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

Answer

D

Explanation

`2AB_(2) (g) Leftrightarrow 2AB(g)+B_(2) (g)`
`{:(,"At t=0"," 2 moles",-,-),(,"At Eqm.",2(1-X)=2,2x,x):}`
Total moles at equilibrium =2+x=2, (x is very small)
`K_(p)=(P_(AB)^(2).P_(B_(2))/(P_(AB_(2))^(2))=((2x)/(2)P)^(2) xx (x/2 P))/(2/2P)^(2)`
`=(x^(2)P^(2). x/2P)/(P^(2))= (x^(3)P)/(2), x=((2K_(p))/(P))^(1//3)`