Predict the sign of ΔS°for the following reaction

Predict the sign of ΔS°for the following reaction: Cu2O(s) + C(s) →2Cu(s) + CO(g)

A. ΔS° ≈ 0
B. Δ S°< 0
C. Δ S°> 0
D. More information is needed to make a reasonable prediction.

I think it's C, as there are two solids going to a solid and a gas, and gas has a higher entropy than a solid. However, because there is still a solid in the product, would the entropy therefore be negligible, and the answer A? I still think C though, am I correct???

You are correct. As long as there is an increase in the number of moles of gas in a reaction, the entropy will almost certainly be positive. Note that in this case, there are 2 moles of solid on the reactant side, and two moles of solid on the product side, so the change in entropy attributable to the solids will be small. On the other hand, there are zero moles of gas on the reactant side, and 1 mole of gaseous reaction products, so this implies a large positive entropy change.

Overall, the entropy change must be positive, and C is the correct answer.