You are viewing a free preview of this lesson.
Subscribe to unlock all 8 lessons in this course and every other course on LearningBro.
Hess's law, formulated by the Swiss-Russian chemist Germain Henri Hess in 1840, states that the enthalpy change for a chemical reaction depends only on the initial and final states of the system, not on the route taken between them. It is a direct consequence of enthalpy being a state function: H depends on the present state of the system (composition, temperature, pressure, phase), not on the history by which that state was reached. Hess's law is the workhorse of A-Level energetics. It lets us compute ΔH for reactions that cannot be measured directly — for example, the formation of carbon monoxide from carbon and oxygen, C(s) + ½O₂(g) → CO(g), cannot be performed cleanly in a calorimeter because the carbon would always combust further to CO₂. By constructing an enthalpy cycle, we obtain ΔfH(CO) indirectly from the easily measurable combustion enthalpies of C and CO. This lesson develops Hess cycles, the formation route, the combustion route, and a sequence of worked examples that drill the sign-and-stoichiometry bookkeeping that examiners reward.
Spec mapping (AQA 7405): This lesson maps to §3.1.4 (energetics — Hess's law and enthalpy cycles). It builds on lesson 0 (calorimetry and the direct measurement of ΔH), develops alongside lesson 1 (mean bond enthalpies — a third route to ΔrH), and is the conceptual foundation for lesson 3 (Born-Haber cycles, which apply Hess's law to ionic-lattice formation). The thermodynamic apparatus extends into §3.1.8 (A2 thermodynamics: entropy, Gibbs free energy) where Hess-like additivity reappears for ΔS and ΔG. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 items include stating Hess's law and giving the formal definitions of standard enthalpy of formation (ΔfH) and standard enthalpy of combustion (ΔᶜH). AO2 calculations require fluent application of two equations: ΔrxnH = ΣΔfH(products) − ΣΔfH(reactants) for the formation route, and ΔrxnH = ΣΔᶜH(reactants) − ΣΔᶜH(products) for the combustion route — note the sign reversal between the two formulas, which is the single most common mark-loss in this topic. AO3 problems demand construction of a custom Hess cycle for an unusual combination of species (e.g. deriving ΔfH(CO) from ΔᶜH(C), ΔᶜH(CO), and ΔfH(CO₂)) and justification of the path-independence on conservation-of-energy grounds.
Key Definition: Hess's law states that the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions (composition, temperature, pressure, phase) are the same.
The deeper justification is that enthalpy H is a state function: the value of H for a system depends only on its current state, not on the path by which that state was reached. The first law of thermodynamics (conservation of energy) demands this — if Hess's law were violated, we could construct a perpetual-motion machine by cycling between two routes with different ΔH and extracting the difference. No such device has ever been built.
Route 1: A ———————————→ D ΔH₁
Route 2: A → B → C → D ΔH₂ + ΔH₃ + ΔH₄
Hess's law: ΔH₁ = ΔH₂ + ΔH₃ + ΔH₄
Hess's original 1840 paper, "Thermochemische Untersuchungen", reported that the heat evolved when sulfuric acid is diluted in two stages equals the heat evolved when the same dilution is performed in a single step. He generalised: the heat evolved in a chemical change is the same whether the reaction occurs in one or several steps. The result long predated the formal first law of thermodynamics (Clausius, 1850s), but Hess intuited the path-independence from his experiments.
A Hess cycle is a closed triangle (or polygon) of reactions whose enthalpy changes must sum to zero around the loop. By identifying a target reaction and two alternative routes — direct, and via a common intermediate (usually elements in their standard states, or combustion products CO₂ + H₂O) — we can solve for the unknown ΔH using the path-independence rule.
Conventions:
The superscript ° denotes standard conditions (often written as ⦵ in modern notation): pressure 100 kPa, with all substances in their thermodynamically most stable form at the temperature specified (usually 298 K). Standard enthalpies of formation and combustion are tabulated at 298 K.
Key Definition: Standard enthalpy of formation, ΔfH°: the enthalpy change when one mole of a substance is formed from its elements in their standard states, all reactants and products being in their standard states. By convention, ΔfH° of an element in its standard state is exactly zero.
Key Definition: Standard enthalpy of combustion, ΔᶜH°: the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions, all reactants and products being in their standard states. Always negative (combustion is exothermic).
When tabulated ΔfH° values are available for every species in a reaction, the reaction enthalpy is given directly by:
ΔrxnH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)
(weighted by stoichiometric coefficients).
