Page 199 - Physical Chemistry

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Chapter 6 In working this problem, we assumed an ideal gas mixture, which is a good
Reaction Equilibrium in Ideal Gas
Mixtures assumption at the T and P of the experiment.

Exercise
If 0.1500 mol of O (g) is placed in an empty container and equilibrium is
2
reached at 3700 K and 895 torr, one finds that 0.1027 mol of O(g) is present.
Find K° and G° for O (g) ∆ 2O(g) at 3700 K. Assume ideal gases. (Answers:
P
2
0.634, 14.0 kJ/mol.)
Exercise
If 0.1500 mol of O (g) is placed in an empty 32.80-L container and equilibrium
2
is established at 4000 K, one finds the pressure is 2.175 atm. Find K° and
P
G° for O (g) ∆ 2O(g) at 4000 K. Assume ideal gases. (Answers: 2.22,
2
26.6 kJ/mol.)

Concentration and Mole-Fraction Equilibrium Constants
Gas-phase equilibrium constants are sometimes expressed using concentrations in-
stead of partial pressures. For n moles of ideal gas i in a mixture of volume V, the par-
i
tial pressure is P n RT/V [Eq. (1.24)]. Defining the (molar) concentration c of
i
i
i
species i in the mixture as
c n >V (6.21)*
i
i
we have
P n RT>V c RT ideal gas mixture (6.22)
i
i
i
Use of (6.22) in (6.7) gives for the ideal-gas reaction aA bB ∆ f F dD
f
f
1c F,eq RT>P°2 1c D,eq RT>P°2 d 1c F,eq >c°2 1c D,eq >c°2 d c°RT f d a b
K° a b (6.23)
P
a
a
1c A,eq RT>P°2 1c B,eq RT>P°2 b 1c A,eq >c°2 1c B,eq >c°2 b P°
3
where c°, defined as c° 1 mol/liter 1 mol/dm , was introduced to make all frac-
tions on the right side of (6.23) dimensionless. Note that c°RT has the same dimensions
as P°. The quantity f d a b is the change in number of moles for the reaction
as written, which we symbolize by n/mol f d a b. Since f d a b is
dimensionless and n has units of moles, n was divided by the unit “mole” in the
definition. For N (g) 3H (g) ∆ 2NH (g), n/mol 2 1 3 2. Defining the
2
3
2
standard concentration equilibrium constant K° as
c
K° q 1c i,eq >c°2 where c° 1 mol>L 1 mol>dm 3 (6.24)
n i
c
i
we have for (6.23)
K° K°1RTc°>P°2 ¢n>mol (6.25)
c
P
Knowing K°, we can find K° from (6.25). K° is, like K°, dimensionless. Since K° de-
P c c P P
pends only on T, and c° and P° are constants, Eq. (6.25) shows that K° is a function of
c
T only.
One can also define a mole-fraction equilibrium constant K :
x
K q 1x i,eq 2 n i (6.26)
x
i
The relation between K and K° is (Prob. 6.7)
x P
K° K 1P>P°2 ¢n>mol (6.27)
P
x