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Energetics is one of the most fundamental topics in A-Level Chemistry. It deals with the energy changes that accompany chemical reactions and physical processes. To tackle this topic with confidence, you need to understand what enthalpy is, how we define different types of enthalpy change, and how to interpret energy profile diagrams. This lesson lays the groundwork for every calculation you will meet later in the course.
Enthalpy (symbol H) is the total energy content of a system at constant pressure. We cannot measure the absolute enthalpy of a substance directly, but we can measure the enthalpy change (ΔH) when a reaction occurs.
ΔH = H(products) − H(reactants)
The sign of ΔH tells us about the direction of energy transfer:
This sign convention is essential — always check whether you are being asked for an exothermic or endothermic process and assign the correct sign.
Common sign error: Students sometimes write a positive value for an exothermic reaction because "the temperature went up." Remember: a temperature rise in the surroundings means the system lost energy, so ΔH is negative.
When we quote enthalpy changes, we do so under standard conditions so that values can be compared fairly. Standard conditions are:
Standard enthalpy changes are denoted with the symbol ΔH° (the superscript ° indicates standard conditions). For example, the standard state of carbon is graphite (not diamond), the standard state of oxygen is O₂(g), and the standard state of water is H₂O(l).
The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
For example, the formation of water:
H₂(g) + ½O₂(g) → H₂O(l) ΔH°f = −286 kJ mol⁻¹
Key points:
| Substance | ΔH°f / kJ mol⁻¹ | Notes |
|---|---|---|
| H₂O(l) | −286 | Liquid water at 298 K |
| H₂O(g) | −242 | Steam — less exothermic because vaporisation is endothermic |
| CO₂(g) | −394 | Very stable product of combustion |
| CH₄(g) | −75 | Methane |
| C₂H₅OH(l) | −277 | Ethanol |
| NH₃(g) | −46 | Ammonia |
| NaCl(s) | −411 | Sodium chloride |
| MgO(s) | −602 | Magnesium oxide |
| O₂(g) | 0 | Element in standard state |
| C(s, graphite) | 0 | Element in standard state |
| Fe(s) | 0 | Element in standard state |
The standard enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen, under standard conditions, with all reactants and products in their standard states.
For example, the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH°c = −890 kJ mol⁻¹
Key points:
| Substance | Formula | ΔH°c / kJ mol⁻¹ |
|---|---|---|
| Hydrogen | H₂(g) | −286 |
| Carbon (graphite) | C(s) | −394 |
| Methane | CH₄(g) | −890 |
| Ethane | C₂H₆(g) | −1560 |
| Propane | C₃H₈(g) | −2220 |
| Methanol | CH₃OH(l) | −726 |
| Ethanol | C₂H₅OH(l) | −1367 |
| Propan-1-ol | C₃H₇OH(l) | −2021 |
The standard enthalpy of neutralisation is the enthalpy change when an acid and a base react to produce one mole of water, under standard conditions.
For a strong acid and strong base:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) ΔH°neut ≈ −57.1 kJ mol⁻¹
Key points:
The standard enthalpy of atomisation is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state, under standard conditions.
For example:
½Cl₂(g) → Cl(g) ΔH°at = +121 kJ mol⁻¹
Na(s) → Na(g) ΔH°at = +107 kJ mol⁻¹
Key points:
The standard enthalpy of atomisation of chlorine is +121 kJ mol⁻¹. This means:
½Cl₂(g) → Cl(g) ΔH = +121 kJ mol⁻¹
To break one mole of Cl–Cl bonds (producing 2 moles of Cl atoms):
Cl₂(g) → 2Cl(g) ΔH = 2 × 121 = +242 kJ mol⁻¹
So the Cl–Cl bond energy is +242 kJ mol⁻¹. Always check whether the question asks for the atomisation enthalpy (one mole of atoms) or the bond energy (one mole of bonds broken).
Energy profile diagrams (also called reaction profiles or enthalpy level diagrams) show how the enthalpy of the system changes as a reaction proceeds.
For an exothermic reaction, the products sit lower than the reactants on the diagram. The difference in height represents ΔH (which is negative). There is a peak between reactants and products called the transition state, and the height from reactants to this peak is the activation energy (Ea).
