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*Revision Materials* 1 Atomic Structure 2 Atoms, molecules and stoichiometry 3 Chemical Bonding 4 States of matter 5 Chemical energetics 6 Electrochemistry 7 Equilibria 8 Reaction kinetics 9 The Periodic Table, chemical periodicity 10 Group 2 11 Group 17 12 Nitrogen and sulfur 13 Organic 14 Hydrocarbons 15 Halogen compounds 16 Hydroxy compounds 17 Carbonyl compounds 18 Carboxylic acids and derivatives 19 Nitrogen compounds 20 Polymerisation 21 Organic synthesis 22 Analytical techniques 23 Chemical energetics 24 Electrochemistry 25 Equilibria 26 Reaction kinetics 27 Group 2 28 Chemistry of transition elements 29 Organic 30 Hydrocarbons 31 Halogen compounds 32 Hydroxy compounds 33 Carboxylic acids and derivatives 34 Nitrogen compounds 35 Polymerisation 36 Organic synthesis 37 Analytical techniques

7 Equilibria

7.1 Chemical equilibria, reversible reactions, dynamic equilibrium 7.2 Brønsted–Lowry theory of acids and bases

Chemical Equilibria: Reversible Reactions, Dynamic Equilibrium

Specification Reference Physical Chemistry: Equilibria 7.1

Quick Notes

  • Reversible reactions go in both directions: forward and reverse.
  • Dynamic equilibrium occurs when the rate of the forward reaction = rate of the reverse reaction, and concentrations of reactants and products remain constant.
  • Le Chatelier’s Principle states that if a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose the change.
  • Factors affecting equilibrium position:
    • Concentration: Increasing reactants shifts equilibrium right; increasing products shifts equilibrium left.
    • Pressure (gases): Increasing pressure shifts equilibrium to the side with fewer moles of gas.
    • Temperature:
      • Increasing temperature shifts equilibrium in endothermic (+ΔH) direction.
      • Decreasing temperature shifts equilibrium in exothermic (−ΔH) direction.
    • Catalysts: Do not shift equilibrium; they increase the rate of both forward and reverse reactions equally.
  • Equilibrium constants (K) are the ratio of products to reactants at equilibrium.
    • Kc uses concentrations
    • Kp uses partial pressures
  • Equilibrium constants only change with temperature.
  • Industrial conditions are chosen to balance yield and rate of production.

Full Notes

Equilibrium and Le Chatelier’s Principle have been outlined in more detail here.
This page is just what you need to know for CIE A-level Chemistry :)

Reversible Reactions

A reversible reaction is one that occurs in both the forward and reverse directions. Reactants can react to form products and then these products can react with each other to reform the reactants.

For Example Haber Process

Forward reaction: N2 + H2 → NH3
Reverse reaction: NH3 decomposes into N2 and H2

CIE A-Level Chemistry diagram of the Haber Process showing the reversible reaction between nitrogen and hydrogen forming ammonia.

Dynamic Equilibrium

If a reversible reaction is left for a period of time in a closed system, eventually the forward reaction will occur at the same rate as the reverse reaction — meaning concentrations of everything will eventually remain constant. This is referred to as dynamic equilibrium.

Le Chatelier’s Principle

Le Chatelier’s Principle states that “If a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose that change.”

Effect of Changing Concentration

Increasing reactant concentration shifts equilibrium right (product concentration increases).
Increasing product concentration shifts equilibrium left (reactant concentration increases).

Example Adding N2 in the Haber Process

CIE A-Level Chemistry diagram of the Haber Process showing the reversible reaction between nitrogen and hydrogen forming ammonia.

Adding more N2 shifts the equilibrium right, producing more NH3.

Effect of Changing Pressure (for Gaseous Equilibria)

Increasing pressure shifts equilibrium towards the side with fewer gas molecules.
Decreasing pressure shifts equilibrium towards the side with more gas molecules.

Example Haber Process molecules

4 moles (N2 + 3H2) ⇌ 2 moles (NH3). Higher pressure shifts equilibrium right, increasing NH3 yield.

Effect of Changing Temperature

Increasing temperature favours the endothermic direction (+ΔH).
Decreasing temperature favours the exothermic direction (−ΔH).

Example Haber Process enthalpy

In the Haber Process, the forward reaction is exothermic (−ΔH).

CIE A-Level Chemistry energy direction diagram for the Haber Process showing exothermic forward reaction.

Increasing temperature shifts equilibrium left, reducing NH3 yield.
Decreasing temperature shifts equilibrium right, increasing NH3 yield.

Effect of a Catalyst on Equilibrium

Catalysts do not shift equilibrium position. They increase the rate of both forward and reverse reactions equally.
Equilibrium is reached faster but at the same position.

Example Iron catalyst in the Haber Process

Speeds up NH3 production but does not affect yield.

Equilibrium Constant Kc

Kc is a constant that expresses the position of equilibrium in terms of concentrations. It applies to homogeneous systems (all substances in the same phase).

