OxfordAQA International GCSE Chemistry: Exothermic and endothermic reactions, and calculating and explaining energy change (9202)

This study guide teaches energy changes in reactions for OxfordAQA International GCSE Chemistry (9202). It follows the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, and covers every statement in section 3.9.1 (a to d) and section 3.9.2 (a to h). The qualification is untiered, so all of it applies to every candidate. Content from any part of the specification can appear on Paper 1 or Paper 2.

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What this unit covers

Statement	What you must be able to do
3.9.1 a	Energy transfer to or from the surroundings; ΔH sign conventions
3.9.1 b	Exothermic reactions, examples and everyday uses
3.9.1 c	Endothermic reactions, thermal decomposition, sports injury packs
3.9.1 d	Reversible reactions: hydrated and anhydrous copper sulfate
3.9.2 a	Compare fuels by simple calorimetry and Q = mcΔT
3.9.2 b	Give energy in J or kJ, per gram or per mole
3.9.2 c	Energy changes of reactions in solution
3.9.2 d	Energy level diagrams
3.9.2 e	Bond breaking and forming; energy calculations
3.9.2 f, g	Exothermic and endothermic in terms of bonds
3.9.2 h	Catalysts and activation energy

Section 6.2 attaches no required practical to 3.9.1 or 3.9.2, but you can still be asked to describe and improve both calorimetry methods.

Per-mole calculations use the mole work in the quantitative chemistry guide.

3.9.1 a: energy, the surroundings and ΔH

Split the situation into the reacting chemicals and the surroundings (the water, the beaker, the air). Energy is conserved, so when a reaction happens energy moves between the two.

ΔH (delta H) is the enthalpy change: the energy change of the chemicals themselves, normally given in kJ/mol.

• If the chemicals lose energy to the surroundings, ΔH is negative. The reaction is exothermic.
• If the chemicals gain energy from the surroundings, ΔH is positive. The reaction is endothermic.

The sign describes the chemicals, not the thermometer: the thermometer rises in an exothermic reaction, yet ΔH is negative.

3.9.1 b: exothermic reactions

An exothermic reaction transfers energy to the surroundings, so the surroundings warm up. Examples you should be able to give:

• Combustion, such as natural gas burning.
• Oxidation, for example iron rusting.
• Neutralisation, such as hydrochloric acid reacting with sodium hydroxide solution.

Everyday uses:

• Self-heating cans (for example coffee) keep two reactants apart until a button is pressed. A common pairing is calcium oxide and water: CaO + H₂O → Ca(OH)₂.
• Hand warmers. Disposable air-activated ones release heat as iron powder oxidises. Reusable ones release heat when a solution crystallises, triggered by flexing a small metal disc.

3.9.1 c: endothermic reactions

In an endothermic reaction, energy is taken in from the surroundings. They cool down, or energy has to be supplied continuously by heating.

• Thermal decompositions are endothermic. Copper carbonate only breaks down into copper oxide and carbon dioxide while you keep heating it: CuCO₃ → CuO + CO₂.
• Sports injury packs that go cold when squeezed keep a solid and water apart. Many use ammonium nitrate, which takes in energy as it dissolves.

3.9.1 d: reversible reactions and energy

In some reactions the products can react to re-form the original reactants. These are reversible reactions, written with the ⇌ sign:

A + B ⇌ C + D

The specification’s example is copper sulfate:

                         endothermic →
hydrated copper sulfate  ⇌  anhydrous copper sulfate + water
       (blue)            ← exothermic        (white)

Heating the blue crystals drives off water and leaves a white powder: endothermic. Adding water to the white powder turns it blue and the mixture gets hot: exothermic.

Key idea: the forward and reverse reactions are opposite types, one exothermic and one endothermic, and the same amount of energy is transferred each way. Only the sign of ΔH changes.

3.9.2 a and b: comparing fuels by simple calorimetry

Use the burning fuel to heat a known mass of water in a glass or metal container, such as a copper can. Weigh the spirit burner before and after, and record the water’s starting and highest temperatures.

