Verified Solution • 2025–26 Session moderate

QJEEM23Chcf40: JEE Main 2023 Chemistry - Class 11 JEEM23Chcf40 Solution: The value of for the reaction at is ___________. (Nearest…

The value of log⁡K\log \mathrm{K} for the reaction A⇌B\mathrm{A} \rightleftharpoons \mathrm{B} at 298 K298 \mathrm{~K} is ___________. (Nearest integer)

Given: ΔH∘=−54.07 kJ mol−1\Delta \mathrm{H}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}

ΔS∘=10 J K−1 mol−1\Delta \mathrm{S}^{\circ}=10 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

(Take 2.303×8.314×298=57052.303 \times 8.314 \times 298=5705 )

Step-by-Step Verified Solution

Given:

\begin{align}\Delta H^0 & = -54.07 \, \text{kJ/mol} = -54070 \, \text{J/mol} \\\\\Delta S^0 & = 10 \, \text{J/K}\cdot \text{mol} \\\\T & = 298 \, \text{K}\end{align}

We find the change in Gibbs free energy ΔG0\Delta G^0 :

\begin{align}\Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\& = -54070 - (10 \times 298) \\\\& = -54070 - 2980 \\\\& = -57050 \, \text{J/mol}\end{align}

Now, we'll use the given expression with the correct constant for RR as 8.314 J/(mol·K):

\begin{align}\Delta G^0 & = -2.303 \times 8.314 \times 298 \times \log K \\\\\log K & = \frac{-57050}{2.303 \times 8.314 \times 298} \\\\& \approx 10\end{align}

So, the answer is log⁡K=10\log K = 10 .

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