Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate
Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate
Michael Zuber
Introduction:
In this experiment 6 different mixtures of known amount of acid, alcohol and/or ester were mixed with dilute hydrochloric acid in a stoppered bottle. The mixture is then set aside for a week to reach equilibrium at room temperature. The mixture is then titrated against a solution of 1 mol.dm-3 NaOH to determine the total amount of acid present in the equilibrium. From this the amount of ethanoic acid can be found and therefore the amount of the other reagents can be found. These values can then be used to find the equilibrium constant for the reaction.
The equation for the reaction is:
Raw Data:
Below is a table of the different volumes of the chemicals initially added in each solution:
Solution
HCl (cm3)
Water (cm3)
Ethyl ethanoate (cm3)
Ethanoic acid (cm3)
Ethanol (cm3)
5
0
5
0
0
2
5
4
0
0
3
5
3
2
0
0
4
5
0
4
0
5
5
0
0
4
6
5
0
0
2
3
Below is a table of the volume of NaOH needed to neutralize the different solutions at equilibrium:
Solution
NaOH (cm3) used to neutralise soltuion (1)
NaOH (cm3) used to neutralize soltuion (2)
NaOH (cm3) used to neutralise soltuion (3)
44.2
43.7
42.9
2
44.5
45.2
3
32.4
32.1
31.1
4
52.9
50.9
51.8
5
22.8
23
6
37.5
36.4
From these values we can find an average amount of 1 mol.dm-3 NaOH needed to neutralize each solution can be found. Below is a table with these values:
Solution
Average amount of NaOH (cm3)
Random Error (±cm3)
Reading Error (±cm3)
Absolute Error (±cm3)
43.6
0.7
0.05
0.7
2
44.9
0.4
0.05
0.4
3
31.9
0.6
0.05
0.7
4
51.9
.0
0.05
.1
5
22.9
0.1
0.05
0.2
6
37.0
0.6
0.05
0.6
By taking an average there will be an inherent random error. This random error is equal to the (max titer - min titer)/2.
Along with the random error is a reading error. This reading error is equal to 1/2 the smallest division of the instrument used to make the measurement. Since we used a biurette the smallest division was 0.1cm3 therefore the reading error is 0.05cm3.
Since both of these errors are of the same units they can be added together to find an absolute error. The random error, reading error and absolute error are all shown in the above table.
From these averages we can determine the amount of moles NaOH needed to neutralize the solution and since both HCl and ethanoic acid react 1 : 1 with NaOH the amount of moles of NaOH reacted will equal the total amount of moles of acid reacted.
Since we know that the concentration of the HCl is 3 mol.dm3 and in each solution 5cm3 of HCl was used, we can find the moles of HCl in the left over acid.
Mole HCl = 0.005 x 3 = 0.015 mol
This value can be subtracted from each ...
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From these averages we can determine the amount of moles NaOH needed to neutralize the solution and since both HCl and ethanoic acid react 1 : 1 with NaOH the amount of moles of NaOH reacted will equal the total amount of moles of acid reacted.
Since we know that the concentration of the HCl is 3 mol.dm3 and in each solution 5cm3 of HCl was used, we can find the moles of HCl in the left over acid.
Mole HCl = 0.005 x 3 = 0.015 mol
This value can be subtracted from each value for the total moles of acid.
The below table shows the total moles of acid, the moles of HCl and the moles of ethanoic acid for each solution:
Solution
Total Moles of Acid
Moles HCl
Moles Ethanoic Acid
0.0436
0.015
0.0286
2
0.0449
0.015
0.0299
3
0.0319
0.015
0.0169
4
0.0519
0.015
0.0369
5
0.0229
0.015
0.0079
6
0.0370
0.015
0.0220
However, since there is an error in the volume of NaOH (as calculated above) there will also be an error in the total moles of acid and therefore in the moles of ethanoic acid. This error will equal the percentage error on the volume of NaOH.
To find the percentage error the absolute error is divided by the average result and this value is multiplied by 100.
This will give us the percentage error on the total moles of acid. Since we consider the HCl to be a standard solution, we also consider it has no error. Therefore the percentage error on the ethanoic acid is equal to the percentage error of the total moles of acid.
Below is a table showing the percentage error on the different moles of ethanoic acid for each solution:
Solution
Moles Ethanoic Acid
Percentage error on Moles (±%)
0.0286
.61
2
0.0299
0.89
3
0.0169
2.20
4
0.0369
2.02
5
0.0079
0.66
6
0.0220
.62
Since one mole of ethanoic acid will react with one mole of ethanol, the above values give the moles of ethanol for each solution.
To find the moles of water and of ethyl ethanoate the initial moles of each chemical must be found. These values must then be subtracted by the moles calculated above.
In solutions 1, 4, 5 and 6 the mass of water is equal to the mass of water in the HCl.
5cm3 of HCl was measured to be 5.28g. In this it was found there was 0.015 mol of HCl. This makes for a mass of 0.5475g of HCl. If this is taken away from the mass of the HCl(aq) than the mass of water can be found.
Mass of water = 5.28 - 0.5475 = 4.73g
Converting this to moles:
Moles water = 4.73/18 = 0.263 moles
Since water has a density of 1 g.cm-3 the amount of cm3 of water added to the initial mixture equals the mass of water added to the initial mixture. Therefore this value must be added to the mass of water and a new mole value will be found.
