(aq) + 7 H 20(l) EO=+ l.33V Overall: Cr20l-(aq) + 3 Cd(s) + 14 H +(aq) � 2 Cr +(aq) + 3 Cd2+(aq) + 7 H 20(l) E:efl =+1.33 -(-0.40)V=+1.73 The reaction is c) Oxidation: Pb(s)2 � Pb2+(aq) + 2 eEO=-0.13V Reduction: Ni \aq) + 2 e- � Ni(s)2 EO=-0.25V E :efl =-0.25 - (-0.13)V= Overall: Pb(s) + N?\aq) � Pb \aq) + Ni(s) Spontaneous reactions have E:efl> O. All three reactions are written as reductions. When two half-reactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction that will result in a positive value of E:efl . Adding (I) and (2) to give a3 spontaneous reaction involves converting ( I ) to oxidation: Oxidation: 2{AI(s) � AI +(aq) + 3 e-} EO �1.66V Reduction: 3 {N204(g) + 2 e- 2 N02-(aq)} EO= 0.867 V E:efl= 0.867 V -(-1.66V)= 3 N 2 04 (g) + 2 AI( s ) � 6 N0 2-(aq) + 2 AI 3 +(aq) Adding ( I ) and (3) to give3a spontaneous reaction involves converting to oxidation: Oxidation: 2 {AI(s) � AI +(aq) + 3 e-} 2 EO �1.66V Reduction: 3 {SO/-(aq) + H 20(l) + 2 e-� S03 3-(aq) + 2 OW(aq)} EO=0.93V AI(s) + 3 SO/-(aq) + 3 H 20(l) � 2 AI \aq) + 3 SO/-(aq) + 6 OW(aq) E,�efl =0.93V - (-1.66V)= Adding (2) and (3) to give a spontaneous reaction involves converting 2 to oxidation: EO= 0.867 V Oxidation: 2 N02-(aq) � N 204(g) + 2 eReduction: SO/-(aq) + H 20(l) + 2 e- SO/-(aq) + 2 0f1(aq) EO= 0.93 V SO/-(aq) + 2 N02-(aq) + H 20(l) � SO/-(aq) + N204(g) + 2 Of1(aq) c)

-0.07 V

not spontaneous

21.33

�

�

�

-0.46 V

not spontaneous. �

spontaneous.

-0.12 V

21.35

�

2.53 V

(J)

2

2.59 V

�

E:efl =0.93V - 0.867 V=0.06 V 21.37

Spontaneous reactions have E:efl> O . All three reactions are written as reductions. When two half-reactions are paired, one half-reaction must be reversed and written as an oxidation. Reverse the half-reaction that will result in a positive value of E:efl . Adding (I) and (2) to give a spontaneous reaction involves converting (2) to oxidation: Oxidation: Pt(s) � Pt2\aq) + 2 eEO= 1.20V Reduction: 2 HCIO(aq) + 2 H\aq) + 2 e- CI2(g) + 22 H 20(l) EO=1.63V 2 HCIO(aq) + Pt(s) + 2 H\aq) CI2(g) + Pt \aq) + 2 H 20(l) �

�

E:efl =1.63V - 1.20V =0.43 V

311

Adding ( I ) and (3) to give a spontaneous reaction involves converting (3) to oxidation: Oxidation: Pb(s) + SO/-(aq) �+ PbS04(s) + 2 eEO = -0.3 1 Reduction: 2 HCIO(aq) + 2 H (aq) + 2 e- CI2(g) + 2 H20(l) EO 1 .63 2 HCIO(aq) + Pb(s) + SO /-(aq) + 2 H\ aq) � CI 2 (g) + PbS04(s) + 2 H 2 0(l) E:
V

V)

=

1.94 V

( )

V V

V

2 1 .39

V V

= 1.51 V

C

would also reduce acid to form H 2 bubbles. C

2 1 .42

Ecell

>

A

>

B.

>

+ + [A j increases and IB I decreases. decreases

=

+ [ A j must

+ equal [B ]

+ [A ]

Yes.

21 .44

>

+ [B ].

cathode

2 1 .45

K.

V V

=

V

5

(

V;

V = 3 X 10 3 5

312

V V

b) Oxidation: 3 { Fe(s) � Fe2 \aq) + 2 e-} EO = -0.44 Reduction: 2{Cr 3 +(aq) + 3 e- � Cr(s)} EO = -0.74 E:ell = E :,tJuxle - E:node = -0.74 - (-0.44 = -0.30 6 electrons are transferred. 6 ( -0.30 V) nP log K= = = -30.4054 (unrounded) 0.0592 0.0592 V K=3.93 1 8 X 1 0-31 = 4 X 10-31

V

V)

V;

--

2 1 .47

Substitute J/C for V. a) /',.Go = -nFEo =-(2 mol e-) (96485 Clmol e-) ( 1 .05 J/C) = -2.026 1 85 x 1 05 =-2.03 X 105 J b) /',.Go = -nFEo = -(6 mol e-) (96485 Clmol e-) (-0.30 llC) = 1 .73673 x 1 05 = 1.73 X 105 J

2 1 .49

Find /',.Go from the fact that /',.Go =-RT In K. Then use /',.Go value to find E:ell from /',.Go = -nFEO. T =(273 + 25) K = 298 K /',.Go = -RT In K = -(8.3 1 4 J/moloK) (298 K) In (5.0 x 1 0 3 ) = -2. 1 1 0 1 959 X 1 04 = -2.1 X 104 J -2. 1 1 0 1 959 x 1 0 4 J /',.G O =0. 2 1 8707 =0.22 V = E =_ nF (I mol e-)(96485 Clmol e-) I llC

(�)

2 1 .5 1

Since this is a voltaic cell, a spontaneous reaction is occurring. For a spontaneous reaction between H 2/H + and Cu/Cu2+ , Cu2+ must be reduced and H 2 must be oxidized: EO = 0.00 Oxidation: H2 (g) � Cu(s) + 2 H+( aq ) + 2 eReduction: Cu2+( aq ) + 2 e- � Cu(s) EO = 0.34 Cu2+( aq ) + H2 (g) � Cu(s) + 2 H+(aq) E:ell = E�alhode - E�node =0.34 V - 0.00 V = 0.34 0.0592 log Q Ecell = E,�ell n [ H+ 0.0592 log Ecell = E :ell n [ Cu 2 + J [ H2 ] For a standard hydrogen electrode [H+] = 1 .0 M and [H2 ] = 1 .0 atm 1 .0 0.0592 log 0.25 = 0.34 2 [ Cu 2 + J t .0

V V

V

r

_

V

V

0.25

_

V - 0.34 V = V

-0.090 =

_

_

1 .0 0.0592 log 2 [ Cu 2+ JI.O

1 .0 0.0592 log 2 [ Cu 2 + JI.O

-3 .04054 = log

[

1 .0 Cu 2 + J 1 .0

Raise each side to

1 9 .1 0876 x 1 0 -4 = -2+ -[Cu ] [Cu2+] = 9. 1 0876 x 1 0 -4 =9 X 10-4 M

313

lOX

21.53

The spontaneous reaction (voltaic cell) involves the oxidation of Co and the reduction ofNi 2 +. EO V Oxidation: Co(s) � Co 2 +Caq) + eEO Reduction: Ni 2 +(aq) + e- � Ni(s) V N i 2 +(aq) + Co(s) � N i(s) + Co2 +(aq)

= -0.28 = -0.25

2

2

= E:""OOe - E:"ode = -0.25 V - (-0.28 V) = 0.03 V Co 2 + ] 0. 0 592 [ equation: a) Use the Nemst EceJl = Ecell - -- I og n [ Ni 2+ ] 20] 0.0592 log -[0.EceJl = 0. 0 3 V 2 [ 0. 8 0 ] = 0.0478209752 V = E:ell

°

--

0.05 V

b) From part (a), notice that an increase in [Co +] leads to a decrease in cell potential. Therefore, the concentration of cobalt ion must increase further to bring the potential down to V. Thus, the new concentrations wi ll be M x (There is a I : I mole ratio) M + x and [Ni 2 +] [Co 2+] +x log x

0. 0 3 = 0.80 = 0. 20 [0.20 ] 0.03 V = 0.03 V 0.0592 2 [0. 8 0 - ] 0.0592 log [0.20 x ] = - -2 [0.80 - x] + x] Raise each side to l Ox = log [0.20 [0.80 - x ] + x] 1 = [0.20 [0.80 - x ] 0.20 x = 0.80 - x x = 0.30 M [N i 2 +] = 0.80 - 0.30 = c) At equilibrium EceJl = 0.00, to decrease the cell potential to 0. 0 0, [Co 2 +] increases and [Ni 2+] decreases. [0.20 x ] 0.00 V = 0.03 V 0.0592 log 2 [0.80 - x] 20 x ] -0.03 V = - 0.0296 log [0.[0.80 - x] 1.0135135 = log [0.[ 0.2800 xx]] x] 10.316052 = [0.20 [ 0.80 ] x = 0.7116332 (unrounded) [Co 2+ ] = 0.20 0.7116332 = 0. 9 116332 = [N?+] = 0. 8 0 - 0.7116332 = 0.08837 = 21.55 The overall cell reaction proceeds t o increase the 0.10 + concentration and decrease the 2.0 + concentration. Therefore, half-cell because it has the lower concentration. = 0. 0 0 V atm) 2H+Caq: 2 eOxidation: Reduction: 2H\aq: 2. 0 2 eatm) EO = 0.00 V 2H+ (aq: atm) 2 (aq H 2 (g: atm) E:ell = 0.00 V n = 2 e_

+

o

.:;.-----3-

o

+

0.50 M

+

_

+

+ -

+ -

+

x

0.91 M 0.09 M

MH A is the anode H2 (g:0.90 � 0.10 My + M) + � H2(g: 0.50 2.0 My + H2 (g:0.90 � W : 0.10 My +

314

MH

EO

0.50

[ H T2 + [H J +

Q for the cell equals

node

PH ( cathode )

cathode

PH ( anode)

( 0. 1 0 / ( 0.50 ) ( 2.0 ) 2 ( 0.90 )

-'...-----'.--'---'- =

0.00 1 3 88889 (unrounded)

0.0592 Ecell = 0.00 V log ( 0.00 1 3 88889 ) = 0.084577 = 0.085 V 2 --

2 1 .57

Electrons flow from the anode, where oxidation occurs, to the cathode, where reduction occurs. The electrons always flow from the anode to the cathode no matter what type of cell.

2 1 .58

A D-sized battery is much larger than an AAA-sized battery, so the D-sized battery contains a greater amount of the cell components. The potential, however, is an intensive property and does not depend on the amount of the cell components. (Note that amount is different from concentration.) The total amount of charge a battery can produce does depend on the amount of cell components, so the D-sized battery produces more charge than the AAA-sized battery.

2 1 .60

The Teflon spacers keep the two metals separated so the copper cannot conduct electrons that would promote the corrosion of the iron skeleton. Oxidation of the iron by oxygen causes rust to form and the metal to corrode.

2 1 .62

Sacrificial anodes are metals with EO less than that for iron, -0.44 so they are more easily oxidized than iron. a) EO(aluminum) = - 1 .66. Yes, except aluminum resists corrosion because once a coating of its oxide covers it, no more aluminum corrodes. Therefore, it would not be a good choice. b) EO(magnesium) = -2. 3 7 V. Yes, magnesium is appropriate to act as a sacrificial anode. c) EO(sodium) = -2. 7 1 V. Yes, except sodium reacts with water, so it would not be a good choice. d) E °(lead) -0. 1 3 V. No, lead is not appropriate to act as a sacrificial anode because its value is too high. e) EO(nickel) = -0.25 V. No, nickel is inappropriate as a sacrificial anode because its value is too high. t) EO(zinc) = -0.76 V. Yes, zinc is appropriate to act as a sacrificial anode. EO(chromium) -0.74 V. Yes, chromium is appropriate to act as a sacrificial anode.

V,

=

g)

2 1 .64

=

3 Cd2 \aq) + 2 Cr(s) -? 3 Cd(s) + 2 Cr3 +(aq) E:.II = -0.40 V - (-0.74 V) = 0.34 V To reverse the reaction requires 0.34 V with the cell in its standard state. A 1 .5 potential, so the cadmium metal oxidizes to Cd2+ and chromium plates out.

