UNIT: 7.7 Equilibrium Concentrations and Calculations

Overview of Equilibrium Calculations

• Focus: Calculating equilibrium concentrations and equilibrium constants (K).
• Methodology: Utilize ICE tables (Initial, Change, Equilibrium) for solving problems.
• Type of problems covered: Type 1 (given initial and one equilibrium condition) and Type 2 (given all initial conditions and the equilibrium constant).

Key Concepts

• ICE Table: This consists of three rows:
  • I: Initial concentrations (or pressures).
  • C: Change in concentrations (or pressures).
  • E: Equilibrium concentrations (or pressures).

Steps for Setting Up an ICE Table

1. Identify the balanced chemical equation.
2. Determine initial concentrations.
   • Convert moles to molarity using volume.
3. Define changes during the reaction.
   • Use variables to represent unknown changes (e.g., -x, +x).
4. Calculate equilibrium concentrations.
   • Use the relationship between changes and equilibrium conditions.
5. Determine the equilibrium constant (K).

Practice Problem: Type 1

• Given: Balanced Reaction: 2SO₂ + O₂ ⇌ 2SO₃
  Initial Conditions:

  • SO₂: 2 moles in a 1L flask → 2M (initial concentration)
  • O₂: 1.5 moles in a 1L flask → 1.5M (initial concentration)
  • SO₃: 3 moles in a 1L flask → 3M (initial concentration)

• Equilibrium Condition: 3.5 moles of SO₃

  • Calculate: Equilibrium concentrations for O₂, SO₂, and K.

Setup:

1. ICE Table
   • Initial:
     • SO₂ = 2M
     • O₂ = 1.5M
     • SO₃ = 3M
   • Change:
     • SO₂ decreases by -x
     • O₂ decreases by -x
     • SO₃ increases by +2x
   • Equilibrium:
     • SO₂ = 2 - x
     • O₂ = 1.5 - x
     • SO₃ = 3 + 2x

Solve for Equilibrium Molarities:

• SO₃ was found to be 3.5M: $3 + 2x = 3.5$
  • Solve for x:
    $2x = 0.5 o x = 0.25$
• Substitute x back:
  • O₂: $1.5 - 0.25 = 1.25M$
  • SO₂: $2 - 0.25 = 1.75M$

Calculate K:

• Use the expression:
  $K = \frac{( ext{SO₃})^2}{( ext{SO₂})^2 imes ( ext{O₂})}$
• Plug in equilibrium concentrations:
  $K = \frac{(1.75)^2}{(1.25)^2}$
• Resulting K value.

Practice Problem: Type 2

Example: Hydrofluoric Acid Synthesis

• Reaction: H₂ + F₂ ⇌ 2HF
• Given: 3 moles H₂, 6 moles F₂, 12 moles HF in a 3L flask
• K at this temperature: 15

Setting Up ICE Table:

1. Initial Concentrations:

   • H₂: $\frac{3}{3} = 1M$
   • F₂: $\frac{6}{3} = 2M$
   • HF: $\frac{12}{3} = 4M$

2. K Expression:

   • $K = \frac{( ext{HF})^2}{( ext{H₂})( ext{F₂})}$

Comparing Q and K:

• Calculate Q using initial concentrations:
  • If Q < K, shift to products.
• Define changes as:
  • H₂ = 1 - x
  • F₂ = 2 - x
  • HF = 4 + 2x

Solving Quadric Equation via Graphing Calculator:

1. Enter K value and Q equation in calculator to find intersection point.
2. Use this to solve for equilibrium concentrations.

Recap:

• The equilibrium constant K only changes with temperature.
• Applying ICE tables through both types of problems aids in understanding equilibrium in reactions.