Determine the equilibrium constant - Kc; of ethanoic acid reacting with ethanol producing an equilibrium to form ethyl ethanoate and water.

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The Determination of an Equilibrium Constant

Determine the equilibrium constant – Kc; of ethanoic acid reacting with ethanol producing an equilibrium to form ethyl ethanoate and water.

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

Following the method as detailed, I conducted experiment 4 and these results were obtained:

To calculate Kc, the concentrations of each reactant must be calculated from the point of equilibrium at which the titration was taken.

Therefore, using 7.75cm3 as the average titre,

Moles = 0.2M x (7.75cm3 / 1000 cm3) = 1.55 x 10-3.

However, this represents the total amount of acid in the system which included ethanoic acid but also the acid catalyst: hydrochloric acid. Therefore this must be taken away from the number of moles of acid to calculate number of moles of ethanoic acid.

Since 25 cm3 of 1.0M HCl was included in the initial 250 cm3 mixture,

Moles = 1 x (25 / 250) = 0.1 moles.

But 1cm3 samples were taken at a time, therefore:

Moles = 0.1 x (1 cm3 / 1000 cm3) = 1 x 10-4.

Taking this away from the total amount of acid = 1.55 x 10-3 – 0.1 x 10-3 = 1.45 x 10-3 (moles of ethanoic acid).

Considering the stoichiometry of the equation, there is equal amounts of ethanol as there is ethanoic acid, therefore there is 1.45 x 10-3 moles of ethanol at equilibrium also. This is because the total amount of acid reacted is equal to the amount of acid, even though it is a weak acid. Because it reacts with a strong base, it means that the acid dissociates completely.

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Now to calculate the number of moles of products;

The number of moles is calculated from multiplying volume by density to calculate mass (which is provided on the enclosed sheet), and then dividing mass by molecular mass. This figure is then divided by 250 to calculate the number of moles in every 1cm3 sample.

Finally, since the equilibrium is shifting to the left during the reaction due to the low concentration of the reactants, the figure 1.45 x 10-3 is added to the reactants and subtracted from the products.

Molecular masses are such:

CH3COOH ...

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