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??