equilibrium constant

Some changes:

1. After adding 8 drops of concentrated sulphuric(VI) acid, titration should be started immediately. But this procedure was missed, V2 (which should be greater than V1) was not accurate. The volume to be subtracted from subsequent titration to correct for the amount of sulphuric (VI) acid present was then found by adding 8 drops of sulphuric (VI) acid to a beaker of water of volume same as before. This procedure was done by Joe. The volume is 0.95 cm3.
2. As the time is not enough, step 5 was not performed.

Result

Calculation

1. A small amount of concentrated sulphuric(VI) acid was added to the reaction mixture at the beginning of the experiment in order to act as a catalyst to speed up the reaction.
2. During refluxing, the mixture may become gases bubbles. Anti-bumping granules were added to the reaction mixture before refluxing in order to prevent the splitting out of the reaction mixture.
3. The refluxing was continued in step 5 until the titre of sodium hydroxide used approaching constant in order to make sure that the reaction had reached equilibrium stage.
4. The equation for the esterification reaction between ethanoic acid and propan-1-ol is:

H2SO4(l)

CH3COOH(aq) + CH3CH2CH2OH(aq)  ⇔  CH3COOCH2CH2CH3(aq) + H2O(l)

5. The equation of the reaction between ethanoic acid and sodium hydroxide is:

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

The volume of NaOH needed for the neutralization with CH3COOH in 1 cm3 sample after refluxing is:

6.1 - 0.95 = 5.15 cm3

mole of CH3COOH ： mole of NaOH = 1：1

mole of NaOH used is:

0.5M × 5.15cm3 × 10-3

= 0.002575 mole

mole of CH3COOH is:

= 0.002575 mole

Therefore, the concentration of ethanoic acid remaining at the end of the reflux is:

= 0.002575 mole per 1 cm3

=2.575 M

6. Concentration of CH3COOH：Concentration of CH3CH2CH2OH

=1：1

Concentration of CH3CH2CH2OH at the end of the reflux is = 2.575 M

Concentration of CH3COOH at the beginning (after procedure 1) is:

= 16.3 cm3 × 10-3 × 0.5M / 10-3 cm3

= 8.15 M

Concentration of CH3COOCH2CH2CH3 after refluxing is:

= 8.15 M - 2.575 M

= 5.575M

Concentration of water after refluxing is:

= 5.575M

7. KC = ([CH3COOCH2CH2CH3] × [H2O])/ ( [CH3COOH] × [CH3CH2CH2OH] )
8. KC = 5.575 ×5.575 / (2.575 × 2.575) = 4.687 mole-1 dm3
9. If the concentration of the sodium hydroxide solution is not known exactly, there will be no effect on the determination of the equilibrium constant for the esterification reaction. The initial concentration of CH3COOH can be found {0.25 / (14.3 + 18.8) × 1000 = 7.55}. The concentration of NaOH can also be found { 7.55 / (1/1000) / (16.3/1000) = 0.46M ≈ 0.5Μ}

Conclusion

The equilibrium constant for the reaction (esterification) between ethanoic acid and propan-1-ol is 4.687 mole-1 dm3

Discussion

1. It is assumed that the volume of 8 drops of concentrated sulphuric(VI) acid is very small which could not greatly affected the volume of the original mixture.
2. It is assumed that the volume of 8 drops of concentrated sulphuric(VI) acid added by Joe is same as that of me.
3. It is assumed that the rate of reaction was slow that when the reaction mixture is cooled down by ice water, KC did not change.
4. After adding of 8 drops of concentrated sulphuric(VI) acid, V2 should be found out immediately. Otherwise, the reaction may have started and the result became incorrect.