Chemical Equilibrium: find concentration from 2 solutions (ICE table)

[doridoridori]

Homework Statement

I've recently encountered this problem https://www.physicsforums.com/threads/need-major-help.117630/
In case the link doesn't open, here's the problem itself:
9. At a certain temperature, T, K for the reaction below is 7.5 liters/mole.

2NO2 <===> N2O4

If 2.0 moles of NO2 are placed in a 2.0-liter container and permitted to react at this temperature, what will be the concentration of N2O4 at equilibrium?

a) 0.39 moles/liter

b) 0.65 moles/liter

c) 0.82 moles/liter

d) 7.5 moles/liter

e) none of these

Relevant Equations

Super stuck on this. All of the deets are given in the question I linked. I'm still getting used to this forum so don't mind please if I accidentally mess something.
Also this is the equation itself:

7.5=x/(1-4x-4x^2)
30x^2-31x+7.5=0
x=0.39(approx.)
x=0.65(also approx.)

Hello! I've recently encountered this problem https://www.physicsforums.com/threads/need-major-help.117630/ and solved it and I'm stuck at choosing between 2 solutions. I don't understand why do we need to plug in 0.39 and 0.65 to (1-2x) and NOT to (1-2x)^2. I mean, we were given 2NO2, not just NO2. I see that in ICE table the result is [1-2x] but then why Keq=N2O4/(NO2)^2 and not just NO2?

 

[Chestermiller]

Those are both correct solutions to your equation, but x=0.65 corresponds to a negative concentration of NO2, which is physically impossible.

 

[doridoridori]

Those are both correct solutions to your equation, but x=0.65 corresponds to a negative concentration of NO2, which is physically impossible.

Hello! Yes, that's essentially what I was asking. Sorry if I made it all unclear! So to know the right answer, we are plugging 0.65 and 0.39 into (1-2x). What I don't understand is why don't we plug it into K=x/(1-2x)^2 instead? I mean, we were given 2 moles of NO2, wouldn't it be reasonable then to calculate for 1-2x squared?

 

[Chestermiller]

Hello! Yes, that's essentially what I was asking. Sorry if I made it all unclear! So to know the right answer, we are plugging 0.65 and 0.39 into (1-2x). What I don't understand is why don't we plug it into K=x/(1-2x)^2 instead? I mean, we were given 2 moles of NO2, wouldn't it be reasonable then to calculate for 1-2x squared?

That is what you do plug into. Try those values and see what you get.