Atkins - Cap 09 - Pares

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CHAPTER 9 
CHEMICAL EQUILIBRIA 
 
 
 
 
9.2 (a) True 
 (b) False. Changing the rate of a reaction will not affect the value of the 
 equilibrium constant; it merely changes how fast one gets to equilibrium. 
 (c) True 
 (d) False. The standard reaction free energy is not 0 at equilibrium. 
 The reaction free energy 
∆ DrG
∆ rG , which is dependent upon the 
 concentrations of the products and reactants, is 0 at equilibrium. 
 
9.4 Cl2 decreases from 2.15 to 2.13 bar, PCl3 decreases from 1.28 to 1.26 bar, 
 and PCl5 increases from 0.02 to 0.04 bar. The shapes for the curves can 
 only be determined accurately if the rate law for the reaction is known (see 
 Chapter 13). 
 
0
0.5
1
1.5
2
2.5
P
re
ss
ur
e 
(b
ar
)
Time
[Cl
2
]
[PCl
3
]
[PCl
5
]
 
247 
9.6 (a) 2
2
2
NO
2
NO O
= PK
P P
; (b) 3 2
5
SbCl Cl
SbCl
= P PK
P
; (c) 2 4
2 2
N H
2
N H
= PK
P P
 
 
9.8 All values should be the same because the same amounts of the substances 
 are present at equilibrium. It doesn’t matter whether we begin with 
 reactants or with products; the equilibrium composition will be the same if 
 the same amounts of materials are used. If different amounts had been 
 used, only (e) would be the same in the two containers. A more detailed 
 analysis follows: 
 H2(g) + Br2(g) 2 HBr(g) →
 The equilibrium constant expression for this system is 
 
2
2 2
[HBr]
[H ] [Br ]
=CK 
 In terms of the change in concentration, x, of H2 and Br2 that has come 
 about at equilibrium, we may write 
 
2 2
2
[2 ] [2 ]
[0.05 ] [0.05 ] [0.05 ]
= =− − −C
x xK
x x x
 (1) 
 in the first container, and in terms of the change in concentration, y, of 
 HBr that has come about at equilibrium in the second container, we may 
 write 
 
2 2
2
[0.10 ] [0.10 ]
2 2 2
− −= =⎡ ⎤ ⎡ ⎤ ⎛ ⎞⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠
C
y yK
y y y
 (2) 
 Because KC is a constant, and because the relative amounts of starting 
 materials are in the ratio of their stoichiometric factors, we must have 
 
2 2
2 2
[2 ] [0.10 ]
[0.05 ]
2
[2 ] [0.10 ]
[0.05 ]
2
−=− ⎛ ⎞⎜ ⎟⎝ ⎠
−=− ⎛ ⎞⎜ ⎟⎝ ⎠
x y
x y
x y
yx
 
248 
 Cross multiplying: 
 
(0.05 )(0.10 )
0.005 0.05 0.10
0.10 2
= − −
= − − +
= −
xy x y
xy y
y x
x xy 
 This may also be seen by solving quadratic equations for Eqs. 1 and 2 for 
 x and y to obtain any value of KC. 
 (a) 2[Br ] 0.05= − x in the first container, 2[Br ] /2= y in the second 
 container. Because /2 0.05= −y x satisfies the conditions of Eq. 3, the 
 concentrations and hence the amounts of Br2 are the same in the two cases. 
 (b) 2[H ] 0.05 /2= − =x y , as above; hence the concentrations of H2 are 
 the same in both systems. 
 (c) Because all concentrations are the same in both cases, this ratio will 
 also be the same. 
 (d) For the same reason as in part (c), this ratio is the same in both cases. 
 (e) This ratio is the equilibrium constant, so it must be the same for both 
 systems. 
 (f) Because all the concentrations and amounts are the same in both cases, 
 and because the volumes and temperatures are the same, the total pressure 
 must be the same: 
 2 2H Br HBr
( )+ += n n n RP
V
T
 
 
9.10 
3 2NH H S
=K P P 
 
For condition 1, 0.307 0.307 0.0942
For condition 2, 0.364 0.258 0.0939
For condition 3, 0.539 0.174 0.0938
= × =
= × =
= × =
K
K
K
 
9.12 (a) 
2
3
3
[Cl ] [ClO ]
[ClO ]
− −
− 
 (b) [CO2] 
249 
 (c) 3
3
[H ][CH COO ]
[CH COOH]
+ −
 
 
9.14 (a) 2 CH4(g) + S8(s) → 2 CS2(l) + 4 H2S(g) 
 
r f 2 f 2
f 4
1 1
1
1
2 (CS , l) 4 (H S, g)
 [2 (CH , g)]
2(65.27 kJ mol ) 4 ( 33.56 kJ mol )
 [2 ( 50.72 kJ mol )]
97.74 kJ mol
− −
−
−
∆ ° = × ∆ ° + × ∆ °
− × ∆ °
= ⋅ + − ⋅
− − ⋅
= + ⋅
G G G
G
 ln∆ ° = −G RT K
 or 
 
1
1 1
18
ln
97 740 J molln 39.4
(8.314 J K mol )(298 K)
8 10
−
− −
−
∆ °= −
+ ⋅= − = −⋅ ⋅
= ×
GK
RT
K
K
 
 (b) CaC2(s) + 2 H2O(l) Ca(OH)→ 2(s) + C2H2(g) 
 
r f 2 f 2 2
f 2 f 2
1 1
1 1
1
(Ca(OH) , s) (C H , g)
 [ (CaC , s) 2 (H O, l)]
( 898.49 kJ mol ) ( 209.20 kJ mol )
 [( 64.9 kJ mol ) 2( 237.13 kJ mol )]
150.1 kJ mol
− −
− −
−
∆ ° = ∆ ° + ∆ °
− ∆ ° + × ∆ °
= − ⋅ + + ⋅
− − ⋅ + − ⋅
= − ⋅
G G G
G G
1
1 1
26
150 100 J molln 60.6
(8.314 J K mol )(298 K)
2 10
−
− −
− ⋅= − = +⋅ ⋅
= ×
K
K
 
 (c) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l) 
 
r f f 2 f 3
1 1
1
4 (NO, g) 6 (H O, l) [4 (NH , g)]
4(86.55 kJ mol ) 6( 237.13 kJ mol ) [4( 16.45 kJ mol )]
1010.8 kJ mol
− −
−
∆ ° = ∆ ° + ∆ ° − ∆ °
= ⋅ + − ⋅ − − ⋅
= − ⋅
G G G G
1−
 
3 1
1 1
177
1010.8 10 J molln 408
(8.314 J K mol )(298 K)
10
−
− −
− × ⋅= − = +⋅ ⋅
≅
K
K
 
250 
 (d) CO2(g) + 2 NH3(g) CO(NH→ 2)2(s) + H2O(l) 
 
r f 2 2 f 2
f 2 f 3
1 1
1 1
1
(CO(NH ) , s) (H O, l)
 [ (CO , g) 2 (NH , g)]
( 197.33 kJ mol ) ( 237.13 kJ mol )
 [( 394.36 kJ mol ) 2( 16.45 kJ mol )]
7.20 kJ mol
− −
− −
−
∆ ° = ∆ ° + ∆ °
− ∆ ° + × ∆ °
= − ⋅ + − ⋅
− − ⋅ + − ⋅
= − ⋅
G G G
G G
3
1 1
7.20 10 Jln 2.91
(8.314 J K mol )(298 K)
18
− −
− ×= − = +⋅ ⋅
=
K
K
 
 
9.16 (a) r
1 1
1
ln K
(8.314 J K mol )(700 K) ln 54
23.2 kJ mol
− −
−
∆ ° = −
= − ⋅ ⋅
= − ⋅
G RT
 (b) r
1 1
1
ln K
(8.314 J K mol )(298 K) ln 0.30
2.98 kJ mol
− −
−
∆ ° = −
= − ⋅ ⋅
= + ⋅
G RT
 
9.18 r f 2
1
(CO , g)
394.36 kJ mol−
∆ ° = ∆ °
= − ⋅
G G 
 3 1
1 1
69
ln
ln
394.36 10 J molln 159.17
(8.314 J K mol )(298 K)
1.3 10
−
− −
∆ ° = −
∆ °= −
− × ⋅= − = +⋅ ⋅
= ×
G RT K
GK
RT
K
K
 
 In practice, no K will be so precise. A better estimate would be . 
 Because Q < K, the reaction will tend to proceed to produce products. 
691 10×
 
9.20 The free energy at a specific set of conditions is given by 
251 
 
5
3 2
r r
r
Cl
r
Cl Cl
1 1
1 1
1
ln
ln ln
ln ln
(8.314 J K mol )(503 K) ln 49
(1.33) (8.314 J K mol )(503 K) ln
(0.22)(0.41)
5.0 kJ mol
− −
− −
−
∆ = ∆ ° +
∆ = − +
∆ = − + ⋅
= − ⋅ ⋅
+ ⋅ ⋅
= − ⋅
G G RT Q
G RT K RT Q
p
G RT K RT
p p
 
 Because r∆ is negative, the reaction will be spontaneous to form 
 products. 
G
 
9.22 The free energy at a specific set of conditions is given by 
 
r r
r
2
r
2 2
1 1
2
1 1
1
ln
ln ln
[HI]ln ln
[H ] [I ]
(8.314 J K mol )(700 K) ln 54
(2.17) (8.314 J K mol )(700 K) ln
(0.16)(0.25)
4.5 kJ mol
− −
− −
−
∆ = ∆ ° +
∆ = − +
∆ = − +
= − ⋅ ⋅
+ ⋅ ⋅
= ⋅
G G RT Q
G RT K RT Q
G RT K RT
 
