Answer:
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 =
