Atkins cap 9 resolução

Prévia do material em texto

CHAPTER 9 
CHEMICAL EQUILIBRIA 
 
 
 
 
9.1 (a) False, Equilibrium is dynamic. At equilibrium, the concentrations of 
reactants and products will not change, but the reaction will continue to 
proceed in both directions. 
(b) False. Equilibrium reactions are affected by the presence of both 
products and reactants. 
(c) False. The value of the equilibrium constant is not affected by the 
amounts of reactants or products added as long as the temperature is 
constant. 
(d) True. 
 
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 1 2 3 4
time
[PCl3] = [Cl2]
[PCl5]
9.3 
 
 
 
 
 
 
 
 
 
 
9.5 (a) C
2
[COCl][Cl] ;
[CO][Cl ]
K = (b) 
2
C
2 2
[HBr] ;
[H ][Br ]
K = (c) 
2 2
2 2
C 2 3
2 2
[SO ] [H O]
[H S] [O ]
K = 
SM-235 
9.7 (a) Because the volume is the same, the number of moles of is larger 
in the second experiment. (b) Because is a constant and the 
denominator is larger in the second case, the numerator must also be 
larger; so the concentration of is larger in the second case. 
(c) Although is the same, will be different, a result 
seen by solving for in each case. (d) Because is a constant, 
 is the same. (e) Because is the same, its 
reciprocal must be the same. 
2O
CK
2O
3
2 3[O ] /[O ]
2
2 2
2 3[O ]/[O ]
CK CK
3
2 3[O ] /[O ]
3
2 3[O ] /[O ]
 
9.9 
2
2 2
[HI]
[H ][I ]C
K = 
2
3 3
2
3 3
2
3 3
(0.0137)for condition 1, 48.8
(6.47 10 ) (0.594 10 )
(0.0169)for condition 2, 48.9
(3.84 10 ) (1.52 10 )
(0.0100)for condition 3, 48.9
(1.43 10 ) (1.43 10 )
C
C
C
K
K
K
− −
− −
− −
= =× ×
= =× ×
= =× ×
 
 
9.11 (a) 
2
3
O
1
P
 
(b) 
2
7
H OP
(c) 2
2 3
NO NO
N O
P P
P
 
 
9.13 To answer these questions, we will first calculate G∆ ° for each reaction 
and then use that value in the expression ln .G RT K∆ ° = − 
(a) 2 2 22 H (g) O (g) 2 H O(g)+ ⎯⎯→
SM-236 
1
r f 2
r
p
1 1
80
2( (H O, g) 2( 228.57 kJ mol ) 457.14 kJ
ln or
ln
457140 Jln 184.42
(8.314 J K mol ) (298.15 K)
1 10
G G
G RT K
G
K
RT
K
K
−
− −
∆ ° = ∆ ° = − ⋅ = −
∆ ° = −
∆ °= −
−= − = +⋅ ⋅
= ×
 
(b) 2 22 CO(g) O (g) 2 CO (g)+ ⎯⎯→
r f 2 f
1 1
1 1
2 (CO , g) [2 (CO, g)]
2( 394.36 kJ mol ) [2( 137.17 kJ mol )]
514.38 kJ
514 380 Jln 207.5
(8.314 J K mol ) (298.15 K)
G G G
K
− −
− −
∆ ° = × ∆ ° − × ∆ °
= − ⋅ − − ⋅
= −
−= − = +⋅ ⋅
 
901 10K = × 
 (c) 3 2CaCO (s) CaO(s) CO (g)⎯⎯→ +
 
r f f 2 f 3
1 1
1 1
(CaO, s) (CO , g) [ (CaCO (s)]
( 604.03 kJ mol ) ( 394.36 kJ mol ) [ 1128.8 kJ mol ]
130.4 kJ
130 400 Jln 52.6
(8.314 J K mol ) (298 K)
G G G G
K
1− − −
− −
∆ ° = ∆ ° + ∆ ° − ∆ °
= − ⋅ + − ⋅ − − ⋅
= +
+= − = −⋅ ⋅
 
 231 10K −= ×
 
9.15 r(a) lnG RT∆ ° = − K
11 1 (8.314 J K mol ) (1200 K) ln 6.8 19 kJ mol− − −= − ⋅ ⋅ = − ⋅ 
 (b) 1 1r
1
ln (8.314 J K mol ) (298 K) ln 1.1 10
68 kJ mol
G RT K 12− − −
−
∆ ° = − = − ⋅ ⋅ ×
= + ⋅
 
 
9.17 First we must calculate K for the reaction, which can be done using data 
from Appendix 2A: 
 1 1r f2 (NO, g) 2 86.55 kJ mol 173.1 kJ molG G
− −∆ ° = × ∆ ° = × ⋅ = ⋅ 
SM-237 
 
1
1 1
ln
ln
173100 J molln 69.9
(8.314 J K mol ) (298 K)
G RT K
GK
RT
K
−
− −
∆ ° = −
∆ °= −
+ ⋅= − = −⋅ ⋅
 
 314 10K −= ×
 Because Q > K, the reaction will tend to proceed to produce reactants. 
 
9.19 The free energy at a specific set of conditions is given by 
r r
r
2
r
2
1 1
2
1 1
-1 1
ln
ln ln
[I]ln ln
[I ]
(8.314 J K mol ) (1200 K) ln (6.8)
(0.98)(8.314 J K mol ) (1200 K) ln
(0.13)
8.3 x 10 kJ mol
G G RT Q
G RT K RT Q
G RT K RT
− −
− −
−
∆ = ∆ ° +
∆ = − +
∆ = − +
= − ⋅ ⋅
+ ⋅ ⋅
= ⋅
 
Because is positive, the reaction will be spontaneous to produce IrG∆ 2. 
 
9.21 The free energy at a specific set of conditions is given by 
r r
r
2
3
r 3
2 2
1 1
2
1 1
3
1
ln
ln ln
[NH ]ln ln
[N ][H ]
(8.314 J K mol ) (400 K) ln (41)
(21)(8.314 J K mol ) (400 K) ln
(4.2) (1.8)
27 kJ mol
G G RT Q
G RT K RT Q
G RT K RT
− −
− −
−
∆ = ∆ ° +
∆ = − +
∆ = − +
= − ⋅ ⋅
+ ⋅ ⋅
= − ⋅
 
Because is negative, the reaction will proceed to form products. rG∆
 
9.23 (a) 2
2
NO Cl 2
2
NOCl
1.8 10
P P
K
P
−= = × 
SM-238 
(3 2)
2 4
12.027 K
12.027 K 12.027 K 1.8 10 4.3 10
500 K
n
C
n
C
TK K
K K
T
∆
−∆
− −
⎛ ⎞= ⎜ ⎟⎝ ⎠
⎛ ⎞⎛ ⎞= = × =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
×
 
(b) 
2CO
167K P= =
(1)
12.027 K 12.027 K 167 1.87
1073 K
n
CK KT
∆ ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
= 
 
9.25 For the reaction written as 2 2 3N (g) 3 H (g) 2 NH (g) Eq.1+ → 
3
2 2
2
NH
3
N H
41
P
K
P P
= = 
(a) For the reaction written as 3 2 22 NH (g) N (g) 3 H (g) Eq. 2→ +
2 2
3
3
N H
2
NH
P P
K
P
= 
This is 
Eq. 1
1 1 0.024
41K
= = 
(b) For the reaction written as 31 2 2 32 2N (g) H (g) NH (g) Eq. 3+ → 
3
2 2
NH
Eq. 3 Eq.11/ 2 3/ 2
N H
41 6.4
P
K K
P P
= = = = 
Note that 1/ 21 Eq. 3 Eq.12Eq. 3 Eq.1and thus .K K= = 
(c) For the reaction written as 2 2 32 N (g) 6 H (g) 4 NH (g) Eq. 4+ → 
3
2 2
4
NH 2 2
Eq. 4 Eq.12 6
N H
41 1.7 10
P
K K
P P
= = = = × 3 
Note that 2Eq. 3 Eq.1Eq. 4 2 Eq.1and thus .K K= =
 
9.27 2 2H (g) I (g) 2 HI(g)+ → 160CK = 
2
2 2
[HI]
[H ][I ]C
K = 
SM-239 
3 2
3
2
(2.21 10 )160
[H ](1.46 10 )
−
−
×= × 
3 2
2 3
5 -
2
(2.21 10 )[H ]
(160) (1.46 10 )
[H ] 2.1 10 mol L
−
−
−
×= ×
= × ⋅ 1
 
 
9.29 3 2
5
PCl Cl
PCl
P P
K
P
= 
3
3
3
PCl
PCl
PCl
(5.43)
25
1.18
(25) (1.18) 5.4
5.43
5.4 bar
P
P
P
=
= =
=
 
 
9.31 (a) 
2 2
2
160HI
H I
pK
p p
= =⋅ 
2(0.10) 0.50
(0.20) (0.10)
Q = = 
(b) the system is not at equilibrium ,Q K≠ ∴
(c) Because more products will be formed. ,Q K<
 
9.33 (a) 
2
63
2
2 2
[SO ] 1.7 10
[SO ] [O ]C
K = = × 
24
23 4
1.0 10
0.500
6.9
1.20 10 5.0 10
0.500 0.500
CQ
−
− −
⎡ ⎤×⎢ ⎥⎣ ⎦= =⎡ ⎤ ⎡ ⎤× ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
 
(b) Because more products will tend to form, which will result 
in the formation of more 
,C CQ K<
3SO .
 
