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ICE chart, Equilibrium question

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the value of Kc for the reaction H2(g) + F2(g) <=> 2HF(g) given that when 0.010 mol H2 and 0.050 mol F2 are added to a 1.00 L container, 0.0125 mol HF is present at equilibrium.



    2. Relevant equations



    3. The attempt at a solution

    I'm confused about how to set up the ICE chart in this scenario. We have

    H2(g) + F2(g) <=> 2HF(g)

    .01....... (.05) ....... 0
    -x ......-x ........... 2x
    .01-x.....(.05-x) ..... (.0125)

    2x = .0125, x = .00625,

    [.0125] ^2 / [.01 - .00625] [ .05-.00625] = 9.3 x 10 ^-6

    Which is the incorrect answer, 9.51 ^-1 is right but I'm not sure how to get there. Any help would be appreciated.
     
    Last edited: Oct 30, 2012
  2. jcsd
  3. Oct 30, 2012 #2

    Borek

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    Staff: Mentor

    0.1 or 0.01?
     
  4. Oct 30, 2012 #3
    Oops, I'm sorry it's .01, .05, Any ideas :D?
     
  5. Oct 30, 2012 #4

    Borek

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    Staff: Mentor

    Check your math. Formula you derived looks OK to me and its value is not 9.3x10-6. Or at least that's not what I got just by keying it into a calculator.
     
  6. Oct 30, 2012 #5
    Ugh...That feeling when you stress out over a question just to type in the right answer 45 minutes later. Thanks you sir!
     
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