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Student Study Guide and Solutions M anual 
for Atkins and Jones's
CHEMICAL PRINCIPLES
The Quest fo r Insight
Fifth Edition
John Krenos
Rutgers. The State University o f New Jersey
Joseph Potenza
Rutgers, The State University o f New Jersey
Carl Hoeger
University o f California, San Diego
Laurence Lavelle
University o f California, Los Angeles
Yinfa Ma
Missouri University o f Science and Technology
IS
W.H. Freeman and Company 
New York
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ISBN-13: 978-1-4292-3135-0 
ISBN-10: 1-4292-3135-1
© 2005, 2008, and 2010 by W.H. Freeman and Company 
A ll rights reserved.
Printed in the United States o f America 
First Printing
W.H. Freeman and Company 
41 Madison Avenue 
New York. N Y 10010
Houndsmills, Basingstoke RG21 6XS England
www.whfreeman.com
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http://www.whfreeman.com
CO N TEN TS
Preface v
Acknowledgments v//
STUDY GUIDE SG_
1 Atoms: The Quantum World 1
2 Chemical Bonds 15
3 Molecular Shape and Structure 25
4 The Properties of Gases 35
5 Liquids and Solids 45
6 Inorganic Materials 57
7 Thermodynamics: The First Law 71
8 Thermodynamics: The Second and Third Laws 85
9 Physical Equilibria 99
10 Chemical Equilibria
11 Acids and Bases 129
12 Aqueous Equilibria I 45
13 Electrochemistry I55
14 Chemical Kinetics I59
15 The Elements: The Main Group Elements 187
16 The Elements: The d Block 217
17 Nuclear Chemistry 233
18 Organic Chemistry 1: The Hydrocarbons 245
19 Organic Chemistry II: Polymers and Biological
Compounds 257
SOLUTIONS MANUAL SM_
Fundamentals 3
1 Atoms: The Quantum World 53
2 Chemical Bonds 75
3 Molecular Shape and Structure 97
4 The Properties o f Gases 119
5 Liquids and Solids I47
6 Inorganic Materials 171
7 Thermodynamics: The First Law 183
8 Thermodynamics: The Second and Third Laws 211
9 Physical Equilibria 239
10 Chemical Equilibria 273
11 Acids and Bases 313
12 Aqueous Equilibria 355
13 Electrochemistry 395
14 Chemical Kinetics 435
15 The Elements: The Main Group Elements 459
16 The Elements: The d Block 479
17 Nuclear Chemistry 499
18 Organic Chemistry 1: The Fiydrocarbons 521
19 Organic Chemistry II: Polymers and Biological
Compounds 543
iii
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SOLUTIONS M ANUAL
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FUNDAMENTALS
A.l (a) chemical; (b) physical; (c) physical
A.3 The temperature of the injured camper, the evaporation and condensation 
of water are physical properties. The ignition o f propane is a chemical 
change.
A.5. (a) Physical; (b) Chemical; (c) Chemical
A.7 (a) intensive; (b) intensive; (c) extensive; (d) extensive
A.9 (a) 1000. grain x
( \ 
1 A'/log rah?
(b) 0.01 batman x
V
(c) lx l 0b mutchkin x
1,000 grain
100 centibatman \ 
1 batman )
= 1 kilograin
= 1 centibatman
1 megamutchkin ] . , , ------ -̂----------- = 1 megamutchkin 10
106 mutchkin J
1 pint 1 quart 0.946 L 1000 mL 
A .ll 1.00 cup x ( ^ ---- ) x ( - 1, ) x ( - ------ - ) x ( — —---- ) = 236 mL
2 cups 2 pints 1 quart 1 L
A .13
. m 
a - — 
V
f 112.32 e
29.27 m L - 23.45 mL
mL 1
cm
- 19.3 g • cm'-'
SM-3
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3.13 The Lewis structures are
F-OO,
(a) The shape of CF,Cl is tetrahedral; all halogen— C— halogen angles should be 
approximately 109.5°. AX4;
(b) TeCl4 molecules will be see saw shaped with Cl— Te— Cl bond angles of 
approximately 90° and 120°. A X (E;
(c) COF2 molecules will be trigonal planar with F—C F and O— C— F angles 
of 120°. AX,;
(d) CH~ ions will be trigonal pyramidal with H— C— H angles of slightly less 
than 109.5°. AX, E
3.15 (a) The angles represented by a and b are expected to be about 120°, while c is
expected to be about 109.5° in 2, 4-pentanedione. All of the angles are expected 
to be about 120° in the acetylacetonate ion.
(b) The major difference arises at the C of the original sp2-hybridized CH2
group, which upon deprotonation and resonance goes to sp2 hybridization with 
only three groups attached (the double-headed arrow is read as “ in resonance 
with” ):
3.17 (a) slightly less than 120; : (b) 180°; (c) 180°; (d) slightly less then
109.5°
H H
SM-99
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7.95 Appendix 2A provides us with the heat o f formation of
h (g) at 298K (+62.44 kJ • mol-1) and the heat capacities of
I, (g) (36.90 J -K '1 • mor1) and I2(s)(54.44 J 'K "1 • mol-1). We can 
calculate theA//sub° at 298K:
1, (s )---- > 1, (g) A//sub0 = +62.44 kJ • mol"1
We can calculate the enthalpy of fusion from the relationship 
= AH
but these values need to be at the same temperature. To correct the value 
for the fact that vve want all the numbers for 298K, we need to alter the 
heat o f vaporization, using the heat capacities for liquid and gaseous 
iodine.
