From C5 Energy Changes you already know exothermic and endothermic reactions, reaction profiles and bond-energy sums. A-Level makes it quantitative and honest about conditions: heat change becomes enthalpy change at constant pressure with defined standard states, calorimetry acquires q = mcΔT and a required practical, Hess’s law lets you calculate enthalpy changes no calorimeter could reach, and bond-energy sums become mean bond enthalpies — with reasons why the three methods disagree.
Enthalpy change
Every reaction redistributes energy between the chemicals and their surroundings:
The sign convention is a rule of the quantity itself, not an afterthought: the sign is part of the value, and an enthalpy change written without one is incomplete.
Enthalpy change (ΔH) — the heat energy change measured under conditions of constant pressure.
Why constant pressure? Because that is how chemistry is actually done — open beakers, flasks and domestic boilers all work at constant atmospheric pressure — so chemists define and tabulate heat that way. And because a heat change depends on the conditions it is measured under, tabulated values are pinned to standard conditions and wear them in the symbol:
One more condition rides along with the symbol: each substance is in its standard state — the state it normally has under those conditions: H2O(l), CO2(g), C(s, graphite).
Two standard enthalpy changes have definitions you must be able to write from memory, every condition present:
Standard enthalpy of combustion (ΔcH⦵) — the enthalpy change when 1 mol of a substance is burned completely in oxygen, with all reactants and products in their standard states under standard conditions.
Standard enthalpy of formation (ΔfH⦵) — the enthalpy change when 1 mol of a compound is formed from its elements, with all substances in their standard states under standard conditions.
The ΔfH⦵ of an element already in its standard state is zero, because forming O2(g) from O2(g) changes nothing. You will lean on that zero constantly in Hess cycles.
The reaction profile you drew at GCSE survives, with upgraded labels: enthalpy on the y-axis, progress of reaction on the x-axis, and ΔH read as the gap between the reactant and product levels. The hump between them is the activation energy, Ea — named on the diagram here, taught properly in Kinetics (3.1.5).
Underneath every ΔH sits a two-line ledger you first met at GCSE: breaking bonds costs energy, making bonds releases it, and the balance decides the sign. It follows from Bonding’s definition of what a bond is — a strong electrostatic attraction — because pulling an attraction apart always costs energy. Changes of state carry enthalpy too — melting and boiling absorb energy to overcome the attractions between particles — which will matter when a calculation mixes H2O(l) with H2O(g). More standard enthalpies — atomisation, lattice enthalpy and the Born–Haber cycle — arrive in Year 2.
- Definitions are all-or-nothing. Recent AQA examiner reports flag answers missing one condition or one noun — “1 mol”, “standard states”, “complete combustion”. Write all four components, every time.
- ΔH carries its sign everywhere. Quoting “890 kJ mol−1” for methane’s combustion without the minus is a different (and wrong) claim.
- “Heat” and “enthalpy change” are not synonyms. ΔH is the heat change at constant pressure — the condition is part of the definition.
GCSE groundwork: C5 Part 1 — exothermic & endothermic and C5 Part 2 — reaction profiles.
🧪 Exam-style questions
Which reaction has an enthalpy change equal to the standard enthalpy of formation of silver iodide, AgI? Tick (✓) one box.
Which reaction has an enthalpy change equal to the standard enthalpy of formation of lithium fluoride, LiF? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers (2020–2024).
Calorimetry & Required Practical 2
Calorimetry is how enthalpy changes are actually measured, and the whole method fits in one sentence: the heat released (or absorbed) by the reaction equals the heat gained (or lost) by something you can measure — usually water, or the reaction solution itself. The instrument is a thermometer; everything else is bookkeeping. The equation:
q = mcΔT
| Term | What it is | The detail that carries marks |
|---|---|---|
| q | heat energy change | comes out in joules — convert to kJ before quoting ΔH |
| m | mass of the substance being heated | the water or the solution — never the fuel burned or the solid added |
| c | specific heat capacity — the energy that warms 1 g of the substance by 1 K | always given in the question (usually 4.18 J g−1 K−1, water) — never recalled |
| ΔT | temperature change | identical in K and °C for a change — use the thermometer reading directly (the kelvin note) |
q is the heat change of the water — but the question asks for a molar enthalpy change. The route from one to the other is always the same three moves: convert q to kilojoules, divide by the moles of the substance the question defines (the limiting reactant, or the fuel burned — moles from solutions do the counting), then attach the sign by hand from the temperature direction:
q (J) — from q = mcΔT
↓ ÷1000 (J → kJ)
q (kJ)
↓ ÷n (the moles the question defines)
the size of ΔH — no sign yet
↓ sign from the temperature direction
ΔH in kJ mol−1
Two set-ups cover every reaction. Reactions in solution — neutralisation, dissolution, displacement — run in an expanded-polystyrene cup, where the solution is both the reaction vessel and the thing being heated. Combustion runs in a spirit burner under a can of water, where the flame heats the water and the fuel is weighed before and after.
