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Lena772

  • one year ago

A(g) + B (g) ↔ C (g) + D (g) Suppose that the concentrations of these gases in an equilibrium mixture were [A] = 0.5 M [B] = 4.0 M [C] = 6.0 M [D] = 8.0 M How many moles of C would have to be added to the mixture to raise the equilibrium concentration of A to 1.0 M?

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  1. Photon336
    • one year ago
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    @Lena772 I feel that for this problem you could set it up and find what the Kc value is because you're given the equilibrium concentrations. \[Kp = \frac{[C][D] }{ [A][B] }\]

  2. Lena772
    • one year ago
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    48/2 Kp=24

  3. Photon336
    • one year ago
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    let's think about this more, we are being told that we need to raise the concentration of A

  4. Photon336
    • one year ago
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    clearly this value for Kp since it's greater than 1 suggests that the equilibrium favors the products.

  5. Lena772
    • one year ago
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    yes should i do an ICE table?

  6. Photon336
    • one year ago
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    yes I think so

  7. Photon336
    • one year ago
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    My thoughts are that you would need to do an ice table, because in order for the concentration of A to increase Q > K and the equilibrium would have to shift to favor the reactants

  8. Photon336
    • one year ago
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    BTW where did you get this question from it's very thought provoking

  9. Lena772
    • one year ago
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    Chem21labs.com is the platform where all my G. Chem II course homework is. @Photon336

  10. Lena772
    • one year ago
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    |dw:1442103625131:dw|

  11. Lena772
    • one year ago
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    Change -x -x -> +x +x?

  12. Photon336
    • one year ago
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    Yea, that's what I was thinking since this reaction is 1:1 and everything's balanced so the changes should be similar

  13. Lena772
    • one year ago
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    x=+0.5 to bring 0.5 M to 1 M ? So 0.5 m added to C to make this change? @Photon336

  14. Photon336
    • one year ago
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    This is what I had initially thought, I don't know why I feel that more needs to be done with this problem

  15. Lena772
    • one year ago
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    I get a few tries for each question. Do you want me to try and submit and let you know?

  16. Photon336
    • one year ago
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    yeah I'm going to work on this myself, my gut tells me though that it's too straightforward

  17. Lena772
    • one year ago
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    Yeah, 0.5 is wrong. :/ @Photon336

  18. Photon336
    • one year ago
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    @sweetburger

  19. Photon336
    • one year ago
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    what is strange about this question is that the concentration of the product goes up. I guess we would have to take the equilibrium concentrations of A,B.C.D as the initial concentrations on our ice table. We have to establish what the new equilibrium concentrations are. I believe that since we found what our Kc is which is 24 this must be constant because were not increasing temperature so Kc shouldn't go up. if the concentration of C goes up because we're adding more moles for that. I think we need to find what Q is and what K is, maybe i'm overcomplicating this. I seriously think that there are two more steps to this problem to getting the answer @Astrophysics do you think you could help?

  20. Lena772
    • one year ago
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    @aaronq

  21. Photon336
    • one year ago
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    @Abhisar this problem completely confused me, if you could provide an answer man that would be great.

  22. Abhisar
    • one year ago
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    Ok, I was trying to think it in a different way, \(\sf A + B <---> C + D\) 0.5 4 6 8 From the above reaction we can say that 1 mole of A reacts to give 1 mole of C or 1 mole of C reacts to give 1 mole of A. So can't we say that in order to increase the concentration of A up to 1 mole we need 0.5 more moles of C

  23. Abhisar
    • one year ago
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    @Photon336

  24. Photon336
    • one year ago
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    I see, but how would we go about setting this up to begin with because the concentration of A ended up increasing. the reaction favors the products, and to my understanding our intitial concentrations are 6,8,0.5 and 4

  25. Lena772
    • one year ago
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    @Abhisar

  26. Abhisar
    • one year ago
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    Ok, that's not very hard. Took me few minutes to figure it out doe :O

  27. Photon336
    • one year ago
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    I missed something the first time around

  28. Abhisar
    • one year ago
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    Let's suppose we add (a) moles of C to increase the amount of A up to 1. Now let's do an ICE table. |dw:1442110714400:dw|

  29. Abhisar
    • one year ago
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    Now you can write a new equilibrium equation and solve it against the older K for a. Getting it?

  30. Photon336
    • one year ago
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    @Abhisar so technically Kc is the same right? because we didn't do anything to the temperature

  31. Abhisar
    • one year ago
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    YEs.

  32. Abhisar
    • one year ago
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    Kc is independent of concentration because change in concentration increases both forward and backward rate and Kc = Kf/Kb

  33. Lena772
    • one year ago
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    so it goes up by 0.5 right? cause i put that and that was wrong

  34. Photon336
    • one year ago
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    Hmm so it would be this right? \[\frac{ 41.25 +7.5a }{ 4.5 } = Kc \]

  35. Abhisar
    • one year ago
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    Yes

  36. Abhisar
    • one year ago
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    Kc is 24. Right?

  37. Photon336
    • one year ago
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    yeah

  38. Abhisar
    • one year ago
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    Answer should be 8.9 @Lena772

  39. Photon336
    • one year ago
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    @Abhisar so why did you put the variable a in front of C and not say D as well This is what I had thought originally Because we already had established equilibrium, If A went up, I thought that both C and D at some point had to go down.. because if the concentration of C went up then the reaction had to have shifted to the reactants. that's what confused me.

  40. Abhisar
    • one year ago
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    Because we only added a moles of C. We never changed the concentration of anything else than C

  41. Lena772
    • one year ago
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    That was right! @Abhisar

  42. Abhisar
    • one year ago
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    c:

  43. Abhisar
    • one year ago
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    Did you get it doe?

  44. Lena772
    • one year ago
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    YES ! :)

  45. Abhisar
    • one year ago
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    Great!!

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