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Partial pressures

  1. Jun 15, 2005 #1
    Just a few questions I'm having trouble with.

    1.Ammonia is formed from nitrogen and hydrogen according to the following equlibrium reaction:

    N2(g)+3H2(g)-->2NH3(g)

    The plant operates close to 700K, at which Kp is 1.00*10^-4 atm-2 and employs the stoiciometric ratio 1:3 of N2:H2. At equilibrium, the partial pressure of NH3 is 50 atm. Calculate the partial pressures of the each reactant.

    2.Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 250c, Kc=1.58*10^-8 M^3 for the following equilibrium:

    NH2COONH4 (s) <---> 2NH3 (g) + CO2 (g)

    If 7.81g of NH2COONH4 is introduced into a 0.500L evacuated container, what is the total partial pressure inside the container at equilibrium at 250c?

    Thanks.
     
    Last edited: Jun 15, 2005
  2. jcsd
  3. Jun 15, 2005 #2
    1. Kp = (Partial Pressure NH3)^2 / ((PP N2) * (PP H2)^3)

    2. Kc = [NH3]^2 * [CO2]

    That should get you started.
     
  4. Jun 15, 2005 #3
    For 1., I got to [N2]*[H2]^3=2.5*10^7 then I just got lost. I know the ratio is 1:3, but that cubed hydrogen has really put me off.
     
  5. Jun 15, 2005 #4
    for ever x atm of N2 there are 3x atm of H2. So (x)(3x)^3=2.5*10^7
     
  6. Jun 15, 2005 #5
    So you get 3x^4=2.5*10^7
    x= 53.73?

    If that's what you meant, apparently it's the wrong answer for N2.
     
  7. Jun 15, 2005 #6
    it would be 27x^4=2.5*10^7, you forgot to cube the 3.
     
  8. Jun 15, 2005 #7
    lol...thanks for that.
     
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