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Dimer-Monomer Percentage

  1. Feb 7, 2010 #1
    URGENT! Dimer-Monomer Percentage

    1. The problem statement, all variables and given/known data
    In the gas phase acetic acid exists as an equilibrium of monomer and dimer molecules.(The dimer consists of two molecules linked through hydrogen bonds.) The equilibrium constant for the dimer-monomer equilibrium (CH3CO2H)2 <=> 2 CH3CO2H has been determined to be 17 at a particular temperature. Assume that acetic acid is present initially at a concentration of .30mol/L at 25 C and that no dimer is present initially. What percentage of acetic acid is converted to dimer?


    2. Relevant equations



    3. The attempt at a solution
    I did an ice table and found that:
    I 0 .3

    C x -x

    E x .3-x


    Then I went 17 = ((.3-x)^2)/x and got the quadratic .09 - 17.6x + x^2 after doing the
    quadratic formula I found x equal to .005115 or 17.59. I figured that .005115 was the only answer that made sense. I divided that concentration by the concentration give and got the percent to be 1.7% dimer but that is incorrect. Can someone point me in the right direction?
     
    Last edited: Feb 7, 2010
  2. jcsd
  3. Feb 7, 2010 #2
    Anything guys?
     
  4. Feb 8, 2010 #3

    chemisttree

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    Ratio of dimer/monomer is 17, right?

    So, 17 = [x]/[0.3-x]2 and you get 0 = x2 - 17.6x + 0.09 from that?

    That's not what I get.
     
  5. Feb 12, 2010 #4

    epenguin

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    Anyway please realise it is not a pedantry to say that that equilibrium constant cannot be 17. It has to have UNITS or it's meaningless.

    If you quote them they will at least tell you if you have got your equation the right way up.
     
  6. Feb 12, 2010 #5

    chemisttree

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    Yep, I've got mine wrong side up. Sorry.

    George3, this part is wrong...
     
    Last edited: Feb 12, 2010
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