The dissociation equilibrium of a gas `AB_2` can be represented as, `2AB_2(g) hArr 2AB (g) +B_2(g)`. The degree of disssociation is 'x' and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant `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

`{:(,2mAB_(2) (g) ,hArr,2 AB (g),+,B_(2)(g)), (" Intial",2"moles",,0,,0), ("After disco.",2-2x,,2x,,2x),(,,,,,"Total"=2+x ):}`,
`p_(AB_(2))=(2-2x)/(2+x)P,p_(AB)= (2x)/(2+x)P, p_(B_(2)) = x/(2+x)P`
`K_(p) = (((2x)/(2+x)P)^(2) (x/(2+x)P))/(((2-2x)/(2+x)P) )=(4x^(3))/((2+x)(2-2x)^(2))P`
` = (4x^(3))/(2(2)^(2))P(" Neglecting x in comparison to 2 ")`
`x^(3)/2P or x^(3) = (2K_(p))/P or x = ((2K_(p))/P)^(1//3)`