Calculate the entropy change in surroundings when `1.00` mol of `H_(2)O(l)` is formed under standard conditions, `Delta_(r )H^(Θ) = -286 kJ mol^(-1)`.

Explanation

`H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), Delta_(f)H^(Θ)=-286 kJ mol^(-1)`
This means that when `1 mol` of `H_(2)O(l)`, is formed, `286 kJ` of heat is relesed. This heat is absorbed by the surroundings, i.e., `q_(surr)=+ 286 kJ mol^(-1)`
`:. DeltaS=q_(surr)/T=(286 kJ mol^(-1))/(298 K)`
`=0.9597 k J K^(-1) mol^(-1)`
`=959.7 J K^(-1) mol^(-1)`