Bromine monochloride is synthesized using the reaction Br2(g)+Cl2(g)↽−−⇀2BrCl(g)𝐾p=1.1×10−4 at 150 KBr2(g)+Cl2(g)↽−−⇀2BrCl(g)�p=1.1×10−4 at 150 K

Question

Bromine monochloride is synthesized using the reaction

Br2(g)+Cl2(g)↽−−⇀2BrCl(g)𝐾p=1.1×10−4 at 150 KBr2(g)+Cl2(g)↽−−⇀2BrCl(g)�p=1.1×10−4 at 150 K

A 203.0 L203.0 L flask initially contains 1.055 kg1.055 kg of Br2Br2 and 1.065 kg1.065 kg of Cl2.Cl2. Calculate the mass of BrClBrCl, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior.

mass of BrClBrCl:

TOOLS

x10^y

g

What is the percent yield of BrCl?BrCl?

percent yield:

Answer 1

Br₂(g) + Cl₂(g) <==> 2BrCl(g) .. Kp = 1.1x10^-4

Initial [Br₂] = 1.055 kg Br₂ x 1000 g / kg x 1mol / 159.8 g = 6.60 mol Br₂ / 203 L = 0.0325 M Br₂

Initial [Cl₂] = 1.065 kg Cl₂ x 1000 g / kg x 1 mol / 70.9 g = 15.02 mol Cl₂ / 203 L = 0.0740 M Cl₂

Br₂(g) + Cl₂(g) <==> 2BrCl(g)

0.0325........0.0740............0..........Initial

-x................-x..................+2x........Change

0.0325-x....0.0740-x.........2x.........Equilibrium

Kp = Kc(RT)^∆n

and since ∆n = 0, Kp = Kc = 1.1x10^-4

Kc = [BrCl]² / [Br₂][Cl₂]

1.1x10^-4 = (2x)² / (0.0325-x)(0.0740-x) = 2x² / (0.0325)(0.0740) = 2x² / 0.00241 (assuming x small)

2x² = 2.65x10^-7

x = 3.64x10^-4 (which is only about 1% of 0.0325 so assumption of x being small was valid).

[BrCl] @ equilibrium = 2x = (2)(3.64x10^-4) = 7.28x10^-4 mols/L

grams BrCl @ equilibrium = 7.28x10^-4 mols / L x 203 L = 0.148 mols x 115.4 g / mol = 17.1 g BrCl

This is the theoretical yield of BrCl.

How can we calculate % yield if we don't know the actual yield?