Solution:
          Â
2 NO(g) ⇔ N2(g) + O2(g)
          Â
0.2702Â Â Â
      0     Â
    0
         Â
          Â
(initial)
(0.2702-x)Â Â Â Â Â 0.5x
      0.5x   Â
          Â
          Â
(Equilibrium, moles)
          Â
(0.2702-x)/2Â Â Â Â Â 0.5x/2
      0.5x/2     Â
(Equilibrium, moles/L)
Equilibrium constant for the above reaction can be given as:
Keq = [N2][O2]/
[NO]2 = 1/0.0153
(0.5x/2)( 0.5x/2)/ [(0.2702-x)/2]2 = 65.35
0.25x2 / (0.2702-x)2 = 65.35
0.25x2 = 4.77
             Â
[neglecting ‘x’ since it is very small at the equilibrium]
x2 = 19.08
x = 4.36