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to satisfy (6.18). If the partial pressures do not satisfy (6.18), the system is not in re- Section 6.2
action equilibrium and its composition will change until (6.18) is satisfied. Ideal-Gas Reaction Equilibrium
For the ideal-gas reaction aA bB ∆ cC dD, the standard (pressure) equilib-
rium constant is given by (6.7).
Since P /P° in (6.13) is dimensionless, the standard equilibrium constant K° is
i
P
dimensionless. In (6.14), the log of K° is taken; one can take the log of a dimension-
P
less number only. It is sometimes convenient to work with an equilibrium constant that
omits the P° in (6.13). We define the equilibrium constant (or pressure equilibrium
constant) K as
P
K q 1P i,eq 2 n i (6.19)
P
i
K has dimensions of pressure raised to the change in mole numbers for the reaction
P
2
as written. For example, for (6.16), K has dimensions of pressure .
P
The existence of a standard equilibrium constant K° that depends only on T is a
P
rigorous deduction from the laws of thermodynamics. The only assumption is that we
have an ideal gas mixture. Our results are a good approximation for real gas mixtures
at low densities.
EXAMPLE 6.1 Finding K° and G°from the equilibrium
P
composition

A mixture of 11.02 mmol (millimoles) of H S and 5.48 mmol of CH was placed
2
4
in an empty container along with a Pt catalyst, and the equilibrium
2H S1g2 CH 1g2 ∆ 4H 1g2 CS 1g2 (6.20)
2
2
4
2
was established at 700°C and 762 torr. The reaction mixture was removed from
the catalyst and rapidly cooled to room temperature, where the rates of the for-
ward and reverse reactions are negligible. Analysis of the equilibrium mixture
found 0.711 mmol of CS . Find K° and G° for the reaction at 700°C.
2
P
Since 0.711 mmol of CS was formed, 4(0.711 mmol) 2.84 mmol of
2
H was formed. For CH , 0.711 mmol reacted, and 5.48 mmol 0.71 mmol
2
4
4.77 mmol was present at equilibrium. For H S, 2(0.711 mmol) reacted, and
2
11.02 mmol 1.42 mmol 9.60 mmol was present at equilibrium. To find
K°, we need the partial pressures P . We have P x P, where P 762 torr and
i
i
P
i
the x ’s are the mole fractions. Omitting the eq subscript to save writing, we have
i
at equilibrium
n H 2 S 9.60 mmol, n CH 4 4.77 mmol, n 2.84 mmol, n CS 2 0.711 mmol
H 2
x H 2 S 9.60>17.92 0.536, x CH 4 0.266, x 0.158, x CS 2 0.0397
H 2
P H 2 S 0.5361762 torr2 408 torr, P CH 4 203 torr,
P 120 torr, P 30.3 torr
H 2 CS 2
The standard pressure P° in K° is 1 bar 750 torr [Eq. (1.12)], and (6.13) gives
P
4
4
1P >P°2 1P >P°2 1120 torr>750 torr2 130.3 torr>750 torr2
K° H 2 CS 2
P
2
2
1P H 2 S >P°2 1P CH 4 >P°2 1408 torr>750 torr2 1203 torr>750 torr2
0.000331
The use of G° RT ln K° [Eq. (6.14)] at 700°C 973 K gives
P
¢G° 38.314 J>1mol K241973 K2 ln 0.000331 64.8 kJ>mol
973