Cebu Institute of Technology – UniversityChemical Engineering Department

A Final Report on Chemical Equilibrium in a Liquid Phase

Kristian Jem Vince Pallar

Engr. Rosario Dangin

I. Abstract

One of the first chemical equilibria to be studied quantitatively was the simple esterification reaction (M, 340-2: Mo, 70-1; G and P, 308).

C2H5OH + CH3COOH -----> CH3COOC2 H5 + H2O

At ordinary temperatures and in the absence of a catalyst, this reaction approaches equilibrium very slowly. Equilibrium may be attained in a reasonable time, either by allowing the reaction mixture to stand at a high temperature, or at room temperature in the presence of a strong acid which acts as a catalyst. The equilibrium constant can be determined by analyzing the equilibrium mixture.

In the present experiment 6M HCl is to be used as a catalyst to avoid the danger of an explosion of the heated sealed tubes. The equilibrium is to be approached from both directions, starting with CH3COOC2 H5 and with the mixture of C2H5OH and CH3COOH. The equilibrium concentrations can be determined by using known weights of the reactants and of the standardized 6M HCl and titrating the total acid present in the equilibrium mixture. The mixtures are to be sealed in small ampules, to avoid loss by volatilization.

The relatively concentrated acid which is used as a catalyst in this experiment, affects the equilibrium constants by changing the activity coefficients of the reactants. A smaller value of the equilibrium constant would be obtained if the reaction were carried out in the absence of the catalyst. It is unnecessary to thermostat the reaction mixture, since the heat of the reaction is practically zero and the equilibrium constant, is therefore, independent of temperature.

II. Objectives

To determine the equilibrium constant.

III. Apparatus and Materials

50 mL burette8 soft glass 25 mL test tubes500 mL of standard 1 N NaOH15 mL each of pure CH3COOC2 H5, glacial CH3COOH, and absolute C2H5OH

IV. Procedure

Part 11. 8 small ampules from clean dry soft glass test tubes are prepared.2. 5 mL of 6 M HCl is pipetted into each of the 8 tubes. 5 mL of distilled

water is added to each of the first 2 tubes. 5 mL of pure CH3COOC2 H5 is added to the third and fourth. 3 mL of glacial CH3COOH and 3 mL of absolute C2H5OH is added to the fifth and sixth tube. 2 mL of glacial CH3COOH and 3 mL of absolute C2H5OH is added to the seventh and eighth.

3. The test tubes are shaken to promote the reaction.Part 24. Two 2 mL samples of glacial CH3COOH is titrated with standard 1 N for

standardization of CH3COOH NaOH using phenolphthalein as an indicator.Part 35. The contents of each tube are analyzed after 3 days.6. 25 mL of standard 1 N NaOH is pipetted into a small Erlenmeyer flask. Its

contents are poured into the flask.7. The excess acid is titrated with 1 N NaOH. Do not attempt to obtain a

permanent endpoint, since its fading is due to the saponification of the ester.

V. Sketch

VI. Data and Results

VII. Computations

Test tube 1

Masswater=5mL x1 g/mL=5 g

MH 2O=5 g x

1mol18.02g.005L

=55.49M

M H+¿= 1N x 19.21ml

25ml=0.7684N∨0.7684M¿

K=¿¿

Test Tube Weight ofEmpty

Tube (g)

Weight of Tube with Solution

(g)

Weight of Solution

Volume of NaOH

used in titration

Kc

1 18.01 29.51 11.50 19.21 1.918x10-4

2 17.92 29.38 11.46 18.95 1.87 x10−4

3 17.60 27.46 9.86 35.89 3.53 x10−3

4 18.25 28.19 9.94 35.40 5.05 x10−3

5 17.48 27.74 10.26 59.60 0.068

6 18.12 28.55 10.43 59.42 0.070

7 17.50 27.74 10.24 40.25 0.023

8 17.38 27.73 10.35 40.19 0.020

Test tube 2

MH 2O=55.49M

MH+¿=1Nx 18.95ml

25ml=0.758M ¿

K=0.758 x 0.75855.492

=1.87 x10−4

Test tube 3

M H 2O=3.91gx 1mol /18g0.005 L

=43 .44M

MCH 3 Ac=5.95gx

1mol88.11g

.005L=13.51M

M H+¿=1Nx 35.89ml

25ml=1.44M ¿

K= 1.44 x 1.4413.51 x 43.44

=3.53 x10−3

Test tube 4

MCH 3 Ac=3.99gx

1mol88.11g

.005L=9.06M

M H+¿=1Nx 35 .40ml

25ml=1.42M ¿

K= 1.42x 1.429.06 x 43.44

=5.05x 10−3

Test tube 5

MHAc=3.210 gx

1mol60g

0.005 L=10.7M

M EtOH=1.79 gx

1mol46g

0.005 L=7.78M

M H+¿=1Nx 59.60ml

25ml=2.38M ¿

K=2.38 x2.387.78 x10.7

=0.068

Test tube 6

MHAc=3.3 gx

1mol60 g

0.005 L=11M

M EtOH=1.7 gx

1mol46 g

0.005 L=7.39M

M H+¿=1Nx 59.42ml

25ml=2.38M ¿

K=2.38 x2.387.39 x11

=0.070

Test tube 7

MHAc=3.6 gx

1mol60g

0.005 L=12M

M EtOH=2.2 gx

1mol46g

0.005 L=9.57M

M H+¿=1Nx 40.25

25ml=1.61M ¿

K=1.61 x1.619.57 x12

=0.023

Test tube 8

MHAc=3.2 gx

1mol60 g

0.005 L=10.67M

M EtOH=2.8 gx

1mol46g

0.005 L=12.17M

M H+¿=1Nx 40.19

25ml=1.61M ¿

K= 1.61 x1.6110.67 x12.17

=0.020

VIII. Data Analysis

Equilibrium constant of each pair is not that far. Therefore, I can assume that the experiment that we conducted is a success. The relatively concentrated acid, which is used as a catalyst in this experiment, affects the equilibrium constant by changing the activity coefficients of the reactants.

IX. Conclusion

The mixtures that are placed in the small vials should be tightly sealed to avoid volatilization. The experiment uses a catalyst for the reaction to occur more rapidly. A smaller value of the equilibrium constant would be obtained if no catalyst was used.