HARD CHEMISTRY QUESTION, HELP!!!!!!

N2O4 >< 2NO2
Keq=11, what will be the molarity of each if we start with 1M N2O4, and then the system goes to equilibrium?

Hint: Write Keq=, then represent things. Some of reactant is taken away, product is added. Quadratic equation required.

If you can, please show work and walk me through please. :D

For any equilibrium question, use an ICE table, where I = initial concentration, C = change in concentration, and E = equilibrium concentration.

So for this equation:
---------N2O4 <----> 2NO2
[I]------1M------------------0

We start with 1M of the reactant and none of the product. Now some of the reactant will form into the product, but right now we don't know how much, so let's call it "x".

--------N2O4 <----> 2NO2
[I]------1M------------------0
[C]----- (-x) ----------- (+2x)

2x of the product will form according to the stoichiometric ratios, one N2O4 makes two NO2. So the equilibrium concentrations are:

--------N2O4 <----> 2NO2
[I]------1M------------------0
[C]----- (-x) ----------- (+2x)
[E]---- (1-x) -------- (2x)

So now we have our equilibrium concentrations (mostly).

Keq is expressed as:

Keq = [NO2eq]^2 / [N2O4eq]

So if we put in the equilibrium concentrations and the value of Keq:

11 = (2x)^2 / (1-x)

And rearrange:

11 (1-x) = (2x)^2
11 - 11x = 4x^2

0 = 4x^2 +11x -11

Since the quadratic is equal to 0, we can use:

x = [ -b ± √(b^2 - 4ac) ] / 2a

To find x.

Then use your value of x to evaluate your equilibrium concentrations:

N2O4 = 1-x
and
NO2 = 2x

Hope this helps!