 Because r∆ is positive, the reaction will proceed to form reactants. G
 
9.24 (a) 3
2 2
2
SO
2
SO O
3.4
P
= =PK
P
 
 C12.027 K
∆⎛ ⎞= ⎜ ⎟⎝ ⎠
n
TK K 
 
(2 3)
212.027 K 12.027 K 3.4 2.8 10
1000 K
−∆ ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
n
CK KT
× 
 (b) 
3 2
2
NH H S 9.4 10
−= = ×K P P 
252 
 
(2 0)
2 412.02 K 12.027 K 9.4 10 1.5 10
297 K
−∆
− −⎛ ⎞⎛ ⎞= = × =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
n
CK KT
× 
 
9.26 For the equation written 2 2 32 SO (g) O (g) 2 SO (g)+ → Eq. 1 
 3
2 2
2
SO 10
2
SO O
2.5 10= = ×PK
P P
 
 (a) For the equation written 12 22SO (g) O (g) SO (g)3+ → Eq. 2 
 3
2 2
SO 10 5
Eq. 2 Eq.11/ 2
SO O
2.5 10 1.6 10= = = × =PK K
P P
× 
 (b) For the equation written 13 2 2SO (g) SO (g) O (g)→ + 2 Eq. 3 
 This equation is the reverse of Eq. 2,
so Eq. 3
Eq. 2 Eq.1
1 1= =K
K K
 
 2 2
3
1/ 2
SO O 6
Eq. 2 10
SO Eq.1
1 1 6.3 10
2.5 10
−= = = = ×
×
P P
K
P K
 
 (c) For the equation written 32 223 SO (g) O (g) 3 SO (g)3+ → Eq. 4 
 This equation is 3/23 Eq. 4 Eq.12 Eq. 1, so× =K K 
 3
2 2
3
SO 3/ 2 10 3/ 2 15
Eq. 2 Eq.13 3/ 2
SO O
(2.5 10 ) 4.0 10= = = × =PK K
P P
× 
 
9.28 H2(g) + Cl2(g) 2 HCl(g) → 85.1 10= ×CK
 
2
2 2
[HCl]
[H ][Cl ]
=CK 
 
3 2
8
3
2
3 2
2 8 3
12 1
2
(1.45 10 )5.1 10
[H ](2.45 10 )
(1.45 10 )[H ]
(5.1 10 )(2.45 10 )
[H ] 1.7 10 mol L
−
−
−
−
− −
×× = ×
×= × ×
= × ⋅
 
253 
9.30 3 2
5
SbCl Cl
SbCl
= P PK
P
 
 2
3
Cl4 (5.02 10 )3.5 10
0.072
−
− ×× = P 
 
2
4
3
Cl 3
(3.5 10 )(0.072) 5.0 10 bar
5.02 10
−
−
−
×= = ××P 
 
9.32 (a) 
2
3
3
2 2
[NH ]
0.278
[N ][H ]
= =CK 
 
2
3
[0.122] 0.248
[0.417][0.524]
= =CQ 
 (b) ≠CQ KC
C
; therefore the system is not at equilibrium. 
 (c) Because , more products will be formed. <CQ K
 
9.34 (a) 
2
3
3
2 2
[NH ]
62
[N ][H ]
= =CK 
 
4 2
3
3 3 3
[1.12 10 ] 2.95 10
[2.23 10 ][1.24 10 ]
−
− −
×= =× ×CQ ×
C
 
 (b) Because , ammonia will decompose to form reactants. >CQ K
 
9.36 2 22 21 1
1.00 g I 0.830 g I
0.003 94 mol I ; 0.003 27 mol I
253.8 g mol 253.8 g mol− −
= =⋅ ⋅ 
 I2(g) U 2 I(g) 
 0.003 94 mol − x 2 x 
 0.003 94 mol 0.003 27 mol− =x
 0.000 67 mol; 2 0.0013 mol= =x x
 
2
4
0.0013
1.00 5.2 10
0.003 27
1.00
−
⎡ ⎤⎢ ⎥⎣ ⎦= =⎡ ⎤⎢ ⎥⎣ ⎦
CK × 
254 
9.38 Pressure (Torr) CO(g) + H2O(g) COU 2(g) + H2(g) 
initial 200 200 0 0 
final 20200 88− 0 88− 88 88 
 Note: Because pressure is directly proportional to the number of moles of 
 a substance, the pressure changes can be used directly in calculating the 
 reaction stoichiometry. Technically, to achieve the correct standard state 
 condition, the Torr must be converted to bar ; however, 
 in this case those conversion factors will cancel because there are equal 
 numbers of moles of gas on both sides of the equation. 
1(750.1 Torr bar )−⋅
 2 2
2
CO H
CO H O
88 88
750.1 750.1 0.62
112 112
750.1 750.1
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
P P
K
P P
 
 
9.40 (a) The balanced equation is Cl2(g) 2 Cl(g) U
 The initial concentration of 122
0.0050 mol Cl
Cl (g) is 0.0025 mol L
2.0 L
−= ⋅ 
 Concentration Cl1(mol L )−⋅ 2(g) 2 Cl(g) U
 initial 0.0025 0 
 change −x +2 x 
 equilibrium 0.0025 − x +2 x 
 
22
3
C
2
2 3
2 3 6
(2 )[Cl] 1.7 10
[Cl ] (0.0025 )
4 (1.7 10 )(0.0025 )
4 (1.7 10 ) (4.25 10 ) 0
−
−
− −
= = = ×−
= × −
+ × − × =
xK
x
x x
x x
 
 
3 3 2
3 3
3 4
(1.7 10 ) (1.7 10 ) 4(4)( 4.3 10 )
2 4
(1.7 10 ) 8.47 10
8
1.3 10 or 8.5 10
− −
− −
− −
− × ± × − − ×= ×
− × ± ×=
= − × + ×
x
x
x
6−
 
255 
 The negative answer is not meaningful, so we choose =x
 The concentration of Cl48.5 10 mol L .−× ⋅ 1− 2 is 
 The concentration of Cl atoms is 
 The percentage decomposition of Cl
40.0025 8.5 10−− ×
0.0017.= 42 (8.5 10 )−× ×
41.7 10−= × 1mol L .−⋅ 2 is given by 
 
48.5 10 100 34%
0.0025
−× × = 
 (b) The balanced equation is Br2(g) U 2 Br(g) 
 The initial concentration of 122
5.0 mol B
Br (g) is 2.5 mol L
2.0 L
−= ⋅r 
 Concentration Br1(mol L )−⋅ 2(g) 2 Br(g) U
 initial 2.5 0 
 change −x +2 x 
 equilibrium 2.5 − x +2 x 
 
22
3
C
2
2 3
2 3 3
(2 )[Br] 1.7 10
[Br ] (2.5 )
4 (1.7 10 )(2.5 )
4 (1.7 10 ) (4.25 10 )
−
−
− −
= = = ×−
= × −
+ × − × =
xK
x
x x
x x 0
 
 
3 3 2
3 1
2 2
(1.7 10 ) (1.7 10 ) 4(4)( 4.25 10 )
2 4
(1.7 10 ) 2.608 10
8
3.3 10 or 3.2 10
− −
− −
− −
− × ± × − − ×= ×
− × ± ×=
= − × + ×
x
x
x
3−
 
 The negative answer is not meaningful, so we choose =x
 The concentration of Br23.2 10 mol L .−× ⋅ 1− 2 is 
 The concentration of Br atoms is 
22.5 3.2 10 2.5.−− × ≈
22 (3.2 10 )−× × = 2 16.4 10 mol L .− −× ⋅ 
 he percentage decomposition of Br2 is given by 
 
23.2 10 100 1.3%
2.5
−× × = 
 (c) At this temperature, Cl2 and Br2 are equally stable since their 
 equilibrium constants are the same. If we had used the same initial 
256 
 amounts of Cl2 and Br2 in parts (a) and (b), then the percent decomposition 
 would have been the same as well. 
 
