The decomposition of ammonium hydrogen sulfide NH4HS(s)⇔NH3+H2S(g) is an endothermic process. A 6.1589-g sample of thesolid is placed in an evacuated 4.000-L vessel at exactly 24∘C. After equilibrium has been established, the totalpressure inside is 0.709 atm. Some solid NH4HS remains in thevessel. (a) What is the Kp for the reaction? (b) What percentage ofthe solid has decomposed? (c) If the volume vessel were doubled atconstant temperature, what would happen to the amount of solid inthe vessel?

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Faiza Fuller · Accepted Answer

a). For the reaction - : NH4HS(s)⇔NH3+H2S(g) initialP 0 0 0 finalP 0+x+x atequilibrium 0 x x k p=P NH3PH2S PT=PNH3+PH2S=0.709 atm 2x=0.709atm x=0.3545atm kp=x2=(0.3545)2=0.1257 b). step1- calculate kc=kp/(0.0821L−atm/K.mol)(297K)?n=0.1257/(0.0821L−atm/K.mol)(297K)2=0.0002114 kc=[NH3][H2S]=Y2 Y=0.0145M [NH3]=[H2S]=0.0145M step2. to calculatemoles moles of NH3=molarity×volume(L) =0.0145M×4.0L=0.05816mol moles of NH3+H2S=0.1163mol moles of NH4HS=6.1589g/(51.13g/mol)=0.1205mol moles of NH4HS decomposed =0.1163mol massof NH4HS decomposed =0.1163mol×51.13g/mol =5.946g percentage of solid decomposed =5.946g6.1589g×100% =96.55% c). When the volume of the vessel is doubled ,pressure of the gaseous product decreases but the amount ofsolid in the vessel remains constant.

The decomposition of ammonium hydrogen sulfideNH4HS(s) <===> NH3 + H2S(g)is an endothermic process.

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