Question

Determine the amount of ester present under equilibrium when 3 moles of ethyl alcohol react with 1 mole of acetic acid, when equilibium constant of the reaction is given as 4.

• A. $0.655$
• B. $0.326$
• C. $0.521$
• D. $0.903$

Answer 1

The correct option is D $0.903$

Given: Initial concentration of $C{H}_{3}COOH$ = $\frac{1}{V}$
Initial concentration of $C_2{H}_{5}OH$ = $\frac{3}{V}$
$C{H}_{3}COOH+C_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_2{H}_5+H_{2}O$
$\frac{1}{V}$$\frac{3}{V}$
$\frac{1-x}{V}\frac{3-x}{V}\frac{x}{V}\frac{x}{V}$
$K_c=4=\frac{\left(\frac{x}{V}\right)\left(\frac{x}{V}\right)}{\left(\frac{1-x}{V}\right)\left(\frac{3-x}{V}\right)}$
$3x^2-16x+12=0$
$x=0.903$
Amount of ester at equilibrium = 0.903 mole