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Equilibrium pressure and Kp

  1. Apr 16, 2010 #1
    A reaction mixture initially contains 2.24 atm Xe and 4.27 atm F2. If the equilibrium pressure of Xe is 0.34 atm, determine Kp for the reaction.

    This question came out of the blue, and all I can think of is that Kp=(PXe)(PF2), but the answer is supposed to be 25.

    I don't get it? Any suggestions? Thanks!
     
  2. jcsd
  3. Apr 16, 2010 #2

    Borek

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    Initial pressure of Xe was 2.24 atm, at equilibrium it was 0.34 atm. What have happened to the rest of Xe?

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    methods
     
  4. Apr 16, 2010 #3
    It must have left the container? I don't know, really :\
     
  5. Apr 16, 2010 #4

    Borek

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    Question asks about reaction equilibrium constant... What reaction?

    --
     
  6. Apr 16, 2010 #5
    Oh, sorry. Xe(g) + 2F2(g) ---> XeF4(g)
     
  7. Apr 16, 2010 #6

    Gokul43201

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    So again, what happened to the rest of the Xe?
     
  8. Apr 16, 2010 #7
    The rest of Xe has gone to the formation of XeF4
     
  9. Apr 16, 2010 #8

    Gokul43201

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    Can you now figure out the partial pressures of all species at equilibrium?

    And take another look at the expression you have for Kp (compare with the equation in post#5).
     
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