1.

When 5.1g of solid NH_4HS is heated at 637K in a closed vessel of volume SL, the given equilibrium establishes: NH_4HS(s)--NH_3(g)+H_2(g)+H_2S(g)+H_2S(g). If 3.06g of NH_4HS remains at equilibrium, then find the value of K_p

Answer»

Solution :Amount of `NH_4HS` DISSOCIATED`=(5.1-3.06)g=2.04g`
2.04g of `NH_4HS=(2.04)/(51)-=0.04mol" of "NH+4HS`
`NH_4HS(s) At equilibrium:0.04 mol0.4 MOL
`K_c=[NH_3][H_2S]=((0.04)/(5))((0.04)/(5))`
`=6.4xx10^(-5)mol^2*L^(-2)`
`K_p=K_c(RT)^(/_\n)=K_c(RT)^2`
`=6.4xx10^(-5)(0.0821xx637)^2=0.175`


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