Question

Without doing any calculations, determine the sign of ? Sₛᵧₛ for each chemical reaction.

a. Mg(s) + Cl₂(g) → MgCl₂(s)

b. 2 H₂S(g) + 3 O₂(g) → 2 H₂O(g) + 2 SO₂(g)

c. 2 O₃(g) → 3 O₂(g)

d. HCl(g) + NH₃(g) → NH₄Cl(s)

Answer 1

The sign of the entropy change ΔS is evaluated based on the phase change and numbers of gas particles involved in reactions. Formation of solid from gases usually decreases entropy, while an increase in the number of gas molecules increases entropy.

Step-by-step explanation:

When predicting the sign of the entropy change (ΔSsys) for a chemical reaction without doing any calculations, consider the phase changes and the number of moles of gaseous reactants and products:

• (a) Mg(s) + Cl₂(g) → MgCl₂(s): Entropy decreases (ΔSsys < 0) since there's a transition from a gaseous state to a solid state, leading to a decrease in disorder.
• (b) 2 H₂S(g) + 3 O₂(g) → 2 H₂O(g) + 2 SO₂(g): The number of moles of gas does not change (5 moles of gas reactants and 4 moles of gas products), so there may be little or no change in entropy; however, the products are more structured and possibly lead to a slight decrease in entropy.
• (c) 2 O₃(g) → 3 O₂(g): Entropy increases (ΔSsys > 0) as there is an increase in the number of gaseous particles, which increases disorder.
• (d) HCl(g) + NH₃(g) → NH₄Cl(s): Entropy decreases (ΔSsys < 0) as gaseous reactants form a solid product, reducing disorder.