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)`

Answer

B

Explanation

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