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 ($K$) by the equation $\Delta G^0 = -RT \ln K$.

Step 2: Detailed Explanation:
1. Identify given values:
$K_P = 1 \times 10^{-6}$.
$T = 227^\circ C = 227 + 273 = 500 \text{ K}$.
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$.
2. Use the formula:
$\Delta G^0 = -2.303 RT \log K_P$
$\Delta G^0 = -2.303 \times 8.3 \times 500 \times \log(10^{-6})$
$\Delta G^0 = -2.303 \times 8.3 \times 500 \times (-6)$
$\Delta G^0 = +2.303 \times 8.3 \times 3000$
$\Delta G^0 = 57344.7 \text{ J mol}^{-1} \approx 57.3 \text{ kJ mol}^{-1}$.

Step 3: Final Answer:
The value of $\Delta G^0$ is $+57.3\text{ kJ mol}^{-1}$.