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I need help with this chem question

  1. Apr 7, 2004 #1
    Gaseous oxides of nitrogen rarely occur as pure substances, but as equilibrium mixtures of them. One of these equilibrium mixtures is nitrogen dioxide and dinitrogen tetroxide. If the density of such a mixture is 2.3 g/l at 74 degrees Celsius and 1.3 atm, calculate the partial pressures of the gasses and Kp for the dissociation of dinitogen tetroxide to nitrogen dioxide.
     
  2. jcsd
  3. Apr 8, 2004 #2
    Let the volume be 1L Then total mass of the system will be

    [tex]n_1*M + 2n_2*M = 2.3 g [/tex]
    Where
    n1 : no of mol of NO2
    n1 : no of mol of N2O4
    M : Molecular mass of NO2

    Also PV=(n1+n2)RT

    So u have two unknown and two equations u can find n1 & n2 from it And Hence Kp or partial pressure
     
  4. Apr 8, 2004 #3

    GCT

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    Using PV=nRT, since you know P and T, you can solve for the solution in the form of moles per liter. For the rest of the problem we will assume a volume of one liter (it does not matter either way since it is only a matter of proportion). Since you have the grams pertaining to this situation you essentially have the "average molar mass of this situation." That is 2.3 grams/the total number of moles (which you are to figure out).


    [molar mass of the nitrogen compound (grams/moles) X N1 (number of nitrogen compound)] + [molar mass of the dinitrogen compound (grams/moles) X N2 (number of the dinitrogen compound)]

    all of this divided by 2

    Solve for the proportion N1/N2, that is proportional to the moles.

    Notice however, that the Kp equation for the reaction pertaining to the addition of the nitrogen to form the dinitrogen should have an exponent of 2 on the pressure of the nitrogen compound; we know that according to the stochiometric equation that there are two moles of the nitrogen compound for every dinitrogen compound.:

    Kp = [partial pressure of dinitrogen compound p]/[partial pressure of nitrogen compound]^2

    Note also that you will need to convert moles into pressure because of the presence of an exponent (with the assumption of 1 L as the volume, volume is insignificant information in this problem).

    --------------
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    Last edited: Apr 8, 2004
  5. Apr 8, 2004 #4
    There's nothing quite like a lungful of NO2 in the morning. It really gives your day a kick.

    :wink:
     
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