9.42 Pressure (bar) 2 BrCl(g) BrU 2(g) + Cl2(g) 
initial 31.4 10−× 0 0 
change 2− x +x +x 
final 31.4 10 2−× − x +x +x 
 
 
2 2Br Cl
2
BrCl
2
3 2 3
2
3 2
( )( )32
(1.4 10 2 ) (1.4 10 2 )
32
(1.4 10 2 )
− −
−
⋅=
= =× − × −
= × −
p p
K
p
x x x
2x x
x
x
 
 
3
3
3
3
32
(1.4 10 2 )
( 32)(1.4 10 2 )
2 32 ( 32)(1.4 10 )
(1 2 32) ( 32)(1.4 10 )
−
−
−
−
=× −
= × −
+ = ×
+ = ×
x
x
x x
x x
x
 
 
3
4
( 32)(1.4 10 )
(1 2 32)
6.4 10
−
−
×= +
= ×
x
x
 
 
2 2
4
Br Cl 6.4 10 bar = 0.64 mbar
−= = ×p p
 3 4 4BrCl 1.4 10 bar 2 (6.4 10 bar) 1.2 10 bar = 0.12 mbar
− − −= × − × = ×p
 The percentage decomposition is given by 
 
4
3
2 (6.4 10 bar) 100 91%
1.4 10 bar
−
−
× × =× 
257 
9.44 (a) concentration of PCl5 initially = 
 
5
1
5 1
2.0 g PCl
208.22 g mol PCl
0.032 mol L
0.300 L
−
−
⎛ ⎞⎜ ⎟⋅⎝ ⎠ = ⋅ 
 Concentration PCl1(mol L )−⋅ 5(g) PClU 3(g) + Cl2(g) 
 initial 0.032 0 0 
 change −x +x +x 
 final 0.032 − x +x +x 
 
2
3 2
C
5
[PCl ][Cl ] ( )( )
[PCl ] (0.032 ) (0.032 )
= = =− −
x x xK
x x
 
 
2
2
2
2
0.61
(0.032 )
(0.61)(0.032 )
(0.61) 0.020 0
(0.61) (0.61) (4)(1)( 0.020)
2 1
(0.61) 0.67 0.03 or 0.64
2 1
=−
= −
+ − =
− ± − −= ⋅
− ±= = + −⋅
x
x
x x
x x
x
x
 
 The negative root is not meaningful, so we choose 10.03 mol L .−= ⋅x 
 13 2[PCl ] [Cl ] 0.03 mol L ;
−= = ⋅ 
 15[PCl ] 0.032 0.03 mol L 0.002 mol L .
1− −= − ⋅ = ⋅ The solution of this 
 problem again points up the problems with following the significant figure 
 conventions for calculations of this type, because we would use only one 
 significant figure as imposed by the subtraction in the quadratic equation 
 solution. The fact that the equilibrium constant is given to two significant 
 figures, however, suggests that the concentrations could be determined to 
 that level of accuracy. Plugging the answers back into the equilibrium 
 expression gives a value of 0.45, which seems somewhat off from the 
 starting value of 0.61. The closest agreement to the equilibrium expression 
 comes from using 13 2[PCl ] [Cl ] 0.0305 mol L
−= = ⋅ and 
258 
 15[PCl ] 0.0015 mol L
−= ⋅ ; this gives an equilibrium constant of 0.62. 
 Rounding these off to two significant figures gives 
 13 2[PCl ] [Cl ] 0.030 mol L
−= = ⋅ and 5[PCl ] 0.0015= , which in the 
 equilibrium expression produces a value of 0.60. The percentage 
 decomposition is given by 
 0.030 100% 94%
0.032
× = 
 
9.46 Starting concentration of 13
0.200 molNH 0.100 mol L
2.00 L
−= = ⋅ 
 Concentration NH1(mol L )−⋅ 4HS(s) NHU 3(g) + H2S(g) 
 initial — 0.100 0 
 change — +x +x 
 final — 0.100 + x +x 
 
C 3 2
4
2 4
2 4
[NH ][H S] (0.100 )( )
1.6 10 (0.100 )( )
0.100 1.6 10 0
( 0.100) ( 0.100) (4)(1)( 1.6 10 )
2 1
0.100 0.1031 0.002 or 0.102
2 1
−
−
−
= = +
× = +
+ − × =
− + ± + − − ×= ⋅
− ±= = + −⋅
K x x
x x
x x
x
x
 
 The negative root is not meaningful, so we
choose 3 12 10 mol L .− −= × ⋅x 
 
1 3
3
1
3 1
2
[NH ] 0.100 mol L 2 10 mol L
 0.102 mol L
[H S] 2 10 mol L
1− − −
−
− −
= + ⋅ + × ⋅
= ⋅
= × ⋅
 
 Alternatively, we could have assumed that x << 0.100, in which case 
 40.100 1.6 10−= ×x 3or 1.6 10 .−= ×x 
 
9.48 The initial concentrations of PCl3 and Cl2 are calculated as follows: 
259 
 
1
3 2
1
0.200 mol 0.600 mol[PCl ] 0.0250 mol L ; [Cl ]
8.00 L 8.00 L
0.0750 mol L
−
−
= = ⋅ =
= ⋅
 
 Concentrations PCl1(mol L )−⋅ 5 PClU 3(g) + Cl2(g) 
 initial 0 0.0250 0.0750 
 change +x −x −x 
 final +x 0.0250 − x 0.0750 − x 
 3 2C
5
[PCl ][Cl ] (0.0250 )(0.0750 ) 33.3
[PCl ] ( )
− −= = =+
x xK
x
 
 
2
2
0.100 0.001875 33.3
33.40 0.001875 0
− + =
− + =
x x x
x x
 
 
2
5
33.40 ( 33.40) (4)(1)(0.001 875) 33.40 33.399 89
(2)(1) 2
5.5 10 or 33.4−
+ ± − − + ±= =
= + × +
x 
 The root +33.4 has no physical meaning because it is greater than the 
 starting concentrations of PCl3 and Cl2, so it can be discarded. 
 
 
 
5 1
5 3[PCl ] 5.5 10 mol L ; [PCl ] 0.0250 mol L
− −= × ⋅ = ⋅ 1−
1−
2[Cl ] 0.0750 mol L 5.5 10 mol L
5 15.5 10 mol L 0.0249 mol L ;− −− × ⋅ = ⋅
1 5 1− − −= ⋅ − × ⋅ 10.0749 mol L .−= ⋅ Note that 
 the normal conventions concerning significant figures were ignored in 
 order to obtain a meaningful answer. 
 
9.50 The initial concentrations of N2 and O2 are 
 
1
2 2
1
0.0140 mol 0.240 mol[N ] 0.001 40 mol L ; [O ]
10.0 L 10.0 L
0.0214 mol L
−
−
= = ⋅ =
= ⋅
 
 Concentrations (m L⋅ N1ol )− 2(g) + O2(g) U 2 NO(g) 
 initial 0.001 40 0.0214 0 
 change −x −x + 2x
 final 0.001 40 − x 0.0214 − x + 2x
260 
 
22
2 2
(2 )[NO]
[N ][O ] (0.001 40 )(0.0214 )
= = − −C
xK
x x
 
 
2
5
2
5
2 5
2 5 2
2 5 2 7
2 7 10
(2 )1.00 10
(0.001 40 )(0.0214 )
41.00 10
0.0228 3.0 10
4 (1.00 10 )( 0.0228 3.0 10 )
4 1.00 10 2.28 10 3.0 10
4 2.28 10 3.0 10 0
−
−
−
− −
− −
− −
× = − −
× = − + ×
= × − + ×
= × − × + ×
+ × − × =
x
x x
x
x x
x x x
x x x
x x
5
10−
 
 
7 7 22.28 10 (2.28 10 ) (4)(4)( 3.0 10 )
(2)(4)
− −− × ± × − − ×=x
10−
 
 
7 5
6 62.28 10 6.93 10 8.6 10 or 8.7 10
8
− −
− −− × ± ×= = ×x − × 
 The negative root can be discarded because it has no physical meaning. 
 [NO] 2= =x 6 52(8.6 10 ) 1.7 10 mol L ;1− −× = × ⋅ − the concentrations of N2 
 and O2 remain essentially unchanged at 10.001 39 mol L−⋅ and 
 , respectively. 10.0214 mol L−⋅
 
9.52 The initial concentrations of N2 and H2 are 
 12 2
0.20 mol[N ] [H ] 0.0080 mol L .
25.0 L
−= = = ⋅ 
 At equilibrium, 5.0 % of the N2 had reacted, so 95.0 % of the N2 remains: 
 1 12[N ] (0.950)(0.0080 mol L ) 0.0076 mol L
− −= ⋅ = ⋅ 
 If 5.0 % reacted, then 
 32 3
2
2 mol NH
0.050 0.200 mol N 0.020 mol NH
mol N
× × = formed. 
 The concentration of 4 13
0.020 molNH formed 8.0 10 mol L .
25.0 L
− −= = × ⋅ 
 The amount of 
 22 2
2
3 mol H
H reacted 0.050 0.200 mol N 0.030 mol H
mol N
= × × = 2 used. 
261 
 Concentration of H2 pr
 
esent at equilibrium 
10.200 mol 0.030 mol 0.0068 mol L .
25.0 L
−−= = ⋅
 
2 4 2
3[NH ] (8.0 10 ) 2.7
−×= = =K 23 3
2 2
10
[N ][H ] (0.0076)(0.0068)
×C 
 
Note: The volume of the system is not used because we are given 
 pressures and K. 
Pressures (bar) N2(g) + 3 H2(g) 2 NH3(g) 
9.54 
 U
 initial 0.025 0.015 0 
 change −x 3− x 2+ x 
 final 0.025 − x 0.015 3− x 2+ x 
 3NH 3
N H .0
= =K
P P
2 2
3 0.036(0 25 )(0.015 3 )
=− −
x
x x
 
explicitly will lead to a high-order equation, so first check to 
see if the assumption that 
2 2(2 )P
 Solving this 
 can be use plify the math: d to sim3 0.015<<x
2
3
2 9
0.036
(0.025)(0.015)
4 3.04 10−
=
= ×x 
5
(2 )
2.8 10−= ×
x
x
 Comparing x to 0.015, we see that the approximation was justified. 
At equilibrium, ; the pressures of 
 N2 and H2 remain essentially unchanged. 
 
