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

Explanation

Enthalpy change for the formation of 1 mole of `H_(2)O` (l),.
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l), DeltaH_(f)H^(@) = -286 kJ mol^(-1)`
Energy related in the above reaction, is absorbed by the surroundings.
It means `q_("surr") = +286 kJ mol^(-1)`
`DeltaS = (q_("sur"))/(T) = (+286 kJ mol^(-1))/(298 K) = 09597 kJK^(-1) mol^(-1) = 959.7 JK^(-1) mol^(-1)`