Q22 / 150 · MHT CET 2026 (11 April Shift 2) · Chemistry
The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298$ K ? ($R = 8.314$ J $\text{K}^{-1}\text{mol}^{-1}$)
- Correct answerA$-11.411$ KJ/mol
- B$-5.744$ KJ/mol
- C$-570.584$ KJ/mol
- D$-57.058$ KJ/mol
Solution
The standard Gibbs free energy change is given by the equation:
$\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$
Substituting the given values:
$R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$
$T = 298 \text{ K}$
$K = 100$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log(100)$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times 2$
$\Delta G^{\circ} = -11411.7 \text{ J mol}^{-1}$
$\Delta G^{\circ} \approx -11.411 \text{ kJ mol}^{-1}$
Answer: $-11.411$ KJ/mol
$\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$
Substituting the given values:
$R = 8.314 \text{ J K}^{-1}\text{mol}^{-1}$
$T = 298 \text{ K}$
$K = 100$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times \log(100)$
$\Delta G^{\circ} = -2.303 \times 8.314 \times 298 \times 2$
$\Delta G^{\circ} = -11411.7 \text{ J mol}^{-1}$
$\Delta G^{\circ} \approx -11.411 \text{ kJ mol}^{-1}$
Answer: $-11.411$ KJ/mol
Hence, the correct answer is option: 'A'.