Question

The rate constant of forward and backward reactions for certain hypothetical reaction are $1.1\times {10}^{-2}$ and $1.5\times {10}^{-3}$, respectively. The equilibrium constant of the reaction is $733\times {10}^{-x}$. The value of $x$ is:

Answer 1

Rate of foward reaction, $k_f=1.1\times {10}^{-2}$

Rate of backward reaction, $k_b=1.5\times {10}^{-3}$

$\begin{array}{cc}{k}_{\omega }& =\frac{k_f}{k_b}=\frac{1.1\times {10}^{-2}}{1.5\times {10}^{-3}}=7.33\\ & \\ & =733\times {10}^{-2}\end{array}$

$x=2$