Question

0.1 mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralised by 100 mL of 0.85 N NaOH. If no hydrolysis of ester is supposed to have undergo, find ${K_c}$
A. ${K_c} = 0.006$
B. ${K_c} = 0.031$
C. ${K_c} = 1.34$
D. ${K_c} = 3.1$

Answer 1

Hint: We have to calculate the milliequivalents of the acetic acid left. From the milliequivalents of acetic acid, we will know the millimoles of acetic acid. We have to then calculate the moles of acetic acid left from the millimoles of acetic acid left. From the moles of acetic acid left we can calculate the value of equilibrium constant by dividing the rate of products to the rate of reactants.

Formula used: We can calculate the equilibrium constant using the formula,
${{\text{K}}_{\text{c}}}{\text{ = }}\dfrac{{{\text{[Rate}}\,{\text{of}}\,{\text{products]}}}}{{{\text{[Rate}}\,{\text{of}}\,{\text{reactants]}}}}$

Complete step by step answer: We know that in a chemical reaction, chemical equilibrium is the state in which the rates of forward reaction and the reverse reaction are equal. The result is that the concentration of the reactants and the products remains unchanged.
When the coefficients in a balanced equation are multiplied with a factor n, the equilibrium expression is raised to the ${{\text{n}}^{{\text{th}}}}$ power.
Given data contains,
Moles of acetic acid are 0.1 mol.
Moles of ethyl alcohol are 0.1 mol.
Volume of sodium hydroxide is 100mL.
Normality of sodium hydroxide is 0.85N.
Acetic acid and ethyl alcohol are reacted with each other to form ethyl acetate and water. We can write the chemical equation for this ester hydrolysis reaction as,

We can write the equilibrium constant expression as,
$
  {K_c} = \dfrac{{{\text{[Rate}}\,{\text{of}}\,{\text{products]}}}}{{{\text{[Rate}}\,{\text{of}}\,{\text{reactants]}}}} \\
    \\
 $
Equilibrium constant for the given reaction can be written as,
${K_c} = \dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right]\left[ {{H_2}O} \right]}}{{\left[ {C{H_3}COOH} \right]\left[ {C{H_3}C{H_2}OH} \right]}}$
We can write the ICE (Initial, Change, Equilibrium) table,
$ {\text{Before}}\,{\text{rxtn}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{0}}{\text{.1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {0.1 - x} \right)\,\,\,\,\,\,\,\,\,\,\,\left( {0.1 - x} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\, \\
 $
Let us now calculate the milliequivalents of acetic acid left.
We can calculate the milliequivalents of acetic acid left using the milliequivalents of sodium hydroxide used.
Milliequivalents of acetic acid left=milliequivalents of sodium hydroxide left
Let us calculate milliequivalents of sodium hydroxide used using the volume and normality of sodium hydroxide.
Milliequivalents of sodium hydroxide used=${\text{Volume $\times$ Normality}}$
Milliequivalents of sodium hydroxide used=${\text{100}}\,{\text{mL $\times$ 0}}{\text{.85}}\,{\text{N}}$
Milliequivalents of sodium hydroxide used=${\text{85}}\,{\text{mEq}}$
The milliequivalents of sodium hydroxide used is ${\text{85}}\,{\text{mEq}}$, therefore the milliequivalents of acetic acid left is also ${\text{85}}\,{\text{mEq}}$.
We have to know that 1mEq is equal to 1mmol.
Therefore, ${\text{85}}\,{\text{mEq}}$ is equal to 85mmol.
From millimoles of acetic acid, we calculate the moles of acetic acid left as,
Moles of acetic acid=$\dfrac{{{\text{Millimoles}}}}{{{\text{1000}}}}$
Moles of acetic acid=$\dfrac{{{\text{85}}\,{\text{mmol}}}}{{{\text{1000}}}}$
Moles of acetic acid=${\text{0}}{\text{.085}}\,{\text{mol}}$
The moles of acetic acid is ${\text{0}}{\text{.085}}\,{\text{mol}}$.
Let us now calculate the value of x in the ICE table.
$
  \left( {0.01 - x} \right) = 0.085 \\
    \\
 $
$x = 0.015$
We can calculate the equilibrium constant as,
$
  {K_c} = \dfrac{{{\text{[Rate}}\,{\text{of}}\,{\text{products]}}}}{{{\text{[Rate}}\,{\text{of}}\,{\text{reactants]}}}} \\
  {K_c} = \dfrac{{\left[ {C{H_3}COO{C_2}{H_5}} \right]\left[ {{H_2}O} \right]}}{{\left[ {C{H_3}COOH} \right]\left[ {C{H_3}C{H_2}OH} \right]}} \\
 $
$
  {K_c} = \dfrac{{{x^2}}}{{{{\left( {0.1 - x} \right)}^2}}} \\
  {K_c} = \dfrac{{{{\left( {0.015} \right)}^2}}}{{{{\left( {0.085} \right)}^2}}} \\
  {K_c} = 0.031 \\
 $
The equilibrium constant is $0.031$.
So, the correct answer is “Option B”.

Note: We know that the equilibrium constant is equal to the rate constant of the forward reaction divided by the rate constant of the reverse reaction. When two or more reactions are summed to give another, the equilibrium constant for the reaction is a product of the equilibrium constants of the equations added.