When the system is at equilibrium, it contains NO 2 at a pressure of 0.817 atm, and N 2 O 4 at a pressure of 0.0667 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

Pressure of NO2 = 1.533 atm

Pressure of N2O4 = 0.2344

Explanation:

Step 1: Data given

Pressure of NO2 = 0.817 atm

Pressure of N2O4 = 0.0667 atm

Step 2: The balanced equation

N2 O4 ⇌ 2NO2

Step 3: calculate the total pressure

Total pressure = 0.817 atm + 0.0667 atm

Total pressure = 0.8837 atm

Step 4: Calculate the value of Kp

Kp = p (NO2) ² / p (N2O4)

Kp = 0.817²/0.0667

Kp = 10.0

Step 5: Calculate

For the second equilibrium, since the volume of the container is halved, the total pressure will be doubled: 2*0.8837 = 1.7674 atm

The partial pressures of N2O4 and NO2 at the equilibrium will be at equilibrium will be 1.7674-x atm and x atm respectively.

Kp = 10.0 = p (NO2) ² / p (N2O4)

10.0 = X² / (1.7674 - X)

X = 1.533

Pressure of NO2 = 1.533 atm

Pressure of N2O4 = 0.2344