For an endothermic reaction, the products sit higher than the reactants. ΔH is positive. There is still an activation energy peak — the height from reactants to the transition state.
Activation energy (Ea) is the minimum energy that colliding particles must possess for a reaction to occur. Even exothermic reactions need an initial input of activation energy to get started. A catalyst lowers the activation energy by providing an alternative reaction pathway, but it does not change ΔH.
An endothermic reaction has ΔH = +178 kJ mol⁻¹ and Ea(forward) = +253 kJ mol⁻¹.
The products sit 178 kJ mol⁻¹ above the reactants. The transition state sits 253 kJ mol⁻¹ above the reactants. For the reverse (exothermic) reaction, the activation energy is measured from the products down to the transition state:
Ea(reverse) = 253 − 178 = 75 kJ mol⁻¹
This makes sense: the reverse reaction is exothermic, so it needs less activation energy.
Understanding enthalpy changes has direct practical importance:
| Mistake | Correction |
|---|---|
| Forgetting the sign on ΔH | Exothermic = negative; endothermic = positive. Always state the sign. |
| Using H₂O(g) instead of H₂O(l) | Under standard conditions (298 K), water is a liquid. |
| Writing 2 moles of product for ΔH°f | Formation must produce exactly 1 mole of product. |
| Saying ΔH°f of O₂ is zero "because it is a gas" | It is zero because O₂(g) is the standard state of oxygen, not because of its physical state. |
| Confusing atomisation enthalpy with bond energy | Atomisation produces 1 mol of gaseous atoms; bond energy breaks 1 mol of bonds (which may produce 2 mol of atoms). |
Understanding these definitions precisely is crucial. Exam questions frequently test whether you can write correct equations for standard enthalpy changes (correct coefficients, correct state symbols, exactly one mole where required) and whether you understand the sign conventions. Pay close attention to:
Edexcel 9CH0 specification Topic 8 (Energetics I) and Topic 13 (Energetics II) require precise definitions of standard enthalpy changes including ΔH°_combustion, ΔH°_formation, ΔH°_atomisation, ΔH°_neutralisation, ΔH°_solution, ΔH°_lattice formation/dissociation, and ΔH°_hydration (refer to the official specification document for exact wording). Topic 8 introduces enthalpy changes, calorimetry and Hess cycles; Topic 13 layers Born–Haber cycles, lattice energy, entropy and Gibbs free energy on top. Definitions are tested on Paper 1 (advanced inorganic and physical chemistry) and Paper 3 (synoptic), and are an unavoidable foundation for any quantitative item across the linear assessment. Standard conditions of 100 kPa, 298 K and standard states must be quoted explicitly when the question demands a definition. The Edexcel data booklet provides standard enthalpy values for selected substances; structural definitions and equations must be written from memory.
Question (8 marks):
(a) State the meaning of standard enthalpy of formation. (2)
(b) Write equations, including state symbols, for the reaction whose enthalpy change is the standard enthalpy of formation of: (i) ethanol, C₂H₅OH(l), and (ii) sodium chloride, NaCl(s). (2)
(c) The standard enthalpies of formation of CO₂(g) and H₂O(l) are −394 and −286 kJ mol⁻¹. ΔH°_c[C₂H₅OH(l)] = −1367 kJ mol⁻¹. Calculate ΔH°_f[C₂H₅OH(l)]. (4)
Solution with mark scheme:
(a) M1 — "the enthalpy change when one mole of a compound is formed from its constituent elements". A1 — "with all reactants and products in their standard states under standard conditions (100 kPa, 298 K)". Both clauses are required for full marks; quoting the temperature and pressure earns the second mark only when both appear.
(b) (i) B1 — 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). The use of graphite (not just C(s)) and the half-coefficient on O₂ are both expected. (ii) B1 — Na(s) + ½Cl₂(g) → NaCl(s). State symbols and the half-coefficient must both appear.
(c) Combustion of ethanol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l).
By Hess's law: ΔH°_c = 2·ΔH°_f[CO₂] + 3·ΔH°_f[H₂O] − ΔH°_f[C₂H₅OH] − 3·ΔH°_f[O₂].
M1 — applying Hess correctly with formation values for products minus reactants.
