Question

Dinitrogentetraoxide partially decomposes according to the following equilibrium:

N2O4 (g) -> 2NO2 (g)
<-

A 1.00-L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol ofN2O4
remains. Keq for this reaction is ?.

hOw can i set up this problem?

Answer 1

Set up an ICE chart, substitute into the Keq expression, and solve for Keq. You want to convert moles to M (but that's easy since it's a 1 L flask.)
initial:
N2O4 = 0.400 M
NO2 = 0

change:
NO2 = +2x
N2O4 = -x

equilibrium:
NO2 = 2x
N2O4 = 0.400-x and the problem tells you this is 0.0055

Now do a little reasoning. Solve 0.400-x=0.0055 for x.
Multiply that by 2 to find NO2 at equilibrium.
Then substitute into Keq expression and solve for Keq.
Check my work.
0

Answer 2

whats the keq expression??'
sorry we're learning about ice charts tomorrow but the teacher gave us a practice worksheet for the coming quiz and i wanted to see if i could figure out some of the problems early.

Answer 3

i did everything you said and got
0.789 im at the step of setting up the keq expression

Answer 4

sorry to have gotten into this post, im solving a similar question and wanted to know how to do the final step

Answer 5

Your 0.789 is 2x and that is (NO2).
(N2O4) = 0.0055 from the problem.
Keq expression is
Keq = (NO2)^2/(N2O4)

If 0.400-x = 0.0055, then x = 0.3945
So (N2O4) = 0.0055
and NO2 = 2x = 0.3945*2 = 0.789

Now substitute into Keq above to obtain:
Keq = (0.789)^2/(0.0055) = ??
about 113 or so.

Answer 6

I got the problem from my book, and it says the answer is 0.87?