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

Answer

B

Explanation

`{:(,2AB_(2)(g),hArr,2AB(g),+,B_(2)(g)),("Initial",2 mol,,0,,0),("At eqm.",2-2x,,2x,,x):}`
Total no. of moles at eqm. `=2+x`
`p_(AB_(2))=(2-2x)/(2+x)P,pAB=(2x)/(2+x)P`
`pB_(2)=(x)/(2+x)P`
`:. K_(p)=(((2x)/(2+x)P)^(2)((x)/(2+x)P))/(((2-2x)/(2+x)P))=(4x^(3))/((2+y)(2-x)^(2))P`
`=(4x^(3))/(2(2)^(2))P` (Neglecting x as compared to 2)
`=(x^(3))/(2)P ` or `x^(3)=(2K_(p))/(P)` or `x=((2K_(p))/(P))^(1//3)`