25.0 cm3 of 1.00 mol dm−3 HCl is mixed with 25.0 cm3 of 1.00 mol dm−3 NaOH; the temperature rises by 6.8 K. Take c = 4.18 J g−1 K−1 and the density of both solutions as 1.00 g cm−3. Find the molar enthalpy of neutralisation.
Step 1 — m is the combined solution, the thing being heated:
m = 25.0 + 25.0 = 50.0 g (1.00 g cm−3)
Step 2 — heat released, then straight into kJ:
q = mcΔT = 50.0 × 4.18 × 6.8 = 1421 J = 1.42 kJ
Step 3 — moles of the reactant the question defines:
n(HCl) = 1.00 × 0.0250 = 0.0250 mol
Step 4 — divide, then attach the sign from the temperature direction (a rise → exothermic):
ΔH = −1.42 ÷ 0.0250 = −56.8 kJ mol−1 (3 sf)
Sig-fig decision: quote no more than 3 sf here — and since the least accurate value is ΔT at 2 sf, −57 kJ mol−1 is the safest answer when a question demands “an appropriate number of significant figures”.
That answer has a name: the enthalpy of neutralisation — the enthalpy change when an acid reacts with an alkali to form 1 mol of water. For any strong acid with any strong alkali it comes out almost identical, ≈ −57 kJ mol−1, because both are fully ionised and the only chemistry happening is the same reaction every time: H+(aq) + OH−(aq) → H2O(l). A weak acid gives a less exothermic value — part of the heat is spent ionising the acid first.
Interactive — run the calorimeter
Burning 0.450 g of ethanol (M = 46.0) raises the temperature of 100.0 g of water by 30.0 K. Find ΔcH and compare it with the data-book value, −1367 kJ mol−1.
Step 1 — heat gained by the water (m is the water, not the fuel), converted to kJ:
q = mcΔT = 100.0 × 4.18 × 30.0 = 12 540 J = 12.5 kJ
Step 2 — moles of fuel burned:
n = 0.450 ÷ 46.0 = 9.78 × 10−3 mol
Step 3 — divide and attach the sign (temperature rose → exothermic), carrying unrounded values:
ΔH = −12.54 ÷ (9.783 × 10−3) = −1280 kJ mol−1 (3 sf — the least precise measurements were 3 sf)
The data book says −1367 kJ mol−1. Our value is less exothermic, and each flaw explains the direction: heat escaped to the surroundings and the apparatus instead of the water; combustion may have been incomplete (a yellow, sooty flame releases less energy); and ethanol that evaporated between weighings counted as burned without releasing any heat.
Interactive — run the numbers: calorimetry
Naming a flaw is one mark; giving its direction is the other — and for endothermic reactions the reasoning reverses. Recent AQA examiner reports flag both: heat-“loss” explanations written where heat gain applies, and error answers that never say which way the value moves.
| Flaw or assumption | What it does to the measurement | Direction of the error in ΔH |
|---|---|---|
| Heat lost to the surroundings (exothermic reaction) | ΔT smaller than it should be → q too small | measured value less exothermic (smaller magnitude) |
| Heat absorbed by the apparatus (can, thermometer, cup) | some q never reaches the water/solution | less exothermic |
| Incomplete combustion (sooty flame, CO formed) | less energy released per gram of fuel | less exothermic |
| Fuel evaporates between weighings | mass loss counted as “burned” released no heat → n too big | less exothermic |
| Solution’s c assumed equal to water’s | q mis-estimated in proportion | state it as an assumption — no direction without data |
| Endothermic reaction in the same cup | the cold mixture gains heat from the surroundings → the temperature fall is too small | measured value less endothermic — the reversal that catches people out |
RP2 is this section done with your own hands — one enthalpy change, measured properly. Practical skills are examined across all three papers.
