Question

An unknown liquid has a boiling point of 653.79 ºC, and the enthalpy change for the conversion of this liquid to a gas is ΔHvap = 77.81 kJ/mol. What is the entropy change for vaporization, ΔSvap, in J/(K•mol)?

Answer 1

T boil = 653.79 °C = 926.94 K

∆Hvap = 77.81 kJ/mol

∆Svap =∆Hvap/T =77.81kJ/mol /926.94 K =77810 J/mol /926.94 K= 83.94 J/(K·mol)