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Equilibrium constant

  1. Jul 19, 2005 #1
    N2(g) + 2O(g) <--> 2NO2(g)

    At a certain temperature, the follwoing are the equilibrium concentrations for the above reaction:

    [N2] = 8.0 mol/L
    [O2] = 2.0 mol/L
    [NO2] = 4.0 mol/L

    Determine the equilibrium constant.

    My answer..

    Ke = [NO2]^2 / [N2][O2]^2

    Ke = [4.0]^2 / [8.0][2.0]^2

    Ke = 0.5

  2. jcsd
  3. Jul 19, 2005 #2
    I believe your calculations are correct. Latex is friendly though =)
  4. Jul 19, 2005 #3
    Another question having to due with equilibrium...

    The Claus Process is an industrial process used to remove toxic hydrogen sulfide gas during the processing of crude oil. The chemicle equation for this process is:

    2 H2S(g) + SO2(g) <--> 3 H2O(g) + Heat

    Use Le Chatliers Principle to describe why the following changes favour the removal of H2S.

    a) Removing sulfur as soon as it forms
    b) Cooling the reaction chamber

    I think it has something to due with the temperature..and either the products or reactants being favored...not sure tho.
  5. Jul 19, 2005 #4
    I'm not sure what you mean by a); since there is no sulfur on the right side of the equation.

    I don't want to give you the answer, but for b), look at the equation. Is it exothermic or endothermic? What would it mean? Try to think of temperature as an actual "molecular species", or in this case, a product. Try to apply your knowledge dealing with the removal or addition of a product in this case to predict which way the system would shift. I hope that leads you to the right answer.
  6. Jul 19, 2005 #5
    oh sorry got the equation wrong.

    2 H2S(g) + SO2(g) <--> 3 S(s) + 2 H2O(g) + Heat
  7. Jul 19, 2005 #6
    yes well, I was too lazy to actually fix your equation haha :P
    do you understand or do you have further questions?
  8. Jul 20, 2005 #7


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    The best way to explain these changes is to offer an explanation in relevance to relative rate constants.
  9. Jul 21, 2005 #8

    remember that heat can be considered as one of the products in this reaction. so cooling the reaction chamber means that you are absorbing heat from the right side of the reaction, and so the reaction will shift to the right by L'Chatlier's principle in order to restore the heat in equilibrium.
  10. Jul 21, 2005 #9


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    I don't particularly favor quetzal's and atremis' approach of "treating heat as a reactant/product" as a means of first understanding Le Chatelier - this approach provides no physical insight. And I strongly recommend GCT's approach of deriving the effects on the forward and reverse reaction rates. Once you have completely understood the reasons though, you may use the first method, as a quick 'n' easy tool.
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