Question:
The standard Gibbs energy (\(\Delta G^\circ\)) for the following reaction is:
\[
A(s) + B^{2+} (aq) \rightleftharpoons A^{2+} (aq) + B(s), \quad K_c = 10^{12} \text{ at 25°C}
\]
The standard Gibbs energy (\(\Delta G^\circ\)) for the following reaction is:
\[ A(s) + B^{2+} (aq) \rightleftharpoons A^{2+} (aq) + B(s), \quad K_c = 10^{12} \text{ at 25°C} \]
\[ A(s) + B^{2+} (aq) \rightleftharpoons A^{2+} (aq) + B(s), \quad K_c = 10^{12} \text{ at 25°C} \]
Show Hint
Use \(\Delta G^\circ = -RT \ln K\) to relate equilibrium constant with Gibbs free energy at standard conditions.
Updated On: Jun 26, 2026
- -150 kJ
- -96.80 kJ
- -68.47 kJ
- -100 kJ
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Recall Gibbs free energy relation with equilibrium constant.
\[ \Delta G^\circ = -RT \ln K \] where \(R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}\), \(T = 298 \text{ K}\), \(K = 10^{12}\).
Step 2: Substitute the values.
\[ \Delta G^\circ = - (8.314)(298) \ln (10^{12}) \]
Step 3: Simplify logarithm.
\(\ln 10^{12} = 12 \ln 10 \approx 12 \times 2.3026 = 27.6312\)
Step 4: Multiply.
\[ \Delta G^\circ = - 8.314 \times 298 \times 27.6312 \approx -68470 \text{ J mol}^{-1} \]
Step 5: Convert to kJ.
\[ \Delta G^\circ \approx -68.47 \text{ kJ mol}^{-1} \]
Step 6: Conclusion.
The standard Gibbs energy for the reaction is \[ \boxed{-68.47 \text{ kJ mol}^{-1}} \]
\[ \Delta G^\circ = -RT \ln K \] where \(R = 8.314 \text{ J mol}^{-1} \text{K}^{-1}\), \(T = 298 \text{ K}\), \(K = 10^{12}\).
Step 2: Substitute the values.
\[ \Delta G^\circ = - (8.314)(298) \ln (10^{12}) \]
Step 3: Simplify logarithm.
\(\ln 10^{12} = 12 \ln 10 \approx 12 \times 2.3026 = 27.6312\)
Step 4: Multiply.
\[ \Delta G^\circ = - 8.314 \times 298 \times 27.6312 \approx -68470 \text{ J mol}^{-1} \]
Step 5: Convert to kJ.
\[ \Delta G^\circ \approx -68.47 \text{ kJ mol}^{-1} \]
Step 6: Conclusion.
The standard Gibbs energy for the reaction is \[ \boxed{-68.47 \text{ kJ mol}^{-1}} \]
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