Final answer:
The sign of ΔS° for the reaction SiCl₄(g) + 2Mg(s) → 2MgCl₂(s) + Si(s) is negative, as the reaction involves a decrease in the number of gas moles, leading to a decrease in entropy.
Step-by-step explanation:
What is the Sign of ΔS° for the Reaction?
The question asks about the sign of the standard entropy change (ΔS°) for the given reaction: SiCl₄(g) + 2Mg(s) → 2MgCl₂(s) + Si(s). To determine the sign of ΔS°, we must consider the change in the number of moles of gas, since gases contribute substantially more to entropy than solids or liquids due to their higher randomness and disorder. In the reaction, we see a decrease from 1 mole of gas to none, as the products are all solids. This reduction in the number of moles of gas indicates a decrease in entropy, hence ΔS° would be negative.