Consider the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g) \quad K=1.0 \times 10^{-37} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Cl}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\).

To calculate ΔG° for the reaction at 25°C, use the relationship between the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K): $$ \Delta G^{\circ}=-RT\ln K $$ where R is the gas constant (\(8.314 \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1}\)) and T is the temperature in Kelvin. Remember, to convert the temperature from Celsius to Kelvin, simply add 273.15 to the temperature in Celsius: $$ T(K)=25^{\circ} \mathrm{C} + 273.15=298.15 \mathrm{K} $$ Now, plug the temperature and equilibrium constant into the equation and solve for ΔG°: $$ \Delta G^{\circ}=-8.314\ \mathrm{J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times 298.15\ \mathrm{K} \times \ln \left( 1.0 \times 10^{-37}\right) $$ $$ \Delta G^{\circ}=1087.53\ \mathrm{J} \cdot \mathrm{mol}^{-1} $$ Thus, the ΔG° for the reaction at 25°C is \(1087.53\ \mathrm{J} \cdot \mathrm{mol}^{-1}\).

In order to calculate ΔGf° for Cl(g) at 25°C, we first find the reaction for the formation of 1 mole of Cl(g) from its standard state: $$ \frac{1}{2}\mathrm{Cl}_2(g) \rightarrow \mathrm{Cl}(g) $$ Notice that this reaction is exactly half of the given reaction, so to find the ΔGf° for Cl(g), we divide the ΔG° calculated in part (a) by 2: $$ \Delta G_{\mathrm{f}}^{\circ} =\frac{\Delta G^{\circ}}{2} = \frac{1087.53\ \mathrm{J} \cdot \mathrm{mol}^{-1}}{2} = 543.76\ \mathrm{J} \cdot \mathrm{mol}^{-1} $$ Therefore, the ΔGf° for Cl(g) at 25°C is \(543.76\ \mathrm{J} \cdot \mathrm{mol}^{-1}\).