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Atomic Ratio and Ratio of Mole Fractions

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Hi folks! :smile:

    This is just a conceptual question that has arisen during some reading. At one point the author states that for the reaction:

    [itex]CO_2 \leftrightharpoons X_{CO}CO + X_{CO_2}CO_2 + X_{O_2}O_2[/itex]

    where X is the mole fraction of each component at equilibrium, that we can relate the ratio of oxygen atoms:carbon atoms to the mile fractions by the following.

    [tex]\frac{\text{No. carbon atoms}}{\text{No. oxygen atoms}}=\frac{1}{2}=\frac{X_{CO}+X_{CO_2}}{X_{CO}+X_{CO_2}+X_{O_2}}[/tex]


    Now I can see that what they have essentially written is

    [tex]\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}
    =
    \frac{\text{mole fractions of everything with carbon in it}}{\Sum\text{mole fractions of everything with oxygen in it}}[/tex]

    Now intuitively this makes sense to me and I can dig it! :smile: BUT, I would like to make the math work to prove it to myself, but I cannot seem to figure it out :confused:

    3. The attempt at a solution

    This is what I did to try to "prove" it. Since Xi = Ni / NT where NT is the total number of moles in the mixture at equilibrium, I can write

    [tex]
    \frac{X_{CO}+X_{CO_2}}{X_{CO}+X_{CO_2}+X_{O_2}} =
    \frac{N_{CO}/N_T+N_{CO_2}/N_T}{N_{CO}/N_T+N_{CO_2}/N_T+N_{O_2}/N_T} =
    \frac{N_{CO}+N_{CO_2}}{N_{CO}+N_{CO_2}+N_{O_2}}
    [/tex]


    I am just not sure where to go from here? I thought about writing each Ni as something like: NCO = (NC + NO), but did not get too far.

    Any thoughts?
     
    Last edited: Jun 29, 2010
  2. jcsd
  3. Jun 29, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    And not

    [tex]\frac{\text{No. oxygen atoms}}{\text{No. carbon atoms}}=\frac{2}{1}[/tex]?

    After all that's the ratio in CO2 which is present initially.

    I will try to get back to the problem, have to do something else at the moment. Seems like your approach should lead to the correct conclusion.
     
    Last edited by a moderator: Aug 13, 2013
  4. Jun 29, 2010 #3

    Borek

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    Staff: Mentor

    How many atoms of everything in 1 mole of CO2?

    And you probably need to use stoichiometry of the reaction:

    2CO2 <-> 2CO + O2
     
    Last edited by a moderator: Aug 13, 2013
  5. Jun 29, 2010 #4
    Yes Borek; my mistake. I have edited to reflect the correct problem.

    So if me let ai be the number of atoms in the ith species and NAis Avagadros number, we have for 1 mole of CO2:

    [tex]a_{\text{1mol}_{CO_2}} = \text{1mol}_C*N_A\frac{\text{atoms}} + 2*\text{1mol}_O*N_A\frac{\text{atoms}}{\text{mol}} = 3N_A \, \text{atoms}[/tex]

    I am can presumably do this for all remaining species and find that the ration is indeed 1/2. Now that I think about it, if I really wanted to *prove* it in general I should have let the subscripts be dummy variables as well as the species names...but that is for another time. When I get the time I will do this on paper since I am still not sure where I would use the stoichiometry of the rxn.
     
    Last edited: Jun 29, 2010
  6. Jun 29, 2010 #5

    Borek

    User Avatar

    Staff: Mentor

    For example - to combine amounts of CO and O2 present. If you assume initially there was CO2 only, you can also calculate XCO2 knowing XCO (or XO2)
     
    Last edited by a moderator: Aug 13, 2013
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