For the water gas reaction,
`C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)`
the standard Gobbs free energy of reaction (at `1000K)` is `-8.1 kJ mol^(-1)`. Calculate its equilibrium constant.

Explanation

We know, `K - antilog ((-DeltaG^(Theta))/(2.3030 RT)) …(i)`
Given that, `DeltaG^(Theta) =- 8.1 kJ mol^(-1)`
`R = 8.314 xx 10^(-3) k J K^(-1)mol^(-1)`
`T = 1000K`
Substituting these values in equation (i), we get
`K = antilog [(+(8.1))/(2.303 xx8.314xx 10^(-3)xx1000)]`
`K = 2.65`