If Kp for the reaction N 2 O 4   2 NO 2   is 0.66 then what is the equilibrium pressure of N2O4. (Total pressure at equilibrium is 0.5 atm)
(1) 0.168                   (2) 0.322                   (3) 0.1                    (4) 0.5

Dear Student,

The given reaction is :
                                                                        N2O42NO2   
                                                      at t = 0       1               0                     initial mole
                                                    at eqm        1 - x            2x                  moles at eqm 
                           
 Mole fraction at                             eqm      1-x1+x           2x1+x 
As the total pressure at eqm = 0.5 atm.
So the partial pressure of N2O4 = ​ 1-x1+x× 0.5 atm 
The partial pressure of NO2 = 2x1+x×0.5 atm 
Now, as the Kp = [p(NO2)]2[p(N2O4)]0.66 = 2x1+x2x1+x × 0.5 ×0.5 1-x1+x× 0.5 0.66 = 4x2 × 0.5 1-x20.66 - 0.66x2 = 2x22.66x2 = 0.66 x = 0.662.66   = 0.5 

Now put the value of x in partial pressure equation of ​N2O4 = 
                 = 1-0.5 1+0.5× 0.5= 0.5 × 0.51.5 = 0.166 atm 

So, the correct option is (1)

Regards.
 

  • 22
What are you looking for?