Question

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

Answer 1

[BrCl] = 0,13M

Step-by-step 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!