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Topic: Equilibrium constant of esterification  (Read 10587 times)

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dolphinsiu

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Equilibrium constant of esterification
« on: April 21, 2005, 03:23:00 PM »
CH3COOH+CH3CH2CH2OH<=>CH3CO2CH2CH2CH3+H2O

From my experimental result:
Add 25 cm3 of distilled water and pipette 1 cm3 of the mixture (CH3COOH+CH3CH2CH2OH)from pear-shaped glass into conical flask
Average volume of 0.5 M NaOH reacted = 15.37cm3


After adding 8 drops of 0.1 M HCl into pear-shaped glass, then add 25 cm3 of distilled water and pipette 1 cm3 of mixture from pear-shaped glass into conical flask
Average volume of 0.5 M NaOH reacted = 16.87cm3

After reflux about 1 hour,
Average volume of 0.5 M NaOH reacted = 5.7cm3

What is the equilibrium constant of esterification?
My attempting answer:
CH3COOH+NaOH=>CH3COONa+H2O
No. of mol of NaOH=
No. of mol of CH3COOH = 15.37 X 0.5/1000=0.00077 mol
Volume of HCl = (16.87-15.37) X 1000 / 0.1 =7.5 cm3

After reflux,
No. of mol of NaOH=
No. of mol of CH3COOH = 5.70X 0.5/1000=0.000285 mol

                 CH3COOH+CH3CH2CH2OH<=>CH3CO2CH2CH2CH3+H2O

AT initial            7.7M          7.7 M
AT Eqm             2.85M         2.85M                     4.85M            4.85M

Kc= 4.85^2/2.85^2=2.895

Is it right?
But is the initial concentration of propan-1-ol equal to the initial concentration of enthanoic acid??
« Last Edit: April 22, 2005, 03:22:08 PM by dolphinsiu »

Offline Donaldson Tan

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Re:Equilibrium constant of esterification
« Reply #1 on: April 27, 2005, 05:56:17 PM »
does your experimental value agrees with literature value?
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