3
5 5
NH 2 2.8 10 bar 5.6 10 bar
− −= × × = ×P 
262 
9.56 Concentrations 
 CH1(mol L )−⋅ 3COOH + C2H5OH U CH3COOC2H5 + H2O 
 initial 0.024 0.059 0 0.015 
 change −x −x +x +x 
 final 0.024 − x 0.059 − x +x 0.015 + x 
 
3 2 5 2
3 2 5
2
2
2
2
2 2
2
2
[CH COOC H ][H O] ( )(0.015 )
[CH COOH][C H OH] (0.024 )(0.059 )
0.015
0.083 0.0014
0.0154.0
0.083 0.001 42
4.0 0.332 0.005 68 0.015
3.0 0.347 0.005 68 0
( 0.347) ( 0.347) (4
+= = − −
+= − +
+= − +
− + = +
− + =
− − ± − −=
C
x xK
x x
x x
x x
x x
x x
x x x x
x x
x
)(3.0)(0.005 68) 0.347 0.228
(2)(3.0) 6.0
0.0958 or 0.0198
+ ±=
=x
 
 The root 0.0958 is meaningless because it is larger than the initial 
 concentration of acetic acid and ethanol, so the value 0.0198 is chosen. 
 The equilibrium concentration of the product ester is, therefore, 
 . The numbers can be confirmed by placing them into the 
 equilibrium expression: 
10.0198 mol L−⋅
 3 2 5 2
3 2 5
[CH COOC H ][H O] (0.0198)(0.015 0.0198) 4.1
[CH COOH][C H OH] (0.024 0.0198)(0.059 0.0198)
+= = − −CK = 
 This is reasonably good agreement, given the nature of the calculation. 
 Given that the KC value is reported to only two significant figures, the best 
 report of the concentration of ester will be 0.020 1mol L−⋅ . 
 
9.58 5
3 2
PCl
PCl Cl
= PK
P P
 
 
2
2
4
Cl
1.3 103.5 10
(9.56)
×× =
P
 
263 
 
2
2
4
Cl 4
1.3 10 3.9 10 bar
(9.56)(3.5 10 )
−×= = ××P 
 
9.60 We use the reaction stoichiometry to calculate the amounts of substances 
 present at equilibrium: 
 Amounts (mol) CO(g) + H2O(g) COU 2(g) + H2(g) 
 initial 1.000 1.000 0 0 
 change −x −x +x +x 
 final 1.000 − x 1.000 − x 0.665 +x 
 (a) Because x = 0.665 mol, there will be (1.000 0.665− ) mol = 0.335 mol 
 CO; 0.335 mol H2O; 0.665 mol H2. The concentrations are easy to 
 calculate because V = 10.00 L: 
 1 12 2 2[CO] [H O] 0.0335 mol L ; [CO ] [H ] 0.0665 mol L
− −= = ⋅ = = ⋅ 
 (b) 
2
2 2
2
2
[CO ][H ] (0.0665) 3.94
[CO][H O] (0.0335)
= = =CK 
 
9.62 The initial concentration of 12
0.100 molH S 0.0100 mol L
10.0 L
−= = ⋅ 
 The final concentration of 12
0.0285 molH 0.002 85 mol L
10.0 L
−= = ⋅ 
 Concentrations 2 H1(mol L )−⋅ 2S(g) U 2 H2(g) + S2(g) 
 initial 0.0100 0 0 
 change 2− x 2+ x +x 
 final 0.0100 2− x 0.002 85 +x 
 Thus, 1 12 0.002 85 mol L or 0.001 42 mol L− −= ⋅ = ⋅x x 
 At equilibrium: 
264 
 
1 1
2
1
2
1
2
2
4
2
[H S] 0.0100 mol L 0.002 85 mol L 0.0072 mol L
[H ] 0.002 85 mol L
[S ] 0.001 42 mol L
(0.002 85) (0.001 42) 2.22 10
(0.0072)
1− − −
−
−
−
= ⋅ − ⋅ = ⋅
= ⋅
= ⋅
= = ×CK
 
 
9.64 The initial concentration of 15
0.865 molPCl 1.73 mol L
0.500 L
−= = ⋅ 
 Concentrations (m L PCl1ol )−⋅ 5(g) PClU 3(g) + Cl2(g) 
 initial 1.73 0 0 
 change −x +x +x 
 final 1.73 − x +x +x 
 
3 2
5
2
2
[PCl ][Cl ]
[PCl ]
( )( )1.80
(1.73 ) (1.73 )
(1.80)(1.73 )
=
= =− −
− =
CK
x x x
x x
x x
 
 
2
2
1.80 3.114 0
1.80 (1.80) (4)(1 3.114) 1.80 3.96
(2)(1) 2
1.08 or 2.88
+ − =
− ± − − − ±= =
= + −
x x
x
x
 
 The negative root is not physically meaningful and can be discarded. The 
 concentrations of PCl3 and Cl2 are, therefore, 1.08 1mol L−⋅ at equilibrium, 
 and the concentration of PCl5 is 1.73 1mol L−⋅ 11.08 mol L−− ⋅ 
 . These numbers can be checked by substituting back into 
 the equilibrium expression: 
10.65 mol L−= ⋅
 
3 2
5
2
[PCl
][Cl ]
[PCl ]
(1.08) 1.79
(0.65)
=
=
CK
 
 which compares well to KC (1.80). 
265 
9.66 We use the ideal gas relationship to find the initial concentration of HCl(g) 
 at 25 C:°
 
1 1
1 atm1.00 bar (4.00 L)
1.013 25 bar
0.176 mol
(0.082 06 L atm K mol )(273 K)− −
=
⎛ ⎞×⎜ ⎟⎝ ⎠= =⋅ ⋅ ⋅
PVn
RT
n
 
 10.176 mol[HCl] 0.0147 mol L
12.00 L
−= = ⋅ 
 2 HCl(g) + I1Concentrations (mol L )−⋅ 2(s) 2 HI(g) + ClU 2(g) 
 initial 0.0147 — 0 0 
 change 2− x — 2+ x +x 
 final 0.147 2− x — 2+ x +x 
 
2
2
2
2
34
2
[HI] [Cl ]
[HCl]
(2 ) ( )1.6 10
(0.0147 2 )
−
=
× = −
CK
x x
x
 
 Because the equilibrium constant is very small, we will assume that x << 
 0.0147: 
 
3
34
2
41.6 10
(0.0147)
−× = x 
 
34 2
3 (1.6 10 )(0.0147)
4
−×=x 
 
34 2
133
(1.6 10 )(0.0147) 2.1 10
4
−
−×= =x × 
 At equilibrium: 
 13 13[HI] 2 2.1 10 4.2 10− −= × × = ×
 132[Cl ] 2.1 10
−= ×
 1[HCl] 0.0147 mol L−= ⋅ 
 
266 
9.68 initial [NH3] 
1
1
25.6 g
17.03 g mol
0.301 mol L
5.00 L
−
−
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ⋅ 
 2 NH1Concentrations (mol L )−⋅ 3(g) U N2(g) + 3 H2(g) 
 initial 0.301 0 0 
 change 2− x +x 3+ x
 final 0.301 2− x +x 3+ x
 
3
2 2
2
3
3
2
4
2
[N ][H ]
[NH ]
( )(3 ) 0.395
(0.301 2 )
27 0.395
(0.301 2 )
=
=−
=−
CK
x x
x
x
x
 
 
4
2
2
27 0.395
(0.301 2 )
3 3 0.628
(0.301 2 )
=−
=−
x
x
x
x
 
 
2
2
2
3 3 (0.628)(0.301 2 ) 0.189 1.26
5.20 1.26 0.189 0
1.26 (1.26) (4)(5.20)( 0.189) 1.26 2.34
(2)(5.20) 10.4
0.105 or 0.396
= − = −
+ − =
− ± − − − ±= =
= + −
x x x
x x
x
x
 
 The negative root is discarded because it is not physically meaningful. 
 Thus, at equilibrium we should have 
 1 12 2 3[N ] 0.105 mol L ; [H ] 0.315 mol L ; [NH ] 0.091 mol L .
− −= ⋅ = ⋅ = 1−⋅
 
9.70 2 HCl(g) → + 2H (g) 2Cl (g)
 2 2H Cl 342
HCl
3.2 10−= = ×P PK
P
 (Eq. 1) 
 At equilibrium, . (Eq. 2) 
2 2H Cl HCl
3.0 bar= + + =TotalP P P P
267 
 And, since the atomic ratio , Cl:H is 1:3
 2
2
HCl Cl
HCl H
2moles Cl 1
moles H 3 2
+= = +
n n
n n
 
 
2 2
2 2
2 2
HCl H HCl Cl
H HCl Cl
H HCl Cl
2 3( 2 )
3
 at , constant so
3 (Eq. 3)
+ = +
= +
∝
= +
gas gas
n n n n
n n n
P n T V
P P P
 We now have three equations in three unknowns, so we can rearrange and 
 substitute to find each partial pressure. 
 Substituting equation 3 into equation 2 gives 
 
2Cl HCl
4 2 3.0 b= + =TotalP P P ar
 
2HCl Cl
3.0 bar 2
2
= −P P 
 Substitution back into equation 3 gives 
 
2 2H C
3.0 bar 
2
= +P P l 
 Using these two expressions for partial pressures in equation 1 gives 
 
2 2
2
Cl Cl
34
2
Cl
3.0 
2 3.2 10
3.0 2
2
−
⎛ ⎞+ ⋅⎜ ⎟⎝ ⎠= =
⎛ ⎞−⎜ ⎟⎝ ⎠
P P
K
P
× 
 Since the equilibrium constant is so small, we would expect the partial 
 pressure of chlorine to be very low at equilibrium, so we can neglect it to 
 simplify this expression. 
 
