Worth a shot..

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

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  1. KpayesMD Registered 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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  3. AntonK Technomage Registered Senior Member

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    What have you go so far? Show us your thought process and we can probably help you out in finding your mistake.

    -AntonK
     
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  5. draqon Banned Banned

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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
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  7. weed_eater_guy It 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. aspaan Registered Member

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    is that 4th year undergrad chemistry!?
     
  9. jon950 Registered Member

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    Definitely first year. We did that stuff a little while ago.
     
  10. weed_eater_guy It ain't broke, don't fix it! Registered Senior Member

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    yeah me too, i just forgot some of the formulas

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  11. jon950 Registered 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
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