Bromine monochloride is synthesized using the reaction Br2(g)+Cl2(g)↽−−⇀2BrCl(g)

Question

𝐾p=1.1×10−4 at 150 K A 202.0 L flask initially contains 1.033 kg of Br2 and 1.028 kg of Cl2. Calculate the mass of BrCl, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior.

What is the percent yield of BrCl?

Answer 1

Br₂(g) + Cl₂(g) <==> 2BrCl(g) ... balanced equation

First, we'll convert mass (kg) to moles and then to moles/liter (M).

Next we'll construct an ICE table to find the equilibrium concentrations of each gas.

Finally, we convert the moles of BrCl into grams.

(NOTE: you could also use PV = nRT and solve for P of each reactant and use the ICE table in units of pressure. In the end you would solve for pressure of BrCl and convert to moles and convert to grams)

moles Br₂ = 1.033 kg x 1000 g/kg x 1 mol/159.8 g = 6.464 moles

[Br₂] = 6.464 moles / 202.0 L = 0.03200 M

moles Cl2 = 1.028 kg x 1000 g/kg x 1 mol /70.91 g = 14.50 moles

[Cl₂] = 14.50 mol / 202.0 L = 0.07177 M

Br₂(g) + Cl₂(g) <==> 2BrCl(g)

0.032......0.0718...........0..........I

-x............-x.................+2x........C

0.032-x....0.0718-x......2x........E

Since we are provided with the Kp, but our values are in molar concentrations and not in pressure units, we can convert the Kp to Kc as follows:

Kp = Kc(RT)^∆n where ∆n is change in number of moles = 2 - 2 = 0 and anything raised to the zero power =1,

so Kp = Kc in this case.

1.1x10^-4 = [BrCl]² / [Br₂][Cl₂]

1.1x10^-4 = (2x)² /(0.032-x)(0.0718-x)

Simplifying and applying the 5% rule, we have...

1.1x10^-4 = (2x)² / (0.032)(0.0718)

2x² = 2.5x10^-7

x = 3.5x10^-4

So [BrCl] = 2 x 3.5x10^-4 = 7.0 x10^-4 M

7.0x10^-4 mol/L x 202 L = 0.1414 moles BrCl

mass BrCl = 0.1414 moles BrCl x 115 g/mol = 16.3 g

(PLEASE CHECK THE MATH)