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Homework Help: Gas problem

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data
    1. A flask initially contains only NOBr gas. Once heated to a temperature T, 34.0 % of the original gas decomposes via the following equation to give a total pressure of 0.25 atm at equilibrium:

    2NOBr(g) [tex]\Leftrightarrow[/tex] 2NO(g) + Br2(g)

    Given a KC of 3.94 x 10-4 I need to find the temperature. I've already gotten the KP and initial concentration as being 9.66x10-6 and 0.213675 atm respectively. I didn't see any equations for this in my notes...so how can one go about this business?
     
    Last edited: Oct 3, 2009
  2. jcsd
  3. Oct 3, 2009 #2

    Borek

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    You have to calculate number of moles of all gases present. Seems to me like 34% and Kp should be enough for that, just remember that pressure is directly proportional to n.

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    methods
     
  4. Oct 3, 2009 #3
    Well I can assume that it's a total of 1L for volume. I do have the partial pressures for all three gasses, so I can probably just work from those. I'll post again once I try this out.
     
  5. Oct 3, 2009 #4
    Ok, I got nowhere.

    I used PV/RT = n to try and get the mols, which failed for obvious reasons.

    So I tried to use the Kc to solve for it:

    3.94X10-4 = [Br2][NO]2 / [NOBr]2

    Can I just throw X in for each of the concentrations and solve? If so then x is equal to the Kc...that seems wrong.
     
  6. Oct 3, 2009 #5

    Borek

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    If there were initially n moles of NOBr, how many moles will be present after 34% reacted? How many moles of both products? Can you build Kp from these values?

    Hm, now I started to wonder if you have enough information for that approach, but it is 2 am here and my thinking is a little bit blurred. CU tomorrow.

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  7. Oct 4, 2009 #6
    From other parts of the question, I calculated the initial pressure of NOBr, as well as the final equlibrium pressures of all the gasses, as well as the Kp.
     
  8. Oct 4, 2009 #7

    Borek

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    What 'other parts'?

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  9. Oct 4, 2009 #8
    Oh, there was an A and B part to that question. This final one I'm stuck on is Part C
     
  10. Oct 4, 2009 #9
    The initial pressure of NOBr was .214 atm, the Kp I found was 9.66x10-6.
     
  11. Oct 4, 2009 #10

    Borek

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    Was there any other data given in these 'other' parts?

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  12. Oct 4, 2009 #11
    1. A flask initially contains only NOBr gas. Once heated to a temperature T, 34.0 % of the original gas decomposes via the following equation to give a total pressure of 0.25 atm at equilibrium:

    2NOBr(g) [tex]\leftrightarrow[/tex] 2NO(g) + Br2(g)

    a) Determine the original pressure of NOBr in the flask.
    b) What is the value of Kp at this temperature T?
    c) If the value of Kc at this temperature T is 3.94 x 10-4, determine the temperature T.

    That is the whole question.
     
  13. Oct 4, 2009 #12

    Borek

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  14. Oct 4, 2009 #13
    That formula is on my formula sheet...ah well. So if [tex]\Delta[/tex]n is 1 then the answer is a decimal...

    9.66x10-6 = 3.94x10-4(0.08206 T)1

    T=.298778534 oK

    Is that right?
     
  15. Oct 4, 2009 #14

    Borek

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    Can't be - 0.3K is so close to absolute zero you can be sure all substances will be liquid, not gaseous. Even helium.

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  16. Oct 4, 2009 #15
    So what went wrong? I see nothing else that failed. My calculations for the Kp and initial pressure look right.

    Can you check my answers please?
     
  17. Oct 4, 2009 #16

    Borek

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    Pressure looks OK, but Kp seems way too small to me.

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    methods
     
  18. Oct 4, 2009 #17
    Ugh...lemme re-run it.

    I ended up with 9.65x10-6 again...
     
  19. Oct 4, 2009 #18

    Borek

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    Show.

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  20. Oct 4, 2009 #19
    Kp = [Br2][NO]2 / [NOBr]2

    Kp = (0.003638)(0.007275)2 / (0.14124)2

    Kp = 9.66 x10-6
     
  21. Oct 4, 2009 #20

    Borek

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    No, where did you get these values from.

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    methods
     
  22. Oct 4, 2009 #21
    The preceding part using an ICE table and some math.

    Hopefully this attachment can get approved soon...

    It's part 1a.
     

    Attached Files:

  23. Oct 4, 2009 #22

    Borek

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    Unfortunately it is 1 am here and tomorrow is Monday, so I have to go to sleep.

    ICE table sounds OK, but values are wrong. From the data given in question [NO]/[NOBr] = 0.34/(1-0.34) - they should be comparable when it comes to value, you have them differ almost 100 times.

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  24. Oct 4, 2009 #23
    I don't get it...
     
  25. Oct 5, 2009 #24

    Borek

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    Show your ICE table and how you calculated - for example - 0.007275.

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  26. Oct 5, 2009 #25
    I re-did it this morning with some hints from my Prof:

    2NOBr [tex]\leftrightarrow[/tex] 2NO + Br2

    I 2x / 0 / 0
    C -2x(.34) / +2x(.34) / +.34x
    E 2x(.66)/ +2x(.34) / .34x

    The total pressure was .25 atm, so:

    0.25 = PNOBr + PNO + PBr

    0.25 = 2x(.66) + 2x(.34) + .34x

    X = .02136752137

    Then I just plugged it into the expression for each gasses concentration, and shoved those values into the Kp equation to get .1926125012.

    That I plugged into Kp=Kc(RT)[tex]\Delta[/tex]n to get a temperature of 5957.399661oK... Is something wrong here?
     
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