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Equilibrium of Reaction of Gases

  1. Dec 14, 2013 #1

    Qube

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    Gold Member

    1. The problem statement, all variables and given/known data

    Nitrous oxide reacts with ozone to form dinitrogen pentoxide. There are four moles of each reactant and the volume of the container in which this reaction takes place is 2 L. T is constant. The equilibrium constant for this reaction is 1. How many moles of product are present when the reaction achieves equilibrium?

    2. Relevant equations


    Balanced equation: [itex]3N_{2}[/itex][itex]O[/itex] + [itex]4O_{3}[/itex] <--> [itex]3N_{2}[/itex][itex]O_{5}[/itex]

    3. The attempt at a solution

    Wrong method:

    [itex]K_{c}[/itex][itex]=[/itex][itex]\frac{[N2O5]^{3}}{[N2O]^{3}[O3]^{4}}[/itex]

    All the components are gaseous, so they can go into the equation. First, however, we have to convert moles to concentrations. So we divide moles by volume (2 L). Also the equilibrium constant is equal to 1. Then we can plug in everything to the equation:

    [itex]1[/itex][itex]=[/itex][itex]\frac{[N2O5]^{3}}{[2]^{3}[2]^{4}}[/itex]

    From here we can solve for the concentration of dinitrogen pentoxide. [itex]1[/itex] [itex]=[/itex][itex]\frac{[N2O5]^{3}}{128}[/itex]

    We get the concentration of dinitrogen pentoxide as the third root of 128, and we must multiply this value by 2 liters to get the number of moles of dinitrogen pentoxide since by dimensional analysis the liters cancel out (moles/liter * liter = moles).

    New and hopefully correct method!

    Let's surmise that x moles of product (dinitrogen pentoxide) were made.

    From the balanced equation we can see that the production of x moles of product consumes x moles of nitrous oxide and (4/3)x moles of ozone.

    Now, we have the equilibrium, rather than the initial, number of moles.

    Equilibrium number of moles:

    Nitrous oxide: 4 - x moles
    Ozone: 4 - (4/3)x moles
    Dinitrogen pentoxide: x moles (as I surmised).

    From here we can plug stuff into our equilibrium constant equation! Volume is constant so we'll just plug in the number of moles.

    [itex]\frac{x^{3}}{(4-x)^{3}(4-4x/3)^{4}}[/itex] [itex]=1[/itex]


    Questions:

    1) I first thought nitrous oxide referred to nitrogen monoxide (NO). I was wrong. Is nitrous oxide the scientific name or just a common name? I'm suspecting it's just a common name because I can't think of a rule for "-ous" compounds outside acids (e.g. sulfuorous acid).

    2) I'm still learning equilibrium here and I'm wondering if my work is correct. I'm pretty sure it's correct after consulting a textbook (which actually refers to N2O as dinitrogen monoxide). Is my answer correct? More importantly, is my process and logic correct?

    3) Are the initial moles of reactant equal to the equilibrium moles of reactant? Is this relevant? I'm suspecting what I did above is wrong because I didn't use the equilibrium moles but instead the initial moles (I just re-read the textbook).
     
    Last edited: Dec 14, 2013
  2. jcsd
  3. Dec 14, 2013 #2

    Borek

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    Staff: Mentor

    That would be correct if the concentrations of reactants after the reaction were 2. They are not - some of the reactants were consumed.

    Note: you are doing strange things to Latex. No need to treat each part separately, combine everything together and use one set of tex tags:

    [tеx]1=\frac{[N_2O_5]^{3}}{[2]^{3}[2]^{4}}[/tеx]

    to get

    [tex]1=\frac{[N_2O_5]^{3}}{[2]^{3}[2]^{4}}[/tex]

    itex tags are for inline formulas (like [itex]e=\sum_0^\infty\frac 1 {n!}[/itex]).

    You can always right-click a formula and select "show math as tex commands" to see how it was entered. Or you can quote the message to see the [itex]\LaTeX[/itex] code.
     
  4. Dec 14, 2013 #3

    Qube

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    Gold Member

    Gotcha! I fixed it in the problem above while preserving the incorrect work so other people can hopefully see what I did wrong and avoid the mistake of assuming the initial concentrations to be the equilibrium concentrations!
     
  5. Dec 14, 2013 #4

    Borek

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    Staff: Mentor

    This looks much better, but you can't ignore the volume. It won't cancel out.

    (BTW, your latex is still slightly messed up - why two separate expressions, each in its own tags, instead of a single expression? Compare our LaTeX tutorial.)
     
  6. Dec 14, 2013 #5

    Qube

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    Gold Member

    You're right on both points; I'll make sure to look at more LaTeX and learn it. I really appreciate the power of LaTex, and I've been lazy by using two separate LaTex expressions instead of combining them.

    Also, yes, volume won't cancel out here; volume is cubed on top and ... raised to the seven power in the denominator. If the degrees of the numerator and denominator were the same, however, then volume would cancel out, I think, correct?

    [itex]\frac{(x/2)^{3}}{[(4-x)/2]^{3}[(4-4x/3)/2]^{4}} = 1 [/itex]

    (Only used one pair of latex tags above! I'm getting it! :) )
     
    Last edited: Dec 14, 2013
  7. Dec 15, 2013 #6

    Borek

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    Staff: Mentor

    Yes.

    Better, but you are still using itex tax for an expression that is put on its own line. This is perhaps not a serious error, but itex generates smaller formulas ([itex]\frac{(\frac x 2)^{3}}{[\frac{4-x} 2]^{3}[\frac {4-\frac{4x}{3}} 2]^{4}} = 1 [/itex]), as it is designed to fit in text. tex tags on the other hand generate images that are larger and easier to read:

    [tex]\frac{(\frac x 2)^{3}}{[\frac{4-x} 2]^{3}[\frac {4-\frac{4x}{3}} 2]^{4}} = 1 [/tex]
     
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