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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?
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?

This Question is Closed

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@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. \[Kp = \frac{[C][D] }{ [A][B] }\]

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0let's think about this more, we are being told that we need to raise the concentration of A

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0clearly this value for Kp since it's greater than 1 suggests that the equilibrium favors the products.

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0yes should i do an ICE table?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0My 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
 one year ago
Best ResponseYou've already chosen the best response.0BTW where did you get this question from it's very thought provoking

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Chem21labs.com is the platform where all my G. Chem II course homework is. @Photon336

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Change x x > +x +x?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Yea, that's what I was thinking since this reaction is 1:1 and everything's balanced so the changes should be similar

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0x=+0.5 to bring 0.5 M to 1 M ? So 0.5 m added to C to make this change? @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0This is what I had initially thought, I don't know why I feel that more needs to be done with this problem

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0I get a few tries for each question. Do you want me to try and submit and let you know?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0yeah I'm going to work on this myself, my gut tells me though that it's too straightforward

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 0.5 is wrong. :/ @Photon336

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0what 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?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@Abhisar this problem completely confused me, if you could provide an answer man that would be great.

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.2Ok, 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

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I 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

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.2Ok, that's not very hard. Took me few minutes to figure it out doe :O

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I missed something the first time around

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.2Let'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
 one year ago
Best ResponseYou've already chosen the best response.2Now you can write a new equilibrium equation and solve it against the older K for a. Getting it?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@Abhisar so technically Kc is the same right? because we didn't do anything to the temperature

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.2Kc is independent of concentration because change in concentration increases both forward and backward rate and Kc = Kf/Kb

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0so it goes up by 0.5 right? cause i put that and that was wrong

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Hmm so it would be this right? \[\frac{ 41.25 +7.5a }{ 4.5 } = Kc \]

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.2Answer should be 8.9 @Lena772

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2Because we only added a moles of C. We never changed the concentration of anything else than C

Lena772
 one year ago
Best ResponseYou've already chosen the best response.0That was right! @Abhisar
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