Hess's Law from Combustion Data

Spec Mapping — OCR H432 Module 3.2.1 — Hess's law (combustion cycle), covering the construction of Hess cycles in which the common intermediates are the combustion products (CO $_2$ (g) + H $_2$ O(l)), the algebraic shortcut $\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$ , comparison of formation-route and combustion-route cycles, and application to organic-chemistry reactions for which only bomb-calorimetric combustion data are available (refer to the official OCR H432 specification document for exact wording).

Lesson 3 established that any reaction enthalpy can be calculated from tabulated $\Delta_f H^\ominus$ data via the formation cycle. But many organic compounds do not have direct $\Delta_f H^\ominus$ measurements — graphite and hydrogen do not combine to form methane in a calorimeter, glucose does not assemble cleanly from its constituent elements, and ethanol cannot be made directly from C(s) and H $_2$ (g) at standard conditions. What we can measure for almost every organic compound is its enthalpy of combustion $\Delta_c H^\ominus$ , because bomb calorimetry burns the substance to completion in excess oxygen and the temperature rise gives $q$ directly with $\pm 0.1\%$ precision. This lesson develops the parallel Hess-cycle technique that uses combustion products (CO $_2$ (g) + H $_2$ O(l)) as the common intermediates instead of elements. You will see why the sign convention inverts compared with the formation cycle — $\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$ rather than the products-minus-reactants of Lesson 3 — and why this inversion confuses examination candidates routinely. Four worked examples cover the formation of methane from combustion data, the hydrogenation of ethene, the fermentation of glucose, and the formation of propan-1-ol.

Key Equation: for any reaction where all species can be combusted (with the conventional exception of O $_2$ and the combustion products themselves): $\Delta_r H^\ominus = \sum \nu_i \Delta_c H^\ominus(\text{reactants}) - \sum \nu_j \Delta_c H^\ominus(\text{products})$ Note the reversal compared to the formation formula. CO $_2$ (g), H $_2$ O(l), O $_2$ (g) and other non-combustible species have no $\Delta_c H^\ominus$ and contribute zero.

Why combustion cycles?

Bomb calorimetry, introduced in Lesson 2, achieves $\pm 0.1\%$ precision for $\Delta_c H^\ominus$ on essentially any organic substance — provided it can be ignited in oxygen and burns to completion. The substance is weighed (often $\sim 1$ g), placed in a steel "bomb" pressurised to $\sim 25$ atm with pure O $_2$ , ignited electrically, and the calibrated heat capacity of the apparatus converts the observed temperature rise into a quoted $\Delta_c H^\ominus$ . The OCR data booklet tabulates $\Delta_c H^\ominus$ for dozens of organic compounds for exactly this reason: they are directly measurable.

If we want the enthalpy change for an organic reaction such as the hydrogenation of ethene, $\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6$ , we can construct a Hess cycle in which the common intermediate is the set of combustion products. Each species in the reaction has a well-tabulated $\Delta_c H^\ominus$ ; the cycle pivots through CO $_2$ (g) and H $_2$ O(l) and gives the target $\Delta_r H^\ominus$ within experimental precision.

Constructing the combustion cycle

The geometry of the combustion cycle is inverted compared to the formation cycle. Both reactants and products combust downwards to the common combustion products (rather than forming upwards from elements):

graph TD
  R[Reactants] -->|ΔrH° = ?| P[Products]
  R -->|Σ ΔcH° reactants| C[Combustion products: CO₂ + H₂O]
  P -->|Σ ΔcH° products| C

To trace the indirect route from reactants → products, we must:

Combust the reactants (descending arrow, $+\sum \Delta_c H^\ominus(\text{reactants})$ ).
Reverse-combust the products back from combustion products to products (ascending arrow, $-\sum \Delta_c H^\ominus(\text{products})$ ).

Adding the two steps:

$\Delta_r H^\ominus = +\sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$

Hence:

$\boxed{\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})}$

Note the reversal compared to the formation cycle. Many candidates mix the two formulae — always draw the arrows first, derive the formula from the geometry, and then substitute.

