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
(a)`(2K_(p)//p)`
(b)`(2K_(p)//p)^(1//3)`
(c)`(2K_(p)//p)^(1//2)`
(c)`(K_(p)//p)`

Options

A. `(2K_t//p)^(1//2)`
B. `K_p//p`
C. `2k_p//p`
D. `(2k_t//p)`

Answer

D

Explanation

`{:(, 2AB_(2 (g)) hArr 2AB_((g)) + B_(2(g))), ("Initial" ,2 " " 0 " " 0), ("Equilibrium", 2(1-x) " "2x " "x):}`
Moles at equilibrium = 2(1 -x ) + 2x + x
= 2- 2x + 2 x + x = x + 2 .
`K_(p) = ([P_(AB)]^(2)[P_(B_(2))])/([P_(AB_(2))]) = (((2x)/(x + 2) xxp)^(2)((x)/(2+x) xxp))/([(2(1-x))/(x+2) xx p]^(2))`
`= ((4x^(3))/(x+2) xx p)/(4(1-x)^(2)) = (4x^(2) xx p)/(2) xx (1)/(4)`
`x = ((2K_(p))/(p))^(1//3) " " `(as 1-x = 1 , 2 + x = 2)