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transcript
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
∆ rG
∆ 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
Pre
ssur
e (b
ar)
Time
[Cl2]
[PCl3]
[PCl5]
247
9.6 (a) 2
2
2NO2
NO O
=P
KP P
; (b) 3 2
5
SbCl Cl
SbCl
=P P
KP
; (c) 2 4
2 2
N H2
N H
=P
KP 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 xKx 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
− −= =
⎡ ⎤ ⎡ ⎤ ⎛ ⎞⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠
Cy 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 yx y
x yyx
248
Cross multiplying:
(0.05 )(0.10 )0.005 0.05 0.100.10 2
= − −= − − += −
xy x yxy yy 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
VT
9.10 3 2NH H S=K P P
For condition 1, 0.307 0.307 0.0942For condition 2, 0.364 0.258 0.0939For condition 3, 0.539 0.174 0.0938
= × == × == × =
KKK
9.12 (a) 2
33
[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 41 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 GG
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
−
− −
−
∆ °= −
+ ⋅= − = −
⋅ ⋅
= ×
GKRT
K
K
(b) CaC2(s) + 2 H2O(l) Ca(OH)→ 2(s) + C2H2(g)
r f 2 f 2 2
f 2 f 21 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 GG 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 G1−
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 GG G
3
1 1
7.20 10 Jln 2.91(8.314 J K mol )(298 K)
18
− −
− ×= − = +
⋅ ⋅=
K
K
9.16 (a) r1 1
1
ln K(8.314 J K mol )(700 K) ln 5423.2 kJ mol
− −
−
∆ ° = −
= − ⋅ ⋅
= − ⋅
G RT
(b) r1 1
1
ln K(8.314 J K mol )(298 K) ln 0.302.98 kJ mol
− −
−
∆ ° = −
= − ⋅ ⋅
= + ⋅
G RT
9.18 r f 21
(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 KGK
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
Clr
Cl Cl
1 1
1 1
1
lnln 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 QG RT K RT Q
pG 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
r2
r2 2
1 1
21 1
1
lnln 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 QG 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
2SO2
SO O
3.4P
= =P
KP
C12.027 K
∆⎛ ⎞
= ⎜ ⎟⎝ ⎠
nTK K
(2 3)
212.027 K 12.027 K 3.4 2.8 101000 K
−∆ ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
n
CK KT
×
(b) 3 2
2NH H S 9.4 10−= = ×K P P
252
(2 0)
2 412.02 K 12.027 K 9.4 10 1.5 10297 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
2SO 102
SO O
2.5 10= = ×P
KP P
(a) For the equation written 12 22SO (g) O (g) SO (g)3+ → Eq. 2
3
2 2
SO 10 5Eq. 2 Eq.11/ 2
SO O
2.5 10 1.6 10= = = × =P
K KP 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. 3Eq. 2 Eq.1
1 1= =K
K K
2 2
3
1/ 2SO O 6
Eq. 2 10SO Eq.1
1 1 6.3 102.5 10
−= = = = ××
P PK
P K
(c) For the equation written 32 223 SO (g) O (g) 3 SO (g)3+ → Eq. 4
This equation is 3/23Eq. 4 Eq.12 Eq. 1, so× =K K
3
2 2
3SO 3/ 2 10 3/ 2 15
Eq. 2 Eq.13 3/ 2SO O
(2.5 10 ) 4.0 10= = = × =P
K KP P
×
9.28 H2(g) + Cl2(g) 2 HCl(g) → 85.1 10= ×CK
2
2 2
[HCl][H ][Cl ]
=CK
3 28
32
3 2
2 8 3
12 12
(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 P
KP
2
3Cl4 (5.02 10 )
3.5 100.072
−−
×× =
P
2
43
Cl 3
(3.5 10 )(0.072) 5.0 10 bar5.02 10
−−
−
×= = ×
×P
9.32 (a) 2
33
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
33
2 2
[NH ]62
[N ][H ]= =CK
4 2
33 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 I0.003 94 mol I ; 0.003 27 mol I
253.8 g mol 253.8 g mol− −= =⋅ ⋅
I2(g) 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
0.00131.00 5.2 10
0.003 271.00
−