By Hess: ΔrH° = Σ ΔcH°(reactants) − Σ ΔcH°(products) = −1697 − (−1560) = −137 kJ mol⁻¹

Memory aid — F-UP / C-DOWN

A useful mnemonic for keeping the two cycles distinct:

F-UP: Formation cycle — arrows go UP from elements to compounds → formula is products minus reactants.
C-DOWN: Combustion cycle — arrows go DOWN from species to combustion products → formula is reactants minus products.

When in doubt, draw the arrows on the page first; the formula follows from "go around the cycle the way the arrows allow, flipping signs whenever you reverse an arrow".

Worked example 1 — enthalpy of formation of methane (from combustion data)

Q: Use the following $\Delta_c H^\ominus$ data to calculate $\Delta_f H^\ominus(\text{CH}_4\text{(g)})$ :

Substance	$\Delta_c H^\ominus$ / kJ mol $^{-1}$
C(s, graphite)	$-393.5$
H $_2$ (g)	$-285.8$
CH $_4$ (g)	$-890.3$

Target equation: $\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$ , so $\Delta_r H^\ominus = \Delta_f H^\ominus(\text{CH}_4)$ .

Using $\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$ :

$\Delta_f H^\ominus(\text{CH}_4) = [\Delta_c H^\ominus(\text{C}) + 2 \times \Delta_c H^\ominus(\text{H}_2)] - [\Delta_c H^\ominus(\text{CH}_4)]$ $= [-393.5 + 2 \times (-285.8)] - [-890.3] = [-393.5 + (-571.6)] - [-890.3]$ $= -965.1 - (-890.3) = -965.1 + 890.3 = \mathbf{-74.8 \text{ kJ mol}^{-1}}$

Matches the literature value exactly. This is the historical method by which $\Delta_f H^\ominus(\text{CH}_4)$ entered the data books — methane cannot be formed directly from graphite and hydrogen in a calorimeter, but all three substances have $\Delta_c H^\ominus$ measurable in a bomb.

Worked example 2 — hydrogenation of ethene

Q: Calculate $\Delta_r H^\ominus$ for $\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \rightarrow \text{C}_2\text{H}_6\text{(g)}$ given:

Substance	$\Delta_c H^\ominus$ / kJ mol $^{-1}$
C $_2$ H $_4$ (g)	$-1411$
H $_2$ (g)	$-286$
C $_2$ H $_6$ (g)	$-1560$

$\Delta_r H^\ominus = [\Delta_c H^\ominus(\text{C}_2\text{H}_4) + \Delta_c H^\ominus(\text{H}_2)] - [\Delta_c H^\ominus(\text{C}_2\text{H}_6)]$ $= [-1411 + (-286)] - [-1560] = -1697 + 1560 = \mathbf{-137 \text{ kJ mol}^{-1}}$

Modestly exothermic — consistent with the fact that catalytic hydrogenation of alkenes is exothermic enough that industrial margarine plants must cool their reactors continuously to maintain selectivity.

Worked example 3 — fermentation of glucose

Q: For the fermentation reaction $\text{C}_6\text{H}_{12}\text{O}_6\text{(s)} \rightarrow 2\text{C}_2\text{H}_5\text{OH(l)} + 2\text{CO}_2\text{(g)}$ , calculate $\Delta_r H^\ominus$ given $\Delta_c H^\ominus(\text{glucose}) = -2803$ and $\Delta_c H^\ominus(\text{ethanol}) = -1367$ kJ mol $^{-1}$ . (CO $_2$ (g) is already a combustion product, so it does not contribute to the cycle.)

$\Delta_r H^\ominus = [\Delta_c H^\ominus(\text{glucose})] - [2 \times \Delta_c H^\ominus(\text{ethanol}) + 2 \times 0]$ $= [-2803] - [2 \times (-1367)] = -2803 - (-2734) = -2803 + 2734 = \mathbf{-69 \text{ kJ mol}^{-1}}$

Modestly exothermic — which is why fermentation tanks in industrial bioethanol plants must be continuously cooled. If left uncontrolled, the temperature rise inhibits the yeast above $\sim 35$ °C and the fermentation stalls.