V supplies more than enough

2 1 .66

The oxidation number of nitrogen in the nitrate ion, N03-, is +5 and cannot be oxidized further since nitrogen has only five electrons in its outer level. In the nitrite ion, N02-, on the other hand, the oxidation number of nitrogen is + 3 , so it can be oxidized to the +5 state.

2 1 .68

Iron and nickel are more easily oxidized than copper, so they are separated from the copper in the roasting step and conversion to slag. In the electrorefining process, all three metals are oxidized into solution, but only Cu2+ ions are reduced at the cathode to form Cu(s).

2 1 . 70

a) At the anode, bromide ions are oxidized to form bromine (Br2 ). b) At the cathode, sodium ions are reduced to form sodium metal (Na).

2 1 . 72

P ossible reductions: Cu2+(aq) + 2 e- -? Cu(s ) EO = +0.34 EO = -2.90 Ba2+(aq) + 2 e- -? Ba(s ) EO = -1 .66 Ae \aq) + 3 e- -? Al(s ) 2 H 20(l) + 2 e- -? H 2(g) + 2 OIr(aq) E = -I with overvoltage Copper can be prepared by electrolysis is its aqueous salt since its reduction half-cell potential is more positive than the potential for the reduction of water. The reduction of copper is more spontaneous than the reduction of water. Since the reduction potentials of Ba2+ and A1 3 + are more negative and therefore less spontaneous than the reduction of water, these ions cannot be reduced in the presence of water since the water is reduced instead.

V

315

V V V

Possible oxidations: 2Br-(aq) Br2(l) + 2 e= +1. 0 7 V = V withitsovervol tage half-cell potential is more 2 H 20(l) can0be2(g)prepared + 4 H+(aq) + 4 eby electrolysis of its aqueous salt1. 4because reduction negative than the potential for the oxidation of water with overvoltage. The more negative reduction potential for Br- indicates that its oxidation is more spontaneous than the oxidation of water. a) Possible oxidations: 2 H 0(l) 0 (g) + 4 H\aq) + 4 e= 1. 4 V with overvoltage = 2. 8 7 2 F-2 F 2(g) +2 2 eSince the reduction potential of water is more negative than the reduction potential for F-, the oxidation of water is more spontaneous than that ofF. The oxidation of water produces oxygen gas and hydronium ions at the anode. Possible reductions: = -1 V with overvoltage 2 H20(l) + 2 e- Hz(g) + 2 OW(aq) Li +(aq) + e- Li(s) = -3. 05 V Since the reduction potential of water is more positive than that of Lt, the reduction of water is more spontaneous than the reduction ofLi+. The reduction of water produces gas and at the cathode. b) Possible oxidations: 2 H zO(l) Oz(g) + 4 W(aq) + 4 e= 1 .4 V with overvoltage The oxidation of water produces oxygen gas and hydronium ions at the anode. The SO/- ion cannot oxidize as S is already in its highest oxidation state in SO/-. Possible reductions: = -1 with overvoltage 2 Hz zO(l) + 2 e- Hz(g) + 2 OW(aq) = -0.14 V + 2 en + Sn(s) S (aq) = -0.63 V (approximate) SO/-(aq) + 4 W(aq) + 2 e- SOz(g) + 2 H 2 0(l) The potential for sulfate reduction is estimated from the Nernst equation using standard state concentrations and 7 pressures for all reactants and products except W, 7which i n pure water i s x 10= 0.20 V - (0. 0 592 / 2)2+log [ I ! x 10- )4] = -0. 6288 = -0.63 V The most easily reduced ion is Sn with the most positive reduction potential, so forms at the cathode. Mgz+ + 2 e- Mg Mg )[ 2 mol e- = 2.92883587 = 2.93 a) (35. 6 g Mg ) 24.mol 3 1 g Mg 1 mol Mg C 5 b) ( 2. 92883587 mol e- ) ( 96485 mol e_ ) = 2.8258872 x 105 = 2.83 10 x 105 C _I_h_ ) � = 31. 3 987 = 31.4 A c) 2.8258872 2.50 h 3600 s 5{ Ra2+ + 2e- Ra In the reduction of radium ions, Ra2+, to radium metal, the transfer of two electrons occurs. l mol e- 1 mol Ra ( 226 g Ra ) = 0.25180 = 0.282 (215 C ) 96485 C 2 mol e- 1 mol Ra Zn2+ + 2 e- Zn Zn ) 2 mol e- 9648 5 Time = (85. 5 g Zn) ( 65.I mol 4 1 g Zn 1 mol Zn 1 mol e- 23. 0 A 5{ = 1 . 0966901 x 104 = 1.10 104 �

Bromine

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21. 8 2 a) The sodium sulfate makes the water conductive, so the current will flow through 7the water to complete the+ circuit, increasing the rate of electrolysis. Pure water, which contains very low (10- concentrations of H and conducts electricity very poorly. b) The reduction of has a more positive half-potential (- I V) than the reduction ofNa+ (-2.71 V); the more spontaneous reduction of water will occur instead of the less spontaneous reduction of sodium ion. The oxidation of is the only oxidation possible because SO/- cannot be oxidized under these conditions. In other words, it is easier to reduce than Na+ and easier to oxidize H 20 than SO/-. 21. 8 3 a) Calculate amount of chlorine gas from stoichiometry, then use ideal gas law to find volume of chlorine gas. 2: I Na: C I2 mole ratio 2NaCI(l) 2Na(l) + CI2(g) 3 Na mol C1 2 Moles Cl2 (30. 0 kg Na ) [ 10I kgg ] ( 22.I mol 99 g Na )(12 mol Na ) 652. 4 5759 mol Cl2 (unrounded) (652. 4 5759 mol Cl 2 ) ( 0. 0 821 molatm ( (273 + 580. ) K ) ) nRT K · V (1. 0 atm ) V 4. 5 692 l04 b) 2Cl- Cl2 + 2 eTwo moles of electrons are passed through the cell for each mole of CI2 produced. �) Coulombs (652. 4 5759 mol CI2 ) [ I2molmolCIe-2 ] ( 96485 I mol e 1.2 59047 x 108 c) Current is charge per time with the amp unit equal to C / s. (125904741 C ) (� 75 C ) 1. 6 787 106 21.84 Zn2+ + 2 e- Zn s ) ( 24 h ) (2.00 day ) [ 1 mol e- ][ I mol zn ]( 65. 4 1 g zn ) Mass (0.755A ) 7:A ( 3600 I h 1 day 96485 C 2 mol e- 1 mol Zn 44.222678 g 21. 8 7 From the current 70. 0% of the moles of product will be copper and 30. 0% zinc. Assume a current of exactly 100 coulombs. The amount of current used to generate copper would be (70. 0% / 100%) (100 C) 70. 0 C, and the amount of current used to generate zinc would be (30. 0% / 100%) (l00 C) 30. 0 C. The half-reactions are: Cu2\aq) 2 e- Cu(s) and Zn2+(aq) 2 e- Zn(s). 1 mol e- ] [ I mol Cu ]( 63.55 g c u ) 0. 023052806 g Cu (unrounded) Mass copper (70.0 C ) [ 96485 C 2 mol e- 1 mol Cu mol e- ][ 1 mol zn ]( 65.4 1 zn ) 0. 0 10168938 g Zn (unrounded) Mass zinc (30.0 C ) [ I96485 C 2 mol e- 1 mol Zn 0.023052806 g cu Mass% copper (0.023052806 0. 0 10168938 ) g Sample ] x 100% 69. 3 907 M)

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The reaction is: Au3 +(aq) + e- -? Au(s) a) Find the volume of gold needed to plate the earring and then use density to find the mass and moles of gold needed. The volume of the gold is the volume of a cylinder (see the inside back cover of the text) V = m2 h 2 I cm 1 0-3 m cm cm3 mm) V= I mm 1 0_2 m

,, 5. 00 (0. 20 0. 3 9269908 2 I mol Au J 0. 0 3847255 mol Au (unrounded) ( 0. 3 9269908 cm3 ) 19.3I cmg �u ) 197. 0 g Au 3 mol e- 96485 C � ( 1 ) ( I h ) ( 1 day Time ( 0. 0 38472 55 mol AU ) I mol Au 1 mol e- 7: 0.0 1 0 A 3600 s 24 h 12.88897

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b) The time required doubles once for the second earring of the pair and doubles again for the second side, thus it will take four times as long as one side of one earring. Time = days) = = 5.2 x lOt d ays c) Start by multiplying the moles of gold from part (a) by four to get the moles for the earrings. Convert this moles to grams, then to troy ounces, and finally to dollars. 1 97.0 g AU I Troy ounce = $3 10 mol AU ) = Cost = I mol Au g Troy Ounce

(4) (12.88897

51.5558887

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The half-reactions and the cell reaction are: Zn(s) + �� -? ZnO(s) + �� + H­ Ag2 0(s) + lhQffi + H- -? Ages) + �� Zn(s) + Ag20(S) -? ZnO(s) + Ag(s) The key is the moles of zinc. From the moles of zinc, the moles of electrons and the moles of Ag2 0 may be found. Only 80% of the initial amount of zince reacts: % I mol Zn Moles Zn = g Zn) = mol Zn (unrounded) 1 00% g Zn The 80% is assumed to have two significant figures.

2

a) Time = =

mol

( 80 zn ) [ 2

mol 1 mol Zn

I

I mol e-

rnA

1 0-3 A

day h

h

s

= 9 1 d ays

b) Mass Ag = =

c) Cost =

2

) ( 65.4 1 ) 0.195688732 e- ) [ 96485 C ) ( �] [ �) ( 1 )( 1 ) _ ( ) (0.1956 88732 24 4. 8 3600 7: 91. 0 54337 ) ( 95% ) ( 2 ) ( 107.91 ) (0.1956887326 zn ) ( 1 44.45224 ) ( $5.50 ) ( 91.054337 ) 0.0820198 ( 44. 45224 ( 95% ) ( (16. 0

21.92

31.10

)(

mol Ag 20 I mol Zn

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1 00%

mol Ag I mol Ag20

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1 00%

I troy oz 3 1 . 1 0 g Ag

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$ 0.08/day

This problem deals with the stoichiometry of electrolysis. The balanced oxidation half-reaction for the chlor-alkali process is given in the chapter: C r(aq) -? CIz(g) + eUse the Faraday constant, F, ( l F = C / mol el and the fact that I mol of CIz produces mol e- or F, to convert coulombs to moles of C1 2 . 7: g CI 2 I kg mol e- 1 mol CI 2 Mass CI2 = X 1 0 4 A ) A Ih C I mol CI 2 kg mol e 1 03 g

2

2

=

2

(3 699.9689

96485 ( 3600 S (8 h ) 1 96485

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=7

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1 02

poun d s

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21.94 a) Aluminum half-reaction: AI3+(aq) + 3 e- AI(s), so n 3. Remember that I A I Cis. I mol Al (3 mol e- 96485 C � I Time (1 000 kg A1) (� l kg l( 26. 9 8 g AI J I mol Al l( I mol e- l 5{ ( 100,000 A J 1. 0 728502 x 105 The molar mass of aluminum limits the significant figures. b)andMultiply the time by the current and voltage, remembering that 1 A 1 Cis (thus 100,000 A is 100,000 Cis) I V 1 J/C (thus, 5.0 V 5.0 J/C). Change units of J to kW·h. C ) ( 5.0 J )( � ) ( I kW · h ) 1 . 4900698 104 (1. 0 728502 x 105 s) ( 100,000 kW. h s C 103 J 3. 6 x 103 kJ c) From part (b), the 1 .5 x 104 kW h calculated is per 1000 kg of aluminum. Use the ratio of kW·h to mass to find kW'hllb and then use efficiency and cost per kW'h to find cost per pound. 104 kW h l( I kg ( 0. 90 cents J( 100% ) 6. 7 57686 Cost ( 1. 4900698 1000 kg Al 2. 205 Ib J I kW h 90. % 21. 9 6 Statement: metal D + hot water reaction Conclusion: D reduces water. Statement: D + salt no reaction. Conclusion: D does not reduce salt, so E reduces D salt. is better reducing agent than D. Statement: D + F salt reaction Conclusion: D reduces F salt. D is better reducing agent than F. If E metal and F salt are mixed, the salt would be reduced producing F metal because has the greatest reducing strength of the three metals is stronger than D and D is stronger than F). The ranking of increasing reducing strength is 21. 9 7 Substitute J/C for V. a) Cell I: Oxidation number (O. N . ) ofH from 0 to + I , so I electron lost from each of 4 hydrogens for a total of 4 electrons. Oxygen O. N . goes from 0 to -2 indicating that 2 electrons are gained by each of the two oxygens for a total of 4 electrons. There is a transfer in the reaction. The potential given in the problem allows the calculation of !)'G: !)'Go -nFEo -(4 mol e-) (96485 Clmol e-) (1 . 2 3 J /C) -4. 7 47062 x 10 5 J Cell I I : I n Pb(s) PbS04, O. N . ofPb goes from 0 to +2 and in Pb02 PbS04, O. N . goes from +4 to +2. There is a transfer of in the reaction. !)'Go -nFEo -(2 mol e-) (96485 Clmol e-) (2. 04 J/C) -3. 9 36588 x 105 J Cell I I I : O. N . of each of two Na atoms changes from 0 to + I and O. N . of Fe changes from +2 to O. There is a transfer of in the reaction. !)'Go -nFEo -(2 mol e-) (96485 Clmol e-) (2.35 J/C) -4.534795 x 10 5 J 32. 0 0 g ° 2. 0 16 g H b) Cell I : Mass of reactants (2 mol H 2 ) ( I mol H 2 2 J (I mol 0 2 ) ( I mol 0 2 2 J 36. 0 32 g (unrounded) -4.74 7062 105 J ( � kJ/g Reactant Mass ( 36. 0 32 g ) 103 J ) = -13.1 7457 Cell I I : Mass of reactants 2 g Pb0 2 J (2 mol H 2 S0 4 ) ( 98. 09 g H 2 S04 ( I mol Pb ) ( 207. 2 g Pb J ( I mol Pb0 2 ) ( 239. I mol H 2 S0 4 J I mol Pb0 2 1 mol Pb 642.58 g (unrounded) -3. 9365 8 8 105 J ( � w kJ /g Reactant Mass ( 642.58 g ) 103 J ) -0. 6 12622 �

