From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K},\) at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \qquad \begin{array}{c} \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} \end{array} $$

Using the given values of \(\Delta H^{\circ}=-844 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K}\), and the formula \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\), we can calculate \(\Delta G^{\circ}\) for reaction (a) at 298 K: \[\Delta G^{\circ}=-844\mathrm{~kJ}-(298\mathrm{~K})(-165 \mathrm{~J} / \mathrm{K})\] First, convert the value of \(\Delta S^{\circ}\) from J/K to kJ/K: \[\Delta S^{\circ} = -165 \times 10^{-3} \mathrm{~kJ} / \mathrm{K}\] Now, calculate \(\Delta G^\circ\): \[\Delta G^{\circ}=-844\mathrm{~kJ}-(298\mathrm{~K})(-0.165 \mathrm{~kJ} / \mathrm{K})\] \[\Delta G^{\circ}=-844\mathrm{~kJ}+49.17\mathrm{~kJ}\] \[\Delta G^{\circ}=-794.83\mathrm{~kJ}\] Since \(\Delta G^{\circ}<0\), the reaction is spontaneous at 298 K, and we don't need to find the temperature at which it becomes spontaneous.

Using the given values of \(\Delta H^{\circ}=572 \mathrm{~kJ}\) and \(\Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K}\), and the formula \(\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}\), we can calculate \(\Delta G^{\circ}\) for reaction (b) at 298 K: \[\Delta G^{\circ}=572\mathrm{~kJ}-(298\mathrm{~K})(179 \mathrm{~J} / \mathrm{K})\] First, convert the value of \(\Delta S^{\circ}\) from J/K to kJ/K: \[\Delta S^{\circ} = 179 \times 10^{-3} \mathrm{~kJ} / \mathrm{K}\] Now, calculate \(\Delta G^\circ\): \[\Delta G^{\circ}=572\mathrm{~kJ}-(298\mathrm{~K})(0.179 \mathrm{~kJ} / \mathrm{K})\] \[\Delta G^{\circ}=572\mathrm{~kJ}-53.322\mathrm{~kJ}\] \[\Delta G^{\circ}=518.678\mathrm{~kJ}\] Since \(\Delta G^{\circ}>0\), the reaction is not spontaneous at 298 K.

To find the temperature at which reaction (b) becomes spontaneous, we must solve \(\Delta G^{\circ} = 0\) for the temperature: \[0 = \Delta H^{\circ} - T\Delta S^{\circ}\] Rearrange the equation to solve for T: \[T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}}\] Using the given values for reaction (b) (\(\Delta H^{\circ}=572 \mathrm{~kJ}\) and \(\Delta S^{\circ}=0.179 \mathrm{~kJ} / \mathrm{K}\)), calculate the temperature: \[T = \frac{572 \mathrm{~kJ}}{0.179 \mathrm{~kJ} / \mathrm{K}}\] \[T = 3196.65 \mathrm{~K}\] Thus, the reaction becomes spontaneous at 3196.65 K.