Objective To find out the equilibrium constant, Kc, for the reaction below, using acid hydrolysis:

Determining an Equilibrium Constant

Objective
To find out the equilibrium constant, Kc, for the reaction below, using acid hydrolysis:

Principle
Although the hydrolysis of ethyl ethanoate is very slow, by using dilute hydrochloric acid as catalyst, the above equilibrium can be attained in 48 hours. After 48 hours, the reaction mixture can then be titrated with standard sodium hydroxide solution. Finally, the equilibrium concentrations of four components below and hence, Kc of hydrolysis of CH3COOCH2CH3 can then be calculated:

Chemicals
2M HCl, 1.0335M NaOH, ethyl ethanoate, phenolphthalein indicator

Apparatus
5 small reagent bottles, 5ml pipette & filler, burette, measuring cylinder, conical flask, electronic balance, stand, white tile

Procedure ── Preparation, allow reaching equilibrium position
         1.> Label 5 reagent bottles & their stoppers as 1A, 1B, 2, 3 and 4.
         2.> Weigh each reagent bottles with their stoppers and record their corresponding masses in Table 3.
         3.> Pipette 2M HCl into each bottles, using measuring cylinders to transfer ethyl ethanoate into bottles 2,3,4 and water into bottles 3,4 according to the
               amount shown in Table 1.

                                  Table 1 shows the amount of chemicals added:

         4.> Record the total mass before and after each addition of chemical into Table 3.
         5.> Set them aside for at least 48 hours and shake the bottles occasionally.

Procedure ── Titration (48 hours later)
         1.> Rinse and fill a burette with standardized NaOH solution.
         2.> Pour the contents of bottle 1A into a conical flask and rinse the bottle with deionized water.
         3.> Add 2 - 3 drops of phenolphthalein indicator and titrate the mixture with standardized NaOH solution.
         4.> Repeat steps 2 - 3 for other regent bottles. The results are recorded in Table 2.

                            Table 2 shows the results of titration:

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Calculation & Results
         Below, it shows how the values in individual items in Table 3 can be obtained:
                   1.> Since HCl(aq) and NaOH(aq) react in equimolar amounts, so,

                   2.> Total amount of acid at equilibrium in bottles 2,3 & 4
                         = no. of moles of HCl(aq) added + CH3COOH(l) formed
                         = concentration of NaOH(aq) x volume of NaOH(aq) added
                   3.> Equilibrium amount of CH3COOH(l)
                         = total amount of acid at equilibrium - amount of HCl(aq) added
                         = the value obtained in (2) - that in (1)
                   4.> Equilibrium amount of CH3CH2OH(l)
                         = equilibrium amount of CH3COOH(l)
                         = the value obtained in (3).
                   5.> Mass of pure HCl in the mixture
                         = amount of HCl(l) x molar mass of HCl (36.5 gmol-1)
                         = the value obtained in (1) x 36.5
                   6.> Mass of water in HCl(aq)
                         = mass of HCl(aq) added - mass of pure HCl
                         = mass of HCl(aq) added - value obtained in (5)
                   7.> Initial amount of CH3COOCH2CH3(l)


                   8.> Equilibrium amount of CH3COOCH2CH3(l)
                         = initial amount of CH3COOCH2CH3(l) - equilibrium amount of CH3COOH(l)
                         = value obtained in (7) - value obtained in (3)
                   9.> Initial amount of H2O(l)


                10.> Equilibrium amount of H2O(l)
                        = initial amount of H2O(l) - equilibrium amount of CH3COOH(l)
                        = value obtained in (9) - value obtained in (3)
                11.> Equilibrium constant, Kc



              Table 3 summarizes all the data analysed:

= 0.21

Precaution
       The deionized water used to rinse the reagent bottle and the time used for titration should be as least as possible to avoid the equilibrium position shifting to right (∵ by Le Chatelier's principle, the increase in [H2O(l)] promote the equilibrium shifts rightward). More CH3COOH(l) is formed and hence alters the Kc.
         The reagent bottles containing ethyl ethanoate should be sealed besides stopped to avoid vapourization of CH3COOCH2CH3(l) (which will decreases [CH3COOCH2CH3(l)]) and hence affects Kc.
         Besides, the reagent bottles should be kept at constant temperature to avoid altering Kc.
         Furthermore, the hands touching the reagent bottles should be dry since this would affect the mass of chemicals obtained and hence Kc.
         Also, we should avoid shaking too fast as this will increase kinetic energy of chemicals and affect Kc.

Discussion
         In this experiment, even if the concentration of sodium hydroxide solution is not known exactly, this would have no effect on the value of Kc calculated, as the errors tend to cancel out in 4 equilibrium expressions.
         Since this hydrolysis is kinetically slow (even in the presence of catalyst), it is possible to measure the [CH3COOH(l)] by titration with standard NaOH (which removes CH3COOH & catalyst H+) but without significantly disturbing the equilibrium position. Besides, the catalyst H+ does not affect the equilibrium position and the percentage yield, it only provides an alternative path with lower activation energies for both forward and backward reactions, thus speed up the time required for equilibrium to achieve.
         Since the amount of water formed is not much more than CH3COOCH2CH3, CH3COOH and CH3CH2OH, i.e. it's not constant and [H2O(l)] should be included in the equilibrium equation.
         The Kc obtained in this experiment has no unit since it's canceled out in 4 equilibrium expressions.
         The errors of Kc obtained may due to the hands touching the reagent bottles are wet, hence the mass of chemicals added seems to be a little bit more. This can be discovered in that the mass of HCl(aq) added to 5 reagent bottles are different (but they should be the same！)
         The Kc obtained in bottle 4 seems to be much larger than others. It may due to the prolong titrating the mixtures or the increase in [H2O(l)] promotes more CH3COOH(l) forming. Certainly, it may also due to the poor titration techniques.

Conclusion
Since Kc cannot be easily affected by physical factors, except for temperature. Hence, Kc of a certain reaction can be obtained by many methods, e.g. electrochemical method and colorimetrical method. The one used in this experiment is titrimetric one since the hydrolysis is kinetically slow.

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Brady Smith

23rd of May 2012

This is a very descriptive, well through out practical. It contains advanced mathematical analysis for an A2 course and draws good conclusions This piece of work is 5 stars out of 5

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