SM-240 
9.35 1
1.90 g HI 0.0148 mol HI
127.91 g mol−
=⋅ 
2 22 HI(g) H (g) I⎯⎯→ + 
0.0172 mol 2 x− x x 
2
3
2 2
0.0148 mol 0.0172 mol 2
0.0012 mol
0.0012 0.0012 0.0012
2.00 2.00 2.00 0.0066 or 6.6 10
0.0148 0.0148
2.00 2.00
C
x
x
K −
= −
=
⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦= = =⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
×
 
 
9.37 4 2 2 4 2 21
4 2 2
25.0 g NH (NH CO ) 0.320 mol NH (NH CO )
78.07 g mol NH (NH CO )−
=⋅ 
42
21
2
4
3 2 3
4
24 4
2 8
3 2
0.0174 g CO 3.95 10 mol CO
44.01g mol CO
2 mol NH are formed per mol of CO , so mol NH 2 3.95 10
7.90 10
7.90 10 3.95 10[NH ] [CO ] 1.58 10
0.250 0.250C
K
−
−
−
−
− −
−
= ×⋅
= × ×
= ×
⎛ ⎞ ⎛ ⎞× ×= = = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
 
 
9.39 (a) The balanced equation is: 2Cl (g) Cl(g)�
The initial concentration of 122
0.0020 mol ClCl (g) is 0.0010 mol L
2.0 L
−= ⋅ 
Concentration → 2 C 1(mol L )−⋅ 2Cl (g) l(g)
initial 0.0010 0 
change x− 2 x+ 
equilibrium 0.0010 x− 2 x+ 
SM-241 
2 2
7
C
2
2 7
2 7 10
7 7 2
7 5
6 6
[Cl] (2 ) 1.2 10
[Cl ] (0.0010 )
4 (1.2 10 ) (0.0010 )
4 (1.2 10 ) (1.2 10 ) 0
(1.2 10 ) (1.2 10 ) 4(4) ( 1.2 10 )
2.4
(1.2 10 ) 4.4 10
8
5.510 or 5.5 10
xK
x
x x
x x
x
x
x
−
−
− −
− −
− −
− −
= = = ×−
= × −
+ × − × =
− × ± × − − ×=
− × ± ×=
= − × + ×
10−
1−
1
 
The negative answer is not meaningful, so we choose 
 The concentration of is essentially 
unchanged because The concentration of 
atoms is 
65.5 10 mol L .x −= + × ⋅ 2Cl
60.0010 5.5 10 0.0010.−− × ≅
Cl 6 52 (5.5 10 ) 1.1 10 mol L .− −× × = × ⋅ − The percentage 
decomposition of is given by 2Cl
65.5 10 100 0.55%
0.0010
−× × = 
(b) The balanced equation is: 2F (g) 2 F(g)�
The problem is worked in an identical fashion to (a) but the equilibrium 
constant is now 41.2 10 .−× 
The initial concentration of 122
0.0020 mol FF (g) is 0.0010 mol L
2.0 L
−= ⋅ 
Concentration 1(mol L )−⋅ 2F (g) → 2 F(g)
initial 0.0010 0 
change x− 2 x+ 
equilibrium 0.0010 x− 2 x+ 
SM-242 
2 2
4
2
2 4
2 4 7
4 4 2
4 3
4 4
[F] (2 ) 1.2 10
[F ] (0.0010 )
4 (1.2 10 ) (0.0010 )
4 (1.2 10 ) (1.2 10 ) 0
(1.2 10 ) (1.2 10 ) 4(4) ( 1.2 10 )
2.4
(1.2 10 ) 1.4 10
8
1.9 10 or 1.6 10
C
xK
x
x x
x x
x
x
x
−
−
− −
− −
− −
− −
= = = ×−
= × −
+ × − × =
− × ± × − − ×=
− × ± ×=
= − × + ×
7−
1−
1−
1−
 
The negative answer is not meaningful, so we choose 
 The concentration of is 
 The concentration of F atoms is 
 The percentage decomposition of 
is given by 
41.6 10 mol L .x −= + × ⋅ 2F
4 40.0010 1.6 10 8 10 mol L .− −− × = × ⋅
4 42 (1.6 10 ) 3.2 10 mol L .− −× × = × ⋅ 2F 
41.6 10 100 16%
0.0010
−× × = 
(c) is more stable. This can be seen even without the aid of the 
calculation from the larger equilibrium constant for the dissociation for 
compared to 
2Cl
2F 
2Cl .
 
9.41 Concentration (bar) 2 HBr(g) → 2 2H (g) Br (g)+ 
initial 31.2 10−× 0 0 
change 2 x− x+ x+ 
final 31.2 10 2 x−× − x+ x+ 
2 2
2
H Br
HBr
p p
K
p
⋅= 
SM-243 
2 2
2
11
3 2 3 2
2
11
3 2
6
3
6 3
6 6 3
8
8
( ) ( )7.7 10
(1.2 10 2 ) (1.2 10 2 )
7.7 10
(1.2 10 2 )
8.8 10
(1.2 10 2 )
(8.8 10 ) (1.2 10 2 )
2(8.8 10 ) (8.8 10 ) (1.2 10 )
1.1 10
1.1 10 bar; the pressure ofH Br
x x x
x x
x
x
x
x
x x
x x
x
p p
−
− −
−
−
−
−
− −
− − −
−
−
× = =× − × −
× = × −
= ×× −
= × × −
+ × = × ×
≅ ×
= = ×
2 2
HBr is essentially unaffected
by the formation of Br and H .
 
The percentage decomposition is given by 
8
3
3
2 (1.1 10 bar) 100 1.8 10 %
1.2 10 bar
−
−
−
× × = ×× 
 
9.43 (a) Concentration of initially 5PCl
5
1
5 1
1.0 g PCl
208.22 g mol PCl
0.019 mol L
0.250 L
−
−
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ⋅ 
Concentration + 1(mol L )−⋅ 5PCl (g) → 3PCl (g) 2Cl
initial 0.019 0 0 
change x− x+ x+
final 0.019 x− x+ x+
2
2 2
5
[PCl ][Cl ] ( ) ( )
[PCl ] (0.019 ) (0.019 )C
x x xK
x x
= = =− − 
2
2
2 2
2 2
1.1 10
(0.019 )
(1.1 10 ) (0.019 )
(1.1 10 ) 2.1 10 0
−
−
− −
= ×−
= × −
+ × − × =
x
x
x x
x x 4
 
 
SM-244 
2 2 2
2
(1.1 10 ) (1.1 10 ) (4)(1)( 2.1 10 )
2.1
(1.1 10 ) 0.031 0.010 or 0.021
2.1
− −
−
− × ± × − − ×=
− × ±= = + −
x
x
4−
 
The negative root is not meaningful, so we choose 10.010 mol L .x −= ⋅ 
1 1
3 2 5[PCl ] [Cl ] 0.010 mol L ; [PCl ] 0.009 mol L .
− −= = ⋅ = ⋅ 
(b) The percentage decomposition is given by 
0.010 100 53%
0.019
× = 
 
