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Kp values and partial pressure

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    • Thread Starter

    Hello everyone,

    I am stuck on the last question of the worksheet attached. I keep on getting a value of 22.5 KPa, but the answer is 80 KPa.

    I got my answer algebraically - at the start the moles of hydrogen and iodine are X, and at eqm there's 1.5 moles of iodine. Therefore, the change in iodine is -(X-1.5), which is the same for the hydrogen) This also implies HI has increased by 2(x-1.5), giving that many moles at eqm, along with X-(X-1.5) moles of hydrogen.

    Are my terms correct - because then I put it in the Kp formula.

    Can anyone help?
    Attached Files
  1. File Type: doc Topic 11 Exercise 4 - Kp (1).doc (30.5 KB, 11 views)

    Nice Q, made me think for a moment.

    Your mathematics are all fine. The thing you're missing is the information given to you by the Kp value.

    Kp = pp(HI)2 / (pp(H2) x pp(I2))

    = (Ptot x n(HI)/ntot)2 / ((Ptot x n(H2)/ntot) x (Ptot x n(H2)/ntot))

    The 2x Ptot and ntot cancel out top and bottom, leaving you with

    = n(HI)2 / n(H2) x n(I2)

    At equilibrium, we know what n(H2) and n(I2) are, so can work out n(HI) and the rest falls into place.
    • Thread Starter

    Thanks for the help.
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