The following equilibria are given by `:`
`N_(2)+3H_(2) hArr 2NH_(3), K_(1)`
`N_(2)+O_(2) hArr 2NO,K_(2)`
`H_(2)+(1)/(2)O_(2) hArr H_(2)O, K_(3)`
The equilibrium constant of the reaction `2NH_(3)+(5)/(2)O_(2)hArr 2NO +3H_(2)O` in terms of `K_(1) , K_(2)` and `K_(3)` is

Options

A. ` K_(1) K_(3)^(3) //k_(2)`
B. ` K_(2) K_(3)^(3) //K_(1)`
C. `K_(2) K_(3) //K_(1) `
D. `K_(2)^(3) K_(3) // K_(1)`

Answer

B

Explanation

(i)`N_(2) + 3 H_(2) hArr 2 NH_(3), K_(1) ([NH_(3)]^(2))/([N_(2)] [ H_(2)]^(3))`
(ii) `N_(2) +O_(2) hArr 2 NO, K_(2) = ([NO]^(2))/([N_(2)] [ O_(2)])`
(iii) ` H_(2) +1/2 O_(2) hArr H_(2)O, K_(3)= ([H_(2)O])/([H_(2)] [ O_(2)]^(1//2))`
Aim : ` 2NH_(3)+ 5/2 O_(2) hArr 2 NO + 3 H_(2)O,`
`K = ([NO]^(2)[H_(2)O]^(3))/([NH_(3)]^(2) [O_(2)]^(5//2))`
Thus, `K = K_(2) xx K_(3)^(3)//K_(1)`