Zinc sulfide can be oxidized by oxygen gas according to the reaction,2 ZnS(s)+ O₂(g)$\leftrightarrows$ 2 ZnO(s)+ 2 S(s).If $\Delta$H $\circ$ = -289.0 kJ/mol and $\Delta$S $\circ$ = -169.4 J/K^.mol,what is $\Delta$G $\circ$ for this reaction at 25.00 $\circ$ C?

Question

Quizplus · Accepted Answer

The Gibbs free energy change (ΔG°) can be calculated using the equation ΔG° = ΔH° - TΔS°. Here, ΔH° = -289.0 kJ/mol, ΔS° = -169.4 J/K·mol = -0.1694 kJ/K·mol, and T = 298 K (25°C). Substituting these values gives ΔG° = -289.0 kJ/mol - (298 K)(-0.1694 kJ/K·mol) = -289.0 kJ/mol + 50.5 kJ/mol = -238.5 kJ/mol.

Zinc Sulfide Can Be Oxidized by Oxygen Gas According to the Reaction,2