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99 g Na ) + ( I mol FeCI 2 ) ( 126. 7 5 g FeCI 2 ) Cell III: Mass of reactants (2 mol Na ) ( 22.1 mol I mol FeCI 2 Na 172. 7 3 g (unrounded) w -4.534795 105 Jl( � ) -2. 625366 [ 172. 7 3 g 103 Reactant Mass Cell I has the highest ratio (most energy released per gram) because the reactants have very low mass while Cell II has the lowest ratio because the reactants are very massive. 2l . 9 8 Examine each reaction to determine which reactant is the oxidizing agent by which reactant gains electrons in the reaction. From reaction between U 3+ + Cr2+3+ Cr2+ + 2U+4+, find that Cr2+3+ oxidizes U3+ . From reaction between Fe + Sn Sn + Fe , find that4+Sn oxidizes2Fe.+ From the fact that no reaction occurs between Fe and U , find that Fe oxidizes U3+. From reaction between Cr23+ + Fe 2 Cr2+ + Fe23++, find that Cr32++ oxidizes Fe.2 From reaction between Cr + +2+Sn + Sn + Cr , find that Sn oxidizes Cr +. 3+ 2+ Notice that nothing oxidizes Sn, so Sn must be the strongest oxidizing agent. Both Cr and Fe oxidize U3+, so U4+ must be the weakest oxidizing agent. Cr3+ oxidizes iron, so Cr3+ is a stronger oxidizing agent than Fe2+ . The half-reactions in order from strongest to weakest oxidizing agent: Sn32+\aq) + 2 e- Sn(s) Cr2+(aq) + e- Cr2+(aq) Fe4 (aq) + 2 e- 3 Fe(s) U +(aq) + e- U \aq) 21.101 Place the elements in order of increasing (more positive) Reducing agent strength: Li > Ba > Na > AI > Mn > Zn > Cr > Fe > Ni > Sn > Pb > Cu > Ag > H g> Au Metals with potentials lower than that of water (-0. 8 3 can displace hydrogen from water by reducing the hydrogen in water. These can displace H 2 from water: Li, Ba, Na, AI, and Mn Metals with potentials lower than that of hydrogen (0. 0 0 can displace hydrogen from acids by reducing the in acid. These can displace H from acid: Li, Sa, Na, AI, Mn, Zn, Cr, Fe, Ni, Sn, and Pb Metals with potentials above 2that of hydrogen (0. 0 0 cannot displace (reduce) hydrogen. These cannot displace H 2 : Cu, Ag, Hg, and Au 21.102 a) Use the stoichiometric relationships found in the balanced chemical equation to find mass of A1203 . Assume that 1 metric ton AI is an exact number. 2 AI203 (in Na3AIF6) + 3 C(gr) 4 AI(l) + 3 CO2(g) AI ) ( 2 mol AI 2 0 3 )( 10l. 9 6 g AI 2 0 3 ) [ 1 kg ] [_1 t_ ] mass AI203 (I t AI ) [ 1013 tkg ] [ 101 kg3 g ] ( 26.1 mol 9 8 g AI 4 mol AI I mol AI 2 03 103 g 103 kg 1.8895478 metric tons AI203 Therefore, are consumed in the production of 1 ton of pure AI. b) Use a ratio of3 mol C: 4 mol Al to find mass of graphite consumed. 1 t ] � ]( 1 mol AI )( 3 mol C )( 12. 0 1 g C)[ �] [ _ mass C (I t AI) [� ][ 3 3 1 t 1 kg 26.98 g AI 4 mol AI 1 mol C 10 g 10 kg 0.3338584 Therefore, are consumed in the production of 1 ton of pure AI, assuming 100% efficiency. c) The percent yield with respect to Ah03 is because the actual plant requirement of l.89 tons AI203 equals the theoretical amount calc ulated in part (a). d) The amount of graphite used in reality to produce 1 ton of AI is greater than the amount calculated in (b). In other words, a 100% efficient reaction takes only 0. 3 339 tons of graphite to produce a ton of AI, whereas real production requires more graphite and is less than 100% efficient. Calculate the efficiency using a simple ratio: (0. 4 5 t) (x) (0. 3 338584 t) (100%) x 74. 19076 =

=

max

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=

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= 1 .890 1 .890 tons of AI 203

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=

0.3339 tons C 0.3339 tons of C =

1 00%

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74 %

320

W

e) For every 4 moles of Al produced, 3 moles of CO2 are produced. Al 3 mol C O 4 moles C = ( I t AI ) [ 1013 tkg )[ 10I kg3 g )( 26.1 mol 9 8 g Al )( 4 mol Al2 ) = 2.7798 10 mol CO2 (unrounded) The problem states that 1 atm is exact. Use the ideal gas law to calculate volume, given moles, temperature and pressure. ( 2.7798 x 104 mol C O 2 )( 0. 0 8206L atm/mol K ) ( ( 273 + 960. ) K 10-3 m 3 ) V = nRT P = l �m IL = 2.81260] x 103 = 21.103 a) The reference half-reaction is: Cu2\aq) + 2 e- Cu(s) = 0. 3 4 Before the addition of the ammonia, = O. The addition of ammonia lowers the concentration of copper ions through the formation of the complex Cu(NH 3)/+. The original copper ion concentration is and the copper ion concentration in the solution containing ammonia is The Nernst equation is used to determine the copper ion concentration in the cell containing ammonia. = 0. 0 592n log ] 0.129 = 0. 0 0 0. 0 5922 log [Cu2+ [Cu2+ ] X

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[CU2+] original"

[C u 2+] . mmonia '

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V

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] 0.129 = 0. 05922 log [Cu2+ [ 0. 0 100 ] ] (0.129 (-2 /0. 0 592) log [Cu2+ [ 0. 0 100 ] V

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V

original

ammonia angma I

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ammonia

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onglna I .

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ammonia ongIna I .

.

Cu2+ ] 4.3842154 10-5 = [[ 0.0100 ] 10-7 (unrounded) 4.3842154 This is the concentration of the=copper ion thatx is not in the complex. The concentration of the complex and of the uncomplexed ammonia must be determined before may be calculated. The original number of moles of copper and the original number of moles of ammonia are found from the original volumes and molarities: . . moles of copper = ( 0. 0 100 mol CU(N03 h ) [ I mol Cu2+ )[ 10-3 L ) ( 90 . 0 m L ) OngInal 1 mol Cu(N03 h 1 mL L 4 = 9.00 x 10- mol . . moles ofammoma. = ( 0. 5 00 mol NH 3 )[ 10-3 L ) ( 10. 0 mL ) = 5. 00 x I0-3 mol NH 3 OngInal I mL L Determine the moles of copper still remaining uncomplexed. Remaining mol of copper = [ 4. 3 842154 x L10-7 mol Cu2+ ) [ 10-I mL3 L ) ( 100. 0 mL ) = 4.3842154 x 10-8 mol Cu ammonia

X

onglnaI .

.

M

[C u 2+] ammonia

Kr

--

C u2+

321

The difference between the original moles of copper and the copper ion remaining in solution is the copper in the complex (= moles of complex). The molarity of the complex may now be found. Moles copper in complex = (9.00 x 10-2 4 -4.3842154 x 10-8) mol Cu2+ = 8.9995615 10-4 mol Cu + (unrounded) 2 4 mol Cu + ) 1 mol CU (NH 3 )�+ (-1 mL ) Mo Ianty. f comp I ex = ( 8. 9995615100.102 0 mL I mol Cu + 10-3 L 3 = 8. 9995615 10- MCu(NH )/+ (unrounded) The concentration of the remaining ammonia is 3found as follows: mol N�3 ) ( 5. 0 0 x 10-3 mol NH 3 ) - ( 8. 9995615 10-4 mol c U 2 + ) ( 14 mol Cu + (-1 mL ) Molarity of ammonia = 100.0 mL 10-3 L = is:0.014001753 Mammonia (unrounded) The Kr equilibrium Cu2\aq) + 4 NH 3(aq) Cu(NH 3)/+(aq) )�+ J - [ [8. 9995615 x 1O-3 J 3 Kr - [CuCu(NH = 5. 34072 10" = [ 2 + ] [NH J 4. 3 842154 x 1O-7 J [0. 0 14001753 t b) The Kr wi II be used to determine the new concentration of free copper ions. Moles uncomplexed ammonia before the addition of new ammonia = (0.014001753 mol NH3 1 L)3 (10-3 Lli mL) (100. 0 mL) = 0.001400175 mol NH 3 Moles ammonia added = 5. 00 x 10- mol NH3 (same as original mol of ammonia) From the stoichiometry: 4 NH3(aq) Cu2\aq) Cu(NH + 3)/+(aq) 0.0014001753 mol Initial moles: 4. 3 842154 x 10-8 mol 8.9995615 x 10-4 mol Added moles: 5. 0 0 x 10mol Cu2+ is limiting: -(4.3842154 x 10-8 mol) -4 (4. 3 842154 10-8 mol) +(4. 3 842154 x 10-8 mol) 0 After the reaction: 0. 006400 mol 9. 00000 10-4 mol Determine concentrations before equilibrium: [Cu2+] = 0 [NH3] = (0. 006400 mol NH3 / 1 1O. 0 mL) (1 mL 1 10-3 L) = 0.0581818 MNH [Cu(NH3=)/+]0. 0=08181818 x 10-4 mol CU(NH3)/+ 1 1 10.0 mL) (1 mL 1 10-3 L) 3 (9.00000MCu(NH )/+ Now allow the system2 to come to equilibrium:3 Cu(NH 3)/+(aq) 4 NH3(aq) Cu \aq) + 0. 0 581818 0. 0 08181818 Initial molarity: 0 Change: +x +4x x Equilibrium: [ x 0. 0 581818 + 4 x 0. 008181818 - x )�+ J [0. 008181818 - x] = 5.34072 x 10 " 3 Kr- [ CuCu(NH 2 + J [NH 3 J [x][O. 0 581818 4x]4 Assume -x and +4x are negligible when compared to their associated numbers: [0. 0 08181818] = 5. 34072 x 10 " Kr= [x][O. 0 581818 t 2 x = [Cu +] = 1. 3 369 x 10-9 Cu2+ X

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X

0

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X

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X

X

X

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+

M

322

5.3

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1 0- 1 1

Use the Nernst equation to determine the new cell potential: [ Cu 2 + ] ammonia E= log [ Cu 2 + ]

