25. Consider the following reaction. 2N*O_{2}(g) = N_2O_4(g)

Question

Consider the following reaction.

2N*O_{2}(g) = N_2O_4(g)

When the system is at equilibrium, it contains NO_{2} at a pressure of 0.810 atm, and N_{2}O_{4} at a pressure of 0.0656 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?

P NO_2 =

atm

PN_2O_4

atm

Answer 1

2NO₂(g) <==> N₂O₄(g)

Kp = (N₂O₄) / (NO₂)²

Kp = 0.0656 / (0.810)² = 0.100

When the volume is reduced to half, the partial pressures will double (P ∝ 1/V). Thus, new pressures will be:

Pressure NO₂ = 2 x 0.810 atm = 1.62 atm

Pressure N₂O₄ = 2 x 0.0656 atm = 0.131 atm

According to LeChatelier's Principle, the equilibrium will shift to the side with fewer number of moles of gas. In this case, it will shift to the right (product side). Set up an ICE table to reflect this shift:

2NO₂(g) <==> N₂O₄(g)

1.62...............0.131.........Initial

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

1.62 - 2x..........0.131+x...Equilibrium

Kp = 0.100 = (N₂O₄) / (NO₂)²

0.100 = 0.131 + x / (1.62 - 2x)²

From here, solve for x (using the quadratic equation) and then plug that value into the "equilibrium" line of the ICE table and solve for partial pressures of each gas.