1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl_(2)` to the extent of 50% `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` Thus `K_(p)` is `:`A. 2.5B. 1.67C. 0.5D. 2

1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl

1.	1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl_(2)` to the extent of 50% `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` Thus `K_(p)` is `:`A. 2.5B. 1.67C. 0.5D. 2
Answer» Correct Answer - A `{:(,PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),("Initial",1"mol",,,,),("At eqm.",1-0.5,,0.5,,0.5):}` Total no. of moles at eqm. `=1.5` Moles of , `PCl_(5)=0.5` `PCl_(3)=0.5` `Cl_(2)=0.5` Initially `PV=nRT` `5xxV=1xxRT` `V=(RT)/(5)` At eqm. `PV=1.5 RT` `(PRT)/(5)=1.5RT` `P=7.5 "atm"` `:. p_(PCl_(5))=(0.5)/(1.5)xx7.5=2.5"atm"` `p_(PCl_(3))=2.5 "atm"` `p_(Cl_(2))=2.5"atm"` `K_(p)=(p_(PCl_(3))xxp_(Cl_(2)))/(p_(PCl_(5)))=(2.5xx2.5)/(2.5)=2.5"atm"`

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1 mole of `PCl_(5)` taken at 5 atm, dissociates into `PCl_(3)` and `Cl_(2)` to the extent of 50% `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)` Thus `K_(p)` is `:`A. 2.5B. 1.67C. 0.5D. 2

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