0.00 V 0. 05922 V - 0. 0 5922 V log [ 1. 3[0.3690 100]10-9 J 0. 203467 c) The first step will be to do a stoichiometry calculation of the reaction between copper ions and hydroxide ions. - ) [ 10-3 L ) Moles O W ( 0.500 mol NaOH )[ 1 mol OH (I O.o ) 5. 0 0 10-3 mol OW L I mol NaOH 1 mL The initial moles of copper ions were determined earlier: 9. 0 0 x 10-4 mol Cu2+ The reaction: + 2 0W(aq)3 Cu2+ (aq ) Cu(OHMs) 9. 0 0 x 10-4 mol Initial moles: 5.00 x 10- mol Cu2+ is limiting: -(9. 0 0 x 10-4 mol) -2 (9. 0 0 x 10-4 mol) 0 0. 0032 mol After the reaction: Determine concentrations before equilibrium: [Cu2+ ] 0 [NH 3 ] (0. 0 032 mol OH- / 100. 0 mL) (1 mL / 10-3 L) 0. 0 32 M O W Now allow the system to come to equilibrium: + 2 0W(aq) Cu(OHMs) Cu2+(aq ) Initial molarity: 0. 0 32 0. 0 Change: +x +2x x 0. 0 32+ 2 x Equilibrium: Ksp [Cu2+ ] [OW] 2 2. 2 10-20 Ksp [x][0. 0 32 + 2 ] 2 2. 2 10-20 Assume 2x is negli § ible compared to 0. 0 32 Ksp [x ] [ 0. 0 32 ] 2 2.2 x 10- 0 x [Cu2+] 2.1487375 10-1 7 2.1 10-1 7 _

original

X

E= E=

= 0.20 V

mL

=

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Use the Nernst equation to determine the new cell potential: [ Cu 2 + J hydroxide E log [ Cu 2 + J

0.00 V 0. 05922 V x 1 O- J - 0.05922 V log [ 2.1487375 [0. 0 100] 0.4 34169 d) Use the Nernst equation to determine the copper ion concentration in the half-cell containing the hydroxide ion. [ Cu 2 + J 5 0 V 2 0. f 0. 0 0 V - � log _

=

original 17

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=

E

=

=

0.43 V

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Cu 2 +

[ Cu 2 + J

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original

0. 3 40 0. 0 5922 V I og [0. 0 100] [ Cu 2 + J (0. 340 V) (-2 /0. 0 592) g [0. 0 100] =

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=

hydroxide

hydroxide

10

323

- 1 1 .486486

3 .262225 6

=

X

[

10 g

1 0-1 2

Cu 2 +

=

]

hydroxide

[ 0. 0 1 00 ]

[ C u 2 + ] hydroxide [ 0. 0 1 00 ]

[ C u 2+hydrox i de = 3 .2622256 X 1 0-1 4 M (unrounded) Now use the Ksp relationship : Ksp [ C u2+] [O Hl 2 . 2 X 1 0-20 Ksp = [ 3 .2622256 X 1 0-1 4 ] [OHlZ = 2 . 2 x 1 0-20 [OHlZ = 6 . 743 862 x 1 0-7 [OHl = 8 .2 1 2 1 x 1 0-4 = 8 . 2 X 1 0-4 M OI1 = 8.2 x 1 0-4 M NaOH =

2 1 . 1 05

=

The half-reactions are (from the Appendix): E = 0.00 V Oxidaton: H z (g) � 2 H +(aq) + 2 eReduction: E = 0 . 80 V 2 (Ag+(aq) + 1 e- � Ag(s)) Ece ll = 0 . 8 0 V - 0.0 V = 0 . 8 0 V 2 Ag+ (aq) + H (g) � 2 Ag(s) + 2 H+(aq) Overall : The hydrogen ion concentration 2can now be found from the Nernst equation. E = EO

_

0. 0592 V 2

0.9 1 5 V = 0.80 V

0.9 1 5 V

_

_

log Q 0 . 0592 V 2

0.80 V =

log

0 . 0592 V

_

2

/

1 . 3 0276

x

= log 1 0 -4

=

[H

+

[ Ag + r PH2

log

(0.9 1 5 V - 0 . 8 0 V) (-2 0.0592 V)

-3 . 8 85 1 3 5

[H+ r [H

+

r

[ 0 . 1 00 ] 2 ( 1 .00 )

= log

[ H: r

[ 0. 1 00 ] ( 1 . 00 )

r

[ 0 .0 1 00 ]

[H + r

[ 0 .0 1 00 ]

[ H+] = 1 . 1 4 1 3 85 1 X 1 0-3 M (unrounded) pH = -log [ H+] = -log ( 1 . 1 4 1 3 8 5 1 x 1 0-3)

=

2 .94256779

324

=

2 .94

Chapter 22 The Transition Elements and Their Coordination Compounds FOLLOW-UP PROBLEMS

22. 1

Plan: Locate the element on the periodic table and use its position and atomic number to write a partial electron configuration. Add or subtract electrons to obtain the configuration for the ion. Partial electron configurations do not include the noble gas configuration and any filledjsublevels, but do include outer level electrons (n-Ievel) and the n- l level d orbitals. Remember that ns electrons are removed before n-l electrons. Solution: a) Ag is in the fifth row of the periodic table, so n = S , and it is in group I B( l l ). The partial electron configuration of the element is Ss 1 4d1 0. For the ion, Ag+ , remove one electron from the Ss orbital. The partial electron configuration of Ag+ is 4dl O • b) Cd is in the fifth row and group 2B( 1 2), so the partial electron configuration of the element is Si4i o. Remove two electrons from the Ss orbital for the ion. The partial electron configuration of Cd2+ is 4dl O • c) Ir is in the sixth row and group The partial electron configuration of the element is 6isd7 . Remove two electrons from the 6s orbital and one from Sd for ion Ir3+ . The partial electron configuration of Ir3 + is 5,1 . Check: Total the electrons to make sure they agree with configuration. a) Ag+ should have 47 - 1 = 4 6 e-. Configuration has 36 e- (from Kr configuration) plus the 1 0 electrons in 4d1 o. This also totals 46 e-. b) Cd2+ should have 48 - 2 = 4 6 e-. Configuration is the same as Ag+, so 46 e- as well. c) 1r3 + should have 77 - 3 = 7 4 e-. Total in configuration is S4 e- (noble gas configuration) plus 1 4 e- ( in/­ sublevel) plus 6 e- (in the partial configuration) which equals 74 e-.

8B(9).

22.2

Plan: Name the compound by following rules outlined in section 22.2. Use Table 22.6 for the names of ligands and Table 22.7 for the names of metal ions. Write the formula from the name by breaking down the name into pieces that follow naming rules. Solution : a) The compound consists of the cation [Cr(H 2 0)sBr] 2- and anion cr. The metal ion in the cation is chromium and its charge is found from the charges on the ligands and the total charge on the ion. total charge = (charge on Cr) + S(charge on H 2 0) + (charge on Br-) -2 = (charge on Cr) + S(O) + (-1 ) charge on Cr = +3 The name of the cation will end with chromium(I I I) to indicate the oxidation state of the chromjum ion since there is more than one possible oxidation state for chromium. The water ligand is named aqua (Table 22.6). There are five water ligands, so pentaaqua describes the (H 2 0)s ligands. The bromide ligand is named bromo. The ligands are named in alphabetical order, so aqua comes before bromo. The name of the cation is pentaaquabromochromium(I I I). Add the chloride for the anion to complete the name: pentaaquabromochromiu m(III) chloride.

b) The compound consists of a cation, barium ion, and the anion hexacyanocobaltate(IIl). The formula of the anion consists of six (from hexa) cyanide (from cyano) ligands and cobalt (from cobaltate) in the +3 oxidation state (from (I I I)). Putting the formula together gives [Co(CN)6r+. To find the charge on the complex ion, calculate from charges on the ligands and the metal ion: total charge = 6( charge on CN-) + (charge on cobalt ion) = 6(- 1 ) + (+3 ) = -3 The formula of complex ion is [CO(CN )6] 3-. Combining this with the cation, Ba2+ , gives the formula for the compound: Ba3[Co(CN)6h. The three barium ions give a +6 charge and two anions give a -6 charge, so the net result is a neutral salt.

32S

22.3

Plan: The given complex ion has a coordination number of 6 (en is bidentate), so it will have an octahedral arrangement of ligands. Stereo isomers can be either geometric or optical. For geometric isomers, check if the ligands can be arranged either next to (cis) or opposite (trans) each other. Then, check if the mirror image of any of the geometric isomers is not superimposable. If mirror image is not superimposable, the structure is an optical isomer. Solution: The complex ion contains two NH} ligands and two chloride ligands, both of which can be arranged in either the cis or trans geometry. The possible combinations are \ ) cis-NH} and cis-Cl, 2) trans-NH} and cis-C!, and 3) cis­ NH} and trans-Cl. Both NH} and Cl trans is not possible because the two bonds to ethylenediamine can be arranged only in this cis-position, which leaves only one set of trans positions. +

+

NH3

J 1

CI H2N " , / ( / Co , H2N Cl NH3

Cl

"'-.. 1 ,/ NH3 ( / Co " H2N 1 NH3

+

H2N

Cl

trans-NH} cis-Cl cis-NH} trans-CI cis-NH ) cis-Cl The mirror images of the second two structures are superimposable since two of the ligands, either ammonia or chloride ion, are arranged in the trans position. When both types of ligands are in the cis arrangement, the mirror image is not a superimposable optical isomer' +

+

cis-NH} cis-Cl There are four stereo isomers of [Co(NH}M en)Cl 2 t

cis-NH} cis-Cl

22.4

Plan: Compare the two ions for oxidation state of the metal ion and the relative ability of ligands to split the d-orbital energies. Solution: The oxidation number of vanadium in both ions is the same, so compare the two ligands. Ammonia is a stronger } field ligand than water, so the complex ion [V(NH3)6f+ absorbs visible light of higher energy than [V(H 2 0)6] + absorbs.

22.5

Plan: Determine the charge on the manganese ion in the complex ion and the number of electrons in its d-orbitals. Since it is an octahedral ligand, the d-orbitals split into three lower energy orbitals and two higher energy orbitals. Check the spectrochemical series for whether CW is a strong or weak field ligand. If it is a strong field ligand, fill the three lower energy orbitals before placing any electrons in the higher energy d-orbitals. If it is a weak field ligand, place one electron in each of the five d orbitals, low energy before high energy, before pairing any electrons. After filling orbitals, count the number of unpaired electrons. The complex ion is low-spin if the ligand is a strong field ligand and high-spin if the ligand is weak field. Solution: Figure the charge on manganese: total charge = (charge on Mn) + 6(charge on CW) charge on Mn -3 - [6(- 1 )] +3 The electron configuration of Mn is [Ar]4i3tf; the electron configuration of Mn3 + is [Ar]3d'. The ligand is CW, which is a strong field ligand, so the four d electrons will fill the lower energy d-orbitals before any are placed in the higher energy d-orbitals: =

=

326

H igher energy d-orbitals

IT]

It . I t it I are unpaired in [Mn(CN)6] 3-. The complex is

L ower energy d-orbitals Two electrons

ligand.

low spin

since the cyanide ligand is a strong field

END--OF-CHAPTER PROBLEMS

22. 1

a) All transition elements in Period 5 will have a "base" configuration of [Kr]5i, and will differ in the number of d electrons (x) that the configuration contains. Therefore, the general electron configuration is li2i2p 63s23p 64i3dlo4p 6 Si4tf .

b) A general electron configuration for Period 6 transition elements includes/sublevel electrons, which are lower in energy than the d sublevel. The configuration is li2i2l3i3l4i3dt 04lsi4dtOSp6 6i4j4Stf . 22.4

a) One would expect that the elements would increase in size as they increase in mass from Period 5 to 6. Because there are 1 4 inner transition elements in Period 6, the effective nuclear charge increases significantly. As effective charge increases, the atomic size decreases or "contracts." This effect is significant enough that Zr4+ and Hr+ are almost the same size but differ greatly in atomic mass. b) The size increases from Period 4 to 5, but stays fairly constant from Period 5 to 6. c) Atomic mass increases significantly from Period 5 to 6, but atomic radius ( and thus volume) hardly increases, so Period 6 elements are very dense.

22.7

a) A paramagnetic substance is attracted to a magnetic field, while a diamagnetic substance is slightly repelled by one. b) Ions of transition elements often have unfilled d-orbitals whose unpaired electrons make the ions paramagnetic. Ions of main group elements usually have a noble gas configuration with no partially filled levels. When orbitals are filled, electrons are paired and the ion is diamagnetic. c) The d-orbitals in the transition element ions are not filled, which allows an electron from a lower energy d orbital to move to a higher energy d-orbital. The energy required for this transition is relatively small and falls in the visible wavelength range. All orbitals are filled in a main-group element ion, so enough energy would have to be added to move an electron to a higher energy level, not just another orbital within the same energy level. This amount of energy is relatively large and outside the visible range of wavelengths.