⎡ ⎤⎢ ⎥⎣ ⎦= =⎡ ⎤⎢ ⎥⎣ ⎦
CK ×
254
9.38 Pressure (Torr) CO(g) + H2O(g) CO2(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 88750.1 750.1 0.62112 112
750.1 750.1
⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
P PK
P P
9.40 (a) The balanced equation is Cl2(g) 2 Cl(g)
The initial concentration of 122
0.0050 mol ClCl (g) is 0.0025 mol L
2.0 L−= ⋅
Concentration Cl1(mol L )−⋅ 2(g) 2 Cl(g)
initial 0.0025 0
change −x +2 x
equilibrium 0.0025 − x +2 x
223
C2
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
−
−
− −
= = = ×−
= × −
+ × − × =
xKx
x xx x
3 3 2
3 3
3 4
(1.7 10 ) (1.7 10 ) 4(4)( 4.3 10 )2 4
(1.7 10 ) 8.47 108
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) 2 Br(g)
The initial concentration of 122
5.0 mol BBr (g) is 2.5 mol L
2.0 L−= ⋅
r
Concentration Br1(mol L )−⋅ 2(g) 2 Br(g)
initial 2.5 0
change −x +2 x
equilibrium 2.5 − x +2 x
223
C2
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 )
−
−
− −
= = = ×−
= × −
+ × − × =
xKx
x xx 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 ) 2.608 108
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) Br2(g) + Cl2(g)
initial 31.4 10−× 0 0
change 2− x +x +x
final 31.4 10 2−× − x +x +x
2 2Br Cl2
BrCl
2
3 2 3
2
3 2
( )( )32(1.4 10 2 ) (1.4 10 2 )
32(1.4 10 2 )
− −
−
⋅=
= =× − × −
=× −
p pK
p
x x x2x x
xx
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 )
−
−
−
−
=× −
= × −
+ = ×
+ = ×
xx
x x
x x
x
3
4
( 32)(1.4 10 )(1 2 32)
6.4 10
−
−
×=
+
= ×
x
x
2 2
4Br 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 =
51
5 1
2.0 g PCl208.22 g mol PCl
0.032 mol L0.300 L
−−
⎛ ⎞⎜ ⎟⋅⎝ ⎠ = ⋅
Concentration PCl1(mol L )−⋅ 5(g) PCl3(g) + Cl2(g)
initial 0.032 0 0
change −x +x +x
final 0.032 − x +x +x
2
3 2C
5
[PCl ][Cl ] ( )( )[PCl ] (0.032 ) (0.032 )
= = =− −
x x xKx 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 0.642 1
=−
= −
+ − =
− ± − −=
⋅− ±
= = + −⋅
xx
x xx 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 L2.00 L
−= = ⋅
Concentration NH1(mol L )−⋅ 4HS(s) NH3(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 0.1022 1
−
−
−
= = +
× = +
+ − × =
− + ± + − − ×=
⋅− ±
= = + −⋅
K x x
x xx x
x
x
The negative root is not meaningful, so we choose 3 12 10 mol L .− −= × ⋅x
1 33
1
3 12
[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 PCl3(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 33.333.40 0.001875 0
− + =
− + =
x x xx 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 15 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) 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
xKx x
25
25
2 5
2 5 2
2 5 2 7
2 7 10
(2 )1.00 10(0.001 40 )(0.0214 )
41.00 100.0228 3.0 10
4 (1.00 10 )( 0.0228 3.0 10 )4 1.00 10 2.28 10 3.0 104 2.28 10 3.0 10 0
−
−−
− −
− −
− −
× =− −
× =− + ×
= × − + ×
= × − × + ×
+ × − × =
xx xx
x xx x xx x xx 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 8.7 108
− −− −− × ± ×
= = ×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 NH0.050 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 HH 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
initial 0.025 0.015 0
change −x 3− x 2+ x
final 0.025 − x 0.015 3− x 2+ x
3NH3
N H .0= =K
P P2 2
3 0.036(0 25 )(0.015 3 )
=− −
xx 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 5NH 2 2.8 10 bar 5.6 10 bar− −= × × = ×P
262
9.56 Concentrations
CH1(mol L )−⋅ 3COOH + C2H5OH 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 52
2
2
2
2 2
2
2
