Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions. (a) \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\) (b) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\)

Short answer

Question: Predict the sign of the standard entropy change (∆S°) for each of the following reactions and provide a brief explanation. a) 2 Na(s) + Cl2(g) → 2 NaCl(s) b) 2 NO(g) + O2(g) → 2 NO2(g) c) C2H4(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(l) d) NH4NO3(s) + H2O(l) → 2 NH3(g) + O2(g) Answer: a) The ∆S° for reaction (a) is negative because there is a conversion of a gaseous reactant (chlorine) to a solid product (sodium chloride). b) The ∆S° for reaction (b) is close to zero but slightly negative due to a slight decrease in molecular complexity. c) The ∆S° for reaction (c) is negative because there is a decrease in the number of gaseous particles and an increase in the number of liquid particles. d) The ∆S° for reaction (d) is positive because there is an increase in the number of gaseous particles and a decrease in solid and liquid particles.

Step-by-step solution

Reaction (a)

The given reaction is: \(2 \mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NaCl}(s)\). Comparing the reactants and products, the gaseous reactant (chlorine) is converted into a solid product (sodium chloride), and the total number of particles per mole is the same on both sides. The reduction of gaseous particles (lower entropy) outweighs the loss of two solid sodium particles. Therefore, the entropy change (\(\Delta S^{\circ}\)) is negative.

Reaction (b)

The given reaction is: \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\). The number of gaseous particles is the same on both sides (3 molecules). There is no significant change in molecular complexity. Therefore, the entropy change (\(\Delta S^{\circ}\)) is close to zero but slightly negative, as molecular complexity decreases slightly.

Reaction (c)

The given reaction is: \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\). There is a decrease in the number of gaseous particles (from 4 to 2) and an increase in the number of liquid particles (from 0 to 3). Since the increase in liquid particles is not enough to offset the decrease in gaseous particles, the entropy change (\(\Delta S^{\circ}\)) is negative.

Reaction (d)

The given reaction is: \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g)\). This reaction shows an increase in the number of gaseous particles (from 0 to 3) and a decrease in the number of solid and liquid particles. Since the increase in gaseous particles increases the entropy significantly more than the decrease in solid and liquid particles, the entropy change (\(\Delta S^{\circ}\)) is positive.