Exothermic and Endothermic Reactions: Revision Notes

Condensed for the final weeks. For the full explanation, use the Exothermic and Endothermic Reactions study guide.

Tier note. Only the basic idea of exothermic/endothermic (energy in vs out, temperature rising vs falling) and reading a simple reaction pathway diagram are 0620 Core. Naming ΔH and its sign convention, activation energy, drawing/labelling a full pathway diagram, explaining ΔH via bond breaking/making, and the bond-energy calculation are all 0620 Extended, 5070 required. 5070 has no Core/Extended split, so all of it is required for O Level.

The two types

	Exothermic	Endothermic
Energy	Released to surroundings	Absorbed from surroundings
Temperature of surroundings	Rises	Falls
ΔH sign	Negative	Positive
Products vs reactants	Products lower in energy	Products higher in energy
Examples	Combustion, neutralisation, respiration, most displacement	Thermal decomposition, photosynthesis, dissolving ammonium salts

The sign convention is from the system’s point of view: exothermic loses energy, so ΔH is negative — even though the thermometer reading goes up. It helps to picture the reacting chemicals as “the system” and everything else (the water bath, the container, the air) as “the surroundings” — ΔH always describes what happens to the system’s own stored energy, not what a thermometer placed in the surroundings records.

Reaction pathway diagrams

EXOTHERMIC

        Ea
       /‾‾\
      /    \
reactants   \
             \____ products

ΔH NEGATIVE (products lower in energy than reactants)

ENDOTHERMIC

              Ea
            /‾‾\
           /    \
          /      \____ products
reactants/

ΔH POSITIVE (products higher in energy than reactants)

Label every diagram with: reactants, products, activation energy Eₐ (reactants to the peak), and ΔH (reactants to products).

Activation energy — the minimum energy colliding particles need to react. It is always measured from the reactants up to the peak of the curve, never from the peak down to the products — a frequent source of a lost mark when labelling a diagram under time pressure.

Bond energy calculations

Breaking bonds is endothermic (energy in). Making bonds is exothermic (energy out).

dH = energy to BREAK bonds  -  energy RELEASED making bonds
   = sum(bonds broken)      -  sum(bonds formed)

Negative result -> exothermic
Positive result -> endothermic

Worked example: H₂ + Cl₂ → 2HCl, given H–H 436, Cl–Cl 242, H–Cl 431 kJ/mol.

Broken:  436 + 242            = 678
Formed:  2 x 431              = 862
dH = 678 - 862 = -184 kJ/mol   -> EXOTHERMIC

Calorimetry — measuring ΔH experimentally

Not on the 0620/5070 syllabus. Calorimetry, specific heat capacity and Q = mcΔT calculations are not part of either syllabus’s chemical energetics content — the only quantitative energetics outcome required at this level is the bond-energy calculation above. This section is included only as background for progression to AS Chemistry (9701), and will not be examined at IGCSE/O Level.

Q = m x c x dT

Q  = energy transferred (J)
m  = mass of the SOLUTION, not just the solute (g)
c  = specific heat capacity (J/g/degC), often 4.18 for a dilute aqueous solution
dT = temperature change (degC)

Worked example. 50.0 cm³ of acid is neutralised by 50.0 cm³ of alkali (equal volumes), and the temperature rises by 6.8 °C. Take the specific heat capacity as 4.18 J g⁻¹ °C⁻¹ and the density as 1.00 g cm⁻³.

total volume = 50.0 + 50.0 = 100.0 cm^3 (acid + alkali, both warmed)
mass = 100.0 g (density 1.00 g/cm^3, so 100.0 cm^3 -> 100.0 g)
Q = 100.0 x 4.18 x 6.8 = 2842.4 J  (~2.84 kJ)

Why the experimental value is usually lower in magnitude than the accepted value: heat is lost to the surroundings, the thermometer and the container during the reaction, so less of the energy released ends up warming the measured solution than the true reaction actually releases; an open or poorly insulated cup (rather than a proper calorimeter) loses more heat still, in the same direction.

A separate simplifying assumption (not itself a reason for the value being lower): the specific heat capacity of the solution is taken to be that of pure water, 4.18 J g⁻¹ °C⁻¹, even though a dilute solution’s true value is slightly different — this is a source of small error, but not one that pushes the answer in a predictable direction.

Catalysts on the diagram

CORE (0620) · REQUIRED (5070) — a catalyst increases the rate of reaction and is chemically unchanged at the end of it.

(0620 Extended, 5070 required) A catalyst lowers Eₐ by providing an alternative pathway. It does not change ΔH — the reactant and product levels are unchanged. Draw a second, lower curve between the same two levels, starting and ending at exactly the same points as the original curve. (This is rates-of-reaction content, topic 6.2, applied here to the energetics diagram.)

Exam traps

• Exothermic = negative ΔH, even though the surroundings get hotter. Most-missed mark in the topic.
• Bond breaking absorbs energy; bond making releases it. Reversing these flips the whole answer.
• Eₐ is measured from the reactants to the peak, not from the peak to the products.
• A catalyst changes Eₐ only, never ΔH.
• Include the sign and units (kJ/mol) in the final answer.
• (Background only, not a 0620/5070 outcome — AS 9701 progression) Using the mass of the solute rather than the total mass of the solution in Q = mcΔT.
• (Background only, not a 0620/5070 outcome — AS 9701 progression) Forgetting that a calorimetry experiment’s measured energy change is usually smaller in magnitude than the accepted (data-book) value, because of heat loss to the surroundings, the thermometer and the container.

Self-test

1. A reaction raises the temperature of the water bath. Is ΔH positive or negative?
2. Is bond breaking exothermic or endothermic?
3. Calculate ΔH for a reaction where 950 kJ is needed to break bonds and 1120 kJ released forming them.
4. What does a catalyst change on a reaction pathway diagram?
5. Give two examples of endothermic processes.

Background only — questions 6 and 7 use calorimetry, which is not a 0620 or 5070 outcome and is included only for progression to AS Chemistry 9701:

6. 25.0 g of water rises in temperature by 4.0 °C. Using c = 4.18 J g⁻¹ °C⁻¹, calculate the energy transferred.
7. Give two reasons why a calorimetry experiment usually underestimates the true energy change.

Answers: 1. Negative — the reaction releases energy to the surroundings, so it is exothermic. 2. Endothermic — energy must be supplied to break a bond. 3. 950 − 1120 = −170 kJ/mol, exothermic. 4. It lowers the activation energy by providing an alternative pathway; the reactant and product energy levels, and therefore ΔH, are unchanged. 5. Any two: thermal decomposition of calcium carbonate, photosynthesis, dissolving ammonium nitrate, reaction of citric acid with sodium hydrogencarbonate. 6. (Background) Q = 25.0 × 4.18 × 4.0 = 418 J. 7. (Background) Any two: heat lost to the surroundings during the reaction, heat absorbed by the thermometer and container rather than staying in the solution, and extra heat lost through an open or poorly insulated cup rather than a proper calorimeter — all of which mean less energy is measured than the reaction actually released.

For the full worked explanation with additional detail, see the Exothermic and Endothermic Reactions study guide; for exam-style questions with full mark schemes, see the Exothermic and Endothermic Reactions practice questions.