9.45 Starting concentration of 13
0.400 molNH 0.200 mol L
2.00 L
−= = ⋅ 
Concentration 1(mol L )−⋅ → 4NH HS(s) 3 2NH (g) + H S(g)
initial — 0.200 0 
change — x+ x+
final — 0.200 x+ x+
3 2
4
2 4
2 4
[NH ][H S] (0.200 ) ( )
1.6 10 (0.200 ) ( )
0.200 1.6 10 0
( 0.200) ( 0.200) (4) (1) ( 1.6 10 )
2.1
0.200 0.2016 0.0008 or 0.2008
2.1
CK x x
x x
x x
x
x
−
−
−
= = +
× = +
+ − × =
− + ± + − − ×=
− ±= = + −
 
The negative root is not meaningful, so we choose 4 18 10 mol Lx − −= × ⋅ 
(note that in order to get this number we have had to ignore our normal 
significant figure conventions). 
1 4 1
3[NH ] 0.200 mol L 8 10 mol L 0.200 mol L
1− − −= + ⋅ + × ⋅ = ⋅ − 
4 1
2[H S] 8 10 mol L
− −= × ⋅ 
Alternatively, we could have assumed that 
4 40.2, the 0.200 1.6 10 , 8.0 10 .x x x− −<< = × = × 
SM-245 
 
9.47 The initial concentrations of and are calculated as follows: 5PCl 3PCl
1 1
5 3
0.200 mol 0.600 mol[PCl ] 0.0500 mol L ; [PCl ] 0.150 mol L
4.00 L 4.00 L
− −= = ⋅ = = ⋅ 
Concentrations 1(mol L )−⋅ 5PCl → 3 2PCl (g) Cl (g)+ 
initial 0.0500 0.150 0 
change x− x+ x+
final 0.0500 – x 0.150 + x x+
3 2
5
2
2
2
[PCl ][Cl ] (0.150 ) ( ) 33.3
[PCl ] (0.0500 )
0.150 (33.3) (0.0500 )
33.45 1.665 0
33.45 (33.45) (4) (1) ( 1.665) 33.4 33.6 0.0497
(2) (1) 2
C
x xK
x
x x x
x x
x
+= = =−
+ = −
+ − =
− ± − − − ±= = =
1
1
 
The negative root has no physical meaning and so it can be discarded. 
1 1 4
5
1 1
3
1
2
[PCl ] 0.0500 mol L 0.497 mol L 3 10 mol L
[PCl ] 0.150 mol L 0.0497 mol L 0.200 mol L
[Cl ] 0.0497 mol L
− − −
− −
−
= ⋅ − ⋅ = ×
= ⋅ + ⋅ = ⋅
= ⋅
−
−
⋅
 
 
9.49 The initial concentrations of and are equal at 0.114 
because the vessel has a volume of 1.00 L. 
2N 2O
1mol L−⋅
Concentrations 1(mol L )−⋅ 2N (g) + → 2 2O (g) NO(g)
initial 0.114 0.114 0 
change x− x− 2 x+ 
final 0.114 – x 0.114 – x 2 x+ 
SM-246 
2 2
2
2 2
2
5
2
2
5
2
[NO] (2 ) (2 )
[N ][O ] (0.114 ) (0.114 ) (0.114 )
(2 )1.00 10
(0.114 )
(2 )1.00 10
(0.114 )
C
x xK
2
x x x
x
x
x
x
−
−
= = =− − −
× = −
× = −
 
3
3
4
4
4
(2 )3.16 10
(0.114 )
2 (3.16 10 ) (0.114 )
2.00316 3.60 10
1.8 10
[NO] 2 2 1.8 10 3.6 10 ;
x
x
x x
x
x
x
−
−
−
−
4− −
× = −
= × −
= ×
= ×
= = × × = × the concentrations of and 
remain essentially unchanged at 0.114 
2N 2O 
1mol L .−⋅ 
 
9.51 The initial concentrations of and are: 2H 2I
1 1
2 2
0.400 mol 1.60 mol[H ] 0.133 mol L ; [I ] 0.533 mol L
3.00 L 3.00 L
− −= = ⋅ = = ⋅ 
Concentrations 1(mol L )−⋅ 2H (g) + Æ 2I (g) 2 HI(g)
initial 0.133 0.533 0 
change x− x− 2 x+ 
final 0.133 x− 0.533 x− 2 x+ 
At equilibrium, 60.0% of the had reacted, so 40.0% of the 
remains: 
2H 2H 
1 1
1 1
1
(0.400)(0.133 mol L ) 0.133 mol L
0.133 mol L (0.400) (0.133 mol L )
0.080 mol L
x
x
x
− −
− −
−
⋅ = ⋅ −
= ⋅ − ⋅
= ⋅
 
at equilibrium: [ 1 12H ] 0.133 mol L 0.080 mol L 0.053 mol L
1− − −= ⋅ − ⋅ = ⋅ 
SM-247 
1 1
2
1 1
2 2
2 2
[I ] 0.533 mol L 0.080 mol L 0.453 mol L
[HI] 2 0.080 mol L 0.16 mol L
[HI] 0.16 1.1
[H ][I ] (0.053) (0.453)C
K
− −
− −
= ⋅ − ⋅ = ⋅
= × ⋅ = ⋅
= = =
1−
 
 
9.53 Initial concentrations of CO and are given by 2O
1
3 1
2
1
2 4 1
2
0.28 g CO
28.01g mol CO
[CO] 5.0 10 mol L
2.0 L
0.032 g O
32.00 g mol O
[O ] 5.0 10 mol L
2.0 L
−
− −
−
− −
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ×
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ×
⋅
⋅
 
Concentration 1(mol L )−⋅ 2 CO(g) + → 2O (g) 22 CO (g)
initial 35.0 10−× 45.0 10−× 0 
change 2 x− x− 2 x+ 
final 35.0 10 2 x−× − 45.0 10 x−× − 2 x+ 
2
2
2
2
2
3 2 4
2
2 5
[CO ]
[CO] [O ]
(2 )
(5.0 10 2 ) (5.0 10 )
4
(4 0.020 2.5 10 ) (5.0 10 )
CK
x
x x
x
4x x x
− −
− −
=
= × − × −
= − + × × −
 
2
3 2 5 8
2 3 2 5
2 3 2 5 8
3 2 5 8
5
5 1
2
40.66
4 0.022 3.5 10 1.25 10
4 (0.66) ( 4 0.022 3.5 10 1.25 10 )
6.064 0.022 3.5 10 1.25 10
0 4 6.04 3.5 10 1.25 10
4.3 10
[CO ] 8.6 10 mol L ; [CO] 4.9 10
x
x x x
x x x x
x x x x
x x x
x
− −
− −
− −
− −
−
− − −
= − + − × + ×
= − + − × + ×
= − + − × + ×
= − − − × + ×
= ×
= × ⋅ = × 3 1
4 1
2
mol L
[O ] 4.6 10 mol L
−
− −
⋅
= × ⋅
8
 
 
SM-248 
 
9.55 Concentrations 
 + + 1(mol L )−⋅ 3CH COOH 2 5 3 2 5C H OH CH COOC H→ 2H O
 initial 0.32 6.30 0 0 
 change x− x− x+ x+
 final 0.32 x− 6.30 x− x+ x+
2
3 2 5 2
2
3 2 5
2
2
2 2
2
2
[CH COOC H ][H O] ( ) ( )
[CH COOH][C H OH] (0.32 ) (6.30 ) 6.62 2.02
4.0
6.62 2.02
4.0 26.48 8.08
3.0 26.48 8.08 0
( 26.48) ( 26.48) (4) (3.0) (8.08) 26.48 24.58
(2) (3.0) 6.0
8.51
C
x x xK
x x x x
x
x x
x x x
x x
x
x
= = =− − − +
= − +
− + =
− + =
− − ± − − + ±= =
= or 0.317
 
The root 8.51 is meaningless because it is larger than the concentration of 
acetic acid and ethanol, so the value 0.317 is chosen. The equilibrium 
concentration of the product ester is, therefore, 0.317 1mol L−⋅ . The 
numbers can be confirmed by placing them into the equilibrium 
expression: 
3 2 5 2
3 2 5
[CH COOC H ][H O] (0.317) (0.317) 5.6
[CH COOH][C H OH] (0.320 0.317) (6.300 0.317)C
K = = − − = 
Note: This number does not appear to agree well with the given value of 
 = 4.0. CK
If 0.316 is used, the agreement is better, giving a quotient of 4.1. If 0.315 
is used, the quotient is 3.3. The better answer is thus 0.316 The 
discrepancy is caused by rounding errors in places that are really beyond 
the accuracy of the measurement. Given that is only given to two 
significant figures, the best report of the concentration of ester would be 
1mol L .−⋅
CK
SM-249 
0.32 , even though this value will not satisfy the equilibrium 
expression as well as 0.316 
1mol L−⋅
1mol L .−⋅ 
 