[CH COOC H ][H O] ( )(0.015 )[CH COOH][C H OH] (0.024 )(0.059 )
0.0150.083 0.0014
0.0154.00.083 0.001 42
4.0 0.332 0.005 68 0.0153.0 0.347 0.005 68 0
( 0.347) ( 0.347) (4
+= =
− −
+=
− +
+=
− +
− + = +
− + =
− − ± − −=
Cx xK
x x
x xx x
x xx x
x x x xx x
x)(3.0)(0.005 68) 0.347 0.228
(2)(3.0) 6.00.0958 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
=P
KP P
2
24
Cl
1.3 103.5 10(9.56)
×× =
P
263
2
24
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) CO2(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 22
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 L10.0 L
−= = ⋅
The final concentration of 12
0.0285 molH 0.002 85 mol L10.0 L
−= = ⋅
Concentrations 2 H1(mol L )−⋅ 2S(g) 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 12
12
12
24
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 L0.500 L
−= = ⋅
Concentrations (m L PCl1ol )−⋅ 5(g) PCl3(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 xx 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
52
[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)− −
=
⎛ ⎞×⎜ ⎟
⎝ ⎠= =⋅ ⋅ ⋅
PVnRT
n
10.176 mol[HCl] 0.0147 mol L12.00 L
−= = ⋅
2 HCl(g) + I1Concentrations (mol L )−⋅ 2(s) 2 HI(g) + Cl2(g)
initial 0.0147 — 0 0
change 2− x — 2+ x +x
final 0.147 2− x — 2+ x +x
22
2
234
2
[HI] [Cl ][HCl]
(2 ) ( )1.6 10(0.0147 2 )
−
=
× =−
CK
x xx
Because the equilibrium constant is very small, we will assume that x <<
0.0147:
3
342
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 g17.03 g mol
0.301 mol L5.00 L
−−
⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ⋅
2 NH1Concentrations (mol L )−⋅ 3(g) N2(g) + 3 H2(g)
initial 0.301 0 0
change 2− x +x 3+ x
final 0.301 2− x +x 3+ x
32 2
23
3
2
4
2
[N ][H ][NH ]
( )(3 ) 0.395(0.301 2 )
27 0.395(0.301 2 )
=
=−
=−
CK
x xx
xx
4
2
2
27 0.395(0.301 2 )
3 3 0.628(0.301 2 )
=−
=−
xx
xx
2
2
2
3 3 (0.628)(0.301 2 ) 0.189 1.265.20 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 xx 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 P
KP
(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 1moles H 3 2
+= =
+
n nn 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 HCl4 2 3.0 b= + =TotalP P P ar
2HCl Cl
3.0 bar 22
= −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 Cl34
2
Cl
3.0 2 3.2 103.0 2
2
−
⎛ ⎞+ ⋅⎜ ⎟⎝ ⎠= =⎛ ⎞−⎜ ⎟⎝ ⎠
P PK
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
Cl34
2
34 34Cl
HCl H
3.0 2 3.2 103.0
2(1.5)(3.2 10 ) 4.8 10 bar
1.5 bar
−
− −
⎛ ⎞ ⋅⎜ ⎟⎝ ⎠≈ = ×⎛ ⎞⎜ ⎟⎝ ⎠
= × = ×
= =
PK
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) 122HgO(s) Hg(l) O (g)+
r f1
r1
r1
r 221 1 1 11
r 21 1
1 1r
[ (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
HS S S S
S
S
At 298 K:
1 1r(298 K)
1
r(298 K)
90.83 kJ (298 K)(108.30 J K )/(1000 J kJ )
58.56 kJ molln
− −
−
∆ ° = − ⋅ ⋅
= ⋅∆ ° = −
G
G RT K
r(298 K)
1
ln
58 560 J 23.6(8.314 J K )(298 K)−
∆ °= −
= − = −⋅
GK
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 f1 1
r1
r
r1 1 1 1
r1 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
HS S S
S
S
At 298 K:
1r(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)−
∆ °= −
= − = −⋅
GK
RT
85 10−= ×K
At 373 K:
1r(373 K)
1 1
1
32.88 kJ mol
(373 K)( 29.2 J K mol )/(1000 J kJ )43.8 kJ mol
−
− − −
−
∆ ° = ⋅
− − ⋅ ⋅ ⋅
= ⋅
G1
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( )l 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⎛ ⎞∆
= −⎜⎝ ⎠
rK HK 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
vapH rH .
