Hess's Law

Spec Mapping — OCR H432 Module 3.2.1 — Hess's law, covering the statement of Hess's law as a consequence of the first law of thermodynamics, the construction of enthalpy cycles connecting reactants and products via a common set of intermediates (elements in their standard states), the algebraic shortcut $\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$Δr​H⊖=∑Δf​H⊖(products)−∑Δf​H⊖(reactants) with the convention $\Delta_f H^\ominus(\text{element}) = 0$Δf​H⊖(element)=0, and the application of these techniques to reactions for which direct calorimetric measurement is impossible or impractical (refer to the official OCR H432 specification document for exact wording).

Many of the most important chemical reactions cannot be performed cleanly in a calorimeter. The enthalpy of formation of methane — $\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$C(s)+2H2​(g)→CH4​(g) — has never been measured directly, because graphite and hydrogen do not combine spontaneously at room temperature, and any high-temperature attempt produces a soup of hydrocarbons rather than pure methane. The hydration of anhydrous copper(II) sulfate cannot be measured cleanly because the solid does not absorb water uniformly in a calorimeter. Yet both values appear in every OCR data booklet, accurate to three significant figures. The trick that delivers them is Hess's law: a 19th-century insight by the Russian–Swiss chemist Germain Hess, who showed in 1840 that the total enthalpy change accompanying a sequence of reactions depends only on the initial and final states, not on the path taken. This lesson develops Hess's law as a direct consequence of the first law of thermodynamics, walks you through the construction of enthalpy cycles using $\Delta_f H^\ominus$Δf​H⊖ data, derives the algebraic shortcut $\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$Δr​H⊖=∑Δf​H⊖(products)−∑Δf​H⊖(reactants), and applies it to four worked examples covering ammonia formation, methane combustion, limestone decomposition, and copper-sulfate hydration. Lesson 4 will then treat the parallel formulation using $\Delta_c H^\ominus$Δc​H⊖ data.

Key Equation: for any reaction, when standard enthalpies of formation are known for every species: $\Delta_r H^\ominus = \sum \nu_i \Delta_f H^\ominus(\text{products}) - \sum \nu_j \Delta_f H^\ominus(\text{reactants})$Δr​H⊖=∑νi​Δf​H⊖(products)−∑νj​Δf​H⊖(reactants) where $\nu$ν is the stoichiometric coefficient. Elements in their standard states contribute zero. This is the single most useful equation in Module 3.2.1.

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Statement of Hess's law

Hess's law: the total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions (and hence the initial and final states) are the same.

This is a direct consequence of the first law of thermodynamics (conservation of energy) combined with the fact that enthalpy is a state function — a property of the current state of the system, depending only on $T$T, $p$p, and the substances present, not on the history by which that state was reached. Two routes from the same starting materials to the same products must therefore yield identical total enthalpy changes; otherwise we could couple them into a perpetual-motion machine that creates or destroys energy by going round the cycle, in violation of the first law.

Geometrically: any closed path round a Hess cycle integrates to zero. Equivalently, if $A \rightarrow B$A→B has enthalpy change $\Delta H_1$ΔH1​ and $B \rightarrow C$B→C has $\Delta H_2$ΔH2​, then $A \rightarrow C$A→C (direct) has $\Delta H_1 + \Delta H_2$ΔH1​+ΔH2​, and $C \rightarrow A$C→A has $-(\Delta H_1 + \Delta H_2)$−(ΔH1​+ΔH2​).

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Enthalpy cycles

A Hess cycle is a diagram with two routes from reactants to products:

• Route 1 (direct): Reactants $\xrightarrow{\Delta_r H^\ominus = ?}$Δr​H⊖=?​ Products
• Route 2 (indirect): Reactants $\rightarrow$→ Intermediates $\rightarrow$→ Products

By Hess's law:

$\Delta H(\text{direct}) = \sum \Delta H(\text{indirect route})$ΔH(direct)=∑ΔH(indirect route)

The intermediates are chosen for convenience: usually either (i) the constituent elements in their standard states, or (ii) the combustion products (CO$_2$2​, H$_2$2​O) of all organic species in the cycle. The first case is treated below; the second is the subject of Lesson 4.

Cycle using enthalpies of formation

When the intermediates are the elements in their standard states, the connecting arrows are enthalpies of formation:

graph TD
  R[Reactants] -->|ΔrH° = ?| P[Products]
  E[Elements in standard states] -->|Σ ΔfH° reactants| R
  E -->|Σ ΔfH° products| P

Each arrow leaving the elements points upwards towards a compound — these are all formation arrows (by definition $\Delta_f H^\ominus$Δf​H⊖ describes the formation of one mole of compound from the elements).

Deriving the algebraic shortcut

To convert reactants → products via the elements, we must:

1. Go backwards from reactants to elements: this is the reverse of forming the reactants, so the enthalpy change is $-\sum \Delta_f H^\ominus(\text{reactants})$−∑Δf​H⊖(reactants) (sign flipped).
2. Then go forwards from elements to products: this is the formation of the products, so the enthalpy change is $+\sum \Delta_f H^\ominus(\text{products})$+∑Δf​H⊖(products).