2
2
2
Cl
34
2
34 34
Cl
HCl H
3.0 
2 3.2 10
3.0 
2
(1.5)(3.2 10 ) 4.8 10 bar
1.5 bar
−
− −
⎛ ⎞ ⋅⎜ ⎟⎝ ⎠≈ = ×
⎛ ⎞⎜ ⎟⎝ ⎠
= × = ×
= =
P
K
P
P P
 
 
268 
9.72 (a) According to Le Chatelier’s principle, an increase in the partial 
 pressure will shift the equilibrium to the left, increasing the partial 
 pressure of . 
2CO
4CH
 (b) According to Le Chatelier’s principle, a decrease in the partial 
 pressure of will shift the equilibrium to the left, decreasing the 
 partial pressure of CO
4CH
2. 
 (c) The equilibrium constant for the reaction is unchanged, because it is 
 unaffected by any change in concentration. 
 (d) According to Le Chatelier’s principle, a decrease in the concentration 
 of H2O will shift the equilibrium to the right, increasing the concentration 
 of CO2. 
 
9.74 The questions can all be answered qualitatively using Le Chatelier’s 
 principle: 
 (a) Adding a reactant will promote the formation of products; the amount 
 of HI should increase. 
 (b) I2 is solid and already present in excess in the original equilibrium. 
 Adding more solid will not affect the equilibrium so the amount of Cl2 will 
 not change. 
 (c) Removing a product will shift the reaction toward the formation of 
 more products; the amount of Cl2 should increase. 
 (d) Removing a product will shift the reaction toward the formation of 
 more products; the amount of HCl should decrease. 
 (e) The equilibrium constant will be unaffected by changes in the 
 concentrations of any of the species present. 
 (f) Removing the reactant HCl will cause the reaction to shift toward the 
 production of more reactants; the amount of I2 should increase. 
 (g) As in (e), the equilibrium constant will be unaffected by the changes 
 to the system. 
 
269 
9.76 Increasing the total pressure on the system by decreasing its volume will 
 shift the equilibrium toward the side of the reaction with fewer numbers of 
 moles of gaseous components. If the total number of moles of gas is the 
 same on the product and reactant sides of the balanced chemical equation, 
 then changing the pressure will have little or no effect on the equilibrium 
 distribution of species present. (a) decrease in the amount of NO2; 
 (b) increase in the amount and concentration of NO; (c) decrease in the 
 amount of HI; (d) decrease in the amount of SO2; (e) increase in the 
 amount and concentration of NO2. (Note that it is not possible to predict 
 whether or not the concentration will increase, decrease, or remain the 
 same in cases where the amount of the substance is decreasing. More 
 specific information about the extent of the changes in the amount and the 
 volume is needed in those cases.) 
 
9.78 (a) The partial pressure of SO3 will decrease when the partial pressure of 
 SO2 is decreased. According to Le Chatelier’s principle, a decrease in the 
 partial pressure of a reactant will shift the equilibrium to the left, 
 decreasing the partial pressure of the products, in this case SO3. 
 (b) When the partial pressure of SO2 increases, the partial pressure of O2 
 will decrease. According to Le Chaltelier’s principle, an increase in the 
 partial pressure of a reactant shifts the equilibrium toward products, 
 decreasing the partial pressure of the other reactant, O2. 
 
9.80 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 reactions given, raising 
 the temperature should favor products in (a) and reactants in (b) and (c). 
 
9.82 Even though numbers are given, we do not need to do a calculation to 
 answer this qualitative question. Because the equilibrium constant is larger 
 at lower temperatures 24 10(4.0 10 at 298 K vs. 2.5 10× × at 500 K, see
270 
 , more products will be present at the lower 
 temperature. Thus we expect more SO
Table 9.2, page 334)
3 to be present at 25ºC than at 500 
 K, assuming that no other changes occur to the system (the volume is 
 fixed and no reactants or products are added or removed from the vessel). 
 
9.84 To answer this question, we must calculate Q: 
 
2 2
2 2
[ClF] (0.92) 15
[Cl ][F ] (0.18)(0.31)
= =Q = 
 Because Q ≠ K, the system is not at equilibrium, and because Q < K, the 
 reaction will proceed to produce more products so ClF will tend to form. 
 
9.86 (a) 1 22HgO(s) Hg(l)
O (g)+U 
 
r f
1
r
1
r
1
r 22
1 1 1 11
r 2
1 1
1 1
r
[ (HgO, s)]
[ 90.83 kJ mol ]
90.83 kJ mol
(Hg, l) (O , g) [ (HgO, s)]
76.02 J K mol 205.14 J K mol
 [70.29 J K mol ]
108.30 J K mol
−
−
− − − −
− −
− −
∆ ° = − ∆ °
∆ ° = − − ⋅
∆ ° = + ⋅
∆ ° = ° + × ° − °
∆ ° = ⋅ ⋅ + × ⋅ ⋅
− ⋅ ⋅
∆ ° = ⋅ ⋅
H H
H
H
S S S S
S
S
 
 At 298 K: 
 
1 1
r(298 K)
1
r(298 K)
90.83 kJ (298 K)(108.30 J K )/(1000 J kJ )
 58.56 kJ mol
ln
− −
−
∆ ° = − ⋅ ⋅
= ⋅
∆ ° = −
G
G RT K
 
r(298 K)
1
ln
58 560 J 23.6
(8.314 J K )(298 K)−
∆ °= −
= − = −⋅
G
K
RT 
 116 10−= ×K
 At 373 K: 
 
1 1
r(373 K)
1
90.83 kJ (373 K)(108.3 J K )/(1000 J kJ )
50.4 kJ mol
− −
−
∆ ° = − ⋅ ⋅
= ⋅
G
271 
 1
8
50 400 Jln 16.3
(8.314 J K )(373 K)
8 10
−
−
= − = −⋅
= ×
K
K
 
 (b) propene (C3H6, g) → cyclopropane (C3H6, g) 
 
r f f
1 1
r
1
r
r
1 1 1 1
r
1 1
r
(cyclopropane, g) [ (propene, g)]
53.30 kJ mol [(20.42 kJ mol )]
32.88 kJ mol
(cyclopropane, g) [ (propene, g)]
237.4 J K mol [266.6 J K mol ]
29.2 J K mol
− −
−
− − − −
− −
∆ ° = ∆ ° − ∆ °
∆ ° = ⋅ − ⋅
∆ ° = + ⋅
∆ ° = ° − °
∆ ° = ⋅ ⋅ − ⋅ ⋅
∆ ° = − ⋅ ⋅
H H H
H
H
S S S
S
S
 At 298 K: 
 
1
r(298 K)
1 1
1
r(298 K)
32.88 kJ mol
(298 K)( 29.2 J K mol )/(1000 J kJ )
41.58 kJ mol
ln
−
− − −
−
∆ ° = ⋅
− − ⋅ ⋅ ⋅
= ⋅
∆ ° = −
G
G RT K
1
 
r(298 K)
1
ln
41 580 J 16.8
(8.314 J K )(298 K)−
∆ °= −
= − = −⋅
G
K
RT 
 85 10−= ×K
 At 373 K: 
 
1
r(373 K)
1 1
1
32.88 kJ mol
 (373 K)( 29.2 J K mol )/(1000 J kJ )
43.8 kJ mol
−
− − −
−
∆ ° = ⋅
− − ⋅ ⋅ ⋅
= ⋅
G
1
 1
7
43 800 Jln 14.1
(8.314 J K )(373 K)
7 10
−
−
= − = −⋅
= ×
K
K
 
 
9.88 We can think of the vaporization of a liquid as an equilibrium reaction, 
 e.g., for substance A, A( ) A( )Ul g . This reaction has an equilibrium 
 constant expression A=K P . Since is the pressure of the gas above a AP
272 
 liquid, it is the equilibrium vapor pressure of that liquid. Therefore, the 
 vapor pressure varies with temperature in the same way that this 
 equilibrium constant varies with temperature. 
 The van’t Hoff equation, 2
1 1
1 1ln
⎛ ⎞∆= −⎜⎝ ⎠
D
rK H
K R T T2
⎟ , (see 16, page 354), 
 gives the temperature dependence of K. To make this equation fit the 
 special case of vapor pressure all we need to do is substitute for K 
 and ∆ for∆
P
D
vapH
D
rH . 
 2
1 1
1 1ln
∆ ⎛ ⎞= −⎜ ⎟⎝ ⎠
D
vapHP
P R T T2
 
 This relationship is known as the Clausius-Clapeyron equation (see 
 equation 4, page 285). Since 0∆ >DvapH , vapor pressure increases with 
 temperature. 
 
