Search any question & find its solution
Question:
Answered & Verified by Expert
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 $R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} ; 2.303 \times 8.314 \times 298=5705$ )
Options:
Solution:
2232 Upvotes
Verified Answer
The correct answer is:
10
10
$$
\begin{aligned}
A & \rightleftharpoons B \\
\Delta G^{\circ} & =\Delta H^{\circ}-T \Delta S^{\circ} \\
\Delta G^{\circ} & =-2303 R T \log K \\
\log K & =\frac{\Delta H^{\circ}-T \Delta S^{\circ}}{-2303 R T} \\
& =\frac{-54.07 \times 10^3-298 \times 10}{-2303 \times 8.314 \times 298} \\
& =\frac{-57050}{-5705}=10
\end{aligned}
$$
\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}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.