Applications of Equilibrium Constants

Key Features of Equilibrium Constants 🔑

  1. Only works when reactant/product concentrations stop changing (at equilibrium!).
  2. Value does NOT depend on starting concentrations.
  3. Depends on temperature 🌡️ – one unique K per reaction at a given temp.
  4. For reverse reaction: Kreverse = 1/Kforward.
  5. If you multiply the reaction by a number (n), new K = (original K)n.

1. Predicting How Far a Reaction Goes (Extent) 📏

K tells you if products or reactants dominate at equilibrium:

  • If Kc > 103: Lots of products! Reaction almost “complete”.
    Example: H₂(g) + Cl₂(g) ⇌ 2HCl(g), Kc = 4.0 × 10³¹ (at 300K)
  • If Kc < 10⁻³: Lots of reactants left. Reaction barely happens.
    Example: N₂(g) + O₂(g) ⇌ 2NO(g), Kc = 4.8 × 10⁻³¹ (at 298K)
  • If 10⁻³ < Kc < 10³: Mix of both reactants and products.
    Example: H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 57.0 (at 700K)

2. Predicting Reaction Direction (Which Way?) 🔄

Use the Reaction Quotient (Q)! Calculate Q the same way as K, but with any concentrations (not just equilibrium).

For a reaction: aA + bB ⇌ cC + dD

\[ Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]

Compare Q to Kc:

  • If Q < Kc: Reaction goes forward (→ products).
  • If Q > Kc: Reaction goes backward (← reactants).
  • If Q = Kc: Already at equilibrium! ⚖️

Example: For H₂(g) + I₂(g) ⇌ 2HI(g) (Kc = 57.0 at 700K):
If [H₂] = 0.10 M, [I₂] = 0.20 M, [HI] = 0.40 M, then
\[ Q_c = \frac{(0.40)^2}{(0.10)(0.20)} = 8.0 \] Since Qc (8.0) < Kc (57.0), reaction moves forward.

3. Calculating Equilibrium Concentrations 🧮

5 Steps to Success:

  1. Write the balanced equation.
  2. Make a table:
    • Initial concentrations
    • Change in concentrations (use x based on reaction ratios)
    • Equilibrium concentrations
  3. Plug equilibrium concentrations into K expression. Solve for x (use quadratic formula if needed).
  4. Find equilibrium concentrations using x.
  5. Check your answer by plugging back into K equation.

Example (N₂O₄ dissociation):
Start: 13.8g N₂O₄ (0.15 mol) in 1L at 400K. Total equil. pressure = 9.15 bar.
\[ \text{Initial pressure of } N_2O_4 = 4.98 \text{ bar} \]
At equilibrium: PN₂O₄ = (4.98 – x) bar, PNO₂ = 2x bar
\[ 9.15 = (4.98 – x) + 2x \implies x = 4.17 \text{ bar} \]
∴ PN₂O₄ = 0.81 bar, PNO₂ = 8.34 bar
\[ K_p = \frac{(8.34)^2}{0.81} = 85.87 \]

4. Link Between K, Q, and Gibbs Energy (G) ⚡

Gibbs energy change (ΔG) tells us if a reaction is spontaneous:

  • ΔG < 0: Spontaneous (→ forward)
  • ΔG > 0: Non-spontaneous (← reverse favored)
  • ΔG = 0: At equilibrium

The magic equation connecting it all:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
At equilibrium (ΔG = 0, Q = K):
\[ \Delta G^\circ = -RT \ln K \]
\[ K = e^{-\Delta G^\circ / RT} \]

Key takeaway:

  • If ΔG° < 0, then K > 1 (products favored).
  • If ΔG° > 0, then K < 1 (reactants favored).

Example 1: ΔG° = 13.8 kJ/mol for glucose phosphorylation. At 298K:
\[ \Delta G^\circ = -RT \ln K_c \implies K_c = 3.81 \times 10^{-3} \]

Example 2: For sucrose hydrolysis (Kc = 2 × 10¹³ at 300K):
\[ \Delta G^\circ = -RT \ln K_c = -7.64 \times 10^4 \text{ J/mol} \]

NEET Must-Knows! 💡

  1. Predict direction using Q vs. K: Master the comparison (Q < K → forward; Q > K → reverse).
  2. Extent from K magnitude: Big K (≫1) = product-favored; Small K (≪1) = reactant-favored.
  3. Calculate equilibrium concentrations: Use the ICE table method + solve for x (quadratic alert!).
  4. ΔG° links to K: ΔG° = -RT ln K (Negative ΔG° means K >1).