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March 24, 2017

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Starting with 0.3500mol Co(g) and 0.05500mol COCL2(g) in a 3.050 L flask at 668K, how many moles of Cl2 are present at equilibrium?

Co(g) = Cl2(g) <-> COCL2(g) Kc = 1.2x10^3

  • Chemistry - ,

    Do you mean Co or CO. Surely you meant CO. You must understand that CO is carbon monoxide and Co is cobalt. Also, I assume the = sign should be a + sign.

    Convert to concns in molarity.
    (CO) = 0.3500/3.050 = ??
    (COCl2) = 0.05500/3.050 = yy
    .............CO + Cl2 ==> COCl2
    initial..0...??....0.......yy
    change.....+x......+x......-x
    equil....0.??+x...x........yy-x

    Substitute the ICE chart I've prepared and substitute into Kc expression, solve for x. That will give you (Cl2) in moles/L, then multiply by L to convert to moles.

  • Chemistry - ,

    When I converted everything I got

    1.2x10^3 = (0.02 + x) / (0.12 - x)(x)
    = (0.02 + x) / (0.12x - x^2)

    1.2x10^3(0.12x - x^2) = 0.02 + x
    144x - 1.2x10^3x^2 = 0.02 + x

    Then I brought x to the other side and I got

    145x - 1.2x10^3x^2 = 0.02

    Then I get confused with the x's and don't know how to continue solving.

    Thanks so much for the previous help though

  • Chemistry - ,

    You're rounding far too early. I may make an error by carrying too many places when doing these quadratic equations but I like to carry more than I'm allowed and round at the end.
    0.3500/3.050 = 0.11475M = (CO)
    0.05500/3.050 = 0.01803M = (COCl2)

    First, note that K = 1.29E3 or 1290 and not 1200.
    ............CO + Cl2 ==> COCl2
    initial.0.11475...0.......0.01803
    change.....+x......+x......-x
    equil..0.11475+x...x.......0.01803-x

    Kc = 1290 = (COCl2)/(CO)(Cl2)
    1290 = (0.01803-x)/(0.11475+x)(x)
    0.01803-x = 1290(0.11475+x)(x)
    0.01803-x = 1290x^2 + 148.03x
    1290x^2 +149.03x -0.01803 = 0
    x = 0.0001208 M = (Cl2)
    Then convert to moles and round to the appropriate number of s.f.

  • Chemistry - ,

    OK thanks so much, I appreciate it.

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