${\mathrm{CH}}_3{\mathrm{COOH}}_{\left(\mathrm{l}\right)} + {\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{l}\right)} \leftrightharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_{5\left(\mathrm{l}\right)} + {\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}$ In the above reaction, one mole of each of acetic acid and alcohol are heated in the presence of little conc. ${\mathrm{H}}_2{\mathrm{SO}}_4.$ On equilibrium being attained

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{COOH}}_{\left(\mathrm{l}\right)}+{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{OH}}_{\left(\mathrm{l}\right)}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOC}}_2{{\mathrm{H}}_5}_{\left(\mathrm{l}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)} \\ \mathrm{Initially}: 1\mathrm{mol} 1\mathrm{mol} 0 0 \\ \mathrm{at} \mathrm{Equil}. : 1-\mathrm{x} \mathrm{x} \mathrm{x} \mathrm{x} \\ \mathrm{where}, \mathrm{x}=\mathrm{degree} \mathrm{of} \mathrm{dissociation}, \mathrm{It} \mathrm{never} \mathrm{be} >1 \end{gathered}$$

Hence In presence of little ${\mathrm{H}}_2{\mathrm{SO}}_4$ (as catalyst) about 2/3 mole of each of ${\mathrm{CH}}_3\mathrm{COOH} \mathrm{and} {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$ react to form 2/3 mole of the product at equilibrium.