Adding the two steps:

$\Delta_r H^\ominus = -\sum \Delta_f H^\ominus(\text{reactants}) + \sum \Delta_f H^\ominus(\text{products})$Δr​H⊖=−∑Δf​H⊖(reactants)+∑Δf​H⊖(products)

Rearranged into the standard form:

$\boxed{\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})}$Δr​H⊖=∑Δf​H⊖(products)−∑Δf​H⊖(reactants)​

Each $\Delta_f H^\ominus$Δf​H⊖ must be multiplied by the stoichiometric coefficient of that species in the balanced equation. Elements in their standard states have $\Delta_f H^\ominus = 0$Δf​H⊖=0 and contribute nothing to the sum (but they must still appear in the bookkeeping — leaving them out and forgetting the zero is a classic error).

By Hess: ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants) = −92.2 − 0 = −92.2 kJ mol⁻¹

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Worked example 1 — formation of ammonia (Haber process)

Q: Calculate $\Delta_r H^\ominus$Δr​H⊖ for $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$N2​(g)+3H2​(g)→2NH3​(g), given $\Delta_f H^\ominus(\text{NH}_3\text{(g)}) = -46.1$Δf​H⊖(NH3​(g))=−46.1 kJ mol$^{-1}$−1.

A:

$\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$Δr​H⊖=∑Δf​H⊖(products)−∑Δf​H⊖(reactants) $= [2 \times \Delta_f H^\ominus(\text{NH}_3)] - [\Delta_f H^\ominus(\text{N}_2) + 3 \times \Delta_f H^\ominus(\text{H}_2)]$=[2×Δf​H⊖(NH3​)]−[Δf​H⊖(N2​)+3×Δf​H⊖(H2​)] $= [2 \times (-46.1)] - [0 + 3 \times 0] = -92.2 - 0 = \mathbf{-92.2 \text{ kJ mol}^{-1}}$=[2×(−46.1)]−[0+3×0]=−92.2−0=−92.2 kJ mol−1

The Haber reaction is moderately exothermic. Notice how the stoichiometric coefficients (2 for NH$_3$3​) multiply the formation enthalpies — a single $-46.1$−46.1 kJ mol$^{-1}$−1 becomes $-92.2$−92.2 kJ for the equation as written. The terms for N$_2$2​ and H$_2$2​ are zero because both are elements in their standard states.

Worked example 2 — combustion of methane via formation enthalpies

Q: Given $\Delta_f H^\ominus(\text{CH}_4) = -74.8$Δf​H⊖(CH4​)=−74.8, $\Delta_f H^\ominus(\text{CO}_2) = -393.5$Δf​H⊖(CO2​)=−393.5 and $\Delta_f H^\ominus(\text{H}_2\text{O(l)}) = -285.8$Δf​H⊖(H2​O(l))=−285.8 kJ mol$^{-1}$−1, calculate $\Delta_c H^\ominus(\text{CH}_4)$Δc​H⊖(CH4​).

A:

Target equation: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l)

$\Delta_r H^\ominus = [\Delta_f H^\ominus(\text{CO}_2) + 2 \times \Delta_f H^\ominus(\text{H}_2\text{O(l)})] - [\Delta_f H^\ominus(\text{CH}_4) + 2 \times \Delta_f H^\ominus(\text{O}_2)]$Δr​H⊖=[Δf​H⊖(CO2​)+2×Δf​H⊖(H2​O(l))]−[Δf​H⊖(CH4​)+2×Δf​H⊖(O2​)] $= [(-393.5) + 2 \times (-285.8)] - [(-74.8) + 2 \times 0]$=[(−393.5)+2×(−285.8)]−[(−74.8)+2×0] $= [-393.5 + (-571.6)] - [-74.8 + 0] = -965.1 - (-74.8) = -965.1 + 74.8 = \mathbf{-890.3 \text{ kJ mol}^{-1}}$=[−393.5+(−571.6)]−[−74.8+0]=−965.1−(−74.8)=−965.1+74.8=−890.3 kJ mol−1

Within 0.3% of the literature value $-890.4$−890.4 kJ mol$^{-1}$−1. The double-negative move ($- (-74.8) = +74.8$−(−74.8)=+74.8) is the commonest arithmetic mistake in this kind of question — work through it slowly.

