Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cu}^{2+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

Short answer

Question: Predict the sign of ΔS° (entropy change) for the following reactions: a) H₂(g) + Ni²⁺(aq) → 2 H⁺(aq) + Ni(s) b) Cu(s) + 2 H⁺(aq) → H₂(g) + Cu²⁺(aq) c) N₂O₄(g) → 2 NO₂(g) Answer: a) The ΔS° for reaction (a) is expected to be negative due to the decrease in the number of moles of gases and the formation of a solid. b) The ΔS° for reaction (b) is expected to be positive due to the increase in the number of moles of gases. c) The ΔS° for reaction (c) is expected to be positive due to the increase in the number of moles of gases.

Step-by-step solution

Reaction (a): \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\)

In this reaction, one mole of gas is consumed to produce no moles of gas. The decrease in the number of moles of gases indicates a decrease in disorder. Additionally, a solid forms which contributes to a more organized state. Therefore, the entropy change (ΔS°) for this reaction is expected to be negative.

Reaction (b): \(\mathrm{Cu}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cu}^{2+}(a q)\)

In this reaction, a solid reactant forms one mole of gas. The increase in the number of moles of gases leads to an increase in disorder. Thus, the entropy change (ΔS°) for this reaction is expected to be positive.

Reaction (c): \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

In this reaction, one mole of gas is transformed into two moles of gas. The increase in the number of moles of gases leads to an increase in disorder. Therefore, the entropy change (ΔS°) for this reaction is expected to be positive.