Consider the following reaction.

Question

2NO2(g)⇌N2O4(g)

When the system is at equilibrium, it contains NO2 at a pressure of 0.724 atm , and N2O4 at a pressure of 0.0524 atm . The volume of the container is then reduced to half its original volume. What is the pressure of each gas after equilibrium is reestablished?

𝑃NO2 =
atm
𝑃N2O4=
atm

Answer 1

..................2NO2(g) ⇌ N2O4(g)
E............0.724 atm........0.0524 atm
Kp = p(N2O4)/p^2(NO2) = 0.0524/(0.724)^2 = 0.1 (rounded to 0.1)
Reducing the volume to 1/2 means pN2O4 becomes 0.0524 x 2 = 0.105 atm and pNO2 = 0.724 x 2 = 1.45 atm. Then, Le Chatelier's Principles tells us that when pressure is increased in a system at equilibrium that the equilibrium will shift to the side with fewer moles. So it will shift to the right producing more N2O4 at the expense of NO2.
.........................2NO2(g) ⇌ N2O4(g)
I.........................1.45...............0.105
C.......................-2x.....................+x
E.....................1.45-2x................0.105+x
Plug the E line into the Kp expression and solve for x, then evaluate 1.45 - 2x and 0.105 + x. Note that I have rounded here and there so you should start from the beginning and use your numbers and not mine.
Post your work if you get stuck.