Full chapter
Chemical Energetics: Thermochemistry
Read reaction energy, measure heat and use supplied cycles without losing signs or stoichiometric factors.
A-Level 8873, revised syllabus (2026-2027)
Breaking needs energy; forming releases it
A reaction can release energy overall and still need an initial barrier to be crossed.
Chemical reactions rearrange bonds. Breaking bonds absorbs energy; forming bonds releases energy. The enthalpy change is the net difference. At constant pressure, an exothermic reaction transfers heat to the surroundings and has negative ΔH; an endothermic reaction absorbs heat from them and has positive ΔH.
Read the barrier and the overall drop
Reactants are at zero, the barrier is 120 kJ mol-1 above them, and products are 80 kJ mol-1 below them. Forward activation energy is 120, reaction enthalpy is -80, and reverse activation energy is 200.
Read ΔH = H(products) - H(reactants). Read the forward activation energy from the reactant level to the barrier, not from the page baseline. In the figure, the reverse reaction starts at -80, so its barrier is 120 - (-80) = 200 kJ mol-1. A catalyst changes the pathway and barrier, but not the reactant/product levels or ΔH.
Worked example
Construct an endothermic profile
Sketch a reaction with Delta H = +30 kJ mol-1 and forward activation energy 80 kJ mol-1.
- Label the vertical axis enthalpy and the horizontal axis reaction progress. Choose the reactants as a zero reference.
- Place products 30 kJ mol-1 above the reactants and the barrier 80 kJ mol-1 above them.
- Join the levels with a smooth rise to the barrier and fall to the products. Label Delta H from reactants to products and Ea from reactants to the barrier.
The product level is higher than the reactant level. The reverse activation energy is 80 - 30 = 50 kJ mol-1.
Name the exact process and amount
An enthalpy value belongs to a stated equation, physical states and conditions.
A standard enthalpy change refers to species in their standard states at a stated temperature, commonly 298 K, with standard pressure 100 kPa. Aqueous standard data use a specified standard concentration, conventionally 1 mol dm-3 at this level. Standard does not by itself mean a reaction occurs at 0 degrees C. Always retain the states and temperature supplied with the data.
| Term | One-mole reference | Illustration |
|---|---|---|
| Enthalpy change of reaction | The reaction as its equation is written | Doubling the equation doubles its enthalpy change. |
| Standard enthalpy of formation | One mole of compound from its elements in their standard states | C(graphite,s) + O2(g) → CO2(g) |
| Standard enthalpy of combustion | One mole of substance completely burned in oxygen under standard conditions | CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) |
| Enthalpy of neutralisation | One mole of water formed when an acid reacts with a base | H+(aq) + OH-(aq) → H2O(l) |
| Bond energy | One mole of specified gaseous covalent bonds broken | H2(g) → 2H(g); positive |
| Lattice energy | One mole of ionic solid formed from separated gaseous ions | Na+(g) + Cl-(g) → NaCl(s); negative |
The formation enthalpy of an element in its standard state is zero by convention. This is not a claim that the element contains no energy. For standard formation of water, H2(g) + ½O2(g) → H2O(l) forms exactly one mole; fractional coefficients are therefore useful.
Lattice energy becomes greater in magnitude when ionic charges increase or ion radii decrease: the opposite charges attract more strongly at shorter separation. MgO, with Mg2+/O2-, has a much more negative lattice energy than NaCl, with Na+/Cl-. For equal charges, a smaller ion generally makes lattice formation more exothermic.
Check your understandingIf a data table instead gives a positive value for separating an ionic lattice into gaseous ions, how should you use it?Think it through, then reveal the answer
Convert a temperature change into heat per mole
First find the surroundings heat, then attach the opposite sign to the reaction.
For a solution approximated as water, q = mcΔT, where m is the mass of the heated solution, c its specific heat capacity and ΔT its temperature change. If c is in J g-1 K-1, use grams and a temperature difference in K or degrees C. The reaction heat is approximately -q when heat exchange with other surroundings and the container is neglected.
Worked example
Neutralisation calorimetry
Mix 50.0 cm3 of 1.00 mol dm-3 HCl and 50.0 cm3 of 1.00 mol dm-3 NaOH. The temperature rises 6.70 K. Assume density 1.00 g cm-3 and c = 4.18 J g-1 K-1.
- Total solution mass = (50.0 + 50.0) × 1.00 = 100.0 g. Use both solutions, not only the acid.
- q(solution) = 100.0 × 4.18 × 6.70 = 2800.6 J = 2.8006 kJ.
- Each reactant supplies 0.0500 mol, forming 0.0500 mol water.
- Enthalpy per mole water = -2.8006/0.0500 = -56.012 kJ mol-1.
The measured enthalpy of neutralisation is -56.0 kJ mol-1, to three significant figures, under the stated approximation.
Worked example
A simple fuel-heating experiment
Burning 0.460 g of a fuel with molar mass 46.0 g mol-1 raises 200 g water by 12.0 K. Estimate its molar combustion enthalpy with c = 4.18 J g-1 K-1.
- q(water) = 200 × 4.18 × 12.0 = 10032 J = 10.032 kJ.
- Fuel burned = 0.460/46.0 = 0.0100 mol.
- Estimated enthalpy = -10.032/0.0100 = -1003.2 kJ mol-1.
About -1.00 × 103 kJ mol-1. This simple apparatus measurement need not equal a standard data-book value.
Heat lost to the room, heat absorbed by the container and incomplete combustion commonly make the measured temperature rise too small, giving an exothermic value that is less negative than the desired value. Insulation and a lid reduce heat loss; stirring reduces temperature gradients. In a temperature-time experiment, extrapolating the cooling trend back towards the mixing time can estimate the temperature rise before heat loss, provided the method is justified by the data.
