Example 16.8: Calculation of \(ΔG298°\)
Consider the decomposition of yellow mercury(II) oxide.
\(\ce{HgO(s,yellow)}\)\(\ce{->}\)\(\ce{Hg(l)}\)\(\ce{ + }\)\(\ce{1/2O2(g)}\)\(\ce{ }\)
Calculate the standard free energy change at room temperature,
\(ΔG298°\), using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution
The required data are available in
Appendix G and are shown here.
Compound | \(\mathrm{ΔGf°(kJ/mol)}\) | \(\mathrm{ΔHf°}\) (kJ/mol) | \(\ce{S298}\) (J/(K mol)) |
| \(\ce{HgO}\) (s, yellow) | -58.43 | -90.46 | 71.13 |
| \(\ce{Hg(l)}\) | 0 | 0 | 75.9 |
| \(\ce{O2(g)}\) | 0 | 0 | 205.2 |
(a) Using free energies of formation
\(ΔG298° = \mathrm{∑νGS298°} \cdot \mathrm{products} - ∑νΔG298° \cdot \mathrm{reactants}\)
\(ΔG298°_{\mathrm{\ce{->}}}\) \(= -ΔG298°_{\mathrm{\ce{HgO(s,yellow)}}}\)
\(\ \ \ =58.43\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
(b) Using enthalpies and entropies of formation:
\(ΔH298° = ∑νΔH298° \cdot \mathrm{products} - ∑νΔH298° \cdot \mathrm{reactants}\)
\(ΔH298°_{\mathrm{\ce{->}}}\) \(= -ΔH298°_{\mathrm{\ce{HgO(s,yellow)}}}\)
\(\ \ \ =90.46\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
\(ΔS298° = ∑νΔS298° \cdot \mathrm{products} - ∑νΔS298° \cdot \mathrm{reactants}\)
\(ΔS298°_{\mathrm{\ce{O2(g)}}}\) \(= 205.2\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(ΔS298°_{\mathrm{\ce{HgO(s,yellow)}}}\) \(= 71.13\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(ΔS298°_{\mathrm{\ce{->}}}\) \(= ΔS298°_{\mathrm{\ce{Hg(l)}}} + \frac{1 }{ 2} \cdot ΔS298°_{\mathrm{\ce{O2(g)}}} - ΔS298°_{\mathrm{\ce{HgO(s,yellow)}}}\)
\(\ \ \ =75.9\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} + \frac{1 }{ 2} \cdot 205.2\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} - 71.13\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(\ \ \ =75.9\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} + 102.600\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} - 71.13\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(\ \ \ =178.50\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}} - 71.13\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(\ \ \ =107.4\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(ΔG° = ΔH° - T \cdot ΔS°\)
\(ΔG°_{\mathrm{\ce{->}}}\) \(= ΔH298°_{\mathrm{\ce{->}}} - T \cdot ΔS298°_{\mathrm{\ce{->}}}\)
\(\ \ \ =90.46\ \frac{\mathrm{kJ}}{\mathrm{mol}} - 298.15\ \mathrm{K} \cdot 107.4\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)
\(\ \ \ =90.46\ \frac{\mathrm{kJ}}{\mathrm{mol}} - 32012.\ \frac{\mathrm{J}}{\mathrm{mol}}\)
\(\ \ \ =58.45\ \frac{\mathrm{kJ}}{\mathrm{mol}}\)
Both ways to calculate the standard free energy change at
25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.