9.57 
2
2 2
[BrCl]
[Cl ][Br ]C
K = 
2
2
2
1
2
(0.145)0.031
(0.495)[Br ]
(0.145)[Br ] 1.4 mol L
(0.495) (0.031)
−
=
= = ⋅
 
 
9.59 We can calculate changes according to the reaction stoichiometry: 
Amount (mol) CO(g) + 23 H (g) → 4CH (g) + 2H O(g)
initial 2.00 3.00 0 0 
change x− 3 x− x+ x+
final 2.00 x− 3.00 3 x− 0.478 x+ 
According to the stoichiometry, 0.478 mol = x; therefore, at equilibrium, 
there are 2.00 mol – 0.478 mol = 1.52 mol CO, 3.00 – 3(0.478 mol) = 1.57 
mol and 0.478 mol To employ the equilibrium expression, we 
need either concentrations or pressures; because is given, we will 
choose to express these as concentrations. This calculation is easy because 
V = 10.0 L: 
2H , 2H O.
CK
1 1
2 4
1
2
4 2
3 3
2
[CO] 0.152 mol L ; [H ] 0.157 mol L ; [CH ] 0.0478 mol L ;
[H O] 0.0478 mol L
[CH ][H O] (0.0478) (0.0478) 3.88
[CO][H ] (0.152) (0.157)C
K
− −
−
= ⋅ = ⋅ =
= ⋅
= = =
1−⋅
 
 
9.61 First, we calculate the initial concentrations of each species: 
1
2
1 1
2 3
0.100 mol[SO ] [NO] 0.0200 mol L ;
5.00 L
0.200 mol 0.150 mol[NO ] 0.0400 mol L ; [SO ] 0.0300 mol L
5.00 L 5.00 L
−
− −
= = = ⋅
= = ⋅ = = ⋅
 
SM-250 
We can use these values to calculate Q in order to see which direction the 
reactions will go: 
(0.0200) (0.0300) 0.75.
(0.0200) (0.0400)
Q = = Because Q < the reaction will proceed 
to produce more products. 
,CK
Concentration 1(mol L )−⋅ 
 + NO(g) + 2SO (g) 2NO (g) → 3SO (g)
initial 0.0200 0.0400 0.0200 0.0300 
change x− x− x+ x+
final 0.0200 x− 0.0400 x− 0.0200 x+ 0.0300 x+ 
3
2 2
2
2
[NO][SO ]
[SO ][NO ]
(0.0200 ) (0.0300 )85.0
(0.0200 ) (0.0400 )
0.0500 0.000 600
0.0600 0.000 800
CK
x x
x x
x x
x x
=
+ += − −
+ += − +
 
2 2
2 2
2
2
85.0( 0.0600 0.000 800) 0.0500 0.000 600
85.0 5.10 0.0680 0.0500 0.000 600
84.0 5.15 0.0674 0
( 5.15) ( 5.15) (4) (84.0) (0.0674) 5.15 1.97
(2) (84.0) 168
x x x x
x x x x
x x
x
− + = + +
− + = + +
− + =
− − ± − − + ±= =
 
0.0424x = + or 0.0189+
The root 0.0424 is not meaningful because it is larger than the 
concentration of . The root of choice is therefore 0.0189. 2NO
At equilibrium: 
1 1
2
1 1
2
1 1
1 1
3
[SO ] 0.0200 mol L 0.0189 mol L 0.0011 mol L
[NO ] 0.0400 mol L 0.0189 mol L 0.0211 mol L
[NO] 0.0200 mol L 0.0189 mol L 0.0389 mol L
[SO ] 0.0300 mol L 0.0189 mol L 0.0489 mol L
1
1
1
1
− − −
− − −
− − −
− − −
= ⋅ − ⋅ = ⋅
⋅= ⋅ − ⋅ =
= ⋅ + ⋅ = ⋅
= ⋅ + ⋅ = ⋅
 
To check, we can put these numbers back into the equilibrium constant 
expression: 
SM-251 
3
2 2
[NO][SO ]
[SO ][NO ]
(0.0389) (0.0489) 82.0
(0.0011) (0.0211)
CK =
=
 
Compared to = 85.0, this is reasonably good agreement given the 
nature of the calculation. We can check to see, by trial and error, if a better 
answer could be obtained. Because the value is low for the 
concentrations we calculated, we can choose to alter x slightly so that this 
ratio becomes larger. If we let x = 0.0190, the concentrations of NO and 
 are increased to 0.0390 and 0.0490, and the concentrations of 
and are decreased to 0.0010 and 0.0200 (the stoichiometry of the 
reaction is maintained by calculating the concentrations in this fashion). 
Then the quotient becomes 91.0, which is further from the value for 
than the original answer. So, although the agreement is not the best with 
the numbers we obtained, it is the best possible, given the limitation on the 
number of significant figures we are allowed to use in the calculation. 
CK
CK
3SO 2SO
2NO
CK
 
9.63 (a) The initial concentrations are: 
1 1
5 3
1
2
1.50 mol 3.00 mol[PCl ] 3.00 mol L ; [PCl ] 6.00 mol L ;
0.500 L 0.500 L
0.500 mol[Cl ] 1.00 mol L
0.500 L
− −
−
= = ⋅ = = ⋅
= = ⋅
 
First calculate : Q
5
3 2
[PCl ] 3.00 0.500
[PCl ][Cl ] (6.00) (1.00)
Q = = = 
Because the reaction is not at equilibrium. ,Q K≠
(b) Because Q < the reaction will proceed to form products. ,CK
(c) 
Concentrations 1(mol L )−⋅ + 3PCl (g) 2Cl (g) → 5PCl (g)
initial 6.00 1.00 3.00 
SM-252 
change x− x− x+
final 6.00 x− 1.00 x− 3.00 x+ 
5
3 2
2
2
2
2
2
[PCl ]
[PCl ][Cl ]
3.00 3.000.56
(6.00 ) (1.00 ) 7 6.00
(0.56) ( 7 6.00) 3.00
0.56 3.92 3.36 3.00
0.56 4.92 0.36 0
( 4.92) ( 4.92) (4) (0.56) (0.36) 4.92 4.48
(2) (0.56) 1.12
CK
x x
x x x x
x x x
x x x
x x
x
=
+ += =− − − +
− + = +
− + = +
− + =
− − ± − − + ±= =
 
x = 9.2 or 0.07 
Because the root 9.2 is larger than the amount of or available, it 
is physically meaningless and can be discarded. Thus, x = 0.071 
giving 
3PCl 2Cl
1mol L ,−⋅
1 1
5
1 1
3
1 1
2
[PCl ] 3.00 mol L 0.07 mol L 3.07 mol L
[PCl ] 6.00 mol L 0.07 mol L 5.93 mol L
[Cl ] 1.00 mol L 0.07 mol L 0.93 mol L
− −
− −
− −
= ⋅ + ⋅ =
= ⋅ − ⋅ =
= ⋅ − ⋅ = ⋅
1
1
1
−
−
−
⋅
⋅ 
The number can be checked by substituting them back into the equilibrium 
constant expression: 
5
3 2
?
[PCl ]
[PCl ][Cl ]
(3.07) 0.56
(5.93) (0.93)
0.56 0.56
CK =
=
=�
 
 
9.65 Pressures (bar) 2 HCl(g) + → 2H (g) 2Cl (g)
 initial 0.22 0 0 
 change 2 x− x+ x+ 
 final 0.22 2 x− x+ x+ 
SM-253 
2 2H Cl
2
HCl
34
2
( ) ( )3.2 10
(0.22 2 )
P P
K
P
x x
x
−
=
× = −
 
Because the equilibrium constant is small, assume that x << 0.22 
2
34
2
2 34
34 2
18
3.2 10
(0.22)
(3.2 10 ) (0.22)
(3.2 10 ) (0.22)
3.9 10
x
x
x
x
−
−
−
−
× =
= ×
= ×
= ± ×
2 
The negative root is notphysically meaningful and can be discarded. x is 
small compared to 0.22, so the initial assumption was valid. The pressures 
at equilibrium are 
2 2
18
HCl H Cl0.22 bar; =3.9 10 barP P P
−= = × 
The values can be checked by substituting them into the equilibrium 
expression: 
2 2
18 18
34?
2
34 34
18
HCl H Cl
(3.9 10 ) (3.9 10 ) 3.2 10
(0.22)
3.1 10 3.2 10
0.22 bar; 3.9 10 barP P P
− −
−
− −√
−
× × = ×
× = ×
= = = ×
 
The numbers agree very well for a calculation of this type. 
 
