Question

the equilibrium constant for the esterification reaction of acidic acid and ethyl alcohol at 100 degree celsius is 4 then what percentage of alcohol has been esterified​

Answer 1

Answer:

reaction of acidic acid and ethyl alcohol at 100 degree celsius is 4 then what percentage of alcohol has been esterified

Answer 2

Given info : the equilibrium constant for the esterification reaction of acidic acid and ethyl alcohol at 100 degree celsius is 4.

To find : the percentage of alcohol has been esterified is ..

solution : esterification reaction is,

CH3COOH + C2H5OH => CH3COOC2H5 + H2O

at eql 1 - x 1 - x x x

now equilibrium constant, k = [H2O][CH3COOC2H5][CH3COOH][C2H5OH]

⇒4 = x × x/(1 - x) × (1 - x)

⇒4 = x²/(1 - x)²

⇒2 = x/(1 - x)

⇒2 - 2x = x

⇒x = 2/3 = 0.667

value of x percentage = 66.7 %

Therefore the percentage of Alcohol has been esterified, is 66.7 %