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Pressure and Equilibrium of Decomposition

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data
    SO3 decomposes on heating:
    2SO2(g) equilibrium.png 2SO2(g) + O2(g)
    A sample of pure SO3 was heated to a high temperature T. At equilibrium, the mole ratio of SO2 to
    SO3 was 0.152 and the total pressure was 2.73 atm. If the piston was pushed in to cut the volume in half what is the new pressure?



    2. Relevant equations
    I tried using the ratio to get the pressures on the left. 0.414 and 2.32 respectively.Also tried multiplying 2.73 by 2 but don't think this is the way.



    3. The attempt at a solution

    See above. I got to "an endpoint" with the problem but feel like it's wrong want to learn a general approach. Sorry these are my weakspot!
     
  2. jcsd
  3. Sep 11, 2015 #2

    Bystander

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    Can you write an equilibrium constant for the reaction?
     
  4. Sep 11, 2015 #3
    With or with out the numbers I found that I was using that ratio they gave me wrong.
     
  5. Sep 11, 2015 #4

    Borek

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    You didn't found that, you were told that ;)
     
  6. Sep 11, 2015 #5
    Which means it was found out. ;) Like some of this makes sense some of it doesn't. It's not a typical ICE problem nor do I think it really applies in this instance? But im lost so..
     
  7. Sep 11, 2015 #6

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    "ICE" is what? Intro. Chem. Eng.?
     
  8. Sep 11, 2015 #7
    Initial CHange equilibrium.
     
  9. Sep 11, 2015 #8

    Bystander

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    "ICE? RICE?" Oh, brother --- there's no way to break this to you gently --- there is one thread in the forum on "ICE," --- and, the method seems to be something of a joke/farce/disaster --- looking at "ICE/RICE" in Wiki, I'm inclined to agree. https://www.physicsforums.com/threads/rice-tables.170018/#post-1329914

    Have you had any conventional chemistry training/instruction on reactions and equilibrium constants? What they are and what they can do for you? I'd really rather not get tangled up in an edu-babble religious ritual trying to make a failed instructional approach make sense to you if we can find some way around it.
     
  10. Sep 12, 2015 #9

    Lol yes I have I am actually studying for the MCAT and am doing problems. Have a bachelor in mS in bio. LOOONG time since I took chem and analytical.

    I felt that ICE wasnt useful.

    To come clean. I looked at the solution and had issues with how to arrive at the molarities. I know I- is 0.1M I also "know" that I2 is 0.0013M but why use this at all? Is there a reason we use the saturation molarity in water only AND that of it in KI? this loses me.

    Im pretty sound on the constant.
     
  11. Sep 12, 2015 #10

    epenguin

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    Have threads got mixed up? I know there was another one about iodine precipitating. #1 is about sulphur oxides and oxygen.

    You ask about a general approach. Do you know/remember the law of chemical equilibrium, what an equilibrium constant is? Even if you don't, look it up and find It's nothing very difficult. Then what is the relation relating concentrations of these three substances in this case? Already asked in #2!

    Then mass conservation laws mean that however they vary in the closed vessel when you vary the pressure or volume, neither the number of sulphur atoms, nor the number of oxygen atoms varies. Here you start from SO3 (g) (please correct your formula) and the result is that once you know the concentration, we'll call it C0 of your starting SO3, and the concentration, call it x, of any of the gases at equilibrium, you easily know concentrations of all three, in terms of x and C0. (It will be a bit simpler if you let x be [O2].)

    So rewrite your equilibrium equation in terms of x and C0. Solve it. Then when you halve the volume what do you know that changes how?
     
  12. Sep 13, 2015 #11
    Equilibrium conc of:

    SO3: 2.73-2x
    SO2: 2x
    O2 : x

    2x/2.73-2x=0.152

    x=0.180

    Eventually I get to the point where K= 0.00415

    Am I in the ballpark here?
     
  13. Sep 13, 2015 #12

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    Start with the reaction: 2SO3 = 2SO2 + O2.

    K = [SO2]2[O2]/[SO3]2

    If I give you that much, can you get where you need to go?
     
  14. Sep 13, 2015 #13
    Excuse me but i got that already.
     
  15. Sep 14, 2015 #14

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    Good.
    You have three gases contributing to the total pressure. You have a mole ratio for two of them, and a stoichiometric ratio for a different pair.
     
  16. Sep 14, 2015 #15

    epenguin

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    Not enough working shown. You needed to write explicitly the equation of #12, and then express it In terms of different quantities as I suggested - explicitly. I think your expression lacks the necessary brackets and in the end there should be x2 terms in it, even x3, and where is C0?
     
    Last edited: Sep 14, 2015
  17. Sep 14, 2015 #16
    That's the value of the ratio given in the problem.

    2.73-2x is the equilibrium value (or so I hope) for SO3. 2x is that of SO2. I set this ratio SO2: SO3 equal to 0.152 to get x and subsequently the new equilibrium values. Then I use these to get k by using the equilibrium expression.

    Hopefully this is OK and makes more sense?
     
  18. Sep 14, 2015 #17

    Borek

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    Lasciate ogni speranza.

    2.73 is a sum of three pressures - that of SO3, that of SO2 and that of O2. If 2x is SO2, and 2.73-2x is SO3, sum of their pressures is 2.73-2x+2x=2.73 - so where is the oxygen?

    No matter what question you solve, all the time you are ignoring the fact amounts of different substances is combined because of the stoichiometry of the reaction. You make the same mistake over and over.
     
  19. Sep 14, 2015 #18

    epenguin

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    If you set out your working explicitly and used the formula quoting and using the formula given you in #12 we could help you more and you could help yourself. It is not clear if you use the square of the ratio. This squared ratio equals something invovling three parameters, C, x and K. No idea how you got any of them. You need to get C from info you are given, then x from stoichiometry, only then can you get K from the ratio given, only after that can you get a new value for new C.

    (As far as I can see that will involve solving a nasty equation but all the more we need explicit clarity; we'll talk about that if we get there.)
     
    Last edited: Sep 14, 2015
  20. Sep 14, 2015 #19
    I know how to apply the equilibrium formula but that ratio is throwing me.
     
    Last edited: Sep 14, 2015
  21. Sep 14, 2015 #20

    Borek

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    Assume partial pressures of gases to be pSO2, pSO3 and pO2. Express the ratio and total pressure using these unknowns. What does the reaction equation tell you about the pSO2 and pO2? Are they linked somehow? Can you express this dependency with a third equation?
     
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