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Chemistry Question on Spontaneity

Calculate the entropy change in surroundings when 1.00 mol1.00\ mol of H2O(l)H_2O(l) is formed under standard conditions. ∆fHΘ=–286 kJmol−1∆_fH^Θ= –286\ kJ mol^{-1}.

Answer

It is given that 286 kJmol–1286\ kJ mol^{–1} of heat is evolved on the formation of 1 mol1\ mol of H2O(l)H_2O(l). Thus, an equal amount of heat will be absorbed by the surroundings.
qsurr=+286 kJmol–1q_{surr} = +286\ kJ mol^{–1}
Entropy change (∆Ssurr)(∆S_{surr}) for the surroundings = qsurr7\frac {q_{surr}}{7}
= 286 kJmol−1298 K\frac {286\ kJ mol^{-1}}{298\ K}
∴ ∆Ssurr=959.73 Jmol–1K–1∆S_{surr}= 959.73\ J mol^{–1} K^{–1}