For the following equilibrium; Br2 (g) + Cl2 (g) ⇄ 2BrCl (g)
the Kc = 7.00 at 127oC. If 0.450 mol of Br2 (g) and 0.450 mol of Cl2 (g) are introduced into a 3.00 L container at 400 K, what will the equilibrium concentration of BrCl(g) be?
I assume you want the concns at 127 C.
M = mols/L = 0.450/3 = 0.150M
............Br2 + Cl2 ==> 2BrCl
I.........0.150..0.150......0
C............x.....-x......2x
E......0.150-x...0.150-x....2x
Substitute into Kc expression and solve for x and 2x.
To find the equilibrium concentration of BrCl (g) in the given equilibrium, we can use the concept of the equilibrium constant, Kc, and the stoichiometry of the balanced equation.
Step 1: Set up the balanced equation: Br2 (g) + Cl2 (g) ⇄ 2BrCl (g)
Step 2: Write the expression for the equilibrium constant, Kc, using the concentrations of gases:
Kc = [BrCl]^2 / ([Br2] * [Cl2])
Step 3: Plug in the given values:
Kc = 7.00
[Br2] = 0.450 mol / 3.00 L = 0.150 M (concentration of Br2)
[Cl2] = 0.450 mol / 3.00 L = 0.150 M (concentration of Cl2)
Step 4: Substitute the values into the equation and solve for [BrCl]:
7.00 = [BrCl]^2 / (0.150^2 * 0.150) (substituting values into the Kc equation)
7.00 = [BrCl]^2 / 0.003375
[BrCl]^2 = 7.00 * 0.003375 (cross-multiplication)
[BrCl]^2 = 0.023625
[BrCl] = √(0.023625)
[BrCl] = 0.153 M
Thus, the equilibrium concentration of BrCl (g) in the 3.00 L container will be 0.153 M.