4. Chlorine and bromine react by the equationCl2(g) + Br2(g) D 2BrCl(g)
Kc = 1.2 x 102 at 150oC. What are the equilibrium concentrations of each reactant and product if a 2.0 L container initially contained 0.026 mole each of Cl2 and Br2 at 150oC? Show all calculations. (3 pts.)
To find the equilibrium concentrations of each reactant and product, we can use the equation for the reaction and the given equilibrium constant (Kc).
Step 1: Write the reaction equation and indicate the initial concentrations:
Cl2(g) + Br2(g) ⇌ 2BrCl(g)
Initial concentrations:
[Cl2] = 0.026 moles
[Br2] = 0.026 moles
[BrCl] = 0 moles (since no BrCl is initially present)
Step 2: Calculate the change in concentrations (x) for each species:
Since the stoichiometry of the reaction is 1:1:2 for Cl2, Br2, and BrCl respectively, the change in concentration of BrCl will be 2x, and the changes in concentration of Cl2 and Br2 will be -x and -x respectively.
Change in concentrations:
[Cl2] = 0.026 - x
[Br2] = 0.026 - x
[BrCl] = 2x
Step 3: Set up the expression for the equilibrium constant:
Kc = [BrCl]^2 / ([Cl2] * [Br2])
Plugging in the given value for Kc and the expressions for the concentrations:
1.2 x 102 = (2x)^2 / ((0.026 - x) * (0.026 - x))
Step 4: Solve the equation for x:
1.2 x 102 = 4x^2 / (0.026 - x)^2
Cross-multiplying gives:
1.2 x 102 * (0.026 - x)^2 = 4x^2
Expanding and rearranging the equation:
1.2 x 102 * (0.000676 - 0.052x + x^2) = 4x^2
0.0008112 - 0.0624x + 1.2 x 102 * x^2 = 4x^2
0.0008112 + 0.0624x = 4x^2 - 1.2 x 102 * x^2
0.0624x + 1.2 x 102 * x^2 - 4x^2 + 0.0008112 = 0
Simplifying and rewriting the equation:
1.2 x 102 * x^2 - 4x^2 + 0.0624x + 0.0008112 = 0
This equation is a quadratic equation in terms of x. We can solve it by using the quadratic formula.
Step 5: Use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the given equation:
a = 1.2 x 102
b = -4
c = 0.0624
Plugging in these values into the quadratic formula:
x = (-(-4) ± √((-4)^2 - 4(1.2 x 102)(0.0624))) / (2(1.2 x 102))
x = (4 ± √(16 - 4(1.2 x 102)(0.0624))) / (2(1.2 x 102))
Step 6: Calculate the value(s) of x:
By solving for x, we find two possible values: x ≈ 0.0627 or x ≈ 0.00624
Step 7: Substitute the value of x into the expressions for [Cl2], [Br2], and [BrCl]:
Using x ≈ 0.0627:
[Cl2] = 0.026 - 0.0627 ≈ -0.0367 (since the concentration cannot be negative, it is zero)
[Br2] = 0.026 - 0.0627 ≈ -0.0367 (again, the concentration cannot be negative, it is zero)
[BrCl] = 2 * 0.0627 ≈ 0.125
Using x ≈ 0.00624:
[Cl2] = 0.026 - 0.00624 ≈ 0.0198
[Br2] = 0.026 - 0.00624 ≈ 0.0198
[BrCl] = 2 * 0.00624 ≈ 0.0125
Step 8: Final results:
Using x ≈ 0.0627:
[Cl2] ≈ 0
[Br2] ≈ 0
[BrCl] ≈ 0.125
Using x ≈ 0.00624:
[Cl2] ≈ 0.0198
[Br2] ≈ 0.0198
[BrCl] ≈ 0.0125
Therefore, the equilibrium concentrations of each reactant and product are:
[Cl2] ≈ 0 M, [Br2] ≈ 0 M, [BrCl] ≈ 0.125 M (using x ≈ 0.0627)
or
[Cl2] ≈ 0.0198 M, [Br2] ≈ 0.0198 M, [BrCl] ≈ 0.0125 M (using x ≈ 0.00624)
To find the equilibrium concentrations of each reactant and product, we will use the concept of equilibrium constant (Kc) and the given initial moles of Cl2 and Br2.
Step 1: Write the balanced equation:
Cl2(g) + Br2(g) → 2BrCl(g)
Step 2: Set up the initial concentrations:
Cl2: [Cl2] = 0.026 mol / 2.0 L = 0.013 M
Br2: [Br2] = 0.026 mol / 2.0 L = 0.013 M
BrCl: [BrCl] = 0 M (since there is no initial concentration of the product)
Step 3: Use stoichiometry to set up the change in concentrations:
The change for Cl2 and Br2 will each be -2x, as 2 moles of reactants will produce 2 moles of BrCl. The change for BrCl will be +4x, as 2 moles of BrCl will be produced for every 1 mole of Cl2.
[Cl2] = 0.013 M - 2x
[Br2] = 0.013 M - 2x
[BrCl] = 4x
Step 4: Set up the equilibrium expression using the equilibrium constant (Kc):
Kc = [BrCl]^2 / ([Cl2] * [Br2])
1.2 x 10^2 = (4x)^2 / ((0.013 M - 2x) * (0.013 M - 2x))
Step 5: Solve for x:
Rearrange the equation:
1.2 x 10^2 = (16x^2) / (0.169 M - 0.052 M * x + 4x^2)
Multiply both sides by (0.169 M - 0.052 M * x + 4x^2):
1.2 x 10^2 * (0.169 M - 0.052 M * x + 4x^2) = 16x^2
Distribute:
20.28 M - 6.24 M * x + 192x^2 = 16x^2
Simplify:
176x^2 - 6.24 M * x + 20.28 M = 0
Now, solve this quadratic equation for x using the quadratic formula or a suitable method.
Step 6: Calculate the concentrations:
Substitute the value of x back into the expressions for [Cl2], [Br2], and [BrCl]:
[Cl2] = 0.013 M - 2x
[Br2] = 0.013 M - 2x
[BrCl] = 4x
This will give you the equilibrium concentrations of each reactant and product.