Question

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

Answer 1

Answer:

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