At temperature T, a compound `AB_(2)(g)` dissociates according to the reaction `2 AB _(2)(g) hArr 2 AB (g) + B _(2) (g)` with degree of dissociation `alpha`, which is small compared with unity. The expression for K, in terms of `alpha` and the total pressure `p_(1)` is

Options

A. `(p _(1)alpha ^(3))/( 2) `
B. `(p _(1) alpha ^(2))/( 3)`
C. `(p _(1) alpha ^(2))/( 2)`
D. `(p _(1) alpha)/(3)`

Answer

A

Explanation

For the given equilibrium, the equilibrium constanttration is
`2 AB _(2) (g) hArr 2 AB (g) + B _(2) (g) `
Equilibrium concentration `c (1-alpha )" " alpha alpha " "(c alpha )/(2)`
`therefore K _(P) = ((P _(B _(2))) (P _(AB )) ^(2))/( ( P _(AB _(2)))^(2)) .` Substituting for `P _(AB), P _(AB _(2))and P _(H_(2)),` in terms of mole fraction and `P _(T),` we get
`K _(P) = (( c alpha //2) xx ( c alpha ) ^(2))/( [ (1- alpha ) ]^(2)) xx (Pr )/([c (1 + alpha //2 )]) = ( alpha ^(3) xx P _(T))/( 2 (1- alpha ) ^(2) (1+ (alpha )/(2)))`
As ` alpha ` is small compared with unity, so `1 - alpha = 1 and 1 + alpha = 1,` therefore, ` K _(P) = (alpha ^(3) xx Pr)/( 2)`