Q34 / 120 · VITEEE 2013 · Chemistry
Consider the following reaction,
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$. For the conversion of chair form to boat form $\mathrm{e}^{-E a / R T}=$ $4.35 \times 10^{-8} \mathrm{~m}$ at $298 \mathrm{~K}$ with pre-exponential factor of $10^{12} \mathrm{~s}^{-1}$. Apparent rate constant $(=k A / k B)$ at $298 \mathrm{~K}$ is
The reaction is of first order in each diagram, with an equilibrium constant of $10^{4}$. For the conversion of chair form to boat form $\mathrm{e}^{-E a / R T}=$ $4.35 \times 10^{-8} \mathrm{~m}$ at $298 \mathrm{~K}$ with pre-exponential factor of $10^{12} \mathrm{~s}^{-1}$. Apparent rate constant $(=k A / k B)$ at $298 \mathrm{~K}$ is
- A$4.35 \times 10^{4} \mathrm{~s}^{-1}$
- Correct answerB$4.35 \times 10^{8} \mathrm{~s}^{-1}$
- C$4.35 \times 10^{-8} \mathrm{~s}^{-1}$
- D$4.35 \times 10^{12} \mathrm{~s}^{-1}$
Solution
$\begin{aligned}
& \mathrm{K}_{\mathrm{B}}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}} \\
&=10^{12} \times 4.35 \times 10^{-8} \\
&=4.35 \times 10^{4} \mathrm{~s}^{-1}
\end{aligned}$
Also equilibrium constant,
$\mathrm{k}=\frac{\mathrm{k}_{\mathrm{A}}}{\mathrm{k}_{\mathrm{B}}}=10^{4}$
$\therefore \mathrm{k}_{\mathrm{A}}=\mathrm{k}_{\mathrm{B}} \times 10^{4}=4.35 \times 10^{8} \mathrm{~s}^{-1}$
Hence, the correct answer is option: 'B'.