0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculate the concentration of Cl_(2) at equilibrium, if the equilibrium constant for the dissociation of PCl_(5) is 0.0414.

0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculat

1.	0*1 mole of PCl_(5) is vaporised in a litrevessel at 260^(@)C.Calculate the concentration of Cl_(2) at equilibrium, if the equilibrium constant for the dissociation of PCl_(5) is 0.0414.
Answer» Solution : `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc. ",0.1 "mole",,0,,0),(" Conc. at EQM. (moles/litres)",(0.1 -x),,x,,x):}` Applying the law of chemical equilibrium, we get`K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]` Here `K_(c) = 0.0414 " (Given) "` `:. 0.0414 = (x XXX)/((0.1-x)) or x^(2)/(0.1 - x) = 0.0414 or x^(2) + 0.0414 x - 0.00414 = 0` `(x= (-0.0414 pm sqrt((0.0414)^(2) - 4 xx 1xx (-0.00414)))/2 ""[ "Using the formula " x = (-b pm sqrt(b^(2) - 4ac))/(2A)]` ( The negative value of x is meaningless and hence is rejected ) Thus, the concentration of `Cl_(2) " at equilibrium will be " 0.0468 mol L^(-1)`