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

`{:(,2AB_(2)(g),hArr,2AB(g)+,B_(2)(g)),("Initial",1,,0,0),("moles",,,,),("At equil.",2(1-x),,2x,x):}`
where, x = degree of dissociation
Total moles at equilibrium
`=2-2x+2x+x=(2+x)`
So, `p_(AB_(2))=(2(1-x)p)/((2+x)), p_(AB)=(2x p)/((2+ x))`
`p_(B_(2))=(x p)/((2+x))`
`K_(p)=((p_(AB))^(2)(p_(B_(2))))/((p_(AB_(2)))^(2))=(((2x p)/(2+x))^(2)[((x)/(2+x))p])/([((2(1-x))/((2+x)))p]^(2))`
`=(4x^(3)p^(3))/((2+x)^(3))xx((2+ x)^(2))/(p^(4)4(1-x)^(2))=(x^(3)p)/((2+x)(1-x)^(2))`
`=(x^(3)p)/(2) " " [because x lt lt lt 1 "and" 2]`
`x=((2K_(p))/(p))^(1//3)`
so, `(1-x)~~1`
`(2+x)~~2`