22.8

a) V: li2i2p63i3p64i3� b) Y: li2i2p63 s23p64i3dt04p6si 4d' c) Hg: [Xe]6i4l'4SdtO

22. 1 0

Check: 2 + 2 + 6 + 2 + 6 + 2 + 3 23 eCheck: 2 + 2 + 6 + 2 + 6 + 2 + 1 0 + 6 + 2 + 1 Check: 54 + 2 + 14 + 1 0 80 e=

=

39 e­

=

Transition metals lose their s orbital electrons first in forming cations. a) The two 4s electrons and one 3d electron are removed to form Sc 3 + : Sc: [Ar]4i3i ; Sc 3 + : [Ar] or li2i2p63 s2 3p6 . There are no u npaired electrons. b) The single 4s electron and one 3d electron are removed to form Cu2+ : Cu: [Ar]4s 1 3dI O; ci+ : [Ar]3tf. There is one unpaired electron . c) The two 4s electrons and one 3d electron are removed to form Fe3 + : Fe: [Ar]4S2 3Ji; Fe 3 + : [Ar]3tf. There are five unpaired electrons since each of the five d electrons occupies its own orbital. d) The two 5s electrons and one 4d electron are removed to form Nb 3+ : Nb: [Kr]5s 24d3 ; Nb 3 + : [Kr]4d2• There are two unpaired electrons.

327

22. 12 The elements in Group 6B(6) exhibit an oxidation state of +6. These elements include . Sg (Seaborgium) is also in Group 6B(6), but its lifetime is so short that chemical properties, like oxidationW states within compounds, are impossible to measure. 22.14 Transition elements in their lower oxidation states act more like metals. The oxidation state of chromium in CrF2 is +2 and in CrF6 is +6 (use -1 oxidation state of fluorine to find oxidation state of Cr). exhibits greater metallic behavior than CrF6 because the chromium is in a lower oxidation state in CrF2 than in CrF6. 22.16 Oxides increases. The oxidation state of chromiumof transition in CrO) ismetals +6 andbecome in CrO less is +2,basicbased(oronmorethe acidic) -2 oxidasatioxidation on state ofstate oxygen. The oxide of the higher oxidation state, produces a more acidic solution. 22.19 The coordination number indicates the number of ligand atoms bonded to the central metal ion. The oxidation number represents the number of electrons lost to form the ion. The coordination number is unrelated to the oxidation number. 22. 2 1 Coordination number of two indicates geometry. Coordination number offour indicates either or geometry. Coordination number of six indicates geometry. 22. 24 a) The oxidation state of nickel is found from the total charge on the ion (+2 because two CI- charges equals -2) and the charge on ligands: charge on nickel +2 - 6(0 charge on water) +2 Name nickel as nickel(II) to indicate oxidation state. Ligands are six (hexa-) waters (aqua). Put together with chloride anions to give b) The cation is [Cr(en))]n+ and the anion is CI04-, the perchlorate ion (see Chapter 2 for naming polyatomic ions). The charge on the cation is +3 to make a neutral salt in combination with 3 perchlorate ions. The ligand is ethylenediamine, which has 0 charge. The charge of the cation equals the charge on chromium ion, so chromium(IIl) is included in the name. The three ethylenediamine ligands, abbreviated en, are indicated by the prefix tris because the name of the ligand includes a numerical indicator, di-. The complete name is c) The cation is K+ and the anion is [Mn(CN)6] 4-. The charge of 4- is deduced from the four potassium ions in the formula. The oxidation state ofMn is - {6(-1)} +2. The name ofCN" ligand is cyano and six ligands are represented by the prefix hexa. The name of manganese anion is manganate(II). The -ate suffix on the complex ion is used to indicate that it is an anion. The full name of compound is 22. 2 6 The charge of the central metal atom was determined in 22. 24 because the Roman numeral indicating oxidation state is part of the name. The coordination number, or number of ligand atoms bonded to the metal ion, is found by examining the bonded entities inside the square brackets to determine if they are unidentate, bidentate, or polydentate. a) The Roman numeral "II" indicates a oxidation state. There are 6 water molecules bonded to Ni and each ligand is un identate, so the coordination number is b) The Roman numeral "III" indicates a oxidation state. There are 3 ethylenediamine molecules bonded to Cr, but each ethylenediamine molecule contains two donor N atoms (bidentate). Therefore, the coordination number is c) The Roman numeral indicates a oxidation state. There are 6 unidentate cyano molecules bonded to Mn, so the coordination number is 22. 2 8 a) The cation is K+, potassium. The anion is [Ag(CN)2r with the name dicyanoargentate (I) ion for the two cyanide ligands and the name of silver in anions, argentate(I). The Roman numeral indicates the oxidation number on Ag. O. N . for Ag I - {2(-I)} + 1 since the complex ion has a charge of - I and the cyanide ligands are also - I .+ The complete name is b) The cation is Na , sodium. Since there are two + 1 sodium ions, the anion is [CdCI 4] 2- with a charge of 2-. The anion is the tetrachlorocadmate(II) ion. With four -1 chloride ligands, the oxidation state of cadmjum is +2 and the name of cadmium in an anion is cadmate. The complete name is Cr, Mo, and

C r F2

C r03,

linear

tetrahedral

square planar

octahedral

=

=

hexa a q u a n ic kel(II) chloride.

tris( ethylened iamine ) c h ro m i u m ( I I I ) perchlorate . -4

=

potassium hexacyanomanganate ( I I ) .

+2

6.

+3

6.

"II"

+2

6.

(1)

= -

=

potassium dicyanoargentate(I).

sodi u m tetrachlorocad mate(I I ) .

328

c) The cation is [Co(NH 3 MH 2 0)Br] 2+. The 2+ charge is deduced from the 2Br- ions. The cation has the name tetraamineaquabromocobalt(III) ion, with four ammonia ligands (tetraammine), one water ligand (aqua) and one bromide ligand (bromo). The oxidation state of cobalt is +3: 2 - {4(0) + 1(0) + The oxidation state is indicated by following cobalt in the name. The anion is Br-, bromide. The complete name is I (-I ) } .

(m),

tetraa mmineaqu abro mocobalt( I I I ) bromide.

22. 3 0 a) The cation is tetramminezinc ion. The tetraammine indicate four NH3 ligands. Zinc has an oxidation state of +2, so the charge on the cation is +2. The anion is SO/-. Only one sulfate is needed to make a neutral salt. The formula of the compound is b) The cation is pentaamminechlorochromium(lII) ion. The ligands are NH from pentaammine, and one chloride from chloro. The chromium ion has a charge of +3, so the complex ion3 has a charge equal to +3 from chromium, plus 0 from ammonia, plus -1 from chloride for a total of +2. The anion is chloride, Two chloride ions are needed to make a neutral salt. The formula of compound is c) The anion is bis(thiosulfato )argentate(I). Argentate(I) indicates silver in the + 1 oxidation state, and / bis(thiosulfato) indicates 2 thiosulfate ligands, S -. The on thesalt.anion s +1 plusof2(-2) to equalis -3. 0 The cation is sodium, Na+ . Three sodium ions are2 needed tototal makecharge a neutral The iformula compound [Zn (NH3)4] S04.

5

[ C r(NH3)sC I ] CI2 .

cr.

N a 3 [Ag(S203)z] '

23. 3 2 Coordination compounds act like electrolytes, i. e . , they dissolve in water to yield charged species. However, the complex ion itself does not dissociate. The "number of individual ions per formula unit" refers to the number of would form per 2coordination compound upon dissolution in water. a)ionsThethatcounter ion is S04.4 -, so the complex ion is [Zn(NH 3)4f+ . Each ammine ligand is unidentate, so the coordination number is Each molecule dissolves in water to form one SO/- ion and one [Zn(NH 3 )4] 2+ ion, so 2 form per formula unit. b) The counter ion is so the complex ion is [Cr(NH 3 )sCI] 2+ . Each ligand is unidentate, so+ the coordination number is Each molecule dissolves in water to form two ions and one [Cr(N H 3 )sCl f ion, so form per formula unit. + c) The counter ion is Na , so the complex ion is [Ag(S203 h] 3-. Assuming that+ the thiosulfate ligand is unidentate, the4coordination number is 2. Each molecule dissolves in water to form 3 Na ions and one [Ag(S203)2] 3- ion, so form per formula unit. 22. 3 4 Ligands that form linkage isomers have two different possible donor atoms. a) The nitrite ion because it can bind to the metal ion through either the nitrogen or one of the oxygen atoms - both have a lone pair of electrons. Resonance Lewis structures are [:'.O=== N -O:] 6. :].. .. [:o-N=== .. b) Sulfur dioxide molecules because both the sulfur and oxygen atoms can bind metal ions because they both have lone pairs. · ·0 S O·. . c) Nitrate ions have. three oxygen atoms, all with a lone pair that can bond to the metal ion, but all of the oxygen atoms are equivalent, so there are The nitrogen does not have a lone pair to use to bind metal ions. : O N O: ions

c r,

cr

6.

3

ions

fo r m s lin kage isomers

-

� ..

for m linkage isomers

.

====

====

no linkage isomers.

:O-N=== O..· .. : 0I :

..

[

.. .. II .. . 0.. -

329

-

j

..

ions

22.36

a) Platinum ion, Pt2+, is a r1 ion so the ligand arrangement is square planar. A and geometric isomer exist for this complex ion: H H H H C H3",-NI "' Br CH3",-NI "' Br Pt/ Pt/ CH3-N/ '" Br Br/ '"N--H H/ HI ICH3"'H Isomer exist because the mirror imagesisomer No optical isomers of both compounds are superimposable on the original molecules. In general, a square planar molecule is superimposable on its mirror image. b) A and geometric isomer exist for this complex ion. No optical isomers exist because the mirror images of both compounds are superimposable on the original molecules. H H H H", I ", I F H--N F H --N '" / '" / Pt Pt / ""/ Cl Cl '"N-- H H --N H/ HI IH "' H Isomerc isomers exist for this molecule,isomer c) Three geometri although they are not named or because all the ligands are different. A different naming system is used to indicate the relation of one ligand to another. H H H H H H ", I ", I ", I F H --N F H--N Cl H --N '" / '" / '" Pt/ Pt Pt / ""/ "'" F Cl / ""-O H H Cl O H -- O HI HI HI Types of isomers for coordination compounds are coordination isomers with different arrangements ofligands and counterions, linkage isomers with different donor atoms from the same ligand bound to the metal ion, geometric isomers with differences in ligand arrangement relative to other ligands, and optical isomers with mirror images that are not superimposable. a) Platinum ion, Pt2+, is a dB ion, so the ligand arrangement is square planar. The ligands are 2 and 2 Br-, so the arrangement can be either both ligands or both ligands to form geometric isomers. cis

trans

CIS

cis

trans

trans

CIS

cis

trans

--

--

22.38

trans

trans

cis

2-

330

cr

b) The complex ion can form linkage isomers with the N02 ligand. Either the N or an 0 may be the donor. 22NH3 NH3 I / NH3 I / NH3 H3N-Cr ONO H3N;;Cr--N02 / H3N 1 H3N 1 NH3 NH3 --

c) In the octahedral arrangement, the two iodide ligands can be either trans to each other, 1800 apart, or cis to each other, 900 apart. 2+ 2+ NH3 1 /1 H3N-Pt--I H3N/ 1 NH3 22.40 a) Four empty orbitals of equal energy are "created" to receive the donated electron pairs from four ligands. The four orbitals are hybridized from an s, two p, and one d-orbital from the previous n level to form 4 dsi orbitals. b) One s and three p-orbitals become four Sp3 hybrid orbitals. 22.43 a) The crystal field splitting energy is the energy difference between the two sets of d-orbitals that result from the bonding of ligands to a central transition metal atom. b) In an octahedral field of ligands, the ligands approach along the and axes. The d and d orbitals are located along the and axes, so ligand interaction is higher in energy than the other orbital-ligand interactions. The other orbital-ligand interactions are lower in energy because the dxy, dyz, and dxz orbitals are located between the and axes. c) In a tetrahedral field ofligands, the ligands do not approach along the and axes. The ligand interaction is greater for the dxy, dyz. and dxz orbitals and lesser for the d and d orbitals. The crystal field splitting is reversed, and the dxy, dyz, and dxz orbitals are higher in energy than the d and d orbitals. 22.45 If � is greater than electrons will preferentially pair spins in the lower energy d-orbitals before adding as unpaired electrons to the higher energy d-orbitals. If � is less than electrons will preferentially add as unpaired electrons to the higher d-orbitals before pairing spins in the lower energy d-orbitals. The first case gives a complex that is low-spin and less paramagnetic than the high-spin complex formed in the latter case. 22.47 To determine the number of d electrons in a central metal ion, first write the electron configuration for the metal atom. Examine the formula of the complex to determine the charge on the central metal ion, and then write the ion's configuration. a) Electron configuration ofTi: [Ar]4i3tf Charge on Ti: Each chloride ligand has a -I charge, so Ti has a +4 charge {+4 + 6(-1) } 2- ion. Both of the 4s electrons and both 3d electrons are removed. Electron configuration ofTi4+: [Ar] Ti4+ has no d electrons. b) Electron configuration of Au: [Xe]6s'4j45io Charge on Au: The complex ion has a -1 charge ([ AuCI n since K has a + 1 charge. Each chloride ligand has a -I charge, so Au has a +3 charge {+ 3 + 4(-1)} 1- ion. The4 6s electron and two d electrons are removed. Electron configuration of Au3+: [Xe]4/45Jl Au3+ has electrons. x,

x,

y,

y,

x,

y,

z

x

z

z

x,

x

2

-

y, y

2

z

z

x

Epairing,

2

-

y

2

z

2

2

-

y

2

z

2

Epairing"

