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. `( (2 K _(P))/( P ))^(1/2)`
B. `(( 3 K_(P))/( P ^(2)))^(1/2)`
C. ` (( 2 K _(P))/( P ))^(1/3)`
D. `(3 K _(P))/(P ^(2))`

Answer

C

Explanation

`2 AB _(2) (g) hArr 2 AB (g) + B _(2) (g)`
`{:(1,0, 0,"mole before dissociation"),( (1-x), x, (x)/(2),"mole alfer dissociation"):} `
Total mole at equilibrium `=1 - x + (x)/(2) = 1 + (x)/(2) . ` Now `K _(P) = (n _(B _(2)) xx n _(AB))/( ( N _(AB _(2)))) xx ((P )/( sum n )) ^( Delta n )`
`K _(P) = ((x )/(2) xx (x ) ^(2))/((1- x) ^(2)) xx ((P)/( 1 + (x)/(2) ) ) ^(1) or K _(P) = (x ^(3) P )/(2) because ` is small `therefore 1- x ~~ 1 and 1 + (x)/(2) ~~ 1 or root (3) ((2K _(P))/( P ))`