What is the equilibrium partial pressure of N2O4 here?

Question

2NO2(g) ↔ N2O4(g) is an equilibrium reaction at 25oC, where Kp = 8.77 atm^-1.
If the equilibrium partial pressure of NO2(g) is 0.250 atm, what is the equilibrium partial pressure of N2O4(g) in atm ?

Any help would be greatly appreciated!!!

Answer 1

Firstly, you must write the expression of Kp:
Kp=(pN2O4)/(pNO2)^2
Then you just have to determine the partial pressure of N2O4 from this expression:
pN2O4=(Kp)*(pNO2)^2
Then you must replace Kp with 8.77 and pNO2 with 0.250
pN2O4=(8.77)*(0.250)^2
So pN2O4=0.548atm
Hope this helped.

Answer 2

It's 0.25 squared x 8.77.