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8873 / 2027
Chemical Energetics: Thermochemistry

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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)

01

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.

Illustrative values relative to the reactants. Reaction progress is not time, and the vertical axis is not temperature.

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 exampleConstruct an endothermic profile

Sketch a reaction with Delta H = +30 kJ mol-1 and forward activation energy 80 kJ mol-1.

Try it first. Then show the worked stepsHide the worked steps
  1. Label the vertical axis enthalpy and the horizontal axis reaction progress. Choose the reactants as a zero reference.
  2. Place products 30 kJ mol-1 above the reactants and the barrier 80 kJ mol-1 above them.
  3. 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.
Answer

The product level is higher than the reactant level. The reverse activation energy is 80 - 30 = 50 kJ mol-1.

Watch out for this

A student says: "The thermometer reading rose, so the reaction gained energy and its delta H is positive." What is wrong with this?

Show the answer

The temperature of the surroundings rises during an exothermic reaction. Their heat gain is positive, but the reacting system loses that energy, so its delta H is negative.

Check your understanding

A reaction has delta H = -40 kJ mol-1 and a forward activation energy of 60 kJ mol-1. What is the activation energy of the reverse reaction?

  1. 60 kJ mol-1, the same as the forward barrier
  2. 20 kJ mol-1, the forward barrier minus delta H
  3. 100 kJ mol-1, from the products to the peak
Show the answer

100 kJ mol-1, from the products to the peak

The peak is 60 kJ mol-1 above the reactants, and the products are 40 kJ mol-1 below the reactants. So the peak is 60 + 40 = 100 kJ mol-1 above the products.

02

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.

Required enthalpy terms
TermOne-mole referenceIllustration
Enthalpy change of reactionThe reaction as its equation is writtenDoubling the equation doubles its enthalpy change.
Standard enthalpy of formationOne mole of compound from its elements in their standard statesC(graphite,s) + O2(g) → CO2(g)
Standard enthalpy of combustionOne mole of substance completely burned in oxygen under standard conditionsCH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Enthalpy of neutralisationOne mole of water formed when an acid reacts with a baseH+(aq) + OH-(aq) → H2O(l)
Bond energyOne mole of specified gaseous covalent bonds brokenH2(g) → 2H(g); positive
Lattice energyOne mole of ionic solid formed from separated gaseous ionsNa+(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 understanding

A data table gives a positive value for separating an ionic lattice into gaseous ions. How should you use it for H1 lattice energy?

  1. Halve it, since the value covers both cation and anion.
  2. Use it as given, since lattice energy is quoted as a positive value.
  3. Change its sign, since H1 lattice energy is for forming the lattice.
Show the answer

Change its sign, since H1 lattice energy is for forming the lattice.

Separating the lattice is the reverse of forming it, so the enthalpy has the same size but the opposite sign. Read the equation, not just the name.

03

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 exampleNeutralisation 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Total solution mass = (50.0 + 50.0) × 1.00 = 100.0 g. Use both solutions, not only the acid.
  2. q(solution) = 100.0 × 4.18 × 6.70 = 2800.6 J = 2.8006 kJ.
  3. Each reactant supplies 0.0500 mol, forming 0.0500 mol water.
  4. Enthalpy per mole water = -2.8006/0.0500 = -56.012 kJ mol-1.
Answer

The measured enthalpy of neutralisation is -56.0 kJ mol-1, to three significant figures, under the stated approximation.

Worked exampleA 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.

Try it first. Then show the worked stepsHide the worked steps
  1. q(water) = 200 × 4.18 × 12.0 = 10032 J = 10.032 kJ.
  2. Fuel burned = 0.460/46.0 = 0.0100 mol.
  3. Estimated enthalpy = -10.032/0.0100 = -1003.2 kJ mol-1.
Answer

About -1.00 × 103 kJ mol-1. This simple apparatus measurement need not equal a standard data-book value.

Try it step by step

Work it out in steps: Heat gained by the water, q = mc delta T, then Moles of fuel burned, then Enthalpy of combustion.

