 #1
doridoridori
 2
 0
 Homework Statement:

I've recently encountered this problem https://www.physicsforums.com/threads/needmajorhelp.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 2.0liter 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/(14x4x^2)
30x^231x+7.5=0
x=0.39(approx.)
x=0.65(also approx.)
Hello! I've recently encountered this problem https://www.physicsforums.com/threads/needmajorhelp.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 (12x) and NOT to (12x)^2. I mean, we were given 2NO2, not just NO2. I see that in ICE table the result is [12x] but then why Keq=N2O4/(NO2)^2 and not just NO2?