A 1 L reaction vessel contained1 mole each of solid NH_(4)HS, NH_(30 and H_(2)S at a temperature of 150^(@)C . When the decompositionof NH_(4)HS was carried out , equilibrium is established K_(p) value at that temperature is 100. Calculate the equilibrium partial pressures at which 60 per centdissociation of NH_(4)HS takes place at a lower temperature where K_(p) value is equal to 200 "atm"^(2) .

A 1 L reaction vessel contained1 mole each of solid NH_(4)HS, NH_(30 a

1.	A 1 L reaction vessel contained1 mole each of solid NH_(4)HS, NH_(30 and H_(2)S at a temperature of 150^(@)C . When the decompositionof NH_(4)HS was carried out , equilibrium is established K_(p) value at that temperature is 100. Calculate the equilibrium partial pressures at which 60 per centdissociation of NH_(4)HS takes place at a lower temperature where K_(p) value is equal to 200 "atm"^(2) .
Answer» SOLUTION :`NH_(4)HS_((s)) hArr NH_(3(g)) + H_(2)S_((g))` `1-x""1+x""1+x` `K_(p)=P_(NH_(3)) xx P_(H_(2)S)` Total no. of moles at EQUILIBRIUM `=(1+x)+(1+x)=2x+2` Partial pressure=MOLE fraction `xx`P Mole fraction of `NH_(3)=(1+x)/(2x+2)` `therefore ` Partial pressure of `NH_(3)=(1+x)/(2x+2)xxP` Mole fraction of `H_(2)S=(1+x)/(2x+2)` `K_(p)=((1+x)/(2x+x)xxP)((1+x)/(2x+2)xxP)` `K_(p)=((1+x)^(2))/((2x+2)^(2))=100` `((1+x)^(2)P^(2))/(4(1+x)^(2))=100 implies P^(2)=100xx4` `P=sqrt(400)=200 ` atm `P_(H_(2)S)=((1+x)20)/(2(x+1))=10`atm `P_(NH_(3))=10` atm per cent DISSOCIATION of `NH_(3) and H_(2)S` corresponds to moles each of `NH_(3) and H_(2)S` `K_(p)=P_(NH_(3))xxP_(H_(2)S)` `P_(NH_(3))=(1.6)/(3.2)xxp,P_(H_(2)S)=(1.6)/(3.2)xxP` `K_(P)=((1.6)xxP^(2))/((3.2)^(2))=400` `P^(2)=(200xx(3.2)^(2))/((1.6)^(2))=400` `P=sqrt(400)=20` atm