A 6.0-l vessel was found to contain 1.0 mol brcl3, 2.0 mol br2 and 6.0 mol cl2. what is the equilibrium constant, kc, for this equilibrium mixture for the reaction 2brcl3 (g) ? br2 (g) + 3cl2 (g) ?

The equilibrium constant (Kc) for this equilibrium mixture for the reaction

2 BrCl3 ⇔ Br2 + 3 Cl2 is 11.94

calculation

Kc = (concentration of the product raised to their coefficients / concentration of the reactant raised to their coefficient)

Kc={ (Br2) Cl2) ^2} / (BrCl3) ^2

find the concentration of the reactant and products

=moles/volume in liters

BrCl3 = 1.0 / 6.0 = 0.167M

Br = 2.0 / 6.0 = 0.333 M

Cl2=6.0/6.0 = 1.00 M

Kc is therefore = { (0.333) (1.00) ^3 / (0.167) ^2} = 11.94