OCR A Level Chemistry A: Enthalpy changes (H432)

This study guide teaches 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. The section sits in Module 3, Periodic table and energy, and has eight lettered learning outcomes, 3.2.1(a) to (h), all covered below. H432 is the full A level. According to the specification’s assessment overview, Module 3 is examined in Periodic table, elements and physical chemistry (01) and Unified chemistry (03). PAG3, Enthalpy determination, is linked to outcomes (e) and (h).

Every H432 topic is linked from the OCR chemistry hub; record your progress on the printable checklist. For a quick check of your starting point, try the free diagnostics.

Outcomes in 3.2.1

Outcome	Skill tested	Specification guidance
3.2.1(a)	Explain exothermic (∆H negative) and endothermic (∆H positive) changes	–
3.2.1(b)	Draw enthalpy profile diagrams comparing reactants and products	M3.1
3.2.1(c)	Explain activation energy qualitatively, using profile diagrams	Minimum energy required for a reaction to take place; M3.1
3.2.1(d)	Use standard conditions and states, ∆rH, ∆fH, ∆cH and ∆neutH	Definitions for formation, combustion and neutralisation only; 100 kPa and a stated temperature, 298 K
3.2.1(e)	Find ∆H directly from experimental results with q = mc∆T	M0.0, M0.2, M2.2 to M2.4; PAG3
3.2.1(f)	(i) Explain average bond enthalpy; (ii) explain ∆H sign by bond breaking and making; (iii) calculate with bond enthalpies	No formal definition; actual values may differ from averages; see 2.2.2(f)
3.2.1(g)	Use Hess’ law cycles with (i) ∆cH, (ii) ∆fH and (iii) unfamiliar cycles	No definition of Hess’ law; unfamiliar cycles provided; HSW2
3.2.1(h)	Describe direct and indirect techniques for finding ∆H	Includes the enthalpy changes in 5.2.1(c); M3.1, M3.2; PAG3; HSW4

3.2.1(a)-(c): Exothermic, endothermic and activation energy

You cannot measure enthalpy, H, itself, only the change: ∆H = H(products) − H(reactants).

• Exothermic: energy passes to the surroundings, which warm up. Products have less enthalpy than reactants, so ∆H is negative. Example: combustion.
• Endothermic: energy is taken from the surroundings, which cool down. Products have more enthalpy, so ∆H is positive. Example: thermal decomposition of a carbonate.

Activation energy, Ea, is the minimum energy required for a reaction to take place. This is why an exothermic reaction may still need a flame to start it.

Worked example 1: drawing enthalpy profiles

Sketch a profile for the burning of methane, an exothermic reaction.

1. Draw a vertical axis labelled enthalpy and a horizontal axis labelled progress of reaction.
2. Draw a horizontal line for CH₄(g) + 2O₂(g) on the left. Draw a lower line for CO₂(g) + 2H₂O(l) on the right.
3. Join them with a curve that rises to a hump before falling to the product level.
4. Draw ∆H as a downward arrow from the reactant level to the product level (∆H negative).
5. Draw Ea as an upward arrow starting at the reactant line and ending at the peak.

For an endothermic reaction the product line sits higher and the ∆H arrow points up. Ea is always measured from the reactants.

3.2.1(d): Standard conditions and enthalpy definitions

Standard conditions: 100 kPa and a stated temperature, usually 298 K. The symbol ⦵ shows a value measured under standard conditions, for example ∆fH⦵.

Standard state means the physical state under standard conditions: H₂O(l), O₂(g), C(s).

Term	Meaning
∆rH, enthalpy change of reaction	The enthalpy change for the amounts in a stated equation
∆fH, enthalpy change of formation	The enthalpy change, under standard conditions, for making 1 mol of a compound from its constituent elements, each in its standard state
∆cH, enthalpy change of combustion	The enthalpy change when 1 mol of a substance is burned completely in oxygen, under standard conditions, all species in standard states
∆neutH, enthalpy change of neutralisation	The enthalpy change when an acid and a base react to form 1 mol of water, under standard conditions, all species in standard states

Learn the wording for ∆fH, ∆cH and ∆neutH. Two consequences: ∆fH of an element in its standard state is zero, and each definition fixes 1 mol of something, so fractions are often needed.

