Question

In an experiment, $onemole$ of acetic acid and $onemole$ of alcohol were allowed to react until equilibrium was established. The equilibrium mixture was found to contain $2/3mole$ of the ester. Calculate the equilibrium constant of the reaction.

Answer 1

${CH}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons {CH}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$
Inital no. of moles $1$ $1$ $0$ $0$
No.of moles at $e{q}^m$, $(1-\frac{2}{3})$ $(1-\frac{2}{3})$ $\frac{2}{3}$ $\frac{2}{3}$
Active conc. $\frac{1}{3V}$ $\frac{1}{3V}$ $\frac{2}{3V}$ $\frac{2}{3V}$
$V$ is the total volume in litres.
Applying the law of mass action,
$K_c=\frac{\left[ester\right]\left[water\right]}{\left[acid\right]\left[alcohol\right]}=\frac{\frac{2}{3V}\times \frac{2}{3V}}{\frac{1}{3V}\times \frac{1}{3V}}=4$