`A_(2)(g)` and `B_(2)(g)` having partial pressures `60mm` of `Hg` & `45mm` of `Hg` respectively, are present in a closed uessel. At equilibrium, partial pressure of `AB(g)` is `28mm` of `Hg`. If all measurements are made under similar condition, then calculate percentage of dissociation of `AB(g)`.

Answer

72

Explanation

`A_(2_g)+B_(2_(g)hArr2AB_((g))`
`{:("at"t=0,60,42,0),("at"eq.,(60-x),(42-x),(2x)):}`
(Partial pressures)
`rArr2x=2B rArr x=14`
`K_(P)=((P_(AB)^(2))/((P_(A_(2)))(P_(B_(2)))))=((28)^(2))/((46)(28))=(14)/(23)`
For `Deltan_(g)=0 , K_(P)=K_(C_(1)={:(14),(23):}`
Let degree of dissociation for `AB` is `x`, then
`2AB(g)hArrA_(2_((g))+B_(2_((g))(K_(C_(2)))=(1)/(K_(C_(1)))=(23)/(14))`
`1 0 0`
at `t=0 (1-x) x x`
`(((x)/(2))((x)/(2)))/((1-x)^(2))=(23)/(14)to alpha=0.719`
Hence percentage of dissociatin`=0.719xx100=72%`