Chemistry Calculating mass in equilibrium reaction

Question

Bromine monochloride is synthesized using the reaction

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

A 204.0 L204.0 L flask initially contains 1.011 kg1.011 kg of Br2Br2 and 1.137 kg1.137 kg of Cl2.Cl2. Calculate the mass of BrClBrCl, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior.

Answer 1

Probably best solved using an ICE table.

[Br₂] = 1.011 kg Br₂ x 1000 g / kg x 1 mol / 159.8 g = 6.327 moles Br2 / 204.0 L = 0.0310 M Br₂

[Cl₂] = 1.137 kg Cl₂ x 1000 g / kg x 1 mol / 70.9 g = 16.04 moles Cl2 / 204.0 L = 0.0786 M Cl₂

Br₂(g) + Cl₂(g) <==> 2BrCl(g)

0.031 ........0.0786.............0.............Initial

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

0.031-x......0786-x............2x...........Equilibrium

Kp = [BrCl]² / [Br₂][Cl₂]

1.1x10^-4 = (2x)² / (0.031-x)(0.0786-x) .. we can assume x is small compared to 0.031, and ignore it

1.1x10^-4 = 4x² / (0.031)(0.0786) = 4x² / 2.43x10^-3

4x² = 2.68x10^-7

x = 2.59x10^-4 (which is only about 1% of 0.031 so above assumption was valid)

To find grams of BrCl at equilibrium, we plug in the concentration @ equilibrium and then convert to grams:

2x = (2)(2.59x10^-4) = 5.18x10^-4 mol / L x 115 g / mol = 0.0596 g / L x 204 L = 12.2 g (not corrected for s.f.)