Equilibrium Question 550
Question: The rate constant for forward and backward reactions 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 $ CH_3COOC_2H_5+H_2O $ ⇌ $ CH_3COOH $ $ +C_2H_5OH $
[AIIMS 1999]
Options:
A) 4.33
B) 5.33
C) 6.33
D) 7.33
Show Answer
Answer:
Correct Answer: D
Solution:
$ 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 $ .
BETA
