Question

One mole of ethyl alcohol was treated with one mole of acetic acid at 25 degree C . 2/3 of the acid changes into Ester at equilibrium. The equilibrium constant for the reaction of hydrolysis of Ester will be

Answer 1

Kc = [2/3][2/3] / [1/3][1/3]
Kc = [4/9] / [ 1/9]
Kc = 4/9 X 9/1
Kc = 4

NB Remember the acid and the alcohol react on a one for one basis
So if 2/3 ester is formed then 2/3 water is formed. leaving unreacted 1/3 acid and 1/3 alcohol.

Answer 2

Explanation:

The given reaction will be as follows.

                       $CH_{3}CH_{2}OH + CH_{3}COOH \rightarrow CH_{3}COOOCH_{2}CH_{3} + H_{2}O$

Before Equilibrium :       1               1                        0

At equilibrium :   $1 -\frac{2}{3}$           $1 -\frac{2}{3}$                          $\frac{2}{3}$                             $\frac{2}{3}$

Hence, equilibrium constant for this reaction will be as follows.

                    K = $\frac{[CH_{3}COOOCH_{2}CH_{3}]}{[CH_{3}CH_{2}OH][CH_{3}COOH]}$

                       = $\frac{[2/3] times [2/3]}{[1/3] \times [1/3]}$

                       = 4

Thus, we can conclude that the equilibrium constant for the reaction of hydrolysis of Ester will be 4.

Kc = [2/3][2/3] / [1/3][1/3]

Kc = [4/9] / [ 1/9]

Kc = 4/9 X 9/1

Kc = 4

NB Remember the acid and the alcohol react on a one for one basis

So if 2/3 ester is formed then 2/3 water is formed. leaving unreacted 1/3 acid and 1/3 alcohol.