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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
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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
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Verified Answer
The correct answer is:
$4.35 \times 10^{8} \mathrm{~s}^{-1}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{B}}=\mathrm{Ae}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}} \\
&=10^{12} \times 4.35 \times 10^{-8} \\
&=4.35 \times 10^{4} \mathrm{~s}^{-1}
\end{aligned}$
Also equilibrium constant,
$\mathrm{k}=\frac{\mathrm{k}_{\mathrm{A}}}{\mathrm{k}_{\mathrm{B}}}=10^{4}$
$\therefore \mathrm{k}_{\mathrm{A}}=\mathrm{k}_{\mathrm{B}} \times 10^{4}=4.35 \times 10^{8} \mathrm{~s}^{-1}$
& \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,
$\mathrm{k}=\frac{\mathrm{k}_{\mathrm{A}}}{\mathrm{k}_{\mathrm{B}}}=10^{4}$
$\therefore \mathrm{k}_{\mathrm{A}}=\mathrm{k}_{\mathrm{B}} \times 10^{4}=4.35 \times 10^{8} \mathrm{~s}^{-1}$
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