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1. The equilibrium constant for th acid-catalysed reaction between ethanol and ethanoic acid to form ethanoate and water can be determined in a laboratory.

C2H5OH (l) + CH3COOH (l) = CH3COOC2H5 (l) + H2O

In an experiment, 20cm3 of ethanoic acid, 20cm3 of ethanol and 30cm3 of ethy ethanoate are mixed together.

10cm3 of 0.5mol dm3 sulphuric acid is then added. This provides the necessary acid catalyst and 10cm3 of water.

1cm3 of the mixture is then pipetted into a small conical flask. 15cm3 of water is added to increase the volume and the misture is then titrated against a 0.25mol dm3 solution of sodium hydroxide. It is found that 14.5cm3 of NaOH is required to neutralise the acid in the mixture.

a) given the following densities calculate the amount (in moles) of ethanol, ethanoic acid, ethyl ethanoate and water that were present at the start of the experiment.
Densities:
C2H5OH = 0.79 g cm-3
CH3COOH = 1.05 g cm-3
CH3COOH = 0.92 g cm-3
H20 = 1 g cm-3

I have answered this question as: volume * density = mass and then mass/mr = moles

C2H5OH = 0.34 mol
CH3COOH = 0.35 mol
CH2COOC2H5 = 0.31 mol
H2O = 0.56 mol

PLEASE will you help me with the following questions:

b) use the titration result to detemine the amount (in moles) of ethanoic acid that is present in the 1cm3 sample taken after equilibrium had been established

c) use your answer from part b to detemine the amount, in moles, of ethanoic acid present in the full 80cm3 of the equilibrium mixture

d) from your answer to part c calculate the amount, in moles, of ethanol, ethyl ethanoate and water present in the equilibrium mixture.

e) calculate the value of the equilibrium constant for this reaction.

Any help would be massively appreciated, thanks! x x x
2. (Original post by ilovepsych_)
The equilibrium constant for th acid-catalysed reaction between ethanol and ethanoic acid to form ethanoate and water can be determined in a laboratory.

C2H5OH (l) + CH3COOH (l) = CH3COOC2H5 (l) + H2O

In an experiment, 20cm3 of ethanoic acid, 20cm3 of ethanol and 30cm3 of ethy ethanoate are mixed together.

10cm3 of 0.5mol dm3 sulphuric acid is then added. This provides the necessary acid catalyst and 10cm3 of water.

1cm3 of the mixture is then pipetted into a small conical flask. 15cm3 of water is added to increase the volume and the misture is then titrated against a 0.25mol dm3 solution of sodium hydroxide. It is found that 14.5cm3 of NaOH is required to neutralise the acid in the mixture.

a) given the following densities calculate the amount (in moles) of ethanol, ethanoic acid, ethyl ethanoate and water that were present at the start of the experiment.
Densities:
C2H5OH = 0.79 g cm-3
CH3COOH = 1.05 g cm-3
CH3COOH = 0.92 g cm-3
H20 = 1 g cm-3

I have answered this question as: volume * density = mass and then mass/mr = moles

C2H5OH = 0.34 mol
CH3COOH = 0.35 mol
CH2COOC2H5 = 0.31 mol
H2O = 0.56 mol

PLEASE will you help me with the following questions:

b) use the titration result to detemine the amount (in moles) of ethanoic acid that is present in the 1cm3 sample taken after equilibrium had been established

c) use your answer from part b to detemine the amount, in moles, of ethanoic acid present in the full 80cm3 of the equilibrium mixture

d) from your answer to part c calculate the amount, in moles, of ethanol, ethyl ethanoate and water present in the equilibrium mixture.

e) calculate the value of the equilibrium constant for this reaction.

Any help would be massively appreciated, thanks! x x x
your moles of water is incorrect. You should appreciate that 0.5M sulphuric acid is mostly water as well, hence you have 20ml = 20g = 1.11 mol, unless I have misunderstood the wording.

Do you add 10ml of water AND 10ml of sulphuric acid?

Use the value for sodium hydroxide to calculate the total moles of H+ present in the sample. Subtract the H+ due to the sulphuric acid and you are left with the H+ due to ethanoic acid. From there you can find he moles of ethanoic acid and hence the change in moles of ethanoic acid. You then use the equilibrium stoichiometry to find the change in moles of all of the other components and hence the final moles of all of the other components. From this you get the equilibrium constant.

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Updated: March 14, 2013
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