For the reaction `N_(2)(g)+O_(2)(g) hArr 2NO(g)`, the equilibrium constant is `K_(1)`. The equilibrium constant is `K_(2)` for the reaction
`2NO(g)+O_(2) hArr 2NO_(2)(g)`
What is `K` for the reaction
`NO_(2)(g) hArr 1/2 N_(2)(g)+O_(2)(g)`?

Options

A. `(1)/((K_(1)K_(2)))`
B. `(1)/((2K_(1)K_(2)))`
C. `(1)/((4K_(1)K_(2)))`
D. `(1/((K_(1)K_(2))))^(1//2)`

Answer

D

Explanation

`N_(2) + O_(2)hArr2NO,K_(1) = ([NO]^(2))/([N_(2)] [O_(2)])` …….(1)
`2NO + O_(2)hArr2NO_(2),K_(2) = ([NO_(2)]^(2))/([NO]^(2)[O_(2)])` …..(2)
Aim `NO_(2) rarr (1)/(2) N_(2) + O_(2), K_(c) = ([NO_(2)]^(1//2)[O_(2)])/([NO]^(2))` ....(3)
To get aim reverse eq. `(2)xx(1)/(2) +` eqn. `(1)xx(1)/(2)`
By observation `K_(c) = sqrt((1)/(K_(1) K_(2)))`