The energy transferred to the water is:

Q = m c ΔT
Q  = energy transferred (J)
m  = mass of water heated (g)
c  = specific heat capacity of water (4.2 J/g °C)
ΔT = temperature change (°C)

Energy is measured in joules (J) or kilojoules (kJ), where 1 kJ = 1000 J. To compare fuels fairly, divide by how much fuel burned: kJ per gram (kJ/g) or kJ per mole (kJ/mol).

Worked example 1: two alcohol fuels

Each fuel heats 200 g of water in a copper can. Use c = 4.2 J/g °C.

Fuel	Mass burned	Water start	Water highest
Propan-1-ol, C₃H₇OH (Mr = 60)	0.84 g	20.5 °C	41.0 °C
Butan-1-ol, C₄H₉OH (Mr = 74)	0.74 g	19.8 °C	41.6 °C

Propan-1-ol

• ΔT = 41.0 − 20.5 = 20.5 °C
• Q = 200 × 4.2 × 20.5 = 17 220 J = 17.2 kJ
• Per gram: 17.22 ÷ 0.84 = 20.5 kJ/g
• Moles burned: 0.84 ÷ 60 = 0.014 mol, so per mole: 17.22 ÷ 0.014 = 1230 kJ/mol

Butan-1-ol

• ΔT = 41.6 − 19.8 = 21.8 °C
• Q = 200 × 4.2 × 21.8 = 18 312 J = 18.3 kJ
• Per gram: 18.312 ÷ 0.74 = 24.7 kJ/g
• Moles burned: 0.74 ÷ 74 = 0.010 mol, so per mole: 1830 kJ/mol (3 significant figures)

Conclusion: butan-1-ol released more energy per gram and per mole. Comparing raw temperature rises would mislead, because different masses burned.

Why the values come out low. Energy also heats the can and the air; some fuel burns incompletely, leaving soot; some alcohol evaporates from the wick. A draught shield, a lid, a short flame-to-can distance and stirring all help. For a fair comparison, keep the mass of water, the container and the flame-to-can distance the same for every fuel.

3.9.2 c: reactions in solution

Mix the reagents in an insulated container, usually a polystyrene cup with a lid, and measure the temperature change of the solution. The insulation cuts energy exchange with the air. The method suits neutralisation reactions and solids reacting with water.

Treat dilute solutions as water: 1 cm³ = 1 g, c = 4.2 J/g °C. Use the total volume of solution.

Worked example 2: neutralisation

40.0 cm³ of 1.50 mol/dm³ nitric acid is added to 40.0 cm³ of 1.50 mol/dm³ potassium hydroxide solution in a polystyrene cup. The temperature rises by 10.2 °C. Calculate the energy change per mole of water formed.

• Mass of solution = 40.0 + 40.0 = 80.0 g
• Q = 80.0 × 4.2 × 10.2 = 3427 J = 3.427 kJ
• Moles of acid = 1.50 × (40.0 ÷ 1000) = 0.0600 mol, and HNO₃ + KOH → KNO₃ + H₂O, so 0.0600 mol of water forms
• Energy per mole = 3.427 ÷ 0.0600 = 57.1 kJ/mol
• The temperature rose, so the reaction is exothermic: ΔH = −57.1 kJ/mol

Worked example 3: a solid dissolving

5.05 g of potassium nitrate, KNO₃ (Mr = 101), dissolves in 50.0 cm³ of water. The temperature falls from 21.4 °C to 13.1 °C.

• ΔT = 21.4 − 13.1 = 8.3 °C (a fall)
• Q = 50.0 × 4.2 × 8.3 = 1743 J = 1.743 kJ (use the mass of water)
• Moles of KNO₃ = 5.05 ÷ 101 = 0.0500 mol
• Energy per mole = 1.743 ÷ 0.0500 = 34.9 kJ/mol
• The solution cooled, so energy was taken in: ΔH = +34.9 kJ/mol

3.9.2 d: energy level diagrams

An energy level diagram plots energy against the progress of the reaction. It shows:

• the relative energies of reactants and products, as two horizontal levels;
• the activation energy, the minimum energy needed for the reaction to happen, measured upwards from the reactants to the highest point of the curve;
• the overall energy change, ΔH, from the reactant level to the product level.

