For the equilibrium below at 400 K, Kc = 7.0.
Br2(g) + Cl2(g) reverse reaction arrow 2 BrCl(g)
If 0.90 mol of Br2 and 0.90 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl?
Br2 + Cl2 ==> 2BrCl
Kc = (BrCl)^2/(Br2)(Cl2)
Do an ICE chart.
initial:
Br2 = 0.9 mol/1 L = 0.9M
Cl2 = 0.9 M
BrCl = 0
change:
Br2 = -x
Cl2 = -x
BrCl = +2x
equilibrium:
Br2 = 0.9-x
Cl2 = 0.9-x
BrCl = +2x
To find the equilibrium concentration of BrCl, we will need to use the given equilibrium constant (Kc) and the initial concentrations of Br2 and Cl2.
Let's denote the initial concentration of Br2 as [Br2]₀ and the initial concentration of Cl2 as [Cl2]₀.
Given:
[Br2]₀ = 0.90 mol
[Cl2]₀ = 0.90 mol
Kc = 7.0
Since we know the stoichiometry of the reaction is 1:1:2 (Br2 : Cl2 : BrCl), we can determine the change in concentration for each species.
Let's denote the change in concentration of Br2 as Δ[Br2], the change in concentration of Cl2 as Δ[Cl2], and the change in concentration of BrCl as Δ[BrCl].
At equilibrium, the concentration of Br2 will be [Br2]₀ - Δ[Br2], the concentration of Cl2 will be [Cl2]₀ - Δ[Cl2], and the concentration of BrCl will be 2(Δ[BrCl]).
To find the equilibrium concentrations, we need to set up an expression for the equilibrium constant (Kc) in terms of these concentrations.
Kc = ([BrCl]equilibrium)² / ([Br2]equilibrium * [Cl2]equilibrium)
Since the concentrations of Br2 and Cl2 are given as initial concentrations, we can express them in terms of [Br2]₀ and [Cl2]₀.
[Br2]equilibrium = [Br2]₀ - Δ[Br2]
[Cl2]equilibrium = [Cl2]₀ - Δ[Cl2]
Now we can substitute these expressions into the Kc equation:
Kc = ([BrCl]equilibrium)² / ([Br2]₀ - Δ[Br2]) * ([Cl2]₀ - Δ[Cl2])
Since the stoichiometry of the reaction is 1:1:2, we can express Δ[Br2] and Δ[Cl2] in terms of Δ[BrCl]:
Δ[BrCl] = 2Δ[BrCl]
Δ[Br2] = -Δ[BrCl]
Δ[Cl2] = -Δ[BrCl]
Substituting these expressions into the Kc equation, we get:
Kc = ([BrCl]equilibrium)² / ([Br2]₀ + Δ[BrCl]) * ([Cl2]₀ + Δ[BrCl])
Now we can simplify further:
Kc = ([BrCl]equilibrium)² / ([Br2]₀ + 2Δ[BrCl]) * ([Cl2]₀ + Δ[BrCl])
Since the equilibrium concentration of BrCl is 2Δ[BrCl], we can express it as:
([BrCl]equilibrium) = 2Δ[BrCl]
Substituting this in, we get:
Kc = (2Δ[BrCl])² / ([Br2]₀ + 2Δ[BrCl]) * ([Cl2]₀ + Δ[BrCl])
We are given that Kc = 7.0, so we can substitute this value as well:
7.0 = (2Δ[BrCl])² / ([Br2]₀ + 2Δ[BrCl]) * ([Cl2]₀ + Δ[BrCl])
Now we can solve this equation to find the value of Δ[BrCl], which will give us the equilibrium concentration of BrCl.
Unfortunately, without knowing the initial concentration of BrCl or any additional information, we cannot determine the exact equilibrium concentration of BrCl in this case.
To find the equilibrium concentration of BrCl, we need to make use of the equilibrium constant expression (Kc) and apply it to the given initial concentrations of Br2 and Cl2.
The equilibrium constant expression for the given reaction is:
Kc = [BrCl]^2 / ([Br2] * [Cl2])
Where [BrCl] represents the concentration of BrCl at equilibrium, [Br2] represents the concentration of Br2 at equilibrium, and [Cl2] represents the concentration of Cl2 at equilibrium.
Given that Kc = 7.0 and the initial concentrations of Br2 and Cl2 are both 0.90 mol in a 1.0 L container, we can set up the equilibrium constant expression as follows:
7.0 = ([BrCl]^2) / (0.90 * 0.90)
Now we can solve for [BrCl].
First, let's rearrange the equation:
7.0 * (0.90 * 0.90) = [BrCl]^2
6.3 = [BrCl]^2
Taking the square root of both sides, we find:
[BrCl] = √6.3
Calculating that, we get:
[BrCl] ≈ 2.51 mol/L
Therefore, the equilibrium concentration of BrCl will be approximately 2.51 mol/L.
Br2 + Cl2 ==> 2BrCl
Kc = (BrCl)^2/(Br2)(Cl2)
Do an ICE chart.
initial:
Br2 = 0.9 mol/1 L - 0.9M
Cl2 = 0.9 M
BrCl = 0
change:
BrCl = +x
Cl2 = -x
Br2 = -x
equilibrium:
Br2 = 0.9-x
Cl2 = 0.9-x
BrCl = +x
Plug the equilibrium concns into the Kc expression and solve for x.