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

`H_(2(g)) + (1)/(2) to H_(2) O_((l)) " "Delta_(f) H= -286 "kJ/mol" -286 "kJ"` heat is released when one mole `H_2 O` formed. So heat is absorbed by surrounding.
`q_("surr") = +286 "kJ/mol"`
`therefore Delta S= (q_("surr"))/( T) = (286)/( 286)`
`= 0.9597 "kJ/mol"= 259.7 "J/K mol"`