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college chem

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N2O4 --->2NO2
a 1 liter flask is charged with .4 mols of N2O4. at equilibrium at 373 k, .0055 mols of N2O4 remain. what is the Kc for this reaction?

please explain to me how to get the rate law for this and solving

  • college chem -

    N2O4 ==> 2NO2
    Set up an ICE chart.
    initial:
    N2O4 = 0.4
    NO2 = 0

    change:
    N2O4 = -x
    NO2 = +2x

    equilibrium: (Just add initial to change to get equilibrium.)
    N2O4 = 0.4 - x = 0.4-x
    NO2 = 0 + 2x = 2x

    The problem gives us that the equilibrium for N2O4 = 0.0055.
    0.4- x = 0.0055
    x = 0.4-0.0055 = ??

    Now go to NO2. The change must be 2x and the total must be
    NO2 = 2x

    Now substitute those numbers into Kc expression.
    Kc = (NO2)2/(N2O4)

  • college chem -

    1.43e2

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