# 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?

##### 1 Answer
Sep 28, 2014

${K}_{p} = 0.11 A t {m}^{2}$

Start by writing down the system which is at equilibrium:

$N {H}_{4} H {S}_{\left(s\right)} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(g\right)} + {H}_{2} {S}_{\left(g\right)}$

The solid reactant has zero partial pressure

So ${K}_{p}$= ${P}_{N {H}_{3}} . {P}_{{H}_{2} S}$

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

${P}_{N {H}_{3}} + {P}_{{H}_{2} S} = 0.66 A t m$

So ${P}_{N {H}_{3}} = 0.33 A t m$

and ${P}_{{H}_{2} S} = 0.33 A t m$

since they are formed in a 1:1 ratio

So

${K}_{p} = {\left(0.33\right)}^{2} = 0.11 A t {m}^{2}$