Question

Bromine monochloride is synthesized using the reaction Br2(g) + Cl2(g) = 2 BrCl(g) Kp = 1.1 x 10-4 at 150 K A 203.0 L flask initially contains 1.070 kg of Br, and 1.115 kg of Cl. Calculate the mass of BrCl, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior. mass of BrCl: g What is the percent yield of BrCl? percent yield: %

Answer 1

The given reaction is

         Br2 (g) + Cl2(g) $\rightleftharpoons$ 2 BrCl (g)

The reactants and products are all gaseous species.

$\Delta$n (g) = moles of gaseous products - moles of gaseous reactants = 2 - ( 1+1) = 0

Hence Kp = Kc = 1.1 x 10^-4 at 150 K

Mass of Br2 = 1.070 Kg = 1070 g

moles of Br2 = 1070 / 159.8 moles

Concentration of Br2 = moles / Volume = 1070 / [159.8 x 203] [ since volume = 203 L]

initial conc of Br2= 0.03298 M

Similarly

Mass of Cl2 = 1.115 Kg = 1115 g

moles of Cl2 = 1115 / 70.9 moles

Concentration of Cl2 = moles / Volume = 1115 / [70.9 x 203] [ since volume = 203 L]

initial conc of Cl2 = 0.07747 M

1)

Let 2x be the equilibrium concentration of BrCl. Then by ICE table

+ Cl2 ► 2Brci 0.07747 0 Br2 Initial 0.03298 Change -X Equilibrium 0.03298 - X -X +2x 0.07747-X 2x

Kp = Kc = [2x]^2 / [0.03298-x] [0.07747-x] = 1.1 x 10^-4

         since x is small we can approximate[ 0.03298 -x] as 0.03298 and [0.07747 - x ] as 0.07747

Therefore 4x^2 = 0.03298 x 0.07747 x 1.1 x 10^-4

                     = 28.1 x 10^-8

               x^2 = 7.025 x 10^-8

                  x = 2.65 x 10^-4

Thus the concentration of BrCl = 2x = 2 x 2.65 x 10^-4 M = 5.3 x 10^-4 M

Hence the equilibrium conc of BrCl in the mixture = 5.3 x 10^-4 M

Therefore total moles present in 203 Litres = 203 x 5.3 x 10^-4 moles

                                                              = 0.1076 moles

Hence the mass of BrCl in the mixture of gases = 0.1076 x 115.35 = 12.41 g

Thus the mass of BrCl in the mixture at equilibrium is 12.41 g

2)

Now since the limiting reagent is Br2, if the reaction goes to 100% completion, the product formed is

         [ 2 x 0.03298] M of BrCl

        = 0.06596 M of BrCl

Therefore moles of BrCl in 203L of the gas = 0.06596 x 203 = 13.39 moles

Hence the mass of BrCl = 13.39 moles x 115.35 g / mol = 1545 g

Thus the theoretical amount of BrCl = 1545 g.

Hence the percentage yield is 12.41 x 100 / 1545 = 0.8%

Thus the % yield is 0.8%