CHEMISTRY 201X: Homework #9
Due
: Monday, December 3, 2007
Equilibrium Applications
(1) (6 pts)
Snow and ice sublime spontaneously when the partial pressure of water vapor is
below the equilibrium vapor pressure of ice.
At 0
°
C the vapor pressure of ice is 0.0060 atm (the
triple point of water).
Taking the enthalpy of sublimation of ice to be 50.0 kJ/mol, calculate the
partial pressure of water vapor below which ice will sublime spontaneously at
−
15
°
C.
Use the ClausiusClapeyron equation for equilibrium between H
2
O(s) and H
2
O(g):
⎛
⎞
⎛
⎞
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
sub
1
1
2
2
1
∆
H
p
p
1
1
50,000
1
1
ln
=

;
ln
=

p
R
T
T
0.006
8.314
273.15
258.15
p
1
= 0.0017 atm
(2) (14 pts)
At 1200 K in the presence of solid carbon, an equilibrium mixture of CO(g) and
CO
2
(g) (called “producer gas”) contains 98.3 mole percent CO and 1.69 mole percent CO
2
(g)
when the total pressure is 1 atm.
The reaction is
CO
2
(g) + C(graphite)
ZZX
YZZ
2 CO(g)
(a)
(2 pts) Calculate
p
(CO) and
p
(CO
2
) (in atm).
p(CO) = x(CO)(1 atm) = 0.983 atm
p(CO
2
) = x(CO
2
)(1 atm) = 0.0169 atm
(b)
(2 pts) Calculate the equilibrium constant
K
p
at 1200 K from these data.
2
2
p
2
p (CO)
(0.983)
K
=
=
= 57.2
p(CO )
(0.0169)
(c)
(2 pts) Calculate
∆
G
°
(in kJ) from these data.
∆
G
°
=
−
(8.314 J/K)(1200 K)ln(57.2) =
−
40370 J =
−
40.4 kJ
(d)
(4 pts) Use thermodynamic data to calculate
∆
G
°
(in kJ) and
K
p
at 1200 K.
Compare
with your results in parts (b) and (c).
∆
H
°
= 2(
−
110.5 kJ)
−
[(
−
393.5 kJ) + (0)] = +172.5 kJ
∆
S
°
= 2(197.9 J/K)
−
[(213.6 J/K) + (5.69 J/K)] = +176.5 J/K
∆
G
°
= (+172.5 kJ)
−
(1200 K)(+0.1765 kJ/K) =
−
39.3 kJ
K
p
= exp[
−
(
−
39300 J)/(8.314 J/K)(1200 K)] = exp(3.939) = 51.4
This estimate relies on assuming that
∆
H
°
and
∆
S
°
remain constant with
temperature, which is strictly incorrect. The equilibrium constant at 1200 K is
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