For the vaporization of a liquid at a given pressure,

Question

Quizplus · Accepted Answer

At low temperatures, the enthalpy change (H) for vaporization is positive (it takes energy to vaporize the liquid), and the entropy change (S) for vaporization is also positive (because the liquid has greater freedom of motion as a gas). This results in a positive Gibbs free energy change (G), so (F1F1F1S1F1F1F10G) is positive at low temperatures. 
At high temperatures, the positive entropy change becomes dominant over the positive enthalpy change, resulting in a negative Gibbs free energy change (G). Therefore, (F1F1F1S1F1F1F10G) is negative at high temperatures. At a certain temperature, the enthalpy and entropy changes cancel out, resulting in a zero Gibbs free energy change.

For the Vaporization of a Liquid at a Given Pressure Δ\DeltaΔ

For the Vaporization of a Liquid at a Given Pressure $\Delta$