Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))is

Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one

1.	Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))is
Answer» 0.13 `MOLL^(-1)` 0.05 `molL^(-1)` 0.065 `molL^(-1)` 0.1 `molL^(-1)` Solution : `K_(c)=([PCl_(3)](Cl_(2)])/([PCl_(3)])=0.13`