1,(1) at )84.3°C-----■+ 1, (g) at 184.3°C Atfv>p0 = +41,96 k.l • mol 1
From Section 6.22, we find the following relationship 
&H,,U +&CPJ ( T 2 - T t)
= A + 3M7, 5,° + (CPJ ( I2,g) - CP,„°(12>I)) (T2 - T t)
=+41.96 kJ-mor'
+ (36.90 J • K~' • mol-1 -80.7 J -K '1 •mor')(298K-475.5K)
= +49.73 kJ-mol'1 
So, at 298K:
+62.44 kJ ■ mol'1 - AH(J + 49.73 kJ • mol-1 
AHtJ =+12.71 kJ-mo!''
7.97 First, we’ ll w7ork out how much heat was required to raise the temperature 
of the water sample: 9 = gCAT = (150g) (4.18J.°CV)(5.00*C) = 31353 
Since both the water and ice samples were at 0.00° C and the water sample 
took 0.5 h to get to 5.00° C, we can estimate that the ice took 10.0 h to 
melt. If 3135 J of heat were transferred in 0.5 h, then the amount of heat 
transferred in 10.0 h is,
SM-202
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There is a correlation between the bond dissociation energy and the free 
energy of formation of the atomic species, but the relationship is clearly 
not linear.
For the heavier three halogens, there is a trend to decreasing free energy of 
formation of the atoms as the element becomes heavier, but fluorine is 
anomalous. The F—F bond energy is lower than expected, owing to 
repulsions of the lone pairs of electrons on the adjacent F atoms because 
the F—F bond distance is so short.
10.109 (a) Using the thermodynamic data in Appendix 2A:
Br,(g) —» 2 Br(g)
AG° = 2(82.40 kJ • mol-1) - 3.11 kJ - mol*' =161.69 kJ • mol’ 1
For equilibrium constant calculations, this is reasonably good agreement 
with the value obtained from part (a), especially if one considers that A/ / 0 
will not be perfectly constant over so large a temperature range.
(b) We will use data from Appendix 2A to calculate the vapor pressure of 
bromine:
120
e
100
9
&
9
O E 60
oco
4)4* 20
0
0 10 20 30 40 50 60
atomic number
= 4.5x10'*
SM-305
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(b) £°(Cc4*/Ce^) = + 1.61 V (anode)
£°(Mn04 “/Mn2*) = +1.51 V (cathode)
Because £°(Cc‘u /Ce3+) > f°(Mn04 */Mn24), the reaction doesnot favor 
products.
(c) £°(Pb4+/Pb2+) = +1.67 V (anode)
E°( Pb2’ /Pb) = -0.13 V (cathode)
Because £°(Pb4+/Pb2+) > E°(Pb24 /Pb), the reaction docs not favor 
products.
(d) £°(NOj “/NO,/H+) = +0.80 V (cathode)
£°(Zn2* /Zn) = -0.76 V (anode)
Because £0(N03 “/NO, /H+) > E°(Zn2* /Zn), the reaction favors 
products.
£°cc|, = +0.80 V - (-0.76 V) = +1.56 V 
NOj “ is the oxidizing agent.
13.33 (a) We want E° for T]3+ <ac0 + 3 e“ -+ Tl(s), n - 3. 
This reaction is the reverse of the formation reaction: 
Tl(s) -» Tl3+(aq) + 3 e"
Therefore, for the T1" /T1 couple, Â = “215 kJ'mo> ’
AG°r =-ij/r£° = -215kJ-mol*1
£0_ aG°. - -2-15xlQ? J-moP1 _ |07>|3V 
- n F -3 x 9.65 x 104 C-moP1
(b) Using the potential from part (a) and the potential from Appendix 2B 
for the reduction of Tl* t we can decide whether or not Tl* will 
disproportionate in solution. The equation of interest is 
3 Tl+(aq) -» 2 Tl(s) + Tl^aq).
The half-reactions to combine arc:
SM-407
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Because we are not given v, it is easiest to make a relative comparison by 
taking the ratio of v for the C—D molecule versus v for the C—H 
molecule:
1 k
2* m - p
_L (Z *Z m --D
2*VlV-H
j n cmu 
mt + mu
" c mo
mc + mn
(12.011) (1.0078)
12.011 + 1.0078
(12.011) (2.0140)
12.011 + 2.0140
( 12.105N 
,13.019, 
v 24.190v 
,14.025,
0.73422
We would thus expect the vibrational frequency for the C—D bond to be 
approximately 0.73 times the value for the C—H bond (lower in energy).
17.57 To determine the effective half-life we need to determine the effective rate 
constant, kE. This constant is equal to the sum of the biological rate
constant (/:B) and the radioactive decay rate constant (£R), both of which
can be obtained from the respective half-lives:
+/tR + - ^ - = 1.56x 10 2 d'1
(effective) =
90.0 d 87.4 d 
In 2 
1.56x10 2 d-i
= 44.4 d
17.59 Remember to convert g to kg.
(a) E = mcz =(1.0x 10~J kg)(3.00x 10s m s'1)2
= 9 .0xl013 kg• m2 • s~2 = 9 .0x l0 ,J J
(b) £ = me1 =(9.109x10 -” kg)(3.00 x 10s m -s1)2
= 8.20 x 1 O'’4 kg • m2 • s'2 = 8.20 x 10 'u J
(c) E = me1
£ = (1.0x1 O'’5 kg) (3.00 x 10s m • s '1 )2 = 90. kg ■ m2 • s'2 = 90. J
\
SM-509
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