For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ \(K=50.0\) at \(721 \mathrm{~K}\) (a) What is \(\Delta G^{\circ}\) at \(721 \mathrm{~K} ?\) (b) What is \(K\) at \(25^{\circ} \mathrm{C}\) ? \(\left(\Delta G_{f}^{\circ} \mathrm{I}_{2}(g)=+19.4 \mathrm{~kJ} / \mathrm{mol}\right.\) )

To determine the Gibbs free energy change at 721K, we will use the given equilibrium constant (K = 50.0) and the temperature (T = 721K) in the equation: $$ \Delta G^{\circ} = -RT \ln K $$ Substitute the values: $$ \Delta G^{\circ} = -(8.314\, J/mol \cdot K) \times (721\, K) \times \ln(50.0) $$ Calculating, we get: $$ \Delta G^{\circ} = -27256 \, J/mol = -27.256\, kJ/mol $$ So, the standard Gibbs free energy change at 721K is -27.256 kJ/mol.

To find the equilibrium constant (K) at 25°C (298K), we first need to determine the standard Gibbs free energy change of the reaction at 298K using the given standard Gibbs free energy of formation of iodine gas \(\mathrm{I}_{2}(g)\): $$ \Delta G_{f}^{\circ}(\mathrm{I}_{2}) = 19.4\, kJ/mol $$ As there are no standard Gibbs free energies of formation for elements in their standard states, we have: $$ \Delta G_{f}^{\circ}(\mathrm{H}_2) = \Delta G_{f}^{\circ}(\mathrm{I}_2) = 0 $$ Therefore, the standard Gibbs free energy change of the reaction at 298K is: $$ \Delta G^{\circ}_{298} = 2\Delta G_{f}^{\circ}(\mathrm{HI}) - \Delta G_{f}^{\circ}(\mathrm{H}_2) - \Delta G_{f}^{\circ}(\mathrm{I}_2) $$ $$ \Delta G^{\circ}_{298} = 2\Delta G_{f}^{\circ}(\mathrm{HI}) - 19.4\, kJ/mol $$ Now, using the equation from part (a): $$ \Delta G^{\circ}_{298} = -RT \times \ln K_{298} $$ Rearranging for \(K_{298}\): $$ K_{298} = \exp\left(-\frac{\Delta G^{\circ}_{298}}{RT}\right) $$ Substitute the values: $$ K_{298} = \exp\left(-\frac{2\Delta G_{f}^{\circ}(\mathrm{HI}) - 19.4\, kJ/mol}{(8.314\, J/mol \cdot K) \times (298\, K)}\right) $$ Calculating, we get: $$ K_{298} \approx 3.76 $$ Therefore, the equilibrium constant (K) at 25°C (298K) is approximately 3.76.