# Worth a shot..

Discussion in 'General Science & Technology' started by KpayesMD, Mar 3, 2006.

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1. ### KpayesMDRegistered Member

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Can anyone figure this out? I can't think straight. Its definitely getting late!! aghhhh *Pulls hair out* Hi all

A mixture of 0.2000 mol of Co2, 0.10000 mol of H2, and 0.1600 mol of H20 is placed in a 2 L vessel. The following equilibrium is established at 500k:

CO2 + H2 <---> CO + H2O

a.) calculate the initial partial pressures of CO2, H2, and H2O
b.) at equilibrium H2O = 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO

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-AntonK

5. ### draqonBannedBanned

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freacking ass wants us to do the stoichiometry job for him...no
remember: k=products/reactants, PV=nRT, PiVi=PfVf, n(total)=(V/RT)(P(total))
were n is moles, and P total is made of P1+P2...

Last edited: Mar 3, 2006

7. ### weed_eater_guyIt ain't broke, don't fix it!Registered Senior Member

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i think we'd need to know an equilibrium constant to figure this one out, but then again i was pretty rusty at chem. good luck!

8. ### aspaanRegistered Member

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is that 4th year undergrad chemistry!?

9. ### jon950Registered Member

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Definitely first year. We did that stuff a little while ago.

10. ### weed_eater_guyIt ain't broke, don't fix it!Registered Senior Member

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yeah me too, i just forgot some of the formulas

11. ### jon950Registered Member

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For:
a) You want to do P = nRT/V for all of them and remember to use the atm gas constant R = 0.082058.
b) You might want to set up a chart like so:

.................CO2 + H2 <---> CO + H2O
Initial:.......4.10...2.05........0......3.28
Change:.......-x......-x........+x.......+x
Equilibrium:....?........?.........?......3.51

You know that the H2O partial pressure is 3.51 at equilibrium, thus subtracting the original value should give you the variable x... It should be easy from there.

Last edited: Mar 11, 2006