Determination of the Equilibrium Constant for Esterification - Ethanoic Acid and Propan-1-ol

GCSE Science

Name: Chan Lo Ho

Class: 6A Class no.: 4

Date:

EQ3 Determination of the Equilibrium Constant for Esterification – Ethanoic Acid and Propan-1-ol

Objectives

To determinate the Equilibrium Constant for Esterification of ethanoic acid and propan-1-ol

Theory

Esterification is a reaction that combines an alcohol with an organic acid, with a water molecule is being taken out, and ester is formed. A concentrated acid catalyst speeds up the esterifications. This time H2SO4 is used.

Esterificaiton is a slow and reversible reaction. The equation for the reaction between an acid RCOOH and an alcohol R’OH (where R and R’ can be the same or different) is:

At equilibrium, the concentrations of both reactants and products are constant.

The number values for “K” are got from experiments by measuring the equilibrium concentrations. And the number value tells the equilibrium ratio of products to reactants.

The value of K is large when the products dominate the mixture and vice versa. The expression for an equilibrium reaction is determined by the coefficients in the balanced equation.

Procedures

0.25mol of glacial ethanoic acid and 0.25mol of propan-1-ol were mixed into a clean, dry 50cm3 pear shaped single neck flask. 1.00 cm3 of the mixture was pipetted to a 250 cm3 conical flask which was containing about 25 cm3 distilled water and 2 drops of phenolphthalein indicator solution. Then the new mixture was titrated to the end point with 0.5M NaOH.

The flask was then be swirled. 8 drops of concentrated H2SO4 was then added to the remainder of the acid-alcohol solutions while the flask was being swirl. A 1.00 cm3 sample was titrated out immediately. And process another titration with 0.5M NaOH.

Then some anti-bumping granules are added to the flask and it was attach to a water cooled reflux condenser. The condenser was then heat under reflux for 1 hour. The flask and its contents was cooled in an iced water bath. Then 1.00 cm3 sample was pipetted out from the flask and was titrated with the 0.5M NaOH.

The flask was continuous heated under reflux for an additional 1/2 hour, and was cooled after 1/2 hour. 1.00 cm3 sample was again pipetted out to go under titration with 0.5M NaOH. The refluxing process was repeated until the two titrations results’ different was within 0.2 cm3.

Results and Calculations

Volume of ethanoic acid used:

0.25 x 60.05 / 1.05 = 14.3 cm3

Volume of propan-1-ol used:

0.25 x 60.1 / 0.8 = 18.8 cm3

Evaluate the equilibrium constant for the esterification, assuming that all solutions are ideal.

(i) Give the equation for the esterification reaction between ethanoic acid ...

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Results and Calculations

Volume of ethanoic acid used:

0.25 x 60.05 / 1.05 = 14.3 cm3

Volume of propan-1-ol used:

0.25 x 60.1 / 0.8 = 18.8 cm3

Evaluate the equilibrium constant for the esterification, assuming that all solutions are ideal.

(i) Give the equation for the esterification reaction between ethanoic acid and propan-1ol.

(ii) How many moles of unreacted CH3COOH remain at the end of the refluxing?

(iii) Calculate the concentrations of the other species present at equilibrium.

(iv) Give the equilibrium expression for the reaction.

(v) Calculate the equilibrium constant for the reaction.

Ans.:

(i). CH3COOH (aq) + CH3CH2CH2OH (aq) H2O (l) + CH3COOCH2CH2CH3 (aq)

(ii). CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

The average volume of NaOH used = (5.75+5.6)/2

= 5.675 cm3

No. of moles NaOH = 0.504 x 5.675 /1000

= 2.8602 x 10-3 mol

= No. of moles of CH3COOH remains

(iii).

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

Start: 6.552 x 10-3 6.552 x 10-3 0 0

At eqm:2.8602 x 10-3 2.8602 x 10-3 3.6918 x 10-3 3.6918 x 10-3

[CH3COOH] at equilibrium = 2.8602 / 1 x 10-3

= 2.8602 mol dm-3

= [CH3CH2CH2OH]

[CH3COOH] at start = (0.504 x 13/1000) / 1 x 10-3

= 6.552 mol dm-3

[CH3COOCH2CH2CH3 (aq)] at equilibrium = 6.552 - 2.8602

= 3.6918 mol dm-3

(iv). Kc = [CH3COOCH2CH2CH3][H2O] / [CH3COOH] [CH3CH2CH2OH]

(v). Kc =3.6918 /2.8602 x 2.8602 = 0.4513 dm3 mol-1

Questions

1. What is the purpose of adding a small amount of concentrated H2SO4 to the reaction mixture at the beginning of the experiment?

