(i) Mole of solid NH4​HS taken initially =513.06​=0.06 At equilibrium NH4​HS(s)⇌0.018NH3​( g)​+0.018H2​ S(g)​Kc​p(NH3​)Kp​​=(20.018​)2=8.1×10−5=20.018×0.082×300​=0.22 atm=(0.22)2=4.84×10−2​(ii) Addition of solid NH4​HS will have no effect on equilibrium.