The equilibrium for the dissociation of `XY_(2)` is given as,
`2XY_(2)(g)hArr2XY(g)+Y_(2)(g)`
if the degree of dissociation x is so small compared to one. Show that
`2K_(p)=PX^(3)` where P is the total pressure and `K_(P)` is the dissociation equilibrium constant of `XY_(2)`.

Explanation

The equilibrium for the dissociation of `XY_(2)` is given as,
`2XY_(2)(g)hArr2XY(g)+Y_(2)(g)`

Total no. of moles `=1-x+x+(X//2)`
`=1+(X//2)~=1`
`[because"Given that"xlt lt1,-x~=1" and "1+(X//2)~=1]`
`K_(P)=((P_(XY))^(2)(P_(Y_(2))))/((P_(XY_(2)))^(2))=((X/1xxP)^(2)((X//2)/(1)xxP))/((1/1xxP)^(2))`
`K_(P)=(X^(3)P)/(2)`
`rArr2K_(P)=X^(3)P`