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)`

Explanation

`N_(2) (g)+O_(2) (g) Leftrightarrow 2NO(g), K_(1)=([NO]^(2))/([N_(2)] [O_(2)])`
`2NO(g)+O_(2) (g) Leftrightarrow 2NO_(2) (s), K_(2)=([NO_(2)]^(2))/([NO]^(2) [O_(2)])`
`N_(2) (g)+2O_(2) (g) Leftrightarrow 2NO_(2) (g), K_(1) xx K_(2)`
Now, `2NO_(2) (g) Leftrightarrow N_(2) (g)+2O(g), (1)/(K_(1) xx K_(2))`
Thus, `NO_(2) (g) Leftrightarrow 1/2 N_(2) (g)+O_(2) (g)`
`K=sqrt((1)/K_(1)) xx sqrt(1/K_(2)), K=[(1)/(K_(1) xx K_(2)]^(1//2)`