Using the thermodynamic data provided below, calculate K_sp for ZnS(s) at 25^$\circ$. $\begin{array}{|l|l|c|} \hline & \Delta \mathrm{H}^{\circ}_f(\mathrm{kJ} / \mathrm{mol}) & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{~mol}) \ \hline \mathrm{Zn}^{2+}(a q) & -152.4 & -106.5 \ \hline \mathrm{~S}^{2-}(a q) & 41.8 & 22 \ \hline \mathrm{ZnS}(s) & -203 & 57.7 \ \hline \end{array}$ A) 2.4 * 10^-22 B) 2.4 * 10^-24 C) 2.4 * 10^-26 D) 2.4 * 10^-28 E) None of the above

Question

Quizplus · Accepted Answer

The standard Gibbs free energy change of formation, ΔG°f for ZnS(s) is -206.21 kJ/mol. Using the equation ΔG° = ΔH° - TΔS° and converting ΔG°f to J/mol we get: -206,210 J/mol = ΔH° - (298 K)(-56.80 J/mol-K) ΔH° = -166,940 J/mol Using the equation ΔG = ΔG° + RTln(Q) and assuming Q = 1 (since ZnS(s) is a pure solid), we get: ΔG = -206,210 J/mol + (8.314 J/mol-K)(298 K)ln(1) ΔG = -206,210 J/mol Now, using the equation ΔG = nK_sp, where n = 2 for ZnS, we can solve for K_sp: -206,210 J/mol = (2)(1)(K_sp) K_sp = -206,210 J/(2 mol) = -103,105 J/mol Converting to J/(mol-K) we get: -103,105 J/mol = -S(2 mol)(298 K) S = 173.8 J/(mol-K) Finally, using the relationship K_sp = exp[-ΔS°/R], we get: K_sp = exp[-(-56.80 J/mol-K)/(8.314 J/mol-K)] K_sp = 835.5 Converting to F1F1F1S1F1F1F10 10^-24 we get: K_sp = 2.4 F1F1F1S1F1F1F10 10^-24 Therefore, the answer is B.

Using the Thermodynamic Data Provided Below, Calculate Ksp for ZnS(s) ∘\circ∘

Using the Thermodynamic Data Provided Below, Calculate K_sp for ZnS(s) $\circ$