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# A2 Chem equilibria question help! watch

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1. So I need a bit of help with this question!
To be honest I'm not really sure where I'm going wrong
I'll explain what I did then hopefully that makes it clearer what I've done wrong!!

The question:
In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction. N2(g) + 3H2(g) == 2NH3(g)

Here's what I did:
1 - work out how much ammonia was produced, admittedly I wasn't too sure how to do this, so just did as follows:
(9+27) x 2/3 = 24

2 - Work out the concentration of each reactant/product, so:
9 ÷ 10 = 0.9
27 ÷ 10 = 2.7
24 ÷ 10 = 2.4

3 - Put these into the Kc equation as follows:
Kc = (2.4)2 over (0.9)(2.7)3

Well, that gives me 0.33 mol-2dm6, but the answer should apparently be 6.6 (same units) and I've tried a few things but can't seem to get to that!

If anybody could do this and (even briefly!) explain how they got there, I'd be really grateful!!

Thank you

P.S For anybody who wondered/noticed, this question was from a worksheet on www.a-levelchemistry.co.uk!
2. (Original post by Kugelmugel)
So I need a bit of help with this question!
To be honest I'm not really sure where I'm going wrong
I'll explain what I did then hopefully that makes it clearer what I've done wrong!!

The question:
In an experiment, 9.0 moles of nitrogen and 27 moles of hydrogen were placed into a vessel of volume 10 dm3 and allowed to reach equilibrium. It was found that two thirds of the nitrogen and hydrogen were converted into ammonia. Calculate Kc for the reaction. N2(g) + 3H2(g) == 2NH3(g)

Here's what I did:
1 - work out how much ammonia was produced, admittedly I wasn't too sure how to do this, so just did as follows:
(9+27) x 2/3 = 24

2 - Work out the concentration of each reactant/product, so:
9 ÷ 10 = 0.9
27 ÷ 10 = 2.7
24 ÷ 10 = 2.4

3 - Put these into the Kc equation as follows:
Kc = (2.4)2 over (0.9)(2.7)3

Well, that gives me 0.33 mol-2dm6, but the answer should apparently be 6.6 (same units) and I've tried a few things but can't seem to get to that!

If anybody could do this and (even briefly!) explain how they got there, I'd be really grateful!!

Thank you

P.S For anybody who wondered/noticed, this question was from a worksheet on www.a-levelchemistry.co.uk!
Two thirds of the nitrogen and hydrogen were used up - so you can say that two thirds of each of those things were used up.

So figure out how many moles of nitrogen or hydrogen were used and how many moles of ammonia that will make.

The other bit is to make sure that you take away the number of moles used up from the original amount of hydrogen/ammonia when calculating Kc.
3. Hello Basically,
2/3 of the nitrogen and hydrogen reacted. So you only have 1/3 of nitrogen and hydrogen left in the mixture at the end. So do 9 moles X 1/3 and you get 3. Do the same with H so times 27 by 1/3 to get 9. Then, 12 moles of NH3 are produced because you need to look at the ratio. I added up the two moles of H2 and N2 to get 36 moles- out of this 36 moles , 2/3 became NH3 so you have 24. HOWEVER, the ratio of reactants to products is 4:2 so you need to divide by two to get 12 moles of NH3

4. At the start you have 9 moles of N2 and 27 moles of H2

At equilibrium you have 3 moles of N2, 9 moles of H2 and 12 moles of NH3 and the total number of moles is 3+9+12=24

So the concentration of H2 = 3/24. N2 = 9/24 and NH3 = 12/24
5. Thank you so much! Would anyone mind posting their working leading up to the answer though, because I'm using what you said, but getting an answer that's even more wrong! O_o

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