Question

Using the heats of vaporization for H2O ΔHvap at 25 ∘C = 44.02 kJ/mol, ΔHvap at 100 ∘C = 40.67 kJ/mol, calculate the entropy change for the vaporization of water at 25 ∘C and at 100 ∘C.

Answer 1

ΔHvap at 25 ∘C = 44.02 kJ/mol

temperature = 25 + 273 = 298 K

ΔS change at 25^oC = ΔH vap / T

                 = 44.02 x 10^3 / 298

                 = 147.72 J/K.mol

ΔS change at 25^oC = 147.72 J/K.mol

ΔHvap at 100 ∘C = 40.67 kJ/mol

temperature = 100 + 273 = 373 K

ΔS change at 100^oC = ΔH vap / T

                        = 40.67 x 10^3 / 373

                    = 109 J/K.mol

ΔS change at 100^oC = 109 J/K.mol