9.67 (a) To determine on which side of the equilibrium position the conditions 
 lie, we will calculate Q : 
 
1 1
2
1
3
33
2 2
2
0.342 mol 0.215 mol[CO] 0.114 mol L ; [H ] 0.0717 mol L ;
3.00 L 3.00 L
0.125 mol[CH OH] 0.0417 mol L
3.00 L
[CH OH] 0.0417 71.1 10
[CO][H ] (0.114) (0.0717)
Q
− −
−
= = ⋅ = =
= = ⋅
= = = ×
⋅
 
SM-254 
 Because Q > the reaction will proceed to produce more of the 
 reactants, which means that the concentration of methanol will decrease. 
,CK
 (b) 
 Concentrations 1(mol L )−⋅ CO(g) + 22 H (g) → 3CH OH(g)
 initial 0.114 0.0717 0.0417 
 change x+ 2 x+ x− 
 final 0.0114 x+ 0.0717 2 x+ 0.0417 x− 
 22
0.0417 1.1 10
(0.0114 ) (0.0717 2 )C
xK
x x
−−= =+ + ×
3−
−
 
 
2 2
3 2 2
2 3 3 2 4 6
2 3 3 2
0.0417 (1.1 10 ) (0.114 ) (0.0717 2 )
(1.2 10 1.1 10 ) (4 0.287 5.14 10 )
4.4 10 8.0 10 4.0 10 6.2 10
0 4.4 10 8.0 10 1.00 0.0417
x x x
x x x
x x x
x x x
−
− −
− − −
− −
− = × + +
= × + × + + ×
= × + × + × + ×
= × + × + −
 This equation can be solved approximately, simply by inspection: it is 
 clear that the x term will be very much larger than the 3x and the 2x terms, 
 because their coefficients are very small compared to 1.00. This leads to a 
 prediction that 10.0417 mol Lx −= ⋅
1mol L ;
 to within the accuracy of the data. 
 Essentially all of the will react, so that 
 0.156
3CH OH
1 1[CO] 0.114 mol L 0.0417 mol L− −= ⋅ + ⋅ = −⋅ 
 1 12[H ] 0.0717 mol L 2 (0.0417 mol L ) 0.155 mol L .
1− − −= ⋅ + ⋅ = ⋅ The 
 mathematical situation is odd in that clearly a 3[CH OH] 0= will not 
 satisfy the equilibrium constant. Knowing that the methanol concentration 
 is very small compared to the CO and concentrations, we can now 
 back-calculate to get a concentration value that will satisfy the equilibrium 
 expression: 
2H
 22 1.1 10(0.156) (0.155)C
yK −= = ×
4
 
 2 2(1.1 10 ) (0.156) (0.155) 2.6 10y − −= × = ×
SM-255 
Alternatively, the cubic equation can be solved with the aid of a graphing 
calculator like the one supplied on the CD accompanying this book. 
 
9.69 Since reactants are strongly favored it is easier to push the reaction as far 
to the left as possible then start from new initial conditions. 
 Pressures (bar) 2 HCl(g) + → 2H (g) 2Cl (g)
 original 2.0 1.0 3.0 
 new initial 4.0 0 2.0 
 change 2 x− x+ x+ 
 final 4.0 2 x− x+ 2.0 x+ 
 
2 2H Cl
2
HCl
34
2 2
33
-2 -1 -1
33
33
( ) (2.0 ) ( ) (2.0)3.2 10
(4.0 2 ) (4.0) 16
2.6 10 bar
nRT (1 mole)(8.314 10 L bar K mol )(298K)V= =
p 2.6 10 bar
9.7 10 L
P P
K
P
x x x x
x
x
−
−
−
=
+× = ≈ =−
= ×
×
×
= ×
 
 
9.71 (a) According to Le Chatelier’s principle, an increase in the partial 
pressure of will result in creation of reactants, which will decrease 
the partial pressure. 
2CO
2H
 (b) According to Le Chatelier’s principle, if the CO pressure is reduced, 
the reaction will shift to form more CO, which will decrease the pressure 
of 2CO .
 (c) According to Le Chatelier’s principle, if the concentration of CO is 
 increased, the reaction will proceed to form more products, which will 
 result in a higher pressure of 2H .
 (d) The equilibrium constant for the reaction is unchanged, because it is 
 unaffected by any change in concentration. 
SM-256 
9.73 (a) According to Le Chatelier’s principle, increasing the concentration of 
NO will cause the reaction to form reactants in order to reduce the 
concentration of NO; the amount of water will decrease. 
 (b) For the same reason as in (a), the amount of will increase. 2O
 (c) According to Le Chatelier’s principle, removing water will cause the 
 reaction to shift toward products, resulting in the formation of more NO. 
 (d) According to Le Chatelier’s principle, removing a reactant will cause 
 the reaction to shift in the direction to replace the removed substance; the 
 amount of should increase. 3NH
 (e) According to Le Chatelier’s principle, adding ammonia will shift the 
 reaction to the right, but the equilibrium constant, which is a constant, will 
 not be affected. 
 (f) According to Le Chatelier’s principle, removing NO will cause the 
formation of more products; the amount of will decrease. 3NH
 (g) According to Le Chatelier’s principle, adding reactants will promote 
the formation of products; the amount of oxygen will decrease. 
 
9.75 As per Le Chatelier’s principle, whether increasing the pressure on a 
 reaction will affect the distribution of species within an equilibrium 
 mixture of gases depends largely upon the difference in the number of 
 moles of gases between the reactant and product sides of the equation. If 
 there is a net increase in the amount of gas, then applying pressure will 
 shift the reaction toward reactants in order to remove the stress applied by 
 increasing the pressure. Similarly, if there is a net decrease in the amount 
 of gas, applying pressure will cause the formation of products. If the 
 number of moles of gas is the same on the product and reactant side, then 
 changing the pressure will have little or no effect on the equilibrium 
 distribution of species present. Using this information, we can apply it to 
 the specific reactions given. The answers are: (a) reactants; 
 (b) reactants; (c) reactants; (d) no change (there is the same number 
 of moles of gas on both sides of the equation); (e) reactants 
SM-257 
9.77 (a) If the pressure of NO (a product) is increased, the reaction will shift to 
 form more reactants; the pressure of should increase. 3NH
 (b) If the pressure of (a reactant) is decreased, then the reaction will 
 shift to form more reactants; the pressure of should increase. 
3NH
2O
 
9.79 If a reaction is exothermic, raising the temperature will tend to shift the 
reaction toward reactants, whereas if the reaction is endothermic, a shift 
toward products will be observed. For the specific examples given, (a) and 
(b) are endothermic (the values for (b) can be calculated, but we know that 
it requires energy to break an X—X bond, so those processes will all be 
endothermic) and raising the temperature should favor the formation of 
products; (c) and (d) are exothermic and raising the temperature should 
favor the formation of reactants. 
 
9.81 Even though numbers are given, we do not need to do a calculation to 
answer this qualitative question. Because the equilibrium constant for the 
formation of ammonia is smaller at the higher temperature, raising the 
temperature will favor the formation of reactants. Less ammonia will be 
present at higher temperature, assuming no other changes occur to the 
system (i.e., the volume does not change, no reactants or products are 
added or removed from the container, etc.). 
 
