Question

One mole of pure ethyl alcohol was treated with one mole of pure acetic acid at${25^ \circ }C$. One third of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be
A.$\dfrac{1}{4}$
B. $2$
C. $3$
D. $4$

Answer 1

Pure ethyl alcohol is also known as Ethanol, is an organic chemistry compound. It is simple alcohol with formula ${C_2}{H_6}O$, $C{H_3} - C{H_2} - OH$ or ${C_2}{H_5}OH$. Acetic acid or ethanoic acid is a colourless kind of liquid organic compound with the chemical formula $C{H_3}COOH$.
Formula used: Chemical equilibrium formula: ${K_C} =$ $\dfrac{{\left( C \right) \times \left( D \right)}}{{\left( A \right) \times \left( B \right)}}$ in the equation of the form$A + B \to C + D$.

Complete step by step answer:
In this question given, the equation is written as, ${C_2}{H_5}OH + C{H_3}COOH \rightleftharpoons C{H_3}COO{C_2}{H_5} + {H_2}O$
Where, ${C_2}{H_5}OH$ is ethyl alcohol, $C{H_3}COOH$ is acetic acid, $C{H_3}COO{C_2}{H_5}$ is ester and ${H_2}O$ is water as a side product. Now, initially when time $t = 0$, ethyl alcohol and acetic acid have one mole each and the product that is ester and water has not produced any mole. At equilibrium, $x$mole dissociates from ethyl alcohol and acetic acid and becomes as follows,
At equilibrium the volume $v$ at, ethyl alcohol $\left( {{C_2}{H_5}OH} \right)$ produces $1 - x$ mole
Acetic acid $\left( {C{H_3}COOH} \right)$ produces $1 - x$ mole
Ester $\left( {C{H_3}COO{C_2}{H_5}} \right)$ produces $x$ mole
Water $\left( {{H_2}O} \right)$ produces $x$ mole
So therefore, we will apply chemical equilibrium formula. On applying the formula,
${K_C} = \dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right] \times \left[ {{H_2}O} \right]}}{{\left[ {{C_2}{H_5}OH} \right] \times \left[ {C{H_3}COOH} \right]}}$
As it is mentioned in the question, one third of the acid changes into ester at equilibrium, so therefore$x = \dfrac{1}{3}$. Now the equation becomes,
${K_C} = \dfrac{{\left( {\dfrac{1}{3}V} \right) \times \left( {\dfrac{1}{3}V} \right)}}{{\left( {\dfrac{2}{3}V} \right) \times \left( {\dfrac{2}{3}V} \right)}}$, Here v stands for the volume of substances at equilibrium
Now, as we can see after further solving the equation, $3$ gets cancelled, $v$ gets cancelled. All that is left is,
${K_C} = \dfrac{{1 \times 1}}{{2 \times 2}}$
$\therefore$ ${K_C} = \dfrac{1}{4}$
So, the right option is (A).

Note:
The word dissociation that came across the solution means, breaking up of a compound into smaller constituents that are usually capable of recombination under other conditions. The concentration of reactants and products which shows no net charge over the time is called equilibrium.