The dissociation equilibrium of a gas AB, can be represented as The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant K,, and total pressure p is

Options

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

Answer

B

Explanation

Initial moles At equilibrium `underset(""_(2)^(1)(1-x))(2AB_(2)(g)) hArr underset(""_(2)^(0))(2AB(g))+underset(""_(x)^(0))(B_(2)(S))`
The moles at equilibrium `=2-2x+2x+x=(2+x)`
So, `P_(AB_(2))=(2(1-x)P)/((2+x)), P_(AB)=(2xp)/((2+x))`
`P_(AB_(2))=(xp)/((2+x))`
`K_(P)=((P_(AB))^(2)(P_(B_(2))))/(P_(AB_(2)))=(((2xp)/(2x+))^(2)((x)/(2+x))p)/([(2(1-x))/(2+x)p])=(x^(3)p)/((2+x)(1-x)^(2))`
`=(x^(3)p)/(2)[ :. x lt lt lt 1 and 2SO_(2), (1-x)~~ 1 and (2+x)~~2.1]`
`x=((2K_(P))/(P))^(1//3)`