The equilibrium constant, `K_(p)` for the reaction :
`A hArr 2B`
is related to degree of dissociation `alpha` of A and total pressure P as :

Options

A. `(4alpha^(2)P)/(1-alpha^(2))`
B. `(4alpha^(2)P^(2))/(1-alpha^(2))`
C. `(4alpha^(2)P^(2))/(1-alpha)`
D. `(4alpha^(2)P)/(1-alpha)`

Answer

A

Explanation

`{:(,A,hArr,2B),("Initial conc.",1,,0),("Equal conc.",1-alpha,,2alpha):}`
(If `alpha` is degree of dissociation)
Total no. of moles `=1-alpha+2alpha=1+alpha`
`p(A)=(1-alpha)/(1+alpha)`. P (P is total pressure)
`p(B)=(2alpha)/(1+alpha).P`
`K_(p)=(((2alpha)/(1+alpha)P)^(2))/(((1-alpha)/(1+alpha)P))`
`=(4a^(2)P)/((1+alpha)(1-alpha))=(4a^(2)P)/(1-a^(2))`