Reactants ————ΔrxnH°————→ Products
↘ ↗
Σ ΔfH°(reactants) Σ ΔfH°(products)
↘ ↗
Elements in
standard states
Going round the loop: down from reactants to elements is −ΣΔfH°(reactants); up from elements to products is +ΣΔfH°(products); across from reactants to products is ΔrxnH°. The closure condition gives the formula.
Calculate ΔrxnH° for: 2CO(g) + O₂(g) → 2CO₂(g)
Given: ΔfH°(CO) = −111 kJ mol⁻¹, ΔfH°(CO₂) = −394 kJ mol⁻¹, ΔfH°(O₂) = 0 (element).
ΔrxnH° = ΣΔfH°(products) − ΣΔfH°(reactants)
ΔrxnH° = [2 × (−394)] − [2 × (−111) + 1 × 0]
ΔrxnH° = (−788) − (−222) = −566 kJ mol⁻¹
This is the molar enthalpy of combustion of 2 mol CO, so ΔᶜH°(CO) = −566/2 = −283 kJ mol⁻¹.
Calculate ΔrxnH° for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Given: ΔfH°(CH₄) = −74.8 kJ mol⁻¹, ΔfH°(CO₂) = −393.5 kJ mol⁻¹, ΔfH°(H₂O(l)) = −285.8 kJ mol⁻¹.
Products: ΔfH°(CO₂) + 2 × ΔfH°(H₂O) = (−393.5) + 2(−285.8) = −965.1 kJ mol⁻¹
Reactants: ΔfH°(CH₄) + 2 × ΔfH°(O₂) = (−74.8) + 2(0) = −74.8 kJ mol⁻¹
ΔrxnH° = (−965.1) − (−74.8) = −890.3 kJ mol⁻¹
This matches the experimentally measured ΔᶜH°(CH₄), confirming the cycle.
Calculate ΔrxnH° for: C₂H₄(g) + H₂(g) → C₂H₆(g)
Given: ΔfH°(C₂H₄) = +52.5 kJ mol⁻¹, ΔfH°(C₂H₆) = −84.7 kJ mol⁻¹, ΔfH°(H₂) = 0.
ΔrxnH° = ΣΔfH°(products) − ΣΔfH°(reactants)
ΔrxnH° = [−84.7] − [+52.5 + 0] = −137.2 kJ mol⁻¹
Exothermic — a weaker C═C π bond is replaced by stronger σ bonds (C—C and two new C—H).
When tabulated ΔᶜH° values are available — particularly common for organic compounds, whose combustion can be measured by bomb calorimetry — the reaction enthalpy is:
ΔrxnH° = Σ ΔᶜH°(reactants) − Σ ΔᶜH°(products)
Note the sign reversal compared with the formation route. This is the single most common error in A-Level energetics. The mnemonic is: formation builds up from elements, so products − reactants; combustion takes down to CO₂ + H₂O, so reactants − products (you "go down twice", and the second negative flips the sign).
Reactants ————ΔrxnH°————→ Products
↘ ↙
Σ ΔᶜH°(reactants) Σ ΔᶜH°(products)
↘ ↙
Combustion
products
(CO₂ + H₂O)
Calculate ΔfH°(C₂H₆) given ΔᶜH°(C₂H₆) = −1560 kJ mol⁻¹, ΔᶜH°(C, graphite) = −394 kJ mol⁻¹, ΔᶜH°(H₂) = −286 kJ mol⁻¹.
Target: 2C(s) + 3H₂(g) → C₂H₆(g), which is ΔfH°(C₂H₆).
Cycle: from 2C + 3H₂ on the left, route via combustion: 2C → 2CO₂ (ΔH = 2×−394) and 3H₂ → 3H₂O (ΔH = 3×−286), giving products 2CO₂ + 3H₂O. From C₂H₆ on the right, combustion goes to 2CO₂ + 3H₂O (ΔH = −1560). Both paths end at the same point.
ΔfH°(C₂H₆) = [2 × ΔᶜH°(C) + 3 × ΔᶜH°(H₂)] − ΔᶜH°(C₂H₆)
ΔfH°(C₂H₆) = [2(−394) + 3(−286)] − (−1560)
ΔfH°(C₂H₆) = [−788 − 858] − (−1560) = −1646 + 1560 = −86 kJ mol⁻¹
(Literature value: −84.7 kJ mol⁻¹, close agreement.)
