`N_(2)O_(4)(g)` is dissociated to an extent of `20%` at equilibrium pressure of `1.0` atm and `57^(@)C`. Find the percentage of `N_(2)O_(4)` at `0.2` atm and `57^(@)C`.

Answer

A, D

Explanation

`{:(,N_(2)O_(4)(g), hArr,2NO_(2)(g)),(underset(("moles"))(t=0),a,,0),(t=t_("eq"),3-aalpha,,2aalpha):}`
`K_(p)=(P_(NO_(2))^(2))/(P_(N_(2)O_(4)))=(((2aalpha)/(a+aalpha).P)^(2))/((a-aalpha)/(a+aalpha)P)=(4alpha^(2)P)/(1-alpha^(2))`
`K_(p)` will remain same as is constant `rArr (4alpha_(1)^(2)P_(1))/(1-alpha_(1)^(2))=(4alpha_(2)^(2)P_(2))/(1-alpha_(1)^(2))`
`[alpha_(1)=0.2, P_(1)=1.0 "atm" P_(2)=0.2 "atm"]`
`rArr alpha_(2)=0.41`