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