Question

Suppose a 500.mL flask is filled with 1.9mol of Br2 , 0.90mol of OCl2 and 2.0mol of BrOCl . The following reaction becomes possible: Br2 (g)+OCl2 (g) +BrOCl (g)=BrCl (g) The equilibrium constant K for this reaction is 0.307 at the temperature of the flask. Calculate the equilibrium molarity of BrOCl . Round your answer to two decimal places.

final answer in M 2 decimal places

Answer 1

Volume of flask = 500 mL = 0.5 L

Let x mole each reactant is reacting.

Correct reaction is represented below:

Br2 (g) + OCl2 (g) --> BrOCl (g) + BrCl (g)

(1.9 – x) + (0.9 – x) --> (2 + x) + x

[Br2] = (1.9 – x)mol / 0.5 L = 2 (1.9 – x) M

[OCl2] = (0.9 – x)mol / 0.5 L = 2 (0.9 – x) M

[BrOCl] = (2 + x)mol / 0.5 L = 2 (2 + x) M

[BrCl] = x mol / 0.5 L = 2x M

Equilibrium constant, K = [BrCl] [BrOCl] / ([Br2] [OCl2]) = 0.307

2 x (2 (2 + x)) / (2 (1.9 – x) 2 (0.9 – x)) = 0.307

Solving we get,

x = 0.176 mol

Equilibrium molarity of BrOCl, M = 2 (2 + x)

M = 4.35 M