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Le Chatelier and Equilibrium

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    At certain pressure and temperature in a 2 L closed flask, equilibrium reaction undergoes such as :

    N2 (g) + O2 (g) ↔ 2NO (g)

    Each concentration of the substances in equilibrium is 0.8 M.
    If we add 2 mole of NO2 gas into the flask, then the concentration of N2 gas in the new equilibrium is ...

    A. 1.13 M
    B. 1.80 M
    C. 2.20 M
    D. 2.60 M
    E. 2.80 M

    2. Relevant equations

    Kc = products of right molarity / products of left molarity

    3. The attempt at a solution

    At initial equilibrium
    Kc = (0.8)^2 / (0.8)(0.8) = 1

    Change in mole = 2 mole => Change in molarity = 2 mole / 2 liter = 2 M

    At final equilibrium
    Kc = (1.8)^2 / (x^2)
    1 = (1.8)^2 / (x^2)
    x = 1.8 M

    However, the answer key is A. 1.13 M :frown:
    Please help what I missed
     
  2. jcsd
  3. Apr 11, 2016 #2

    DrClaude

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    Staff: Mentor

    Do you mean NO?
     
  4. Apr 11, 2016 #3

    Borek

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    Staff: Mentor

    Nope, concentration of NO at equilibrium is not 1.8.

    This is best done using an ICE table.
     
  5. Apr 11, 2016 #4
    NO2 is what written in the problem. Maybe, it's a typo from the problem-maker.

    Okay, I'll try

    2nqze3m.png

    Kc = (1.8-2x)^2/(0.8+x)^2
    1 = (1.8-2x)^2/(0.8+x)^2

    Using calculator, x = 0.3333 M
    N2 concentration = 0.8+0.33 = 1.13 M

    Thanks for all your help! ICE table is really a reliable method! :smile:
     
  6. Apr 11, 2016 #5

    Borek

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    Staff: Mentor

    Remember ICE is just an easy and convenient way of keeping track of the stoichiometry, it doesn't add anything new to the way we solve equilibrium problems.
     
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