IB Group 4 Internal Assessment Chemistry Biology Physics
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Introduction:
The equilibrium constant could be calculated by using the formula. Where A + B -> C + D
In determining the equilibrium constant for an esterification reaction,
The equilibrium constant, KC , would be;
The research question would be; determine the KC values for different mixtures.
Table 1. Uncertainties of Equipments
Table 2. Mixtures and Its Properties
Table 3. Mass of Each of the Liquids
Results:
Table 4. Raw Data – Titres against 0.9531M of NaOH
Table 5. Qualitative Data
Table 6. Processed Data - Rearrangements of Data
Table 7. Processed Data – Average Titration Value for Each Mixture
Mixture 1; average and uncertainty calculation
Average = 44.65ml
Uncertainty = 0.60ml
Calculations:
Balanced equation for the esterification reaction is;
Calculate the number of moles of ethyl ethanoate, ethanoic acid, ethanol and water present in the original mixture.
Mixture 1, 2, 3, 4 and 5;
Since there were no ethanoic acid present in the mixture, the amount in moles of ethanoic acid present in the original mixture is 0.
Mixture 1 contained 5ml of ethyl ethanoate, hence the amount in moles of ethyl ethanoate present in the original mixture is;
Moles of ethyl ethanoate; calculating uncertainty
0.052 ± 2.3x10-4 mol of ethyl ethanoate was present
Moles of water are 0.212±0.0009mol.
Volume of water = 5 – (0.032x36.45)
= 3.83g
Calculating uncertainty,
Uncertainty = 0.9531 x (33.24x10-3±0.15x10-3)
= 0.032±0.451%
= 5±0.15 – (0.032±0.451% x 36.45)
= 5±0.15 – 1.1664±0.00526
= 3.83ml±0.16ml
Mixture 2 contained 1ml of water and 4ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and water present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.041 ± 2.3x10-4 mol of ethyl ethanoate was present.
Moles of water are 0.268±0.0017mols
Mixture 3 contained 2ml of water and 3ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and water present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.031 ± ...
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= 5±0.15 – (0.032±0.451% x 36.45)
= 5±0.15 – 1.1664±0.00526
= 3.83ml±0.16ml
Mixture 2 contained 1ml of water and 4ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and water present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.041 ± 2.3x10-4 mol of ethyl ethanoate was present.
Moles of water are 0.268±0.0017mols
Mixture 3 contained 2ml of water and 3ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and water present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.031 ± 2.3x10-4 mol of ethyl ethanoate was present.
Moles of water are 0.323±0.0017mols
Mixture 4 contained 3ml of water and 2ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and water present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.021 ± 2.3x10-4 mol of ethyl ethanoate was present.
Moles of water are 0.378±0.0017mols
Mixture 5 contained 1ml of ethanol and 4ml of ethyl ethanoate; hence the amount in moles of ethyl ethanoate and ethanol present in the original mixture are;
Moles of ethyl ethanoate; calculating uncertainty
0.041 ± 2.3x10-4 mol of ethyl ethanoate was present.
Moles of ethanol; calculating uncertainty
0.017 ± 4.4x10-4 mol of ethanol was present.
Moles of water are 0.212±0.0009mol.
Mixture 6 and 7;
Both of the mixture has 1ml of ethanoic acid, hence the amount in moles of ethanoic acid present in the original mixture is;
Moles of ethanoic acid; calculating uncertainty
Therefore, there was 0.017mol±3.3x10-4mol of ethanoic acid present in the original mixture for mixture 6 and mixture 7.
Mixture 6 contained 4ml ethyl ethanoate; hence the amount in moles of ethyl ethanoate present in the original mixture is;
Moles of ethyl ethanoate; calculating uncertainty
0.041 ± 2.3x10-4 mol of ethyl ethanoate was present. Moles of water are 0.212±0.0009mol.
Mixture 7 contained 4ml of ethanol; hence the amount in moles of ethanol present in the original mixture is;
Moles of ethanol; calculating uncertainty
0.068 ± 4.4x10-4 mol of ethanol was present
Moles of water are 0.212±0.0009mol.
Mixture 8 contained 2ml of ethanoic acid and 3ml of ethanol; hence the amount in moles of ethanoic acid and ethanol present in the original mixture are;
Moles of ethanoic acid; calculating uncertainty
0.035mol±3.3x10-4mol of ethanoic acid was present
Moles of ethanol; calculating uncertainty
0.051 ± 4.4x10-4 mol of ethanol was present
Moles of water are 0.212±0.0009mol.
Table 8. Summary of Calculation Question – Moles in Mixture
Calculate the amount in moles of ethanoic acid present in the equilibrium mixture and hence the amount of ethanol, ethyl ethanoate and water in the equilibrium mixture.
Mixture 1 contains 5ml of HCl and 5ml of ethyl ethanoate. There is no water present in this mixture; however, since HCl is in aqueous state, there is water present in the HCl solution.
