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`

Options

A. 6.65 atm
B. 0.665 atm
C. 0.0665 atm
D. 66.5 atm

Answer

B

Explanation

`NH_(4) HS hArr NH_3 +H_2S`
`0.5 + x" " x`
or x (0.5 + x) = 0.11
or `x^2 + 0.5 x -0.11 =0`
or `x=-0.5pmsqrt((0.25+0.44))/(2)`
`=0.165`
`:. P_(NH_3) = 0.5 +0.165 =0.665.`