Worked example 2: writing definition equations

• ∆fH of propanone: 3C(s) + 3H₂(g) + ½O₂(g) → CH₃COCH₃(l)
• ∆cH of butan-2-ol: CH₃CH(OH)CH₂CH₃(l) + 6O₂(g) → 4CO₂(g) + 5H₂O(l)
• ∆neutH for sulfuric acid with sodium hydroxide: ½H₂SO₄(aq) + NaOH(aq) → ½Na₂SO₄(aq) + H₂O(l)

3.2.1(e): Finding ∆H from experimental results

In calorimetry, the reaction changes the temperature of a known mass of water or solution. Use:

q = mc∆T

• q: energy transferred, in J
• m: mass of the water or solution heated, in g (take 1.00 cm³ of a dilute solution as 1.00 g)
• c: specific heat capacity, 4.18 J g⁻¹ K⁻¹ for water
• ∆T: temperature change in K (a 1 °C change equals a 1 K change)

Then ∆H = −q ÷ n, with q in kJ and n the amount of the substance in the definition. A temperature rise means energy was released, hence the minus sign.

Worked example 3: enthalpy change of combustion

A spirit burner containing butan-2-ol (M = 74.0 g mol⁻¹) is used to heat 175 g of water. Its mass falls by 0.444 g while the water warms by 13.4 K.

1. q = 175 × 4.18 × 13.4 = 9802 J = 9.802 kJ
2. n(butan-2-ol) = 0.444 ÷ 74.0 = 0.00600 mol
3. ∆cH = −9.802 ÷ 0.00600 = −1630 kJ mol⁻¹ (3 s.f.)

The answer is less exothermic than a data value because of heat loss and incomplete combustion.

Worked example 4: enthalpy change of neutralisation

35.0 cm³ of 1.20 mol dm⁻³ KOH(aq) is mixed with 35.0 cm³ of 1.20 mol dm⁻³ HNO₃(aq) in a polystyrene cup. The temperature rises by 8.2 K.

1. m = 35.0 + 35.0 = 70.0 g of solution
2. q = 70.0 × 4.18 × 8.2 = 2399 J = 2.399 kJ
3. n(H₂O) = 0.0350 × 1.20 = 0.0420 mol
4. ∆neutH = −2.399 ÷ 0.0420 = −57.1 kJ mol⁻¹

Both solutions are warmed, so m uses the total volume.

3.2.1(f): Average bond enthalpies

Bond enthalpy is the energy needed to break 1 mol of a given bond in gaseous molecules. Breaking bonds is endothermic; making bonds is exothermic. A bond’s strength depends on its surroundings: C–H in methane is not identical to C–H in ethanol. Tables therefore give average values over many compounds, and the actual value in a given molecule may differ.

A reaction is exothermic when making new bonds releases more energy than breaking old ones takes in, and endothermic when the reverse is true.

∆H = Σ(energy to break the reactant bonds) − Σ(energy released forming the product bonds)

Worked example 5: ∆H from bond enthalpies

Calculate ∆H for 2NH₃(g) + 3Cl₂(g) → N₂(g) + 6HCl(g). Use these values (kJ mol⁻¹): N–H 391, Cl–Cl 243, N≡N 945, H–Cl 432.

1. Broken: 6 N–H + 3 Cl–Cl = 6(391) + 3(243) = 2346 + 729 = 3075 kJ
2. Made: 1 N≡N + 6 H–Cl = 945 + 6(432) = 945 + 2592 = 3537 kJ
3. ∆H = 3075 − 3537 = −462 kJ mol⁻¹

Worked example 6: finding a bond enthalpy

For H₂(g) + Br₂(g) → 2HBr(g), ∆H = −103 kJ mol⁻¹. H–H is 436 and Br–Br is 193 kJ mol⁻¹. Find the H–Br bond enthalpy.

1. −103 = (436 + 193) − 2x
2. 2x = 629 + 103 = 732
3. x = 366 kJ mol⁻¹

Bond enthalpy answers differ from ∆fH-based values because averages are not exact for your molecules, and because they apply only to gases, so changes of state for liquids or solids are left out.

