If 0.25 mol of br2 and 0.55mol of cl2 are introduced into a 3.0-l container at 400 k, what will be the equilibrium concentration of brcl ? ? br2(g)+cl2(g) 2brcl at 400 k, kc= 7.0

[BrCl] = 0,13M

Explanation:

For the reaction:

Br₂(g) + Cl₂(g) ⇄ 2BrCl

kc = [BrCl]²/[Br₂] [Cl₂] (1)

If initial [Br₂] and [Cl₂] are 0,25mol/3,0L=0,083M and 0,55mol/3,0L=0,18M, the concentrations in equilibrium are:

[Br₂] = 0,083M - x

[Cl₂] = 0,18M - x

[BrCl] = 2x

Replacing in (1)

7 = (2x)²/ (0,083 - x)(0,18 - x)

7 = 4x²/ x²- 0,263x + 0,01494

7x² - 1,841x + 0,10458 = 4x²

3x² - 1,841x + 0,10458 = 0

Solving for x:

x = 0,0633. As [BrCl] = 2x. [BrCl] = 0,13M

I hope it helps!

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