Determination of the equilibrium constant for esterification

AS and A Level Science

Chemistry TAS Report

Experiment Number : 9
Date : 17/12/2007
Title : Determination of the equilibrium constant for esterification

Aims/Objective :

To determine the equilibrium constant of esterification.

Introduction / Theory:

Esterification is the reaction of a carboxylic acid with an alcohol.

This experiment is an esterification reaction between ethanoic acid and propan-1-ol when heated:

CH3COOH(aq) + CH3CH2CH2OH(aq) ⬄ CH3COOCH2CH2CH3(aq) + H2O(l)

The formation of propyl ethanoate is particularly well-suited to the determination of the equilibrium constant. The reaction is slow enough at room temperature so the order of mixing, temperature fluctuations over the reaction time and even a final titration with a strong base only have little effect on the reaction. Since this is a homogeneous reaction with the same number of moles of reactants and products, the equilibrium constant (Kc) is generally expressed in terms of molarity or can be calculated in terms of moles alone which is more convenient.

Because the reaction is very slow at room temperature, it is sped up by addition of catalyst (concentrated sulphuric(VI) acid ). But the catalyst does not take part in the overall reaction.

The analysis of the equilibrium mixtures is based on a simple titration with standardized NaOH. Since the initial amounts of all materials are known and the overall changes in the reaction can be reflected by the determination of the final amount of acid. The concentrated H2SO4 catalyst remains unchanged and once an amount of acid is subtracted out, there is only ethanoic acid. If less acid is detected than original addition, the reaction has moved in the forward direction. If more acid is detected, the reaction has moved in the reverse direction. Since the stoichiometric ratios in the reaction are all unity, the loss or gain in ethanoic acid can be used to figure the loss or gain in everything else.