ICE TABLE
N2O4 ⇋ 2NO2
Initial 1 mol 0
change x 2x
Equilibrium. 1-x 2x
Mole fraction of NO2 = 2x/(1-x+2x) => 2x/(x+1) = ½
[since NO2 is 50% of the total volume ]
solving we get, 4x = x+1
3x = 1
x = 1/3
Since volume of the vessel is 1L
The concentration equilibrium constant
Kc = > (2x)^2/(1-x) = (2/3)^2/(1-1/3) =(4/9)/(2/3) = 4*3/9*2 = 2/3=0.66
Option A