Bromine monochloride is synthesized using the reaction:
Br2(g) + Cl2(g) <> 2 BrCl (g)
Kp = 1.1 x 10^− 4 at 150 K
1. A 199.0 L flask initially contains 0.979 kg of Br 2 and 1.075 kg of Cl 2 . Calculate the mass of BrCl 2 , in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior.

11.72 grams

Explanation:

Let the equilibrium concentration of BrCl be y

Initial concentration of Br2 = number of moles ÷ volume = (0.979×1000/160) ÷ 199 = 0.031 M

Initial concentration of Cl2 = (1.075×1000/71) ÷ 199 = 0.076 M

From the equation of reaction

1 mole of Br2 reacted with 1 mole of Cl2 to form 2 moles of BrCl

Therefore, equilibrium concentration of Br2 = (0.031 - 0.5y) M while that of Cl2 = (0.076 - 0.5y) M

Kp = [BrCl]^2/[Br2][Cl2]

1.1×10^-4 = y^2/(0.031 - 0.5y)(0.076 - 0.5y)

y^2/0.002356-0.0535y+0.25y^2 = 0.00011

y^2/0.00011 = 0.002356-0.0536y+0.25y^2

9090.9y^2-0.25y^2+0.0536y-0.002356 = 0

9090.65y^2+0.0535y-0.002356 = 0

The value of y must be positive and is obtained using the quadratic formula

y = [-0.0535 + sqrt(0.0535^2 - 4×9090.65×-0.002356)] ÷ 2(9090.65) = 9.2025/18181.3 = 0.00051 M

Mass of BrCl = concentration×volume×MW = 0.00051×199×115.5 = 11.72 grams