- #1

doridoridori

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- 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?