9.83 To answer this question we must calculate Q: 
 
2 2
3
3 3
2 2
[NH ] (0.500) 0.0104
[N ][H ] (3.00) (2.00)
Q = = = 
 Because the system is not at equilibrium and because the 
reaction will proceed to produce more products. 
,Q K≠ ,Q K<9.85 Because we want the equilibrium constant at two temperatures, we will 
 need to calculate and rH∆ ° rS∆ ° for each reaction: 
SM-258 
 (a) 4 3NH Cl NH (g) HCl(g)→ +
 r f 3 f f 4(NH , g) (HCl, g) (NH Cl, s)H H H H∆ ° = ∆ ° + ∆ ° − ∆ °
 1 1r ( 46.11 kJ mol ) ( 92.31 kJ mol ) ( 314.43 kJ mol )H
1− − −∆ ° = − ⋅ + − ⋅ − − ⋅ 
 1r 176.01 kJ molH
−∆ ° = ⋅ 
 r 3 4(NH , g) (HCl, g) (NH Cl, s)S S S S∆ ° = ° + ° − °
 1 1 1 1 1 1r 192.45 J K mol 186.91 J K mol 94.6 J K molS
− − − − − −∆ ° = ⋅ ⋅ + ⋅ ⋅ − ⋅ ⋅
 1 1r 284.8 J K molS
− −∆ ° = ⋅ ⋅
 r rG H T S∆ ° = ∆ ° − ∆ °r
 At 298 K: 
 
1 1
r(298 K)
1
r(298 K)
176.01 kJ (298 K) (284.8 J K )/(1000 J kJ )
91.14 kJ mol
ln
G
G RT K
− −
−
∆ ° = − ⋅ ⋅
= ⋅
∆ ° = −
 
 
r(298 K)
1
ln
91140 J 36.8
(8.314 J K ) (298 K)
G
K
RT
−
∆ °= −
= − = −⋅
 
 
161 10
At 423 K:
K −= ×
 
1 1
r(423 K)
1
r(423 K)
176.01 kJ (423 K) (284.8 J K )/(1000 J kJ )
55.54 kJ mol
ln
G
G RT K
− −
−
∆ ° = − ⋅ ⋅
= ⋅
∆ ° = −
 
r(423 K)
1
ln
55 540 J 15.8
(8.314 J K )(423 K)
G
K
RT
−
∆ °= −
= − = −⋅
 
 71 10K −= ×
 (b) 2 2 2 2H (g) D O(l) D (g) H O(l)+ → +
SM-259 
 
r f 2 f 2
1 1
r
1
r
r 2 2 2 2
1 1 1 1
r
1 1 1
(H O, l) [ (D O, l)]
( 285.83 kJ mol ) [ 294.60 kJ mol ]
8.77 kJ mol
(D , g) (H O, l) [ (H , g) (D O, l)]
144.96 J K mol 69.91 J K mol
[130.68 J K mol 75.94 J K
H H H
H
H
S S S S S
S
− −
−
− − − −
− − −
∆ ° = ∆ ° − ∆ °
∆ ° = − ⋅ − − ⋅
∆ ° = ⋅
∆ ° = ° + ° − ° + °
∆ ° = ⋅ ⋅ + ⋅ ⋅
− ⋅ ⋅ + ⋅ 1
1 1
r
mol ]
8.25 J K molS
−
− −
⋅
∆ ° = ⋅ ⋅
 At 298 K: 
 
1 1 1
r(298 K)
1
8.77 kJ mol (298 K) (8.25 J K mol )/(1000 J kJ )
6.31 kJ mol
G − − −
−
∆ ° = ⋅ − ⋅ ⋅ ⋅
= ⋅
1−
K r(298 K) lnG RT∆ ° = −
 r(298 K)ln
G
K
RT
∆ °= − 
 1
6310 J 2.55
(8.314 J K ) (298 K)−
= − = −⋅ 
 
27.8 10
At 423 K:
K −= ×
 
1 1 1
r(423 K)
1
8.77 kJ mol (423 K) (8.25 J K mol )/(1000 J kJ )
5.28 kJ mol
G − − −
−
∆ ° = ⋅ − ⋅ ⋅ ⋅
= ⋅
1−
 1
5280 Jln 1.50
(8.314 J K )(423 K)
0.22
K
K
−= − = −⋅
=
 
 
9.87 ( ) n cK RT K∆= 
 2
1 1
1 1ln
° ⎛ ⎞∆= − −⎜ ⎟⎝ ⎠
rK H
K R T T2
 
 
( )
( )
2 2
1 21 1
1 1ln
∆ °
∆
⎛ ⎞ ⎛ ⎞∆⎜ ⎟ = − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
n
c r
n
c
RT K H
R T TRT K
 
SM-260 
 
22
1 1 1
2 2
1 1 2
1 1ln ln
1 1ln ln
°
°
⎛ ⎞⎛ ⎞ ⎛∆∆ + = − −⎜ ⎟⎜ ⎟ ⎜⎝ ⎠ ⎝⎝ ⎠
⎛ ⎞ ⎛ ⎞ ⎛∆= − − − ∆⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝⎝ ⎠
c r
c
c r
c
KT Hn
T K R T
K H T
n
K R T T
2
1
⎞⎟⎠
⎞⎟⎠
T
T
 
 
9.89 
 
ln∆ ° = − = ∆ ° − ∆ °G RT K H T S
-1 -1 15
298
4 -1
-1 -1 14
303
4 -1
298 303
(8.3145JK mol )(298.15K) ln(9.9197 10 )
 7.9933 10 Jmol
(8.3145JK mol )(303.15K) ln(1.4689 10 )
 8.0283 10 Jmol
(298.15K) [ (303.15K
−
−
∆ ° = − ×
= ×
∆ ° = − ×
= ×
∆ ° − ∆ ° = ∆ ° − ∆ ° − ∆ ° −
G
G
G G H S H
2 -1
-1 -1
) ]
3.5097 10 Jmol (5.00K)
70.194JK mol
∆ °
− × = ∆ °
∆ ° = −
S
S
S
 The sign is negative indicating that the product ions create more order in 
 the system probably because of their interaction with solvating water 
 molecules. 
 
9.91 (a) According to Le Chatelier’s principle, adding a product should cause a 
 shift in the equilibrium toward the reactants side of the equation. 
 (b) Because there are equal numbers of moles of gas on both sides of the 
 equation, there will be little or no effect upon compressing the system. 
 (c) If the amount of is increased, this will cause the reaction to shift 
 toward the formation of products. 
2CO
 (d) Because the reaction is endothermic, raising the temperature will favor 
 the formation of products. 
 (e) If the amount of is removed, this will cause the reaction to 
 shift toward the formation of products. 
6 12 6C H O
 (f) Because water is a liquid, it is by definition present at unit 
 concentration, so changing the amount of water will not affect the 
 reaction. As long as the glucose solution is dilute, its concentration can be 
 considered unchanged. 
SM-261 
 (g) Decreasing the concentration of a reactant will favor the production of 
 more reactants. 
 
9.93 (a) In order to solve this problem, we will manipulate the equations with 
 the known so that we can combine them to give the desired overall 
 reaction: 
’sK
 First, reverse equation (1) and multiply it by 12 : 
 12 2 22H (g) O (g) H O(g)+ → 4 11 1/ 21(1.6 10 )K −= × (4) 
 Multiply equation (2) by 12 also: 
 12 22CO (g) CO(g) O (g)→ + (5) 1/ 2105 (1.3 10 )K −= ×
 Adding equations (4) and (5) gives the desired reaction. The resultant 
 equilibrium constant will be the product of the K’s for (4) and (5): 
 2 2 2CO (g) H (g) H O(g) CO(g)+ → +
1/ 210
5 11
1.3 10 2.9
1.6 10
K
−
−
⎛ ⎞×= ⎜ ⎟×⎝ ⎠
= (3) 
 (b) To obtain the K value for a net equation from two (or more) others, 
 the K’s are multiplied, but r’sG∆ ° are added: 
 (1) 2 22 H O(g) 2 H (g) O (g)+� 2
11−
2
10
 
r
1 1
2 1
ln
(8.314 J K mol ) (1565 K) ln 1.6 10
3.2 10 kJ mol
G RT K
− −
−
∆ ° = −
= − ⋅ ⋅ ×
= + × ⋅
 (2) 22 CO (g) 2 CO(g) O (g)→ +
 
r
1 1
2 1
ln
(8.314 J K mol ) (1565 K) ln 1.3 10
3.0 10 kJ mol
G RT K
− − −
−
∆ ° = −
= − ⋅ ⋅ ×
= + × ⋅
 The corresponding values for (4) and (5) are 
 
2 21
r(4) 2
2 21
r(5) 2
(3.2 10 kJ) 1.6 10 kJ
(3.0 10 kJ) 1.5 10 kJ
G
G
∆ ° = − × = − ×
∆ ° = × = × 
 Summing these two values will give 
SM-262 
 2 2r(3) 1.6 10 kJ 1.5 10 kJ 10 kJ molG
1−∆ ° = − × + × = − ⋅ 
 
1
r
1 1
10 000 J molln 0.8
(8.314 J K mol ) (1565 K)
GK
RT
−
− −
∆ ° − ⋅= − = − = +⋅ ⋅ 
 2K = 
 This value is in reasonable agreement with the one obtained in (a), given 
 the problems in significant figures and due to rounding errors. 
 
