Question

An equilibrium mixture contains N2O4, (P= 0.27atm ) and NO2 (P= 1.1 atm ) at 350 K. The volume of the container is doubled at constant temperature.

A) Calculate the equilibrium pressure of

N2O4 when the system reaches a new equilibrium.

B) Calculate the equilibrium pressure of

NO2 when the system reaches a new equilibrium.

Answer 1

N2O2 ------> 2NO2

Kp = P²_NO2/P_N2O4

    = (1.1)² /0.27 = 4.48

According to Boyle's law volume is double pressure is half.

P_NO2 = 1.1/2 = 0.55atm

P_N2O4   = 0.27/2 = 0.135 atm

N2O2 ------> 2NO2

0.55           0.135     initial

-x              +2x    change

0.55-x      0.135+2x at equlibrium

Kp = P²_NO2/P_N2O4

4.48 =( 0.135+2x)²/0.55-x

2.464-4.48 x= 0.018225 + 4x² + 0.54x

4x² + 5.02x-2.445775 =0

x = 0.375

P_NO2 = 0.135+2x = 0.135+2*0.375 = 0.885 atm

P_N2O4 =0.55-x = 0.55-0.375 = 0.175 atm

cheking new pressure

Kp = P²_NO2/P_N2O4

      = (0.885)²/0.175   = 4.48 so it is correct answer