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Equilibrium Constant for esterification

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Introduction

Chemistry Experiment 15 (a) Title: Equilibrium Constant for esterification (b) Aim: To find out the equilibrium constant for esterification of ethanoic acid and propan-1-ol. (c) Theory: In the presence of conc. H2SO4 and under reflux, ethanoic acid undergoes a reversible reaction with propan-1-ol to form ester and water. As equilibrium is reached, CH3COOH(l)+CH3CH2CH2OH(l) CH3COOCH2CH2CH3(l)+H2O(l) Kc = By titrating the reaction mixture with standard alkali solution before and after refluxing, the quantity of the acid used, and thus the quantities of the alcohol used as well as of the ester and water formed can be determined. Knowing the quantities or molar concentrations of the species present in the equilibrium mixture, the equilibrium constant, Kc can be calculated. Set-up: (d) Result: Titration of 8 drops of H2SO4 against NaOH Titration 1 (trial) 2 Final burette reading (cm3) 15.1 30.2 Initial burette reading (cm3) 0 15.1 Volume of NaOH used (cm3) ...read more.

Middle

of moles of CH3COOH reacted in 1 cm3 of mixture = 8.75 x 10-3 mol - 3.65 x 10-3 = 5.10 x 10-3 mol No. of moles of CH3CH2CH2OH in 1 cm3 of reaction mixture after heating = No. of moles of H+ from 1 cm3 of CH3COOH = 3.65 x 10-3 mol No. of moles of CH3COOCH2CH2CH3 in 1 cm3 of reaction mixture after heating = No. of moles of H+ reacted in 1 cm3 of CH3COOH = 5.10 x 10-3 mol No. of moles of H2O in 1 cm3 of reaction mixture after heating = No. of moles of CH3COOCH2CH2CH3 = 5.10 x 10-3 mol Therefore, Kc = (5.10 x 10-3 x 5.10 x 10-3) (3.65 x 10-3 x 3.65 x 10-3) = 1.95 When comparing with the theoretical value (Kc=3.00), Percentage error = [(1.95 - 3.00) / 3.00] x 100% = - 35% (e) ...read more.

Conclusion

3. If the NaOH needed for the end-point of titration is constant, it indicates that the concentration of the reacting substances do not have further changes, meaning that the esterification process has completed and reached an equilibrium. 4. (i) The number of moles of NaOH required for complete neutralization with H2SO4 and also the CH3COOH in the reaction mixture will not be accurate, causing errors when estimating the number of moles of the 8 drops of the added H2SO4 and also that of the CH3COOH reacted. Hence, the equilibrium constant obtained will be inaccurate. (ii) Since the actual volume of the reaction mixture was lower than the expected one, the concentration of the former would be higher than the latter. Since the volume after mixing the reactants is used to determine the concentration of the 4 reactants, the equilibrium constant obtained will be inaccurate. (g) Conclusion: The equilibrium constant (Kc) for the reaction CH3COOH(l)+CH3CH2CH2OH(l) CH3COOCH2CH2CH3(l)+H2O(l) is 1.95, with an percentage error of -35% ?? ?? ...read more.

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