Dynamic Equilibrium, Le Chatelier’s Principle and Kc

Learning Objectives (OCR Spec 3.2.3)

By the end of this lesson you should be able to:

• Explain what is meant by dynamic equilibrium in a closed system
• State and apply Le Chatelier’s principle to predict the effect of changes in concentration, pressure and temperature
• Define and write the expression for Kc for a homogeneous equilibrium
• Perform calculations using Kc
• Describe how compromise conditions are chosen in industrial processes such as the Haber and Contact processes

Dynamic Equilibrium

Consider a reversible reaction in a closed container:

A + B ⇌ C + D

Initially only A and B are present, so only the forward reaction occurs. As C and D build up, the reverse reaction starts and its rate increases. As A and B are consumed, the forward rate decreases. Eventually the forward and reverse rates become equal, and the concentrations of all four species stop changing.

This is dynamic equilibrium.

Key Features of Dynamic Equilibrium

1. The reaction occurs in both directions (forward and reverse).
2. The rate of the forward reaction equals the rate of the reverse reaction.
3. The concentrations of reactants and products remain constant (but not necessarily equal).
4. It is reached only in a closed system (where matter cannot enter or leave).
5. It can be approached from either side (pure reactants or pure products).

"Dynamic" means nothing has stopped - molecules are still constantly colliding, reacting and separating - but the net rate of change is zero.

Le Chatelier’s Principle

"When a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to partially oppose that change."

This principle lets us predict qualitatively how an equilibrium mixture responds to changes in concentration, pressure or temperature. It does not say by how much the equilibrium shifts - it says in which direction.

Effect of Concentration

If we increase the concentration of a reactant, the system shifts to the right (towards products) to consume the extra reactant and restore equilibrium. If we remove a product, the system shifts to the right to replace it.

Example: N2 + 3H2 ⇌ 2NH3

• Add more N2 → position shifts right, more NH3 formed.
• Remove NH3 → position shifts right, more NH3 produced to replace it.
• Add more NH3 → position shifts left, NH3 is consumed.

Importantly, Kc is unchanged by concentration changes at constant temperature. Only the position of equilibrium (the relative amounts) shifts.

Effect of Pressure (Gas Equilibria)

Increasing the pressure of a gas-phase equilibrium shifts the position towards the side with fewer moles of gas, to reduce the pressure. Decreasing the pressure shifts it towards the side with more moles.

Example: N2(g) + 3H2(g) ⇌ 2NH3(g)

• Left side: 1 + 3 = 4 moles of gas
• Right side: 2 moles of gas
• Increasing pressure shifts position to the right (fewer moles), producing more NH3.

Example: H2(g) + I2(g) ⇌ 2HI(g)

• Left and right sides both have 2 moles of gas.
• Pressure changes have no effect on position of equilibrium.

Kc is also unchanged by pressure changes at constant temperature.

Effect of Temperature

Temperature affects equilibria differently from concentration or pressure - it actually changes the value of Kc. The direction of the shift depends on whether the forward reaction is exothermic or endothermic.

• If the forward reaction is exothermic (ΔH < 0), increasing temperature shifts the position to the left (reverse, endothermic direction), and Kc decreases.
• If the forward reaction is endothermic (ΔH > 0), increasing temperature shifts the position to the right and Kc increases.

Example: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = −92 kJ mol⁻¹ (exothermic forward).

• Lower temperature favours product formation (higher Kc).
• But at low temperature the rate is very slow (Boltzmann), so we compromise.

Example: 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = −196 kJ mol⁻¹ (exothermic).

• Lower temperature favours SO3 production.
• In practice 450 °C is used - a compromise.

Decision Tree for Le Chatelier

graph TD
  A[What is being changed?] --> B[Concentration]
  A --> C[Pressure]
  A --> D[Temperature]
  A --> E[Catalyst]
  B --> B1[Shift AWAY from the added species]
  C --> C1[Shift towards FEWER moles of gas]
  D --> D1{Forward exothermic?}
  D1 -->|Yes| D2[Higher T: shift LEFT, Kc decreases]
  D1 -->|No| D3[Higher T: shift RIGHT, Kc increases]
  E --> E1[No shift in position; rate increased both ways]

Effect of a Catalyst

A catalyst does not change the position of equilibrium. It speeds up the forward and reverse reactions equally, so equilibrium is reached faster, but the final concentrations (and Kc) are unchanged. This is why catalysts appear on both sides of an energy profile diagram.

The Equilibrium Constant Kc

For a homogeneous equilibrium aA + bB ⇌ cC + dD, the equilibrium constant Kc is:

Kc = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)

where each concentration is the equilibrium concentration in mol dm⁻³. The powers come from the stoichiometric coefficients.

Units of Kc

The units depend on the overall change in moles (Δn = products − reactants exponents). For N2 + 3H2 ⇌ 2NH3:

Kc = [NH3]² / ([N2][H2]³) Units: (mol dm⁻³)² / [(mol dm⁻³)(mol dm⁻³)³] = 1 / (mol dm⁻³)² = mol⁻² dm⁶ or dm⁶ mol⁻².

For H2 + I2 ⇌ 2HI: Kc = [HI]² / ([H2][I2]), units cancel → no units.

Meaning of Kc

A large Kc (≫ 1) means the equilibrium lies to the right (products favoured at equilibrium). A small Kc (≪ 1) means the equilibrium lies to the left (reactants favoured). Kc = 1 means roughly equal concentrations at equilibrium.