Question

The equilibrium constant $K_P$ of a homogenous equilibrium reaction is $1 \times 10^{-6}$ at $227^\circ C$. What is the value of $\Delta G^0$ of the reaction at the same temperature? ($R = 8.3\text{JK}^{-1}\text{mol}^{-1}$)}

• A. 26.3 kJ mol-1
• B. +57.3 kJ mol-1
• C. -57.3 kJ mol-1
• D. -26.3 kJ mol-1

Hint:

Since $K_P < 1$, the log term is negative, making $\Delta G^0$ positive. This indicates that the reaction is non-spontaneous in the forward direction under standard conditions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The standard Gibbs free energy change (\( \Delta G^0 \)) is related to the equilibrium constant by a logarithmic relation. A small equilibrium constant (\( K<1 \)) implies a positive \( \Delta G^0 \) (non-spontaneous under standard conditions).
Step 2: Key Formula or Approach:
\( \Delta G^0 = -2.303 RT \log K_P \).
Temperature must be in Kelvin (\( T = 227 + 273 = 500 \text{ K} \)).
Step 3: Detailed Explanation:
Given: \( K_P = 10^{-6} \), \( T = 500 \text{ K} \), \( R = 8.3 \text{ J/Kmol} \).
\[ \Delta G^0 = -2.303 \times 8.3 \times 500 \times \log(10^{-6}) \] \[ \Delta G^0 = -2.303 \times 8.3 \times 500 \times (-6) \] \[ \Delta G^0 = 2.303 \times 8.3 \times 3000 \] \[ \Delta G^0 = 19.11 \times 3000 = 57330 \text{ J/mol} \] Convert to kJ/mol: \( \frac{57330}{1000} \approx +57.3 \text{ kJ/mol} \).
Step 4: Final Answer:
The value of \( \Delta G^0 \) is +57.3 kJ mol-1.