A curved line joins reactants to products, rising over the activation energy “hill” first.

Energy (kJ)
 345 |           ___            uncatalysed peak
     |          /   \
 280 |         / _-_ \          catalysed peak
     |        / /   \ \
 210 |_______/_/     \ \        reactants
     |                \ \
  95 |                 \_\______ products
     |
     +--------------------------------
                progress of reaction

Worked example 4: reading a profile

Reactants sit at 210 kJ, the uncatalysed peak at 345 kJ and the products at 95 kJ.

• Activation energy = 345 − 210 = 135 kJ
• Overall energy change = 95 − 210 = −115 kJ. Products sit below reactants, which makes it exothermic.
• With a catalyst the peak is 280 kJ, so the activation energy is 280 − 210 = 70 kJ. ΔH stays at −115 kJ.

For an endothermic reaction, the product level sits above the reactant level and ΔH is positive.

3.9.2 e, f and g: bond breaking and bond making

During a reaction:

• energy must be supplied to break bonds (an endothermic step);
• energy is released when bonds form (an exothermic step).

Exothermic overall (3.9.2 f): the energy released forming new bonds is greater than the energy needed to break the existing bonds. Endothermic overall (3.9.2 g): the energy needed to break existing bonds is greater than the energy released forming new bonds.

Activation energy links to bond breaking: bonds must start to break before new ones form, which needs an energy input even in an exothermic reaction.

To calculate the energy transferred from bond dissociation energies you are given:

energy change = energy to break bonds − energy released forming bonds

A negative answer means exothermic; a positive answer means endothermic.

Worked example 5: burning ammonia (exothermic)

4NH₃ + 3O₂ → 2N₂ + 6H₂O. Bond energies to use, in kJ/mol: N–H 391, O=O 498, N≡N 945, O–H 464.

• Bonds broken: 4 NH₃ contain 12 N–H, plus 3 O=O. 12 × 391 + 3 × 498 = 4692 + 1494 = 6186 kJ
• Bonds formed: 2 N≡N plus 6 H₂O containing 12 O–H. 2 × 945 + 12 × 464 = 1890 + 5568 = 7458 kJ
• Energy change = 6186 − 7458 = −1272 kJ for the equation as written, so exothermic. That is −318 kJ per mole of ammonia.

Worked example 6: decomposing methanol (endothermic)

CH₃OH → CO + 2H₂. Bond energies to use, in kJ/mol: C–O 358, C–H 413, O–H 464, H–H 436, C≡O 1077.

• Bonds broken: 3 C–H + 1 C–O + 1 O–H = 1239 + 358 + 464 = 2061 kJ
• Bonds formed: 1 C≡O + 2 H–H = 1077 + 872 = 1949 kJ
• Energy change = 2061 − 1949 = +112 kJ/mol, so endothermic, as you would expect for a thermal decomposition.

Picture each displayed formula and count every bond, then multiply by the coefficients.

3.9.2 h: catalysts

With a catalyst, the reaction follows a different pathway whose activation energy is lower. On an energy level diagram, draw the catalysed curve with a lower peak between the same reactant and product levels. A catalyst does not change ΔH.

Common errors

• Giving ΔH a positive sign because the temperature went up.
• Putting the fuel’s mass into Q = mcΔT instead of the water’s.
• Using only one solution’s volume when two solutions are mixed.
• Dividing joules by moles but writing the unit as kJ/mol.
• Counting bonds in one molecule but forgetting the coefficient, such as the 4 in 4NH₃.
• Calculating “formed − broken”, which flips the sign.

Where next

Use the revision notes, then the practice questions. For covalent bonds, see the bonding guide.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. This guide covers section 3.9.1 Exothermic and endothermic reactions and section 3.9.2 Calculating and explaining energy change.