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

Answer

D

Explanation

`2AB_2 hArr 2AB + B_2`
`{:(,1,0,0),("At eqm." ,1 - x ,x ,x//2):}`
`K_p = ((P_(AB))^2(P_(B_2)))/((P_(AB))^2), `
`implies K_p=([(x)/(1+x//2)xxP]^2 [(x//2)/(1+x//2)xxP])/([(1-x)/(1+x//2)xxP]^2)`
`implies K_p=(x^3P)/(2)`
`implies x = ((2K_p)/(P))^(1//3)`