The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right) \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) 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?

First, let's examine the given reaction: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ We can observe that a solid substance (crystalline hydrate) is losing water and converting into its anhydrous form along with water molecules in the gaseous state. The process is spontaneous and has a positive enthalpy change (ΔH°).

We know that for a spontaneous reaction, the Gibbs free energy change (ΔG°) is negative. Taking the relationship between Gibbs free energy change, enthalpy change, and entropy change into account: $$ \Delta G^{\circ} = \Delta H^{\circ} - T \times \Delta S^{\circ} $$ Given that ΔH° is positive and ΔG° is negative for the spontaneous reaction, we can infer that ΔS° must be positive to make the right side of the equation negative: $$ \Delta G^{\circ} = (+) - T \times (+) $$ Thus, the sign of ΔS° at room temperature is positive.

In the given scenario, a large amount of water vapor is already present in the closed vessel. The entropy change (ΔS°) is mainly dependent on the difference in the number of moles of gaseous products and reactants and the arrangement of particles in each state. Since the water is in the gaseous state in both the reaction and the vessel, the presence of excess water vapor would not change the nature of the reaction or the arrangement of the particles. Therefore, ΔS° remains the same in this case, and it is still positive at room temperature. In summary: (a) The sign of ΔS° at room temperature is positive. (b) The presence of a large amount of water vapor in the closed vessel does not change ΔS° for this reaction at room temperature.