The reaction, PCI_(5)hArrPCI_(3)+CI_(2) is started in a five litre container by taking one mole of PCI_(5). If 0.3 mol PCI_(5) is there at equilibrium, concentration of PCI_(3) "and" K_(c) will respectively be:

The reaction, PCI_(5)hArrPCI_(3)+CI_(2) is started in a five litre con

1.	The reaction, PCI_(5)hArrPCI_(3)+CI_(2) is started in a five litre container by taking one mole of PCI_(5). If 0.3 mol PCI_(5) is there at equilibrium, concentration of PCI_(3) "and" K_(c) will respectively be:
Answer» `0.14,(49)/(150)` `0.12,(23)/(100)` `0.07,(23)/(100)` `20,(49)/(150)` Solution :`{:(PCI_(3),hArr,PCI_(3),+,CI_(2),),(1,,0,,0,"Initial mole"),((1-0.7)/(5),,(0.7)/(5),,(0.7)/(5),"Conc. at equilibrium"):}` Total mole of `PCI_(3)=0.7` Concentration`=0.14` `K_(c)=(X^(2))/((1-x)V)=(0.7xx0.7)/(0.3xx5)=(49)/(150)`