In the reaction `AB_(2)(l)+3X_(2)(g)hArrAX_(2)(g)+2BX_(2)(g)+2BX_(2)(g)Delta H= -270` kcal per mol. of `AB_(2)(l)`.
the enthalpies of formation of `AX_(2)(g) & BX_(2)(g)` are in the ratio of `4:3` and have opposite sign. The value of `Delta H_(f)^(0)(AB_(2)(l))= +30` kcal/mol. Then

Options

A. `DeltaH_(f)^(0)(AX_(2))= -96 kcal//mol`
B. `DeltaH_(f)^(0)(BX_(2))= +480 kcal//mol`
C. `K_(p)=K_(c) & Delta H_(f)^(0)(AX_(2))= + 480 kcal//mol`
D. `K_(p)=K_(c)RT & DeltaH_(f)^(0)(AX_(2))+DeltaH_(f)^(0)(BX_(2))= -240 kcal//mol`

Answer

C

Explanation

`AB_(2)(l)+3X_(2)(g)hArrAX_(2)(g)+2BX_(2) " "DeltaH= -270 kcal`
`Delta H_(F)^(o)(AX_(2))+2DeltaH_(F)^(o)(BX_(2))-Delta H_(F)^(o)(AB)_(2)= -270`
`DeltaH_(F)^(o)(AX_(2))+2 DeltaH_(F)^(o)(BX_(2))= -240 rArr 4x-6x= -240`
`rArr -2x= -240 " "rArr 4x-6x= -240`
`Delta H_(F)^(o)(AX_(2))=4xx120" "rArr x=120`
`DeltaH_(F)^(o)(AX_(2))=4xx120=480` Kcal/mole
`K_(p)=K_(c)(RT)^(Delta n g)`
`Delta n_(g)=0` So, `K_(p)=K_(c)`