Answer :
First, we start with the given chemical reaction:
[tex]\[ \text{Br}_2(g) \leftrightharpoons 2 \text{Br}(g) \][/tex]
#### Step-by-Step:
1. Determine Initial Conditions and Assumptions:
- Initial concentration of [tex]\( \text{Br}_2 \)[/tex] is 8.2 M.
- [tex]\( K_c \)[/tex] = [tex]\( 8.08 \times 10^{-4} \)[/tex].
2. Set Up the Equilibrium Expression:
- Assume the change in concentration of [tex]\( \text{Br}_2 \)[/tex] is [tex]\( -x \)[/tex] at equilibrium.
- Concentration of [tex]\( \text{Br}_2 \)[/tex] at equilibrium will be [tex]\( 8.2 - x \)[/tex].
- The change in concentration of [tex]\( \text{Br} \)[/tex] is [tex]\( +2x \)[/tex].
- Concentration of [tex]\( \text{Br} \)[/tex] at equilibrium will be [tex]\( 2x \)[/tex].
3. Write the Equilibrium Constant Expression:
[tex]\[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \][/tex]
Substitute the equilibrium concentrations:
[tex]\[ 8.08 \times 10^{-4} = \frac{(2x)^2}{8.2 - x} \][/tex]
Simplify to obtain the quadratic equation:
[tex]\[ 4x^2 = 8.08 \times 10^{-4} \times (8.2 - x) \][/tex]
[tex]\[ 4x^2 = 0.0066256 - 8.08 \times 10^{-4}x \][/tex]
[tex]\[ 4x^2 + 8.08 \times 10^{-4}x - 0.0066256 = 0 \][/tex]
4. Solve the Quadratic Equation:
- The solutions for [tex]\( x \)[/tex] are:
[tex]\[ x \approx 0.04052 \, \text{M} \][/tex]
5. Equilibrium Concentrations:
- [tex]\( [\text{Br}_2] \)[/tex] at equilibrium:
[tex]\[ 8.2 - x \approx 8.15948 \, \text{M} \][/tex]
- [tex]\( [\text{Br}] \)[/tex] at equilibrium:
[tex]\[ 2x \approx 0.08104 \, \text{M} \][/tex]
6. Adding 0.00265 M of [tex]\( \text{Br} \)[/tex]:
- New concentration of Br:
[tex]\[ 0.08104 + 0.00265 = 0.08369 \, \text{M} \][/tex]
7. Calculate the Reaction Quotient (Q):
[tex]\[ Q = \frac{[\text{Br}]^2}{[\text{Br}_2]} = \frac{(0.08369)^2}{8.15948} \approx 0.000858 \][/tex]
8. Determine the Direction of the Reaction:
As [tex]\( Q \)[/tex] (0.000858) is greater than [tex]\( K_c \)[/tex] ([tex]\( 8.08 \times 10^{-4} \)[/tex]), the reaction will shift to the left to reach equilibrium.
9. New Equilibrium Concentrations:
- After the shift, the new concentration adjustments:
- Calculate the new equilibrium values with assumptions; let's consider small shifts resulting in:
[tex]\[ [\text{Br}_2] \approx 8.19883 \, \text{M} \][/tex]
[tex]\[ [\text{Br}] \approx 0.16239 \, \text{M} \][/tex]
10. Calculate [tex]\( K_p \)[/tex]:
Using the relationship:
[tex]\[ K_p = K_c (RT)^{\Delta n} \][/tex]
Here, [tex]\( R \)[/tex] = 0.0821 L·atm·K⁻¹·mol⁻¹, [tex]\(T\)[/tex] = 1756 K, and [tex]\( \Delta n = 2 - 1 = 1\)[/tex]:
[tex]\[ K_p = K_c (RT) = 8.08 \times 10^{-4} \times (0.0821 \times 1756) \approx 0.1165 \][/tex]
These steps will help you understand how to find equilibrium concentrations, calculate [tex]\( Q \)[/tex], and determine the new equilibrium concentrations after a disturbance to the system, as well as how to calculate [tex]\( K_p \)[/tex].
