Equilibria: dynamic equilibrium, Le Chatelier’s principle, equilibrium constants

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Physical Chemistry – Equilibria (Cambridge AS & A Level Chemistry 9701)

1. Dynamic Equilibrium

  • Definition: A reversible reaction in a closed system reaches a state where the forward and reverse processes continue to occur simultaneously.
  • Macroscopic view: The concentrations (or partial pressures) of all species appear constant with time.
  • Microscopic view: Molecules are constantly colliding and reacting, but the rate of the forward reaction equals the rate of the reverse reaction.
  • Closed system: No matter can enter or leave the reaction mixture.
  • Pure solids and liquids: Do not appear in the equilibrium‑constant expression because their activities are constant (≈ 1).
  • Temperature dependence: The equilibrium constant (K) is a constant **only at a fixed temperature**.

2. Le Chatelier’s Principle

When a system at equilibrium is subjected to a disturbance (a “stress”), it responds by shifting in the direction that counteracts the change. The direction of shift can be predicted for each type of stress.

2.1 Types of Stress, Expected Shift and Effect on K

Stress Applied Shift of Equilibrium Microscopic Rationale Effect on K
Increase in concentration of a reactant Right (towards products) More reactant molecules collide → forward reaction accelerates until the excess is used up. No change
Decrease in concentration of a reactant Left (towards reactants) Fewer reactant molecules → reverse reaction produces more reactant. No change
Increase in concentration of a product Left Excess product is removed by the reverse reaction. No change
Decrease in concentration of a product Right System produces more product to replace what was lost. No change
Increase in total pressure (or decrease in volume) – gases only Towards side with fewer moles of gas (Δn < 0) Higher pressure favours the side occupying less volume. No change
Decrease in total pressure (or increase in volume) – gases only Towards side with more moles of gas (Δn > 0) Lower pressure is alleviated by forming more gas molecules. No change
Increase in temperature – exothermic reaction (ΔH < 0) Left (towards reactants) Heat is treated as a product; adding heat drives the reverse reaction. K decreases
Increase in temperature – endothermic reaction (ΔH > 0) Right (towards products) Heat is treated as a reactant; adding heat drives the forward reaction. K increases
Decrease in temperature – exothermic reaction Right Removing heat favours the forward (heat‑producing) direction. K increases
Decrease in temperature – endothermic reaction Left Removing heat favours the reverse (heat‑absorbing) direction. K decreases
Addition of a catalyst No shift in position Catalyst lowers the activation energy for both forward and reverse reactions equally. No change

2.2 Temperature – the Only Stress That Changes K

Because K is derived from the standard Gibbs free energy (ΔG° = –RT ln K), any change in temperature alters ΔG° and therefore the magnitude of K. All other stresses affect the position of equilibrium but leave K unchanged.

2.3 Industrial Applications (Syllabus Requirement)

Process Net Reaction Le Chatelier Optimisation
Haber Process (NH3 synthesis) N2(g) + 3 H2(g) ⇌ 2 NH3(g) ΔH = –92 kJ mol⁻¹
  • High pressure (≥ 150 atm) – shifts right because Δn = –2.
  • Moderately high temperature (≈ 500 °C) – a compromise: higher T speeds the reaction but reduces K; the optimum balances rate and yield.
  • Iron catalyst – increases both forward and reverse rates without changing K.
Contact Process (SO3 production) 2 SO2(g) + O2(g) ⇌ 2 SO3(g) ΔH = –198 kJ mol⁻¹
  • High pressure (≥ 1 atm) – shift right because Δn = –1.
  • Temperature around 450 °C – high enough for a reasonable rate, low enough that K remains large (exothermic).
  • V₂O₅ catalyst – accelerates the reaction without altering K.

3. Equilibrium Constants

3.1 General Form – Concentration (Kc)

For a reversible reaction

aA + bB ⇌ cC + dD

the equilibrium constant in terms of molar concentrations is

Kc = \(\displaystyle\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\)

Square brackets denote concentrations in mol L⁻¹. Pure solids and liquids are omitted.

