Lena772
  • Lena772
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?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Photon336
  • Photon336
@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. \[Kp = \frac{[C][D] }{ [A][B] }\]
Lena772
  • Lena772
48/2 Kp=24
Photon336
  • Photon336
let's think about this more, we are being told that we need to raise the concentration of A

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More answers

Photon336
  • Photon336
clearly this value for Kp since it's greater than 1 suggests that the equilibrium favors the products.
Lena772
  • Lena772
yes should i do an ICE table?
Photon336
  • Photon336
yes I think so
Photon336
  • Photon336
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
Photon336
  • Photon336
BTW where did you get this question from it's very thought provoking
Lena772
  • Lena772
Chem21labs.com is the platform where all my G. Chem II course homework is. @Photon336
Lena772
  • Lena772
|dw:1442103625131:dw|
Lena772
  • Lena772
Change -x -x -> +x +x?
Photon336
  • Photon336
Yea, that's what I was thinking since this reaction is 1:1 and everything's balanced so the changes should be similar
Lena772
  • Lena772
x=+0.5 to bring 0.5 M to 1 M ? So 0.5 m added to C to make this change? @Photon336
Photon336
  • Photon336
This is what I had initially thought, I don't know why I feel that more needs to be done with this problem
Lena772
  • Lena772
I get a few tries for each question. Do you want me to try and submit and let you know?
Photon336
  • Photon336
yeah I'm going to work on this myself, my gut tells me though that it's too straightforward
Lena772
  • Lena772
Yeah, 0.5 is wrong. :/ @Photon336
Photon336
  • Photon336
@sweetburger
Photon336
  • Photon336
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?
Lena772
  • Lena772
@aaronq
Photon336
  • Photon336
@Abhisar this problem completely confused me, if you could provide an answer man that would be great.
Abhisar
  • Abhisar
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
Abhisar
  • Abhisar
@Photon336
Photon336
  • Photon336
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
Lena772
  • Lena772
@Abhisar
Abhisar
  • Abhisar
Ok, that's not very hard. Took me few minutes to figure it out doe :O
Photon336
  • Photon336
I missed something the first time around
Abhisar
  • Abhisar
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|
Abhisar
  • Abhisar
Now you can write a new equilibrium equation and solve it against the older K for a. Getting it?
Photon336
  • Photon336
@Abhisar so technically Kc is the same right? because we didn't do anything to the temperature
Abhisar
  • Abhisar
YEs.
Abhisar
  • Abhisar
Kc is independent of concentration because change in concentration increases both forward and backward rate and Kc = Kf/Kb
Lena772
  • Lena772
so it goes up by 0.5 right? cause i put that and that was wrong
Photon336
  • Photon336
Hmm so it would be this right? \[\frac{ 41.25 +7.5a }{ 4.5 } = Kc \]
Abhisar
  • Abhisar
Yes
Abhisar
  • Abhisar
Kc is 24. Right?
Photon336
  • Photon336
yeah
Abhisar
  • Abhisar
Answer should be 8.9 @Lena772
Photon336
  • Photon336
@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.
Abhisar
  • Abhisar
Because we only added a moles of C. We never changed the concentration of anything else than C
Lena772
  • Lena772
That was right! @Abhisar
Abhisar
  • Abhisar
c:
Abhisar
  • Abhisar
Did you get it doe?
Lena772
  • Lena772
YES ! :)
Abhisar
  • Abhisar
Great!!

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