Solve this:

1. 
One mole of N2O4 (g) in a 1 L flask decornposes to attain the equilibrium N2O4 (g)  2NO2(g) 
At the equilibrium the mole fraction of NO2 is 1/2 Hence Kc will be:
(1) 1/3        (2) 1/2         (3) 2/3        (4) 1

                        N2O42NO2Initial:             1              -At eqm           1-x        2xTotal moles at eqm = 1-x+2x= 1+x  Mole fraction of NO2 = Number of moles of NO2Total moles = 2x1+xSo 2x1+x = 124x=1+x3x=1x=1/3 = 0.333At eqm          N2O4= 1-x = 1-0.33=0.67   [N2O4]=nv= 0.671=0.67    NO2=2x =2 (0.33)=0.66[NO2]=nv= 0.661=0.66Kc=[NO2]2[N2O4] =0.6620.67 =0.66 =23Thus option 3 is correct

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