Question

Bromine monochloride is synthesized using the reaction Br_(g) + Cl2(g) = 2 BrCl(g) Kp = 1.1 x 10-4 at 150 K A 190.0 L flask initially contains 1.078 kg of Br, and 1.068 kg of Cl.. Calculate the mass of BrCl,, in grams, that is present in the reaction mixture at equilibrium. Assume ideal gas behavior. mass of BrCI: g What is the percent yield of BrCl? percent yield:

Answer 1

Moles of Br2 = 1078 g/(160 g/mol) = 6.7375 mol

Moles of Cl2 = 1068 g/(71 g/mol) = 15.0422 mol

Here, Kp = Kc.(RT)^$\Delta$n = Kc.(RT)^2-2 = Kc

i.e. Kc = Kn

Kn = n(BrCl)²/n(Br2).n(Cl2)

i.e. 1.1*10^-4 = (2x)²/(6.7375-x)(15.0422-x)

i.e. 1.1*10^-4 = 4x²/(101.347 - 21.78 x + x²)

i.e. 3.99989x² + 0.0024 x - 0.01115 = 0

i.e. x = 0.0525

i.e. The no. of moles of BrCl = 2*0.0525 = 0.105 mol

i.e. The mass of BrCl = 0.105 mol * 115.5 g/mol = 12.1275 g

The percent yield of BrCl = {0.105/(2*6.7375)}*100 = 0.78%