Check your understandingIn the neutralisation example, what happens if only 50 g is used for m?Think it through, then reveal the answer
Follow the arrows in the supplied cycle
The total enthalpy change depends on the initial and final states, not the route.
Hess' Law allows two routes between the same initial and final states to be equated. Read a supplied cycle by reversing signs for reversed arrows and multiplying energy values when an equation is scaled. The H1 calculation scope uses supplied simple cycles, formation/combustion/neutralisation data and bond energies; constructing a Born-Haber cycle is not an H1 requirement.
A supplied methane formation cycle
Elements and methane at the top both burn to the same products below. The left route releases 394 plus twice 286 kJ; the right combustion releases 890 kJ. The top arrow is the unknown methane formation enthalpy.
Worked example
A formation enthalpy that is difficult to measure directly
Use the supplied cycle and combustion values C = -394, H2 = -286 and CH4 = -890 kJ mol-1.
- Direct left route: -394 + 2(-286) = -966 kJ.
- Top then right route: Delta Hf(CH4) + (-890).
- Equate them: Delta Hf - 890 = -966.
ΔHf(CH4) = -76 kJ mol-1. The multiplier two belongs to the two moles of H2 burned.
The equivalent algebraic rule is ΔH(reaction) = sum of product formation enthalpies - sum of reactant formation enthalpies, each multiplied by its coefficient. When using combustion data instead, both sides lead down to common combustion products, so the subtraction direction differs. Following the actual arrows prevents memorised-sign errors.
Neutralisation data are quoted per mole of water formed. If a supplied cycle step neutralises two moles of water, use twice the listed neutralisation enthalpy. A reverse neutralisation arrow has the opposite sign. Check species, amount and physical states before treating two routes as having the same endpoint.
A supplied cycle using neutralisation data
The direct neutralisation route releases 53.6 kJ. An alternative route first separates HA into H+ and A-, then neutralises H+ with OH- to release 57.1 kJ. Equal endpoints allow the unknown first step to be calculated.
Worked example
Use neutralisation data without constructing a new cycle
In the supplied cycle, find the enthalpy for HA(aq) → H+(aq) + A-(aq).
- The direct route is HA + OH- to A- + H2O, with Delta H = -53.6 kJ for one mole of HA.
- Along the alternative route, HA first forms H+ and A-. The OH- is unchanged at this stage, so this arrow is the requested dissociation process.
- The second arrow is H+ + OH- to H2O, releasing 57.1 kJ. A- is unchanged in that step.
- Equate routes: -53.6 = Delta H(dissociation) + (-57.1). Therefore Delta H(dissociation) = -53.6 + 57.1 = +3.5 kJ mol-1.
Dissociation is endothermic by +3.5 kJ mol-1 for this supplied example. Some of the neutralisation energy is used in dissociation, so the net weak-acid neutralisation is less exothermic. The conclusion follows from these data; do not assume every weak acid has the same dissociation enthalpy.
Worked example
Average bond energies provide another route
Estimate ΔH for H2(g) + Cl2(g) → 2HCl(g), given H-H = 436, Cl-Cl = 243 and H-Cl = 431 kJ mol-1.
- Break one H-H and one Cl-Cl bond: energy in = 436 + 243 = 679 kJ.
- Form two H-Cl bonds: energy out = 2 × 431 = 862 kJ.
- Delta H = energy for bonds broken - energy released by bonds formed.
ΔH = 679 - 862 = -183 kJ per mole of reaction as written, or -91.5 kJ per mole of HCl formed. Average bond-energy estimates may differ from measured values because the molecular environment matters.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Step | Remember |
|---|---|
| Sign | Exothermic reaction negative; its surroundings gain heat. |
| One mole of what? | Formation: compound; combustion: fuel; neutralisation: water; lattice: solid. |
| Calorimetry | q = mcDeltaT; use all heated solution mass; reaction heat is -q. |
| Hess cycle | Reverse arrow → reverse sign; scaled equation → scaled enthalpy. |
| Bond energies | Broken minus formed, using gas-phase bonds. |
| Lattice comparison | Higher charges/smaller ions → greater magnitude, more negative formation energy. |
Scope and references
Learning outcomes and sources
6. Chemical Energetics: Thermochemistry (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
6(a) Explain exothermic and endothermic changes.
- Bond breaking/forming, heat transfer and sign of enthalpy.
6(b) Construct and interpret energy profiles.
- Reaction enthalpy, activation energy and link to kinetics.
6(c) Define the specified energy terms.
- (i) Reaction enthalpy, standard conditions, formation, combustion and neutralisation.
- (ii) Bond energy as positive bond breaking.
- (iii) Lattice energy as negative gaseous-ion-to-solid formation.
Name the exact process and amountBreaking needs energy; forming releases it
6(d) Calculate experimental enthalpy changes.
- q = mcDeltaT; heat sign, reacting amount and assumptions.
6(e) Explain lattice-energy magnitude.
- Effects of ionic charge and radius, qualitative only.
6(f) Use supplied Hess cycles and bond energies.
- Supplied simple cycles; formation/combustion/neutralisation scope.
- (i) Indirect determination, including formation from combustion.
- (ii) Average bond energies; cycle construction not required.
- SEAB H1 Chemistry 8873, 2026 revision
Official scope: section 6, printed pages 16-17. All lettered outcomes and nested requirements checked.
- SEAB H1 Chemistry 8873, 2027
Section 6: same substantive outcomes as the revised 2026 course.
- NJC Energetics H1 Lecture Notes Student (Grail)
Consulted printed p. 8 on neutralisation/dissociation and p. 19 on supplied Hess routes. Added a new illustrative cycle and independently verified +3.5 kJ mol-1 calculation with different data; H1 cycle-construction exclusion retained.