PCl_5, PCl_3 and Cl_2 are at equilibrium at 500 K and having concentration 1.59 M PCl_3, 1.59 M Cl_2 and 1.41 M PCl_5. Calculate K_c for the reaction , PCl_5 hArr PCl_3 + Cl_2 .

PCl_5, PCl_3 and Cl_2 are at equilibrium at 500 K and having concentra

1.	PCl_5, PCl_3 and Cl_2 are at equilibrium at 500 K and having concentration 1.59 M PCl_3, 1.59 M Cl_2 and 1.41 M PCl_5. Calculate K_c for the reaction , PCl_5 hArr PCl_3 + Cl_2 .
Answer» SOLUTION :`{:("Equilibrium reaction :",PCl_5 hArr, PCl_3 + ,Cl_2),("Equilibrium CONCENTRATION:",1.41 M, 1.59 M, 1.59 M):}` Equilibrium concentration is mol `L^(-1)` means molarity (M). `therefore K_c=([PCl_3][Cl_2])/([PCl_5])` `=((1.59 M)(1.59 M))/(1.41 M)` = 1.7930 M Additional calculation -1 : Equilibrium constant of reverse reaction. `K._c=1/K_c=1/1.793 = 0.5577 M^(-1)` Additional calculation -2: Calculation `K_p` . R=0.0831 bar L `mol^(-1) K^(-1), K_p = K_c(RT)^(Deltan)` Where, `K_c`= 1.7930 mol `L^(-1)` R=0.0831 bar L `mol^(-1) K^(-1)` T=500 K `Deltan`=(1+1) -1 =+1 `therefore K_p`=(1.7930 mol `L^(-1)` ) [(0.0831 bar L `mol^(-1) K^(-1)` ) (500 K) ] = 74.499 bar `approx` 74.5 bar Thus, `K_p gt K_c` because `Deltan gt 0`