Worked example 4 — formation of propan-1-ol

Q: Calculate $\Delta_f H^\ominus$ for propan-1-ol: $3\text{C(s)} + 4\text{H}_2\text{(g)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_3\text{H}_7\text{OH(l)}$ , given $\Delta_c H^\ominus(\text{C}) = -394$ , $\Delta_c H^\ominus(\text{H}_2) = -286$ , and $\Delta_c H^\ominus(\text{C}_3\text{H}_7\text{OH(l)}) = -2021$ kJ mol $^{-1}$ .

$\Delta_f H^\ominus = [3 \times \Delta_c H^\ominus(\text{C}) + 4 \times \Delta_c H^\ominus(\text{H}_2) + \tfrac{1}{2} \times 0] - [\Delta_c H^\ominus(\text{C}_3\text{H}_7\text{OH})]$ $= [3 \times (-394) + 4 \times (-286) + 0] - [-2021]$ $= [-1182 + (-1144) + 0] - [-2021] = -2326 + 2021 = \mathbf{-305 \text{ kJ mol}^{-1}}$

Oxygen contributes zero because O $_2$ has no $\Delta_c H^\ominus$ — it is the combustion agent, not a fuel. This is a frequent trap: candidates sometimes look up an " $\Delta_c H^\ominus$ of O $_2$ " and find nothing tabulated, then guess at a value rather than recognising that O $_2$ does not combust.

Formation vs combustion cycle — side-by-side comparison

Feature	Formation cycle	Combustion cycle
Common intermediate	Elements in standard states	Combustion products (CO $_2$ (g) + H $_2$ O(l))
Arrow direction (re: cycle base)	UP from elements to compounds	DOWN from substances to combustion products
Formula	$\Delta_r H^\ominus = \sum \Delta_f H^\ominus(\text{products}) - \sum \Delta_f H^\ominus(\text{reactants})$	$\Delta_r H^\ominus = \sum \Delta_c H^\ominus(\text{reactants}) - \sum \Delta_c H^\ominus(\text{products})$
Elements with $\Delta H = 0$	Elements in standard state (Lesson 3)	O $_2$ (g), CO $_2$ (g), H $_2$ O(l) (already at the combustion-products level)
Best for	Inorganic reactions; well-tabulated $\Delta_f H^\ominus$	Organic reactions; bomb-calorimetric $\Delta_c H^\ominus$
Mnemonic	F-UP, products − reactants	C-DOWN, reactants − products
Mechanism of indirect route	Decompose reactants to elements; form products from elements	Combust reactants to CO $_2$ + H $_2$ O; reverse-combust products from CO $_2$ + H $_2$ O

When both data sets are available, the two cycles give identical results — they are alternative geometric realisations of the same underlying Hess principle.

Decision flowchart — choosing the cycle

flowchart TD
  A[Inspect the question's data table] --> B{What data given?}
  B -->|ΔfH° for every species| C[Use FORMATION cycle - Lesson 3]
  B -->|ΔcH° for every species| D[Use COMBUSTION cycle - this lesson]
  B -->|Mix of ΔfH° and ΔcH°| E[Use whichever can fill all gaps;<br/>sometimes need both]
  B -->|Mean bond enthalpies only| F[Use BOND ENTHALPY method - Lesson 5]
  C --> G[Σ ΔfH°(prod) − Σ ΔfH°(react)]
  D --> H[Σ ΔcH°(react) − Σ ΔcH°(prod)]
  E --> I[Construct ad-hoc cycle from given data]
  F --> J[Σ bonds broken − Σ bonds formed]

When in doubt, draw the cycle. The geometry of the cycle determines the formula, not the other way round.

Hess’s Law from Combustion Data