Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. 
= –286 kJ mol–1.
It is given that an energy of 286 kJ is released per mole, when water is formed under standard conditions. This means 286 kJ of energy is absorbed by the surroundings in this process. So,
q = - ∆H = 286 kJ mol-1
∆S = q/T
⇒ ∆S = 286 kJ/ 298 K
⇒ ∆S = 959.73 J mol-1K-1
T = Temperature = 298K
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