JEE Advanced · Chemistry · 6. Thermodynamics (C)
The value of \(\log _{10} \mathrm{~K}\) for a reaction \(A \rightleftharpoons B\) is (Given: \(\Delta_{\mathrm{r}} H_{298 \mathrm{~K}}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{r}} S_{298 \mathrm{~K}}^{\circ}\)\(=10 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) and \(R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; 2.303 \times 8.314 \times 298\)\(=5705\) )
- A 5
- B 10
- C 95
- D 100
Answer & Solution
Correct Answer
(B) 10
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
A & \rightleftharpoons B \\
\Delta G^{\circ} & =\Delta H^{\circ}-T \Delta S^{\circ} \\
\Delta G^{\circ} & =-2303 R T \log K \\
\log K & =\frac{\Delta H^{\circ}-T \Delta S^{\circ}}{-2303 R T} \\
& =\frac{-54.07 \times 10^3-298 \times 10}{-2303 \times 8.314 \times 298} \\
& =\frac{-57050}{-5705}=10
\end{aligned}
\)
\begin{aligned}
A & \rightleftharpoons B \\
\Delta G^{\circ} & =\Delta H^{\circ}-T \Delta S^{\circ} \\
\Delta G^{\circ} & =-2303 R T \log K \\
\log K & =\frac{\Delta H^{\circ}-T \Delta S^{\circ}}{-2303 R T} \\
& =\frac{-54.07 \times 10^3-298 \times 10}{-2303 \times 8.314 \times 298} \\
& =\frac{-57050}{-5705}=10
\end{aligned}
\)
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