AQA A Level Chemistry

Revision Notes

8.1.6 Determination of Kc

Determination of Kc

  • Determination of Kc for an esterification reaction
  • 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 the other substances can be deduced and Kc determined
  • A number of reaction mixtures are made up and analysed so that an average value for Kc can be found
  • The reaction is slow, so the reaction mixtures are made up and allowed to reach equilibrium in stoppered flasks over the course of a week

Steps in the procedure

  • Using burettes or graduated pipettes, various reaction mixtures are made up and left inside small stoppered flasks
  • At the end of a week the contents of each flask are titrated against 1.0 mol dm-3 NaOH

Volume- composition mixtures for an esterification reaction table

 

 

 

 

 

 

  • Some mixtures are made for the forward reaction and some are for the backward reaction
  • A ‘blank’ flask with only the catalyst in (concentrated HCl) is titrated against NaOH, so the volume of base needed to neutralise the acid catalyst in each flask can be subtracted

Practical tips

Specimen Results

  • Titration results from experiment 5:

Analysis

  • You need the concentrations of the four components at equilibrium
  • You can use the density of ethanol, ethanoic acid, ethyl ethanoate and water to calculate the mass of each liquid used
  • For example, the mass of 4 cm3 of ethanol is:

0.789 g cm-3  x  4 g  = 3.156 g

  • From the molar mass you can find the number of initial moles of ethanol:

moles of ethanol =  3.156 / 46.1 = 0.0685 mol

  • The total volume of the mixture is 10 cm3 or 0.01 dm3, so you can calculate the initial concentration:

0.0685 / 0.01 = 6.85 mol dm-3

  • The results of the titration tell you the number of moles of ethanoic acid at equilibrium and from this you can deduce the moles of ethanol, ethyl ethanoate and water
  • For example, if 23.19 cm3 of 1.0 mol dm3 NaOH were needed to react with CH3COOH then the moles of CH3COOH
  • at equilibrium would be:

23.19/1000    x   1.0  = 0.02319 mol   

  • If you started with 0.100 mol of ethanoic acid, then you can deduce the other equilibrium amounts

 

  • These are moles rather than concentrations, so you would need to divide by the volume in dm3 in order to find the concentrations

 

 

 

 

 

 

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