The dissociation equilibrium of a gas `AB_(2)` can be represented as
`2AB_(2)(g)hArr 2AB(g)+B_(2)(g)`
The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium contant `K_(p)` and total pressure p is

Options

A. `(2K_(p)//P)^(1//2)`
B. `(K_(p)//P)`
C. `(2K_(p)//P)`
D. `(2K_(p)//P)^(1//3)`

Answer

D

Explanation

`underset(2(1-x))underset(2)(2AB_(2)(g))hArrunderset(2x)underset(0)(2AB(g))+underset(x)underset(0)(B_(2)(g))underset("at eq.")underset("initially")(.)`
Total amount of moles at equilibrium
`2=x(1-x)+2x+x=2+x`
`K_(p)=([P_(AB)]^(2)[P_(B_(2))])/([P_(AB_(2))]^(2))`
`K_(p)=([(2x)/(2+x)xxP]^(2)xx[(x)/(2+x)xxP])/([(2(1-x))/(2+x)xxP]^(2))`
`K_(p)=((4x^(3))/(2+x)xxP)/(4(1-x)^(2))`
`K_(p)=(4x^(3)xxP)/(2)xx(1)/(4)("As "1-x ~=1 and 2+x~=2)`
`x=(8K_(p)//4P)^(1//3)=(2K_(p)//P)^(1//3)`