3.2.1(g): Hess’ law and enthalpy cycles

Hess’ law: going from the same reactants to the same products, every route gives the same total ∆H. It follows from conservation of energy (HSW2).

Draw the reaction across the top and a third box below, linked to both sides. Go the long way round: add a ∆H when you follow its arrow, subtract it when you go against it.

Worked example 7: ∆rH from ∆cH values

Find ∆rH for C₃H₆(g) + H₂(g) → C₃H₈(g). Use ∆cH (kJ mol⁻¹): C₃H₆(g) −2058, H₂(g) −286, C₃H₈(g) −2219.

Both sides burn to the same products, 3CO₂(g) + 4H₂O(l), so both arrows point down to that box.

∆rH = Σ∆cH(reactants) − Σ∆cH(products) = (−2058 + −286) − (−2219) = −125 kJ mol⁻¹

Worked example 8: ∆rH from ∆fH values

Find ∆rH for CS₂(l) + 3O₂(g) → CO₂(g) + 2SO₂(g). Use ∆fH (kJ mol⁻¹): CS₂(l) +89, CO₂(g) −394, SO₂(g) −297.

Here the arrows point up from the elements, C(s) + 2S(s) + 3O₂(g), to both sides.

∆rH = Σ∆fH(products) − Σ∆fH(reactants) = (−394 + 2(−297)) − (+89 + 0) = −988 − 89 = −1077 kJ mol⁻¹

O₂ counts as zero. The ∆cH and ∆fH formulas run in opposite orders; drawing the cycle stops you mixing them up.

Worked example 9: an unfamiliar cycle

Find ∆H for Na₂CO₃(s) + 10H₂O(l) → Na₂CO₃·10H₂O(s). Dissolving each solid in water gives the same solution. Use these values: dissolving anhydrous Na₂CO₃, −25.0 kJ mol⁻¹; dissolving Na₂CO₃·10H₂O, +66.0 kJ mol⁻¹.

Route: anhydrous solid → solution (−25.0), then solution → hydrated solid (the reverse of dissolving it, −66.0).

∆H = −25.0 − 66.0 = −91.0 kJ mol⁻¹

3.2.1(h): Techniques for direct and indirect determination

Direct methods (PAG3) measure the temperature change of the reaction you want.

• Solutions (neutralisation, or dissolving a solid for an enthalpy change of solution, 5.2.1(c)): use a lidded polystyrene cup in a beaker. Measure the volumes, or the solid’s mass and the water’s volume. Record the start temperature, mix, stir and read the temperature at regular intervals.
• Combustion: weigh a capped spirit burner, use it to heat a known mass of water in a metal can, stir, then reweigh. Use a draught shield.

Temperature-time graphs (M3.1, M3.2). Cooling starts before the reaction finishes, so the highest reading is too low. Plot temperature (y) against time (x), fit a straight line to the falling readings and extend it backwards to the moment of mixing. For example, a solution at 21.4 °C is mixed at 3 minutes. Readings at 4, 5, 6 and 7 minutes are 30.8, 30.5, 30.2 and 29.9 °C, a steady fall of 0.3 K per minute. Extrapolating back gives 31.1 °C at 3 minutes, so ∆T = 9.7 K, not 9.4 K.

Indirect methods combine two measurable reactions in a Hess cycle, for example worked example 9, or a thermal decomposition found from two reactions with excess acid. An enthalpy change of hydration (5.2.1(c)) is also found indirectly, from a measured enthalpy change of solution and the lattice enthalpy.

Main errors: heat loss, ignoring the container’s heat capacity, incomplete combustion (soot) and fuel evaporating from the wick.

Common errors

• Putting the fuel’s mass into q = mc∆T in place of the water’s.
• Forgetting that a temperature rise gives a negative ∆H.
• Dividing by moles of acid instead of moles of water for ∆neutH.
• Using non-standard states such as H₂O(g) in a definition equation.
• Missing a bond: draw out every structure and count each bond.

Next steps

Condense this with the Enthalpy changes revision notes, then test yourself with the Enthalpy changes practice questions. Bond strength first appears in Bonding and structure, and mole calculations in Amount of substance.

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.