The Student Room Group

Kc Chemistry A2 Rates Question

The equilibrium below was established by allowing 2 moles of ethanol and 1 mole of ethanoic acid to react at 25 degrees Celsius. At equilibrium the mixture contained 0.845 moles of ethyl ethanoate. Calculate Kc at this temperature.

CH3COOH + CH3CH2OH -> CH3COOCH2CH3 + H2O

So obviously work out Kc. But then can I just sub the values of the moles in since no volume was given or do I need to work out the Mr etc.
Original post by EmmaBxoxo
The equilibrium below was established by allowing 2 moles of ethanol and 1 mole of ethanoic acid to react at 25 degrees Celsius. At equilibrium the mixture contained 0.845 moles of ethyl ethanoate. Calculate Kc at this temperature.

CH3COOH + CH3CH2OH -> CH3COOCH2CH3 + H2O

So obviously work out Kc. But then can I just sub the values of the moles in since no volume was given or do I need to work out the Mr etc.


I think no volume was given because no matter what volume you have, in the Kc equation they'll cancel eachother out. So you can just sub in the moles.

You could see for yourself that if you assign a volume V, then you'll end up with V^2/V^2 which is one, and then the mole stuff.
Reply 2
Original post by SeanFM
I think no volume was given because no matter what volume you have, in the Kc equation they'll cancel eachother out. So you can just sub in the moles.

You could see for yourself that if you assign a volume V, then you'll end up with V^2/V^2 which is one, and then the mole stuff.


Ah right thanks :smile:

Just one last question - is it necessary to calculate the amount of moles of H2O?

Thanks x
Original post by EmmaBxoxo
Ah right thanks :smile:

Just one last question - is it necessary to calculate the amount of moles of H2O?

Thanks x


Yes, you'll need it for the calculation :smile: I *think* it's equal to the number of moles of ethyl ethanoate.
Reply 4
Original post by SeanFM
Yes, you'll need it for the calculation :smile: I *think* it's equal to the number of moles of ethyl ethanoate.


Awesome, I did that in the first place :smile:

Are you doing this paper this year.
Original post by EmmaBxoxo
Awesome, I did that in the first place :smile:

Are you doing this paper this year.


I did it last year, I'm in my first year of uni now :colondollar:
Reply 6
Original post by SeanFM
I did it last year, I'm in my first year of uni now :colondollar:


Ah the june 2014 f324 paper was horrible, haven't looked at the paper for f325 yet.

Any advice to do well in a2 chem :smile:
Original post by EmmaBxoxo
Ah the june 2014 f324 paper was horrible, haven't looked at the paper for f325 yet.

Any advice to do well in a2 chem :smile:


I did AQA but I think every June 2014 paper was a nightmare. I'm not great at Chem but I'd suggest making lots of notes (maybe on flashcards if that works for you), using the textbook and the revision guide and just doing past papers over and over. Good luck!
Reply 8
Original post by SeanFM
I did AQA but I think every June 2014 paper was a nightmare. I'm not great at Chem but I'd suggest making lots of notes (maybe on flashcards if that works for you), using the textbook and the revision guide and just doing past papers over and over. Good luck!


Thank you Sean will do.

repped
Reply 9
Original post by EmmaBxoxo
But then can I just sub the values of the moles in since no volume was given or do I need to work out the Mr etc.


Using V in the equation c = n/V

Writing out the expression for Kc, you'll end up with two V terms on the top and two below. The V terms cancel out.
Reply 10
Original post by Qeebo
Hi Emma, I was wondering if you had the 2014 june ocr chemistry of material paper.
Thank you x


Hey I don't I'm afraid. It was the hardest of all past ocr papers current spec imo. You could take a look at the examiners report though :smile:

Original post by Pigster
Using V in the equation c = n/V

Writing out the expression for Kc, you'll end up with two V terms on the top and two below. The V terms cancel out.


Yes because they are all the same order, thanks :smile:
Reply 11
Does anyone know what Kc works out to be?
Original post by EmmaBxoxo
Yes because they are all the same order, thanks :smile:


The exponents in the Kc expression are not orders (as in the rate equation).

Orders are values determined by experiments and relate to the number of each particle in the rate determining step.

The exponents in the Kx expression are the reacting ratios in the overall equation, i.e. the stoichiometry.

When there are as many particles on the left as there are on the right, then the V terms will cancel out. c.f. when pressure doesn't affect gas equilibrium position.
Original post by SLTLucy
Does anyone know what Kc works out to be?


Kc= 3.99 (to 2d.p) with no units
Original post by SLTLucy
Does anyone know what Kc works out to be?


4 I think?
Reply 15
Original post by ethanoylchloride
Kc= 3.99 (to 2d.p) with no units



Original post by buxtonarmy
4 I think?


Thank you to both of you! Reassuring as I also got this!
Reply 16
Original post by Pigster
The exponents in the Kc expression are not orders (as in the rate equation).

Orders are values determined by experiments and relate to the number of each particle in the rate determining step.

The exponents in the Kx expression are the reacting ratios in the overall equation, i.e. the stoichiometry.

When there are as many particles on the left as there are on the right, then the V terms will cancel out. c.f. when pressure doesn't affect gas equilibrium position.


Yes, that makes sense. Thanks for the clarity.

I got 3.99 too.
Reply 17
Original post by ethanoylchloride
Kc= 3.99 (to 2d.p) with no units


Yes I got this too, thanks for the reassurance :smile:

Quick Reply

Latest