At a certain temperature, the equilibrium constant, 𝐾c, is 6.01×10−3 for the reaction

Question

At a certain temperature, the equilibrium constant, 𝐾c, is 6.01×10−3 for the reaction

Cl2(g)↽−−⇀2Cl(g)

A. If 3.21 g Cl2 is placed in a 2.50 L flask at this temperature, what are the equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M
B. Following the establishment of equilibrium in part A, the volume of the flask is suddenly increased to 4.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M
C. Following the establishment of equilibrium in part A, the volume of the flask is instead suddenly decreased to 1.00 L while the temperature is held constant. What are the new equilibrium concentrations of Cl2 and Cl?
[Cl2]=
M
[Cl]=
M

Answer 1

A. (Cl2) = M = moles/L where moles = g/molar mass = 3.21/71 = 0.045 mols, then 0.045/02.50 L = 0.018M.

..................Cl2(g)↽−−⇀2Cl(g)
I ..................0.018............0
C...................-x.................2x
E................0.018-x............2x
Substitute the E line into the Kc expression and solve for x = (Cl^-) and evaluate 0.0108-x = (Cl2)
For B: See the previous problem.
For C: See the previous problem.
Note: This is a Kc problem and the previous problem was a Kp. For this one decreasing volume increaes concn and increasing volume decreases concn.
Post your work if you get stuck.

Answer 2

To solve this problem, we'll use the given equilibrium constant, initial amounts, and the ideal gas law to find the equilibrium concentrations of Cl2 and Cl in each scenario.

A. Initial Conditions:
- Initial amount of Cl2 = 3.21 g
- Volume of the flask = 2.50 L

1. Convert the mass of Cl2 to moles:
- Molar mass of Cl2 = 2*35.45 g/mol = 70.90 g/mol
- Moles of Cl2 = 3.21 g / 70.90 g/mol = 0.04526 mol

2. Use the ideal gas law to find the initial concentration of Cl2:
- Concentration = (moles of Cl2) / (volume of the flask)
- [Cl2] = 0.04526 mol / 2.50 L = 0.018104 M

3. According to the balanced chemical equation, the stoichiometry of Cl2 to Cl is 1:2. So, the initial concentration of Cl is twice that of Cl2:
- [Cl] = 2 * [Cl2] = 2 * 0.018104 M = 0.036208 M

Therefore, at equilibrium:
[Cl2] = 0.018104 M
[Cl] = 0.036208 M

B. In this scenario, the volume of the flask is suddenly increased to 4.00 L while the temperature is held constant. However, the amount of substance remains the same.

Since the number of moles of Cl2 and Cl remains the same, the new equilibrium concentrations will be the same as the initial concentrations:
[Cl2] = 0.018104 M
[Cl] = 0.036208 M

C. In this scenario, the volume of the flask is suddenly decreased to 1.00 L while the temperature is held constant. Again, the amount of substance remains the same.

Using the ideal gas law, we can determine the new equilibrium concentrations:
[Cl2] = (moles of Cl2) / (new volume of the flask)
[Cl] = (moles of Cl) / (new volume of the flask)

1. Moles of Cl2 and Cl remain the same, so we can directly plug in the values from part A:
[Cl2] = 0.018104 M
[Cl] = 0.036208 M

2. Calculate the new equilibrium concentrations:
[Cl2] = 0.04526 mol / 1.00 L = 0.04526 M
[Cl] = 0.09052 mol / 1.00 L = 0.09052 M

Therefore, the new equilibrium concentrations of Cl2 and Cl in scenario C are:
[Cl2] = 0.04526 M
[Cl] = 0.09052 M

Answer 3

To solve this problem, we can use the equilibrium expression and the given equilibrium constant to calculate the equilibrium concentrations of Cl2 and Cl.

The equilibrium constant expression for the given reaction is:
Kc = [Cl]^2 / [Cl2]

A. For the initial conditions, we are given:
[Cl2] initial = 3.21 g / (70.91 g/mol) = 0.0452 mol
[Cl] initial = 0 mol (since no Cl has been formed yet)

To find the equilibrium concentrations, we can assume that the change in concentration of Cl2 is -2x and the change in concentration of Cl is +x (as per the stoichiometry of the balanced equation), where x is the change in concentration at equilibrium.

This gives us the following concentrations at equilibrium:
[Cl2] = [Cl2] initial - 2x
[Cl] = [Cl] initial + x

Since we are given the equilibrium constant Kc = 6.01×10^(-3), we can substitute the equilibrium concentrations into the equilibrium constant expression:

6.01×10^(-3) = ([Cl]^2) / [Cl2]
6.01×10^(-3) = ([Cl] initial + x)^2 / ([Cl2] initial - 2x)

Simplifying the equation and rearranging, we get a quadratic equation:

6.01×10^(-3) * ([Cl2] initial - 2x) = ([Cl] initial + x)^2

Solve the quadratic equation to find the value of x. Then use this value to calculate the equilibrium concentrations [Cl2] and [Cl].

B. For part B, the volume is suddenly increased to 4.00 L while the temperature is held constant. Since no additional Cl2 or Cl is added or removed, the number of moles of each substance remains the same. Therefore, the equilibrium concentrations remain the same as in part A.

C. For part C, the volume is suddenly decreased to 1.00 L while the temperature is held constant. Again, since no additional Cl2 or Cl is added or removed, the number of moles of each substance remains the same. Therefore, the equilibrium concentrations remain the same as in part A.

Please note that to solve the quadratic equation, you may use methods such as factoring, completing the square, or using the quadratic formula.