Pressure of each gas after equilibrium

Question

Consider the following reaction.

2NO2(g)⇌N2O4(g)

When the system is at equilibrium, it contains NO2 at a pressure of 0.835 atm, and N2O4 at a pressure of 0.0697 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?

Answer 1

2NO₂(g) <==> N₂O₄(g)

Kp = (N₂O₄) / (NO₂)² = (0.0697) / (0.835)²

Kp = 0.09997 = 0.1

If the volume is halved, the pressures will be doubled, giving us

Pressure NO₂ = 2 x 0.835 atm = 1.67 atm

Pressure N₂O₂ = 2 x 0.0697 atm = 0.1394 atm

Le Chatelier's principal tells us that the equilibrium will shift to the side with fewer moles of gas, so in this case it will shift to the right (product side). Set up and ICE table...

2NO₂(g) <==> N₂O₄(g)

1.67..................0.1394...........Initial

-2x....................+x.................Change

1.67-2x............ 0.1394+x......Equilibrium

Kp = 0.10 = (0.1394 + x) / (1.67-2x)²

From here, you need to solve for x (algebra) and then plug that value into the "equilibrium" line of the ICE table and solve for partial pressures of each gas.