It also acts as a positive catalyst to speed up the reaction by providing an alternate pathway with lower activation energy. It was used to correct the volume of CH3COOH.

2. Why do we need to add anti-bumping granules to the reactant mixtures before refluxing?

It is used to prevent the solution from boiling vigorously during heating. Also propan-1-ol is a volatile solution; it will easily catch fire if it is boiled vigorously.

3. Why should we keep on refluxing until the volume of the NaOH needed for complete titration is constant (or more or less constant)?

It can help us to make sure that the reaction has reached the equilibrium position. If the volume of NaOH needed is not constant, it means the concentration of ethanoic acid is still decreasing and, instant of the same; the forward rate is larger than the backward rate. We need to keep on refluxing so the forward and backward rates become the same.

4. Explain what effect, if any, each of the following would have on your determination of the equilibrium constant for the esterification reaction:

(i). The concentration of the sodium hydroxide solution is not known exactly.

The remained CH3COOH is then cannot be calculated as the NaOH is used to titrate with it, and we cannot find the concentration, so that the equilibrium constant cannot be determined without the concentration of Sodium Hydroxide.

(ii). 100 cm3 of ethanoic acid and 100 cm3 of propan-1-ol yield 190 cm3 of solution, which yields 190 cm3 of product at equilibrium.

The equilibrium constant remains unchanged. It is because the volume of the ethanoic acid and propan-1-ol are used to calculate the concentration of ethanoic acid and propan-1-ol respectively. When the concentration of the ethanoic acid and propan-1-ol are increased, the concentration of propyl ethanoate and water will also be increased. As a result, there will have no change to the equilibrium constant.

Discussions

A summary of the main result

The results of the 1st and the 2nd titration are not accurate. The volume of NaOH used to react with the reaction mixture is too large because it is at the beginning of the reaction. As the volume is used to calculate the concentration of CH3COOH and CH3CH2CH2OH, the larger the volume of NaOH, the larger the concentration of CH3COOH and CH3CH2CH2OH will be calculated. Although the volume of NaOH used in the 1st and the 2nd titration is larger than literature volume, both values are still sensible. This is because the volume of the 2nd titration is slightly larger than that of the 1st titration. As about 8 drops of concentrated sulphuric acid was added to the reaction mixture after the 1st titration and before the 2nd, the concentration of acid content in the mixture is larger. So a larger amount of NaOH was used to react with the acid.

The results in the 3rd and the 4th titration are also reliable because the volume of NaOH used in the 4th titration is slightly larger than that in the 3rd titration and the different between two values are within 0.2 cm3. There exists a small difference because the mixture that was titrated with NaOH in the 4th titration which had been refluxed for 1/2 hour is more than that in the 3rd titration. The reflux condenser used to reduce the loss of mixture due to evaporation through condensation however the loss still existed. Besides, CH3COOH would further react with CH3CH2CH2OH in the above 1/2 hour. Then the concentration of acid content in the mixture would decrease; so a smaller amount of NaOH was used for the titration.

An examination of the errors in the results

The errors in the 1st and the 2nd titration were known by the change in colour of the solution. The colour changed from colourless to deep pink (nearly purple in colour). In this case, we knew that the reaction was further away from the equivalence point and too much NaOH solution was added. Therefore the results of two titrations were not accurate.

Comparison of results with available literature value

The equilibrium constant in this experiment is different from the literature value. This is because the results in the 3rd and the 4th titrations are not accurate and the volume of NaOH used in both titrations is not accurate. And these inaccurate values are used to calculate the concentration of the species in the mixture. And the concentration values are used to calculate the equilibrium constant. When the concentrations of the species are inaccurate, the value of the equilibrium constant will also be inaccurate.

Defects in the experiment and suggested improvements

When we process the titration in this experiment, the NaOH solution should be added slowly to the reaction mixture. If there was a large proportion of area of the solution turned to pink colour when the NaOH solution is added and then back to colourless, the NaOH should be added slowly by adding half drop of it.

The end of the report