## vera_ewing one year ago Chem question

1. Rushwr

Hey what is the answer u got ?

2. Rushwr

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

3. vera_ewing

Wait, is the reaction at equilibrium?

4. sweetburger

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

5. Australopithecus

6. vera_ewing

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

7. Rushwr

$N _{2}O _{4} => 2NO _{2}$

8. Rushwr

So first write and equation for Kc.

9. Australopithecus

http://www.800mainstreet.com/7/0007-007-Equi_exp_k.html see this and see the section: "What does an equilibrium constant tell"

10. vera_ewing

Okay, and then what?

11. sweetburger

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

12. Rushwr

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

13. vera_ewing

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

14. Rushwr

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

15. Australopithecus

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

16. Rushwr

I get 4.5 x 10^-2

17. vera_ewing

How? Can you show me your work?

18. sweetburger

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

19. vera_ewing

Ohhh I see. I messed up.

20. vera_ewing

Thanks so much.

21. sweetburger

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

22. Australopithecus

Check out that page I linked

23. Rushwr

$Kc = 0.15^{2}/ 0.5$

24. Rushwr

So it's gonna be D !

25. Australopithecus

and the image I posted

26. sweetburger

yes I agree it should be D.