The following equilibrium are given :
`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_(2).K_(3)`
B. `K_(1)K_(2)//K_(3)`
C. `K_(1)K_(3)^(2)//K_(2)`
D. `K_(2)K_(3)^(3)//K_(1)`

Answer

D

Explanation

`K_(1)=([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
`K_(2)=([NO]^(2))/([N_(2)][O_(2)])`
`K_(3)=([H_(2)][O_(2)]^(1//2))/([H_(2)O])`
`K=([NO]^(2)[H_(2)O]^(3))/([NH_(3)]^(2)[O_(3)]^(3//2))=(K_(2)K_(3)^(2))/(K_(1))`.