For the reaction,
`N_(2)O_(4)(g)hArr 2NO_(2)(g)`
the reaction connecting the degree of dissociation `(alpha)` of `N_(2)O_(4)(g)` with eqilibrium constant `K_(p)` is
where `P_(tau)` is the total equilibrium pressure.

Options

A. `alpha=(K_(p)//P)/(4+K_(P)//P)`
B. `alpha=(K_p)/(4+k_(P))`
C. `alpha=(K_(p)//P)/(4+K_(p)//P)^(1//2)`
D. `alpha=(K_(P))/(4+K_(P))^(1/2)`

Answer

C

Explanation

`N_(2)O_(4) (g) Leftrightarrow 2NO_(2) (g)`
`1-alpha " "2alpha , n_("total")=1+alpha`
`P_(N_(2)O_(4))=(1- alpha)/(1+alpha) P " "P_(NO_(2))=(2 alpha)/(1+alpha) P`
`K_(p)=(P^(2) NO_(2))/(P_(N_(2)O_(4))=((2 alpha)/(1+alpha) P)^(2) ((1+alpha)/(1-alpha )1/P)) =(4 alpha^(2))/(1-alpha^(2)) P`
Rearranging we get `alpha =(K_(p) //p)/(4+K_(p) //p)^(1//2)`