2
1 1
1 1ln∆ ⎛ ⎞
= −⎜ ⎟⎝ ⎠
vapHPP R T T2
This relationship is known as the Clausius-Clapeyron equation (see
equation 4, page 285). Since 0∆ >vapH , vapor pressure increases with
temperature.
9.90 Recall ln and 10−∆ = − = ∆ − ∆ = pKr r rG RT K H T S K
and 2
1 1
1 1ln⎛ ⎞∆
= −⎜⎝ ⎠
rK HK R T T2
⎟ , the van’t Hoff equation (see equation 16,
page 354).
Rearranging and plugging in values,
13.8330 1 1
215.136
1
1 2
4 1
1
10 8.314 J K molln ln1 1101 1
293 K 303 K
(3.000)(7.381 10 J mol )221.4 kJ mol
− −
−
−
−
⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞ ⋅ ⋅
∆ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠ ⎜ ⎟−⎜ ⎟−⎜ ⎟ ⎝ ⎠⎝ ⎠= × ⋅
= ⋅
rK RHK
T T
−
3 11 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
−− − −
− −
∆∆ = +
× ⋅= ⋅ ⋅ +
= ⋅ ⋅
rr
HS R K
T
273
The sign on is positive because of the randomization that occurs as
deuterons dissociate and redistribute throughout the liquid.
∆ rS
From the results above, the “autoprotolysis” constant of heavy water at
25°C is
2
1 1 2
3 115.136
2 1 1
1 1ln
221.4 10 J mol 1 1ln ln10293 K 298 K8.314 J K mol
33.327
−−
− −
⎛ ⎞∆= −⎜ ⎟
⎝ ⎠⎛ ⎞× ⋅ ⎛ ⎞= −⎜ ⎟⎜ ⎟⋅ ⋅ ⎝ ⎠⎝ ⎠
= −
rK HK 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 31
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:
1r(298 K)
1 1
1
285.4 kJ mol
(298 K)(137.56 J K mol )/(1000 J kJ )326.4 kJ mol
−
− − −
−
∆ ° = − ⋅
− ⋅ ⋅
= − ⋅
G1⋅
°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
GK
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
12 4 2
1
2.50 g 0.330 g92.02 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 NO2(g)
initial 0.0136 0.003 59
change −x 2+ x
final 0.0136 − x 0.003 59 2+ x
275
22
2 42
3
3 2
5 3 2 2
2 2 5
[NO ][N O ]
(0.003 59 2 )4.66 100.0136
(4.66 10 )(0.0136 ) (0.003 59 2 )6.34 10 4.66 10 4 1.44 10 1.29 104 1.91 10 5.05 10 0
−
−
− − −
− −
=
+× =
−
× − = +
× − × = + × + ×
+ × − × =
K
xx
x xx x x
x x
�
5−
1
Solving by using the quadratic equation gives 31.89 10 .−= ×x
1 3 1 1
2 41 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:
23?
3 3
(0.0074) 4.66 100.012
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 cis-2-butene K1
2-methyl propene trans-2-butene K2
cis-2-butene trans-2-butene K3
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 11
1 12
1 13
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
rr
11
1 11 1
12
2 21 1
13
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)
−
− −
−
− −
−
− −
∆ °∆ ° = − =
−∆ ° ⋅
= = − = − =− ⋅ ⋅
∆ ° ⋅= = − = − =− ⋅ ⋅
∆ ° − ⋅= = −− ⋅ ⋅
GG RT K K
RTG
K KRTG
K KRTG
KRT 31.17; 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 Kcis
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 VKcis
V
9.98 A 2 B + 3 C
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 PK
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) 78.18.56
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 nRTmPV RTM
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
− −
−
=
⋅ ⋅ ⋅=
⋅=
mRTPMV
278
1 2
1 2
23.59 atm298 K 400 K
=
=
P PT 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)+
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) 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 ) 4127
4.88 2 413 3
4.88 2 3 12333.3 2 4.88 0
0.354
−=
−=
−=
− =
+ − ==
xx x
xx
xx
x xx x
x
= 4.17 bar or 4.11 atm; 3NHP
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 mol2(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
− −
− − − −
− −
− −
− −
− −
∆ ° = − ⋅ = − ⋅
∆ ° = ⋅ ⋅ − ⋅ ⋅
+ ⋅ ⋅
= − ⋅ ⋅
∆ ° = − ⋅ ⋅
− − ⋅ ⋅
= − ⋅
HS
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) 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
