Equilibrium Constant for esterification

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Chemistry Experiment 15

  1. 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

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Titration of final reaction mixture against NaOH

No. of moles of CH3COOH in 1 cm3 of reaction mixture before heating

=         [(density x volume) / molar mass] / volume

=        [(1.05 gcm-3 x 15 cm3) / (12x2+1x4+16x2)]

=        8.75 x 10-3 mol

        2 NaOH + H2SO4                 Na2SO4 + 2 H2O

No. of moles of H+ from H2SO4 in 1 cm3  of reaction mixture before heating (8 drops)

=        [(No. of moles of NaOH reacted / 2) x 2] / volume

=        {[(15.1/1000) x 0.5] / 2} x 2 / 30

=        2.52 x 10-4 mol

=        No. of moles of NaOH of 1 cm3  reacted ...

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