Question

The rate of disappearance of $A$ in the reaction at equilibrium $A\rightleftharpoons B$ is given by :
$-\frac{d\left[A\right]}{dt}=2\times {10}^{-2}\left[A\right]-4\times {10}^{-3}\left[B\right]$ at $300K$.

The equilibrium constant $K_c$ is :

Answer 1

The rate of disappearance of $A$ in the reaction at equilibrium $A\rightleftharpoons B$ is given by
$-\frac{d\left[A\right]}{dt}=k_f\left[A\right]-k_b\left[B\right]=2\times {10}^{-2}\left[A\right]-4\times {10}^{-3}\left[B\right]$ at $300K$.

The equilibrium constant, $K_c=\frac{K_f}{K_b}=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}=5$