For the reaction   $N_2{O}_4\rightleftharpoons 2N{O}_{2\left(g\right)}$, the degree of dissociation at equilibrium is 0.2 at 1 atm. Then calculate  $\frac{1}{K_p}$

Step 1: Start with the dissociation reaction and initial pressures

The reaction is:

N2O4 ⇌ 2NO2

Let the initial pressure of N₂O₄ be P₀. After dissociation, the partial pressures will be:

• P_N2O4 = P₀(1−α)
• P_NO2 = 2P₀α

Step 2: Total pressure at equilibrium

The total pressure is the sum of the partial pressures:

Ptotal = PN2O4 + PNO2 = P0(1−α) + 2P0α = P0(1 + α)

Step 3: Relate P₀ to the total pressure

Given that P_total = 1 atm,

P0(1 + α) = 1

P0 = 1 / (1 + α)

Substitute α = 0.2:

P0 = 1 / (1 + 0.2) = 1 / 1.2 ≈ 0.833 atm

Step 4: Calculate partial pressures at equilibrium

Now that we have P₀ = 0.833 atm:

• P_N2O4 = P₀(1−α) = 0.833 × (1 − 0.2) = 0.833 × 0.8 = 0.666 atm
• P_NO2 = 2P₀α = 2 × 0.833 × 0.2 = 0.333 atm

Step 5: Expression for K_p

The equilibrium constant K_p is:

Kp = (PNO2)² / PN2O4

Substitute the values:

Kp = (0.333)² / 0.666 = 0.111 / 0.666 ≈ 0.167 atm

Kp ≈ 0.167 atm