Bromine and chlorine gas react to give bromine monochloride (BrCl) according to
Br2(g)+Cl2(g)⇌2BrCl(g)
Given initial concentrations [Br2]=1.50 mol L-1, [Cl2]=0.90 mol L-1 and [BrCl]=0 and K=7 determine (to two decimal places) the equilibrium concentration of BrCl in mol L-1
7=(x)^2/(1.5-x)(.9-x)
do the algebra, get a quadratic , use the quadratic equation to solve for x.
To determine the equilibrium concentration of BrCl (Bromine monochloride), we can use the given equilibrium constant (K) and the initial concentrations of Br2 and Cl2.
First, we need to set up an ICE table (Initial - Change - Equilibrium) to track the changes in concentrations during the reaction:
Equation: Br2(g) + Cl2(g) ⇌ 2BrCl(g)
Initial: 1.50 0.90 0
Change: -x -x +2x
Equilibrium:1.50-x 0.90-x 2x
Where x represents the change in concentration for both Br2 and Cl2. Since 1 molecule of Br2 reacts with 1 molecule of Cl2 to form 2 molecules of BrCl, the change in concentration for BrCl is +2x.
Since the equilibrium concentration of BrCl is given as 0, we can assume that the value of x is very small compared to the initial concentrations of Br2 and Cl2. Therefore, we can approximate 1.50 - x as 1.50 and 0.90 - x as 0.90.
Now, we can substitute the equilibrium concentrations into the equilibrium expression and solve for x:
K = [BrCl]^2 / ([Br2] * [Cl2])
7 = (2x)^2 / (1.50 * 0.90)
Simplifying the equation, we have:
7 = 4x^2 / (1.35)
4x^2 = 7 * 1.35
4x^2 = 9.45
x^2 = 9.45 / 4
x^2 = 2.3625
x ≈ √(2.3625)
x ≈ 1.54
Since x represents the change in concentration for both Br2 and Cl2, the equilibrium concentration of BrCl can be calculated as:
2x = 2 * 1.54 = 3.08 mol L^-1
Therefore, the equilibrium concentration of BrCl is approximately 3.08 mol L^-1.
To determine the equilibrium concentration of BrCl in mol L-1, we can use the equilibrium expression and the given initial concentrations.
The equilibrium expression for the reaction is:
K = [BrCl]² / ([Br2] * [Cl2])
Given:
[Br2] = 1.50 mol L-1
[Cl2] = 0.90 mol L-1
[BrCl] = 0
K = 7
Let's substitute the given values into the equilibrium expression:
7 = (0)² / (1.50 * 0.90)
To solve for [BrCl], we rearrange the equation:
(0)² = 7 * 1.50 * 0.90
0 = 9.45
Since we obtained 0 = 9.45, it means that the reaction will not proceed to form any appreciable concentration of BrCl. Therefore, the equilibrium concentration of BrCl is 0 mol L-1.
It is important to note that the equilibrium expression and the given initial concentrations were used to determine the equilibrium concentration of BrCl.