The following equilibria are given by
rho`N_(2)+3H_(2)hArr2NH_(3):K_(1)`
`N_(2)+O_(2)hArr2NO,K_(2)`
`H_(2)+1/2O_(2)hArrH_(2)O,K_(3)`
The equlibrium constant of the reaction
`2NH_(3)+5/2O_(2)hArr2NO+3H_(2)O`
in terms of `K_(1) and K_(3)` is

Explanation

Let solubility of `AgBr` be `xM`. Thus, `[Br^(Θ)] = xM`, but `[Ag^(o+)] != xM` since it will react with `NH_(3)` to form a complex and thus, its concentration will be decied by the disscoiation of complex. So, let `[Ag^(o+)] = yM`.
`AgBr hArr Ag^(o+)(aq) +Br^(Θ) (aq)`
`rArr K_(sp) = [Ag^(o+)] [Br^(-)] = yx = 5.0 xx 10^(-13)`
Since the formation constant `(K_(f))` of the complex is very high, assume that whole of `Ag^(o+)` formed is consumed.
`{:(Ag^(o+)+,2NH_(3)rarr,Ag(NH_(2))_(2).^(oplus),,),(x,0.4,,,),(,0.4-2x,x,,):}`
`{:(Ag(NH_(3))_(2)^(o+)hArr,Ag^(o+)(aq),+2NH_(3),(K_(d)=5.0xx10^(-13)),,),(x,,0.4-2x,,),(x-y,,y0.4-2x+2y,,):}`
Thus, `K_(d) = ([Ag^(oplus)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(oplus)])=(y(0.4-2x+2y)^(2))/(x-y) =6xx10^(-8)`
Assuming `x-y ~~` since `K_(d)` is low and `x lt lt 0.4`, we get :
`K_(d) = (y(0.4)^(2))/(x)`
Solving for `x` :
`x = 1.15 xx 10^(-3)M`
(Verify the approximation yourself).