9.90 Recall ln and 10−∆ = − = ∆ − ∆ =D D D pKr r rG RT K H T S K
 and 2
1 1
1 1ln
⎛ ⎞∆= −⎜⎝ ⎠
D
rK H
K R T T2
⎟ , the van’t Hoff equation (see equation 16, 
 page 354). 
 Rearranging and plugging in values, 
 
13.8330 1 1
2
15.136
1
1 2
4 1
1
10 8.314 J K molln ln
1 1101 1
293 K 303 K
(3.000)(7.381 10 J mol )
221.4 kJ mol
− −
−
−
−
⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞ ⋅ ⋅∆ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠ ⎜ ⎟−⎜ ⎟−⎜ ⎟ ⎝ ⎠⎝ ⎠
= × ⋅
= ⋅
D
r
K RH
K
T T
−
 
 
3 1
1 1 15.136
1 1
ln
221.4 10 J mol(8.314 J K mol ) ln(10 )
293 K
465.9 J K mol
−
− − −
− −
∆∆ = +
× ⋅= ⋅ ⋅ +
= ⋅ ⋅
D
D r
r
HS R K
T
 
273 
 The sign on is positive because of the randomization that occurs as 
 deuterons dissociate and redistribute throughout the liquid. 
∆ DrS
 From the results above, the “autoprotolysis” constant of heavy water at 
 25°C is 
 
2
1 1 2
3 1
15.136
2 1 1
1 1ln
221.4 10 J mol 1 1ln ln10
293 K 298 K8.314 J K mol
33.327
−
−
− −
⎛ ⎞∆= −⎜ ⎟⎝ ⎠
⎛ ⎞× ⋅ ⎛ ⎞= −⎜ ⎟⎜ ⎟⋅ ⋅ ⎝ ⎠⎝ ⎠
= −
D
rK H
K R T T
K + 
 15D, 298 K D, 298 K3.359 10 or 14.474
−= × =K pK
 Since for regular water < H, 298 K 14.00=pK D, 298 K 14.474=pK , 
 substitution of D for H causes reactants, i.e., the associated molecules, to 
 be favored more in heavy water than in regular water even though we 
 expect D and H to be chemically equivalent. We can suggest that this 
 observation is another example of the kinetic isotope effect where the 
 zero-point energy of the O-D bond is lower than that of the O-H bond (see 
 the solutions to Exercises 17.55 and 17.85 for further discussion). If bond 
 dissociation is the rate-limiting step in the mechanism, and if only the 
 vibrational energy of the ground state of the water molecules is changed 
 appreciably, then the activation barrier for dissociation is bigger in heavy 
 water since the zero-point energy is lower due to the heavier mass of D. 
 This increase in the activation energy for bond dissociation causes the rate 
 of dissociation to decrease while the rate of association is not changed 
 appreciably. The position of equilibrium then favors the associated 
 molecules more than it does in the molecules. 
2D O
2H O
 
9.92 (a) r f 3
1
1
[2 (O , g)]
[2(142.7 kJ mol )]
285.4 kJ mol
−
−
∆ ° = − ∆ °
= − ⋅
= − ⋅
H H
274 
 
r 2 3
1 1 1 1
1 1
3 (O , g) [2 (O , g)]
3(205.14 J K mol ) [2(238.93 J K mol )]
137.56 J K mol
− − − −
− −
∆ ° = ° − °
= ⋅ ⋅ − ⋅ ⋅
= ⋅ ⋅
S S S
 At 298 K: 
 
1
r(298 K)
1 1
1
285.4 kJ mol
 (298 K)(137.56 J K mol )/(1000 J kJ )
326.4 kJ mol
−
− − −
−
∆ ° = − ⋅
− ⋅ ⋅
= − ⋅
G
1⋅
°G or directly from ∆ values f
 
r f 3
1
1
2 (O , g)
2(163.2 kJ mol )
326.4 kJ mol
−
−
∆ ° = − ∆ °
= − ⋅
= − ⋅
G G
 (b) r(298 K)
r(298 K)
1
57
ln
ln
326 390 J 132
(8.314 J K )(298 K)
10
−
∆ ° = −
∆ °= −
−= − = +⋅
=
G RT K
G
K
RT
K
 The equilibrium constant for the decomposition of ozone to oxygen is 
 extremely large, making this process extremely favorable. Note: The 
 presence of ozone in the upper atmosphere is largely due to kinetic factors 
 that inhibit the decomposition; however, in the presence of a suitable 
 catalyst, this process should occur extremely readily. 
 
9.94 
1 1
1
2 4 2
1
2.50 g 0.330 g
92.02 g mol 46.01 g mol
[N O ] 0.0136 mol L ; [NO ]
2.00 L 2.00 L
0.003 59 mol L
− −
−
−
⎛ ⎞ ⎛⎜ ⎟ ⎜⋅ ⋅⎝ ⎠ ⎝= = ⋅ =
= ⋅
⎞⎟⎠ 
 N1Concentration (mol L )−⋅ 2O4(g) 2 NOU 2(g) 
 initial 0.0136 0.003 59 
 change −x 2+ x 
 final 0.0136 − x 0.003 59 2+ x 
275 
 
2
2
2 4
2
3
3 2
5 3 2 2
2 2 5
[NO ]
[N O ]
(0.003 59 2 )4.66 10
0.0136
(4.66 10 )(0.0136 ) (0.003 59 2 )
6.34 10 4.66 10 4 1.44 10 1.29 10
4 1.91 10 5.05 10 0
−
−
− − −
− −
=
+× = −
× − = +
× − × = + × + ×
+ × − × =
K
x
x
x x
x x x
x x
�
5−
1
 
 Solving by using the quadratic equation gives 31.89 10 .−= ×x
 
1 3 1 1
2 4
1 3 1
2
[N O ] 0.0136 mol L 1.89 10 mol L 0.012 mol L
[NO ] 0.003 59 mol L 2 (1.89 10 mol L ) 0.0074 mol L
− − − −
− − −
= ⋅ − × ⋅ = ⋅
= ⋅ + × ⋅ = −⋅ 
 These numbers can be checked by substituting them into the equilibrium 
 expression: 
 
2
3?
3 3
(0.0074) 4.66 10
0.012
4.6 10 4.66 10
−
− −√
= ×
× = ×
 
 
9.96 We can write an equilibrium expression for the interconversion of each 
 pair of isomers: 
 2-methyl propene U cis-2-butene K1
 2-methyl propene U trans-2-butene K2
 cis-2-butene trans-2-butene KU 3 
 We need to calculate only two of these values in order to determine the 
 ratios. 
 The K’s can be obtained from the r∆ °G ’s for each interconversion: 
 
1 1
1
1 1
2
1 1
3
65.86 kJ mol 58.07 kJ mol 7.79 kJ mol
62.97 kJ mol 58.07 kJ mol 4.90 kJ mol
62.97 kJ mol 65.86 kJ mol 2.89 kJ mol
− −
− −
1
1
1
−
−
− − −
∆ ° = ⋅ − ⋅ = ⋅
∆ ° = ⋅ − ⋅ = ⋅
∆ ° = ⋅ − ⋅ = − ⋅
G
G
G
 
276 
 
r
r
1
1
1 11 1
1
2
2 21 1
1
3
3 1 1
ln ; ln
7790 J molln 3.14; 0.043
(8.314 J K mol )(298 K)
4900 J molln 1.98; 0.14
(8.314 J K mol )(298 K)
2890 J molln
(8.314 J K mol )(298 K)
−
− −
−
− −
−
− −
∆ °∆ ° = − = −
∆ ° ⋅= = − = − =− ⋅ ⋅
∆ ° ⋅= = − = − =− ⋅ ⋅
∆ ° − ⋅= = −− ⋅ ⋅
G
G RT K K
RT
GK K
RT
G
K K
RT
G
K
RT 3
1.17; 3.2= + =K
 
 Each K gives a ratio of two of the isomers: 
 1 2 3
[ -2-butene] [ -2-butene] [ -2-butene]; ;
[2-methylpropene] [2-methylpropene] [ -2-butene]
= = =cis trans transK K K
cis
 
 We will choose to use the first two K’s to calculate the final answer: 
 [ -2-butene] 0.043
[2-methylpropene]
=cis [ -2-butene] 0.14
[2-methylpropene]
=trans 
 If we let the number of moles of 2-methylpropene = 1, then there will be 
 0.14 mol trans-2-butene and 0.043 mol cis-2-butene. The total number of 
 moles will be 1 mol + 0.14 mol + 0.043 mol = 1.18 mol. The percentage 
 of each will be 
 
1 mol% 2-methylpropene 100% 85%
1.18 mol
0.043 mol% -2-butene 100% 4%
1.18 mol
0.14 mol% -2-butene 100% 12%
1.18 mol
= × =
= × =
= × =
cis
trans
 
 (The sum of these should be 100% but varies by 1%, due to limitations in 
 the use of significant figures/rounding conventions.) 
 The relative amounts are what we would expect, based upon the free 
 energies of formation (the more negative or less positive value 
 corresponding to the thermodynamically more stable compound), which 
 indicate that the most stable compound is the 2-methylpropene followed 
 by trans-2-butene with cis-2-butene being the least stable isomer. We can 
 use K3 as a check of these numbers: 
277 
 
?
3
0.14
[ -2-butene]3.2
0.043[ -2-butene]
3.2 3.3
= = =
≅
trans VK
cis
V
 
 
9.98 A 2 B + 3 C U
 initial 10.00 atm 0 atm 0 atm 
 change − +2 x +3 x x
 final 10.00 − x +2 x +3 x 
 The equilibrium expression is 
2 3×= B C
A
P P
K
P
 
 We can also write total 10.00 2 3 10.00 4 15.76= − + + = + =P x x x x 
 x = 1.44 atm 
 PA = 8.56 atm; PB = 2.88 atm; PC = 4.32 atm 
 
2 3
1 1
1
(2.88) (4.32) 78.1
8.56
ln (8.314 J K mol )(298 K)ln(78.1)
10.8 kJ mol
− −
−
= =
∆ ° = − = − ⋅ ⋅
= − ⋅
K
G RT K 
 
