The following equilibrium constants are given :
` N_(2) + 3 H_(3) hArr 2 NH_(3) , K_(1)`
` N_(2) + O_(2) hArr 2 NO , K_(2)`
` H_(2) + 1/2 O_(2) hArr H_(2) O , K_(3)`
The equilibrium constant for the oxidation of `NH_(3)` by oxygen to give NO is :

Options

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

Explanation

For equilibrium `K_(1)=([NH_(3)]^(2))/([N_(2)] [H_(2)]^(3)) ........(i)`
`K_(2)=([NO]^(2))/([N_(2)] [O_(2)]) ..........(ii)`
`K_(3)=([H_(2)O])/([H_(2)] [O_(2)]^(1//2))......(iii)`
For a reaction, [oxidation of `NH_(3)` by oxygen to given NO]
`2NH_(3) (g) +5/2 O_(2) (g) Leftrightarrow 2NO(g)+ 3H_(2)O (g)`
`K=([NO]^(2) [H_(2)O]^(3))/([NH_(3)]^(2) [O_(2)]^(5//2)) ..........(iv)`
For getting K, we must do
`K_(3)^(3)=([H_(2)O]^(3) [O_(2)]^(3//2)) .........(v)`
from equation (i), (ii) and (v) `K=(K_(2) xx K_(3)^(3))/(K_(1))`