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Calculate the entropy change in the surroundings when 1.00 mol of H2O (l) is formed under standard conditions: Δ fH0 = - 286 kj mol-1. |
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Answer» H2 (g) + ½ O2 (g) ---> H2O (g), Δf H0 = -286kj mol-1 This means that when 1 mol of H2O (l), is formed , 286 kj of heat is released. This heat is absorbed by the surroundings, i.e., q rev = +286 kj mol-1 . So ΔS = qsurr /T =286KJmol-1 /298K =0.9597kjK-1 mol-1 = 959.7 JK-1 mol |
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