Suppose a 250.mL flask is filled with 0.70mol of Br2, 0.20mol of OCl2 and 1.1mol of BrOCl. The following reaction becomes possible: Br2gOCl2g BrOClgBrClg The equilibrium constant K for this reaction is 5.53 at the temperature of the flask. Calculate the equilibrium molarity of BrOCl. Round your answer to two decimal places.

4.97M

Explanation:

Based on the reaction:

Br₂(g) + OCl₂(g) → BrOCl(g) + BrCl(g)

And K of this reaction is defined as:

K = 5.53 = [BrOCl] [BrCl] / [Br₂] [OCl₂]

Where [] are equilibrium concentrations

Initial concentrations are:

Br₂(g) = 0.70mol / 0.25L =2.8M

OCl₂(g) = 0.20mol / 0.25L = 0.8M

BrOCl(g) = 1.1mol /  0.25L = 4.4M

BrCl(g) = 0

When the equilibrium is reached, some Br₂(g) and OCl₂(g) react producing more BrOCl(g) and BrCl(g), that is:

Br₂(g) =2.8M - X

OCl₂(g) = 0.8M - X

BrOCl(g) = 4.4M + X

BrCl(g) = X

Replacing in K expression:

5.53 = [4.4 + X] [X] / [2.8-X] [0.8-X]

5.53 = 4.4X + X² / 2.24 -3.6X + X²

12.3872 - 19.908X + 5.53X² = 4.4X + X²

12.3872 - 24.308X + 4.53X² = 0

Solving for X:

X = 0.57M. Right solution.

X = 4.8M. False solution. Produce negative concentrations

That means equilibrium concentration of BrOCl is:

BrOCl(g) = 4.4M + X = 4.4M + 0.57M =

answer; natural resources;