At temperature T, a compound `AB_(2)(g)` dissociates according to the reaction `2AB_(2)(g) hArr 2AB(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. `(p_(T) alpha^(3))/(2)`
B. `(p_(T) alpha^(2))/(3)`
C. `(p_(T) alpha^(2))/(2)`
D. `(p_(T) alpha)/(3)`

Answer

A

Explanation

For the given equilibrium , the equilibrium concentration is
`{:(,2AB_(2)(g),hArr,2AB(g),+,B_(2)(g)),("Equilibrium concentration",c(1-alpha),,c alpha,,):}`
`therefore K_(p) = ((pB_(2))(p_(AB))^(2))/((p_(AB_(2))^(2))`. Substituting for `p_(AB), p(AB_(2))` and `p_(B_(2))`, in therms of mole fraction and `P_(T)`, we get
`K_(p) = ((c alpha//2)(c alpha)^(2))/([c(1-alpha)]^(2)) xx (p_(T))/([c(1+alpha//2)]) = (alpha^(3) xx p_(T))/(2(1-alpha)^(2)(1+(alpha)/(2)))`
As `alpha` is amll compared with unity , so `1- alpha ~= ` and `1 + (alpha)/(2) = 1`, therefore , `K_(p) = (alpha^(3) xx p_(T))/(2)`