`0.1 mol` of `PCl_(5)` is vaporised in a litre vessel at `260^(@)C`. Calculate the concentration of `Cl_(2)` at equilibrium, if the equilibrium constant for the dissociation of `PCl_(5)` is `0.0414`.

Explanation

`{:(,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 1 xx (-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)`