The cooling correction deserves its own drawing, because it is the RP2 skill written papers actually probe. A reaction is never instantaneous and the cup is never perfect: the mixture starts cooling the moment the reaction starts, so the biggest temperature you ever read is already an underestimate. The fix: record the temperature every minute before mixing, mix at a known time, keep recording through the peak and the cool-down — then extrapolate the cooling line back to the moment of mixing and read ΔT there.
Interactive — your turn: extrapolate the cooling line
- m is the mass being heated. In Zn + CuSO4(aq), m is the solution’s mass — not the zinc’s, and never solution + solid.
- Joules to kilojoules, visibly. Recent AQA examiner reports flag J→kJ slips and missing minus signs on exothermic answers — build both conversions into the written route, as their own lines.
- Sig figs follow the least accurate measurement — usually the thermometer. State the decision rather than defaulting to whatever the calculator shows.
- Error answers need a direction, not just a flaw — and for endothermic reactions the vessel gains heat from the surroundings, so the standard heat-loss sentence is the wrong way round.
GCSE groundwork: C5 Part 1 — measuring temperature changes.
🧪 Exam-style questions
The temperature changed from 21.8 °C to 19.2 °C during a calorimetry experiment. The uncertainty of each reading of the thermometer is ±0.1 °C. What is the percentage uncertainty in the temperature change? Tick (✓) one box.
State the meaning of the term enthalpy change as applied to a chemical reaction.
Show answer
The heat energy change measured at constant pressure. 1 mark
A student determines the enthalpy change for the reaction between calcium carbonate and hydrochloric acid.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
The student measures 50 cm3 of 1.00 mol dm−3 hydrochloric acid with a measuring cylinder into a glass beaker, weighs 2.50 g of solid calcium carbonate on a watch glass and tips it into the acid, stirs with a thermometer and records the maximum temperature reached. Explain how the experimental method and apparatus can be improved to provide more accurate data, and describe how the data from the improved method can be used to determine an accurate value for the temperature change.
Show answer
Apparatus. Measure the acid with a burette or pipette rather than a measuring cylinder; use an insulated polystyrene cup (with a lid) rather than a glass beaker; reweigh the watch glass after adding the solid to find the mass actually added; use powdered calcium carbonate so it reacts quickly. 2 marks
Temperature readings. Record the temperature for a few minutes before adding the solid, then at regular intervals (e.g. every minute) for several minutes after adding it. 2 marks
Finding ΔT. Plot a graph of temperature against time and extrapolate the cooling line back to the moment of addition; read the temperature change there. 2 marks
Marked by levels of response — all three stages (apparatus, timed readings, extrapolation graph) must appear for the top band. The extrapolation is the same cooling-correction move as the graph above.
In a different experiment, 50.0 cm3 of 0.500 mol dm−3 hydrochloric acid are reacted with 50.0 cm3 of 0.500 mol dm−3 sodium hydroxide.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ΔH = −57.1 kJ mol−1
The initial temperature of each solution is 18.5 °C. Calculate the maximum final temperature of the reaction mixture. Assume the specific heat capacity of the mixture, c = 4.18 J K−1 g−1, and its density is 1.00 g cm−3.
Show answer
n(HCl) = 0.500 × 0.0500 = 0.0250 mol
Moles of the reactant (acid = alkali here) 1 mark
q = 57.1 × 0.0250 = 1.4275 kJ = 1427.5 J
Heat released = ΔH × moles 1 mark
ΔT = qmc, with m = 50.0 + 50.0 = 100 g
Combined solution mass is what is heated 1 mark
ΔT = 1427.5 ÷ (100 × 4.18) = 3.4 °C
Temperature change 1 mark
final temperature = 18.5 + 3.4 = 21.9 °C
Added to the initial temperature (must be a rise — the reaction is exothermic) 1 mark
Suggest how, without changing the apparatus, the experiment in part (c) could be improved to reduce the percentage uncertainty in the temperature change.
Show answer
Increase the concentration of the solutions, so the temperature change is larger and the fixed ±0.1 °C uncertainty is a smaller fraction of it. 1 mark
State the meaning of the term standard enthalpy of combustion.
