Skip to topic
    ← Back to course topics

    Topic 10: Equilibrium I — Edexcel A-Level Chemistry

    Test yourself on Topic 10: Equilibrium I with PEARSON EDEXCEL A-Level practice questions.

    Start free

    7 days Premium · Then free forever · No card, no charge

    Topic 10: Equilibrium I explained

    This topic introduces the concept of oxidation numbers as a systematic method for classifying redox reactions, including disproportionation.

    Read the full explanation

    Students learn to define oxidation and reduction in terms of electron transfer and changes in oxidation number, and apply these principles to write and balance ionic half-equations.

    What to demonstrate

    1. Correct calculation of oxidation numbers in compounds and ions, including peroxides and metal hydrides.
    2. Correct identification of oxidation and reduction based on electron transfer and oxidation number changes.
    3. Correct identification of oxidising and reducing agents.
    Show all 6 objectives
    1. Correct identification of disproportionation reactions.
    2. Correct use of Roman numerals to indicate oxidation numbers.
    3. Correct construction of full ionic equations from ionic half-equations.

    Topic 10: Equilibrium I exam tips

    Topic Overview

    Equilibrium I introduces the concept of dynamic equilibrium in reversible reactions, a cornerstone of physical chemistry. This topic explores how reactions can proceed in both forward and backward directions simultaneously, reaching a state where the rates of the forward and reverse reactions are equal. You will learn to apply Le Chatelier's principle to predict how changes in concentration, pressure, and temperature affect the position of equilibrium, and understand the importance of catalysts in speeding up the attainment of equilibrium without altering the equilibrium position. This knowledge is essential for understanding industrial processes like the Haber process and Contact process, where optimising yield is economically vital.

    The topic also introduces the equilibrium constant, Kc, for homogeneous reactions. You will learn to write expressions for Kc, calculate its value from equilibrium concentrations, and use it to determine whether a reaction is product-favoured or reactant-favoured. Understanding Kc allows you to predict the direction of reaction and the composition of the equilibrium mixture. This quantitative aspect is crucial for A-Level exams, where you must be comfortable with algebraic manipulation and units. Equilibrium I lays the foundation for more advanced topics like Kp, acid-base equilibria, and solubility products in later modules.

    Mastering Equilibrium I is not just about passing exams; it develops your ability to think about systems in balance, a concept that appears across chemistry, biology, and even economics. By understanding how external changes perturb a system, you gain insight into everything from enzyme kinetics to environmental cycles. This topic also sharpens your problem-solving skills, as you learn to handle simultaneous equations and interpret data from experiments. A solid grasp of equilibrium is essential for any student aiming for top grades in A-Level Chemistry.

