Chem 2

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The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4?
N2O4(g) 2 NO2(g)

  • Chem 2 -

    First Kc must determined.
    .........N2O4 ==> 2NO2
    Kc = (NO2)^2/(N2O4).
    Substitute the equilibrium numbers and solve for Kc. Then set up an ICE cart for the remainder of the problem.

    ............NO2 ==> 2NO2
    initial....0.55......0.25
    add....................0.1
    change......+x........-2x
    new equil.0.55+x.....0.26-2x

    Kc = (NO2^2/(N2O4)
    Substitute the new equilibrium line and solve for x, then evaluate NO2 and N2O4.

  • Chem 2 -

    I was hoping DrBob222 could have been more vague and unclear in his response.

  • Chem 2 -

    0.25+0.1 is not 0.26, but 0.35

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