At temperature T, a compound `AB_2(g)` dissociation according to the reaction, `2AB _2(g) hArr 2AB(g) +B_2(g)` with degree of dissociation, `alpha`, which is small compared to unity . Deduce the expression for `alpha` in terms of the equilibrium constant `K_p` and the total pressure P.

Options

A. `(2K_P//P)^(1//2)`
B. `K_P//P`
C. `2K_P//P`
D. `(2K_P//P)^(1//3)`

Explanation

`2AB_(2) (g) Leftrightarrow 2AB(g)+B_(2) (g)`
`{:(,t_(0),2,0,0),(,t_(eq),2(1-x),2x,"x total moles =2+x"):}`
`"Partial pressure "(2(1-x))/(2+x) xx P" "(2x)/(2+x) xx P" "(x)/(2+x) xx P`
`K_(p)=p_(AB)^(2) xx p_(B_(2))/(p_(AB_(2)))= ([((2)/((2+x))P)]^(2) [(xP)/((2+x))])/([(2(1-x))/(2+x) xx P])`
`=(4x^(3) xx P)/(4(2+x) (1-x)^(2)`
`(2+x)=2, (1-x)=1`
`K_p=(Px^(3))/(2)`
`therefore x=(2K_(p)//P)^(1//3)`