At temperature T, a compound `AB_(2)(g)` dissociates according to the reaction
`2AB_(2)(g)hArr2AB(g)+B_(2)(g)`
with degree of dissociation `alpha`, which is small compared with unity. The expression for `K_(p)` in terms of `alpha` and the total pressure `P_(T)` is

Options

A. `(Px^3)/2`
B. `(Px^2)/3`
C. `(Px^3)/3`
D. `(Px^2)/2`

Answer

A

Explanation

`2AB_(2(g)) hArr2AB_((g)) +B_(2(g))`
`{:("Initially",1,0,0),("At equilibrium", (1–x) ,x, x//2):}`
Total no. of moles at equilibrium
`= (1-x) + x+ (x)/2 = (2+x)/2`
Partial pressure = mole fraction `xx` total pressure
Applying `K_p=(P_(AB)^2xxP_(B_2))/(P_(AB_2)^2)`
`=(((x)/((2+x)/2)xxP)xx(((x)/(2))/((2+x)/(2))xxP))/((((1-x)/(2+x))/(2)+P))=(px^3)/((2+x)(1-x)^2)`
Since `x lt lt 1` so `(1 - x)^2` can be neglected and `(2 + x)` can be taken as 2.
`:. K_p = (px^3)/2`