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Find the equilibrium concentration

  1. Apr 19, 2009 #1
    1. The problem statement, all variables and given/known data
    At 300 K, kp = 1.5 x 1018 for the reaction 3NO --> N2O + NO2

    If 0.030 mol of NO were placed in a 1.00 L vessel and and equilibrium was established, find the equilibrium concentrations of NO, N2O and NO2.


    2. Relevant equations



    3. The attempt at a solution
    First, I converted kp to kc using the formula kc = kp (RT)(change in mols of gas)

    I get kc = (1.5 x 10^18)((0.0821)(300K))^(-1) = 6.08 x 10^16

    Then, using an ICE chart, I found the mass action expression is equal to:

    x2 / (0.030 - 3x)3

    kc should be equal to the mass action expression.

    But I can't figure out how to solve for x. There don't seem to be any simplifications I can make.
     
  2. jcsd
  3. Apr 20, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    Looks to me like equilibrium is shifted faaaar to the right. What does it tell you about amount of products?
     
  4. Apr 20, 2009 #3
    That the amount of products is very big compared to the amount of reactants. The reaction goes pretty much to completion.

    So, I shouldn't even do an ICE chart, I should just use stoich?

    At what point is kc big enough that I can assume the reaction goes to completion?
     
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