Final answer:
The standard entropy change for the reaction MgCl₂(s) + H₂O(l) → MgO(s) + 2HCl(g) is -150 J/(mol·K). So, the correct answer is (b).
Step-by-step explanation:
The standard entropy change for a reaction is determined by taking the difference between the total standard entropies of the products and the total standard entropies of the reactants. In this case, we have:
MgCl₂(s) + H₂O(l) → MgO(s) + 2HCl(g)
The standard entropies at 25°C are:
S°(MgCl2) = 89.3 J/(mol·K)
S°(H2O) = 69.9 J/(mol·K)
S°(MgO) = 26.8 J/(mol·K)
S°(HCl) = 186.9 J/(mol·K)
With the given standard entropy values for the substances involved, we can determine the standard entropy change for the reaction by applying the formula.
S°reaction = [2 × S°(H2O)] + [2 × S°(HCl)] - [S°(MgCl2) + S°(MgO)]
S°reaction = [2 × 69.9 J/(mol·K)] + [2 × 186.9 J/(mol·K)] - [89.3 J/(mol·K) + 26.8 J/(mol·K)]
S°reaction = -150 J/(mol·K)
The negative sign indicates that the reaction results in a decrease in entropy, consistent with the fact that the reactants involve a solid and a liquid, while the products include a solid and a gas.