Question

When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when two-thirds of the acid and alcohol are consumed. The equilibrium constant of the reaction will be:
(A) 0.4
(B) 4
(C) 40
(D) 0.04

Answer 1

The concentration of both products and reactants can be calculated from the given fraction of reactants consumed at equilibrium in the question. We can then calculate the equilibrium constant for the given reaction by dividing the concentration of products by concentration of reactants.

Complete Solution:
-As we know, equilibrium constant is the ratio of the product concentration that is raised to the powers of their coefficient to the products of the reactant concentration that is raised to powers of their coefficient.
-Now let's consider the given reaction. Since its given as equimolar ratio,let one mole of acetic acid ( $C{{H}_{3}}COOH$) and ethyl alcohol ( ${{C}_{2}}{{H}_{5}}OH$) mixed together. ${{C}_{2}}{{H}_{5}}OH$ We can write the equation as follows
${{C}_{2}}{{H}_{5}}OH+C{{H}_{3}}COOH\quad \to C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$

-This is an esterification reaction. $OH$ from ethyl alcohol and $H$ from acetic acid condenses and water molecules gets eliminated. Thus, an ester is formed.
- In the question it's given that the equilibrium is attained when two-thirds of the acid and alcohol are consumed. Hence, we can write
$Fraction\text{ }of\text{ }both\text{ }consumed=\dfrac{2}{3}$
Therefore, the number of moles of ethyl alcohol( ${{C}_{2}}{{H}_{5}}OH$) and acetic acid ($C{{H}_{3}}COOH$)left
= $1-\dfrac{2}{3}$
= $\dfrac{1}{3}$
The number of moles of ester ( $C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}$) and water at equilibrium
= $\dfrac{2}{3}$
Therefore, the equilibrium constant ( ${{K}_{c}}$) for the reaction can be written as
${{K}_{c}}=\dfrac{\left[ {{H}_{2}}O \right]\left[ C{{H}_{3}}COO{{C}_{2}}{{H}_{5}} \right]}{\left[ {{C}_{2}}{{H}_{5}}OH \right]\left[ C{{H}_{3}}COOH \right]}$

By substituting the values, we got,
${{K}_{c}}=\dfrac{\left[ \dfrac{2}{3} \right]\times \left[ \dfrac{2}{3} \right]}{\left[ \dfrac{1}{3} \right]\times \left[ \dfrac{1}{3} \right]} =4$

So, the correct answer is “Option B”.

Note: We can also represent the answer in an easier way as follows, $\begin{matrix} {} & {{C}_{2}}{{H}_{5}}OH+ & C{{H}_{3}}COOH\to & C{{H}_{3}}COO{{C}_{2}}{{H}_{5}} & +{{H}_{2}}O \\\ Initial & 1\text{ }mol & 1\text{ }mol & \- & \- \\\ Equilibrium & \dfrac{1}{3}mol & \dfrac{1}{3}mol & \dfrac{2}{3}mol & \dfrac{2}{3}mol \\\ \end{matrix}$ Therefore, ${{K}_{c}}=\dfrac{\left[ \dfrac{2}{3} \right]\times \left[ \dfrac{2}{3} \right]}{\left[ \dfrac{1}{3} \right]\times \left[ \dfrac{1}{3} \right]} =4$