9.95 Pressure Æ + 5PCl (g) 3PCl (g) 2Cl (g)
 initial n
 change nα− nα+ nα+ 
 final (1 )n α− nα+ nα+ 
 
2 2 2
2 2
2
( ) ( )
(1 ) (1 ) (1 )
(1 ) (1 )
1
(1 ) 1 1 1
α α α α
α α α
α α α α
α
2α α α
α α α α
= = =− − −
= − + + = +
= +
⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟− + − −⎝ ⎠ ⎝ ⎠
n n n nK
n n
P n n n n
Pn
n P PK
 
 
2 2
2 2
2 2
2
(1 )α α
α α
α α
α
α
− =
− =
= +
= +
= +
K P
K K P
K P K
K
P K
K
P K
 
 (a) For the specific conditions 4.96K = and 0.50P = bar, 
 4.96 0.953
0.50 4.96
α = =+ 
 (b) For the specific conditions 4.96K = and 1.00P = bar, 
 4.96 0.912
1.00 4.96
α = =+ 
SM-263 
 
9.97 (a) If K = 1.00, then G∆ °must be equal to 0 ( ln ).G RT K∆ ° = − 
 (b) This can be calculated by determining the values for H∆ ° and 
 at 25 C.S∆ ° °
 
 
1 1
1
393.51 kJ mol [( 110.53 kJ mol ) ( 241.82 kJ mol )]
41.16 kJ mol
H − −
−
∆ ° = − ⋅ − − ⋅ + − ⋅
= − ⋅
1−
1 1 1 1
1 1 1 1
1 1
130.68 J K mol 213.74 J K mol
 [197.67 J K mol 188.83 J K mol ]
42.08 J K mol
− − − −
− − −
− −
∆ ° = ⋅ ⋅ + ⋅ ⋅
− ⋅ ⋅ + ⋅ ⋅
= − ⋅ ⋅
S
−
1
 
 
1 1 1( 41.16 kJ mol ) (1000 J kJ ) ( 42.08 J K mol ) 0
978 (or 705 C)
G T
T K
− − − −∆ = − ⋅ ⋅ − − ⋅ ⋅ =
= °
 (c) CO(g) + + 2H O(g) → 2CO (g) 2H (g)
 10.00 bar 10.00 bar 5.00 bar 5.00 bar 
 change x− x− x+ x+ 
 net 10.00 x− 10.00 x− 5.00 x+ 5.00 x+ 
 2.50 barx =
 All pressures are equal to 7.50 bar. 
 (d) First, check Q to determine the direction of the reaction: 
 Q = (10.00) (5.00) 2.08
(6.00) (4.00)
= 
 Because Q is greater than 1, the reaction will shift to produce reactants. 
 CO(g) + + 2H O(g) → 2CO(g) 2H (g)
 6.00 bar 4.00 bar 10.00 bar 5.00 bar 
 change x+ x+ x− x−
 net 6.00 x+ 4.00 x+ 10.00 x− 5.00 x− 
SM-264 
 
2 2 2
2 2
CO(g) H O(g) CO (g) H (g)
(10.00 ) (5.00 ) 1
(6.00 ) (4.00 )
(10.00 ) (5.00 ) (6.00 ) (4.00 )
15.00 50.0 10.00 24.0
25.00 26.0
1.04 bar
7.04 bar; 5.04 bar; 8.96 bar; 3.96 bar
x x
x x
x x x x
x x x x
x
x
P P P P
− − =+ +
− − = + +
− + = + +
=
=
= = = =
 
 
9.99 (a) These values are easily calculated from the relationship 
 ln K. For the atomic species, the free energy of the reaction will be 
G R∆ ° = − T
1
2 of 
 this value because the equilibrium reactions are for the formation of two 
 moles of halogen atoms. 
 The results are 
 Bond Dissociation Energy G∆ ° 
 Halogen 1(kJ mol )−⋅ 1(kJ mol )−⋅
 fluorine 146 19.2 
 chlorine 230 47.8 
 bromine 181 42.8 
 iodine 139 5.6 
 (b) 
 
 
 
 
0
10
20
30
40
50
60
0 50 100 150 200 250 300
bond dissociation energy (kJ⋅mol-1)
fr
ee
 e
ne
rg
y 
of
r
io
m
at
 fo
n
(k
J ⋅m
ol
-1
)
SM-265 
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. 
 
0
20
40
60
80
100
120
0 10 20 30 40 50 60
atomic number
fr
ee
 e
ne
rg
y 
of
 d
is
so
ci
at
io
n 
(k
J ⋅m
ol
-1
)
 
 
 
 
 
 
 
 
 
 
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. 
 
9.101 (a) K<1 but will increase at higher temperature because S∆ >0. 
 (b) K>1 
 (c) K>1 
 (d) K>1 
 
9.103 (a) Use the relationship ln 2
1 2
1 1 .K H
K R T T
⎛ ⎞∆ °= − −⎜⎝ ⎠1 ⎟
 The value of H∆ ° is 
 first obtained by using the two points given in Table 9.1. 
 ln 
3
5
1.7 × ∆ 10 1 1
3.4 10 1200 1000
H
R
−
−
° ⎛ ⎞= − −⎜ ⎟× ⎝ ⎠
 2 11.96 10 kJ molH −∆ ° = × ⋅ 
SM-266 
 We then use this value plus one of the known equilibrium constant points 
 in the equation 
 ln
1
298
5 1 1
196 000 J mol 1 1
3.4 10 8.314 J K mol 298 1000
K −
− − −
⎛ ⎞⋅ ⎛ ⎞= − −⎜ ⎟⎜ ⎟× ⋅ ⋅ ⎝ ⎠⎝ ⎠
 
 29298 2.6 10K
−= ×
 (b) Using the thermodynamic data in Appendix 2A: 
 
2
1 1
/ 29
Br (g) 2 Br(g)
2(82.40 kJ mol ) 3.11 kJ mol 161.69 kJ mol
e 4.5 10G RT
G
K
1− − −
−∆ ° −
→
∆ ° = ⋅ − ⋅ = ⋅
= = ×
 
 For equilibrium constant calculations, this is reasonably good agreement 
 with the value obtained from part (a), especially if one considers that H∆ ° 
 will not be perfectly constant over so large a temperature range. 
 (c) We will use data from Appendix 2A to calculate the vapor pressure of 
 bromine: 
 
2 2
1
Br (l) Br (g)
3.11 kJ mol
e 0.285G /RT
G
K
−
−∆ °
→
∆ ° = ⋅
= =
 The vapor pressure of bromine will, therefore, be 0.285 bar or 0.289 atm. 
 Remember that because the standard state for the thermodynamic 
 quantities is 1 bar, the values in K will be derived in bar as well. 
(d) 
2
2 2
Br(g) Br(g)29
Br (l)
4.5 10
0.285 bar
P P
P
−× = = 
2 15 15
Br(g) 3.6 10 bar or 3.6 10 atmP
− −= × × 
 (e) Use the ideal gas law: 
 1 1(0.289 atm) (0.0100 mol) (0.082 06 L atm K mol ) (298 K)
0.846 L or 846 mL
PV nRT
V
V
− −
=
= ⋅ ⋅ ⋅
=
 
9.105 First, we calculate the equilibrium constant for the conditions given. 
SM-267 
 
2
3
(23.72) 41.0,
(3.11) (1.64)
K = =
3
3
 which corresponds to the reaction written as 
 2 2N (g) 3 H (g) 2 NH (g)+ →
 We then set up the table of anticipated changes upon introduction of the 
 nitrogen: 
 2 2N (g) 3 H (g) 2 NH (g)+ →
 initial 4.68 bar 1.64 bar 23.72 bar 
 change x− 3 x− 2 x+ 
 total 4.68 x− 1.64 3x− 23.72 2 x+ 
 
2
3
(23.72 2 )41.0
(4.68 ) (1.64 3 )
x
x x
+= − − 
 The equation can be solved using a graphing calculator, other computer 
 software, or by trial and error. The solution is 0.176.x = The pressures of 
 gases are 
 
2
2
3
N
H
NH
4.33 bar or 4.39 atm
1.11 bar or 1.12 atm
24.07 bar or 24.39 atm
P
P
P
=
=
=
 
50
650
1250
250 300 350
T / K
K
9.107 
 
 
 
 
 
 
 
 
 
 
 
