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!!!
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- Thread Starter
- 16-06-2005 20:31
- 16-06-2005 20:50
is the anser given in the question, or did you get it from the markscheme?
- 16-06-2005 20:53
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.
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.
- Thread Starter
- 17-06-2005 10:04
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!!!
- 17-06-2005 11:26
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.