Show answer
The enthalpy change when 1 mol of a substance burns completely in oxygen … 1 mark
… with all substances in their standard states (at a stated temperature and 100 kPa). 1 mark
A student determines the enthalpy of combustion of propan-1-ol (CH3CH2CH2OH, Mr = 60.0). Burning 0.497 g of propan-1-ol raises the temperature of 150 g of water from 21.2 °C to 35.1 °C. Calculate a value, in kJ mol−1, for the enthalpy of combustion in this experiment. The specific heat capacity of water is 4.18 J K−1 g−1.
Show answer
q = mcΔT = 150 × 4.18 × 13.9 = 8715 J = 8.72 kJ
Heat gained by the water (ΔT = 35.1 − 21.2 = 13.9 K) 1 mark
n = 0.497 ÷ 60.0 = 8.28 × 10−3 mol
Moles of fuel burned 1 mark
ΔH = −8.72 ÷ (8.28 × 10−3) = −1050 kJ mol−1
Divide and attach the sign (must be negative; at least 2 sf) 1 mark
The enthalpy of combustion determined experimentally is less exothermic than that calculated using enthalpies of formation. Give one possible reason for this, other than heat loss.
Show answer
Any one of: incomplete combustion; evaporation of the fuel; the experiment is not carried out under standard conditions. 1 mark
A bomb calorimeter is a container of fixed volume used for accurate determination of the heat change during combustion: the fuel is mixed with pure oxygen, ignited, and the temperature change recorded. Its total heat capacity, Ccal, is found using a fuel of known heat change. Igniting 2.00 g of hexane (Mr = 86.0) gives a temperature change ΔT of 12.4 °C; under these conditions 1.00 mol of hexane releases 4154 kJ. Using q = CcalΔT, calculate Ccal in kJ K−1.
Show answer
n(hexane) = 2.00 ÷ 86.0 = 0.02326 mol
Moles of hexane 1 mark
q = 4154 × 0.02326 = 96.6 kJ
Heat released 1 mark
Ccal = 96.6 ÷ 12.4 = 7.79 kJ K−1
q ÷ ΔT 1 mark
The experiment is repeated with 2.00 g of octane (Mr = 114.0) and the temperature change recorded is 12.2 °C. Calculate the heat change, in kJ mol−1, for octane in this combustion.
Show answer
q = CcalΔT = 7.79 × 12.2 = 95.0 kJ
Heat released, from part (a)’s Ccal 1 mark
n(octane) = 2.00 ÷ 114.0 = 0.01754 mol → 95.0 ÷ 0.01754 = −5420 kJ mol−1
Per mole, with the sign 1 mark
State why the heat change calculated from the bomb calorimeter experiment is not an enthalpy change.
Show answer
The pressure is not constant in the sealed, fixed-volume bomb — an enthalpy change is the heat change measured at constant pressure. 1 mark
The thermometer used to measure the 12.2 °C change in part (b) has an uncertainty of ±0.1 °C in each reading. Calculate the percentage uncertainty in this use of the thermometer, and suggest one change that decreases it while using the same thermometer.
Show answer
percentage uncertainty = 2 × 0.112.2 × 100 = 1.6%
Two readings, each ±0.1 °C 1 mark
Use a bigger mass of fuel, so ΔT is greater. 1 mark
Source: AQA A-Level Chemistry past papers (2020–2024).
Hess’s law
Calorimetry has a hard limit: it can only price reactions you can actually run, cleanly and completely, in a cup or under a can. Methane refuses to assemble itself from carbon and hydrogen; sodium hydrogencarbonate decomposes far too reluctantly to track with a thermometer. Yet the data book quotes an enthalpy change for both. The escape is Hess’s law — the tool that turns enthalpy changes you can measure into the ones you can’t.
Hess’s law — the enthalpy change of a reaction is independent of the route taken, provided the initial and final states are the same.
Why must it be true? Because energy is conserved. And the working method is drawn cycles, not memorised formulas — the reaction across the top, the data book’s reference substances at the base, and arrows you read like a map:
Two cycles cover nearly every question, and each collapses into a formula worth having as a check on the drawing.
From formation data, the elements sit at the base — every compound in the cycle is built up from them:
ΔH = ΣΔfH⦵(products) − ΣΔfH⦵(reactants)
ΔfH⦵/kJ mol−1: C3H8(g) −105; CO2(g) −394; H2O(l) −286. Calculate the standard enthalpy of combustion of propane.
