Equilibrium Ques 14
14. 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
[1995]
$CH_3 COOC_2 H_5+H^{+} \rightleftharpoons$ $CH_3 COOH+C_2 H_5 OH$ is
(a) $4.33$
(b) $5.33$
(c) $6.33$
(d) $7.33$
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Answer:
Correct Answer: 14.(d)
Solution:
- (d) Rate constant of forward reaction $(K_f)$ $=1.1 \times 10^{-2}$ and rate constant of backward reaction $(K_b)=1.5 \times 10^{-3}$ per minute. Equilibrium constant $(K_c)$
$=\frac{K_f}{K_b}=\frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}}=7.33$
BETA
