Question

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?

Answer 1

`NH_(4)HS(s)hArrNH_(3)(g)+H_(2)S(g)`
moles of `NH_(4)HS=3.06/51=0.06`
Dehree of dissociation `=0.3`
At equilibrium,
`[NH_(3)(g)]=(0.3xx0.06)/2,[H_(2)S(g)]=0.3/2xx0.06`
`K_(c )=[NH_(3)(g)][H_(2)S(g)]=(0.3xx0.06xx0.3xx0.06)/(2xx2)`
`=8.1xx10^(-5)`
Now applying,
`K_(p)=K_(c )(RT)^(Deltan)=8.1xx10^(-5)xx(0.082xx300)^(2)=0.049`
Since `NH_(4)HS` is solid, so it causes no change in equilibrium.