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Finding the composition of a mixture at equlibrium

  1. Oct 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Initially a mixture contains 0.723 mol each of N2 and O2 in an 7.20 L vessel. Find the composition of the mixture when equilibrium is reached at 3900°C. For the reaction below, the Kc = 0.0123 at 3900°C.
    N2(g) + O2(g) 2 NO(g)



    2. Relevant equations
    I.C.E Tables (not an equation but relevant) and the quadtratic formula

    x=-b +or- sq root(b^2 -4ac) / 2a


    3. The attempt at a solution

    I did my I.C.E Table
    Conc.(M) N2 + O2 --> 2 NO
    Start 0.6025 0.6025 0
    Change -x -x 2x
    Eq. 0.6025-x 0.6025-x 2x

    Then I plugged that into my Kc formula
    Kc=[products]/[reactants]
    0.0123=[2x]^2/[0.6025-x][0.6025-x]

    I then rearranged that to get a quadratic equation to solve for x (this is where I think I made a mistake)
    -324x^2 -1.205x +0.3630=0
    I plugged it into the quadratic formula and
    got 0.03166 for x, which i plugged into the equations from the EQ. line in my I.C.E table, however when i plugged those values into the Kc formula, i didn't come out with the Kc value I was given (0.0123)

    Can anyone help?
     
  2. jcsd
  3. Oct 15, 2008 #2
    nevermind, I was (stupidly) using a value from another question, but my method was right.
     
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