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A complex equilibrium question

  1. Dec 30, 2014 #1
    • Member warned about not using the homework template
    Kc1 = (5.8*2/5)^2 / (14/5)(1.4/5) = 6.865

    6.865 = [HI]^2 / (45/253.8/100)(0.5/2/100)
    [HI] = 5.52x10^-3 M
    mass no of HI = [HI] x (126.9+1) x 100 = 70.6g
    is it correct?
    And how to do 3bii
    I GOT 3bi Kc2 = [HCl(g)]^2/ / Kc(1) [H2(g)] [Cl2(g)]
    20141228_f511cddd68ca462e6de7NFWjy7I2WqFy.png
     
  2. jcsd
  3. Dec 30, 2014 #2

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    Think about it --- you've got 45 g I2 to start --- what's the largest mass of HI you can make?
     
  4. Dec 30, 2014 #3
    6.865 = x^2 / ((45/253.8/100)-0.5x)((0.5/2/100)-0.5x)
    This equation is OK?because the mole ratio of H2 to Hi is 1:2
    So [HI]=x=5.33*10-4M
    HI =6.813g?
     
    Last edited: Dec 30, 2014
  5. Dec 30, 2014 #4

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    The 6.865 is okay for an equilibrium constant. The notation you're using for the equation is unconventional. What're the "100s" doing in there?
     
  6. Dec 30, 2014 #5
    100L
    The volume
     
  7. Dec 30, 2014 #6

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    And the HI is not also occupying that volume?
     
  8. Dec 30, 2014 #7
    Yes, but molarity = mole/volume
    Does not volume is necessary?
     
  9. Dec 30, 2014 #8

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    You're using it in the denominator for hydrogen and iodine concentrations. Don't you think you should be using it in the numerator as well? After all, you did it that way when finding the equilibrium constant.
     
  10. Dec 30, 2014 #9
    ((45/253.8)/100)
    ??this way??
     
  11. Dec 30, 2014 #10

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    No, the way you did it for the very first step when you solved for the equilibrium constant.
     
  12. Dec 30, 2014 #11
    You're using it in the denominator for hydrogen and iodine concentrations
    What it mean.....
     
  13. Dec 30, 2014 #12

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    Do you see where you inserted the number "5" for five liters in the very first calculation you did? Do you see it in both numerator, HI, and denominator, H and I?
     
  14. Dec 30, 2014 #13
    For example
    14 is the mole of h2,so 14/5=2.8 which is concentration of h2, right?
    OK then there is 0.5g of h2,
    0.5/2=0.25mol
    0.25/100=1.25*10^-3M
    What wrong with me?
     
  15. Dec 30, 2014 #14

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    So, you will have x moles of HI in 100 liters. Clear?
     
  16. Dec 30, 2014 #15
    Does numerator will not be wrong?
     
  17. Dec 30, 2014 #16

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    If you are using the 100 liter volume in the denominator, you HAVE to include it in the numerator.
     
  18. Dec 30, 2014 #17
    I know what you mean sir
    if I use (x/100)^2, .x will be the mole of HI right?
    But in the numerator,did you see((0.5/2/100)-0.5x)
    if x =mole, minus 0.5x
    this is the mole instead of concentration
     
  19. Dec 30, 2014 #18

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    Okay. We are making progress. You also need to be dividing the 0.5x by 100 if you're going to work with concentration. It is your option or choice whether to express x the amount actually reacting subtracted from the amount you begin with as just moles, or to divide by the volume and handle amounts as concentrations. The one thing you do have to do is treat reactants and products and fraction reacted all the same way, either as number of moles, or as concentrations.
     
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