For the reaction N2O4(g)2NO2(g) , Kp = 0.25 at a certain temperature. ....HELP!?

Question

For the reaction N2O4(g) <--->2NO2(g) , Kp = 0.25 at a certain temperature. If NO2(g) is placed in an otherwise empty container at an initial pressure of 0.050 atm, calculate the equilibrium pressures of NO2(g) and N2O4(g).


If the amount of NO2 that reacts to reach equilibrium is 2x, then the pressures at equilibrium (in terms of x) are

PN2O4 = x
PNO2 = 0.050-2x

Substitute the equilibrium pressures of NO2 and N2O4 (in terms of x) into the equilibrium expression and solve for x.

Answer 1

You've already been given a lot of help here. But you were supposed to write down

Kp = (p NO2) squared/(p N2O4)

So 0.25 = (0.05 - 2x) squared/x

And now all you need is some quadratic work to solve for x.