N₂O₄ at an initial pressure of 2atm. and 300k dissociates to an extent of 20% at the same temperature by the time equilibrium is established. 

K_p for the reaction 2NO₂ $\rightleftharpoons$N₂O₄ is

find K_p for N₂O₄ $\rightleftharpoons$2NO₂ then reverse 

Calculation of K_p for N₂O₄ $\rightleftharpoons$2NO₂

20% of N₂O₄ is decomposed = $\frac{20\times 2}{100}=0.4$

pressure of each component present at equilibrium :

N₂O₄ = 2-0.4=1.6 atm

NO₂ = 2$\times$0.4 = 0.8 atm

K_p=$\frac{{\left[{\mathrm{P}}_{{\mathrm{NO}}_2}\right]}^2}{\left[{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}\right]}=\frac{0.8\times 0.8}{1.6}=0.4 \mathrm{atm}$

For 2NO₂$\rightleftharpoons$N₂O₄

${\mathrm{K}}_{\mathrm{p}}^{\text{'}}=\frac{1}{{\mathrm{K}}_{\mathrm{p}}}=\frac{1}{0.4}=2.5 {\mathrm{atm}}^{-1}$