Worked example 3 — thermal decomposition of calcium carbonate

Q: Calculate $\Delta_r H^\ominus$Δr​H⊖ for the industrial-scale decomposition $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$CaCO3​(s)→CaO(s)+CO2​(g), given:

Substance	$\Delta_f H^\ominus$Δf​H⊖ / kJ mol$^{-1}$−1
CaCO$_3$3​(s)	$-1207$−1207
CaO(s)	$-635$−635
CO$_2$2​(g)	$-394$−394

A:

$\Delta_r H^\ominus = [\Delta_f H^\ominus(\text{CaO}) + \Delta_f H^\ominus(\text{CO}_2)] - [\Delta_f H^\ominus(\text{CaCO}_3)]$Δr​H⊖=[Δf​H⊖(CaO)+Δf​H⊖(CO2​)]−[Δf​H⊖(CaCO3​)] $= [-635 + (-394)] - [-1207] = -1029 + 1207 = \mathbf{+178 \text{ kJ mol}^{-1}}$=[−635+(−394)]−[−1207]=−1029+1207=+178 kJ mol−1

Endothermic, as expected — limestone is heated to $\sim 900$∼900 °C in a cement kiln to produce quicklime (CaO), the basis of every cement clinker in the modern world. The endothermicity is the reason limestone is stable at room temperature: the reverse direction (carbonation of CaO in atmospheric CO$_2$2​) is exothermic, but only the high-$T$T kiln environment makes the forward direction kinetically accessible.

Worked example 4 — hydration of copper sulfate

Q: Calculate $\Delta_r H^\ominus$Δr​H⊖ for $\text{CuSO}_4\text{(s)} + 5\text{H}_2\text{O(l)} \rightarrow \text{CuSO}_4\cdot 5\text{H}_2\text{O(s)}$CuSO4​(s)+5H2​O(l)→CuSO4​⋅5H2​O(s), given $\Delta_f H^\ominus(\text{CuSO}_4\text{(s)}) = -771$Δf​H⊖(CuSO4​(s))=−771, $\Delta_f H^\ominus(\text{H}_2\text{O(l)}) = -286$Δf​H⊖(H2​O(l))=−286, and $\Delta_f H^\ominus(\text{CuSO}_4\cdot 5\text{H}_2\text{O(s)}) = -2280$Δf​H⊖(CuSO4​⋅5H2​O(s))=−2280 kJ mol$^{-1}$−1.

A:

$\Delta_r H^\ominus = [-2280] - [-771 + 5 \times (-286)] = -2280 - [-771 - 1430] = -2280 - [-2201]$Δr​H⊖=[−2280]−[−771+5×(−286)]=−2280−[−771−1430]=−2280−[−2201] $= -2280 + 2201 = \mathbf{-79 \text{ kJ mol}^{-1}}$=−2280+2201=−79 kJ mol−1

This value cannot be measured directly — anhydrous CuSO$_4$4​(s) does not absorb water in a clean calorimetric way; instead it goes through partial hydrates (CuSO$_4 \cdot \text{H}_2\text{O}$4​⋅H2​O, CuSO$_4 \cdot 3\text{H}_2\text{O}$4​⋅3H2​O) before reaching the pentahydrate, and the heat is released over hours. Hess's law lets us bypass the experimental difficulty entirely by combining tabulated formation enthalpies.

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Decision flowchart — which Hess cycle to use?

flowchart TD
  A[Read the question — what data are given?] --> B{ΔfH° values<br/>provided?}
  B -->|Yes| C[Use FORMATION cycle]
  B -->|No| D{ΔcH° values<br/>provided?}
  D -->|Yes| E[Use COMBUSTION cycle - Lesson 4]
  D -->|No| F{Mean bond<br/>enthalpies provided?}
  F -->|Yes| G[Use BOND ENTHALPY method - Lesson 5]
  F -->|No| H[Construct ad-hoc cycle from given data]
  C --> I[Formula: ΣΔfH°(products) − ΣΔfH°(reactants)]
  E --> J[Formula: ΣΔcH°(reactants) − ΣΔcH°(products)]
  G --> K[Formula: Σ bonds broken − Σ bonds formed]

The first step in any Hess calculation is to inspect the data table the question provides — that determines the cycle. A common trap is to memorise one formula and force-fit every problem to it; the correct approach is to derive the formula from the cycle in each case.

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Drawing good Hess cycles — exam technique

OCR examiners routinely penalise candidates who do not draw the cycle clearly. The conventional layout:

1. Write the target equation horizontally across the top with "$\Delta_r H^\ominus = ?$Δr​H⊖=?" above an arrow connecting reactants and products.
2. Write the intermediate state (elements in standard states, or combustion products) horizontally below.
3. Draw two arrows connecting reactants and products to the intermediate. Each arrow should be labelled with the relevant $\Delta H$ΔH (with sign, stoichiometric coefficient, and species).
4. State which formula you are using ($\Sigma \Delta_f H^\ominus$ΣΔf​H⊖ or $\Sigma \Delta_c H^\ominus$ΣΔc​H⊖).
5. Apply Hess's law explicitly: "$\Delta_r H^\ominus = \Delta H(\text{indirect route})$Δr​H⊖=ΔH(indirect route)".
6. Substitute numbers; check sign physically (combustion should be exothermic, dissociations endothermic).

Even a candidate whose arithmetic fails can earn method marks for a correctly-set-up cycle. Showing the cycle is more valuable than rushing to the answer.

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Why Hess's law matters historically