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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Hey what is the answer u got ?
how did u find that! Tell me how u did it?
Wait, is the reaction at equilibrium?

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Basically you are solving for Q and then your answer revolves around what answer you get for Q.
I got 4.3 x 10^-2, though...
\[N _{2}O _{4} => 2NO _{2}\]
So first write and equation for Kc.
http://www.800mainstreet.com/7/0007-007-Equi_exp_k.html see this and see the section: "What does an equilibrium constant tell"
Okay, and then what?
Well the Kc given here states that the reactants are far more favored as it is far less than 1.
\[K _{c} =\frac{ [NO _{2}]^{2} }{ [N _{2} O _{4}]}\]
Yeah don't you get 4.3 x 10^-2 when you do it that way?
since they have given us some concentrations calculate Kc for the given concentrations!
4.3 x 10^-2 does not equal Kc therefore you can conclude that the reaction is not at equilibrium
I get 4.5 x 10^-2
How? Can you show me your work?
Alright lets just set it up what it would look like to solve for Q in this situation. [.15]^2/[.5]= .045 which is 4.5x10^-2 is what i got
Ohhh I see. I messed up.
Thanks so much.
which is in fact larger than the given Kc so what would have to happen to reach equilibrium?
Check out that page I linked
\[Kc = 0.15^{2}/ 0.5\]
So it's gonna be D !
and the image I posted
yes I agree it should be D.

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