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. `(2K_(p)//p)`
B. `(2K_(p)//p)^(1//3)`
C. `(2K_(p)//p)^(1//2)`
D. `(K_(p)//p)`

Answer

B

Explanation

`{:(,2AB_(2)(g),hArr,2AB(g),+,B_(2)(g)),("Initial moles",1,,0,,0),("At equilibrium",2(1-x),,2x,,x):}`
Total moles at equilibrium
`=2-2x+2x+x=(2+x)`
So, `P_(AB_(2))=(2(1-x)p)/((2+x)), P_(AB)=(2xp)/((2+x)) P_(B^(2))=(xp)/((2+x))`
`K_(p)=((P_(AB))^(2)(P_(B_(2))))/((P_(AB))^(2))=(((2xp)/(2+x))^(2)((x)/(2+x))p)/([[2(1-x))/(2+x)p])`
`=(x^(3)p)/((2+x)(1-x)^(2))=(x^(3)p)/(2)`
[`:' x lt lt lt 1` and `2` So, `(1-x) ~~ 1` and `(2+x)~~2`.]
`x=((2K_(p))/(p))^(1//3)`