The amount of moles of water added to each solution initially is shown in the below table.
Solution
Initial Moles water
0.263
2
0.318
3
0.429
4
0.263
5
0.263
6
0.263
To find the moles of ethyl ethanoate, a similar approach must be used. In the solution where no ethy; ethanoate was introduced initially, the amount of moles of ethyl ethanoate produced will equal to the amount of moles of ethanoic acid reacted (due to the 1 : 1 mole ratio of the reaction equation). However, for the solutions where ethyl ethanoate was put into the mixture the amount of moles of ethyl ethanoate will equal the amount of moles initially put in subtracted by the amount of moles of ethanoic acid.
The mass of 5cm3 ethyl ethanoate was measured out to be 4..37g.
Therefore for 5cm3 of ethyl ethanoate there is 4.37/88 = 0.0497 moles.
For 4 cm3 the amount of moles will equal 4/5 x 0.0497 = 0.0397
For 2 cm3: moles = 0.0199
These values are then subtracted by the moles of ethanoic acid in their respective solutions.
The below table shows the initial amount of ethyl ethanoate for each solution:
Solution
Moles Ethyl ethanoate (initial)
0.0497
2
0.0397
3
0.0199
4
0.0397
5
0.0000
6
0.0000
Below is a table with the moles of water and ethyl ethanoate reacted in each solution:
Solution
Moles water reacted
Moles Ethyl ethanoate reacted
0.234
0.0211
2
0.288
0.0099
3
0.412
0.0030
4
0.226
0.0028
5
0.255
0.0079
6
0.241
0.0220
The equation for the equilibrium constant for this reaction is as follows:
To find the concentration we mist simply divide the moles by the total volume. However, since there are equal numbers of variable on the top and on the bottom, the volume cancels out. Therefore we can simply use the moles of each chemical to find the equilibrium constant.
Solution
Moles Ethanoic Acid
Moles water
Moles Ethyl ethanoate
0.0286
0.234
0.0211
2
0.0299
0.288
0.0099
3
0.0169
0.412
0.0030
4
0.0369
0.226
0.0028
5
0.0079
0.255
0.0079
6
0.0220
0.241
0.0220
Since the moles of ethanol equal the moles of ethanoic acid we can simply square the value. Above is a table with all the need values of moles to calculate the equilibrium constant.
From these values we can calculate the equilibrium constant. In the below table the calculated values for the equilibrium constant are shown.
Solution
Kc
6.05
2
3.19
3
4.39
4
0.47
5
32.29
6
0.98
Conclusion:
These values can be averaged to find an average value for the equilibrium constant:
Average = 9.56
However, by doing this there will be an inherent random error. This random error is calculated as done before.
Random error = 15.91
This can be converted into a percentage error.
Percentage error = 166.4%
We must also take into account the percentage errors calculated before.
The average percentage error for the percentage errors before was 1.50%.
Since this percentage error id the same on each value for the moles, it must be multiplied by 4.
This gives us a percentage error of 6%. This must then be added to the percentage error calculated above.
This gives us 170.4%.
This can then be converted into an absolute error.
Absolute error = 16.5
Therefore the Kc = 9.56 ± 16.5
Evaluation:
Accuracy of Apparatus: This experiment was largely subject to the accuracy of the apparatus used. Seeing as very small volumes of liquids were used, if the apparatus used had a large reading error, the error in our calculations would have also been very large therefore making our results very inaccurate. In this experiment, biurettes were used to measure out the volumes of liquids. Therefore the reading error was fairly small making the experiment fairly accurate. This is evident in the small percentage errors varying from 0.9 - 2.2%.
Time Management: Time management was also an issue in this experiment. In order to get a decent amount of results each individual student only prepared two separate flasks with different mixtures in each. This meant that information had to be shared which led to other inaccuracies (which will be explained below). Therefore, the constricting time did not enable each individual student to get his/her own set of results.
Sharing of Data: As explained, in order for a decent amount of results, results had to be shared in between students. This method is very unreliable since no one can be sure that everyone else is doing the experiment properly. This therefore meant that if one student did not follow the instructions properly his/her results could have been very inaccurate and therefore the results for the entire class would have suffered.
Labeling of Liquids: An error in this practical was that not all of the liquids were properly labeled. Students were told where each liquid was in the room but not all of the prepared biurettes were properly labeled with the liquids inside of them students could have easily gotten confused. For example, since both the distilled water and the aqueous hydrochloric acid have the same colorless appearance, a student could easily get confused between the two and mix them up therefore leading to inaccurate mixtures therefore affecting our results.
Improvements:
Accuracy of Apparatus: As explained above, the use of the biurettes is a fairly accurate method of measuring out the liquids. However, more accurate apparatus could be used. For example, electronic flow meters would give even more accurate results.
Time Management: In order to allow students to get there own sets of results more time should have been allotted to this experiment so that each individual student could get there own results.
Sharing of Data: The sharing of data is very useful to get a wider range of values. However, when there is a very limited amount of results it can be more problematic than helpful. Therefore, if the above proposal is used data could still be shared because if one set of results was wrong it would have less of an effect on the overall results and it would stick out more.
Labelling of Liquids: In order to avoid confusion, all biurettes should have been properly labeled with the liquids that they contained.