Question

A 2.00 mol sample of nitrogen dioxide was placed in a 40.0 L vessel. Nitrogen dioxide decomposes according to the equation At a certain temperature, the nitrogen dioxide was 8.6%; that is, for each mol of NO2 before the reaction, (1.000 – 8.6×10-2) mol NO2 remains after dissociation. Calculate the value of KC for this reaction.

Answer 1

You have forgotten the reaction. I looked in internet and found that reaction is:
   2NO2          ----->                2NO +     O2
       2                                            0             0     (initial moles)
2*(1-8.6*10^-2)        2*8.6*10^-2    8.6*10^-2   (at equilibrium number of moles)

Since Volume, V = 40 L
[NO2] = number of moles of NO2 / Volume
              = {2*(1-8.6*10^-2) } / 40
              =0.0457 M
[NO] = number of moles of NO / Volume
              = {2*8.6*10^-2 } / 40
              =4.3*10^-3 M
[O2] = number of moles of O2 / Volume
              = {8.6*10^-2 } / 40
              =2.15*10^-3 M

KC = [NO]^2 [O2] / [NO2]^2
      = (4.3*10^-3 )^2 * (2.15*10^-3 ) / (0.0457 )^2
      =1.9*10^-5
Answer: 1.9*10^-5