Equilibrium - Result Question 20
####21. 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]
$CH_3 COOC_2 H_5+H^{+} \rightarrow$
$CH_3 COOH+C_2 H_5 OH$ is
(a) 4.33
(b) 5.33
(c) 6.33
(d) 7.33
Topic 2: Relation between $K, Q$ and $G$ and Factors Effecting Equilibrium
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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 $(K_c)$
$=\frac{K_f}{K_b}=\frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}}=7.33$
