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(g))+O_(2(g))hArr2NO_((g)),K_(1)`
`(2NO_((g))+O_(2(g))hArr2NO_(2(g)),K_(2))/(N_(2(g))+2O_(2(g))hArr2NO_(2(g)),K=K_(1)xxK_(2))`
`therefore"For"NO_(2(g))hArr1/2N_(2(g))+O_(2(g)),K'[(1)/(K_(1)K_(2))]^(1/2).`