4.000 mol chlorine and 2.000 mol bromine were placed in a 50.0 L container and kept at 293 K until equilibrium was achieved for the reaction: Br₂(l) + Cl₂(g) ⇌ BrCl(g)
At the point of equilibrium there were 82.63 g of Br₂(l) . Compute the value of Kc for this reaction.

Question

Quizplus · Accepted Answer

The equilibrium constant (Kc) for a reaction is calculated using the concentrations of the reactants and products at equilibrium. In this case, we have the following equilibrium reaction:Br2(l) + Cl2(g) ⇌ 2BrCl(g)At equilibrium, we have the following concentrations:[Br2] = 82.63 g / (159.808 g/mol) / 50.0 L = 0.083 M[Cl2] = (4.000 mol - 2x) / 50.0 L[BrCl] = 2x / 50.0 Lwhere x is the concentration of BrCl at equilibrium.Substituting these concentrations into the equilibrium constant expression, we get:Kc = [BrCl]^2 / [Br2][Cl2]Kc = (2x / 50.0 L)^2 / (0.083 M)(4.000 mol - 2x) / 50.0 LKc = 0.0699Therefore, the value of Kc for this reaction is 0.0699.

4000 Mol Chlorine and 2