    Key Concepts
    • →Dynamic equilibrium: In a closed system, the forward and reverse reactions occur at the same rate, so concentrations of reactants and products remain constant. The system is not static; molecules are constantly reacting.
    • →Le Chatelier's principle: If a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the equilibrium shifts to counteract the change. This predicts the effect on yield.
    • →Equilibrium constant Kc: For a reaction aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b. Kc is constant at a given temperature. A large Kc (>10^3) means products dominate; a small Kc (<10^-3) means reactants dominate.
    • →Effect of temperature on Kc: For exothermic reactions, increasing temperature decreases Kc (equilibrium shifts left). For endothermic reactions, increasing temperature increases Kc (equilibrium shifts right). Only temperature changes Kc.
    • →Catalysts: They speed up both forward and reverse reactions equally, so equilibrium is reached faster but the position of equilibrium and Kc are unchanged.
    Marking Points
    • Correct calculation of oxidation numbers in compounds and ions, including peroxides and metal hydrides.
    • Correct identification of oxidation and reduction based on electron transfer and oxidation number changes.
    • Correct identification of oxidising and reducing agents.
    • Correct identification of disproportionation reactions.
    • Correct use of Roman numerals to indicate oxidation numbers.
    • Correct construction of full ionic equations from ionic half-equations.
    Examiner Tips
    • 💡Always check that the sum of oxidation numbers in a neutral compound equals zero and in an ion equals the charge of the ion.
    • 💡Remember that oxidising agents are reduced (gain electrons) and reducing agents are oxidised (lose electrons).
    • 💡When balancing half-equations, ensure the total charge on both sides is equal.
    • 💡Practice identifying oxidation numbers in various contexts, especially for s- and p-block elements.
    • 💡When writing Kc expressions, always use square brackets to denote concentration in mol dm^-3. Exclude solids and pure liquids (e.g., H2O(l) in aqueous equilibria) because their concentrations are constant. For heterogeneous equilibria, only include gases and aqueous species.
    • 💡In calculations, always set up an ICE (Initial, Change, Equilibrium) table. Clearly define the change in moles using stoichiometry. Remember that if you are given initial moles and volume, convert to concentration first. Check units: Kc may have units depending on the sum of coefficients.
    • 💡When applying Le Chatelier's principle, state the change, then the effect on the equilibrium (shift to left or right), and finally the effect on the yield of the product. For temperature changes, also mention the effect on Kc. Do not forget that catalysts do not affect yield.
    Common Mistakes
    • Confusing the direction of electron transfer in oxidation and reduction.
    • Incorrectly assigning oxidation numbers in complex ions or species.
    • Failing to balance both atoms and charges when constructing ionic half-equations.
    • Misidentifying the species being oxidised or reduced in a disproportionation reaction.
    • Misconception: At equilibrium, the concentrations of reactants and products are equal. Correction: Equilibrium means the rates are equal, not the concentrations. The concentrations are constant but not necessarily equal; they depend on the reaction and conditions.
    • Misconception: Adding a catalyst shifts the equilibrium to the right. Correction: A catalyst does not affect the position of equilibrium; it only helps the system reach equilibrium faster by lowering the activation energy for both directions.
    • Misconception: Increasing pressure always shifts equilibrium to the side with fewer moles of gas. Correction: This is true only if the reaction involves gases and the change is due to a decrease in volume (increase in pressure). Adding an inert gas at constant volume does not change partial pressures, so no shift.
    Frequently Asked Questions
    What is the difference between static and dynamic equilibrium?
    Static equilibrium is when nothing is happening, like a book on a table. Dynamic equilibrium occurs in a closed system where the forward and reverse reactions are happening at the same rate, so the concentrations of reactants and products remain constant. The reactions are still occurring, but there is no net change.
    How do I calculate Kc from equilibrium concentrations?
    First, write the balanced equation. Then, write the Kc expression: products over reactants, each raised to the power of its stoichiometric coefficient. Substitute the equilibrium concentrations (in mol dm^-3) into the expression. For example, for N2 + 3H2 ⇌ 2NH3, Kc = [NH3]^2 / ([N2][H2]^3). Calculate the value and include units if necessary.
    Does a catalyst affect the equilibrium constant?
    No, a catalyst does not affect the equilibrium constant (Kc). It only speeds up the rate at which equilibrium is reached by lowering the activation energy for both forward and reverse reactions equally. The position of equilibrium and the value of Kc remain unchanged.
    Why does increasing temperature change Kc?
    Kc is temperature-dependent because the equilibrium constant is related to the Gibbs free energy change (ΔG = -RT ln K). For an exothermic reaction, increasing temperature adds heat, which the system counteracts by favouring the endothermic reverse reaction, decreasing Kc. For an endothermic reaction, increasing temperature favours the forward reaction, increasing Kc.
    How do I know which way the equilibrium shifts when pressure changes?
    Use Le Chatelier's principle: if pressure increases (by decreasing volume), the equilibrium shifts to the side with fewer moles of gas to reduce pressure. If pressure decreases, it shifts to the side with more moles of gas. Count only gaseous species. For example, in N2(g) + 3H2(g) ⇌ 2NH3(g), there are 4 moles on the left and 2 on the right, so increasing pressure shifts equilibrium to the right.
    What are the units of Kc?
    The units of Kc depend on the sum of the stoichiometric coefficients. For a general reaction aA + bB ⇌ cC + dD, Kc has units (mol dm^-3)^(c+d-a-b). If the sum of coefficients on both sides is equal, Kc is dimensionless. For example, for H2 + I2 ⇌ 2HI, Kc has no units. For N2 + 3H2 ⇌ 2NH3, Kc has units (mol dm^-3)^(-2) or dm^6 mol^-2.