8.1.6 Determination of Kc

• The determination of an equilibrium constant can be carried out in a school laboratory using an esterification reaction between ethanol and ethanoic acid in the presence of an acid catalyst
• The equation for the reaction is:

C₂H₅OH (l)   +  CH₃COOH (l)    ⇌   CH₃COOCH₂CH₃ (l) + H₂O (l)

• The composition of the equilibrium is determined by titrating against standard sodium hydroxide solution and deducing the number of moles of acid present at equilibrium
• Once the acid is known, the equilibrium concentrations of the other substances can be deduced and K_c determined
• A number of reaction flasks are made up and analysed so that an average value for K_c can be found

Key steps in the procedure

• 6.0 g of concentrated ethanoic acid and 6.2 g of ethanol are added to a conical flask- this is equivalent to 0.1 mol
• A drop of concentrated sulfuric acid is added and the flask is stoppered and shaken
• A second flask is made up as a 'blank' containing a drop of sulfuric acid and about 20 cm³ of distilled water
• The reaction is slow, so the reaction mixture is allowed to reach equilibrium over the course of a week
• After a week the flask contents are titrated against 1.0 mol dm^-3 sodium hydroxide solution using phenolphthalein as an indicator
• From the difference in the titres of the two flasks, the amount of ethanoic acid at equilibrium can be calculated

Finding Kc for an esterification

Practical tips

• Although you could measure the liquids by volume, for accurate work it is easier to weigh them directly into the reaction flask and you don't have a problem of drops of reactants remaining in measuring cylinders
• Safety spectacles and gloves should be worn as both acids as very corrosive
• The ethanoic acid has a very strong pungent smell so it is best handled in a fume cupboard

Specimen Results

• Volume of 1.0 mol dm^-3 NaOH needed to neutralise the reaction mixture = 32.4 cm³
• Volume of 1.0 mol dm^-3 NaOH needed to neutralise the 'blank' flask = 1.4 cm³

Analysis

• Volume of 1.0 mol dm^-3 NaOH equivalent to ethanoic acid = 32.4 -1.4 = 31.0 cm³
• Amount of 1.0 mol dm^-3 NaOH  = (31.0/ 1000) x 1.0 = 0.0031 moles
• The equation for the reaction is

CH₃COOH    + NaOH  →  CH₃COONa + H₂O  

• The moles of ethanoic acid in the equilibrium mixture is = 0.0031 moles
• The equilibrium compositions will be

Equilibrium Composition Table

• The equilibrium constant can be then calculated: