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An odd equilibrium question

  1. Apr 20, 2014 #1
    ImageUploadedByPhysics Forums1397974725.378461.jpg

    My solution for this question is Kc=1/10

    As the ratio of both forward and backward rate constants never changes unless the temperature changes.

    My teacher told me that the answer is 1/100 because the reaction's products and reactants are doubled, that seems senseless to me, I don't want to go in an argument with him until I make sure whether my answer is right or wrong,
    so I need your help, thanks in advance
     
  2. jcsd
  3. Apr 20, 2014 #2

    Borek

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    Staff: Mentor

    Your teacher is right.

    Have you wrote Kc for both reactions? Name them Kc and Kc', and try to express one using the other. There is no need to guessing here, pure and simple arithmetic.
     
  4. Apr 20, 2014 #3

    I did that, but I'm not convinced that it should be done, because when A,B and C is doubled the reaction should move toward C opposing the change so Kc remains constant

    "I'm sure that you are right but I don't want to blindly accept that answer, I want to know why"
     
  5. Apr 20, 2014 #4
    Here, I hope this helps! ;)
    If you change the stoichiometric numbers, the equilibrium constant (the concentration quotient "Kc" in fact) also changes, because it's dependant on those stoichiometric numbers.
     

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  6. Apr 20, 2014 #5

    Q changes not Kc it's not the same thing
     
  7. Apr 20, 2014 #6
    At equilibrium it's the same thing ;)
     
  8. Apr 20, 2014 #7

    Borek

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    Kc value depends on the way reaction is written.

    That's why it is convenient to follow the convention that properly balanced reaction equation uses set of the lowest possible integer coefficients, it removes ambiguity.
     
  9. Apr 20, 2014 #8
    In a closed container if Iodine reacts with hydrogen to give hydrogen iodide, if we opened the container and doubled the amount of iodine, hydrogen, and hydrogen iodide, can we say that Kc Changes ??
     
  10. Apr 20, 2014 #9

    Borek

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    You are confusing stoichiometric coefficients with concentrations.

    Doubling concentrations doesn't change stoichiometric coefficients, doubling stoichiometric coefficients doesn't change concentrations.
     
  11. Apr 20, 2014 #10
    aA + bB [itex]\rightarrow[/itex] cC

    If we say that a,b and c are the number of molecules
    this means that cC is only produced when aA reacts with bB

    2aA + 2bB [itex]\rightarrow[/itex] 2cC

    doesn't this equation give the same meaning as the first one ??

    :confused:
     
    Last edited: Apr 20, 2014
  12. Apr 21, 2014 #11
    I don't understand the difference between the two equations
     
  13. Apr 21, 2014 #12

    Borek

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    No, they are different.

    Lets say we have a simple reaction

    A <-> B

    and in the mixture at equilibrium concentration of A is 2 M and concentration of B is 3 M.

    For the reaction as written

    [tex]K_c = \frac {}{[A]}[/tex]

    and Kc value is

    [tex]K_c = \frac {[3]}{[2]} = 1.5[/tex]

    Now lets say we want to write the reaction as

    2A <-> 2B

    If so

    [tex]K_c' = \frac {^2}{[A]^2}[/tex]

    Our mixture was at equilibrium, so concentrations are what they are, however, now Kc' is

    [tex]K_c' = \frac {[3]^2}{[2]^2} = 2.25[/tex]

    Just because we decided to use a different reaction equation value of the equilibrium constant changes, and, obviously

    [tex]K_c' = K_c^2[/tex]

    That's just the way it is.
     
  14. Apr 21, 2014 #13



    That means that the rate constant of the forward reaction of the second equation is square the rate constant of the forward reaction of the first equation and same thing with the rate constant of the backward reaction ?
     
  15. Apr 21, 2014 #14

    Borek

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    In terms of kinetic theory of chemical equilibrium - yes. But please remember that the reaction kinetics is never guaranteed to follow the reaction equation, and that the kinetic theory of the equilibrium is not the only theory explaining how the equilibrium works and how it is reached.
     
  16. Apr 21, 2014 #15
    Logically, I can't still visualize why 2A , AB react giving 2C faster than A , B
     
  17. Apr 22, 2014 #16

    Borek

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    Because in reality they don't. What you see is just an artifact of the kinetic theory of equilibrium. It looks as if the reaction kinetic was higher, because the theory blindly assumes reaction kinetic can be calculated from the reaction equation. I told you it is not true, have you read the second phrase of my previous post?
     
  18. Apr 24, 2014 #17

    Theoretically speaking, the reaction is faster ?
     
  19. Apr 24, 2014 #18

    Borek

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    No, real reaction speed (just like the equilibrium) doesn't depend on the way the reaction equation is written.
     
  20. Apr 24, 2014 #19

    Then all the calculations we do is just crap !!!
     
  21. Apr 24, 2014 #20

    Borek

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    You are either trolling, or just wasting my time:

    I told you twice what the problem is. As we know what the problem is, we DON'T calculate kinetics from the overall reaction equation. OTOH equilibrium calculations are perfectly right.

    I can't stop you from doing crappy calculations, but you have been warned.
     
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