Step 1 — the combustion equation, coefficients and all:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Step 2 — each side’s formation total, coefficient × value on its own line:
ΣΔfH⦵(products) = 3(−394) + 4(−286) = −1182 − 1144 = −2326 kJ mol−1
ΣΔfH⦵(reactants) = −105 + 0 = −105 kJ mol−1 (O2 is an element: 0)
Step 3 — products minus reactants:
ΔcH⦵ = −2326 − (−105) = −2221 kJ mol−1
From combustion data, the logic inverts: burn everything. The combustion products sit at the base, the arrows fall into them, and because both routes now end in the same puddle of CO2 and H2O, the cycle closes itself:
ΔH = ΣΔcH⦵(reactants) − ΣΔcH⦵(products)
ΔcH⦵/kJ mol−1: C(s) −394; H2(g) −286; CH4(g) −890. Calculate the standard enthalpy of formation of methane.
Step 1 — the formation equation:
C(s) + 2H2(g) → CH4(g)
Step 2 — each side’s combustion total:
ΣΔcH⦵(reactants) = −394 + 2(−286) = −966 kJ mol−1
ΣΔcH⦵(products) = −890 kJ mol−1
Step 3 — reactants minus products — the product’s combustion is subtracted, the classic sign slip:
ΔfH⦵ = −966 − (−890) = −76 kJ mol−1
Interactive — run the numbers: a Hess cycle, term by term
The reactions AQA names for indirect determination — the thermal decomposition of NaHCO3, the hydration of MgSO4 or CuSO4 — cannot be run in a polystyrene cup: one needs sustained heating, the others are solid-to-solid changes with nothing to stir a thermometer through. Instead you measure two related reactions that a cup handles happily — dissolve each solid in water or acid and record its temperature change — then draw both onto one cycle and close it with Hess’s law. This is RP2 arithmetic feeding a paper method, and a classic written-paper context.
Interactive — build the Hess route
Tap the route’s terms in travel order — reactants → base of the cycle → products:
- Products − reactants for formation data; reactants − products for combustion data. Don’t trust the memorised order under pressure — re-derive it from the arrows each time: with an arrow, add; against it, subtract.
- Coefficients before summing. Recent AQA examiner reports flag missed factor-of-2 values and mishandled signs in energetics cycles — write each coefficient × value as its own line before adding anything.
- State symbols change the numbers. H2O(l) and H2O(g) differ by the enthalpy of vaporisation — a cycle mixing them without noticing gets a different (wrong) answer.
The coefficient arithmetic is Amount of Substance — calculations from equations.
🧪 Exam-style questions
Some enthalpy change data are shown.
C(s) + 2H2(g) → CH4(g) ΔH = −75 kJ mol−1
H2(g) → 2H(g) ΔH = +436 kJ mol−1
What is the enthalpy change, in kJ mol−1, for CH4(g) → C(s) + 4H(g)? Tick (✓) one box.
Two reactions of iron with oxygen are shown.
Fe(s) + ½O2(g) → FeO(s) ΔH = −272 kJ mol−1
2Fe(s) + 1½O2(g) → Fe2O3(s) ΔH = −822 kJ mol−1
What is the enthalpy change, in kJ mol−1, for 2FeO(s) + ½O2(g) → Fe2O3(s)? Tick (✓) one box.
Define standard enthalpy of formation.
Show answer
The enthalpy change when 1 mol of a substance is formed from its elements … 1 mark
… with all substances in their standard states (at 100 kPa and a stated temperature). 1 mark
Silver nitrate(V) is formed when silver nitrate(III) undergoes thermal decomposition.
2AgNO2(s) → Ag(s) + AgNO3(s) + NO(g) ΔH = +56.2 kJ mol−1
The standard enthalpy of formation of AgNO3(s) is −123.0 kJ mol−1 and that of NO(g) is +90.4 kJ mol−1. Determine the standard enthalpy of formation of AgNO2(s).
Show answer
ΔH = ΣΔfH(products) − ΣΔfH(reactants)
+56.2 = [0 + (−123.0) + (+90.4)] − 2x
Ag(s) is an element, so its ΔfH is 0; set up with 2x for the two moles of AgNO2 1 mark
2x = −32.6 − 56.2 = −88.8 → x = −44.4 kJ mol−1
Rearrange and divide by 2 1 mark
Suggest why the enthalpy change for the thermal decomposition of solid silver nitrate(III) is difficult to determine experimentally.