=

=

8d

331

2

c) Electron configuration of Rh: [Kr]Si4d7 Charge on Rh: Each chloride ligand has a -1 charge, so Rh has a +3 charge {+ 3 + The Ss electrons and one 4d electron are removed. Electron configuration of Rh3 + : [Kr]4Ji Rh3 + has 6 d electro n s . 22 . 49

6(-1)}

=

3- ion.

a) Ti: [Ar]4i3cf. The electron configuration of Ti 3 + is [Ar]3i . With only one electron in the d-orbitals, the titanium(llI) ion cannot form high and low spin complexes - all complexes will contain one unpaired electron and have the same spin. Co: [ar]4i3d7 . The electron configuration of C02+ is [Ar]3d7 and will form high and low-spin complexes with 7 electrons in the d-orbital. c) Fe: [Ar]4i3Ji. The electron configuration of Fe2+ is [Ar]3Ji and will form high and low-spin complexes with 6 electrons in the d-orbital . d) Cu: [Ar]4S I 3dIO. The electron configuration of Cu2+ is [Ar]3d 9 , so in complexes with both strong and weak field ligands, one electron will be unpaired and the spin in both types of complexes is identical. Cu2+ cannot form high and low-spin complexes.

b)

(a)

(b)

LL

L_

11 11 L

11 11 11 low

high - spin

(c)

LL 11 L L high - spin

22 . 5 1

(d)

11 11 11 low

-

spin

-

spin

li L . 11 1L 11

To draw the orbital-energy splitting diagram, first determine the number of d electrons in the transition metal ion. Examine the formula of the complex ion to determine the electron configuration of the metal ion, remembering that the ns electrons are lost first. Determine the coordination number from the number of ligands, recognizing that 6 ligands result in an octahedral arrangement and 4 ligands result in a tetrahedral or square planar arrangement. Weak-field ligands give the maximum number of unpaired electrons (high-spin) while strong-field ligands lead to electron pairing (low-spin). a) Electron configuration of Cr: [Ar]4S I 3d5 Charge on Cr: The aqua ligands are neutral, so the charge on Cr is +3 . Electron configuration of Cr3 + : [Ar]3d 3 Six ligands indicate an octahedral arrangement. Using Hund's rule, fill the lower energy t2 g orbitals first, filling empty orbitals before pairing electrons within an orbital.

332

b) Electron configuration ofCu: [Ar]4s i 3di O Charge on Cu: The aqua ligands are neutral, so Cu has a +2 charge. Electron configuration9 ofCu2+: [Ar]3d 9 Four ligands and a d configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry). Use Hund's rule to fill in the 9 d electrons. Therefore, the correct orbital-energy splitting diagram shows one unpaired electron. c) Electron configuration of Fe: [Ar]4s23d 6 Charge on Fe: Each fluoride ligand has a - 1 charge, so Fe has a +3 charge to make the overall complex charge equal to -3 . Electron configuration of Fe3+: [Ar]3d 5 Six ligands indicate an octahedral arrangement. Use Hund's rule to fill the orbitals. is a weak field ligand, so the splitting energy, �, is not large enough to overcome the resistance to electron pairing. The electrons remain unpaired, and the complex is called high-spin. F

(b)

(a)

-L L L 22.53

Figure 22.2 1 describes the spectrum of splitting energy, �. N02- is a stronger ligand than NH3, which is stronger than H20. The energy oflight absorbed increases as �3 increases. [Cr(H 2 0)6 1 3+

22.55

22.60

<

[Cr(NH )6f+ < [Cr(N0 2)6 1 3

violet complex absorbs yellow-green light. The light absorbed by a complex with a weaker ligand would be at a lower energy and higher wavelength. Light oflower energy than yellow-green light is yellow, orange, or red light. The color observed would be blue or green. a) The coordination number of cobalt is 6 . The two cr ligands are unidentate and the two ethylenediamine ligands are bidentate, so a total of 6 ligand atoms are connected to the central metal ion. b) The counter ion is CI-, so the complex ion is [Co(en)2CI2t. Each chloride ligand has a -1 charge and each en ligand is neutral, so cobalt has a +3 charge: +3 + 2(0) + 2(- 1 ) + 1 . c) One mole of complex dissolves in water to yield one mole of [Co(en)2CI2t ions and one mole ofCI- ions. Therefore, each formula unit yields 2 individual ions. d) One mole of complex dissolves to form one mole ofCr ions, which reacts with the Ag+ ion (from AgN 03) to form one mole of AgCl precipitate. [Co(NH3MH20)CI] 2+ tetraammineaquachlorocobalt(IlI) ion 2 geometric isomers 2+ 2+ CI CI H3� I / NH3 H3� I / OH2 Co Co...... ,... .. H3N,..... I ...... NH3 H3N I NH3 NH3 H20 cis CI and NH 3 trans CI and H20 A

=

22.64

333

[Cr(H20)3Br2CI] triaquadibromochlorochromium(III) 3 geometric isomers CI CI H20 I / Br H20 I ....... OH2 "Cr "Cr Br./" I "OH2 H20""""'" I 'Br Br H20 Br's Br's Br's H20 's facial H20's meridional (Unfortunately meridional and facial isomers are not covered in the text. Facial (fac) isomers have three adjacent comers of the octahedron occupied by similar groups. Meridional (mer) isomers have three similar groups around the outside of the complex.) [Cr(NH3MH20)2Br2]+ diamminediaquadibromochromium(lII) ion 6 isomers (5 geometric) NH3 H20 I / Br 'Cr Br./" I "OH2 NH3 Only NH3's are All pairs are NH3 NH3 H20 I / NH3 H20 I / Br 'Cr 'Cr Br./" I "OH2 Br./" I "'NH3 Br OH2 Only H20 's are Only Br's are NH3 NH3 H20 I /NH3 H3N I ....... OH2 'Cr...... 'Cr H;5 I 'Br Br./" I "OH2 Br Br All pairs are These are optical isomers of each other. a) The first reaction shows no change in the number of particles. In the second reaction, the number of reactant particles is greater than the number of product particles. A decrease in the number of particles means a decrease in entropy, while no change in number of particles indicates little change in entropy. Based on entropy change only, the first reaction is favored. b) The ethylenediamine complex is more stable with respect to ligand exchange with water because the entropy change is unfavorable. cis

cis

trans

+

trans

+

trans

+

trans

trans

+

cis .

22.69

334

+

+

Chapter 23 Nuclear Reactions and Their Applications FOLLOW-UP PROBLEMS

23 . 1

Plan: Write a skeleton equation that shows an unknown nuclide, � X , undergoing beta decay, _� � , to form cesium- 1 33, I ;; CS . Conserve mass and atomic number by ensuring the superscripts and subscripts equal one another on both sides of the equation. Determine the identity of X by using the periodic table to identify the element with atomic number equal to Z. Solution: The unknown nuclide yields cesium- 1 3 3 and a � particle: A X --t 1 lJ Cs + 0 A - I I-' 55 Z To conserve atomic number, Z must equal 54. Element is xenon. To conserve mass number, A must equal 1 33 . The identity o f � X i s I ;! X e . The balanced equation is' 13354 Xe --t 13355 Cs + -\0 AtJ Check: A = 1 33 = 1 33 + 0 and Z = 54 = 55 + ( I) so mass is conserved. •

-

,

23.2

Plan: Nuclear stability is found in nuclides with an NIZ ratio that falls within the band of stability in Figure 23.2. Nuclides with an even N and Z, especially those nuclides that have magic numbers, are exceptionally stable. Examine the two nuclides to see which of these criteria can explain the difference in stability. Solution: Phosphorus-3 1 has 1 6 neutrons and 1 5 protons, with an NIZ ratio of 1 .07. P hosphorus-30 has 1 5 neutrons and 1 5 protons, with an NlZ ratio o f 1 .00. An NIZ ratio of 1 .00 when Z < 2 0 could indicate a stable nuclide, but Figure 23.2 indicates a departure from the 1 : 1 line, even with smaller nuclides. 3 1 p has an even N and a slightly higher NIZ ratio, accounting for its stability over 3 0p.

23.3

Plan: Examine the NIZ ratio and determine which mode of decay will yield a product nucleus that is closer to the band. Solution: a) Iron-6 1 has an NlZ ratio of ( 6 1 - 26)/26 = 1 .35, which is too high for this region of the band. lron-6 1 will undergo � decay. °A new NIZ = (6 1 - 27)/27 = 1 .26 2616 Fe --t 2671 C O + -I }J b) Americium-24 1 has Z > 83, so it undergoes a decay. 2;; Am --t 2 �� Np + � He

23.4

Plan: Use the half-life of 24Na to find k. Substitute the value of k, initial activity (Ao), and time of decay (4 days) into the integrated first-order rate equation to solve for activity at a later time (AI)' Solution: k (In 2) 1 tl/2 = ( In 2) 1 1 5 h = 0.0462098 h-I (unrounded) In Ao - In AI = kt In AI = -kt + In Ao = -(0.0462098 h-I) ( 4.0 days) ( 24 h/day) + In ( 2.5 x 1 09 ) = 1 7 .2034 (unrounded) In AI = 1 7.2034 AI = 2.96034 X 1 0 7 = 3.0 X 1 0 7 dis Check: In 4 days, a little more than 6 half-lives ( 6.4) have passed. A quick calculation on the calculator shows that (2 .5 x 1 09 ) 1 2 = 1 .25 X 1 09 ( first half-life). Dividing by 2 again shows 6.25 x 1 0 8 (secon d half-life). Continued division by 2 shows that this value comes to 3.9 x 1 0 7 dis after six divisions by 2. This number is slightly higher than the calculated answer because 6 half-lives have been calculated, not 6.4 half-lives. =

335

23.5

Plan: The wood from the case came from a living organism, so Ao equals 1 5 .3 d/minog. Substitute the current activity of the case (At), Ao and the tll2 of carbon (5730 yr) into the first-order rate expression and solve for t. Solution: ( 1 5 .3 d / minog ) I A 1 1 A 40 1 8.2 4.02 x 1 03 years In In -o t - In -o . ) n g d n m 1 9.4 ( A, k A, i o 730 =

=

[;(,J

= C /s' l")

=

=

Check: Since the activity of the bones is a little more than half of the modem activity, the age should be a little less than the half-life of carbon. 23.6

2 5

Plan: Uranium-23 5 has 92 protons and 1 43 neutrons in its nucleus. Calculate the mass defect (�m) in one 3 U atom, convert to MeV and divide by 235 to obtain binding energy/nucleon. Solution: �m [(92 x mass H atom) + ( 1 43 x mass neutron)] - mass 2 3 5 U atom �m [(92 x 1 .007825 amu) + ( 1 43 x 1 .008665)] - 235.043924 amu �m 1 .9 1 507 1 amu Binding Energy / nucleon [( 1 .9 1 507 1 amu) (93 1 .5 MeV/amu)] / (235 nucleons) 7.59 1 0 1 5 7.59 1 MeV / nucleon The BE/nucleon of 1 2 C is 7.680 MeV/nucleon. The energy per nucleon holding the 23 5 U nucleus together is less than that for 1 2 C (7.59 1 < 7.680), so 23 5 U is less stable than 1 2 c. Check: The answer is consistent with the expectation that uranium is less stable than carbon. =

=

=

=

=

=

END-OF-CHAPTER PROBLEMS

23 . 1

a) Chemical reactions are accompanied by relatively small changes in energy while nuclear reactions are accompanied by relatively large changes in energy. b) Increasing temperature increases the rate of a chemical reaction but has no effect on a nuclear reaction. c) Both chemical and nuclear reaction rates increase with higher reactant concentrations. d) If the reactant is limiting in a chemical reaction, then more reactant produces more product and the yield increases in a chemical reaction. The presence of more radioactive reagent results in more decay product, so a higher reactant concentration increases the yield in a nuclear reaction.

