Q. In a chemical equilibrium, the rate constant of the backward reaction is $7.5 \times 10^{- 4}$ and the equilibrium constant is 1.5. So, the rate constant of the forward reaction is
A) $1.125 \times 10^{- 3}$
B) $2.225 \times 10^{- 3}$
C) $3.335 \times 10^{- 5}$
D) $1.125 \times 10^{- 1}$
Explanation
(a)
$K_{c} = \frac{k_{f}}{k_{b}}$
$\therefore k_{f} = K_{c} \times k_{b} = 1.5 \times 7.5 \times 10^{- 4} = 1.125 \times 10^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.