Question

In an experiment starting with $1$ mole of ethyl of alcohol, $1$ mole of acetic acid and $1$ mole of water at ${100^ \circ }C$ , the equilibrium mixture on analysis shows that $53.4{\text{\% }}$ of the acid is esterified. Calculate the equilibrium constant of the reaction.

Answer 1

Let us first of all have a brief idea about equilibrium constant of a reaction. When a chemical process reaches equilibrium, the equilibrium constant (typically indicated by the symbol ${\text{K}}$) offers information on the relationship between the products and reactants. The ratio of the concentration of products to the concentration of reactants, each raised to their respective stoichiometric coefficients, is the equilibrium constant of concentration of a chemical reaction.

Complete answer:
From the question, we have
$1$ mole of ethyl of alcohol,
$1$ mole of acetic acid,
and $1$ mole of water at ${100^ \circ }C$.

$${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH(l) + }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH(l)}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_2}{{\text{H}}_5}{\text{(l) + }}{{\text{H}}_{\text{2}}}{\text{O}} \\\ {\text{Initial concentration}} \\\ {\text{(in moles) }}1{\text{ 1 0 1}} \\\ {\text{Equilibrium concentration}} \\\ {\text{(in moles) (1 - x) (1 - x) x (1 + x)}} \\\ {\text{ 1 - 0}}{\text{.543 1 - 0}}{\text{.543 0}}{\text{.543 1 + 0}}{\text{.543 }} \\\$$

It is also given in the question that $53.4{\text{\% }}$ of the acid is esterified.
This implies that ${\text{x = }}\dfrac{{53.4}}{{100}}{\text{ = 0}}{\text{.543}}$.
Now, we can apply the mass action law.
So, from mass action law, we have
${{\text{K}}_c} = \dfrac{{{\text{[ester][water]}}}}{{{\text{[acid][alcohol]}}}} = \dfrac{{0.543(1 + 0.543)}}{{(1 - 0.543)(1 - 0.543)}} = 4$
Thus, the required equilibrium constant of the reaction is $4$.
In case the stoichiometry of an equilibrium reaction is changed, the power of the equilibrium constant is likewise changed by the same amount.

Note:
We can note some characteristics of equilibrium constant. It's reaction-specific, and it's set at a constant temperature like the rate of forward and backward reactions is changed equally by a catalyst, hence the value of the equilibrium constant is not affected. Changes in concentration, pressure, temperature, and the presence of inert gases can modify the equilibrium, favouring either forward or backward reaction but not the equilibrium constant.