PCl_(5) is kept in a closed container at a temperature of 250 K the equilibrium concentrations . PCl_(5) , PCl_(3) and Cl_(2) are 0.045 moles L^(-1) , 0.096 moles L^(-1) , 0.096 moles L^(-1) respectively . The value of equilibrium constant for the reaction PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) will be .............

PCl_(5) is kept in a closed container at a temperature of 250 K the eq

1.	PCl_(5) is kept in a closed container at a temperature of 250 K the equilibrium concentrations . PCl_(5) , PCl_(3) and Cl_(2) are 0.045 moles L^(-1) , 0.096 moles L^(-1) , 0.096 moles L^(-1) respectively . The value of equilibrium constant for the reaction PCl_(5) (g) hArr PCl_(3) (g) + Cl_(2) (g) will be .............
Answer» Solution :0.205 `K_(C) = ([PCl_(3)] [ Cl_(2)])/([PCl_(5)]) = (0.096 XX 0.096)/(0.045) = 0.2048 = 0.205`