Partial Pressures when volume is doubled

Question

The partial pressures of an equilibrium mixture of N₂O₄ and NO₂ are P_N2O4 = 0.34 atm and P_NO2 = 1.20 atm at a certain temperature. The volume of the container is doubled. Calculate the partial pressures of the two gases when a new equilibrium is established.

Answer 1

First, we will find the Kp for this reaction

N₂O₄ <==> 2NO₂

Kp = (P_NO2)² / (P_N2O4) = (1.20)² / 0.34 = 4.24

Now, if the volume is doubled, the pressure of each gas will be 1/2 of the original

P_N2O4 = 0.34 atm / 2 = 0.17 atm

P_NO2 = 1.20 atm / 2 = 0.60 atm

Establishing a NEW equilibrium we will have (use an ICE table)

N₂O₄ <==> 2NO₂

0.17...........0.60.........Initial

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

0.17-x......0.60+2x....Equilibrium

Kp = 4.24 = (P_NO2)² / (P_N2O4) = (0.60+2x)² / (0.17-x)

0.72 - 4.24x = 4x² + 2.4x + 0.36

4x² + 6.64x - 0.36 = 0 use quadratic to solve for x

x = 0.053

Partial pressure of the gases is now...

P_N2O4 = 0.17 - 0.053 = 0.12 atm

P_NO2 = 0.60 + (2 x 0.053) = 0.71 atm

Total pressure of the 2 gases = 0.83 atm

(NOTE: be sure to check the math)