Calculate `DeltaG^(0)` for the reaction `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)` at 298K. Given, at 298K `DeltaH_(f)^(0)` for `H_(2)O(l)` is -286 kJ`*mol^(-1)` and the molar entropies `(S^(0))` for `H_(2)(g),O_(2)(g) and H_(2)O(l)` are 130.7, 205.1 and 69.9 `J*K^(-1)*mol^(-1)` respectively.

Explanation

For the given reaction, `DeltaS^(0)=S_(H_(2)O(l))^(0)-[S_(H_(2)(g))^(0)+(1)/(2)S_(O_(2)(g))^(0)]`
`=[69.9-(130.7+(1)/(2)xx205.1)]J*K^(-1)=-163.35J*K^(-1)`
Given that `DeltaH_(f)^(0)[H_(2)O(l)]=-286kJ*mol^(-1)`
So, `DeltaH^(0)=-286kJ` for the given reaction.
We know, `DeltaG^(0)=DeltaH^(0)-TDeltaS^(0)`
`therefore DeltaG^(0)=[-286xx10^(3)-298(-163.35)]J=-237.32kJ`.