9.100 (a) First we calculate the initial pressure of the gas at 127ºC using the 
 ideal gas relationship: 
 
=
=
PV nRT
mPV RT
M
 
 
1 1
1
(10.00 g)(0.082 06 L atm K mol )(298 K)
(17.03 g mol )(4.00 L)
3.59 atm
− −
−
=
⋅ ⋅ ⋅= ⋅
=
mRTP
MV
 
278 
 
1 2
1 2
23.59 atm
298 K 400 K
=
=
P P
T T
P
 
 = 4.82 atm or 4.88 bar (1 atm = 1.01325 bar) 2P
 From Table 9.1 we find that the equilibrium constant is 41 for the equation 
 2 2 3N (g) 3 H (g) 2 NH (g)+ U z
 Although this exercise technically begins with pure ammonia and allows it 
 to dissociate into N2(g) and H2(g), we can use the same expression. 
 pressure (bar) + 3 H2N (g) 2(g) U 2 NH3(g) 
 initial 0 0 4.88 
 change +x 3+ x 2− x
 final +x 3+ x 4.88 2− x 
 
2
3
2
4
2
2
2
(4.88 2 ) 41
( )(3 )
(4.88 2 ) 41
27
4.88 2 41
3 3
4.88 2 3 123
33.3 2 4.88 0
0.354
− =
− =
− =
− =
+ − =
=
x
x x
x
x
x
x
x x
x x
x
 
 = 4.17 bar or 4.11 atm; 
3NH
P
2N
0.354=P bar or 0.349 atm; bar 
 or 1.05 atm 
2H
1.06=P
 Ptotal = 4.11 atm + 0.349 atm + 1.05 atm = 5.51 atm 
 (b) The equilibrium constant can be calculated from ∆ °G at 400 K, 
 which can be obtained from ∆ °H and ∆ °S . The values are 
279 
 
1 1
1 1 1 1
1 1
1 1
1 1
1 1
2( 46.11 kJ mol ) 92.22 kJ mol
2(192.45 J K mol ) [(191.61 J K mol )
 3(130.68 J K mol )]
198.75 J K mol
( 92.22 kJ mol )(1000 J K )
 (400 K)( 198.75 J K mol )
12.72 kJ
− −
− − − −
− −
− −
− −
− −
∆ ° = − ⋅ = − ⋅
∆ ° = ⋅ ⋅ − ⋅ ⋅
+ ⋅ ⋅
= − ⋅ ⋅
∆ ° = − ⋅ ⋅
− − ⋅ ⋅
= − ⋅
H
S
G
1 1mol 12720 J mol− −= − ⋅
 
1 1 1/ 12720 J mol / (8.314 J K mol )(400K)e e
− − −−∆ ° + ⋅ ⋅ ⋅= = =G RTK 46
 This is reasonably good agreement considering the logarithmic nature of 
 these calculations. For comparison, the ∆ °G value calculated from K = 41 
 at 400 K is 12.34 1kJ mol .−⋅ 
 
9.102 (a) pressure (bar) 2 AsH3(g) U 2 As(s) + 3 H2(g) 
 initial 0.52
 change 2− x 3+ x 
 final 0.52 2− x 3+ x 
 At equilibrium, total pressure = 0.64 bar. 
 
0.64 0.52 2 3 0.52
0.12 bar
= − + = +
=
x x x
x
 
 3
2
AsH 0.52 bar 2(0.12 bar) = 0.28 bar
3(0.12 bar) = 0.36 bar
= −
=H
P
P
 (b) Find the number of moles of that decomposed and relate it to 
 the mass of As. 
3AsH
 
1 1
3
3
3
3
3
(0.24 bar)(1.00 L)
(0.0831 L bar K mol )(298 K)
9.69 10 mol AsH decomposed
2 mol As 74.9 g Asmass As = (9.69 10 mol AsH )
2 mol AsH 1 mol As
0.73 g As
− −
−
−
= = ⋅ ⋅ ⋅
= ×
⎛ ⎞⎛ ⎞× ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=
PVn
RT
 
280 
 (c) 2
3
3 3
H
2 2
AsH
(0.36)K 0
(0.28)
= = =P
P
.60 
 
9.104 The free energy change for the reaction is calculated from the equation 
 ln ln ln∆ = ∆ ° + = − +G G RT Q RT K RT Q
 In order to determine the range of Cl2 pressures that we need to examine, 
 we will first calculate the equilibrium pressures of the gases present. 
 Br2(g) + Cl2(g) 2 BrCl(g) U
 initial 1.00 bar 1.00 bar 0 
 change − x −x 2 x 
 final 1.00 − x 1.00 − x 2 x 
 
2 2
2 2
BrCl
2
Br Cl
(2 )0.2
(1.00 )
20.2
1.00
0.183 0.2
= = −
= −
= ≈
P x
P P x
x
x
x
 
 In other words, equilibrium will be reached when BrCl 2 0.36 bar= =P x , at 
 the point on the graph where 0.∆ =G 
-700
-600
-500
-400
-300
-200
-100
0
100
0 0.1 0.2 0.3 0.4 0.5
de
lta
 G
 (k
J 
pe
r m
ol
)
Pressure of BrCl (bar) 
281 
 Notice that the free energy change for the reaction is most negative the 
 farther the system is from equilibrium and that the value approaches 0 as 
 equilibrium is attained. 
 
9.106 The graph is generated for ln K vs. 1/T according to the relationship 
 1ln ∆ ° ∆ °= − ⋅ +H SK
R T R
 
 where ∆ °− H
R
 is the slope and ∆ °S
R
 is the intercept. 
 
0
4
8
12
16
20
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
ln
 K
1/T (1/K)
y = -5398.4x +17.899
R = 0.99213
 
 
1
1 1
5398 K
44.9 kJ mol
17.9
149 J K mol
−
− −
∆ °− = −
∆ ° = ⋅
∆ ° =
∆ ° = ⋅ ⋅
H
R
H
S
R
S
 
 Because the reaction involves breaking only the N—N bond, the enthalpy 
 of reaction should be an approximation of the N—N bond strength. 
 Notice that this value is considerably less than the average 
 value for N—N bonds given in Table 2.2. Therefore it is a very weak 
1163 kJ mol−⋅
 N—N bond. 
282 
9.108 (a) Since we are only using two temperatures, it really is not necessary to 
 make graphs in order to determine ∆ °H and ∆ °S . We can easily 
 substitute into the van’t Hoff equation, 2
1 1
1 1ln
⎛ ⎞∆= −⎜⎝ ⎠
D
rK H
K R T T2
⎟ , for each 
 halogen to get ∆ °H , then use it to find ∆ °S from 
 . A spread sheet will work well. ln∆ ° = ∆ ° − ∆ = −G H T S RT K
 (b) & (c) Results from our spread sheet based on the relationships given 
 above and data from Table 9.2 are given below. The ∆ °S values are for 
 the reactions X2(g) 2 X(g), so we can calculate the
standard molar 
 entropies for the atomic species using data from Appendix 2A and the 
 relationship 
U
m m2 (X, g) (X , g∆ ° = ° − °S S S 2 ).
 
Halogen 
∆ °H 
1kJ mol−⋅
∆ °S 
1 1J K mol− −⋅ ⋅
2(X , g)°S 
1 1J K mol− −⋅ ⋅
(X, g)°S 
1 1J K mol− −⋅ ⋅ 
Fluorine 164 6 202.78 164 12
Chlorine 256 160 223.07 192 
Bromine 195 110 245.46 178 
Iodine 163 152 260.69 206 
 
.110 The general form of this type of equation is 
 
 
9
r r1ln
∆ ° ∆ °⎛ ⎞= − +⎜ ⎟⎝ ⎠
H S
K
R T R
 
 which is easily derived from the relationships 
r r . By inspection, we can see 
 that for the original expression to be valid, 
 
 ∆ ° = − ∆ ° = ∆ ° − ∆ °G RT K G H T Sr rln and
r
r
1 1 1
21 700, which gives
(21 700 K)(8.314 J K mol )/(1000 J kJ ) 180 kJ mol 1− − − −
− = − ∆ °
= + ⋅ ⋅ ⋅ = ⋅
H
R 
 
∆ °H
283 
9.112 (a) 
3 3PH BCl
1= ⋅K P P 
 (b) c
c 1 1
4
( )
19.2
( ) [(0.08314 L bar K mol )(333 K)]
1.48 10
∆
∆ − −
=
= = ⋅ ⋅ ⋅
= ×
n
n 2−
RT
KK
RT
K K 
 (c) The initial pressure of PH3 is: 
 3
1 1
PH
(0.0128 mol)(0.08314 L bar K mol )(333 K)
0.500 L
0.7088 bar
− −⋅ ⋅= =
=
nRTP
V 
 pressure (bar) PH3(g) + BCl3(g) U PH3BCl3(s) 
 initial 0.7088 0 
 change +x +x 
 equilibrium 0.7088+ x x
 