  1. Problem 1. Burning 0.320 g of methanol (M = 32.0 g mol-1) raises the temperature of 150 g of water by 11.5 K. Use c = 4.18 J g-1 K-1. Find the enthalpy of combustion in kJ mol-1 to 3 significant figures.
  2. Problem 2. Burning 0.580 g of propanone (M = 58.0 g mol-1) raises the temperature of 100 g of water by 20.0 K. Use c = 4.18 J g-1 K-1. Find the enthalpy of combustion in kJ mol-1 to 3 significant figures.
  3. Problem 3. Burning 0.230 g of ethanol (M = 46.0 g mol-1) raises the temperature of 100 g of water by 12.0 K. Use c = 4.18 J g-1 K-1. Give each answer to 3 significant figures, with the sign of delta H.
Show the answer

Problem 1.

Heat gained by the water, q = mc delta T: q = 150 x 4.18 x 11.5 = 7211 J = 7.21 kJ, gained by the water.

Moles of fuel burned: n(methanol) = 0.320 / 32.0 = 0.0100 mol.

Enthalpy of combustion: delta H = -7.21 / 0.0100 = -721 kJ mol-1. The sign is negative because burning gave heat to the water.

Problem 2.

Heat gained by the water, q = mc delta T: q = 100 x 4.18 x 20.0 = 8360 J = 8.36 kJ.

Moles of fuel burned: n(propanone) = 0.580 / 58.0 = 0.0100 mol.

Enthalpy of combustion: delta H = -8.36 / 0.0100 = -836 kJ mol-1. Heat lost to the air makes this much smaller in size than the data-book value.

Problem 3.

Heat gained by the water, q = mc delta T: q = 100 x 4.18 x 12.0 = 5016 J, which is 5.02 x 103 J to 3 significant figures.

Moles of fuel burned: n(ethanol) = 0.230 / 46.0 = 0.00500 mol.

Enthalpy of combustion: delta H = -5.016 / 0.00500 = -1003 kJ mol-1, which is -1.00 x 103 kJ mol-1 to 3 significant figures.

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 understanding

In the neutralisation example, 50 g of acid was mixed with 50 g of alkali. What happens if only 50 g is used for m in Q = mc(delta T)?

  1. Nothing changes, because the temperature rise is still the same.
  2. The heat calculated, and the size of delta H, come out half as big.
  3. The heat calculated comes out twice as big, as each solution counts once.
Show the answer

The heat calculated, and the size of delta H, come out half as big.

Both solutions were mixed and warmed together, so the mass heated is about 100 g. Using 50 g halves Q and so halves the molar enthalpy change.

04

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.

Given cycle: carbon is graphite; hydrogen, oxygen, methane and carbon dioxide are gases; water is liquid. Both downward routes consume 2O2. Values are kJ per reaction as drawn.

Worked exampleA 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Direct left route: -394 + 2(-286) = -966 kJ.
  2. Top then right route: Delta Hf(CH4) + (-890).
  3. Equate them: Delta Hf - 890 = -966.
Answer

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.

Supplied illustrative data, in kJ for one mole of HA. All solutes are aqueous and water is liquid at the same stated temperature. Positions show routes, not enthalpy levels; OH- is carried unchanged through the first step.

Worked exampleUse neutralisation data without constructing a new cycle

In the supplied cycle, find the enthalpy for HA(aq) → H+(aq) + A-(aq).

Try it first. Then show the worked stepsHide the worked steps
  1. The direct route is HA + OH- to A- + H2O, with Delta H = -53.6 kJ for one mole of HA.
  2. 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.
  3. The second arrow is H+ + OH- to H2O, releasing 57.1 kJ. A- is unchanged in that step.
  4. Equate routes: -53.6 = Delta H(dissociation) + (-57.1). Therefore Delta H(dissociation) = -53.6 + 57.1 = +3.5 kJ mol-1.
Answer

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 exampleAverage 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.

Try it first. Then show the worked stepsHide the worked steps
  1. Break one H-H and one Cl-Cl bond: energy in = 436 + 243 = 679 kJ.
  2. Form two H-Cl bonds: energy out = 2 × 431 = 862 kJ.
  3. Delta H = energy for bonds broken - energy released by bonds formed.
Answer

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.

