Given
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g),K_(1)`
`N_(2)(g)+O_(2)(g)hArr2NO(g),K_(2)`
`H_(2)(g)+(1)/(2)O_(2)hArrH_(2)O(g),K_(3)`
The equilibrium constant for
`2NH_(3)(g)+(5)/(2)O_(2)(g)hArr2NO(g)+3H_(2)O(g)`
will be

Options

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

Answer

D

Explanation

For equilibrium,
(a) `N_(2)(g) + 3H_(2)(g)hArr2NH_(3)(g)`
`K_(1) = ([NH_(3)]^(2))/([N_(2)][H_(2)]^(3))`
(b) `N_(2)(g) + O_(2)(g)hArr2NO(g)`
`K_(2) = ([NO]^(2))/([N_(2)][O_(2)])`
(c) `H_(2)(g) + (1)/(2)O_(2)(g)hArrH_(2)O(g)`
`K_(3) = ([H_(2)O])/([H_(2)][O_(2)]^(1//2))`
For reaction,
`2NH_(3)(g) + (5)/(2)O_(2)(g)hArr2NO(g) + 3H_(2)O(g)`
`K = ([NO]^(2))/([NH_(3)]^(2)[O_(2)]^(5//2))`
For getting K, we must do
`K_(3)^(3) = ([H_(2)O]^(3))/([H_(2)]^(3)[O_(2)]^(3//2))`
Form eqs. (i) (ii) and (v)
`K = (K_(2)xxK_(3)^(3))/(K_(1))`