Show answer
The solid has to be heated to decompose it, so you cannot measure the temperature change caused by the reaction itself. 1 mark
This is exactly why Hess’s law is needed — the enthalpy change is reached from formation data instead of a thermometer.
Source: AQA A-Level Chemistry past papers (2020–2024).
Bond enthalpies
The third route to ΔH needs no apparatus and no table of formation values — just a price list for bonds. Every reaction is, underneath, bonds breaking and bonds forming (Bonding); if each bond has a known energy price, the enthalpy change is a shopping bill. The catch is in the price list’s small print, and AQA examines the small print.
Mean bond enthalpy — the enthalpy change when 1 mol of a covalent bond is broken, in the gas phase, averaged over many compounds containing that bond.
All four bold components are individually creditworthy — and individually fatal to omit.
Why “mean”? Because a C–H bond is not one thing. The C–H in methane sits in different electronic surroundings from the C–H in chloromethane, and their actual bond enthalpies differ. Tabulated values average across a range of compounds — which makes them portable, and makes every answer built from them an approximation. Remember that trade: it is both what the table is for and where its error comes from.
The calculation itself is a ledger with the page’s colour rule built in: breaking bonds costs energy (positive, red), making bonds releases it (negative, green), and the reaction’s ΔH is the balance:
ΔH ≈ Σ(bond enthalpies broken) − Σ(bond enthalpies formed)
The ≈ is doing honest work — this method estimates. An exothermic reaction is one that forms stronger bonds than it breaks; that single sentence explains every negative ΔH on this page. And the method is strictly gas phase: mean bond enthalpies are defined for gaseous molecules, so using them on liquids or solids quietly imports the enthalpy of the missing state changes as extra error.
Mean bond enthalpies/kJ mol−1: C–H 412; O=O 496; C=O 805 (in CO2); O–H 463. Estimate ΔH for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), and explain why the answer differs from the data-book −890 kJ mol−1.
Step 1 — draw every molecule and count bonds:
broken: 4 × C–H and 2 × O=O · formed: 2 × C=O and 4 × O–H
Step 2 — sum the bonds broken, its own line:
Σ(broken) = 4(412) + 2(496) = 1648 + 992 = 2640 kJ mol−1
Step 3 — sum the bonds formed, its own line:
Σ(formed) = 2(805) + 4(463) = 1610 + 1852 = 3462 kJ mol−1
Step 4 — broken minus formed, then read the sign:
ΔH ≈ 2640 − 3462 = −822 kJ mol−1
The data book’s value is −890 kJ mol−1. Both reasons for the gap earn credit, in full sentences: mean bond enthalpies are averages over many compounds, not the actual bond enthalpies of the bonds in these particular molecules; and the calculation is all-gas — it produced H2O(g), while the standard enthalpy of combustion is defined with H2O(l), which sits lower by the enthalpy of condensation.
Interactive — the bond-enthalpy stepper
-
Step 1 — count the bonds broken (every reactant bond):
-
Step 2 — energy in: Σ(bonds broken), its own line:
kJ -
Step 3 — count the bonds formed (every product bond):
-
Step 4 — energy out: Σ(bonds formed), its own line:
kJ -
Step 5 — broken − formed, and read the sign:
kJ mol−1
- The definition has four components — enthalpy change, 1 mol of bonds, gas phase, averaged over many compounds. Recent AQA examiner reports flag “energy” written for “enthalpy”, a missing “1 mol” and a missing average idea as repeated definition mark-losers.
- Broken − formed, in that order. Sign chaos is the flagged failure mode of these calculations — write the two Σ lines separately, as in the worked example, before subtracting anything.
- The answer is approximate by construction. When a question asks you to comment on the value, say so — and say why: mean values, gas phase only.
GCSE groundwork: C5 Part 2 — bond energy calculations.
🧪 Exam-style questions
Define the term enthalpy change.
Show answer
The heat (energy) change at constant pressure. 1 mark
Propane undergoes complete combustion.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = −2046 kJ mol−1
Mean bond enthalpies/kJ mol−1: C–H 412, C=O 743, O–H 463. The bond enthalpy for O=O is 496 kJ mol−1, and for H2O(l) → H2O(g), ΔH = +41 kJ mol−1. Use these data to calculate a value for the C–C bond enthalpy in propane.
