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

Answer

B

Explanation

Consider the equilibrium expression
`{:(,2AB_(2)(g)hArr,2AB(g),B_(2)(g)),("Initial moles",a,0,0),("Equilibrium moles",a(1-x),ax,ax//2):}`
Total moles `= (a-ax)+ax+ax//2`
`= a+ax//2`
`P_(AB_(2))= (a-ax)/(a+ax//2)P= (a(1-x))/(a(1+x//2))P =(1-x)/(1+x//2)P`
`P_(AB) = (ax)/(a+ax//2)P = (ax)/(a(1+x//2))P = (x)/(1+x//2)P`
`P_(B_(2))=(ax//2)/(a+ax)P = (x//2)/(1+x//2)P`
`K_(P)= (P_(AB)^(2)P_(B_(2)))/(P_(AB_(2))^(2))`
`=(((x)/(1+x//2)P)^(2)((x//2)/(1+x//2)P))/(((1-x)/(1+x//2)P)^(2))`
Neglecting x with respect to 1, we have
`K_(p) = ((xp)^(2)((x)/(2)P))/(P^(2))=(x^(3))/(2)P`
or `x^(3) = (2K_(P))/(P)`
or `x= ((2K_(p))/(P))^(1//3)`