At a particular temperature, Kp 0.25 for the reaction a. A ask containing only N2O4 at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A ask containing only NO2 at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

a.  pNO₂ = 1 atm    pN₂O₄ = 4 atm

b.  pNO₂ = 1 atm    pN₂O₄ = 4 atm

c. It does not matter.

Explanation:

From the information given in this question we know the equilibrium involved is

N₂O₄ (g) ⇄ 2 NO₂ (g)

with Kp  given by

Kp = p NO₂²/ p N₂O₄ = 0.25

We know that if we place 4.5 atm of N₂O₄ is placed in a flask, a quantity x is going to be consumed producing 2x atm of NO₂  and we can setup the following equation:

0.25 =  p NO₂²/ p N₂O₄ =  (2x)² / (4.5 - x)

0.25 x  (4.5 - x) = 4x²

4x² + 0.25 x - 1.125 = 0

after solving this quadratic equation, we get two roots

x₁ = 0.5

x₂ = -0.56

the second root is physically impossible, and the partial pressures for x₁ = 0.5  will be

pNO₂ = 2 x 0.5 atm = 1.0 atm

pN₂O₄ = (4.5 - 0.5) atm = 4.0 atm

Similarly for part b, we get the equilibrium equation

0.25 = (9- 2x)² / x

0.25x = 81 - 36x + 4x²

the roots of this equation are:

x₁ = 5.0625

x₂ = 4

the first root is physically impossible since it will give us a negative partial pressure of N₂O₄ :

p N₂O₄ = 9 - 2(5.0625) = -1.13

the second root give us the following partial pressures:

p N₂O₄ = (9 - 2x4) atm = 1 atm

p NO₂ = 4 atm

The partial pressures are the same, it does not matter from which direction an equilibrium position is reached since what is essential is that the partial pressures of the gasses N₂O₄ and  NO₂ obey the equilibrium equation.