2 mole each of CH₃OH & CH₃COOH are taken and heated in the presence of conH₂SO₄ so that equilibrium is established. If K_C for esterfication process is 4, mass of ester present at equilibrium (in gm) is

Easy

Solution

Stoichiometry CH₃OH+ CH₃COOH
$\large \rightleftharpoons$
CH₃COOCH₃ + H₂O
Initial moles 2 2 - -
Moles at equilibrium 2 - x 2 - x x x
$\large {K_C} = \frac{{\left[ {C{H_3}COO{H_3}} \right]\left[ {{H_2}O} \right]}}{{\left[ {C{H_3}OH} \right]\left[ {C{H_3}COOH} \right]}} = \frac{{{X^2}}}{{{{\left( {2 - x} \right)}^2}}}$
Given K_C = 4
$\large 4 = \frac{{{x^2}}}{{{{\left( {2 - x} \right)}^2}}}$
$\large 2=\frac{x}{2-x}$
4 - 2x = x
3x = 4
$\large \boxed{\;x=\frac{4}{3}}$
$\large {n_{C{H_3}COOC{H_3}}} = \frac{4}{3}$
$\large {W_{C{H_3}COOC{H_3}}} = \frac{4}{3} \times \left( {74} \right)$

$W_{{CH}_{3} {COOCH}_{3}} = 98.6 gr$

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