At temperature T, a compound `AB_(2)(g)` dissociates according to the reaction:
`2AB _(2) (g) hArr 2 AB (g) + B _(2) (g)` with a degree of dissociation ‘x’ which is small compared to unity. Deduce the expression for ‘x’ in terms of the equilibrium constant `K_(P)` and the total pressure P.

Options

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

Answer

C

Explanation

`{:(2AB_(2)(g),hArr,2AB(g),+,B_(2)(g),),(1,,0,,0,"molebefore dissociation"),((1-x),,x,,(x)/(2),"mole after dissociation"):}`
Total mole at equilibrium `= 1-x + x + (x)/(2) = 1 + (x)/(2)`. Now `K_(p) = (n_(B_(2)) xx n_(AB))/((N_(AB_(2)))) xx ((p)/(Sigma n))^(Delta n)`
`K_(p) = ((x)/(2) xx (x)^(2))/((1-x)^(2)) xx ((P)/(1+(x)/(2)))` or `K_(p) = (x^(3)P)/(2) because x` is small `therefore 11-x ~~ 1` and `1 + (x)/(2) ~~ 1` or `x = 3sqrt((2K_(p))/(P))`