The esterification reaction between ethanoic acid and ethanol proceeds as follows:
CH3COOH(org)+CH3CH2OH(org)⇌CH3COOCH2CH3(org)+H2O(org)
A mixture of 1.00 mol of ethanoic acid and 2.00 mol of ethanol was dissolved in an inert organic solvent (designated as “org”) to produce 1.00 dm3 of the solution and heated at 50 °C in the presence of an acid catalyst until equilibrium was reached. The equilibrium mixture contained 1.40 mol of ethanol.