SM-268 
9.109 (a) 
2 2
2
2 4
[NO ] (2.13) 11.2
[N O ] 0.405C
K = = = 
 (b) If is added, the equilibrium will shift to produce more 
 The amount of will be greater than initially present, but less than the 
 present immediately upon making the addition. will not 
 be affected. 
2NO 2 4N O .
2NO
13.13 mol L−⋅ CK
 (c) 
 Concentrations 1(mol L )−⋅ Æ 2 4N O 22 NO
 initial 0.405 3.13 
 change x+ 2 x− 
 final 0.405 x+ 3.13 2 x− 
 
2
2
2
2
2
(3.13 2 )11.2
0.405
(11.2) (0.405 ) (3.13 2 )
11.2 4.536 4 12.52 9.797
4 23.7 5.26 0
( 23.7) ( 23.7) (4) (4) (5.26) 23.7 21.9
(2) (4) 8
5.70 or 0.23
x
x
x x
x x x
x x
x
x
−= +
+ = −
+ = − +
− + =
− − ± − − ±= =
=
 
 At equilibrium 1 12 4[N O ] 0.405 mol L 0.23 mol L 0.64 mol L
1− − −= ⋅ + ⋅ = ⋅ 
 1 12[NO ] 3.13 mol L 2(0.23 mol L ) 2.67 mol L
− −= ⋅ − ⋅ = 1−⋅
 These concentrations are consistent with the predictions in (b). 
 
9.111 To find the vapor pressure, we first calculate G∆ ° for the conversion of 
 the liquid to the gas at 298 K, using the free energies of formation found in 
 the appendix: 
SM-269 
 
2 2 2 2
2 2 2 2
H O(l) H O(g) f(H O(g)) f(H O(l))
1 1
1
D O(l) D O(g) f(D O(g)) f(D O(l))
1 1
1
( 228.57 kJ mol ) [ 237.13 kJ mol ]
8.56 kJ mol
( 234.54 kJ mol ) [ 243.44 kJ mol ]
8.90 kJ mol
G G G
G G G
→
− −
−
→
− −
−
∆ ° = ∆ ° − ∆ °
= − ⋅ − − ⋅
= ⋅
∆ ° = ∆ ° − ∆ °
= − ⋅ − − ⋅
= ⋅
 The equilibrium constant for these processes is the vapor pressure of the 
 liquid: 
 
2 2H O D O
orK P K P= = 
 Using ln K, we can calculate the desired values. G R∆ ° = − T
 For 2H O:
 ln 
1
1 1
8560 J mol 3.45
(8.314 J K mol ) (298 K)
GK
RT
−
− −
∆ ° ⋅= − = − = −⋅ ⋅ 
 bar 0.032K =
 1 atm 760 Torr0.032 bar 24 Torr
1.013 25 bar 1 atm
× × = 
 For 2D O:
 ln 
1
1 1
8900 J mol 3.59
(8.314 J K mol ) (298 K)
GK
RT
−
− −
∆ ° ⋅= − = − = −⋅ ⋅ 
 bar 0.028K =
 1 atm 760 Torr0.028 bar 21 Torr
1.013 25 bar 1 atm
× × = 
 The answer is that D has a lower zero point energy than H. This makes the 
 — “hydrogen bond” stronger than the — hydrogen 
 bond. Because the hydrogen bond is stronger, the intermolecular forces are 
 stronger, the vapor pressure is lower, and the boiling point is higher. 
2D O 2D O 2H O 2H O
 Potential energy curves for the O—H and O—D bonds as a function of 
 distance: 
 energy required to break O—H or O—D bond. E∆ =
SM-270 
9.113 First, we derive a general relationship that relates fG∆ ° values to 
 nonstandard state conditions. We do this by returning to the fundamental 
 definition of and G∆ G∆ ° . 
 f m(products) m(reactants)G G G∆ ° = Σ ° − Σ °
 Similarly, for conditions other than standard state, we can write 
 f m(products) m(reactants)G G G∆ = Σ − Σ
 Because m m ln ,G G RT Q= ° +
 f m i (products) m i (reactants)( ln ) ( ln )G G RT P G RT P∆ = Σ ° + − Σ ° +
 But because all reactants or products in the same system refer to the same 
 standard state, 
 f m(products) m(reactants) (reactants)( lnG G G n RT Q∆ = Σ ° − Σ ° + ∆ )
)f f (reactants)( lnG G n RT Q∆ = ∆ ° + ∆
 The value f , which is the nonstandard value for the conditions of 1 
 atm, 
G∆
 etc. becomes the new standard value. 11 mol L ,−⋅
 (a) 1 1 1 12 22 2 2 2H (g) I (g) HI(g), 1 mol ( mol mol) 0n+ → ∆ = − + = 
 1 atm = 1.013 25 bar 
 
f
1 1
1.70 kJ mol
(0) (8.314 J K mol ) (298 K) (ln 1.013 25)/(1000 J kJ )
1.70 kJ mol
G
1− − −
∆ = ⋅
+ ⋅ ⋅ ⋅
= ⋅
 
 (b) 1 122 2C(s) O (g) CO(g), 1 mol mol moln+ → ∆ = − = 12 
 
f
1 1 1−⋅12
1
137.17 kJ mol
( ) (8.314 J K . mol ) (298 K)(ln 1.013 25)/(1000 J kJ )
137.15 kJ mol
G
− −
−
∆ = − ⋅
+ ⋅
= − ⋅
 
 (c) 1 1 1 12 22 2 2 2C(s) N (g) H (g) HCN(g), 1 mol ( mol mol)
0
n+ + → ∆ = − +
=
 
 1 Torr 31 atm 1.013 25 bar 1.333 10 bar
760 Torr 1 atm
−× × = × 
SM-271 
 
f
1 1 3
1
124.7 kJ mol
(0) (8.314 J K mol ) (298 K) (ln 1.333 10 )/(1000 J kJ )
124.7 kJ mol
G
1− − −
−
∆ = ⋅
+ ⋅ ⋅ × ⋅
= ⋅
−
1
 
 (d) 2 4C(s) 2 H (g) CH (g), 1 mol 2 mol 1 moln+ → ∆ = − = −
 51 Pa 10 bar−=
 
f
1 1 5
1
50.72 kJ mol
( 1) (8.314 J K mol ) (298 K)(ln 10 )/(1000 J kJ )
22.2 kJ mol
G
− − − −
−
∆ = − ⋅
+ − ⋅ ⋅ ⋅
= − ⋅
 
9.115 (a) (i) Increasing the amount of a reactant will push the equilibrium toward 
 the products. More NO2 will form. (ii) Removing a product will pull the 
 equilibrium toward products. More NO2 will form. (iii) Increasing total 
 pressure by adding an inert gas not change the relative partial pressures. 
 There will be no change in the amount of NO2. (b) Since there are two 
 moles of gas on both sides of the reaction, the volume cancels out of the 
 equilibrium constant expression and it is possible to use moles directly for 
 each component. 
 At equilibrium, SO2 = 2.4 moles -1.2 moles = 1.2 moles, SO3 = 1.2 moles 
 and NO = 1.2 moles since the reaction is 1:1:1:1. The original number of 
 moles of NO2 is set to x. 
 K = 3.00 = (1.2)(1.2)/(1.2)(x-1.2) = (1.2)/(x-1.2) 
 x=1.2/3.00 + 1.2 
 x=1.6 moles of NO2
 
9.117 (a) Reversing both reactions then adding them together gives the reaction 
 of interest. 
 NO + O Æ NO2 K’=1/K=1/6.8 x 10-49=1.47 x1048
 NO2 + O2 Æ O3 + NO K”=1/K=1/5.8 x 10-34=1.72 x 1033
 ---------------------------- 
 O2 + O Æ O3 K=K’*K”=2.5 x 1081
SM-272 
 (b) Since the initial mixture is equimolar in the two reactants (i.e. each 
 partial pressure is 2.0 bar) and the equilibrium strongly favors products, 
 we can see that essentially all of both reactants will combine to form an 
 equimolar amount of ozone, or a final equilibrium partial pressure of 2.0 
 bar. Some very small partial pressure, x, of each reactant will be left and 
 must satisfy the equilibrium constant expression: 
 K = 2.5 x 1081 = PO3/(PO2*PO ) = 2.0 bar/(x2) 
 x = 2.8 x 10-41 bar = PO2 = PO and PO3 = 2.0 bar at equilibrium 
 
SM-273 
	CHEMICAL EQUILIBRIA