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.
Express your answer using two significant figures.
Part B
Calculate the equilibrium pressure of NO2 when the
system reaches a new equilibrium.
Express your answer using two significant figures.
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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