Question:
Following method of extracting Zn is based on thermodynamics
A: \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)
B: \(\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\)
If \(\Delta G_f^{\circ}\) (standard free energies of formation, in kJ mol\(^{-1}\)) of ZnS = -205.4, ZnO = -318.0, SO\(_2\) = -300.4, and CO = -137.3. Free energy changes of the above reaction A and B (respectively) will be:
Following method of extracting Zn is based on thermodynamics
A: \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)
B: \(\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\)
If \(\Delta G_f^{\circ}\) (standard free energies of formation, in kJ mol\(^{-1}\)) of ZnS = -205.4, ZnO = -318.0, SO\(_2\) = -300.4, and CO = -137.3. Free energy changes of the above reaction A and B (respectively) will be:
A: \(2\text{ZnS} + 3\text{O}_2 \rightarrow 2\text{ZnO} + 2\text{SO}_2\)
B: \(\text{ZnO} + \text{C} \rightarrow \text{Zn} + \text{CO}\)
If \(\Delta G_f^{\circ}\) (standard free energies of formation, in kJ mol\(^{-1}\)) of ZnS = -205.4, ZnO = -318.0, SO\(_2\) = -300.4, and CO = -137.3. Free energy changes of the above reaction A and B (respectively) will be:
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Negative \(\Delta G\) = spontaneous. Reaction B is non-spontaneous at standard conditions.
Updated On: Apr 23, 2026
- -826.4 kJ, +180.9 kJ
- +826.4 kJ, -180.9 kJ
- -826.4 kJ, -180.9 kJ
- +826.4 kJ, +180.9 kJ
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
\(\Delta G^{\circ}_{\text{rxn}} = \sum \Delta G_f^{\circ}(\text{products}) - \sum \Delta G_f^{\circ}(\text{reactants})\).
Step 2: Detailed Explanation:
For A: \(\Delta G_A = [2(-318.0) + 2(-300.4)] - [2(-205.4) + 3(0)]\)
= \([-636.0 - 600.8] - [-410.8]\) = \([-1236.8] + 410.8 = -826.0\ \text{kJ}\) (approx -826.4).
For B: \(\Delta G_B = [(-137.3) + 0] - [(-318.0) + 0] = -137.3 + 318.0 = +180.7\ \text{kJ}\) (approx +180.9).
Step 3: Final Answer:
Thus, \(\Delta G_A = -826.4\ \text{kJ}\), \(\Delta G_B = +180.9\ \text{kJ}\).
\(\Delta G^{\circ}_{\text{rxn}} = \sum \Delta G_f^{\circ}(\text{products}) - \sum \Delta G_f^{\circ}(\text{reactants})\).
Step 2: Detailed Explanation:
For A: \(\Delta G_A = [2(-318.0) + 2(-300.4)] - [2(-205.4) + 3(0)]\)
= \([-636.0 - 600.8] - [-410.8]\) = \([-1236.8] + 410.8 = -826.0\ \text{kJ}\) (approx -826.4).
For B: \(\Delta G_B = [(-137.3) + 0] - [(-318.0) + 0] = -137.3 + 318.0 = +180.7\ \text{kJ}\) (approx +180.9).
Step 3: Final Answer:
Thus, \(\Delta G_A = -826.4\ \text{kJ}\), \(\Delta G_B = +180.9\ \text{kJ}\).
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