The crystalline hydrate $\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)$ loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

The relationship between Gibbs free energy (\(\Delta G^{\circ}\)), enthalpy (\(\Delta H^{\circ}\)), and entropy (\(\Delta S^{\circ}\)) is given by the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] Since the process is spontaneous, the change in Gibbs free energy (\(\Delta G^{\circ}\)) must be negative. We are given that the change in enthalpy (\(\Delta H^{\circ}\)) is positive for this process. The only way to satisfy a negative value for \(\Delta G^{\circ}\) is if the term \(T\Delta S^{\circ}\) is greater in magnitude than \(\Delta H^{\circ}\) and therefore has a positive value: \[\Delta H^{\circ} - T\Delta S^{\circ} < 0 \Rightarrow T\Delta S^{\circ} > \Delta H^{\circ}\] Since room temperature is a positive value, the change in entropy (\(\Delta S^{\circ}\)) must also be positive to satisfy the inequality. Thus, we can conclude that: (a) \(\Delta S^{\circ}\) is positive at room temperature.

Placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor will not change the intrinsic change in entropy (\(\Delta S^{\circ}\)) as \(\Delta S^{\circ}\) depends on the reaction itself and not on the initial conditions of the system. However, it can potentially affect partial pressures of the water below and above the hydrate, which in turn can affect the spontaneity of the process. Nevertheless, the intrinsic change in entropy at room temperature for the reaction will remain the same. Therefore, we can conclude that: (b) No, placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor does not change the intrinsic entropy change (\(\Delta S^{\circ}\)) of the reaction at room temperature.