Question

For a phase change, H = 31 kJ/mol and sº = 0.093 kJ/(K-mol). What are ∆G and the spontaneity of the phase change at 300 K?

Answer 1

`\begin{gathered} \Delta G=\Delta H-T\Delta S \\ G:Gibb^(\prime)s\text{ }free\text{ }energy\text{ }in\text{ }kJ \\ H:Enthalpy\text{ }kJ \\ T:Temperature\text{ }in\text{ }K \\ S:Entropy\text{ }kJK^(-1) \end{gathered}`

By substituting the given values and solving for the unknown we have:

`\begin{gathered} \Delta G=31\text{ }kJ-(300K*0.093\text{ }kJK^(-1)) \\ \Delta G=+3.1\text{ }kJ \end{gathered}`

The change in Gibb's energy is 3.1kJ and is non-spontaneous.

When we have a +G then the re