When `3.06 g` of solid `NH_(4)HS` is introduced into a two-litre evacuated flask at `27^(@)C, 30%` of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate `K_(c )` and `K_(p)` for the reaction at `27^(@)C`. (ii) What would happen to the equilibrium when more solid `NH_(4)HS` is introduced into the flask?

Explanation

`{:(,NH_(4)HS_((s)),hArr,NH_(3(g)),+,H_(2)S_((g))),("Initial mole",(3.06)/(51),,0,,0),("Mole at eqm.",(3.06)/(51)xx(70)/(100),,(3.06)/(51)xx(30)/(100),,(3.06xx30)/(51xx100)):}`
Given, `V=2 litre`, `T=27^(@)C`, `Deltan=2-0=2`
`K_(c)=[NH_(3)][H_(2)S]=(3.06xx30)/(51xx100xx2)xx(30.6xx30)/(51xx100xx2)`
`=8.1xx10^(-5)mol^(2)litre^(-2)`
Also `K_(p)=K_(c)(RT)^(Deltan)=8.1xx10^(-5)xx(0.082xx300)^(2)`
`=4.90xx10^(-2) atm^(2)`
Addition of more `NH_(4)HS` on this equilibrium will cause on effect because concentration of `NH_(4)HS` is not involved in formulae of `K_(p)` or `K_(c)`.