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

It is given that 286 given that 286 kJ `"mol"^(-1)` of heat is evolved on the formation of 1 mol of `H_(2)O_((l)).` Thus, an equal amount of heat will be absorbed by the surroundings.
`q_("surr")=+286 kJ "mol"^(-1)`
Entropy change `(DeltaS_("surr"))` for the surroundings =`(q_("surr"))/(7)`
`=(286 kJ "mol"^(-1))/(298 J)`
`:. DeltaS_("surr")=957.73 J "mol"^(-1)J^(-1)`