Some acid `NH_(4)HS` is placed in flask containing `0.5 atm` of `NH_(3)`. What would be pressures of `NH_(3)` and `H_(2)S` when equilibrium is reached?
`NH_(4)HS_((s))hArrNH_(3(g))+H_(2)S_((g))`, `K_(p)=0.11`

Answer

`0.17`

Explanation

`{:(,NH_(4) HS_((s)) hArr NH_(3(g)) + H_(2) S_((g))),("Initial pressure"," 0.5 0" ),("Pressure after dissociation"," (0.5 + P) P"):}`
Let pressure equivalent to P atm is developed by `NH_(3)` and `H_(2) S` on dissociation of `NH_(4) HS`, when 0.5 atm of `NH_(3)` is already present
`:. K_(P) = P._(NH_(3)) xx P._(H_(2)S) = (0.5) + P ) P`
or `0.11 = (0.5 + P) P rArr P = 0.1653` [`:. K_(P) =0.11` ]
`:. P._(NH_(3)) = (0.5 + P) = 0.5 + 0.1653 = 0.6653 ` atm
`P_(H_(2)S) = P = 0-.1653 ` atm