AL chem(equilibrium)

2010-05-24 8:08 am

Same number of mole of glaical ethanoic acid and propan-1-ol were put in a clean dry pear-shaped flask. 1.0cm^3 of this mixture require 15.5cm^3 0.50M NaOH to reach the end point. 8 drops of conc. H2SO4 was added to the remainder of the mixture, and titrated with another 1.0cm^3 sample immediately. 17.2cm^3 of 0.50M NaOH solution is required for titration. The mixture was reflux for half hour to reach equilibrium. The flask was cooled. 1.0cm^3 of this mixture require 5.6cm^3 0.50M NaOH to reach the end point.

(a)Calculate the concentration of ethanoic acid
(i)at the beginning of the experiment
(ii)at the end of the reflux

(b)Calculate the concentrations of the other species present at equilibrium

(c)Calculate the equilibrium constant(Kc) for the esterification reaction.

回答 (1)

fooks

2010-05-24 9:10 am

✔ 最佳答案

1.
(a) (i)
Titration after mixing ethanoic acid and propan-1-ol:
CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1
No. of moles of NaOH used = 0.5 x (15.5/1000) = 0.00775 mol
No. of moles of CH3COOH = 0.00775 mol
At the beginning of the experiment,
Concentration of CH3COOH = 0.00775/(1/1000) = 7.75 M

(a)(ii)
Titration after the reflux:
CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1
Volume of NaOH to react with H2SO4 = 17.2 - 15.5 = 1.7 cm³
Volume of NaOH to react with CH3COOH = 5.6 - 1.7 = 3.9 cm³
No. of moles of NaOH to react with CH3COOH = 0.5 x (3.9/1000) = 0.00195 mol
At the end of the reflux,
Concentration of CH3COOH = 0.00195/(1/1000) = 1.95 M

(b)
CH3COOH + PrOH ≒ CH3COOPr + H2O

Initial concentrations:
[CH3COOH]o = [PrOH]o = 7.75 M
[CH3COOPr]o = [H2O]o = 0 M

At equilibrium:
[CH3COOH] = [PrOH] = 1.95 M
[CH3COOPr] = [H2O] = 7.75 - 1.95 = 5.8 M

(c)
Kc
= [CH3COOPr] [H2O] / [CH3COOH] [PrOH]
= (5.8)² / (1.95)²
= 8.85

參考: fooks

👍 0 👎 0