Can you please check if I came up with the right answer for this please.

A 4.30L container had 55.10 moles of N2O4 added to it. When equilibrium was established the concentration of N2O4 was found to be 1.69M.What is the Kc for the following reaction?
N2O4(g)<-->2NO2(g)

Reasoning and workings:
N2O4<-->2No2
12.814M of O
12.814-x2x
12.814-x=1.69
x=11.124M
Kc=[NO2]^2/[N2o4]
=22.248^2/1.69
=292.88M
Hows that ??
Is it right??
If it is i am going to be so happy!!
Thanks andy

I have (N2O4) = 55.10/4.30 = 12.01 M BUT you ended up with the right value for x and the right value for NO2 as well as the right value for K.

Doess this mean I fluked the write answer??

Could you write it out as I have above please with your logical answers so I can go away and study if we are right..I am never comfortable not knowing I have the right workings
Thanks andy

No, you were right. I read my calculator wrong. My eyesight leaves something to be desired and I read 12.81 as 12.01. So all of your numbers are correct. See below for the only picky thing I could find.

N2O4<-->2No2
12.814M of O This should be 12.814 M N2O4, not of O. Everything else looks ok to me.
12.814-x2x
12.814-x=1.69
x=11.124M
Kc=[NO2]^2/[N2o4]
=22.248^2/1.69
=292.88M

Thanks DR BOB,

You had me worried for a second there!!
Thanks for all your great help on this one
Regards Andy

To determine if your answer is correct, let's go through the steps together:

1. First, we need to calculate the initial concentration of N2O4. As given, the initial moles of N2O4 are 55.10 and the volume of the container is 4.30L.

Initial concentration of N2O4 = Moles of N2O4 / Volume of container
= 55.10 moles / 4.30 L
= 12.814 M

2. Next, we need to calculate the equilibrium concentration of N2O4. The equilibrium concentration of N2O4 is given as 1.69M.

Equilibrium concentration of N2O4 = 1.69 M

3. The change in concentration of N2O4 can be calculated by subtracting the equilibrium concentration from the initial concentration.

Change in concentration of N2O4 = Initial concentration - Equilibrium concentration
= 12.814 M - 1.69 M
= 11.124 M

4. Since the stoichiometry of the reaction is 1:2, the change in concentration of NO2 is twice that of N2O4 (i.e., 2 * 11.124 M = 22.248 M).

5. Finally, using the balanced equation, we can calculate Kc by taking the square of the equilibrium concentration of NO2 and dividing it by the equilibrium concentration of N2O4.

Kc = [NO2]^2 / [N2O4]
Kc = (22.248 M)^2 / (1.69 M)
Kc ≈ 293.228 M (rounded to three significant figures)

Therefore, based on the calculations, the correct value of Kc is approximately 293.228 M, which is very close to your answer of 292.88 M. So, congratulations! Your answer is correct.