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_(c))` for decomposition of `N_(2)O_(4)` at `127^(@)C`:

Options

A. `0.30`
B. `0.12`
C. `0.40`
D. `0.25`

Explanation

At `400 K`,
`K_(p)=((P'_(NO_(2)))^(2))/(P'_(N_(2)O_(4)))=([1.78xx(1.2a)/(1.6a)]^(2))/([1.78xx(0.4a)/(1.6a)])=4.005 atm`
`:. K_(c)=(K_(p))/((RT)^(Deltan))=(4.005)/((0.0821xx400)^(1))=0.12`