Substitute: −1367 = 2(−394) + 3(−286) − ΔH°_f[C₂H₅OH] − 0.
M1 — correct substitution; ΔH°_f for O₂(g) recognised as zero.
ΔH°_f[C₂H₅OH] = 2(−394) + 3(−286) − (−1367) = −788 − 858 + 1367 = −279 kJ mol⁻¹.
A1 — correct numerical answer (−279 kJ mol⁻¹). A1 — correct sign (negative; ethanol's formation is exothermic).
Total: 8 marks (M2 A2 B2 + 2A).
Question (6 marks): Define each of the following standard enthalpy changes, and for each write a balanced equation with state symbols.
(a) Enthalpy of atomisation of sodium. (2) (b) Enthalpy of neutralisation. (2) (c) Enthalpy of hydration of Mg²⁺(g). (2)
Mark scheme decomposition by AO:
| Part | Marks | AO | Earned by |
|---|---|---|---|
| (a) | M1 | AO1.1 | "enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state" |
| (a) | A1 | AO2.1 | Na(s) → Na(g), correct state symbols, ΔH°_at positive |
| (b) | M1 | AO1.1 | "enthalpy change when 1 mole of water is formed by neutralisation" under standard conditions |
| (b) | A1 | AO2.1 | H⁺(aq) + OH⁻(aq) → H₂O(l) |
| (c) | M1 | AO1.1 | "enthalpy change when 1 mole of gaseous ions is dissolved in water to form an aqueous solution" |
| (c) | A1 | AO2.1 | Mg²⁺(g) + aq → Mg²⁺(aq) |
Total: 6 marks split AO1 = 3, AO2 = 3. Definition questions reward precision: "1 mole" must appear, the species and conditions must be unambiguous, and equations must use the correct state symbols (in particular (g) for atomisation, (aq) for hydration, (l) for water in neutralisation).
| AO | Typical share on definition questions | Earned by |
|---|---|---|
| AO1 (knowledge) | 60–70% | Stating definitions verbatim from the spec; quoting standard conditions; identifying signs |
| AO2 (application) | 20–30% | Writing balanced equations with correct state symbols; selecting which enthalpy term applies to a given equation |
| AO3 (analysis) | 10–20% | Comparing enthalpies (e.g. why ΔH°_neut for HCN is less exothermic than for HCl) |
Phrasing rewarded by examiners: "one mole", "standard states", "100 kPa and 298 K", "gaseous ions/atoms", "aqueous solution of infinite dilution". Phrasing that loses marks: "the energy of the reaction" (ambiguous — energy is not enthalpy); "at room temperature and pressure" (RTP differs from standard conditions); omitting "(g)" for atoms or ions in atomisation/lattice contexts. Avoid quoting examiner reports verbatim; the Edexcel mark schemes vary in wording but consistently demand "one mole" and "standard states".
Question: Define standard enthalpy of combustion and write a balanced equation for the standard enthalpy of combustion of methane.
Grade A response (~140 words):*
The standard enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen, with all reactants and products in their standard states under standard conditions of 100 kPa and 298 K.
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Examiner commentary: Full marks (3/3). The candidate names "one mole", "complete combustion", "excess oxygen", "standard states" and "standard conditions" with explicit values for pressure and temperature. The equation uses (l) for water (correct under standard conditions, where water is liquid) and balances cleanly. A common Grade B slip is to omit "complete" — a partial combustion gives CO or soot, which is not the defined process — losing the M1.
Question: (a) Define standard enthalpy of formation. (b) Explain why ΔH°_f[O₂(g)] = 0. (c) Write the equation for ΔH°_f of liquid ethanol.
Grade A response (~210 words):*
(a) The standard enthalpy of formation is the enthalpy change when one mole of a substance is formed from its constituent elements, with all reactants and products in their standard states under standard conditions (100 kPa, 298 K).
(b) Oxygen gas, O₂(g), is the standard state of the element oxygen at 100 kPa and 298 K. Forming it from itself requires no chemical change, so by definition the enthalpy change is zero. This is a convention: it provides a baseline against which compound formation enthalpies are measured.