Hence, the amount of pure HCl could be calculated;
Since there were 5ml of HCl solution in the mixture, then the amount of water could be calculated;
Mixture 1 contained 5ml of ethyl ethanoate. To find the number of moles for ethyl ethanoate in equilibrium;
33.24x10-3dm3 of NaOH was needed to react with 10ml of HCl, since in mixture 1 has 5ml of HCl, amount used of NaOH to react with HCl needs to be taken off from the overall amount of NaOH used for the mixture.
Uncertainty Calculations
Uncertainty = (0.9531±0) x (44.65x10-3±1.5x10-4)
= 0.043±0.0034
Uncertainty Calculations
Uncertainty = (44.65x10-3±1.5x10-4) – [(33.24x10-3±1.5x10-4)/2]
= (44.65x10-3±1.5x10-4) – (16.62x10-3±7.5x10-5)
= 28.03x10-3±2.25x10-4
Number of mols of ethanoic acid in equilibrium would be 0.027mol.
Since initial is 0 mols and equilibrium is 0.027, the change is 0.027 mols.
Moving onto water, as shown above, there is no water present in the mixture, but only in the HCl, which is 3.83ml. If 3.83ml is to be converted into moles, then the number of moles at initial state would be;
Uncertainty Calculations
Uncertainty = (0.9531±0)x(28.03x10-3±2.25x10-4)
= 0.0027±2.17x10-5
Because the change is 0.0027±2.17x10-5 mols, the number of moles of water in equilibrium is;
0.212±0.0009mol – 0.0027±2.17x10-5 = 0.185±9.217x10-4
As the initial number of moles of ethyl ethanoate in equilibrium could be calculated;
Because the change is 0.027mols, the number of moles of ethyl ethanoate in equilibrium is;
0.052±2.286x10-4 – 0.027±2.17x10-5 = 0.025±2.50x10-4 mols.
As there are no ethanoic acid and ethanol present in the mixture, both of their numbers of moles in initial state are 0. Also, as written above, the change is 0.027mols. However, because this is the other side of the reaction, the 0.027mols changes into -0.027mols. Hence, the numbers of moles for both ethanoic acid and the ethanol in equilibrium are 0.027±2.17x10-5 mols.
Overall in mixture 1 at equilibrium state, there are 0.027±2.17x10-5 mols of ethanoic acid, 0.027±2.17x10-5 mols of ethanol, 0.025±2.50x10-4 mols of ethyl ethanoate and 0.185±9.217x10-4 mols of water.
Mixture 2 contains 5ml of HCl, 1ml of water and 4ml of ethyl ethanoate. Mixture 2 in an equilibrium state, there are 0.026±2.17x10-5 mols of ethanoic acid, 0.026±2.17x10-5 mols of ethanol, 0.015±2.49x10-4 mols of ethyl ethanoate and 0.241±1.722x10-3 mols of water.
Mixture 3 contains 5ml of HCl, 2ml of water and 3ml of ethyl ethanoate. Mixture 3 in an equilibrium state, there are 0.021±2.17x10-5 mols of ethanoic acid, 0.021±2.17x10-5 mols of ethanol, 0.010±2.49x10-4 mols of ethyl ethanoate and 0.302±1.722x10-3 mols of water.
Mixture 4 contains 5ml of HCl, 3ml of water and 2ml of ethyl ethanoate. Mixture 4 in an equilibrium state, there are 0.016±2.17x10-5 mols of ethanoic acid, 0.016±2.17x10-5 mols of ethanol, 0.005±2.49x10-4 mols of ethyl ethanoate and 0.362±1.722x10-3 mols of water.
Mixture 5 contains 5ml of HCl, 1ml of ethanol and 4ml of ethyl ethanoate. Mixture 5 in an equilibrium state, there are 0.020±2.17x10-5 mols of ethanoic acid, 0.037±4.35x10-4 mols of ethanol, 0.021±2.49x10-4 mols of ethyl ethanoate and 0.192±9.217x10-4 mols of water.
Mixture 6 contains 5ml of HCl, 1ml of ethanoic acid and 4ml of ethyl ethanoate. Mixture 6 in an equilibrium state, there are 0.035±3.55x10-4 mols of ethanoic acid, 0.018±2.17x10-5 mols of ethanol, 0.023±2.49x10-4 mols of ethyl ethanoate and 0.194±9.217x10-4 mols of water.
Mixture 7 contains 5ml of HCl, 1ml of ethanoic acid and 4ml of ethanol. Mixture 7 in an equilibrium state, there are 0.008±3.55x10-4 mols of ethanoic acid, 0.059±4.57x10-4 mols of ethanol, 0.009±2.49x10-4 mols of ethyl ethanoate and 0.221±9.217x10-4 mols of water.
Mixture 8 contains 5ml of HCl, 2ml of ethanoic acid and 3ml of ethanol. Mixture 8 in an equilibrium state, there are 0.015±3.55x10-4 mols of ethanoic acid, 0.031±4.57x10-4 mols of ethanol, 0.020±2.49x10-4 mols of ethyl ethanoate and 0.192±9.217x10-4 mols of water.
