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The change in Gibbs free energy of vapourisation of 1 mol of H2O at 1 atm and 100◦C is (qP is the heat change at constant P)
H2O (l) --> H2O (g)S = 69.91J --> 188.83 Jdh=40.6 kj/moldS = prod - reactantdS = 188.83 J - 69.91 J = 118.92J / K = 0.1189 kJ / K100Celsius .... Kelvin temp = 373KdG = dH - TdSdG = 40.6 kJ - (373K) (0.1189 kJ/K)dG = 40.6 kJ - 44.35 kJdG = – 3.74 kJ
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