OCR A Level Chemistry A: Enthalpy changes (H432) – Revision Notes

These notes condense section 3.2.1, Enthalpy changes, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover all eight learning outcomes, 3.2.1(a) to (h), in Module 3, Periodic table and energy. H432 is the full A level. The explanations behind every point are in the Enthalpy changes study guide.

When you are ready, move on to the Enthalpy changes practice questions. Other H432 topics are on the OCR chemistry hub, and the printable checklist lists every outcome. Use the free diagnostics to see where your gaps are.

3.2.1(a)-(c): Signs, profiles and activation energy

	Exothermic	Endothermic
Sign of ∆H	Negative	Positive
Energy flow	System to surroundings	Surroundings to system
Temperature of surroundings	Rises	Falls
Products on profile	Below reactants	Above reactants
∆H arrow	Points down	Points up

• Activation energy, Ea: the minimum energy required for a reaction to take place.
• On a profile, Ea is measured upwards from the reactant line to the peak, whichever way ∆H goes.
• Label both axes (enthalpy; progress of reaction), both levels, ∆H and Ea.

3.2.1(d): Standard conditions and definitions

• Standard conditions: 100 kPa and a stated temperature, 298 K. Shown by ⦵.
• Standard state: the physical state of a substance under standard conditions, e.g. Br₂(l), Na(s), N₂(g).
• ∆rH: the enthalpy change for the amounts in a stated equation.

Definitions you must be able to write:

Term	Definition	Fixed 1 mol of
∆fH	Enthalpy change, under standard conditions, for making 1 mol of a compound from its constituent elements, each in its standard state	Compound formed
∆cH	Enthalpy change when 1 mol of a substance burns completely in oxygen, under standard conditions, all in standard states	Substance burned
∆neutH	Enthalpy change when an acid and a base react to form 1 mol of water, under standard conditions, all in standard states	Water formed

∆fH of any element in its standard state is zero.

Formulas

Use	Formula	Notes
Energy from calorimetry	q = mc∆T	m = mass of water or solution in g; c = 4.18 J g⁻¹ K⁻¹
Enthalpy change per mole	∆H = −q ÷ n	q in kJ; n of the substance in the definition
Bond enthalpies	∆H = Σ(bonds broken) − Σ(bonds made)	Gaseous species only
Hess with ∆fH	∆rH = Σ∆fH(products) − Σ∆fH(reactants)	Elements count as zero
Hess with ∆cH	∆rH = Σ∆cH(reactants) − Σ∆cH(products)	Opposite order to ∆fH

Method in steps

Calorimetry, 3.2.1(e)

1. Find ∆T (extrapolated if a temperature-time graph is given).
2. q = mc∆T, using the mass being heated, not the reacting solid or fuel.
3. Convert J to kJ by dividing by 1000.
4. Find n of the substance named in the definition.
5. ∆H = −q ÷ n for a temperature rise; +q ÷ n for a fall. Give a sign and units.

Bond enthalpies, 3.2.1(f)

1. Draw out every molecule so you can see each bond.
2. Total the bonds broken in the reactants.
3. Total the bonds made in the products.
4. ∆H = broken − made.
5. To find one unknown bond, put x in the equation and solve.

Hess cycles, 3.2.1(g)

1. Write the target reaction across the top.
2. Put the common third box below: the elements (∆fH, arrows up) or the combustion products (∆cH, arrows down).
3. Multiply each ∆H by its coefficient.
4. Go round the long way: add with the arrow, subtract against it.

Small worked reminders

Dissolving a solid. 2.12 g of LiCl (M = 42.4 g mol⁻¹) dissolves in 60.0 g of water. The temperature rises by 7.4 K. q = 60.0 × 4.18 × 7.4 = 1856 J = 1.856 kJ; n = 2.12 ÷ 42.4 = 0.0500 mol; ∆H = −1.856 ÷ 0.0500 = −37.1 kJ mol⁻¹.

Bond enthalpies. CH₃CH=CH₂(g) + HCl(g) → CH₃CHClCH₃(g). Use these values (kJ mol⁻¹): C=C 612, H–Cl 432, C–C 347, C–H 413, C–Cl 346. Only the changing bonds are needed. Broken: C=C + H–Cl = 612 + 432 = 1044. Made: C–C + one new C–H + C–Cl = 347 + 413 + 346 = 1106. ∆H = 1044 − 1106 = −62 kJ mol⁻¹.

3.2.1(h): Techniques

Direct (PAG3)

• Solutions: polystyrene cup with lid, in a beaker; measure volumes with a measuring cylinder or pipette; stir; read the temperature at regular intervals.
• Dissolving a solid (for enthalpy change of solution, 5.2.1(c)): weigh the solid, add to a measured volume of water, stir and record.
• Combustion: weighed spirit burner under a metal can of water; reweigh after heating; draught shield.

