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Equilibrium Partial Pressure Kp/Kc Question

  1. Dec 9, 2012 #1
    1. The problem statement, all variables and given/known data
    At 100 o C Kc=.078 for the reaction SO2Cl2<-->SO2 + Cl2. In an equilibrium mixture the [SO2CL2]=.0108 M and [SO2]=.052 M.

    What is the partial pressure of Cl2 in the eq. mixture?

    2. Relevant equations
    Kp=Kc(RT)[itex]\Delta[/itex]n
    P=RT/V

    3. The attempt at a solution
    I solve for Kp but I don't know how to find the partial pressure since neither the total pressure nor the volume is given.
     
  2. jcsd
  3. Dec 10, 2012 #2

    Borek

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    Staff: Mentor

    Pressure can't be a function of the volume, which suggests in the end it will cancel out. Would it help to assume 1L total volume?
     
  4. Dec 10, 2012 #3
    I tried that, here's the work:
    P=RT=(.0821)(373)=30.6233 atm

    Moles is M*L so all of the moles equal the concentration. Total number of moles is .0108+.052+.0162 (moles of Cl2)=.079 total moles. So the pCl2= .0162/.079*30.6233=6.27 atm.

    The book says 5 atm so where am I going wrong?
     
  5. Dec 10, 2012 #4

    Borek

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    Staff: Mentor

    List assumptions used to calculate this number.

    But I don't see how the final answer can be 5 atm. Are you sure you have not missed a decimal point?
     
  6. Dec 10, 2012 #5
    I used the ideal gas law, PV=nRT rearranged to solve for P. (I did this in another problem in the same set to get overall pressure). I just set the volume as 1 L.

    I checked all the numbers and the answer in the book and everything seems correct. :S
     
  7. Dec 10, 2012 #6

    Borek

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    Staff: Mentor

    P=nRT/V, what have you used for n?
     
  8. Dec 10, 2012 #7
    Thanks for the help, I did indeed misplace a decimal xD.
     
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