3.2 Partial‑Pressure Form – Kp

When all species are gases, the constant may be written using partial pressures (bar or atm):

Kp = \(\displaystyle\frac{(P_{C})^{c}(P_{D})^{d}}{(P_{A})^{a}(P_{B})^{b}}\)

3.3 Relationship Between Kc and Kp

For gaseous equilibria

Kp = Kc\,(RT)^{\Delta n}

  • R = 0.08314 L bar K⁻¹ mol⁻¹ (or 8.314 J K⁻¹ mol⁻¹ in SI units).
  • T = temperature in kelvin.
  • Δn = (total moles of gaseous products) − (total moles of gaseous reactants).

3.4 Temperature Dependence of the Equilibrium Constant

The value of K is constant only at a given temperature. Changing the temperature changes the standard Gibbs free energy (ΔG°) and therefore the magnitude of K. This is why temperature is the only stress that alters K (see § 2.2).

4. Reaction Quotient (Q) – Predicting the Direction of Shift

Before equilibrium is reached, the reaction quotient Q is calculated with the same expression as K but using the **current** concentrations or partial pressures.

  • If Q < K → forward reaction is favoured (shift right).
  • If Q > K → reverse reaction is favoured (shift left).
  • If Q = K → the system is already at equilibrium.

5. Worked Examples

5.1 Determining Kc from Equilibrium Concentrations

Reaction:

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

At 500 K the equilibrium concentrations are:

  • [N₂] = 0.40 mol L⁻¹
  • [H₂] = 0.60 mol L⁻¹
  • [NH₃] = 0.10 mol L⁻¹

Calculation:

Kc = \(\displaystyle\frac{[NH₃]^{2}}{[N₂][H₂]^{3}} = \frac{(0.10)^{2}}{(0.40)(0.60)^{3}} = \frac{0.0100}{0.0864} \approx 0.12\)

5.2 Using Q to Predict the Shift After Adding NH₃

Extra 0.20 mol L⁻¹ NH₃ is added:

  • New [NH₃] = 0.30 mol L⁻¹
  • [N₂] and [H₂] unchanged.

New reaction quotient:

Qc = \(\displaystyle\frac{[NH₃]^{2}}{[N₂][H₂]^{3}} = \frac{(0.30)^{2}}{(0.40)(0.60)^{3}} = \frac{0.0900}{0.0864} \approx 1.04\)

Since Qc > Kc (1.04 > 0.12), the system shifts **left**, consuming NH₃ and forming more N₂ and H₂ until equilibrium is restored.

5.3 Converting Kc to Kp for the Same Reaction

Given Kc = 0.12 at 500 K:

  • Δn = (2 mol NH₃) − (1 + 3 mol N₂ + H₂) = –2
  • R = 0.08314 L bar K⁻¹ mol⁻¹

Calculation:

Kp = Kc(RT)^{Δn} = 0.12 × (0.08314 × 500)^{‑2} = 0.12 × (41.57)^{‑2} = 0.12 × 5.78 × 10⁻⁴ \(\approx 6.9 × 10^{-5}\)

6. Summary Checklist (Cambridge Syllabus 7.1 & 7.2)

  1. Dynamic equilibrium: forward rate = reverse rate; macroscopic concentrations (or pressures) are constant; pure solids/liquids omitted from expressions; equilibrium constant is temperature‑specific.
  2. Le Chatelier’s principle:
    • System shifts to oppose any stress (concentration, pressure/volume, temperature, catalyst).
    • Only temperature changes the value of K (exothermic ↔ endothermic behaviour).
    • Industrial examples – Haber and Contact processes – illustrate how pressure, temperature and catalysts are chosen to maximise yield.
  3. Equilibrium constants:
    • Kc = \(\displaystyle\frac{[{\text{products}}]^{u}}{[{\text{reactants}}]^{u}}\) (concentration).
    • Kp = \(\displaystyle\frac{(P_{\text{products}})^{u}}{(P_{\text{reactants}})^{u}}\) (partial pressure).
    • Δn = (moles of gaseous products) − (moles of gaseous reactants).
    • Kp = Kc(RT)^{Δn}.
    • Both constants are constant only at a fixed temperature.
  4. Reaction quotient (Q) predicts the direction of shift before equilibrium is reached (Q < K → right, Q > K → left).
  5. Pressure/volume effect for gases: use Δn to decide – increase pressure shifts toward the side with fewer gas moles (Δn < 0), decrease pressure shifts toward the side with more gas moles (Δn > 0).

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