Equilibrium Question 633

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 $ .



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