Question

One mole of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$and one mole of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}}$reacts to produce $\dfrac{{\text{2}}}{{\text{3}}}{\text{mol}}$ of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$. The equilibrium constant is:
A. ${\text{2}}$
B. ${\text{ + 2}}$
C. ${\text{ - 4}}$
D. ${\text{4}}$

Answer 1

Hint: To reach out to the solution of the given problem we will define the moles of the substances before reaction and then after reaction. After that we will use the formula of equilibrium constant.

Complete step by step answer: For the reaction, ${\text{aA + bB}} \rightleftharpoons {\text{cC + dD}}$.
Equilibrium constant is ${{\text{K}}_{\text{C}}}{\text{ = }}\dfrac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}$. Equilibrium constant is nothing but the ratio of the product of the concentration of the products raised to the power of their stoichiometric coefficient to the product of the concentrations of the reactants raised to the power of the stoichiometric coefficients.
For the given reaction, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$.
We will define the number of moles at the start of the reaction.
${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH + }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$.
At start, ${\text{1}}$ ${\text{1}}$ ${\text{0}}$ ${\text{0}}$
At equilibrium, ${\text{1 - }}\dfrac{{\text{2}}}{{\text{3}}}$ ${\text{1 - }}\dfrac{{\text{2}}}{{\text{3}}}$ $\dfrac{{\text{2}}}{{\text{3}}}$ $\dfrac{{\text{2}}}{{\text{3}}}$
$\dfrac{{\text{2}}}{{\text{3}}}$ is the amount of moles of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}$produce in reaction.
Now, let the total volume be VL.
So, the concentration of the reactants:
${\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH] = }}\dfrac{{\text{1}}}{{\text{3}}}{\text{Vmol/L}}$
${\text{[}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH] = }}\dfrac{{\text{1}}}{{\text{3}}}{\text{Vmol/L}}$.
The concentration of the products will be as follows:

${\text{[C}}{{\text{H}}_{\text{3}}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{] = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{Vmol/L}}$
${\text{[}}{{\text{H}}_{\text{2}}}{\text{O] = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{Vmol/L}}$.
Now, after calculating all the required concentrations of the reactants and the products we will substitute that into the expression of equilibrium constant ${{\text{K}}_{\text{C}}}{\text{ = }}\dfrac{{{{{\text{[C]}}}^{\text{c}}}{{{\text{[D]}}}^{\text{d}}}}}{{{{{\text{[A]}}}^{\text{a}}}{{{\text{[B]}}}^{\text{b}}}}}$.
\[\]${{\text{K}}_{\text{C}}}{\text{ = }}\dfrac{{{{{\text{[CH}}{}_{\text{3}}{\text{COO}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{]}}}^{\text{1}}}{{{\text{[}}{{\text{H}}_{\text{2}}}{\text{O]}}}^{\text{1}}}}}{{{{{\text{[C}}{{\text{H}}_{\text{3}}}{\text{COOH]}}}^{\text{1}}}{{{\text{[}}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{OH]}}}^{\text{1}}}}}$.
Substituting the values,
${{\text{K}}_{\text{C}}}{\text{ = }}\dfrac{{\dfrac{{\text{2}}}{{\text{3}}}{\text{Vmol/L \times }}\dfrac{{\text{2}}}{{\text{3}}}{\text{Vmol/L}}}}{{\dfrac{{\text{1}}}{{\text{3}}}{\text{Vmol/L \times }}\dfrac{{\text{1}}}{{\text{3}}}{\text{Vmol/L}}}}$.
${{\text{K}}_{\text{C}}}{\text{ = }}\dfrac{{\dfrac{{\text{4}}}{{\text{9}}}{{\text{V}}^{\text{2}}}{{{\text{(mol/L)}}}^{\text{2}}}}}{{\dfrac{{\text{1}}}{{\text{9}}}{{\text{V}}^{\text{2}}}{{{\text{(mol/L)}}}^{\text{2}}}}}$ (The units will get cancelled. Therefore, the equilibrium constant is a constant with no units.)
${{\text{K}}_{\text{C}}}{\text{ = 4}}{\text{.}}$

So, the correct answer is “Option D”.

Note: The units should be kept in mind while writing the concentrations. Proper units should be used. Moreover, the equation should be balanced and calculations should be carefully done while calculating the equilibrium constant.