Question:

Calculate the value of K for this reaction (using units of mol/L for the concentrations).?

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At a particular temperature 8.0 mol NO2 gas is placed in a 1.0-L container. Over time the NO2 decomposes to NO and O2:

2NO2(g)-------->2NO(g) + O2(g)

At equilibrium the concentration of NO(g) was found to be 3.2 mol/L.

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2 ANSWERS


  1. 2 NO2 <------> 2 NO + O2

    initial concentration

    8.0

    change

    -2x... . .. . . .. . . . +2x. . . +x

    2x = 3.2 M= [NO]

    [O2] = 3.2 / 2 = 1.6M

    [NO2] = 8.0 - 3.2 = 4.8M

    K = (3.2)^2 x 1.6 / (4.8)^2 = 0.711


  2. Actually

    2 NO2 <------> 2 NO + O2

    initial concentration

    8.0

    change

    -2x... . .. . . .. . . . +2x. . . +x

    2x = 3.2 M= [NO]

    [O2] = 3.2 / 2 = 1.6M

    [NO2] = 8.0 - 3.2 = 4.8M

    K = (3.2)^2 x 1.6 / (4.8)^2 = 0.711

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