Question

In an experiment starting with $1$ mole of ethyl of alcohol, $1$ mole of acetic acid and $1$ mole of water at ${100}^{o}C$, the equilibrium mixture on analysis shows that $54.3$% of the acid is esterified. Calculate the equilibrium constant of this reaction.

Answer 1

We start with $1$ mole of ethyl alcohol, $1$ mole of acetic acid and $1$ mole of water.
${CH}_{3}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons {CH}_{3}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$
Initial moles $1$ $1$ $0$ $1$
Equilibrium moles $(1-x)$ $(1-x)$ $x$ $(1+x)$
$1-0.543$ $1-0.543$ $0.543$ $1+0.543$
Note: $54.3$% of the acid is esterified. Hence, $x=\frac{54.3}{100}=0.543$
Applying law of mass action,
$K_c=\frac{\left[ester\right]\left[water\right]}{\left[acid\right]\left[alcohol\right]}=\frac{0.543(1+0.543)}{(1-0.543)(1-0.543)}=4$