Posted by Jane on Monday, October 25, 2010 at 5:42pm.
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?
- Chemistry Class - DrBob222, Monday, October 25, 2010 at 6:22pm
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.
- Chemistry Class - Laynie, Monday, October 25, 2010 at 6:42pm
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.
- Chemistry Class - Laynie, Monday, October 25, 2010 at 6:44pm
i did everything you said adn got
0.789 im at the step of setting up the keq expression
- Chemistry Class - Laynie, Monday, October 25, 2010 at 7:11pm
sorry to have gotten into this post, im solving a similar question and wanted to know how to do the final step
- Chemistry Class - DrBob222, Monday, October 25, 2010 at 8:32pm
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.
- Chemistry Class - Jane, Monday, October 25, 2010 at 9:12pm
I got the problem from my book, and it says the answer is 0.87?
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