Question

Chem question

Answer 1

Hey what is the answer u got ?

Answer 2

how did u find that! Tell me how u did it?

Answer 3

Wait, is the reaction at equilibrium?

Answer 4

Basically you are solving for Q and then your answer revolves around what answer you get for Q.

Answer 5

I got 4.3 x 10^-2, though...

Answer 6

\[N _{2}O _{4} => 2NO _{2}\]

Answer 7

So first write and equation for Kc.

Answer 8

http://www.800mainstreet.com/7/0007-007-Equi_exp_k.html

see this and see the section:

"What does an equilibrium constant tell"

Answer 9

Okay, and then what?

Answer 10

Well the Kc given here states that the reactants are far more favored as it is far less than 1.

Answer 11

\[K _{c} =\frac{ [NO _{2}]^{2} }{ [N _{2} O _{4}]}\]

Answer 12

Yeah don't you get 4.3 x 10^-2 when you do it that way?

Answer 13

since they have given us some concentrations calculate Kc for the given concentrations!

Answer 14

4.3 x 10^-2 does not equal Kc therefore you can conclude that the reaction is not at equilibrium

Answer 15

I get 4.5 x 10^-2

Answer 16

How? Can you show me your work?

Answer 17

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

Answer 18

Ohhh I see. I messed up.

Answer 19

Thanks so much.

Answer 20

which is in fact larger than the given Kc so what would have to happen to reach equilibrium?

Answer 21

Check out that page I linked

Answer 22

\[Kc = 0.15^{2}/ 0.5\]

Answer 23

So it's gonna be D !

Answer 24

and the image I posted

Answer 25

yes I agree it should be D.