Question

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 $e^{-Ea/RT}=4.35\times {10}^{-8}$ at $298K$ with pre-exponential factor of ${10}^{12}{s}^{-1}$. Apparent rate constant $(=k_A)$ at $298K$ is:

• A. $4.35\times {10}^4{s}^{-1}$
• B. $4.35\times {10}^8{s}^{-1}$
• C. $4.35\times {10}^{-8}{s}^{-1}$
• D. $4.35\times {10}^{12}{s}^{-1}$

Answer 1

The correct option is B $4.35\times {10}^8{s}^{-1}$

$b_B=A{e}^{-E_a/RT}$
$={10}^{12}\times 4.35\times {10}^{-8}$
$=4.35\times {10}^4{s}^{-1}$

Also equilibrium constant, $K=\frac{k_A}{k_B}={10}^4$

$\therefore k_A=k_B\times {10}^4=4.35\times {10}^8{s}^{-1}$