# Kc Chemistry A2 Rates Question

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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.

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.

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#2

(Original post by

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.

**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.

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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(Original post by

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.

**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.

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

Thanks x

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#4

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Ah right thanks

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

Thanks x

**EmmaBxoxo**)Ah right thanks

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

Thanks x

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Yes, you'll need it for the calculation I *think* it's equal to the number of moles of ethyl ethanoate.

**SeanFM**)Yes, you'll need it for the calculation I *think* it's equal to the number of moles of ethyl ethanoate.

Are you doing this paper this year.

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#6

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Awesome, I did that in the first place

Are you doing this paper this year.

**EmmaBxoxo**)Awesome, I did that in the first place

Are you doing this paper this year.

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I did it last year, I'm in my first year of uni now

**SeanFM**)I did it last year, I'm in my first year of uni now

Any advice to do well in a2 chem

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#8

(Original post by

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

**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

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(Original post by

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!

**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!

repped

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#10

(Original post by

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.

**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.

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.

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(Original post by

Hi Emma, I was wondering if you had the 2014 june ocr chemistry of material paper.

Thank you x

**Qeebo**)Hi Emma, I was wondering if you had the 2014 june ocr chemistry of material paper.

Thank you x

(Original post by

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.

**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.

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#13

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Yes because they are all the same order, thanks

**EmmaBxoxo**)Yes because they are all the same order, thanks

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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#14

(Original post by

Does anyone know what Kc works out to be?

**SLTLucy**)Does anyone know what Kc works out to be?

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#16

(Original post by

Kc= 3.99 (to 2d.p) with no units

**ethanoylchloride**)Kc= 3.99 (to 2d.p) with no units

(Original post by

4 I think?

**buxtonarmy**)4 I think?

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(Original post by

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.

**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.

I got 3.99 too.

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(Original post by

Kc= 3.99 (to 2d.p) with no units

**ethanoylchloride**)Kc= 3.99 (to 2d.p) with no units

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