JEE MAIN - Chemistry (2023 - 6th April Morning Shift - No. 19)
The value of $$\log \mathrm{K}$$ for the reaction $$\mathrm{A} \rightleftharpoons \mathrm{B}$$ at $$298 \mathrm{~K}$$ is ___________. (Nearest integer)
Given: $$\Delta \mathrm{H}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$$
$$\Delta \mathrm{S}^{\circ}=10 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$
(Take $$2.303 \times 8.314 \times 298=5705$$ )
Explanation
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 $\Delta G^0$:
$$
\begin{align}
\Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\
& = -54070 - (10 \times 298) \\\\
& = -54070 - 2980 \\\\
& = -57050 \, \text{J/mol}
\end{align}
$$
Now, we'll use the given expression with the correct constant for $ R $ 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 $\log K = 10$.
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