PCl_5,PCl_3 and Cl_2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 xx 10^(-3) "mol L"^(-1), 1.2 xx 10^(-3) "mol L"^(-1) and 1.2 xx 10 "mol L"^(-1), respectively. The value of K_c for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) will be

PCl_5,PCl_3 and Cl_2 are at equilibrium at 500 K in a closed container

1.	PCl_5,PCl_3 and Cl_2 are at equilibrium at 500 K in a closed container and their concentrations are 0.8 xx 10^(-3) "mol L"^(-1), 1.2 xx 10^(-3) "mol L"^(-1) and 1.2 xx 10 "mol L"^(-1), respectively. The value of K_c for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) will be
Answer» `1.8xx10^(3) "MOL L"^(-1)` `1.8xx10^(-3)` `1.8xx10^(-3)"mol L"^(-1)` `0.55xx10^4` Solution :For the REACTION, `PCl_(5) hArr PCl_(3) + Cl_2` At 500 K in a closed CONTAINER, `[PCl_5]=0.8 xx 10^(-3) "mol"^(-1)` `[PCl_5]=1.2xx10^(-3) "mol"^(-1)` `[Cl_2]=1.2xx10^(-3) "mol"^(-1)` `K_c=([PCl_3][Cl_2])/([PCl_5])=((1.2xx10^(-3))xx(1.2xx10^(-3)))/((0.8xx10^(-3)))` `=1.8xx10^(-3)`