MCAT General Chemistry · Lesson 6

Equilibrium

3 min read3 sectionsUpdated January 3, 2026

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6.1Equilibrium
6.2Le Chatelier's Principle
6.3Kinetic and Thermodynamic Control

3 sections

6.1 Equilibrium

This section explains the concept of chemical equilibrium, dynamic versus static equilibria, the Law of Mass Action, reaction quotients, and calculations related to equilibrium constants.

Dynamic vs Static Equilibria
- Reversible reactions do not proceed to completion; products can revert to reactants.
- Equilibrium: rate(products) = rate(reactants)
- Static equilibrium: reactions have stopped.
- At equilibrium: entropy is maximized, Gibbs free energy $G$ is minimized.
Law of Mass Action
- At constant temperature, the equilibrium ratio is constant:
  - $K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
- Example: $2A → B + C$ $2 A \to B + C$
  - Rate(forward) = $k_f[A]^2$
  - Rate(reverse) = $k_r[B][C]$
- For gases: $K_c = K_p$
- $K_c$ and $K_{eq}$ interchangeable; $K_c = K_{eq} = k_f/k_r$ .
Reaction Quotient
- $Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ $Q_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ at any point in time.
  - If $Q < K_{eq}$ → reaction shifts forward.
  - If $Q = K_{eq}$ → reaction at equilibrium.
  - If $Q > K_{eq}$ → reaction shifts reverse.
Properties
- Concentrations of pure solids and liquids do not appear in $K$ expressions (activities ≈ 1).
- $K_{eq}$ is temperature dependent.
- Large $K_{eq}$ → favors products; small $K_{eq}$ → favors reactants.
- For reverse reaction: $K_{rev} = 1/K_{fwd}$
Equilibrium Calculations
- Example: 3 mol N₂O₄ in 0.5 L container $N_2O_4 → 2 NO_2$ $N_{2} O_{4} \to 2 N O_{2}$ , $K_{eq} = 6 × 10^{-6}$ $K_{e q} = 6 \times 1 0^{- 6}$ How much $NO_2$ $N O_{2}$ is in the container?
  - $[N_2O_4] = 3\ \text{mol}/0.5\ \text{L} = 6\ M$
  - $K_{eq} = \frac{[NO_2]^2}{[N_2O_4]}$
  - Let $x$ = change in N₂O₄: $6 × 10^{-6} = \frac{(2x)^2}{6}$
  - Solve: $4x^2 = 36 × 10^{-6} → x^2 = 9 × 10^{-6} → x = 3 × 10^{-3} M$
  - $[NO_2] = 2x = 6 × 10^{-3} M$

6.2 Le Chatelier's Principle

Explains how a system at equilibrium responds to changes in concentration, pressure, or temperature to re-establish equilibrium.

Khan Academy Video: Le Chatlier’s Principle

Le Chatlier’s PrinciplePlay video

Stress Responses
- If stress applied, system shifts to relieve it.
- Example: $N_2 + 3H_2 → 2NH_3$ $N_{2} + 3 H_{2} \to 2 N H_{3}$
  - Increase pressure → shifts to side with fewer moles of gas (left)
  - Decrease pressure → shifts to side with more moles (right)
- Always consider moles of gas only.
Temperature Effects
- Endothermic: heat treated as reactant; ↑T → shifts forward
- Exothermic: heat treated as product; ↑T → shifts reverse

6.3 Kinetic and Thermodynamic Control

Discusses how reaction conditions affect product distribution and stability, distinguishing between kinetic and thermodynamic products.

Pathway Control
- Low temperatures → kinetic product dominates
  - Lower activation energy, forms faster, products higher in energy.
- High temperatures → thermodynamic product dominates
  - Higher activation energy, forms slower, products lower in free energy (more stable).
Stability Considerations
- Dependent on torsional strain, angle strain, and non-bonded interactions.
- Thermodynamic product: more substituted double bond → less reactive, more stable.
- Kinetic product: less substituted double bond → forms faster at low temperatures.
- Ring systems can be attacked based on steric and electronic factors.

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