Question

At 25°C in a closed system, ammonium hydrogen sulfide exists as the following equilibrium. NH4HS(s)<--->NH3(g) + H2S(g) When a sample of pure NH4HS(s) is placed in an evacuated reaction vessel and allowed to come to equilibrium at 25°C, total pressure is 0.660 atm. What is the value of Kp?

Answer 1

`K_p=0.11Atm^2`

Start by writing down the system which is at equilibrium:

`NH_4HS_((s))rightleftharpoonsNH_(3(g))+H_2S_((g))`

The solid reactant has zero partial pressure

So `K_p`= `P_(NH_(3)).P_(H_2S)`

Where `P` denotes the partial pressure of each gas.

`P_(NH_(3))+P_(H_2S)=0.66Atm`

So `P_(NH_(3))=0.33Atm`

and `P_(H_2S)=0.33Atm`

since they are formed in a 1:1 ratio

So

`K_p=(0.33)^2=0.11Atm^2`