The entropy of vaporization of water is 109.0 J/mol . K at its normal boiling point.What is the enthalpy of vaporization of water at this temperature?

Question

Quizplus · Accepted Answer

By definition, the enthalpy of vaporization is the amount of energy required to vaporize one mole of a substance at its normal boiling point. It is related to the entropy of vaporization (ΔSvap) by the equation ΔHvap = TΔSvap. Thus, we can find the enthalpy of vaporization of water at its normal boiling point (100°C, or 373.15 K) using the given entropy of vaporization (109.0 J/mol K) as follows:

ΔHvap = TΔSvap = (373.15 K)(109.0 J/mol K) = 40.66 kJ/mol

Therefore, the answer is D (+40.66 kJ/mol).

The Entropy of Vaporization of Water Is 109