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Equality reaction

  1. Oct 26, 2004 #1
    Hi I have simplified my last question in the hope of that someone, can help med answer it.

    The following equality reaction:

    [itex]2 NO_{2} \leftrightharpoons N_{2} O_{4}[/itex]

    Where the total number of mole's of the two substances in equilibrium is
    2,6 x 10^2 mol.

    The reaction takes place in a container with a volume of 0,50 Liters.

    This means that the total concentration of the two substances are
    0,052 mol/liter.

    I need to calculate the concentration of both [itex] [NO_2] [/itex] and
    [itex] [N_{2} O_{4}] [/itex]

    2 NO_2 <----> N_2 O_4
    int | 0,052 + x | 0,052 + x
    change | x | x
    equi | 0,052 + x - (2x) | 0,052 + x - (2x)
  2. jcsd
  3. Oct 26, 2004 #2


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    Gold Member

    Okay, let me say that the total moles are 3x, and [itex]NO_2[/itex] is 2x while [itex]N_2O_4[/itex] is x, which is equal to 8,67.10-3 moles.

    There you can do the calculation to find the corresponding concentrations of the gases in 0.50 liters of a container.
    Last edited: Oct 26, 2004
  4. Oct 26, 2004 #3

    Thank You for Your answer.

    My assumption is accurate then ?

    [itex]K_{c} = \frac{[N_2 O_4]}{[NO_2]^2} = \frac{(0.052 + x - (2x))}{(0.052 +x - (2x))^2} = ???[/itex]


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