Question

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

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

Hint:

The standard Gibbs free energy change can be directly calculated from the equilibrium constant using the formula \( \Delta G^{\circ} = -RT \ln K_c \), where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

Answer 1

The Correct Option is C

Step 1: {Using the Gibbs Free Energy Relation} 
The standard Gibbs free energy change is related to the equilibrium constant \( K_c \) by the equation: \[ \Delta G^{\circ} = -RT \ln K_c \] where: - \( R = 8.314 \, {J/K} \cdot {mol} \) is the universal gas constant. - \( T = 298 \, {K} \) is the standard temperature. - \( K_c = 10^{12} \). 
Step 2: {Substitute Values and Calculate} 
\[ \Delta G^{\circ} = - (8.314 \, {J/K} \cdot {mol}) \times (298 \, {K}) \times \ln(10^{12}) \] \[ \Delta G^{\circ} = - 8.314 \times 298 \times 2.303 \] \[ \Delta G^{\circ} = - 68.47 \, {kJ/mol} \] Thus, the correct answer is (C).