CHAP. 8: CHEMICAL EQUILIBRIUM [CONTENTS] 253
Solution
If we use the subscript 3 to denote∆rGstof the reaction whose equilibrium constant we want to
calculate, we can write
∆rGst 3 = ∆rGst 1 −2∆rGst 2
and by substituting from (8.16) we obtain
lnK 3 = lnK 1 −2 lnK 2
and thus
K 3 =
K 1
K 22
8.4.4 Conversions
- Conversion to another temperature
The dependence of enthalpy, entropy and the Gibbs energy on temperature is given in [3.5.3
through3.5.7]. The temperature dependence of the equilibrium constant is given in [8.2.1].
- Conversion to another pressure
The dependence of enthalpy, entropy and the Gibbs energy on pressure is given in [3.5.3
through3.5.7]. The pressure dependence of the equilibrium constant is given in [8.3.2].
Example
Oxidation of sulphur dioxide
0 =SO 3 (g)−SO 2 (g)−
1
2
O 2 (g),
has the equilibrium constantK= 1.794 (standard state of the gaseous component in the state
of an ideal gas at temperature 1000 K and standard pressure 101.325 kPa). Calculate the
equilibrium constant of the same reaction for the standard state of a pure gaseous component
at the temperature and pressure of the system if the system pressure is 1 kPa. Assume ideal
behaviour.
