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Partial pressures and equilibrium question

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    For the reaction 2HI (g) <-- --> H2 (g) + I2 (g) Kp = .02 at 445 degrees C
    Parital pressures are .05, .01 and .01 respectively. Perform a calc to see which way it goes then figure out the equilibrium partial pressures of the gasses

    3. The attempt at a solution

    Qp = (.01)(.01)/(.05^2) = .0001/.0025 = .04

    Since Qp>Kp reaction goes left

    I .05 .01 .01
    C +2x -x -x
    E .05+2x .01-x .01-x

    .02 = (.01-x)^2/(.05+2x)
    0 = -x^2 + .06x + .0009

    Quadratic equation yields -.0264 and -.0336... neither of which make sense as the final answers are given and they should be:
    .0546, .0077 and .0077 atm
    Which makes x = .0023...

    That was my third time running through this problem and each time I got a different answer... no idea how I screwed this one up. it's only the first problem on an assignment due tomorrow... any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 13, 2009 #2

    Borek

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    Staff: Mentor

    This is OK.

    This is not. Just check your math.
     
  4. Mar 13, 2009 #3
    You're right.. it should be 0= x^2 -.06x -.0009
    Which has roots of 0.072 and -0.012 ...

    But .05 + 2(.072) = 0.194 yet it should equal .0546 atm :(
     
  5. Mar 13, 2009 #4

    Borek

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    Staff: Mentor

    while (x != 0.0023) check_your_math;
     
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