1.

Calculate K_(p) for the equilibrium, NH_(4)HS_((s))iffNH_(3(g))+H_(2)S_((g)) if the total pressure inside the reaction vessel is 1.12 atm at 105^(@)C.

Answer»

`0.56`
`1.25`
`0.31`
`0.63`

Solution :`{:(,NH_(4)HS_((s)),iff,NH_(3(g)),+,H_(2)S_((g))),("LET initial moles",1,,0,,0),("At equilibrium",(1-x),,x,,x):}`
TOTAL GASEOUS moles at equilibrium = `x+x=2x`
We know `K_(p)=p_(NH_(3))xxp_(H_(2)S)`
but partial pressure (p) = mole fraction `xx` total pressure (P)
`K_(p)=((x)/(2x)xxP)((x)/(2x)xxP)=((P)/(2))^(2)=((1.12)/(2))^(2)=0.3136`
`II^(nd)` method : Both `NH_(3)andH_(2)S` have same number of moles at equilibrium so have same mole fraction and thus equal partial pressures.
i.e., `p_(NH_(3))=p_(H_(2)S)=(1.12)/(2)`
`K_(p)=p_(NH_(3))xxp_(H_(2)S)=(1.12)/(2)xx(1.12)/(2)=0.3136`


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