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?

Question

29049 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-28T00:27:17

#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#