InterviewSolution
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InterviewSolution
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A container of volume `V L` contains an equilibrium mixture that consists of `2` mol each of gaseous `PCl_(5), Pcl_(3)` and `Cl_(2)` at `3` atm `T K`. Some `Cl_(2)` is added unitl the volume is double keeping `P` and `T` constant. Calculate moles of `Cl_(2)` added and `K_(p)` for `PCl_(5) rarr PCl_(3)+Cl_(2)` |
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Answer» `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2),),(2,,2,,2,P=3.0 "atm"):}` Total moles `=2+2+2=6` `p_(PCl_(5))=2xx3//6=1, p_(PCl_(3))=1` `K_(p)=((1)(1))/(1)=1 "atm"` When volume is double `(i.e., Vrarr 2V)` at constant `P` and `T` total moles brcome `=12` If volume is double, reaction will more backward at constant `P` and `T` `{:(PCl_(5),hArr,PCl_(3),+,Cl_(2),("Let x mole of" Cl_(2) "us added")),(2,,2,,2+x,("Let y mole of" PCl_(5) "is formed")),(2+y,,2-y,,(2+x)-y,):}` `p_(PCl_(5))=((2+y))/(12)xx3, p_(PCl_(3))=((2-y))/(12)xx3` `p_(Cl_(2))=((2+y)-y)/(12)xx3` `:. K_(p)=1=(((2-y)/(12))xx3xx[((2+x)-y)/(12)]xx3)/(((2+y)/(12))xx3) ...(i)` Also `2+y+2-y+(2+x)-y=12 ...(ii)` From equation (i) and (ii), solve, `y=2//3, xrArr 20//3` |
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