Ch 1 · Exothermic and Endothermic Reactions · IGCSE Chemistry 0620
Topic 1 of 1 in Chemical Energetics — IGCSE Chemistry 0620 — 45 lessons.
Introduction
Strike a match and the surroundings get hotter; dissolve a cold pack and they get colder. Every chemical reaction either gives out or takes in thermal energy, and chemists show this on a reaction pathway diagram. Extended candidates go further, calculating the energy change from the bonds broken and made. Precise wording about the surroundings is what separates full marks from half marks here.
Key ideas — Exothermic and endothermic reactions
An exothermic reaction transfers thermal energy to the surroundings, increasing their temperature. Combustion, neutralisation, respiration, the reaction of metals with acids and the hydration of anhydrous copper(II) sulfate are all exothermic.
🧑🏫 Teacher's explanation
Think of energy like money in a bank account. In an exothermic reaction, the chemical system is 'spending' its energy, throwing it out into the surroundings like cash. Because the surroundings receive this extra energy, they get hotter. In an endothermic reaction, the system is 'borrowing' energy from the surroundings to make the reaction happen. Because the surroundings lose energy, they get colder.
📖 Definitions
🔍 Worked example — Easy
A student reacts magnesium with hydrochloric acid. The temperature of the solution increases from 25 circC to 35 circC. Is this reaction exothermic or endothermic?
🔍 Worked example — Standard
Calculate the enthalpy change (Δ H) for a reaction where the energy of the reactants is 500 kJ/mol and the energy of the products is 300 kJ/mol.
⚠️ Common errors
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🎯 Exam question
Key ideas — Reaction pathway diagrams and activation energy
A reaction pathway diagram plots energy against the progress of the reaction. The reactants start at one level and the products finish at another.
🧑🏫 Teacher's explanation
Think of a reaction like a hiker trying to get to a valley on the other side of a mountain. Even if the destination (the products) is much lower than the starting point (the reactants), the hiker still has to climb over the mountain peak first. That climb represents the 'activation energy'. If the hiker ends up at a lower elevation than they started, they have released potential energy to the surroundings (exothermic). If they end up higher up, they must have absorbed energy from the surroundings to make that climb (endothermic).\n\nIn chemistry, we track this using energy profiles. The vertical axis represents the energy level. If the products are at a lower energy level than the reactants, the difference is the energy released to the surroundings. If the products are higher, the difference is the energy absorbed from the surroundings.
📖 Definitions
🔍 Worked example — Easy
A reaction has a reactant energy level of 100 kJ/mol and a product energy level of 40 kJ/mol. Calculate the enthalpy change, Δ H.
🔍 Worked example — Standard
In an endothermic reaction, the activation energy (E_a) is 50 kJ/mol and the enthalpy change (Δ H) is +30 kJ/mol. Calculate the energy level of the reactants if the products are at 100 kJ/mol.
⚠️ Common errors
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🎯 Exam question
Key ideas — Enthalpy change, ΔH
🧑🏫 Teacher's explanation
Imagine you are playing a video game where you have to spend 'energy points' to perform an action. If you spend points to perform a move, your personal total goes down (negative change). If you perform a move that rewards you with points, your total goes up (positive change). In chemistry, we look at the 'bank account' of the chemicals.
📖 Definitions
🔍 Worked example — Easy
A reaction has an enthalpy change (Δ H) of -280 kJ/mol. Is the reaction exothermic or endothermic? Explain your answer.
🔍 Worked example — Standard
Calculate the enthalpy change (Δ H) for a reaction where the energy of the reactants is 450 kJ/mol and the energy of the products is 120 kJ/mol.
🔍 Worked example — Exam-style
Calculate the enthalpy change (Δ H) for the reaction H_2 + Cl_2 arrow 2HCl using the following bond energies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol.
⚠️ Common errors
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🎯 Exam question
Key ideas — Bond energies
🧑🏫 Teacher's explanation
Think of chemical bonds like springs connecting two balls. To pull the balls apart (breaking a bond), you have to put energy into the system, like stretching a spring. This is endothermic. When the balls snap back together to form a new connection (making a bond), they release that stored energy, like a spring snapping shut. This is exothermic.
📖 Definitions
🔍 Worked example — Easy
Calculate the enthalpy change for the reaction: H_2 + I_2 arrow 2HI. Given bond energies: H--H = 436 kJ/mol, I--I = 151 kJ/mol, and H--I = 299 kJ/mol.
🔍 Worked example — Standard
In a reaction, 500 kJ/mol of energy is absorbed to break the reactant bonds, and 850 kJ/mol of energy is released when the product bonds form. Determine the Δ H and state if the reaction is exothermic or endothermic.
⚠️ Common errors
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🎯 Exam question
Worked example
Question. Methane burns in oxygen:
🔍 Worked example — Easy
A reaction involves breaking one bond with energy 400 kJ/mol and forming one bond with energy 350 kJ/mol. Calculate the enthalpy change Δ H.
🔍 Worked example — Standard
Calculate the enthalpy change for the reaction: H_2 + Cl_2 arrow 2HCl. Given bond energies: H-H = 436 kJ/mol, Cl-Cl = 242 kJ/mol, H-Cl = 431 kJ/mol.
🔍 Worked example — Exam-style
Methane burns in oxygen: CH_4 + 2O_2 arrow CO_2 + 2H_2O. Bond energies (kJ/mol): C-H = 412, O=O = 496, C=O = 743, O-H = 463. Calculate Δ H.
⚠️ Common errors
Common mistakes
Practice
Answers A reaction that takes in thermal energy from the surroundings, decreasing their temperature [B1]; e.g. thermal decomposition of calcium carbonate or photosynthesis [B1]. The minimum energy that colliding particles must have to react [B1]. Products are at a lower energy level than the reactants [B1]. Products drawn above reactants [B1]; E_a arrow from reactant level to the peak [B1]; Δ H arrow from reactant level up to product level, labelled positive [B1]. Bonds broken = 436 + 193 = 629 kJ [M1]; bonds formed = 2 × 366 = 732 kJ [M1]; Δ H = 629 - 732 = -103 kJ/mol [A1]; more energy is released in forming the H–Br bonds than is needed to break the H–H and Br–Br bonds [B1].