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InterviewSolution
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In the dissociation of `PCl_(5)` as `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)` If the degree of dissociation is `alpha` at equilibrium pressure P, then the equilibrium constant for the reaction isA. `K_(p)=(alpha^(2))/(1+alpha^(2)P)`B. `K_(p)=(alpha P^(2))/(1-alpha^(2))`C. `K_(p)=(alpha^(2)P^(2))/(1-alpha^(2))`D. `K_(p)=(alpha^(2)P)/(1-alpha^(2))` |
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Answer» Correct Answer - D `{:(,PCl_(5),hArr,PCl_(3),+,Cl_(2)),("Initial moles",1,,-,,-),("Equilibrium moles :",1-alpha,,alpha,,alpha):}` Total moles `=1- alpha + alpha + alpha = (1+alpha)` Total pressure `=P` `:. P_(PCl_(5))=((1-alpha)P)/((1+alpha))` `P_(PCl_(3))=(alphaP)/((1+alpha)),P_(Cl_(2))=(alphaP)/((1+alpha))` Now `K_(p)=(P_(Cl_(3))xxP_(Cl_(2)))/(P_(PCl_(5)))=(alpha^(2)P^(2))/((1+alpha)(1-alpha)P)` `=(alpha^(2)P)/(1-alpha^(2))` |
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