`N_(2)O_(4)` is 25% dissociated at `37^(@)C` and one atmosphere pressure. Calculate (i) Kp and (ii) the percentage dissociation at 0.1 atm and `37^(@)C` .

Explanation

`N_(2)O_(4)(g)hArr2NO_(2)(g)`
`{:("Initial",1,0),("At equilibrium",(1-alpha),2alpha):}`
Total moles`=(1-alpha)+2alpha=(1+alpha)`
`p_(N_(2)O_(4))=((1-alpha)/(1+alpha))P, p_(NO_(2))=(2alpha)/((1+alpha)).P`
Given, `alpha=0.25` and `P=1 "atm"`
`p_(N_(2)O_(4))=((1-0.25)/(1+0.25))xx1=0.6 "atm"`
`p_(NO_(2))=((2xx0.25)/(1+0.25))xx1=0.4 "atm"`
`K_(p)=((p_(NO_(2)))^(2))/p_(N_(2)O_(4))=(0.4xx0.4)/(0.6)=0.267 "atm"`
ii. Let the degree of dissociation of `N_(2)O_(4)` at `0.1 "atm"` be them
`p_(N_(2)O_(4))=((1-alpha)/(1+alpha))xx0.1` and `p_(NO_(2))=(2alpha)/((1+alpha))xx0.1`
`K_(p)=(((2alpha)/(1+alpha))^(2)xx(0.1)^(2))/(((1-alpha)/(1+alpha))xx0.1)=(4alpha^(2)xx0.1)/((1-alpha)(1+alpha))=(0.4alpha^(2))/((1-alpha^(2)))`
or `0.267=(0.4alpha^(2))/((1-alpha))` or `0.267=0.667 alpha^(2)`
`alpha=0.632`
Hence, dissociation of `N_(2)O_(4)=63.2%`