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

Question

A 207.0 L flask initially contains 1.092 kg of Br2 and 1.088 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) ... Kp = 1.1x10^-4

Moles of Br₂ = 1.092 kg Br₂ x 1000 g / kg x 1 mol Br2 / 159.8 g = 6.833 mols

Moles of Cl₂ = 1.088 kg Cl2 x 1000 g / kg x 1 mol Cl2 / 70.91 g = 15.34 mols

Pressure Br₂: PV = nRT and P = nRT/V

P = (6.833)(0.0821)(150) / 207 = 0.4065 atm

Pressure Cl₂: PV = nRT and P = nRT/V

P = (15.34)(0.0821)(150) / 207

P = 0.9126 atm

Using an ICE table...

Br₂(g) + Cl₂(g <==> 2BrCl(g)

0.4065....0.9126........0...........Initial

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

.4065-x...0.9126-x.....2x.........Equilibrium

Kp = (BrCl)² / (Br₂)(Cl₂)

1.1x10^-4 = (2x)2 / (0.4065)(0.9126) assuming x is small compared to initial pressures

4x² = 4.081x10^-5

x² = 1.020x10^-5

x = 3.19x10^-3 atm (which is small relative to initial pressures so assumption was valid)

I'm out of time, but plug x back into the equilibrium pressures in the ICE table and find pressure of each gas. You can then use ideal gas law to find moles of BrCl and then grams.