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NH(4)HS(s)hArrNh(3)(g)+H(2)S(g) K(P(1) ...

`NH_(4)HS(s)hArrNh_(3)(g)+H_(2)S(g) K_(P_(1)`
`NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) K_(P_(2)`
`2` mol `NH_(4)HS(s)` is taken & `50%` of this is dissociated till at equilibrium in `1` litre container. Find `(K_(P_(2)^(2)))/(K_(P_(1)^(6))` if `0.25` moles of `N_(2)` are found finally.

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