PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed container and their concentrations are 0.8xx10^(-3)mol L^(-1), 1.2xx10^(-3)mol L^(-1) and 1.2xx10^(-3)mol L^(-1) respectively. The value of K_(c) for the reaction PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g) will be

PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed co

1.	PCl_(5), PCl_(3) and Cl_(2) are at equilibrium at 500 K in a closed container and their concentrations are 0.8xx10^(-3)mol L^(-1), 1.2xx10^(-3)mol L^(-1) and 1.2xx10^(-3)mol L^(-1) respectively. The value of K_(c) for the reaction PCl_(5)(g) hArrPCl_(3)(g)+Cl_(2)(g) will be
Answer» `1.8xx10^(-3) ` mol `L^(_1)` `1.8xx10^(-3)` `1.8xx10^(-3) L "mol"^(-1)` `0.55xx10^(4)` Solution :`K_(c)=([PCl_(3)][Cl_(2)])/([PCl_(5)]) = ((1.2xx10^(-3))xx(1.2xx10^(-3)))/(0.8xx10^(-3))=1.8xx10^(-3)`.