Question

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 :
$C{H}_{3}COO{C}_2{H}_5+H_{2}O\rightleftharpoons C{H}_{3}COOH+C_2{H}_{5}OH$

• A. $33.7$
• B. $7.33$
• C. $5.33$
• D. $33.3$

Answer 1

Answer: (B)

$7.33$

Given,

$k_f=1.1\times {10}^{-2},k_b=1.5\times {10}^{-3}$

$K_c=\frac{k_f}{k_b}=\frac{1.1\times {10}^{-2}}{1.5\times {10}^{-3}}=7.33$