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//(4K_1K_2)`
B. `[1//K_1K_2]^(1//2)`
C. `1//(K_1K_2)`
D. `1//(2K_1K_2)`

Answer

B

Explanation

`N_2(g)+O_2(g)hArr2NO(g),K_1` .....(i)
`2NO(g)+O_2(g)hArr2NO_2(g),K_2` .....(ii)
On adding Eqs.(i) and (ii)
`N_2(g)+2O_2(g)hArr2NO_2(g),K=K_1xxK_2` .....(iii)
On drividing (iii) by `(1)/(2) ` and on reversing we get,
`NO_2(g) hArr91)/(2)N_2(g) +O_2(g)`
So, `K=((N_2)^(1//2)(O_2))/(NO_2)`
`K=[(1)/(K_1K_2)]^(1//2)`