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Determination of the equilibrium constant for esterification

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Chemistry TAS Report 1. Experiment Number : 9 2. Date : 17/12/2007 3. Title : Determination of the equilibrium constant for esterification 4. Aims/Objective : To determine the equilibrium constant of esterification. 5. Introduction / Theory: Esterification is the reaction of a carboxylic acid with an alcohol. This experiment is an esterification reaction between ethanoic acid and propan-1-ol when heated: CH3COOH(aq) + CH3CH2CH2OH(aq) <=> CH3COOCH2CH2CH3(aq) + H2O(l) The formation of propyl ethanoate is particularly well-suited to the determination of the equilibrium constant. The reaction is slow enough at room temperature so the order of mixing, temperature fluctuations over the reaction time and even a final titration with a strong base only have little effect on the reaction. Since this is a homogeneous reaction with the same number of moles of reactants and products, the equilibrium constant (Kc) is generally expressed in terms of molarity or can be calculated in terms of moles alone which is more convenient. Because the reaction is very slow at room temperature, it is sped up by addition of catalyst (concentrated sulphuric(VI) ...read more.


2. 1.0 cm3 of the mixture was transferred by pipette to a 250 cm3 conical flask that about 25 cm3 of deionized water and 2 drops of phenolphthalein indicator were contained. The solution was then titrated to the end point with 0.50 M sodium hydroxide solution. 3. The volume (V1 cm3) of titre was recorded. 4. 8 drops of concentrated sulphuric(VI) acid were added to the remainder of the acid-alcohol solution and the flask was swirled continuously. 5. Step 2 was repeated immediately. 6. The volume (V2 cm3) of titre was recorded. 7. A few anti-bumping granules were added to the flask and then it was attached to a water-cooled reflux condenser. 8. The solution was refluxed for 30 minutes. Then, the flask was cooled by an ice bath. 9. Step 2 was repeated again. 10. The volume (V3 cm3) of titre was recorded. 11. The solution was refluxed continuously for additional 20 minutes. Then, the flask was also cooled by an ice bath. 12. ...read more.


= 2.525 M 6. Concentration of propan-1-ol = Concentration of ethanoic acid = 2.525 M Concentration of propyl ethanoate = 0.5 x (V2 - V4) / (1 x 10-3) = 0.5 x [(19.5 - 5.45) x 10-3] / (1 x 10-3) = 7.025 M Concentration of water = Concentration of propyl ethanoate = 7.025 M 7. Kc = [CH3COOCH2CH2CH3(aq)] [ H2O(l) ] [CH3COOH (aq)][ CH3CH2CH2OH (aq)] 8. Kc = (7.025/2.525)2 = 7.74 9. If the concentration of the sodium hydroxide solution is not known exactly, it would not have any effect on the determination of the equilibrium constant for the esterification reaction. Since Kc = [CH3COOCH2CH2CH3(aq)] [ H2O(l) ] [CH3COOH (aq)][ CH3CH2CH2OH (aq)] Kc = { [NaOH]( V2- V4 ) / [NaOH] [V4 - (V2 - V1)]}2 = { ( V2- V4 ) / [V4 - (V2 - V1)] }2 Thus, Kc is not affected by the concentration of sodium hydroxide. 10. There is error in this experiment. (1) Taking reading in titration. Error estimation - When taking initial reading, error is + 0.05 cm3. When taking final reading, error is also + 0.05 cm3. Therefore, error is + 0.1 cm3. ...read more.

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