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Example 16.7: Evaluation of \(ΔG°\) Change from \(ΔH°\) and \(ΔS°\)

Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for \(ΔG°\) say about the spontaneity of this process?

Solution

The process of interest is the following:


The standard change in free energy may be calculated using the following equation:


From Appendix G, here is the data:
Substance

\(\mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}}\) \(= -286.83\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)


\(S°_{\mathrm{\ce{H2O(l)}}}\) \(= 70.0\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


\(\mathrm{ΔHf°}_{\mathrm{\ce{H2O(g)}}}\) \(= -241.82\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)


\(S°_{\mathrm{\ce{H2O(g)}}}\) \(= 188.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


Combining

\(ΔH°\) \(= \mathrm{ΔHf°}_{\mathrm{\ce{H2O(g)}}} - \mathrm{ΔHf°}_{\mathrm{\ce{H2O(l)}}}\)

\(\ \ \ =-241.82\ \frac{\mathrm{kJ}}{\mathrm{mol}} - (-286.83\ \frac{\mathrm{kJ}}{\mathrm{mol}})\)

\(\ \ \ =45.01\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)


\(ΔS°\) \(= S°_{\mathrm{\ce{H2O(g)}}} - S°_{\mathrm{\ce{H2O(l)}}}\)

\(\ \ \ =188.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} - 70.0\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)

\(\ \ \ =118.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


\(T\) \(= 298\ \mathrm{K}\)


\(ΔG°\) \(= ΔH° - T \cdot ΔS°\)

\(\ \ \ =45.01\ \frac{\mathrm{kJ}}{\mathrm{mol}} - 298\ \mathrm{K} \cdot 118.8\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)

\(\ \ \ =45.01\ \frac{\mathrm{kJ}}{\mathrm{mol}} - 35402.\ \frac{\mathrm{J}}{\mathrm{mol}}\)

\(\ \ \ =9.61\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)


At 298 K (25 °C) \(ΔG298°>0\), and so boiling is nonspontaneous (not spontaneous).
Think about it: \(ΔH°\) for boiling water is positive because:
a) The freedom of particles increases
b) The freedom of particles decreases
c) You gain energy when breaking hydrogen bonds
d) You expend energy to break hydrogen bonds