Calculate the equilibrium constant for the following reaction at `298K` and `1` atmospheric pressure:
`C(s) +H_(2)O(l) rarr CO(g) +H_(2)(g)`
Given `Delta_(f)H^(Theta) at 298 K` for `H_(2)O(l) =- 286.0 kJ mol^(-1)`
for `CO(g) =- 110.5 kJ mol^(-1)`
`DeltaS^(Theta)` at `298K` for the reaction `= 252.6 J K^(-1) mol^(-1)`

Answer

`2.72 xx 10^(-18)`

Explanation

`DeltaH^(@)` for the reaction
` = [DeltaH^(@)_(f) (CO) + Delta_(F)H^(@)(H_(2))] - [ DeltaH^(@)_(f)(C) + DeltaH_(f)^(@)(C) + DeltaH_(f)^(@)(H_(2)O)]`
` = [- 110.5 + 0] - [0 + (-286.0)]`
`= 175.5 kJ "mol"^(-1)`
`DeltaG^(@) = DeltaH^(@) - TDeltaS^(@)`
`= 175.5 - 298 xx 252.6 xx 10^(-3)`
` = 100.225 kJ "mol"^(-1)`
Now, `DeltaG^(@) = -2.303 RT "log K`
or `log K = -(DeltaG^(@))/(2.303 RT)`
` = -(100225)/(2.303 xx 8.31xx298) = -17.565`
or `K = 2.72 xx 10^(-18)`