Question

$K_C$ for the esterification reaction : $C{H}_{3}COOH+C{H}_2{H}_{5}OH\rightleftharpoons C{H}_{3}COO{C}_{2}H$ is $4$.If $4mol$ each of acid and alcohol are initially ,what is the equilibrium moles the acid:

• A. $\frac{2}{3}$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. None of these

Answer 1

The correct option is B $\frac{4}{3}$

Given reaction is-
$C{H}_{3}COOH+C_2{H}_{5}OH\leftrightharpoons C{H}_{3}COO{C}_{2}H$
Initially, $4$ $4$ $0$
Finally, $4-x$ $4-x$ $x$
$K_C=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]}{\left[C_2{H}_{5}OH\right]\left[C{H}_{3}COOH\right]}=\frac{x}{(4-x{)}^2}=4$
$⟹x=\frac{8}{3}$
So, number of moles of acid $=4-\frac{8}{3}=\frac{4}{3}$