Kc Chemistry A2 Rates Question

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#1
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.
0
6 years ago
#2
(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.
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#3
(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

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

Thanks x
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6 years ago
#4
(Original post by EmmaBxoxo)
Ah right thanks

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 I *think* it's equal to the number of moles of ethyl ethanoate.
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#5
(Original post by SeanFM)
Yes, you'll need it for the calculation I *think* it's equal to the number of moles of ethyl ethanoate.
Awesome, I did that in the first place

Are you doing this paper this year.
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6 years ago
#6
(Original post by EmmaBxoxo)
Awesome, I did that in the first place

Are you doing this paper this year.
I did it last year, I'm in my first year of uni now
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#7
(Original post by SeanFM)
I did it last year, I'm in my first year of uni now
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
1
6 years ago
#8
(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
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!
1
#9
(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
0
6 years ago
#10
(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.
1
#11
(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

(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
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6 years ago
#12
Does anyone know what Kc works out to be?
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6 years ago
#13
(Original post by EmmaBxoxo)
Yes because they are all the same order, thanks
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.
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6 years ago
#14
(Original post by SLTLucy)
Does anyone know what Kc works out to be?
Kc= 3.99 (to 2d.p) with no units
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6 years ago
#15
(Original post by SLTLucy)
Does anyone know what Kc works out to be?
4 I think?
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6 years ago
#16
(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!
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#17
(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.
0
#18
(Original post by ethanoylchloride)
Kc= 3.99 (to 2d.p) with no units
Yes I got this too, thanks for the reassurance
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