Question

Bromine chloride, BrCl, a reddish covalent gas with properties similar to Cl₂ , may eventually replace Cl₂ as a water disinfectant. One mole of Chlorine gas and one mole of Bromine gas are enclosed in a 5.00 Liter flask and allowed to reach equilibrium at a certain temperature. Show all work                       Cl₂ (g) + Br₂ (g) <-----> 2BrCl (g)      Kc= 4.7 x 10^-2

A) What percent of the Chlorine has reacted at equilibrium?

B) What weight (in grams) of BrCl(g) is present at equilibrium?

C) What would a decrease in volume shift the equilibrium position, if at all?

Answer 1

Initial concentration of Cl2 = initial concentration of Br2 = initial moles / volume in L

                                                                               = 1 mol / 5.00 L = 0.2 M

                           Cl₂ (g) + Br₂ (g) <-----> 2BrCl (g)      Kc= 4.7 x 10^-2

Initial conc            0.2         0.2                   0

Conc                    -c            -c                   +2c

Equb conc         0.2-c        0.2-c                  2c

Kc = [BrCl]2 / ([Cl2][Br2]) = 4.7 x 10^-2

       (2c)² / [(0.2-c)(0.2-c)] = 4.7 x 10^-2

    2c / (0.2-c) = 0.217

         2c = 0.043 - 0.217 c

          c = 0.0196 M

So Equilibrium concentration of Cl2 = 0.2-c = 0.18 M

    Equilibrium concentration of Br2 = 0.2-c = 0.18 M    

Equilibrium concentration of BrCl = 2c = 0.0392 M

A) percent of the Chlorine has reacted at equilibrium = (concentration of chlorine reacted /initial concentration )*100

= [(0.2 -0.18)/0.2]*100

= 10%

B) Equilibrium concentration of BrCl = 2c = 0.0392 M

Equilibrium moles of BrCl = Molarity x volume in L

                                     = 0.0392 M x 5.00 L

                                     = 0.196 moles

C) There is no decrease in volume shift since the number of moles of reactants are equal to the number of moles of products