Question

For the following gaseous reaction: N2O3 --- NO + NO2 the degree of dissociation for N2O3 is found to be .243 at 25 degrees C when initially 2 moles of N2O3 are placed in a 2 liter flask. Kc=.078

If the volume is increased to 4L, what will the new concentration of N2O3 be?

Answer 1

I have the answer in front of me, but not quite sure how to get it. It should be 0.338 mol/L. I understand that Kc= [NO][NO2] (which are the same)/[N2O3]. I also understand that degree of dissociation equals change in mols over initial moles

Answer 2

N2O3 -> NO+ NO2
inital 2 0 0
at eq.,2-2@ 2@ 2@
where @ is degree of dissociation.
Now convert the moles to concentration by dividing by volume.
N2O3=(2-2@)/4=(1-@)/2, NO=2@/4=@/2, NO2=@/2.
Now apply Kc equation
Kc= @^2 /2(1-@). You get a quadratic you can solve it for @ and then N2O3 at eq.,=2(1-@).

Answer 3

Or wait you could make use of the extra data but i'm not sure how. The method above is very tedious(Solving the quadratic and all). There has to be a better method