33
33
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
− −
−
−
= =⋅ ⋅ ⋅
= ×
⎛ ⎞⎛ ⎞× ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
=
PVnRT
280
(c) 2
3
3 3H2 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)
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 2BrCl
2Br Cl
(2 )0.2(1.00 )
20.21.000.183 0.2
= =−
=−
= ≈
P xP P x
xx
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
delta
G (k
J pe
r mol
)
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 SKR T R
where ∆ °−
HR
is the slope and ∆ °SR
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 +17.899R = 0.99213
1
1 1
5398 K
44.9 kJ mol
17.9
149 J K mol
−
− −
∆ °− = −
∆ ° = ⋅∆ °
=
∆ ° = ⋅ ⋅
HR
HSRS
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⎛ ⎞∆
= −⎜⎝ ⎠
rK HK 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 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 SK
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
rr
1 1 1
21 700, which gives
(21 700 K)(8.314 J K mol )/(1000 J kJ ) 180 kJ mol 1− − − −
− = − ∆ °
= + ⋅ ⋅ ⋅ = ⋅
HR
∆ °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−
RTKK
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 L0.7088 bar
− −⋅ ⋅= =
=
nRTPV
pressure (bar) PH3(g) + BCl3(g) PH3BCl3(s)
initial 0.7088 0
change +x +x
equilibrium 0.7088+ x x
3 3PH BCl
1119.2(0.7088 )
= = =⋅ +
KP P x x
=
2 0.7088 0.05208 0+ −x x
20.7088 (0.7088) 4(0.05208)
20.06712 (negative root isn't physically possible)
− ± +=
=
x
So 3
0.70PH 88 0.06712 0.7759 bar= + =P
and 3
3= =c PH 1
PH 1 1
0.7759 bar 0.0280 mol L(0.08314 L bar K mol )(333 K)
−− − = ⋅
⋅ ⋅
PRT
(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 RTKK
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) BCl3(g) + BClBr2(g)
2 BClBr2(g) BBr3(g) + BCl2Br(g)
2 BCl3(g) + BBr3(g) 3 BCl2Br(g)
BCl3(g) + 2 BBr3(g) 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 −∆
=rG
RTK e and
. ∆ = ∑ ∆ −∑ ∆r prods f reacts fG n G m G
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 3 BClBr2 1871exp 1.421
8.314 298.15⎛ ⎞= =⎜ ⎟×⎝ ⎠
K
(2) 2 BCl3 + BBr3 3 BCl2Br 22399exp 2.632
8.314 298.15⎛ ⎞= =⎜ ⎟×⎝ ⎠
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
3BClBr
1 2=P
Kxy
and 2
3BCl Br
2 2=P
Kx y
so , and 2
1/ 3 1/ 3 2 / 3BClBr 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 oBCl BBr BCl BCl Br BClBr BBr+ = + + +P P P P P P
3
2
33
3=
3
where are the initial pressures. Balancing Cl atoms
and Br atoms gives us
3 3
o oBCl BBr 1 bar= =P P
3 3 2
oBCl BCl BCl Br BClBr3 3 2= + +P P P P
3 2 2
oBBr 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 zy
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/ 32 1
3 2
33 2
33(0.975) 2(1.381)(0.975) (1.124)(0.975)0.461
=+ +
=+ +
=
yz K z K z
Then 3 3(0.975) (0.461) 0.428= = =x z y
And finally:
( )2
1/ 32BClBr (1.421)(0.428)(0.461) 0.506= =P
( )2
1/ 32BCl 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) BBr3(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
−
−
−= = =
− =
= ×
= ×
xx
x xx
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
= = = .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 HRate of effusion of HI
127.9 g mol2.016 g mol
7.97
−
−
=
⋅=
⋅
=
MM
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 HRate of effusion of HI
=P MP 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
2HI
H
1/ 2HI H (0.345 0.69) 0.49 bar
=
= ⋅ = × =
PK
P
P K P
2
2
1H2 HI
1HI H
Rate of effusion of H 0.69 bar 127.9 g molRate of effusion of HI 0.49 bar 2.016 g mol
1110 (to 1 sig fig)
−
−
⋅⎛ ⎞= = ⎜ ⎟ ⋅⎝ ⎠
==
P MP M
So hydrogen effuses ten times as fast as hydrogen iodide under these
conditions.
290