Question

# An equilibrium mixture contains N2O4, (P= 0.28 atm ) and NO2 (P= 1.0 atm ) at...

An equilibrium mixture contains N2O4, (P= 0.28 atm ) and NO2 (P= 1.0 atm ) at 350 K. The volume of the container is doubled at constant temperature.

Part A
Calculate the equilibrium pressure of N2O4 when the system reaches a new equilibrium.

Part B
Calculate the equilibrium pressure of NO2 when the system reaches a new equilibrium.

N2O4 -------->2NO2

Initial

0.28 .............1.0

Kp = (1.0)^2/(0.28) = 3.57

Now

When Volume is Doubled.....Pressure will be halved ...at constant temperature .....

therefore

N2O4 ----------------> 2NO2

Initial

0.14..........................0.5

Final

0.14-x........................0.5+2x

Kp = (0.5+2x)^2/(0.14-x)

3.57*(0.14-x) = (0.5+2x)^2

0.4998 - 3.57x = 0.25 + 4x^2 + 2x

4x^2 +5.57x -0.25 =0

solving for x

we get

x =0.0435

a) Equilibrium pressure of N2O4 = 0.14 - x = 0.10 atm

b) Equilibrium pressure of NO2 = 0.5 +2x = 0.59 atm

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