Q. The rate constant for forward reaction and backward reaction of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively. Equilibrium constant for the reaction is $CH_{3}COOC_{2}H_{5} + H_{2}O \rightleftharpoons CH_{3}COOH + C_{2}H_{5}OH$
A) 33.7
B) 7.33
C) 5.33
D) 33.3
Explanation
(b)
$k_{f} = 1.1 \times 10^{- 2},\ k_{b} = 1.5 \times 10^{- 3}$
$K_{c} = \frac{k_{f}}{k_{b}} = \frac{1.1 \times 10^{- 2}}{1.5 \times 10^{- 3}} = 7.33$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.