General formula for Kc:

CIE A-Level Chemistry general reaction used to define the equilibrium constant Kc.

Kc =

CIE A-Level Chemistry expression for Kc in terms of equilibrium concentrations.

[A], [B], [C], [D] are equilibrium concentrations in mol dm−3.
a, b, c, and d are the balancing numbers from the equation.

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Matt’s exam tip

Solids aren’t included in Kc expressions and if water is a solvent as well as a reactant or product, it also isn’t included.

Interpreting Kc

Mole Fractions and Partial Pressure

The total pressure of a gaseous system at equilibrium is directly related to the number of moles of each gas in the mixture. The contribution of a gas to the total pressure is its partial pressure. All partial pressures add to the total pressure.

Partial pressure is calculated using:

CIE A-Level Chemistry formula for partial pressure in terms of mole fraction and total pressure.

The mole fraction of a gas is the moles of that gas divided by the total moles of gas in the mixture. It is calculated using:

CIE A-Level Chemistry formula for mole fraction of a gas in a mixture.

Equilibrium Constant Kp

Kp is the equilibrium constant for gaseous systems; it works like Kc but uses partial pressures instead of concentration. It applies to homogeneous gaseous equilibria.

For the general reaction:

CIE A-Level Chemistry general gaseous reaction used to define the equilibrium constant Kp.

The Kp expression is:

CIE A-Level Chemistry expression for Kp in terms of partial pressures of gases.

where P[A], P[B], P[C], P[D] are the partial pressures of each gas, and a, b, c, d are the stoichiometric coefficients from the balanced equation.
Units depend on the reaction and must be included in answers.

Calculating Kc and Kp

You need to be able to calculate Kc and Kp values based on given data and use them to determine equilibrium quantities.

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Matt’s exam tip

Remember the concentrations and partial pressures in Kc and Kp expressions are the equilibrium values. You may need to calculate these from starting moles — read questions carefully and include correct units for Kc and Kp.


Worked Example (Kc)

Find Kc for the system at 298 K:
CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

  1. Given equilibrium concentrations:
    [CH3COOH] = 0.20 mol dm−3
    [C2H5OH] = 0.20 mol dm−3
    [CH3COOC2H5] = 0.40 mol dm−3
    [H2O] = 0.40 mol dm−3
  2. Kc calculation:
    Kc = [CH3COOC2H5][H2O] / ([CH3COOH][C2H5OH])
    Kc = (0.40 × 0.40) / (0.20 × 0.20) = 4.0
  3. Since Kc > 1, equilibrium favours the products.

Worked Example (Kp)

Find Kp for the equilibrium given total pressure = 400 kPa and equilibrium moles N2O4 = 0.40, NO2 = 0.60:
N2O4(g) ⇌ 2NO2(g)

  1. Mole fractions:
    X(N2O4) = 0.40 / (0.40 + 0.60) = 0.40
    X(NO2) = 0.60 / (0.40 + 0.60) = 0.60
  2. Partial pressures:
    P(N2O4) = 0.40 × 400 = 160 kPa
    P(NO2) = 0.60 × 400 = 240 kPa
  3. Kp expression:
    Kp = (P[NO2]2) / (P[N2O4]) = (2402) / 160 = 360 kPa

Calculating Quantities at Equilibrium

For many questions, you need to determine equilibrium quantities or concentrations from starting amounts.

To do this, we can use ICE tables (see worked example below).

Worked Example (ICE method)

At 298 K, 0.5 moles of N2 and 1.5 moles of H2 are mixed in a sealed container of volume 1.00 dm3. At equilibrium, [NH3] = 0.300 mol dm−3. Calculate the equilibrium concentrations of nitrogen and hydrogen.
N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 1: Set up an ICE table (I = Initial, C = Change, E = Equilibrium)

N2 H2 NH3
I (mol dm−3) 0.500 1.500 0
C (mol dm−3) −x −3x +2x
E (mol dm−3) 0.500 − x 1.500 − 3x 0.300

Step 2: Solve for x using stoichiometry (2x = 0.300) → x = 0.150 mol dm−3

Step 3: Calculate other equilibrium concentrations
[N2] = 0.500 − x = 0.500 − 0.150 = 0.350 mol dm−3
[H2] = 1.500 − 3x = 1.500 − 0.450 = 1.050 mol dm−3
[NH3] = 0.300 mol dm−3 (given)

What Affects the Equilibrium Constant?

Factor Effect on Kc / Kp
Temperature Changes K (value depends on endo/exothermic direction)
Concentration No change
Pressure No change
Catalyst No change

Only temperature affects Kc or Kp. Other changes shift the equilibrium position, but the system adjusts to restore the same constant value.

Industrial Application Examples: The Haber and Contact Processes

Haber Process (NH3 production)

N2 + 3H2 ⇌ 2NH3 (forward reaction exothermic = −ΔH)

Contact Process (H2SO4 production)

2SO2 + O2 ⇌ 2SO3 (forward reaction endothermic = −ΔH)

Summary