Question

When ethyl alcohol and acetic acid mixed together in equimolecular proportions, equilibrium is attained when two-third of 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

Hint: The equilibrium constant can be calculated by calculating the concentrations of the reactants and the products at equilibrium and then taking the ratio of the product concentration and the reactant concentration.

Complete step by step solution:
-Weak electrolytes cannot dissociate completely into their respective ions. So their ionization is less than 100%. Weak acids, weak bases and sparingly soluble salts come under this category.
Eg. Weak acids like formic acid, acetic acid, hydrogen sulfide are weak electrolytes.
Weak bases like ammonia, pyridine and other nitrogen ring bases are weak electrolytes.
-The reaction is feasible only when strong acids and strong bases react to give weak conjugate acids/bases. So, the equilibrium of the reaction proceeds in that direction only where strong electrolytes are converted to weak electrolytes. This is why the dissociation constant is calculated.
-Both ethyl alcohol and acetic acid are weak bases and acid respectively. So they do not dissociate completely. Their reaction can be written as
${{C}_{2}}{{H}_{5}}OH+C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O$
-The equilibrium constant is defined as the ratio of all the products to all the reactants when the reaction reaches its equilibrium condition. For this reaction, the equilibrium constant is defined as ${{K}_{c}}=\dfrac{\left[ {{H}_{2}}O \right]\left[ C{{H}_{3}}COO{{C}_{2}}{{H}_{5}} \right]}{\left[ {{C}_{2}}{{H}_{5}}OH \right]\left[ C{{H}_{3}}COOH \right]}$
-We can see that 1 mole of both the reactants is used to form the products. So their initial moles can be taken as 1. Now, from the question, equilibrium is attained when two-third of acid and alcohol are consumed. Since equal moles were present initially, both the reactants will be consumed in equal amounts.
-Fraction of moles of reactants consumed = $\dfrac{2}{3}$ (given in the question)
So, the number of reactants which remain at equilibrium will be given as $1$-$\dfrac{2}{3}$= $\dfrac{1}{3}$
The amount of both the products formed during the equilibrium will be $\dfrac{2}{3}$as this is the moles being consumed by the reactants.
-Thus, equilibrium constant can be given as
${{K}_{c}}=\dfrac{\left[ {{H}_{2}}O \right]\left[ C{{H}_{3}}COO{{C}_{2}}{{H}_{5}} \right]}{\left[ {{C}_{2}}{{H}_{5}}OH \right]\left[ C{{H}_{3}}COOH \right]}$
= $\dfrac{\left( \dfrac{2}{3} \right)x\left( \dfrac{2}{3} \right)}{\left( \dfrac{1}{3} \right)x\left( \dfrac{1}{3} \right)}$ = 4

Therefore the correct option is (B) 4.

Note: The degree of dissociation of an electrolyte is the fraction of one mole of electrolyte which has dissociated under the given set of conditions. It depends on the nature of the electrolyte, nature of solvent, dilution, pressure and temperature. It is denoted as$\alpha $. The concept of the dissociation constant is valid only for weak electrolytes and not for strong electrolytes as for them, $\alpha =1$.