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?

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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

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