Question

The standard Gibbs energy (\(\Delta G^\circ\)) for the following reaction is:
\[ A(s) + B^{2+} (aq) \rightleftharpoons A^{2+} (aq) + B(s), \quad K_c = 10^{12} \text{ at 25°C} \]

• A. -150 kJ
• B. -96.80 kJ
• C. -68.47 kJ
• D. -100 kJ

Hint:

Use \(\Delta G^\circ = -RT \ln K\) to relate equilibrium constant with Gibbs free energy at standard conditions.

Answer 1

The Correct Option is C

Step 1: Use \(\Delta G^\circ = -2.303RT\log K\).
\(\Delta G^\circ = -2.303 \times 8.314 \times 10^{-3} \text{ kJ} \times 298 \times 12\).

Step 2: Calculate.
\(= -2.303 \times 2.478 \times 12 \approx -68.47\) kJ.
\[ \boxed{-68.47 \text{ kJ}} \]