On decomposition of `NH_(4)HS` , the following equilibrium is estabilished: `NH_(4) HS(s)hArrNH_(3)(g) + H_(2)S(g)` If the total pressure is P atm, then the equilibrium constant `K_(p)` is equal to

Options

A. P atm
B. `P^(2) atm^(2)`
C. `P^(2) // 4 atm^(2)`
D. 2P atm

Answer

C

Explanation

`P_(NH_(3)) = P_(H_(2)S) = (P)/(2)`
Hence `K_(p) = P_(NH_(3))xxP_(H_(2)S)`
`=(P)/(2)xx(P)/(2) = (P^(2))/(4)`