Page 197 - Physical Chemistry

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Chapter 6 and (6.8) becomes
Reaction Equilibrium in Ideal Gas
Mixtures n i
¢G° RT a n ln 1P i,eq >P°2 RT a ln 1P i,eq >P°2 (6.10)
T
i
i i
k
where k ln x ln x was used. The sum of logarithms equals the log of the product:
n n
a ln a ln a ln a p ln a ln 1a a p a 2 ln q a i
n
2
1
i
n
1 2
i 1 i 1
where the large capital pi denotes a product:
n
q a a a p a n (6.11)*
i
1 2
i 1
As with sums, the limits are often omitted when they are clear from the context. Use
of ln a ln a in (6.10) gives
i
i
i
i
¢G° RT ln c q 1P i,eq >P°2 d (6.12)
n i
T
i
We define K° as the product that occurs in (6.12):
P
K° q 1P i,eq >P°2 n i ideal-gas reaction equilib. (6.13)*
P
i
Equation (6.12) becomes
¢G° RT ln K° ideal-gas reaction equilib. (6.14)*
P
y
Recall that if y ln x, then x e [Eq. (1.67)]. Thus (6.14) can be written as
e
K° e ¢G°>RT (6.15)
P
Equation (6.9) shows that G° depends only on T. It therefore follows from (6.15) that
K° for a given ideal-gas reaction is a function of T only and is independent of the pres-
P
sure, the volume, and the amounts of the reaction species present in the mixture: K°
P
K°(T). At a given temperature, K° is a constant for a given reaction. K° is the stan-
P P P
dard equilibrium constant (or the standard pressure equilibrium constant) for the
ideal-gas reaction.
Summarizing, for the ideal-gas reaction 0 ∆ n A , we started with the general
i i i
condition for reaction equilibrium n m 0 (where the n ’s are the stoichiometric
i i i i
numbers); we replaced each m with the ideal-gas-mixture expression m m°
i i i
RT ln (P /P°) for the chemical potential m of component i and found that G°
i i
RT ln K°. This equation relates the standard Gibbs energy change G° [defined by
P
(6.9)] to the equilibrium constant K° [defined by (6.13)] for the ideal-gas reaction.
P
Because the stoichiometric numbers n are negative for reactants and positive for
i
products, K° has the products in the numerator and the reactants in the denominator.
P
Thus, for the ideal-gas reaction
N 1g2 3H 1g2 S 2NH 1g2 (6.16)
2
2
3
we have n 1, n 3, and n 2, so
N 2 H 2 NH 3
2
K° 3P1NH 2 >P°4 3P1N 2 >P°4 1 3P1H 2 >P°4 3 (6.17)
P
3 eq
2 eq
2 eq
2
3P1NH 2 >P°4
3 eq
K° (6.18)
P
3P1N 2 >P°43P1H 2 >P°4 3
2 eq
2 eq
where the pressures are the equilibrium partial pressures of the gases in the reaction
mixture. At any given temperature, the equilibrium partial pressures must be such as