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Question: The rate constant for forward and backward reactions of hydrolysis of ester are \(1.1 \times 10 ^ { ...

The rate constant for forward and backward reactions of hydrolysis of ester are 1.1×10−21.1 \times 10 ^ { - 2 } and 1.5×10−31.5 \times 10 ^ { - 3 } per minute respectively. Equilibrium constant for the reaction,

CH3COOC2H5+H2O\mathrm { CH } _ { 3 } \mathrm { COOC } _ { 2 } \mathrm { H } _ { 5 } + \mathrm { H } _ { 2 } \mathrm { O }⇌ CH3COOH+C2H5OH\mathrm { CH } _ { 3 } \mathrm { COOH } + \mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }is

A

4.33

B

5.33

C

6.33

D

7.33

Answer

7.33

Explanation

Solution

Kf=1.1×10−2K _ { f } = 1.1 \times 10 ^ { - 2 }, Kb=1.5×10−3K _ { b } = 1.5 \times 10 ^ { - 3 };

Kc=KfKb=1.1×10−21.5×10−3=7.33K _ { c } = \frac { K _ { f } } { K _ { b } } = \frac { 1.1 \times 10 ^ { - 2 } } { 1.5 \times 10 ^ { - 3 } } = 7.33