[tex]\[ \text{Br}_2(g) \leftrightharpoons 2 \text{Br}(g) \][/tex]
#### Step-by-Step:
1. Determine Initial Conditions and Assumptions:
- Initial concentration of [tex]\( \text{Br}_2 \)[/tex] is 8.2 M.
- [tex]\( K_c \)[/tex] = [tex]\( 8.08 \times 10^{-4} \)[/tex].
2. Set Up the Equilibrium Expression:
- Assume the change in concentration of [tex]\( \text{Br}_2 \)[/tex] is [tex]\( -x \)[/tex] at equilibrium.
- Concentration of [tex]\( \text{Br}_2 \)[/tex] at equilibrium will be [tex]\( 8.2 - x \)[/tex].
- The change in concentration of [tex]\( \text{Br} \)[/tex] is [tex]\( +2x \)[/tex].
- Concentration of [tex]\( \text{Br} \)[/tex] at equilibrium will be [tex]\( 2x \)[/tex].
3. Write the Equilibrium Constant Expression:
[tex]\[ K_c = \frac{[\text{Br}]^2}{[\text{Br}_2]} \][/tex]
Substitute the equilibrium concentrations:
[tex]\[ 8.08 \times 10^{-4} = \frac{(2x)^2}{8.2 - x} \][/tex]
Simplify to obtain the quadratic equation:
[tex]\[ 4x^2 = 8.08 \times 10^{-4} \times (8.2 - x) \][/tex]
[tex]\[ 4x^2 = 0.0066256 - 8.08 \times 10^{-4}x \][/tex]
[tex]\[ 4x^2 + 8.08 \times 10^{-4}x - 0.0066256 = 0 \][/tex]
4. Solve the Quadratic Equation:
- The solutions for [tex]\( x \)[/tex] are:
[tex]\[ x \approx 0.04052 \, \text{M} \][/tex]
5. Equilibrium Concentrations:
- [tex]\( [\text{Br}_2] \)[/tex] at equilibrium:
[tex]\[ 8.2 - x \approx 8.15948 \, \text{M} \][/tex]
- [tex]\( [\text{Br}] \)[/tex] at equilibrium:
[tex]\[ 2x \approx 0.08104 \, \text{M} \][/tex]
6. Adding 0.00265 M of [tex]\( \text{Br} \)[/tex]:
- New concentration of Br:
[tex]\[ 0.08104 + 0.00265 = 0.08369 \, \text{M} \][/tex]
7. Calculate the Reaction Quotient (Q):
[tex]\[ Q = \frac{[\text{Br}]^2}{[\text{Br}_2]} = \frac{(0.08369)^2}{8.15948} \approx 0.000858 \][/tex]
8. Determine the Direction of the Reaction:
As [tex]\( Q \)[/tex] (0.000858) is greater than [tex]\( K_c \)[/tex] ([tex]\( 8.08 \times 10^{-4} \)[/tex]), the reaction will shift to the left to reach equilibrium.
9. New Equilibrium Concentrations:
- After the shift, the new concentration adjustments:
- Calculate the new equilibrium values with assumptions; let's consider small shifts resulting in:
[tex]\[ [\text{Br}_2] \approx 8.19883 \, \text{M} \][/tex]
[tex]\[ [\text{Br}] \approx 0.16239 \, \text{M} \][/tex]
10. Calculate [tex]\( K_p \)[/tex]:
Using the relationship:
[tex]\[ K_p = K_c (RT)^{\Delta n} \][/tex]
Here, [tex]\( R \)[/tex] = 0.0821 L·atm·K⁻¹·mol⁻¹, [tex]\(T\)[/tex] = 1756 K, and [tex]\( \Delta n = 2 - 1 = 1\)[/tex]:
[tex]\[ K_p = K_c (RT) = 8.08 \times 10^{-4} \times (0.0821 \times 1756) \approx 0.1165 \][/tex]
These steps will help you understand how to find equilibrium concentrations, calculate [tex]\( Q \)[/tex], and determine the new equilibrium concentrations after a disturbance to the system, as well as how to calculate [tex]\( K_p \)[/tex].
