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

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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: Reagent bottle's number 1A 1B 2 3 4 Volume of HCl(aq) ...read more.


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: Reagent bottle's number 1A 1B 2 3 4 Mass of reagent bottle with stopper /g 101.01 69.99 67.71 62.52 66.57 (1) Amount of HCl /mol 0.01039 0.01039 0.01039 0.01039 0.01039 (2) Total amount of acid at eqm. /mol - - 0.03927 0.03597 0.02692 (3) Eqm. amount of CH3COOH(l) /mol - - 0.02888 0.02558 0.01653 (4) Eqm. amount of CH3CH2OH(l) /mol - - 0.02888 0.02558 0.01653 (5) Mass of pure HCl /g 0.3792 0.3792 0.3792 0.3792 0.3792 Mass of bottle after adding HCl(aq) /g 106.11 75.15 72.79 67.62 71.75 Mass of HCl(aq) ...read more.


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. ...read more.

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