A given sample of `N_(2)O_(4)` in a closed vessel shows `20%` dissociation in `NO_(2)` at `27^(@)C` and `760 "torr"`. The sample is now heated upto `127^(@)C` and the analysis of mixture shows `60%` dissociation at `127^(@)C`.
The equilibrium constant `(K_(p))` for the decomposition of `N_(2)O_(4)` at `27^(@)C` is :

Options

A. `0.165`
B. `0.29`
C. `0.523`
D. `0.25`

Explanation

At `300K`, `P_(NO_(2))=(0.4a)/(1.2a)xx1=0.33 atm`
(`P'=P_(T)xx"Mole fraction"`)
`P_(N_(2)O_(4))=(0.8a)/(1.2a)xx1=0.66`
`:. K_(p)=((0.33)^(2))/(0.66)=0.165`