Show answer
Bonds broken: 2(C–C) + 8(C–H) + 5(O=O); bonds made: 6(C=O) + 8(O–H) 1 mark
Bond enthalpies describe gaseous species, but the water is made as a liquid — include the vaporisation of the four waters, 4 × (+41) 1 mark
2(C–C) = 6(743) + 8(463) + 4(41) − 2046 − 8(412) − 5(496) = 504
Total for the two C–C bonds 1 mark
C–C = 504 ÷ 2 = 252 kJ mol−1
Divide by the number of C–C bonds 1 mark
Miss the 4 × (+41) water-vaporisation term and the answer comes out at 170 — a classic dropped mark, because the tabulated bond enthalpies are all gas-phase but this reaction makes liquid water.
Explain why the value given for the O=O bond enthalpy is not a mean value.
Show answer
O2 is the only substance that contains an O=O bond, so the value is a single, exact figure rather than an average taken across a range of compounds. 1 mark
The equation for the complete combustion of gaseous pentan-1-ol is shown.
CH3(CH2)3CH2OH(g) + 7½O2(g) → 5CO2(g) + 6H2O(g) ΔH = −3388 kJ mol−1
Bond enthalpies/kJ mol−1: C–H 412, C–O 360, O–H 463, C=O 805, O=O 496. Use these data to calculate a value for the mean C–C bond enthalpy in pentan-1-ol.
Show answer
Bonds broken: 4(C–C) + 11(C–H) + (C–O) + (O–H) + 7½(O=O); bonds made: 10(C=O) + 12(O–H) 1 mark
−3388 = 4(C–C) + 9075 − 13606, so 4(C–C) = 1143
Broken (apart from C–C) = 11(412)+360+463+7.5(496) = 9075; made = 10(805)+12(463) = 13606 1 mark
C–C = 1143 ÷ 4 = 286 kJ mol−1
Divide by the four C–C bonds 1 mark
Everything here is already gaseous, so — unlike the propane question above — there is no water-vaporisation term to add.
Source: AQA A-Level Chemistry past papers (2020–2024).
Capstone: one reaction, three routes
Everything on this page prices the same kind of thing three different ways, so put them side by side. The reaction: the complete combustion of ethanol, C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l). The data book says ΔcH⦵ = −1367 kJ mol−1. Here is what each route returns, what it needs, and where it goes wrong:
| Calorimetry (RP2) | Hess’s law, formation data | Mean bond enthalpies | |
|---|---|---|---|
| Value returned | −1280 kJ mol−1 (the spirit-burner worked example) | −1369 kJ mol−1 from [2(−394) + 3(−286)] − (−277) | −1279 kJ mol−1 from Σbroken 4719 − Σformed 5998 |
| Data it needs | your own readings — a balance, a thermometer, q = mcΔT | tabulated ΔfH⦵: ethanol(l) −277, CO2(g) −394, H2O(l) −286 | tabulated mean bond enthalpies, plus every molecule drawn out |
| Assumptions built in | all the heat reaches the water; combustion complete; no fuel evaporates | only that the tabulated values are right — states exactly as written | every species gaseous; every bond exactly average |
| Characteristic error — with direction | systematically less exothermic: heat loss, incomplete combustion, evaporation all point the same way | closest to the data book (−1367) — the standard values were themselves measured carefully and corrected | less exothermic here: ethanol and water are really liquids, and no bond in these molecules is exactly average |
Which value would you trust? For the data book — the Hess value: it is built from standard values that were measured carefully and corrected, and states are exactly as defined. For this apparatus, today — the calorimetry value: it honestly reports what your burner and can did, flaws included. For a quick estimate of a gas-phase reaction nobody has tabulated — the bond-enthalpy value, quoted with its ≈. And notice the trap in the numbers: calorimetry (−1280) and bond enthalpies (−1279) nearly agree here — by coincidence, for entirely different reasons. Two wrong answers agreeing is not the same as two right ones.
Interactive — route runner
Phase 1 — which method does this data permit?
Phase 2 — execute it. Tap the moves in order, then give the value:
🧪 Exam-style questions
In an experiment to determine the enthalpy of combustion of limonene (C10H16), burning 1.31 g of limonene increases the temperature of 60.0 g of water in a copper calorimeter by 52.1 °C. The specific heat capacity of water is 4.18 J K−1 g−1. Calculate a value for the enthalpy of combustion, in kJ mol−1, of limonene.
