Question

Predict the sign of \(\Delta S^{\circ}\) for each of the following reactions: (a) \(\mathrm{CCl}_{4}(l)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+4 \mathrm{ClO}_{2}(g)\) (b) \(8 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{S}_{8}(s) \longrightarrow 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g)\) (c) \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}_{2}(s)\) (d) \(2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\)

Short answer

Question: Predict the sign of standard entropy change for the following reactions: (a) CCl4(l) + 5 O2(g) → CO2(g) + 4 Cl2(g) (b) 2 H2O(l) + S8(s) → 16 H2O(g) (c) Br2(l) → Br2(s) (d) 2 NH3(g) → N2(g) + 3 H2(g) Answer: (a) In reaction (a), the number of moles of gas increases, thus making ΔS° positive. (b) In reaction (b), the number of moles of gas increases and there is a change from a more ordered phase to a less ordered phase, making ΔS° positive. (c) In reaction (c), there is a change from a less ordered phase to a more ordered phase, thus making ΔS° negative. (d) In reaction (d), the number of moles of gas decreases, thus making ΔS° negative.

Step-by-step solution

(a) Reaction involving CCl4 and O2

In this reaction, we have 1 mole of liquid and 5 moles of gas reacting to form 1 mole of gas and 4 moles of gas. There is an increase in the moles of gas (5 moles to 6 moles), which means the entropy will increase, making \(\Delta S^{\circ}\) positive.

(b) Reaction involving H2O and S8

Here, we have 8 moles of liquid and 1 mole of solid reacting to form 8 moles of gas and 4 moles of gas. As the number of moles of gas increases (0 moles to 12 moles) and a more ordered phase (liquid/solid) is changing to a less ordered phase (gas), this leads to an increase in entropy, making \(\Delta S^{\circ}\) positive.

(c) Reaction involving Br2

In this reaction, there is a phase change from liquid to solid Br2. Since there is a change from a less ordered phase (liquid) to a more ordered phase (solid), the entropy will decrease, making \(\Delta S^{\circ}\) negative.

(d) Reaction involving NH3

In this reaction, we have 2 moles of gas reacting to form 1 mole of gas and 3 moles of gas. As the total number of moles of gas decreases (3 moles) during the reaction, the entropy will also decrease, making \(\Delta S^{\circ}\) negative.