Table 9. Summary Table for Calculation – Moles in Equilibrium
Calculate the equilibrium concentrations of the water, ethyl ethanoate, ethanoic acid and ethanol in the mixture and hence the equilibrium constant.
Mixture 1 in an equilibrium state, there are 0.025mols of ethanoic acid, 0.025mols of ethanol, 0.027mols of ethyl ethanoate and 0.187mols of water.
To calculate the concentration, each number of moles should be divided with 10ml as mixture 1 is formed of 10ml solution.
Calculating Uncertainty
(0.027±2.17x10-5)/10
= (0.027±0.080%)/10
= 0.0027±2.17x10-6)
(0.0185±9.217x10-5)(0.0025±2.49x10-5)/(0.0027±2.17x10-6)(0.0027±2.17x10-6)
= 0.498% + 0.996% + 0.080% + 0.080%
= 1.654%
= 1.654/100 x 6.34
= ±0.10
Mixture 2 is an equilibrium state; there are 0.015mols of ethanoic acid, 0.015mols of ethanol, 0.026mols of ethyl ethanoate and 0.252 mols of water.
Mixture 3 in an equilibrium state, there are 0.010mols of ethanoic acid, 0.010mols of ethanol, 0.021mols of ethyl ethanoate and 0.313 mols of water.
Mixture 4 in an equilibrium state, there are 0.050mols of ethanoic acid, 0.050mols of ethanol, 0.0157mols of ethyl ethanoate and 0.328 mols of water.
Mixture 5 in an equilibrium state, there are 0.022mols of ethanoic acid, 0.039mols of ethanol, 0.019mols of ethyl ethanoate and 0.190 mols of water.
Mixture 6 in an equilibrium state, there are 0.024mols of ethanoic acid, 0.007mols of ethanol, 0.034mols of ethyl ethanoate and 0.205 mols of water.
Mixture 7 in an equilibrium state, there are 0.009mols of ethanoic acid, 0.068mols of ethanol, 0.008mols of ethyl ethanoate and 0.214 mols of water.
Mixture 8 in an equilibrium state, there are 0.015mols of ethanoic acid, 0.031mols of ethanol, 0.020mols of ethyl ethanoate and 0.192 mols of water.
Table 10. Summary Table for Calculation – Concentration and KC
Conclusion/Discussion:
It is stated in the Chemistry Course Companion by Geoffrey Neuss that the theoretical equilibrium constant of this esterification reaction is 4 at standard room temperature, 298K. In table 10, it shows the concentration of all the substances in each mixture and also shows the equilibrium constant. Calculating the percentage difference between the theoretical value and KC values of each mixtures;
Percentage difference of mixture 1 is 58.5%±0.9%.
Calculating Uncertainty
[(6.34±0.10) – (4±0)]/(4±0)x100 = [(6.34±1.58%) – (4±0%)]/(4±0%)x100
= (2.34±1.58%)/(4±0%)x100
= 0.585±1.58%x100
= 58.5±0.9%
Percentage difference of mixture 2 is 33.8%±0.9%.
Percentage difference of mixture 3 is 71.3%±2.3%.
Percentage difference of mixture 4 is 76.8%±4.3%.
Percentage difference of mixture 5 is 37.3%±2.8%.
Percentage difference of mixture 6 is 82.3%±5.6%.
Percentage difference of mixture 7 is 5.3%±0.4%.
Percentage difference of mixture 8 is 38.0%±2.0%.
Considering these percentage differences, it could be stated that there were numerous errors been carried out throughout the experiment.
In room temperature, the enthalpy change of this esterification reaction is +17.5kJ/mol, an endothermic reaction. It is expected for the reaction to have greater KC value when the reaction was taken place in higher temperature. For example, if the reaction was to be taken place in 60°C environment, then there is more energy provided into the reaction. Hence, KC value will be greater as the added energy is used up and pushes the equilibrium to the right due to Le Chatelier’s equilibrium principal.
During this esterification reaction, HCl was added to the reactants. However, the yield of ester formed in the reaction could be increased when concentrated sulphuric acid was to be added to the reactants. This is because HCl has only one H+ ion, whereas, concentrated sulphuric acid contains two H+ ion. Hence, there are more H+ ions in concentrated sulphuric acid to be used up in the esterification reaction. By adding concentrated sulphuric acid rather than HCl, the yield of ester formed will increase by squared, as there is double the amount of H+ ions in concentrated sulfuric acid.
Improvements:
Table 11. Lists of Improvements
Bibliography:
Neuss, Geoffrey. Chemistry Course Companion. Oxford: Oxford University Press, 2007
Group 1D + Group 2D
Group 1B + Group 1B
Group 3D + Group 4D
Group 3B + Group 4B
Group 5D + Group 6D
Group 5B + Group 6B
Team 6 data was eliminated from calculating the average as it is an outlier
Team 1 and Team 6 data was eliminated from calculating the average as both of them are outliers
Team 1 and Team 6 data was eliminated from calculating the average as both of them are outliers
Team 3 and Team 6 data was eliminated from calculating the average as both of them are outliers
