Consider the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g) \quad K=4.4 \times 10^{-19} $$ (a) Calculate \(\Delta G^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). (b) Calculate \(\Delta G_{i}^{\circ}\) for \(\mathrm{N}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\).

To calculate \(\Delta G^{\circ}\) for the reaction, we will use the formula connecting the equilibrium constant (K) with the standard Gibbs free energy change: $$ \Delta G^{\circ} = -RT \ln K $$ where \(R\) is the gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin (298 K) and \(K\) is the equilibrium constant given as \(4.4 \times 10^{-19}\). Plug in the values and solve for \(\Delta G^{\circ}\): $$ \Delta G^{\circ} = - (8.314 \,\text{J/mol K})(298 \,\text{K}) \ln (4.4 \times 10^{-19}) $$ $$ \Delta G^{\circ} \approx 2.67 \times 10^{2} \,\text{kJ/mol} $$

To calculate the standard Gibbs free energy of formation for N2O, we will use the equation for the standard Gibbs free energy change in terms of the standard Gibbs free energies of formation of the products and reactants: $$ \Delta G^{\circ} = \sum_\text{products}(\text{nG}) - \sum_\text{reactants}(\text{nG}) $$ For the given reaction: $$ \Delta G^{\circ} = 3 \Delta G_{\mathrm{NO}}^{\circ} - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + \Delta G_{\mathrm{NO}_{2}}^{\circ}) $$ We are given the standard Gibbs free energy of formation for NO2 and NO: $$ \Delta G_{\mathrm{NO}_{2}}^{\circ} = 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{NO}}^{\circ} = 87.5 \,\text{kJ/mol} $$ Insert the values and the calculated \(\Delta G^{\circ}\) from part (a): $$ 2.67 \times 10^{2} \,\text{kJ/mol} = 3 (87.5 \,\text{kJ/mol}) - (\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} + 52.1 \,\text{kJ/mol}) $$ Now, solve for \(\Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ}\): $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} = 3 (87.5 \,\text{kJ/mol}) - 2.67 \times 10^{2} \,\text{kJ/mol} - 52.1 \,\text{kJ/mol} $$ $$ \Delta G_{\mathrm{N}_{2} \mathrm{O}}^{\circ} \approx 8.4 \,\text{kJ/mol} $$