`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` .

Answer

(i) `0.266` atm (ii) `63.25%`

Explanation

`N_(2)O_(4)hArr2NO_(2)`
`{:(1,0,,,),(1-.25,.50,,,),(.75,.50,,,n_("total")=1.25):}`
`P_(N_(2)O_(4))=((P_(NO_(2)))^(2))/((P_(N_(2)O_(4))))=(.50//1.25)^(2)/((.75//1.25))=(.50xx.50)/(1.25xx.75)=(4)/(15)=0.266`
At pressure `0.1` atm
`N_(2)O_(4)hArr2NO_(2)`
`1 0 `
`(1-alpha) 2a`
`K_(P)=((1-alpha)/(1+alpha))xx0.1 , P_(NO_(2)=(((2alpha))/(1+alpha))xx0.1`
`K_(P)=((2alpha)/(1+alpha)xx0.1)^(2)/(((1-alpha)/(1+alpha))xx0.1), K_(P)=(4alpha^(2)xx0.1)/((1+alpha)(1-alpha))`
`0.266=(0.1xx4alpha^(2))/(1-alpha^(2))`
`0.665=(1+0.665)alpha^(2). rArr alpha=63.25%`