(c) 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
Examiner commentary: Full marks (6/6). Part (b) is the discriminator: the Grade B candidate often writes "because oxygen is an element" without explaining the convention embedded in the definition. The A* answer pinpoints "standard state" and notes that the value is set by definition. Part (c) earns the A1 by quoting graphite (the standard state of carbon, not diamond) and using fractional ½O₂.
Question: ΔH°_f values: NH₃(g) = −46; H₂O(l) = −286; NO(g) = +90; HNO₃(aq) = −207. Calculate ΔH for: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l), and discuss the implication of the sign for industrial ammonia oxidation.
Grade A response (~310 words):*
By Hess's law, ΔH°_r = ΣΔH°_f(products) − ΣΔH°_f(reactants).
Products: 4(+90) + 6(−286) = +360 − 1716 = −1356 kJ. Reactants: 4(−46) + 5(0) = −184 kJ. (O₂ is in its standard state.)
ΔH°_r = −1356 − (−184) = −1172 kJ mol⁻¹ (per equation as written).
The reaction is strongly exothermic. In the industrial Ostwald process, ammonia is oxidised over a platinum–rhodium catalyst at ~900 °C and ~5 atm. The exothermicity of this step provides much of the heat needed to sustain the process. Le Chatelier reasoning suggests that a high temperature would shift equilibrium back toward reactants, but kinetic considerations (rate of NO formation) and the use of a catalyst at high temperature give an acceptable compromise. The sign also tells engineers that heat must be removed continuously from the converter to prevent runaway.
Examiner commentary: Full marks (9/9). The candidate (i) computes correctly using Hess (M1, M1, A1), (ii) identifies O₂ as having zero ΔH°_f (an easy slip), (iii) connects the sign of ΔH to industrial design (AO3), (iv) addresses both the kinetic and thermodynamic aspects without conflating them. A Grade A candidate scoring 7/9 typically computes correctly but writes only "the reaction is exothermic and useful for industry"; the A* candidate explains why the sign matters in an engineering sense.
Oxbridge interview prompt: "Why is ΔH°_f[C(diamond)] positive? What does this say about the thermodynamic stability of diamond at room temperature, and why does diamond not spontaneously convert to graphite on your finger?"
Energetics is anchored by CP1: Determination of the enthalpy change of a reaction and CP9: Determination of the enthalpy change of an insoluble salt by an indirect method (Hess cycle). For this lesson, CP1 is the relevant practical: students measure the temperature change when, for example, a known mass of an alcohol is burned beneath a calorimeter containing a known mass of water, and compute ΔH°_c via q = mcΔT and n = m/Mr. The experimental value is invariably less exothermic than the data-book value because of heat losses to the surroundings, incomplete combustion (yellow flame, soot), and evaporation of the alcohol before combustion. Students should practise stating these limitations explicitly and quantitatively, using the textbook value as a benchmark. CP9, met later in this course, uses standard enthalpies of formation in a Hess cycle to deduce a value that cannot be measured directly — the connection between today's definitions and tomorrow's calorimetric reasoning.
This content is aligned with the Pearson Edexcel GCE A Level Chemistry (9CH0) specification, Topics 8 and 13 — Energetics I and II. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.
graph TD
A["Enthalpy change ΔH°<br/>defined under<br/>standard conditions"] --> B{"What process?"}
B -->|"Element → 1 mol compound"| C["ΔH°_f<br/>formation<br/>(can be ±)"]
B -->|"1 mol fuel + O₂"| D["ΔH°_c<br/>combustion<br/>(always −)"]
B -->|"Element → 1 mol gaseous atoms"| E["ΔH°_at<br/>atomisation<br/>(always +)"]
B -->|"Acid + base → 1 mol H₂O"| F["ΔH°_neut<br/>(always −)"]
B -->|"Solid → aqueous"| G["ΔH°_sol<br/>(can be ±)"]
B -->|"Gaseous ions → lattice"| H["ΔH°_lat<br/>formation (−)<br/>dissociation (+)"]
B -->|"Gaseous ion → aqueous"| I["ΔH°_hyd<br/>(always −)"]
C --> J["Standard states<br/>100 kPa, 298 K<br/>1 mol product"]
D --> J
E --> J
F --> J
G --> J
H --> J
I --> J
style J fill:#27ae60,color:#fff