JEE MainChemistryModeratePhysical Chemistry

In an experiment starting with 1 mole of ethyl alcohol, 1 mole
of acetic acid and 1 mole of water at TºC, the equilibrium mixture on
analysis shows that 54.3 of the acid is esterfied. Calculate the
equilibrium constant of this reaction.

Show answer and solution

{CH}_{3}COOH(I) + C_{2}H_{5}OH(l) \rightleftharpoons {CH}_{3}{COOC}_{2}H_{5}(l) + H_{2}O(l)

Initial 1 1 0 1
At equilibrium 1 - x 1 - x x 1 + x
1 - 0.543 1 - 0.543 0.543 1 + 0.543

\left( 54.3\%\ of\ 1mole = \frac{1 \times 54.3}{100} = 0.543mole \right)

Hence given x = 0.543 mole
Applying law of mass action

K_{C} = \frac{\lbrack\ ester\ \rbrack\lbrack\ water\ \rbrack}{\lbrack\ acid\ \rbrack\lbrack\ alcohol\ \rbrack} = \frac{0.543 \times 1.543}{0.457 \times 0.457} = 4.0

Source: Grass-G / CH-24 Chemistry Paper 31 Dec..docx

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