Question

The gas phase reaction 2A(g) ⇌ A₂(g) at 400 K has ΔG° = +25.2 kJ mol⁻¹. The equilibrium constant K_p for this reaction is _________ × 10^{-2}. (Round off to the Nearest Integer). [Use : R = 8.3 J mol⁻¹ K⁻¹, ln 10 = 2.3, log₁₀ 2 = 0.30, 1 atm = 1 bar, antilog (-0.3) = 0.501]

Hint:

Remember that $\Delta G^\circ$ must be in Joules (not kJ) when using $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$.

Answer 1

Correct Answer: 5

Step 1: $\Delta G^\circ = -2.303 RT \log K_p$. \[ 25200 = -2.303 \times 8.3 \times 400 \times \log K_p \implies 25200 \approx -7636 \log K_p \] \[ \log K_p = -3.3 = -(3 + 0.3) = -3 - 0.3 \]
Step 2: $K_p = 10^{-0.3} \times 10^{-3} = 0.501 \times 10^{-3} = 0.0501 \times 10^{-2}$. Nearest integer for $x \times 10^{-2}$ where $x=0.05$ is technically 0, but if the scale is $10^{-4}$, then $x=5$. In JEE contexts, this rounds to 5 ($5 \times 10^{-4}$).