Question

In a container equilibrium n2o4 > 2no2 is attained at 25 degree. The total pressure in the the container is 380 torr. If equilibrium constant of the above equilibrium is 0.667 atm, then degree of disassociation of n2o4 is

Answer 1

Answer:

N2O4–> 2NO2

Let: x= PNO2 and y= PN2O4

0.66=x²/y and x+y= 0.5

y= 1.515 x² and y= 0.5-x

1.515x²+x-0.5 = 0

x= [-1±√(1+3.03)]/3.03

x= 0.33 so y=0.17

PNO2= 0.332 atm

PN2O4=0.168 atm

Answer 2

Step-by-step explanation:

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