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» <html><body><p></p>Solution : `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),(" Intial conc. ",0.1 "mole",,0,,0),(" Conc. at <a href="https://interviewquestions.tuteehub.com/tag/eqm-446398" style="font-weight:bold;" target="_blank" title="Click to know more about EQM">EQM</a>. (moles/litres)",(0.1 -x),,x,,x):}` <br/> Applying the law of chemical equilibrium, we get`K_(c) = ([PCl_(3)][Cl_(2)])/([PCl_(5)]` <br/> Here `K_(c) = 0.0414 " (Given) "` <br/> `:. 0.0414 = (x <a href="https://interviewquestions.tuteehub.com/tag/xxx-1463707" style="font-weight:bold;" target="_blank" title="Click to know more about XXX">XXX</a>)/((0.1-x)) or x^(2)/(0.1 - x) = 0.0414 or x^(2) + 0.0414 x - 0.00414 = 0` <br/> `(x= (-0.0414 pm sqrt((0.0414)^(2) - <a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a> xx 1xx (-0.00414)))/2 ""[ "Using the formula " x = (-b pm sqrt(b^(2) - 4ac))/(<a href="https://interviewquestions.tuteehub.com/tag/2a-300249" style="font-weight:bold;" target="_blank" title="Click to know more about 2A">2A</a>)]` <br/> ( The negative value of x is meaningless and hence is rejected ) <br/> Thus, the concentration of `Cl_(2) " at equilibrium will be " 0.0468 mol L^(-1)`</body></html>