Question

At 300 K, \( \Delta_r G^\circ \) for the reaction \( A(g) \rightleftharpoons B(g) \) is \( -11.5 \) kJ mol\(^{-1}\). The equilibrium constant at 300 K is approximately: \((R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1})\)

• A. \( 10 \)
• B. \( 100 \)
• C. \( 1000 \)
• D. \( 25 \)

Hint:

Use the relation \( \Delta G^\circ = -RT \ln K \) to calculate equilibrium constants. If \( \Delta G^\circ \) is negative, \( K \) is greater than 1, favoring product formation.

Answer 1

The Correct Option is B

Step 1: Using the Gibbs Free Energy and Equilibrium Constant Relation 
The standard Gibbs free energy change \( \Delta_r G^\circ \) is related to the equilibrium constant \( K \) by: \[ \Delta_r G^\circ = - RT \ln K \] where: - \( R = 8.314 \) J mol\(^{-1}\) K\(^{-1}\) = \( 8.314 \times 10^{-3} \) kJ mol\(^{-1}\) K\(^{-1}\),
- \( T = 300 \) K,
- \( \Delta_r G^\circ = -11.5 \) kJ mol\(^{-1}\). 
Step 2: Rearranging for \( K \) \[ \ln K = \frac{-\Delta_r G^\circ}{RT} \] Substituting values: \[ \ln K = \frac{-(-11.5)}{(8.314 \times 10^{-3}) \times 300} \] \[ \ln K = \frac{11.5}{2.4942} \] \[ \ln K \approx 4.61 \] Step 3: Finding \( K \) 
Taking the exponent: \[ K = e^{4.61} \] Approximating: \[ e^{4.61} \approx 100 \] Final Answer: The equilibrium constant is approximately \( 100 \), which matches Option (2).