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# A Kc OCR question.. watch

1. i was hoping someone could help me with a Kc question i have come across that i cant seem to get right:

0.204 mol CH3COOC2H5 and 0.645 mol H2O were introduced into a flask, acidified and left until a steady state was reached. a sample was then analysed, and 0.114 mol of CH3COOH was found. calculate Kc for the reaction.

CH3COOC2H5(l) + H2O(l) -> CH3COOH(l) + C2H5OH(l)
0.204 0.645 0.114 x
i dont even know how to go about it, there is no volume of the flask givem. is the question just wrong, or am i missing something?

the answer is given as 0.27

any help would be appreciated!!!
2. is the anser given in the question, or did you get it from the markscheme?
3. Since 0.114 moles of CH3COOH is present in the equilibrium mixture, 0.114 moles of CH3COOC2H5 will have been used up. Hence:

moles CH3COOC2H5 in equilibrium mixture = 0.204 - 0.114 = 0.0900 moles

The same is true for water, hence:

moles H2O in equlibrium mixture = 0.645 - 0.114 = 0.531 moles

If 0.114 moles of CH3COOH is present in the equilibrium mixture, then 0.114 moles of C2H5OH must also be present in the equilibrium mixture.

Therefore:

Kc = 0.114 x 0.114 / 0.531 x 0.0900 = 0.272 (3sf)

It's all to do with molar ratios, which in this case is 1:1.

The volumes cancel out, so it's not required.
4. Okay...i think i get that. i just didnt get what the mols of C2H5OH would be, but i take it if you take amounts from the equation, you assume it is 0.114?

excellent! thanks a lot!!!
5. Yeah, it's all about the equation. The equation basically says: for every mole of ethanoic acid produced, you will also get one mole of ethanol.

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