Discuss the effect of temperature change on the spontaneity of the following reactions at 1 atm. (a) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow 2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) $$ \Delta H^{\circ}=+851.5 \mathrm{~kJ} ; \Delta S^{\circ}=+38.5 \mathrm{~J} / \mathrm{K} $$ (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(l) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g)\) $$ \Delta H^{\circ}=-50.6 \mathrm{~kJ} ; \Delta S^{\circ}=0.3315 \mathrm{~kJ} / \mathrm{K} $$ (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) $$ \Delta H^{\circ}=98.9 \mathrm{~kJ} ; \Delta S^{\circ}=+0.0939 \mathrm{~kJ} / \mathrm{K} $$

For the reaction (a), we have the following values of enthalpy and entropy change: \(\Delta H^{\circ} = +851.5 \mathrm{~kJ}\) and \(\Delta S^{\circ} = +38.5 \mathrm{~J} / \mathrm{K}\) Both enthalpy and entropy change are positive. Let's plug these values into the Gibbs free energy change formula and calculate the critical temperature at which \(\Delta G\) becomes negative: \(\Delta G = 851.5 - T(0.0385)\) For the reaction to be spontaneous, \(\Delta G < 0\). Therefore: \(851.5 - T(0.0385) < 0\) Now, find the critical temperature: \(T > \frac{851.5}{0.0385} \approx 22117 \mathrm{K}\) So, for reaction (a) to be spontaneous, the temperature needs to be greater than 22117 K.

For the reaction (b), we have the following values of enthalpy and entropy change: \(\Delta H^{\circ} = -50.6 \mathrm{~kJ}\) and \(\Delta S^{\circ} = +0.3315 \mathrm{~kJ} / \mathrm{K}\) In this case, enthalpy change is negative and entropy change is positive. Let's plug these values into the Gibbs free energy change formula: \(\Delta G = -50.6 - T(0.3315)\) For the reaction to be spontaneous, \(\Delta G < 0\). Therefore: \(-50.6 - T(0.3315) < 0\) In this case, no matter the value of temperature, the \(\Delta G\) will always be negative, making the reaction spontaneous at all temperatures.

For the reaction (c), we have the following values of enthalpy and entropy change: \(\Delta H^{\circ} = +98.9 \mathrm{~kJ}\) and \(\Delta S^{\circ} = +0.0939 \mathrm{~kJ} / \mathrm{K}\) Both enthalpy and entropy change are positive. Let's plug these values into the Gibbs free energy change formula and calculate the critical temperature at which \(\Delta G\) becomes negative: \(\Delta G = 98.9 - T(0.0939)\) For the reaction to be spontaneous, \(\Delta G < 0\). Therefore: \(98.9 - T(0.0939) < 0\) Now, find the critical temperature: \(T > \frac{98.9}{0.0939} \approx 1053 \mathrm{K}\) So, for reaction (c) to be spontaneous, the temperature needs to be greater than 1053 K. In conclusion, the temperature conditions for each reaction to be spontaneous are: - Reaction (a): spontaneous at T > 22117 K - Reaction (b): spontaneous at all temperatures - Reaction (c): spontaneous at T > 1053 K