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Consider the following equation:

N2O4(g)+ 2NO2(g) Kc= 5.8 x 10^-3

If the initial concentration of N2O4(g)= 0.040 M and the initial concentration of NO2(g) is 0 M, what is the equilibrium of concentration of N2O4(g)?

  • chemistry -

    Did you make a typo, typing + instead of -->

  • chemistry -

    Yes, I meant to right -->...I apologize for typing the question incorrectly.

  • chemistry -

    If so, then
    N2O4 ==> 2NO2

    Set up ICE chart.
    N2O4 = 0.040
    NO2 = O

    NO2 = 2x
    N2O4 = -x

    NO2 = 2x
    N2O4 = 0.040-x

    Set up Keq expression of
    Keq = (NO2)^2/(N2O4), plug in the equilibrium values from above and solve for x. Finally, 0.040-x will be the N2O4 at equilibrium.

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