Show answer
q = mcΔT = 60.0 × 4.18 × 52.1 = 13067 J = 13.07 kJ
Heat gained by the water 1 mark
Mr(C10H16) = 136.0 → n = 1.31 ÷ 136.0 = 9.63 × 10−3 mol
Moles of limonene burned 1 mark
ΔH = −13.07 ÷ (9.63 × 10−3)
Divide, with the sign 1 mark
= −1360 kJ mol−1
Final answer 1 mark
The table shows values, obtained by different methods, for the enthalpy of combustion of a different liquid hydrocarbon.
| Method | Enthalpy of combustion / kJ mol−1 |
|---|---|
| 1 — standard enthalpy of combustion, ΔcH⦵298 | −4194 |
| 2 — calculated from a calorimetry experiment | −1100 |
| 3 — calculated using mean bond enthalpies | −3159 |
Suggest reasons for the differences between the values obtained by Methods 2 and 3 and the value obtained by Method 1.
Show answer
Method 2 (calorimetry) is much less exothermic than Method 1 because heat is lost to the surroundings and absorbed by the copper calorimeter, combustion may be incomplete, and some liquid hydrocarbon may evaporate rather than burn. up to 3 marks
Method 3 (mean bond enthalpies) is less exothermic than Method 1 because mean bond enthalpies are averaged across a range of compounds, not the actual bonds in this molecule, and the all-gas calculation ignores the enthalpy changes of vaporising the liquid fuel and condensing the water. up to 2 marks
This is the page’s capstone in exam form: the data-book standard value (Method 1) is the reference; calorimetry undershoots through heat loss, and bond enthalpies undershoot through averaging and the gas-phase assumption. Award any 5 of the reasons above.
Source: AQA A-Level Chemistry past papers (2020–2024).
- Exothermic: heat released to the surroundings, ΔH negative. Endothermic: heat absorbed, ΔH positive. The sign is part of the value — write it every time.
- Enthalpy change (ΔH) = the heat energy change measured at constant pressure.
- Standard conditions = 100 kPa (not 1 atm) and a stated temperature, usually 298 K — all carried by the ⦵ sign in ΔH⦵; standard state = the substance’s normal state under those conditions.
- ΔcH⦵: 1 mol of substance burned completely in oxygen, all substances in standard states, standard conditions.
- ΔfH⦵: 1 mol of compound formed from its elements, all in standard states, standard conditions — so an element in its standard state has ΔfH⦵ = 0.
- q = mcΔT: m is the mass being heated (the solution or the water — never the fuel or the solid); c is always given; ΔT is the same in K and °C.
- Route to a molar value: q (J) → ÷1000 → ÷n → attach the sign from the temperature direction.
- RP2 — one reaction from: dissolution of KCl or Na2CO3, HCl + NaOH, Zn + CuSO4, combustion of alcohols — weigh by difference, stir, read to the thermometer’s precision.
- Cooling correction: extrapolate the cooling line back to the time of mixing — the corrected ΔT is bigger than any rise you observed.
- Error answers need a direction: heat loss makes an exothermic value less exothermic; an endothermic mixture gains heat from the surroundings, so its value comes out less endothermic.
- Sig figs follow the least accurate measurement; show the J → kJ conversion as its own line.
- Hess’s law: the enthalpy change is independent of the route, provided initial and final states are the same.
- Formation data: ΔH = ΣΔfH⦵(products) − ΣΔfH⦵(reactants). Combustion data: ΔH = ΣΔcH⦵(reactants) − ΣΔcH⦵(products). Derive both from the arrows: with = add, against = subtract.
- Multiply every value by its coefficient before summing — and keep state symbols honest (H2O(l) ≠ H2O(g)).
- Mean bond enthalpy: the enthalpy change when 1 mol of a covalent bond is broken, in the gas phase, averaged over many compounds — four components, all required.
- ΔH ≈ Σ(bonds broken) − Σ(bonds formed); breaking costs (+), making pays (−); exothermic reactions form stronger bonds than they break.
- Bond-enthalpy answers are approximate: the values are averages, and the method assumes everything is a gas — both reasons earn the mark when Hess and bond values disagree.