Colourful Solutions > Practical Scheme of Work > E701 - Determination of an equilibrium constant (HL)

IB Chemistry Shop

Standard level

An equiibrium is a chemical system in which the rate of a forward chemical reaction is equal to the rate of the reverse chemical reaction, and consequently the concentrations of the components remain unchanged.


Background

Esterification and the reverse process, hydrolysis of an ester are the forward and back reactions of the ester equilibrium.

Esterification

CH3COOH(l) + C2H5OH(l) CH3COOC2H5(l) + H2O(l)

Hydrolysis of an ester

CH3COOC2H5(l) + H2O(l) CH3COOH(l) + C2H5OH(l)

The equilibrium constant is given by dividing the multiple of the concentrations of the products raised to the power ot their stoichiometric coefficients, by the multiple of the concentration of the reactants raised to the power of their stoichiometric coefficients.

For the equilibrium:

aA + bB cC + dD

kc = [C]c[D]d/[A]a[B]b

Clearly, the actual value of kc depends on which way round the equilibrium is written. The value of kc for the forward reaction is the reciprocal of the value of kc for the back reaction.

kc(forward) = 1/kc(backward)

For the esterification and the hydrolysis of the ester a catalyst speeds up the attainment of the equilibrium condition, but does not affect the value(s) of the equilbrium constant. Dilute hydrochloric acid is used as the catalyst.

At room temperature the equilibrium is slow to establish, so the reaction mixtures should be prepared a couple of weeks before the analysis.

Analysis is a simple titration to find the total acid in the equilibrium mixture. The catalyst acid remains unchanged, while the hydrolysis reaction increases the total acid in the mixture due to the formation of ethanoic acid.


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