Chemistry

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For the equilibrium below at 400 K, Kc = 7.0.

Br2(g) + Cl2(g) reverse reaction arrow 2 BrCl(g)
If 0.90 mol of Br2 and 0.90 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl?

  • Chemistry -

    Br2 + Cl2 ==> 2BrCl

    Kc = (BrCl)^2/(Br2)(Cl2)
    Do an ICE chart.
    initial:
    Br2 = 0.9 mol/1 L - 0.9M
    Cl2 = 0.9 M
    BrCl = 0

    change:
    BrCl = +x
    Cl2 = -x
    Br2 = -x

    equilibrium:
    Br2 = 0.9-x
    Cl2 = 0.9-x
    BrCl = +x

    Plug the equilibrium concns into the Kc expression and solve for x.

  • Chemistry -

    Br2 + Cl2 ==> 2BrCl

    Kc = (BrCl)^2/(Br2)(Cl2)
    Do an ICE chart.
    initial:
    Br2 = 0.9 mol/1 L = 0.9M
    Cl2 = 0.9 M
    BrCl = 0

    change:
    Br2 = -x
    Cl2 = -x
    BrCl = +2x

    equilibrium:
    Br2 = 0.9-x
    Cl2 = 0.9-x
    BrCl = +2x

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