Try it step by step

Work it out in steps: Energy absorbed breaking bonds, then Energy released forming bonds, then Enthalpy change.

  1. Problem 1. Estimate delta H for CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g). Use C-H = 410, Cl-Cl = 244, C-Cl = 340 and H-Cl = 431 kJ mol-1. Give each answer as a whole number of kJ.
  2. Problem 2. Estimate delta H for C2H4(g) + H2(g) -> C2H6(g). Use C=C = 610, H-H = 436, C-C = 350 and C-H = 410 kJ mol-1. Give each answer as a whole number of kJ.
  3. Problem 3. Estimate delta H for 2H2(g) + O2(g) -> 2H2O(g). Use H-H = 436, O=O = 496 and O-H = 460 kJ mol-1. Give each answer as a whole number of kJ.
Show the answer

Problem 1.

Energy absorbed breaking bonds: Only one C-H and the Cl-Cl bond break: 410 + 244 = 654 kJ absorbed. The other three C-H bonds are unchanged.

Energy released forming bonds: Form one C-Cl and one H-Cl: 340 + 431 = 771 kJ released.

Enthalpy change: delta H = 654 - 771 = -117 kJ mol-1.

Problem 2.

Energy absorbed breaking bonds: Break C=C and H-H: 610 + 436 = 1046 kJ absorbed. The four C-H bonds in ethene stay.

Energy released forming bonds: Form C-C and two new C-H: 350 + 2(410) = 1170 kJ released.

Enthalpy change: delta H = 1046 - 1170 = -124 kJ mol-1.

Problem 3.

Energy absorbed breaking bonds: Break two H-H and one O=O: 2(436) + 496 = 1368 kJ absorbed.

Energy released forming bonds: Form four O-H, two in each water: 4(460) = 1840 kJ released.

Enthalpy change: delta H = 1368 - 1840 = -472 kJ mol-1 for the equation as written.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Standard enthalpy change of formation, ΔHf°The enthalpy change when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K, 100 kPa).
Standard enthalpy change of combustion, ΔHc°The enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions (298 K, 100 kPa).
Standard enthalpy change of neutralisationThe enthalpy change when one mole of water is formed by the reaction of an acid with a base in dilute aqueous solution under standard conditions (298 K, 100 kPa).
Bond energyThe average energy required to break one mole of a particular covalent bond in gaseous molecules.
Lattice energyThe enthalpy change when one mole of a solid ionic compound is formed from its constituent gaseous ions under standard conditions (298 K, 100 kPa). It is always negative.
Hess's lawThe enthalpy change of a reaction depends only on the initial and final states, and not on the route taken.
Thermochemistry checks
StepRemember
SignExothermic reaction negative; its surroundings gain heat.
One mole of what?Formation: compound; combustion: fuel; neutralisation: water; lattice: solid.
Calorimetryq = mcDeltaT; use all heated solution mass; reaction heat is -q.
Hess cycleReverse arrow → reverse sign; scaled equation → scaled enthalpy.
Bond energiesBroken minus formed, using gas-phase bonds.
Lattice comparisonHigher charges/smaller ions → greater magnitude, more negative formation energy.

Scope

Learning outcomes

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

  1. 6(a) Explain exothermic and endothermic changes.

    • Bond breaking/forming, heat transfer and sign of enthalpy.

    Breaking needs energy; forming releases it

  2. 6(b) Construct and interpret energy profiles.

    • Reaction enthalpy, activation energy and link to kinetics.

    Breaking needs energy; forming releases it

  3. 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

  4. 6(d) Calculate experimental enthalpy changes.

    • q = mcDeltaT; heat sign, reacting amount and assumptions.

    Convert a temperature change into heat per mole

  5. 6(e) Explain lattice-energy magnitude.

    • Effects of ionic charge and radius, qualitative only.

    Name the exact process and amount

  6. 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.

    Follow the arrows in the supplied cycle