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`

Answer

0.665 atm , 0.1653 atm

Explanation

`NH_4NS(s) hArr NH_3(g) + H_2S(g)`
Initially 0.5 atm of `NH_3` was present and `NH_4HS(s)` is added to the flask. Let at equilibrium , `NH_4HS` dissociates to give .a. moles of `NH_3` of .a. moles of `H_2S`.
`therefore p_(NH_3)`=(0.5+a)
`p_(H_2S)=a`
`K_p=p_(NH_3)xxp_(H_2S)`
0.11=(0.5 + a) x (a)
`a^2+0.5 a -0.11 =0`
Solving we get , a=0.1653 atm
`therefore p_(NH_3)` = 0.1653+0.5=0.6653 atm
`p_(H_2S)` = 0.1653 atm