Variation Two: Equilibrium Constant Calculation

Variation Two

• Focus: Calculating equilibrium constant $K$ with different initial information.
• Information provided: Initial concentrations of reactants and the concentration of a reactant lost.
  • Differs from previous examples where all ICE table values could be directly filled.

Example: Ethanol and Acetic Acid Reaction

• Reactants:
  • Ethanol ($C<em>2H</em>5OH$)
  • Acetic acid ($CH_3COOH$)
• Initial Information:
  • Amounts given in moles.
  • Volume given in milliliters (mL).
• Conversion:
  • Convert volume to liters (L) for concentration calculations.
  • C = N/V where
    • $C$ is concentration,
    • $N$ is moles, and
    • $V$ is volume.
  • Example: 1 mole in 250 mL ($0.25$ L) gives a concentration of 4 M. $(C = \frac{1 \text{ mol}}{0.25 \text{ L}} = 4 \text{ M})$
• Reaction:
  • Balanced reaction is essential.
  • Note: Acetic acid should have "aq" (aqueous) beside it ($CH_3COOH(aq)$).
  • $C<em>2H</em>5OH(aq) + CH<em>3COOH(aq) \rightleftharpoons CH</em>3COOC<em>2H</em>5(aq) + H_2O(l)$

ICE Table Considerations

• Water ($H_2O$):
  • In liquid form, so its concentration doesn't change significantly.
  • Not included in the ICE table or the mass action expression.
• Initial Concentrations:
  • Ethanol: 4 M
  • Acetic Acid: 4 M
• Temperature:
  • Kept constant at 100°C.
  • Important for maintaining equilibrium.
• Product: Ethyl Acetate ($CH<em>3COOC</em>2H_5$)
  • Initial concentration: 0 M (assumed no initial product).
• Change Row:
  • Determined by the amount of acetic acid consumed at equilibrium.
  • Equilibrium concentration:
    • 0. 25 moles of acetic acid consumed in 0.25 L.
    • Concentration change: 1 M $(\frac{0.25 \text{ mol}}{0.25 \text{ L}} = 1 \text{ M})$

ICE Table Calculations

• Changes based on stoichiometry:
  • Reaction is 1:1:1 (acetic acid : ethanol : ethyl acetate).
  • Acetic acid decreases by 1 M.
  • Ethanol also decreases by 1 M.
  • Ethyl acetate increases by 1 M.
• Equilibrium Concentrations:
  • Ethanol: 4 M (initial) - 1 M (change) = 3 M
  • Acetic Acid: 4 M (initial) - 1 M (change) = 3 M
  • Ethyl Acetate: 0 M (initial) + 1 M (change) = 1 M

Equilibrium Constant Calculation

• Mass Action Expression:
  • $K = \frac{[Products]}{[Reactants]}$
  • $K = \frac{[CH<em>3COOC</em>2H<em>5]}{[C</em>2H<em>5OH][CH</em>3COOH]}$
• Substitute Equilibrium Concentrations:
  • $K = \frac{1}{(3)(3)} = \frac{1}{9}$
• Result:
  • $K ≈ 0.11$