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.

For the water gas reaction, `C(s) +H_(2)O(g) hArr CO(g)+H_(2)(g)` the

1.	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.
Answer» 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`

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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.

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