Posted by Jane on .
i already posted this question, but im still confused on it.
Dinitrogentetraoxide partially decomposes according to the following equilibrium:
N2O4 (g) > 2NO2 (g)
<
A 1.00L flask is charged with 0.400 mol of N2O4. At equilibrium at 373 K, 0.0055 mol ofN2O4
remains. Keq for this reaction is ?.
so my steps:
N2O4 (g) > 2NO2 (g)
Initial conc. 0.400 mol 0 mole (no reaction yet)
Change 0.3945 mol + 2 x 0.3945 mole (based on equation above)
Equilibrium 0.4000.3945 mol 0.7890 mole
0.0055 mol 0.7890
Keq = [NO2]2/[N2O4]
Keq = (0.789)^2/(0.0055) =113
but my book says the answer is 0.87?

chemistry 
DrBob222,
I think the book answer must be wrong. I used 113.2 as Keq and started with 0.4 M N2O4 and zero NO2 and solved backwards. x came out to be 0.3945M which makes equilibrium concn N2O4 = 0.0055 which was our starting number in the problem.
If I use 0.87 as Keq, the quadratic has a solution which working backwards gives x = 0.2056 so (N2O4) at equilibrium = 0.1944 (not 0.0055). The values of 0.2056 for (N2O4) and 0.411 = (NO2) gives Keq = 0.87; however, the problem gives (N2O4) at equilibrium of 0.0055 M and not 0.2056M.
I don't believe 0.87 is correct. 
chemistry 
Jane,
yes, ive asked multiple people and they've all told me the same thing. i'll study the question the way you said, i have a test next week and i wanted to know how to solve problems like this,