posted by Jane .
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?
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.)
N2O4 = 0.400 M
NO2 = 0
NO2 = +2x
N2O4 = -x
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.
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.
i did everything you said and got
0.789 im at the step of setting up the keq expression
sorry to have gotten into this post, im solving a similar question and wanted to know how to do the final step
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.
I got the problem from my book, and it says the answer is 0.87?