A2 OCR Chemistry - unifying concepts question

8

Really stuck with this question, can't seem to get my head round it.

equilibrium equation: CO + 2H2 ----> CH3OH

The equilibrium reaction is normally carried out at 10MPa pressure and 550K, starting with a 1:2 CO :h:

2 mixture. At equilibrium, only 10% CO reacted. Then it gives a table, giving you the following info and asking you to fill in the blanks:

initial amount/mol CO - 1.0 H2 - 2.0 CH3OH - 0.0

equilbrium amount/mol CO - 0.9

mole fraction at equilibrium

partial pressure at equilib/MPa

Does anyone have a good (easy!) way to explain this? Any help greatly appreciated! :biggrin:

Reply 1

Intelligentsia

11

Well the equation is:

CO + 2H2 <---> CH3OH

At Equilibrium only 10% CO REACTED. Which means that you have 90% (0.9)CO REMAINING. But the 10% that REACTED gives you 0.1 CH3OH and also 0.2 H2 were used up (due to the mole ratios 1:2 for CO:2H2 and 1:1 for CO:CH3OH).

The mole fraction is given by no. of moles of that substance/total no. of moles. So at equilibrium its:

For CO = 0.9/ (0.9 + 0.1 + 0.2)
For H2 = 0.2/ (0.9 + 0.1 + 0.2)
For CH3OH = 0.1/ (0.9 + 0.1 + 0.2)

Partial pressure for each substance is also just the Mole fraction x total pressure. In this case it would be the individual mole fractions x 10MPa.

Reply 2

xsbrx

OP

8

Intelligentsia

Well the equation is:

CO + 2H2 <---> CH3OH

At Equilibrium only 10% CO REACTED. Which means that you have 90% (0.9)CO REMAINING. But the 10% that REACTED gives you 0.1 CH3OH and also 0.2 H2 were used up (due to the mole ratios 1:2 for CO:2H2 and 1:1 for CO:CH3OH).

The mole fraction is given by no. of moles of that substance/total no. of moles. So at equilibrium its:

For CO = 0.9/ (0.9 + 0.1 + 0.2)
For H2 = 0.2/ (0.9 + 0.1 + 0.2)
For CH3OH = 0.1/ (0.9 + 0.1 + 0.2)

Partial pressure for each substance is also just the Mole fraction x total pressure. In this case it would be the individual mole fractions x 10MPa.

Thanks for the reply.
For the equilib amount, the markscheme says H2 is 1.8, not 0.2. Did u mean do 2.0-0.2?

Reply 3

Intelligentsia

11

Sorry yes, if you read it I did say 0.2 were USED UP i.e. reacted because the H2: CH3OH ration is 2:1. The equilibrium amount is 2 x the equilibrium amount of CO so you have 1.8. My bad, and obviously the Mole fraction also changes then. I would wait for another confirmation.

Reply 4

xsbrx

OP

8

ok so the mole fraction would be:

For CO = 0.9/ (0.9 + 0.1 + 1.8)
For H2 = 1.8/ (0.9 + 0.1 + 1.8)
For CH3OH = 0.1/ (0.9 + 0.1 + 1.8)
right!?

But this isn't what it comes out to in the markscheme. For mole fraction, they get:
For CO = 0.9/2.8 (same as above)
For H2 = 0.32/0.3
For CH3OH =1.8/2.8

Reply 5

Intelligentsia

11

Hmm.. what year and session is this question from?

Reply 6

xsbrx

OP

8

Intelligentsia

Hmm.. what year and session is this question from?

it's from june 2005, question 2.

Thanks for your help btw, i'll rep you 2moro :biggrin:

Reply 7

Intelligentsia

11

xsbrx

ok so the mole fraction would be:

For CO = 0.9/ (0.9 + 0.1 + 1.8)
For H2 = 1.8/ (0.9 + 0.1 + 1.8)
For CH3OH = 0.1/ (0.9 + 0.1 + 1.8)
right!?

But this isn't what it comes out to in the markscheme. For mole fraction, they get:
For CO = 0.9/2.8 (same as above)
For H2 = 0.32/0.3
For CH3OH =1.8/2.8

I think you're just a lil confused by the mark scheme layout, it's really badly spaced out with no tables or anything. The mark scheme is here:

http://www.pastpaperexams.com/downloads/UnifC/2005%20June%20MS.pdf

Notice it says "0.9/2.8 OR 0.321 OR 0.32/0.3" .. if you divide 0.9/2.8 you get 0.321ish so I think they're all just acceptable versions of the same mole fraction for CO.

Then right next to all that there's a "1.8/2.8 OR 0.643OR 0.64/0.6" again acceptable versions which I believe refers to the H2, and finally "0.1/2.8 OR 0.036 OR 0.04" for the CH3OH.

I think that sorts it all out. :smile:

Cheers for the rep, I'm new round here lol.

Reply 8

xsbrx

OP

8

Intelligentsia

I think you're just a lil confused by the mark scheme layout, it's really badly spaced out with no tables or anything. The mark scheme is here:

http://www.pastpaperexams.com/downloads/UnifC/2005%20June%20MS.pdf

Notice it says "0.9/2.8 OR 0.321 OR 0.32/0.3" .. if you divide 0.9/2.8 you get 0.321ish so I think they're all just acceptable versions of the same mole fraction for CO.

Then right next to all that there's a "1.8/2.8 OR 0.643OR 0.64/0.6" again acceptable versions which I believe refers to the H2, and finally "0.1/2.8 OR 0.036 OR 0.04" for the CH3OH.

I think that sorts it all out. :smile:

Cheers for the rep, I'm new round here lol.

Ooooh yeh i think i was mixing up the answers for H2 and CH30H :redface:

Thanks for the help :smile:

Reply 9

Intelligentsia

11

np.

A2 OCR Chemistry - unifying concepts question

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