Recompute Example 3 using combustion data: ΔᶜH°(C₂H₄) = −1411 kJ mol⁻¹, ΔᶜH°(H₂) = −286 kJ mol⁻¹, ΔᶜH°(C₂H₆) = −1560 kJ mol⁻¹.
ΔrxnH° = ΣΔᶜH°(reactants) − ΣΔᶜH°(products)
ΔrxnH° = [(−1411) + (−286)] − [(−1560)] = −1697 + 1560 = −137 kJ mol⁻¹
Identical (to rounding) to the formation-route answer in Example 3. This is Hess's law in action: two independent routes, one identical answer.
The direct synthesis C(s) + ½O₂(g) → CO(g) cannot be carried out cleanly: carbon burned in limited oxygen produces a mixture of CO and CO₂, and burned in excess oxygen produces only CO₂. So ΔfH°(CO) is not measurable directly — it must be obtained by Hess's law.
Given: ΔᶜH°(C, graphite) = −394 kJ mol⁻¹ (i.e. ΔfH°(CO₂) = −394 kJ mol⁻¹) and ΔᶜH°(CO) = −283 kJ mol⁻¹ (i.e. ΔH for CO + ½O₂ → CO₂).
Cycle:
Sum of path B: −394 + 283 = −111 kJ mol⁻¹.
By Hess: ΔfH°(CO) = −111 kJ mol⁻¹. (Matches the value used in Example 1.)
This is the canonical worked example for the combustion route, and it appears regularly in AQA past papers.
Calculate ΔᶜH° for: C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)
Given: ΔfH°(C₆H₁₂O₆, s) = −1273 kJ mol⁻¹, ΔfH°(CO₂) = −394 kJ mol⁻¹, ΔfH°(H₂O, l) = −286 kJ mol⁻¹.
ΔᶜH° = [6(−394) + 6(−286)] − [(−1273) + 6(0)]
ΔᶜH° = [−2364 − 1716] − [−1273] = −4080 + 1273 = −2807 kJ mol⁻¹
A bond-enthalpy estimate (lesson 1) gives roughly −2800 kJ mol⁻¹, within 10 kJ — but the Hess-law value via ΔfH° is the definitive textbook number because tabulated formation enthalpies use experimental data, whereas bond enthalpies are gas-phase averages that miss the solid-state lattice energy of crystalline glucose.
For non-standard combinations of data — for example, you are given some ΔfH° values and one ΔᶜH° value, and asked for an unrelated reaction — the recipe is:
The discipline of always drawing the cycle — rather than memorising a formula — is what separates A and A* responses. The cycle makes the sign conventions visible.
| Pitfall | Description | Mitigation |
|---|---|---|
| Sign reversal (formation vs combustion) | ΔrxnH = products − reactants for ΔfH, but reactants − products for ΔᶜH | Memorise: "formation builds up; combustion goes down" |
| Forgetting stoichiometric factor | Failing to multiply ΔfH(H₂O) by 2 when 2H₂O appear | Always rewrite each term with its coefficient explicitly |
| Treating ΔfH(element) as non-zero | E.g. quoting ΔfH(O₂) = +498 (which is the O═O bond enthalpy, not formation) | ΔfH of any element in its standard state is exactly 0 |
| Mis-direction of arrows in cycle | Drawing the combustion arrow upward, forgetting to flip the sign | Always orient combustion arrows downward toward CO₂ + H₂O |
| Phase confusion in ΔᶜH | Using H₂O(g) instead of H₂O(l) — the difference is +44 kJ mol⁻¹ per mole of water (enthalpy of vaporisation) | Standard ΔᶜH° uses H₂O(l); check the data table |
| Mixing routes mid-calculation | Switching between formation and combustion data without a cycle diagram | Draw the cycle first; then read off the equation |
Question 1. [13 marks total]
(a) State Hess's law. [2 marks]
(b) Use the following standard enthalpies of formation to calculate the standard enthalpy change for the reaction:
2NH₃(g) + 3N₂O(g) → 4N₂(g) + 3H₂O(l)
Given: ΔfH°(NH₃) = −46 kJ mol⁻¹, ΔfH°(N₂O) = +82 kJ mol⁻¹, ΔfH°(H₂O, l) = −286 kJ mol⁻¹. [4 marks]
Subscribe to continue reading
Get full access to this lesson and all 8 lessons in this course.