Temperature-time graph: extend the best-fit cooling line backwards to the mixing time; this corrects for heat lost while the reaction finishes.

Indirect: measure two reactions that share a common product, then combine them in a Hess cycle. Used when the target reaction is too slow, cannot be controlled, or gives no clean temperature change (hydration of an anhydrous salt; thermal decomposition of a carbonate via reaction with acid). The enthalpy change of hydration of ions (5.2.1(c)) cannot be measured directly; it comes from a cycle combining a measured enthalpy change of solution with the lattice enthalpy.

Why experimental ∆cH values are too small: heat loss to the air, heat absorbed by the can, incomplete combustion, evaporation of fuel, and non-standard conditions.

Must-know distinctions

• ∆H sign vs temperature change: the surroundings warm up when ∆H is negative.
• Average vs actual bond enthalpy: the average comes from many compounds; the actual value in one molecule may be different. This is why bond enthalpy answers differ from ∆fH-based answers.
• ∆fH vs ∆cH cycles: arrows up from elements for ∆fH; arrows down to oxides for ∆cH.
• ∆neutH vs ∆rH for neutralisation: ∆neutH is per mole of water, so a diprotic acid needs ½ in the equation.
• Standard state vs physical state in an experiment: water in a combustion definition is H₂O(l), even if steam forms in the flame.

Quick self-test

1. State the conditions that are meant by “standard conditions”.
2. Write the equation that represents ∆fH of ethanoic acid, CH₃COOH(l).
3. Explain why ∆fH of O₂(g) is zero.
4. Calculate q when 80.0 g of water warms by 5.5 K.
5. A neutralisation forms 0.0250 mol of water and releases 1.40 kJ. Calculate ∆neutH.
6. Burning 0.0100 mol of a fuel releases 13.4 kJ. Calculate ∆cH.
7. Use ∆cH (kJ mol⁻¹) C(s) −394, H₂(g) −286 and CH₃COOH(l) −874 to calculate ∆fH of ethanoic acid.
8. Use ∆fH (kJ mol⁻¹) SiCl₄(l) −687, H₂O(l) −286, SiO₂(s) −911 and HCl(g) −92 to calculate ∆rH for SiCl₄(l) + 2H₂O(l) → SiO₂(s) + 4HCl(g).
9. A reaction breaks bonds totalling 1850 kJ and makes bonds totalling 2010 kJ. Calculate ∆H and state whether it is exothermic or endothermic.
10. Give two reasons why a spirit-burner value of ∆cH is less negative than the data value.
11. On an enthalpy profile, state where the activation energy arrow starts and ends.
12. Explain why the temperature change is found by extrapolation on a temperature-time graph.

Answers

1. 100 kPa and a stated temperature, 298 K.
2. 2C(s) + 2H₂(g) + O₂(g) → CH₃COOH(l)
3. Oxygen is an element in its standard state, so forming it from itself involves no change.
4. q = 80.0 × 4.18 × 5.5 = 1839 J (1.84 kJ)
5. ∆neutH = −1.40 ÷ 0.0250 = −56.0 kJ mol⁻¹
6. ∆cH = −13.4 ÷ 0.0100 = −1340 kJ mol⁻¹
7. ∆fH = [2(−394) + 2(−286)] − (−874) = −1360 + 874 = −486 kJ mol⁻¹
8. ∆rH = [−911 + 4(−92)] − [−687 + 2(−286)] = −1279 + 1259 = −20 kJ mol⁻¹
9. ∆H = 1850 − 2010 = −160 kJ mol⁻¹; exothermic.
10. Any two: heat loss to the surroundings; heat absorbed by the can; incomplete combustion; fuel evaporating from the wick.
11. It starts at the reactant line and ends at the peak (highest energy point).
12. The mixture cools while the reaction is still finishing, so the highest reading is lower than the true value; extrapolating the cooling line to the time of mixing corrects for this.

Where marks are usually lost

• Putting the mass of the solid or fuel into q = mc∆T in place of the water or solution.
• Leaving out the sign or the units on a final ∆H.
• Forgetting to divide q by 1000 before dividing by n.
• Dividing by moles of acid, not moles of water, for a diprotic acid.
• Writing H₂O(g) or omitting state symbols in a definition equation.
• Missing “under standard conditions” or “1 mol” from a written definition.
• Not multiplying ∆fH or ∆cH by the coefficients in the equation.
• Using products minus reactants with ∆cH data.
• Counting bonds in the wrong molecule or forgetting double bonds such as C=O.
• Drawing the Ea arrow from the product level on an exothermic profile.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.2.1 Enthalpy changes, learning outcomes (a) to (h). Published by OCR.