How do I set up this equilibrium pressure question?

Question

An equilibrium mixture contains N2O4, (P= 0.28 atm ) and NO2 (P= 1.0 atm ) at 350 K. The volume of the container is doubled at constant temperature.

Calculate the equilibrium pressure of N2O4 when the system reaches a new equilibrium.

Answer 1

First, we need to know what the overall reaction is. It appears to be the following:

N₂O₄(g) <==> 2NO₂(g)

Step 1: Let's find the equilibrium constant, Kp

Kp = (PNO₂)² / (PN₂O₄) = (1.0)² / (0.28)

Kp = 3.57

Step 2: Find new pressures after doubling the volume

According to Boyle's law, a doubling of the volume will result in the pressure being halved (assuming constant temperature.

New PN₂O₄ = 0.28 atm / 2 = 0.14 atm

New PNO₂ = 1.00 atm / 2 = 0.500 atm

According to LeChatelier, a decrease in pressure will push the equilibrium to the side with more moles of gas. In this case, it will shift to the right (product side), so PN₂O₄ will decrease and PNO₂ will increase.

Let X be the decrease in PN₂O₄

We then have PN₂O₄ = 0.14 atm - X and PNO₂ = 0.500 atm + 2X

Substituting in the Kp expression, we have...

3.57 = (0.5 + 2X)² / (0.14-X)

Use the quadratic formula to solve for X and then equilibrium pressure of N2O4 will be 0.14 - x