Question

A reaction has a forward rate constant $2.3\times {10}^6{s}^{-1}$ and an equilibrium constant of $4.0\times {10}^8.$ What is the rate constant for the reverse reaction ?

• A. $1.1\times {10}^{-15}{s}^{-1}$
• B. $5.75\times {10}^{-3}{s}^{-1}$
• C. $1.7\times {10}^2{s}^{-1}$
• D. $9.2\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is B $5.75\times {10}^{-3}{s}^{-1}$

Solution:

$A\stackrel{For}{\underset{Back}{\rightleftharpoons }}B$

$\therefore$ At equilibrium $K_c=\frac{[B{]}_e}{[A{]}_e}$

$r_f=K_f\left[A\right]=r_r=K_r\left[B\right]$

$\frac{K_f}{K_r}=\frac{[B{]}_e}{[A{]}_e}=K_c$ in term of concentration (equilibrium const).

$K_r=\frac{K_r}{K_c}=\frac{2.3\times {10}^{-6}{s}^{-1}}{4\times {10}^8}=5.75\times {10}^{-3}{s}^{-1}$

$\therefore$ Option $B$ is correct answer.