3 3PH BCl
1119.2
(0.7088 )
= = =⋅ +K P P x x
=
 
 2 0.7088 0.05208 0+ −x x
 
20.7088 (0.7088) 4(0.05208)
2
0.06712 (negative root isn't physically possible)
− ± +=
=
x 
 So 
3
0.70PH 88 0.06712 0.7759 bar= + =P 
 and 3
3
= =c PH 1PH 1 10.7759 bar 0.0280 mol L(0.08314 L bar K mol )(333 K)
−
− − = ⋅⋅ ⋅
P
RT
 (d) K increases as T increases, so the equilibrium position is shifting 
 toward products with the addition of heat. The reaction is endothermic. 
 (e) c
c 1 1
4
( )
26.2
( ) [(0.08314 L bar K mol )(343 K)]
2.13 10
∆
∆ − −
=
= = ⋅ ⋅ ⋅
= ×
n
n
K K RT
KK
RT 2−
 
284 
 (f) has a lone pair of electrons to donate in order to form an adduct 
 with , which can be an electron acceptor. Therefore is a Lewis 
 base while is a Lewis acid. 
3PH
3BCl 3PH
3BCl
 
9.114 (a) The values of the standard free energies of formation are 
 Compound 1f (kJ mol )
−∆ ° ⋅G
 BCl3(g) 387.969−
 BBr3(g) 236.914−
 BCl2Br(g) 338.417−
 BClBr2(g) 287.556−
 (b) A number of equilibrium reactions are possible including 
 2 BCl2Br(g) BClU 3(g) + BClBr2(g) 
 2 BClBr2(g) BBrU 3(g) + BCl2Br(g) 
 2 BCl3(g) + BBr3(g) U 3 BCl2Br(g) 
 BCl3(g) + 2 BBr3(g) U 3 BClBr2(g) 
 (c) BCl3 because it is the most stable as determined by its f∆ °G value. 
 (d) No. Although the equilibrium reaction given in the exercise would 
 require that these two partial pressures be equal, there are other equilibria 
 occurring as in (b) that will cause the values to be unequal. 
 (e) We will solve this problem graphically rather than by writing a 
 computer program to solve a system of simultaneous equations. Standard 
 free energies of formation are only used to get equilibrium constants for 
 two essential reactions from 
−∆
=
D
rG
RTK e and 
 . ∆ = ∑ ∆ −∑ ∆D Dr prods f reacts fG n G m GD
 In order to solve this part of the problem, first we need to figure out the 
 independent reactions between the four species. If we consider BCl3 and 
 BBr3 as pure species in terms of the halogens atoms, then the mixed 
 species BClBr2 and BCl2Br can be made from the pure species by these 
 two reactions: 
285 
 (1) BCl3 + 2 BBr3 U 3 BClBr2 1 871exp 1.4218.314 298.15
⎛ ⎞= =⎜ ⎟×⎝ ⎠K 
 (2) 2 BCl3 + BBr3 U 3 BCl2Br 2 2399exp 2.6328.314 298.15
⎛ ⎞= =⎜ ⎟×⎝ ⎠K 
 All other reactions between the species can be derived from these two 
 essential reactions, so there are only two independent reactions for the 
 system. 
 Let , and 
3BCl
=P x
3BBr
=P y then 2
3
BClBr
1 2=
P
K
xy
 and 2
3
BCl Br
2 2=
P
K
x y
 
 so , and 
2
1/ 3 1/ 3 2 / 3
BClBr 1=P K x y 2 1/ 3 2 / 3 1/ 3BCl Br 2=P K x y
 Atom balances are determined by the initial composition of the system. 
 Balancing B atoms we get 
 
3 3 3 2 2
o o
BCl BBr BCl BCl Br BClBr BBr+ = + + +P P P P P P 3
2
3
3
3=
3
 where are the initial pressures. Balancing Cl atoms 
 and Br atoms gives us 
3 3
o o
BCl BBr 1 bar= =P P
 
3 3 2
o
BCl BCl BCl Br BClBr3 3 2= + +P P P P
 
3 2 2
o
BBr BCl Br BClBr BBr3 2= + +P P P P
 (Notice that the B balance is not an independent condition because it is 
 implied by the Cl and Br balances.) In terms of x and y, the Cl balance and 
 Br balance can be written as 
 Cl balance: 1/ 3 2 / 3 1/ 3 1/ 3 1/ 3 2 / 32 13 2+ +x K x y K x y
 Br balance: 1/ 3 2 / 3 1/ 3 1/ 3 1/ 3 2 / 32 12 3+ + =K x y K x y y 
 Subtracting the second equation from the first equation, dividing the 
 resulting equation by y, and also using this new definition 
 
1/ 3⎛ ⎞ ≡⎜ ⎟⎝ ⎠
x z
y
 
 we arrive at following equation in terms of z: 
 3 1/ 3 2 1/ 32 13 3+ − −z K z K z 0=
286 
 3 23 1.381 1.124 3 0+ − −z z z =
 This cubic equation can be solved graphically by noting where 
 crosses 0. The only real root is 
 . 
3 2( ) 3 1.381 1.124 3= + − −f z z z z
0.975=z
-25
-20
-15
-10
-5
0
5
10
15
20
25
-3 -2 -1 0 1 2 3
z
 
 Substituting 3=x z y back into the Cl balance equation gives 
 
3 1/ 3 2 1/ 3
2 1
3 2
3
3 2
3
3(0.975) 2(1.381)(0.975) (1.124)(0.975)
0.461
= + +
= + +
=
y
z K z K z
 
 Then 3 3(0.975) (0.461) 0.428= = =x z y
 And finally: 
 ( )
2
1/ 32
BClBr (1.421)(0.428)(0.461) 0.506= =P
 ( )
2
1/ 32
BCl Br (2.632)(0.428) (0.461) 0.606= =P
 
287 
 The results can be summarized as: 
 bar 
3BCl
0.428=P
 bar 
3BBr
0.461=P
 bar 
2BCl Br
0.606=P
 bar 
2BClBr
0.506=P
 Note that in part (c) we predicted that BCl3 is the most abundant species in 
 the equilibrated mixture. However, this calculation reveals that it is the 
 least abundant species at equilibrium given the stated initial conditions. 
 The result is simply non-intuitive. What we need to recognize is that the 
 formation reactions and their corresponding values of free energy of 
 reaction describe equilibria between the elements B(s), Cl2(g), and Br2(l) 
 and the boron trihalides under standard conditions. In our mixture the 
 compounds are certainly not in equilibrium with their elements in their 
 standard states, so the prediction in part (c) fails. 
 Also note that in this system, B(s) and Br2(l) cannot exist at same time, 
 because the equilibrium constant of B(s) + 1.5 Br2(l) BBrU 3(g) is very 
 large, while the actual pressure of BBr3(g) is way below this equilibrium 
 constant. B(s) may exist from a microscopic amount of decomposition of 
 BCl3(g), so Br2(l) can not exist. This condition further prevents BCl2Br 
 and BClBr2 from decomposing into their elements, so only BCl3(g) may 
 decompose to a small extent. Clearly, this small amount of Cl2(g) is far 
 below standard conditions. 
 
9.116 (a) Find Q to see if the reaction is at equilibrium. 
 cyclohexane (C) methylcyclohexane (M) 
 [M] 0.100Q 5.00 K 0.140
[C] 0.0200
= = = > = 
 The reaction will proceed to the left to form more reactant. 
 
 
288 
 (b) [C] [M] 
 initial 0.0200 0.100 
 change +x −x 
 equil 0.0200 + x 0.100 − x 
 
2
2
[M] 0.100K 0.140
[C] 0.0200 +
0.100 (0.140)(0.0200 + )
1.140 9.72 10
8.53 10
−
−
−= = =
− =
= ×
= ×
x
x
x x
x
x
 
 [M] = 0.100 − 0.0853 = 0.0147 = 0.015 M 
 [C] = 0.0200
+ 0.0853 = 0.1053 M 
 (c) At equilibrium, [C] = 0.100 M. The total concentration of all species is 
 0.120 M. Therefore, [M] = 0.020 at equilibrium. 
 
50 C
[M] 0.100K 5
[C] 0.020
= = =D .0 
 (d) As the temperature increased, K also increased resulting in the 
 reaction forming more products: C + heat M. Therefore, the 
 reaction is endothermic. 
 
9.118 Refer to Graham’s Law of Effusion, equation 19, page 144, which states 
 that the ratio of the rates of effusion of two gases, both at the same 
 temperature, is equal to the square-root of the inverse ratio of their molar 
 masses. 
 
2
2 HI
H
1
1
Rate of effusion of H
Rate of effusion of HI
127.9 g mol
2.016 g mol
7.97
−
−
=
⋅= ⋅
=
M
M
 
 This treatment suggests that H2 effuses ~8 times faster than HI. However, 
 it is also important to note that the partial pressure of each gas affects the 
 collision frequency of the molecules with the pinhole. Therefore, an 
 expression that takes this dependence into account is 
289 
 2
2
H2 H
HI H
Rate of effusion of H
Rate of effusion of HI
= P M
P M
I 
 The partial pressure of hydrogen is 0.7 atm, or 0.69 bar. The partial 
 pressure of HI can be found from 
 2
2
2
HI
H
1/ 2
HI H (0.345 0.69) 0.49 bar
=
= ⋅ = × =
PK
P
P K P
 
 
2
2
1
H2 HI
1
HI H
Rate of effusion of H 0.69 bar 127.9 g mol
Rate of effusion of HI 0.49 bar 2.016 g mol
11
10 (to 1 sig fig)
−
−
⋅⎛ ⎞= = ⎜ ⎟ ⋅⎝ ⎠
=
=
P M
P M
 
 So hydrogen effuses ten times as fast as hydrogen iodide under these 
 conditions. 
290