23.2

Radioactive decay that produces a different element requires a change in atomic number (Z, number of protons). A mass number (protons + neutrons) �X number of protons (positive charge) Z symbol for the particle X A - Z (number of neutrons) N a) Alpha decay produces an atom of a different element, i.e., a daughter with two less protons and two less neutrons. �X -t �� Y + � He 2 fewer protons, 2 fewer neutrons b) Beta decay produces an atom of a different element, i.e., a daughter with one more proton and one less neutron. A neutron is converted to a proton and � particle in this type of decay. �X -t Z �I Y + _ ?f3 I more proton, 1 less neutron c) Gamma decay does not produce an atom of a different element and Z and N remain unchanged. ( � X * energy rich state), no change in number of protons or neutrons. � X * -t � X + � 'Y d) Positron emission produces an atom of a different element, i.e., a daughter with one less proton and one more neutron. A proton is converted into a neutron and positron in this type of decay. �X -t Z�I Y + +� � I less proton, I more neutron e) Electron capture produces an atom of a different element, i.e., a daughter with one less proton and one more neutron. The net result of electron capture is the same as positron emission, but the two processes are different. � X + _� e -t Z�I Y 1 less proton, I more neutron A different element is produced in all cases except (c). =

336

23 .4

A neutron-rich nuclide decays to convert neutrons to protons while a neutron-poor nuclide decays to convert protons to neutrons. The conversion of neutrons to protons occurs by beta decay: 0A I I o n � I p + -\ .., The conversion of protons to neutrons occurs by either positron decay: l OA I I p � 0 n + I tJ or electron capture: : p + _� e � � n Neutron-rich nuclides, with a high NIZ, undergo � decay. Neutron-poor nuclides, with a low NIZ, undergo positron decay or electron capture.

23.6

In a balanced nuclear equation, the total of mass numbers and the total of charges on the left side and the right side must be equal. a) 2�� U � � He + 239� Th Mass: 234 = 4 + 230; Charge: 92 = 2 + 90 b) 2!� Np + _� e � 2�� U Mass: 232 + 0 = 232; Charge: 93 + (-I ) = 9 2 12 12 O A c) 7 N � I t-' + 6 C Mass: 1 2 = 0 + 1 2; Charge: 7 = I + 6

23.8

a) In other words, an unknown nuclide decays to give Ti-48 and a positron. 48 y � 48 Ti + 0I AI-' 23 22 b) In other words, an unknown nuclide captures an electron to form Ag- I 07. I �� Cd + _� e � I �; Ag c) In other words, an unknown nuclide decays to give Po-206 and an alpha particle. 2�� Rn � 2�: Po + � He

23. 1 0

Look at the NIZ ratio, the ratio of the number of neutrons to the number of protons. If the NIZ ratio falls in the band of stability, the nuclide is predicted to be stable. Also check for exceptionally stable numbers of neutrons and lor protons - the "magic" number of 2, 8, 20, 28, 50, 82 and (N = 1 26). Also, even numbers of protons and or neutrons are related to stability whereas odd numbers are related to instability. (a) 2� O appears stable because its Z (8) value is a magic number, but its NIZ ratio ( 1 2 :8 = 1 .50) is too high and this nuclide is above the band of stability; 2� O is unstable. (b) �� Co might look unstable because its Z value is an odd number, but its NIZ ratio (32:27 1 . 1 9) is in the band of stability, so �� Co appears stable. ( c) � Li appears unstable because its NIZ ratio (6 :3 = 2.00) is too high and is above the band of stability. =

23. 1 2

a) 2�� U : N uclides with Z > 8 3 decay through ex decay. b) The NIZ ratio for i! Cr is (48 - 24)/24 1 .00. This number is below the band of stability because N is too low and Z is too high. To become more stable, the nucleus decays by converting a proton to a neutron, which is positron decay. Alternatively, a nucleus can capture an electron and convert a proton into a neutron through electron capture. c) The N/Z ratio for �� Mn is (50 - 25)/25 = 1 .00. This number is also below the band of stability, so the nuclide undergoes positron decay or electron capture. =

23 . 1 4

Stability results from a favorable NIZ ratio, even numbers of N and/or Z, and the occurrence of magic numbers. The NIZ ratio of �� Cr is ( 52 - 24)/24 = 1 . 1 7, which is within the band of stability. The fact that Z is even does not account for the variation in stability because all isotopes of chromium have the same Z. However, �� Cr has 28 neutrons, so N is both an even number and a magic number for this isotope only.

337

23 . 1 8

N o, it is not valid to conclude that tl/2 equals 1 minute because the number of nuclei is so small (6 nuclei). Decay rate is an average rate and is only meaningful when the sample is macroscopic and contains a large number of nuclei, as in the second case. Because the second sample contains 6 x 1 0 12 nuclei, the conclusion that tll2 = 1 minute is valid.

23 .20

Specific activity of a radioactive sample is its decay rate per gram. Calculate the specific activity from the number of particles emitted per second (disintegrations per second = dps) and the mass of the sample. Specific activity = decay rate per gram. I Ci = 3 .70 x 1 0 1 0 dps 1 .66 x 1 0 6 dPS 1 �g 1 Ci Specific Activity = = 2.8945 X 1 0-2 = 2.89 X 1 0-2 Ci/g 1 .5 5 mg 1 0- g 3 .70 x 1 0 1 0 dps

[

23 .22

][ ][

]

The rate constant, k, relates the number of radioactive nuclei to their decay rate through the equation A = kN. The number of radioactive nuclei is calculated by converting moles to atoms using Avogadro's number. The decay rate is 1 .39 x 1 0 5 d/yr or more simply, 1 .39 x 1 0 5 yr- I (the disintegrations are assumed). Decay rate = -(llN / Llt) = kN 1 .00 x 1 0. 1 2 mOI 6.022 x 1 0 2 3 atom - 1 .39 x 1 0 5 atom =k I mol 1 .00 yr 1 .39 x 1 0 5 atom /yr = k (6.022 x l O " atom) k = ( 1 .39 x 1 0 5 atom /yr) / 6.022 x 1 0 " atom k = 2.30820 X 1 0-7 = 2 3 1 X 1 0-7 y r- I

[

_

][

]

.

23 .24

Radioactive decay is a first-order process, so the integrated rate law is In � = kt Nt 3 To calculate the fraction of bismuth-2 1 2 remaining after 3 .75 x 1 0 h, first find the value of k from the half-life, then calculate the fraction remaining with No set to 1 (exactly). tll2 = 1 .0 1 yr t = 3 .75 X 1 0 3 h k = (In 2) / (tid = (In 2) / ( 1 .0 1 yr) = 0.686284 yr- I (umounded) 1 In _- = (0.686284 yr- I ) (3 .75 x 1 0 3 h) ( I day / 24 h) ( 1 yr / 365 days) Nt 1 In - = 0.293 785958 (unrounded) Nt I - = 1 .34 1 4967 (unrounded) Nt Nt = 0.745436 (unrounded) Multiplying this fraction by the initial mass, 2.00 mg, gives 1 .490872 = 1 .49 mg of bismuth-2 1 2 remaining after 3 .75 x 1 0 2 h.

23 .26

Lead-206 is a stable daughter of 23 8 U. Since all of the 206 Pb came from 23 8 U, the starting amount of 23 8 U was (270 /lmol + 1 1 0 /lmol) = 380 /lmol = No. The amount of 23 8 U at time t (current) is 270 /lmol = Nt. Find k from the first-order rate expression for half-life, and then substitute the values into the integrated rate law and solve for t. tll2 = (In 2) / k k = (In 2) / tll2 = (In 2) / (4.5 x 1 0 9 yr) = 1 .540327 x 1 0- 1 0 yr- I (unrounded) N In o = kt Nt

3 80 /lmol = ( 1 .540327 x 1 0- 10 yr- I )t 270 /lmol 0.34 1 749293 = ( 1 .540327 x 1 0- 1 0 yr- I )t t = (0.34 1 749293) / ( 1 .540327 x 1 0- 10 yr- I ) = 2.2 1 868 x 1 09 = 2.2 X 1 09 y r In

338

23 .28

Use the conversion factor, I Ci = 3 .70 x 1 0 1 0 disintegrations per second (dps). 6 X I 0 - 1 l mCi --1 0 -3 Ci 3 .70 x 1 0 1 0 dP S --. . 60 s 1 000 mL A ctlVlty = = 1 26.0 = 1 x mL 1 mCi 1 Ci I min I .OS7 qt

)[

[

)(

)[

)

)(

10

2

dpm/qt

23 .32

Protons are repelled from the target nuclei due to the interaction of like (positive) charges. Higher energy is required to overcome the repulsion.

23 .33

a)

I � B + � He

--7

�n + X

Since the charge on the left, S + 2 = 7, the charge on the right must = 7 Element X has a = 7 which is nitrogen. The mass is 1 4 on each side. lO B + ' He --7 I n + 1 3 N 2 0 5 7 b) Si + H --7 �� P + X Since the charge on the left, 1 4 + I = I S , the charge on the right must = I S . P-29 already has a charge of I S so X must have = 0, a neutron. Si + H --7 �� P + � n The mass on each side = 30 c) X + ; He --7 2 � n + 2 Cf The charge on the right = 0 + 98 = 98. The charge on the left must = 98. He-4 has a charge of 2, so element X has a charge of 98 - 2 = 96. Element 96 is Cm. 2 Cm + ; He --7 2 � n + 2 Cf The mass of Cm is 242 so that the mass on each side = 244

Z

�� � �� �

Z

��

��

��

23.37

Ionizing radiation is more dangerous to children because their rapidly dividing cells are more susceptible to radiation than an adult' s slowly dividing cells.

23 .38

a) The rad is the amount of radiation energy absorbed per body mass in kg. 3 .3 x 1 0-7 J 2.20 S Ib 1 rad = S .39 X 1 0-7 = 5.4 X 1 0-7 rad 1 kg 1 3S Ib 1 X 1 0-2 J / kg b) Conversion factor is 1 rad = 0.01 Gy (S .39 x 1 0-7 rad) (0.0 1 Gy/ I rad) = S .39 x 1 0-9 = 5.4 X 1 0-9 Gy

23 .40

)[

)(

[

)

[ X ](

a) Convert the given information to units of J /kg. I rad = 0.0 1 J /kg = 0.0 I Gy 6.0 x 1 0 5 � 8.74 x 1 0- 1 4 J /� 0 . 0 1 GY 1 rad = 7.49 1 4 X 1 0- 10 = 7.5 X Dose = 70. kg I rad 0.0 I g

(

)

)(

)(

) [ :r )

=

( )[ )( )[

23 .44

1 0- 1 0 G y

b) Convert grays to rads and multiply rads by RBE to find rems. Convert rems to mrems. I rad I em rem = rads x RBE = 7 .49 1 4 x 1 0- 1 0 GY ( 1 .0) = 7.49 1 4 x 1 O-5 = 7.5 x I0-5 mrem 0.0 1 Gy 1 0 rem c) 1 rem = 0.0 1 Sv Sv (7.S X 1 0-5 mrem) ( 1 0-3 rem / 1 mrem) (0. 0 1 Sv / rem) 7.49 1 4 x 1 0- 1 0 = 7.5 X 1 0- 10 Sv. Use the time and disintegrations per second (8q) to find the number of 60Co atoms that disintegrate, which equals the number of � particles emitted. The dose in rads is calculated as energy absorbed per body mass. I rad 60 S 47S 8q 1 0 3 g l dP S S .OS x 1 0- 14 J 27.0 = 2.447 X 1 0-3 = 2.45 X 1 0-3 ra d min ) Dose = ( . . 1 mill 0.0 I I .S 88 g I kg I Bq 1 dlslllt. g

(

23 .42

)

=

)

( )[

X

]

N AA does not destroy the sample while chemical analyses does. N eutrons bombard a non-radioactive sample, "activating" or energizing individual atoms within the sample to create radioisotopes. The radioisotopes decay back to their original state (thus, the sample is not destroyed) by emitting radiation that is different for each isotope.

