Equilibrium constant

Question

The value of Kc for the following reaction is 0.490 at 429K.
N2O4(g)->2NO2(g)
If a reaction vessel at that temperature initially contains 0.0400M NO2 and 0.0400M N2O4 What is the concentration of NO2 at
equilibrium?

Answer 1

......N2O4(g)<===>2NO2(g)

I....0.04....................0.04....

C....-x....................+2x....

E...0.04-x.............0.04+2x...

Kc = 0.490 = [NO2]^2/[N2O4] = (0.04-2x)2/(0.04-x)

Solve for x and add 2x that number to 0.04 to get [NO2] at equilibrium