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2Zn+O2→2ZnO;ΔG∘=−616 J2\text{Zn} + \text{O}_2 \rightarrow 2\text{ZnO}; \Delta G^\circ = -616\text{ J} 2Zn+S2→2ZnS;ΔG∘=−293 J2\text{Zn} + \text{S}_2 \rightarrow 2\text{ZnS}; \Delta G^\circ = -293\text{ J} S2+2O2→2SO2;ΔG∘=−408 J\text{S}_2 + 2\text{O}_2 \rightarrow 2\text{SO}_2; \Delta G^\circ = -408\text{ J} ΔG∘\Delta G^\circ for the following reaction is: 2ZnS+3O2→2ZnO+2SO22\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2

A

−731 J-731\text{ J}

B

−1317 J-1317\text{ J}

C

−501 J-501\text{ J}

D

+731 J+731\text{ J}

Step-by-Step Solution

Given reactions: (i) 2Zn+O2→2ZnO;ΔG1∘=−616 J2\text{Zn} + \text{O}_2 \rightarrow 2\text{ZnO}; \Delta G_1^\circ = -616\text{ J} (ii) 2Zn+S2→2ZnS;ΔG2∘=−293 J2\text{Zn} + \text{S}_2 \rightarrow 2\text{ZnS}; \Delta G_2^\circ = -293\text{ J} (iii) S2+2O2→2SO2;ΔG3∘=−408 J\text{S}_2 + 2\text{O}_2 \rightarrow 2\text{SO}_2; \Delta G_3^\circ = -408\text{ J}

The target reaction is: 2ZnS+3O2→2ZnO+2SO22\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2 This can be obtained by reversing equation (ii) and adding it to equation (i) and equation (iii): Target reaction = −(ii)+(i)+(iii)-(\text{ii}) + (\text{i}) + (\text{iii}) Therefore, by applying Hess's Law for Gibbs free energy: ΔG∘=−ΔG2∘+ΔG1∘+ΔG3∘\Delta G^\circ = -\Delta G_2^\circ + \Delta G_1^\circ + \Delta G_3^\circ ΔG∘=−(−293 J)+(−616 J)+(−408 J)\Delta G^\circ = -(-293\text{ J}) + (-616\text{ J}) + (-408\text{ J}) ΔG∘=293−616−408=−731 J\Delta G^\circ = 293 - 616 - 408 = -731\text{ J}

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