Bromine monochloride is synthesized using the reaction Br2(g)+Cl2(g)↽−−⇀2BrCl(g) 𝐾p=1.1×10−4 at 150 K. A 196.0 L flask initially contains 1.097 kg of Br2 and 1.041 kg Cl2 .

Question

Bromine monochloride is synthesized using the reaction Br2(g)+Cl2(g)↽−−⇀2BrCl(g) 𝐾p=1.1×10−4 at 150 K. A 196.0 L flask initially contains 1.097 kg of Br2 and 1.041 kg Cl2 . Calculate the mass of BrCl, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior.

Mass of BrCl: ?

What is the percent yield of BrCl: ?

Answer 1

Br₂(g) + Cl₂(g) <==> 2BrCl(g)

molar mass Br2 = 159.8 g / mol

molar mass Cl2 = 70.9 g / mol

molar mass BrCl = 115.4 g / mol

moles Br2 initially present = 1.097 kg x 1000 g / kg x 1 mol / 159.8 g = 6.86 mols

moles Cl2 initially present = 1.041 kg x 1000 g / kg x 1 mol / 70.9 g = 14.68 mols

Converting moles of each gas to pressure using the Ideal gas law, we have...

PV = nRT

P = nRT/V

P_Br2 = (6.86 mol)(0.0821 Latm/Kmol)(150K) / 196 L = 0.4310 atm

P_Cl2 = (14.68 mol)(0.0821 Latm/Kmol) 150K) / 196 L = 0.9924 atm

Plugging these pressures into an ICE table, we have...

Br₂(g) + Cl₂(g) <==> 2BrCl(g)

0.4310...0.9924..........0............Initial

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

0.4310-x...0.9924-x......2x...........Equilibrium

K_p = 1.1x10^-4 = (BrCl)² / (Br₂)(Cl₂)

1.1x10^-4 = (2x)² / (0.4310-x)(0.9924-x) assume x is small and ignore it in the denominator. Simplify to:

1.1x10^-4 = 4x² / (0.4310)(0.9924)

1.1x10^-4 = 4x² / 0.4277

4x² =4.705x10^-5

x² = 1.176x10^-5

x= 3.430x10^-3 (note: this is only ~ 0.8% of 0.4310 and less than that for 0.9924 so above assumption was ok)

Equilibrium pressure of BrCl = 2x = (2)(3.430x10^-3) = 6.860x10^-3 atm

Plugging this back into the ideal gas law, and solving for moles, we have ...

PV = nRT

n = PV / RT = (6.860x10^-3 atm)(196 L) / (0.0821 Latm/Kmol)(150K)

n = 0.0192 moles

Mas BrCl = 0.0192 moles x 115.4 g /mol = 12.60 g