Example \(\PageIndex{1}\): Calculation of \(ΔG^\circ_{298}\)
Consider the decomposition of yellow mercury(II) oxide.
\[\ce{HgO}(s,\,\ce{yellow})⟶\ce{Hg}(l)+\dfrac{1}{2}\ce{O2}(g) \nonumber\]
Calculate the standard free energy change at room temperature, \(ΔG^\circ_{298}\), 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? The required data are available in Table T1 .
Solution
The required data are available in Table T1 and are shown here.
Compound |
\(ΔG^\circ_\ce{f}\:\mathrm{(kJ/mol)}\) |
\(ΔH^\circ_\ce{f}\:\mathrm{(kJ/mol)}\) |
\(S^\circ_{298}\:\textrm{(J/K⋅mol)}\) |
HgO (s, yellow) |
−58.43 |
−90.46 |
71.13 |
Hg(l) |
0 |
0 |
75.9 |
O2(g) |
0 |
0 |
205.2 |
(a) Using free energies of formation:
\[ΔG^\circ_{298}=∑νGS^\circ_{298}(\ce{products})−∑νΔG^\circ_{298}(\ce{reactants}) \nonumber\]
\[=\left[1ΔG^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔG^\circ_{298}\ce{O2}(g)\right]−1ΔG^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\]
\[\mathrm{=\left[1\:mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)\right]−1\: mol(−58.43\: kJ/mol)=58.43\: kJ/mol} \nonumber\]
(b) Using enthalpies and entropies of formation:
\[ΔH^\circ_{298}=∑νΔH^\circ_{298}(\ce{products})−∑νΔH^\circ_{298}(\ce{reactants}) \nonumber\]
\[=\left[1ΔH^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔH^\circ_{298}\ce{O2}(g)\right]−1ΔH^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\]
\[\mathrm{=[1\: mol(0\: kJ/mol)+\dfrac{1}{2}mol(0\: kJ/mol)]−1\: mol(−90.46\: kJ/mol)=90.46\: kJ/mol} \nonumber\]
\[ΔS^\circ_{298}=∑νΔS^\circ_{298}(\ce{products})−∑νΔS^\circ_{298}(\ce{reactants}) \nonumber\]
\[=\left[1ΔS^\circ_{298}\ce{Hg}(l)+\dfrac{1}{2}ΔS^\circ_{298}\ce{O2}(g)\right]−1ΔS^\circ_{298}\ce{HgO}(s,\,\ce{yellow}) \nonumber\]
\[\mathrm{=\left[1\: mol(75.9\: J/mol\: K)+\dfrac{1}{2}mol(205.2\: J/mol\: K)\right]−1\: mol(71.13\: J/mol\: K)=107.4\: J/mol\: K} \nonumber\]
Now use these values in Equation \(\ref{18.5.3}\) to get \(ΔG^o\):
\[ΔG°=ΔH°−TΔS°=\mathrm{90.46\: kJ−298.15\: K×107.4\: J/K⋅mol×\dfrac{1\: kJ}{1000\: J}} \nonumber\]
\[ΔG°=\mathrm{(90.46−32.01)\:kJ/mol=58.45\: kJ/mol} \nonumber\]
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 (since \(ΔG^o > 0\).