Reference no: EM13585599
Determine a value of Kc for the equilibrium which is established when ethanol reacts with ethanoic acid.
Ch3COOH + CH3CH2OH = CH3COOCH2CH3 + H2O
Set up a mixture of ethanoic acid, ethanol, water and dilute HCL(catalyst) leave for a week to come to equilibrium. A control of dilute HCL is set up to.
The mixture is then poured into excess water and then titrated with 0.100 mol dm-3 sodium hydroxide solution. Results give the total concentration of acid in the equilibrium mixture. Titration of the control shows the concentration of the dilute HCL from this the concentration of the ethanoic acid can be determined.
Use the densities and volumes given to calculate the initial amount in moles , for the ethanoic acid , ethanol and water in the equilibrium mixture. Assume 2.0cm3 of 1mol dm3 HCL adds 2.0cm3 of water to the mixture.
Titrate the control:
Pour contents of control and add 100ml of distilled water. Use indicator for pink endpoint and titrate with NaOH solution from burette.
My Control was 25.8ml
Titration of equilibrium mixture:
Use 1.0ml of equilibrium mixture and add 100ml of distilled water into conical flask.
Titrate as done above with NaOH and indicator.
My average of three results was 40.35ml
Table of the initial volumes for each reagent al in cm3
|
Reagent
|
Density/g cm-3
|
Control
|
Mixture
|
|
Glacial ethanoic acid
|
1.05
|
0.0
|
6.0
|
|
Ethanol
|
0.79
|
0.0
|
6.0
|
|
Water
|
1.00
|
18.0
|
6.0
|
|
1 mol dm-3 HCL
|
1.00
|
2.0
|
2.0
|
|
|
|
|
|
|
Total Volume
|
|
20.00
|
20.00
|
|
|
|
|
|
|
|
|
|
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Questions:
Calculate the amount of HCL, in moles, in the control flask using the titration result for the control.
Calculate the total amount of acid in moles, in the equilibrium mixture using the average titre from the titration of the equilibrium mixture.
Subtract the amount of HCL from the total amount of acid in the equilibrium mixture to calculate the amount of ethanoic acid, in moles, remaining in the equilibrium mixture.
Calculate the amount of ethanoic acid, in moles, that has reacted as the equilibrium is established. The amount of moles is the same as the amount of ethyl ethanoate, in moles and the amount of water in moles which have been formed at equilibrium. It is also the amount of alcohol, in moles, which has reacted as the equilibrium is established.
Calculate the amount of ethanol in moles, that remains at equilibrium from the original amount, in moles that was put in the flask
Calculate the amount of water in moles at equilibrium. (Do not forget to include the fact that the equivalent of 8.0cm3 of water was added initially
Calculate the concentration in mol dm-3 of ethanoic acid at equilibrium.
Calculate the concentration in mol dm-3 of ethanol at equilibrium
Calculate the concentration in mol dm-3 of ethanoate at equilibrium
Calculate the concentration in mol dm-3 of water at equilibrium
Use the data to calculate the value of Kc for the equilibrium mixture
(Given that the accepted value for this constant is quoted as 4.00 at 298 K, comment on the result.