InterviewSolution
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InterviewSolution
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The degree of dissociation of `PCl_(5)` at `1` atm pressure is `0.2`. Calculate the pressure at which `PCl_(5)` is dissociated to `50%`? |
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Answer» Correct Answer - A::B::D `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("moles before",1,,0,,0),("dissociation",,,,,),("moles after",1-alpha,,alpha,,alpha),("dissociation",,,,,):}` Given `=0.2` at `1` atm temperature `:. K_(p)=(n_(PCl_(3))xxn_(Cl_(2)))/n_(PCl_(5))xx[P/(Sigman)]^(Deltan)` `=(alpha.alpha)/((1-alpha))[P/(1+alpha)]=(Palpha^(2))/(1-alpha^(2))=(1xx(0.2)^(2))/(1-(0.2)^(2))` `K_(p)=0.0416 "atm"` Again when `alpha` is desired at `0.5, K_(p)` remains constant and thus, `K_(SP)=(Palpha^(2))/(1-alpha^(2))` `0.0416=(Pxx(0.5)^(2))/(1-(0.5)^(2)), :. P=0.1248 "atm"` |
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