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Grade: 12th pass

                        

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)

5 years ago

Answers : (1)

Ashutosh Sharma
askIITians Faculty
181 Points
							H2O (l) --> H2O (g)
S = 69.91J --> 188.83 J
dh=40.6 kj/mol
dS = prod - reactant
dS = 188.83 J - 69.91 J = 118.92J / K = 0.1189 kJ / K

100Celsius .... Kelvin temp = 373K

dG = dH - TdS
dG = 40.6 kJ - (373K) (0.1189 kJ/K)
dG = 40.6 kJ - 44.35 kJ
dG = – 3.74 kJ
5 years ago
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