339

oxygenin theinformal formalddehyde ehydeproduct. comes fromThe methanol because thethe oxygen isaciotoped reactant in the methanol reactant 23.45 The appears oxygen i s otope i n chromi c appears i n waterdehyde. product, not the formaldehyde product. The isotope traces the oxygen in methanol to the oxygen ithen formal 23.48 Apply the appropriate conversions from the chapter or the inside back cover. 6 ev ) a) Energy (0.0 1861 MeV) [ 10I MeV 6 eV ) [ 1.6021 xeV10- 1 9 ) 2 .981 322 10-1 5 b) Energy (0. 0 1861 MeV)[ 101 MeV 23. 5 0 Mass Calculof27 ate �m,IH convert the27 mass deficit to27.MeV,2 11275andamu divide by 59 nucleons. atoms x 1. 0 07825 Mass of32Totalneutrons 32 x59.1.0408665 32.27728 amu 88555 amu mass Mass defect �m 59. 4 88555 -58.933198 0. 5 55357 amurCo 0. 5 55357 glmol 59CO amu (931. 5 MeV] CO ) 59 a) B m· d'mg energy [ 0. 5 55357 59 nucleons 1 amu 8 768051619 5 MeV ] 517.3 150 b) B inding energy [ 0. 5 553571 atoamum 59 CO ) ( 931.I amu c) Use �E �mc2 Binding energy [ 0. 5 55357molg 59 co )[ 101 �gg ) ( 2. 99792 108 m s )2 kg ·I m 10I � J 4.9912845 101 0 andisfissia spontaneous on, radioactiprocess ve partiinclewhis arechemiunstabl tted,ebutnucltheei process leoadiactinvgetopartithecemiles andssion is 23. 5 3 Indiffbotherent.radiRadioactioactive vdecay e decay emi t radi energy.idesFitossibreak on occurs as thelerresul t dofes,hiradigh-energy bombardment ofenergy. nuclei with small particles that cause the nucl i n to smal nucli o acti v e parti c l e s, and Infissia ochain andn reacti orgen, alnucll fissieusoncaneventsleadareto splnotitthetingsame. Thelargecolnucllisioeni ibetween the ofsmalwaysl partito produce cle emit several ed in the the l a of the n a number different products. 23. 5 6 The water serves to slow the neutrons so that they are better able to cause a fission reaction. Heavy water (�H 2 0neutrons or D20)areis avai a betterlablemoderator because iotndoesprocess. not absorb neutrons asdoeswelnotl asoccur light natural water l(:y Hin20)greatdoes, so more to i n i t i a t e the fissi However, D20 abundance, so production of 020 adds to the cost of a heavy water reactor. In addition, if heavy water does absorb a neutron, it becomes i.e., it contains the isotope tritium, �H , which is radioactive. 23.60 a) Use the valCmues giv�enPuin the Heproblem to calculate the mass change (reactant -products) for: �mass (kg) [243. 0614 amu -(4.0026 + 239.0522) amu] [ 1.66054amu10-24 g )[ 101 kgg ) 1.095956 10-29 4 = 1 .861 x 1 0 eV

=

J =

=

= =

=

X

=

= .

=

=

X

=

= 8 . 768 M e V I

= 4.99 128

X

1010

[ j(,]( )

/

S

kJ

2:

2

I mol

+�

X

=

=

X

= 1.1

X

2 1 0- 9

kg

340

nue I eon

= 51 7.3 MeV I atom

tritiated,

-t

J

=

=

2:�

t o- 1 5

=

=

X

X

=

=

=

= 2 .98 1

-

3-

J

b) E = �mc 2 = ( 1 .095956

X

1 0-29 kg )( 2.99792

X

1 0 8 m/s

t

[ %] 1J kg m 0

= 9.84993 x 1 0- 1 3 = 9.8 X t o- 13 J

S2 1 3 23 c) E released = (9.84993 x 1 0- J/reaction) (6.022 x 1 0 reactions/mol) ( 1 kJ / 1 0 3 J ) = 5 .93 1 6 X 1 0 8 = 5.9 X 1 0 8 kJ/mol This is approximately 1 million times larger than a typical heat of reaction. 22.62

Determine k for 1 4C using the half-life (5730 yr): k = (In 2) / tl/2 = (In 2) / (5730 yr) = 1 .2096809 x 1 0-4 yr- I (unrounded) Determine the mass of carbon in 4.38 grams of CaC0 3 : I mol CaC 0 3 1 2.0 1 g C I mol C mass = ( 4.38 g caC0 3 ) = 0.52556 g C (unrounded) 1 mol C 1 00.09 g CaC0 3 1 mol CaC0 3 The activity is: (3.2 dpm) / (0.52556 g C) = 6.08874 dpmog- ' Using the integrated rate law:

(

In

( �: )

6.08874 dpm g- I 1 5 .3 dpm g- I t = 76 1 6.98 = 7.6 x 1 03 y r 23 .65

[

)

= -( l .2096809 X 1 0-4 yr- I )t

Determine how many grams of AgCI are dissolved in 1 mL of solution. The activity of the radioactive Ag+ indicates how much AgCI dissolved, given a starting sample with a specific activity ( 1 75 nCi/g). 1 .25 X 1 0-2 Bq 1 dPS 1 Ci 1 nCi 1 g AgCl Concentration = 1 mL I Bq 3 .70 X 1 0 0 dps 1 75 nCi 1 0-9 Ci = 1 .93050 x 1 0-6 g AgCI/mL (unrounded) Convert glmL to moUL (molar solubility) using the molar mass of AgCI. 1 mol AgC I 1 mL 1 .93050 x 1 0 .6 g AgCI Molarity = 1 43 .4 g AgCI 1 0-3 L mL = 1 .34623 X 1 0-5 = 1 .3 5 X t o-5 M AgCI

)( )[

[

)(

[

23 .66

)

No = 1 5.3 dpm g- 1 (the ratio of I 2 e : 14 C in living organisms)

= -kt In

)(

)(

)(

)[

)

)( )

a) Find the rate constant, k, using any two data pairs (the greater the time between the data points, the greater the reliability of the calculation). Calculate tl/2 using k. In

(�J

= -kt

(

)

495 photons / s = -k(20 h) 5000 photons / s -2.3 1 2635 = -k(20 h) k = 0. 1 1 563 h- ' (unrounded) tll2 = (In 2) / k = (In 2) / (0. 1 1 563 h- I ) = 5.9945 = 5.99 h (Assuming the times are exact, and the emissions have three significant figures.) In

34 1

2. 0

2. 0

b) The percentage of isotope remaining is the fraction remaining after h (Nt where t = h) divided by the initial amount (No), i.e., fraction remaining is N/No. Solve the first order rate expression for NtlNo, and then subtract from to get fraction lost. In

( �J

100%

= -kt

(��) -(0.1 1563 h-I) (2.0 (��) 0.793533 ( �� ] 100% 79.3 533% In

=

=

-0.23126 (unrounded)

(unrounded)

x

=

(unrounded)

100% - 79.3533% 20. 6467%

The fraction lost is

23. 6 8

h) =

=

= 21%

of the isotope is lost upon preparation.

The production rate of radon gas (volume/hour) is also the decay rate of 22 6Ra. The decay rate, or activity, is proportional to the number of radioactive nuclei decaying, or the number of atoms in g of 22 6Ra, using the relationship A = kN. Calculate the number of atoms in the sample, and find k from the half-life. Convert the activity in units of nuclei/time (also disintegrations per unit time) to volume/time using the ideal gas law. 2�: Ra � � He + 2�! Rn k = (In / tl/2 = (In / yr) h/yr)] = x h- I (unrounded) 6 22 amu / atom or g / mol. The mass of Ra is x ) Ra atoms mol Ra = x N= g Ra Ra atoms (unrounded) mol Ra g Ra x 1 0 21 Ra atoms) = x Ra atoms / h (unrounded) A = kN = x 6 22 This result means that x Ra nuclei are decaying into 222 Rn nuclei every hour. Convert atoms of 222 Rn into volume of gas using the ideal gas law. moles = x Ra atoms / h) ( l atom Rn / atom Ra) mol Rn / x 1 0 23 Rn) = x mol Rn / h (unrounded) L atm x mol Rn / h K) mol · V = nRT / P = atm 9 = x = 4.904 X 1 0- L / h Therefore, radon gas is produced at a rate of x Llh. Note: Activity could have been calculated as decay in moles/time, removing Avogadro's number as a multiplication and division factor in the calculation.

1 . 000

2)

2) [(1599 (8766 4. 9451051 10-8 226. 025402 226.2025402 ) 2.6643023 102 1 (1.000 ( 226.1025402 ) [ 6. 022 1 10 3 1 . 3 175254 101 4 (4. 9451051 10-8 h-I) (2.4 6643023 1. 3 18 101 (1 . 3 175254 101 4 1 (1 6. 022 2.1878536 10-10 (2.1878536 10-10 ) ( 0. 08206 K ) (273.15 1 4. 904006 10-9 4. 904 10-9 •

23. 7 5

2 1 2 (3 32 00 106 K) 2. 0709066 10-1 7 ( 23 ) (8.3146. 022 102K)(1. 3 1 . 7 975048 101 7 (2. 00 (2. 99792 108 1.7975048 1017 )[ ) ( 1 . 0078 J ) 2 3 3 1 10 6. 022 10 2. 0709066 10-17 J 1. 4 525903 107

a) Kinetic energy = ] / mv2 = ( / ) m RT/MH) = ( / ) (RT / NA) x J / mol · = Energy = = 2.07 X 1 0- 1 7 J/ato m : H X x atom / mol b) A kilogram of I H will annihilate a kilogram of anti-H ; thus, two kilograms will be converted to energy: x Energy = mc 2 = kg) x m/s) \J / (kg·m 2/s2 )) = J (unrounded) x gH I mol H J I kg H atoms H atoms = I kg H mol H g atoms H x x =

[

X

= 1 .45

X

1 07

H atoms

342

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c) 4 : H

�

� He + 2 � P

�m

=

(Positrons have the same mass as electrons.) [4( 1 .007825 amu)] - [4.00260 amu - 2(0.000549 amu)] 0.027602 amu / � He

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0.027602 g I kg I mol He I mol H ( 1 .4525903 X 1 0 7 H atoms ) 2 3 3 mol He 4 mol H 6.022 x 1 0 H atoms 10 g 1 .6644967 x 10-22 kg (unrounded) Energy (�m)c 2 ( 1 .6644967 x 1 0-22 kg) (2.99792 X 1 0 8 mlS) 2 (J / (kgom2/s2 )) 1 .4959705 x 1 0-5 1 .4960 X 1 0-5 J d) Calculate the energy generated in part (b): 1 .7975048 x 1 0 1 7 J 1 kg 1 .0078 g H 1 mol H Energy 3 . 008 1 789 x 1 0- 10 J 3 1 mol H 1 kg H 6.022 x 1 0 2 3 atoms H 10 g Energy increase ( 1 .4959705 x 1 0-5 - 3 .008 1 789 x 1 0- 10 ) J 1 .4959404 x 1 0-5 1 .4959 X 1 0-5 J e) 3 : H � � He + 1 � P �m [3 ( 1 .007825 amu)] - [3.01 603 amu + 0.000549 amu] 0.006896 amu / � He 0.006896 g / mol � He �m

=

=

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)

0.006896 g 1 kg 1 mol He I mol H ( 1 .4525903 x 1 07 H atoms ) 3 mol He 1 0 g 3 mol H 6.022 x 1 0 2 3 H atoms 5 .5447042 x 1 0-23 kg (unrounded) Energy (�m)c 2 (5.5447042 x 1 0-23 kg) (2.99792 X 1 0 8 mlS) 2 (J / (kgom2/s2 )) 4.983 3 1 64 x 1 0-6 4.983 X 1 0-6 J No, the Chief Engineer should advise the Captain to keep the current technology.

�m

=

=

=

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=

343