1. Tardigrade
  2. Question
  3. Chemistry
  4. The rate constant for forward and backward reaction of hydrolysis of ester are 1.1 × 10-2 and 1.5 × 10-3 per minute respectively. Equilibrium constant for the reaction CH 3 COOC 2 H 5+ H + leftharpoons CH 3 COOH + C 2 H 5 OH is

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Q. The rate constant for forward 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 $CH _3 COOC _2 H _5+ H ^{+} \rightleftharpoons CH _3 COOH + C _2 H _5 OH$ is

Equilibrium

Solution:

Rate constant of forward reaction $(K)=1.1 \times 10^{-2}$ and rate constant of backward reaction $\left(K_b\right)=1.5 \times 10^{-3}$ per minute.
Equilibrium constant $\left(K_c\right)=\frac{K_f}{K_b}=\frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}}=7.33$