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Some solid NH(4)HS is placed in flask co...

Some solid `NH_(4)HS` is placed in flask containing `0.5` atm of `NH_(3)`. What would be the pressure of `NH_(3)` and `H_(2)S` when equilibrium is reached.
`NH_(4)HS(g) hArr NH_(3)(g)+H_(2)S(g), K_(p)=0.11`

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`NH_(3) and H_(2)S ` produced by the decomposition of `NH_(4)HS` will be same . Suppose at equilibrium each has pressure = p atm due to decompistion of ` NH_(4)HS` . Then
` p_(H_(2)S) = p atm `
` p_(NH_(3) ) p + 0*5 " " ( :' NH_(3) " is already present at" 0*5 atm )`
Applying law of chemical equilibrium to the given reaction
`p_(NH_(3)) xx p_(H_(2)S) = K_(p)`
` :. p xx (p + 0*5) = 0*11`
or ` p^(2) + 0*5 p = 0*11`
or `p^(2) + 0*5 p - 0*11 = 0 `
` :. p= -0*5 pm sqrt((0*5)^(2)-4 (-0*11))/2 = (0*5 pm sqrt (0*25 + 0*44))/2 = -(0*5 pm sqrt(0*69))/2`
` ( -0*5 pm 0*83 )/2 = (0*33)/2 = 0*165 ("Neglecting -ve value ")`
` :. p_(H_(2)S) = 0* 165 "atm " , p_(NH_(3)) = 0*665 " atm "`
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