Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. $\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)$ b. $4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$ c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. $\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$ e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

We will go through each of the given reactions and compare the particles and states in the reactants and products. a. SiF6(aq) + H2(g) → 2 HF(g) + SiF4(g) Reactants: 1 aqueous, 1 gas Products: 2 gases In this case, the number of gas particles increased, so the disorder increases. ΔS° will be positive. b. 4 Al(s) + 3 O2(g) → 2 Al2O3(s) Reactants: 4 solids, 3 gases Products: 2 solids Here, the number of gas particles decreased, and the number of solid particles also decreased. As a result, the disorder decreases, and ΔS° will be negative. c. CO(g) + Cl2(g) → COCl2(g) Reactants: 1 gas, 1 gas Products: 1 gas The number of gas particles decreased, so the disorder decreases. ΔS° will be negative. d. C2H4(g) + H2O(l) → C2H5OH(l) Reactants: 1 gas, 1 liquid Products: 1 liquid The number of gas particles decreased, so the disorder decreases. ΔS° will be negative. e. H2O(s) → H2O(l) Reactants: 1 solid Products: 1 liquid In this case, there is a transition from a solid state to a liquid state, which increases the disorder. ΔS° will be positive.

Based on our analysis of each reaction, the reactions with negative ΔS° are: - Reaction b: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) - Reaction c: CO(g) + Cl2(g) → COCl2(g) - Reaction d: C2H4(g) + H2O(l) → C2H5OH(l)