Final answer:
The sign of ΔS is negative for NaCl formation from ions and the formation of Fe₂O₃ from solids and gas due to decreased disorder, while it is positive for the reaction of C₆H₁₄ with oxygen as it leads to an increase in gas particles. The sign of ΔS for the combustion of methane is difficult to predict as there's no significant change in the number of gas molecules.
Step-by-step explanation:
When predicting the sign of entropy change (ΔS) for a chemical reaction, we examine the number of gas particles, distribution of matter, and phase changes involved in the process. Entropy generally increases when substances become more dispersed and there is an increase in the number of gas particles.
- (a) Na+ (aq) + Cl¯(aq) → NaCl(s): Negative ΔS, as ions in the solution combine to form a solid, reducing the dispersal of matter.
- (b) 2Fe(s) + ⅔O₂ (g) → Fe₂O₃ (s): Negative ΔS, as solid and gas react to form a solid, hence reducing the number of gaseous particles.
- (c) 2C₆H₁₄ (l) + 19O₂ (g) → 14H₂O(g) + 12CO₂(g): Positive ΔS, the number of gas molecules increases significantly, indicating increased disorder.
For the balanced chemical equation of the combustion of methane, CH₄(g), the reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Predicting whether ΔS is positive or negative in this scenario is difficult because there is no net change in the number of gas molecules; however, the products may be more disorderly than the reactants, causing a potential increase in entropy.