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Some acid NH(4)HS is placed in flask con...

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`

A

6.65 atm

B

0.665 atm

C

0.0665 atm

D

66.5 atm

Text Solution

Verified by Experts

The correct Answer is:
B
`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.`
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