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For the reaction I2(g) + Br2(g) 2 IBr(g), Kc = 280. at 150.°C. Suppose that 0.570 mol IBr in a 2.50-L flask is allowed to reach equilibrium at 150.°C. What are the equilibrium concentrations of IBr, I2, and Br2?

  • chem -

    M = mols/L = 0.570/2.50L = about 0.228M

    ........I2 + Br2 ==> 2IBr
    I.......0....0......0.228
    C.......x....x.....0.228-2x
    E.......x....x.....0.228-2x

    Substitute the E line into Kc expression and solve.

  • chem -

    do u have to use the quadratic to find x?

  • chem -

    what do i do after i get

    .051984+4x^2 - .912/(x^2)

  • chem -

    .051984+4x^2 - .912x=280(x^2)

  • chem -

    then subtract the 4x^2? so

    .051984-.912x=276x^2

    now what?

  • chem -

    To here is ok.
    .051984+4x^2 - .912x=280(x^2)

    Just rearrange to
    0.05918 + 4x^2 - 0.912x = 280x^2
    Then 280x^2-4x^2 + 0.912x - 0.05198 = 0 and
    276x^2+0.912x-0.05198 = 0 and use the quadratic formula. I have x = 0.0122 or close to that.

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