What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of 1Cl was 0.78 M ? 2I Cl (g) hArr I_(2) (g) + Cl_(2) (g), K_(C) = 0.14

What is the equilibrium concentration of each of the substances in the

1.	What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of 1Cl was 0.78 M ? 2I Cl (g) hArr I_(2) (g) + Cl_(2) (g), K_(C) = 0.14
Answer» Solution :Suppose at EQUILIBRIUM the molar CONCENTRATION of both `I_(2) (g)` and `Cl_(2) (g)` is x mol `L^(-1)` `{:( ,2I Cl (g) , HARR , I_(2) (g) + Cl_(2) (g)) , ("Initial molar conc." , 0.78 M , , 0 "" 0), ("Eqn. molar conc." , (0.78 - 2x) M , , x "" x):}` `K_(C) = ([I_(2) (g)][Cl_(2) (g)])/([I Cl (g)]^(2)) = ( (x) xx (x))/((0.78 - 2x)^(2))` `(x)/((0.78 - 2x)) = (0.14)^((1)/(2)) = 0.374` or `x= 0.374 (0.78 - 2x)` `x = 0.292 - 0.748x ` or `x = 1.748 x = 0.292 , x = (0.292)/(1.748) = 0.167` `[I Cl] = (0.78 - 2 xx 0.167) = (0.78 - 0.334) = 0.446 M` `[I_(2)]= 0.167 M , [Cl_(2)] = 0.167 M`