Discussion Forum

1)

The rate constant for forward reaction and backward reaction of hydrolysics of ester are $1.1 \times 10^{-2}$ and $1.5 \times 10^{-3}$ per minute respectively . Equilibrium constant for the reaction is 

$CH_{3}COOC_{2}H_{5} + H_{2}O \rightleftharpoons CH_{3}COOH+C_{2}H_{5}OH$

Answer:

Option B

Explanation:

given $k_{f}=1.1 \times 10^{-2} , k_{b}=1.5 \times 10^{-3}$

$k_{c}=\frac{k_{f}}{k_{b}}=\frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}}=7.33$