Question

In a chemical equilibrium the rate constant of backward reaction is $3.2 \times 10^{-2}$ and the equilibrium constant is $2^{-5}$. The rate constant of forward reaction is

• A. $1 \times 10^{-3}$
• B. $2 \times 10^{-2}$
• C. $8 \times 10^{-2}$
• D. $4 \times 10^{-2}$

Answer 1

Answer: (A)

In a chemical equilibrium
Equilibrium constant,
$K=\frac{\text{Rate constant of forward reaction}, K_{f}}{\text{Rate constant of backward reaction}, K_{b}}$
$\therefore K_{f}=K_{b}\times K$
$=3.2\times10^{-2}\times2^{-5}$
$=\frac{3.2\times10^{-2}}{32}=1\times10^{-3}$