Chemical Thermodynamics · Lesson 6 of 10

Combustion, Atomisation, and Bond Enthalpy

“Connect fuel energy with the energy required to separate atoms and break chemical bonds.”

Learning Objectives

• Interpret a standard molar enthalpy of combustion with specified product states. • Use combustion data to infer a formation enthalpy. • Distinguish atomisation, bond dissociation, and mean bond enthalpy. • Explain why successive bond dissociations can have different energies. • Estimate gas-phase reaction enthalpy from bonds broken and formed.

Energy released by combustion

Burning a fuel releases energy that can heat a room, power an engine, or drive a rocket. The standard molar enthalpy of combustion is the enthalpy change for complete combustion of one mole of substance, with reactants and products in their stated standard states. Combustion reactions considered here are exothermic, so their enthalpy changes are negative.

Product states are essential. Burning a fuel to liquid water releases more heat than making water vapour, because condensation releases additional energy. For butane the balanced equation is $C_{4} H_{10} (g) + \frac{13}{2} O_{2} (g) \to 4C O_{2} (g) + 5 H_{2} O (l)$ . Using the formation table gives about $- 2877 kJ mo l^{- 1}$ . A value near −2658 kJ/mol instead corresponds approximately to vapour as the water product; attaching it to liquid-water products would mix endpoints.

Glucose oxidation illustrates the link to food energy: $C_{6} H_{12} O_{6} (s) + 6 O_{2} (g) \to 6C O_{2} (g) + 6 H_{2} O (l)$ releases roughly 2802 kJ per mole using the chapter’s example. In the body this overall conversion proceeds through many enzyme-controlled steps. Hess’s law explains why splitting an overall conversion into steps does not create a new total enthalpy between the same endpoints; physiology also controls how energy is transferred and used.

Example — Formation of benzene from combustion data

Problem
One mole of liquid benzene releases 3267.0 kJ when it forms carbon dioxide gas and liquid water. Find its formation enthalpy using −393.5 and −285.83 kJ/mol for those products.

1.Balance combustion: $C_{6} H_{6} (l) + \frac{15}{2} O_{2} (g) \to 6C O_{2} (g) + 3 H_{2} O (l)$ . The reaction enthalpy is $- 3267.0 kJ mo l^{- 1}$ .
2.The product formation sum is $6 (- 393.5) + 3 (- 285.83) = - 3218.49 kJ mo l^{- 1}$ . Oxygen has zero formation enthalpy in its reference state.
3.Write $- 3267.0 = - 3218.49 - Δ_{f} H^{\circ} (C_{6} H_{6}, l)$ .
4.Rearrange to get $Δ_{f} H^{\circ} = + 48.51 kJ mo l^{- 1}$ . The positive sign follows from reversing the combustion reaction in the Hess route.
5.Small differences from a rounded table value such as +49.0 arise from the data used; the sign must remain positive for this dataset.

Making separated gaseous atoms

Before analysing bonds, specify a common separated-particle endpoint: isolated atoms in the gas phase. Atomisation is conversion of a substance to those gaseous atoms. Its quoted enthalpy must be tied to the written amount of substance and the number of atoms produced. For a metal, converting solid metal into gaseous atoms is both atomisation and sublimation.

Definition

Atomisation enthalpy

The enthalpy change for converting the stated amount of a substance completely into gaseous atoms. For example, $C H_{4} (g) \to C (g) + 4H (g)$ atomises one mole of methane; $Na (s) \to Na (g)$ produces one mole of sodium atoms.

For $H_{2} (g) \to 2H (g)$ , breaking one mole of H–H bonds gives two moles of H atoms and needs about 435 kJ using the rounded example. If a datum instead refers to $\frac{1}{2} H_{2} (g) \to H (g)$ , its value is half as large. This distinction prevents a factor-of-two error when constructing cycles.

Example — An atomisation amount

Problem
The enthalpy for $H_{2} (g) \to 2H (g)$ is 435.0 kJ per mole of hydrogen molecules. How much is needed to produce 0.50 mol of H atoms?

1.Each mole of hydrogen molecules produces two moles of atoms.
2.To obtain 0.50 mol of atoms, dissociate $0.50/2 = 0.25 mol$ of molecules.
3.The energy is $(0.25) (435.0) = 108.75 kJ$ .
4.State the chemical equation associated with an atomisation value; the word alone does not specify the amount basis.

Individual bonds and mean values

Breaking a covalent bond requires energy, whereas making the corresponding bond from separated fragments releases energy. Bond dissociation enthalpy refers to a specified gaseous molecule and gaseous products. A polyatomic molecule can be dissociated in stages, and each stage starts with a different species.

Definition

Bond dissociation enthalpy

The enthalpy change for breaking one mole of a specified covalent bond in gaseous species to give specified gaseous fragments. For a diatomic molecule dissociated into atoms, it equals the atomisation change for that written dissociation.

Methane has four equivalent C–H bonds in the original molecule, but after one H atom is removed the remaining species is methyl, not methane. Further removals change the electronic structures again. The sequence therefore has different dissociation enthalpies even though the original bonds were equivalent. A mean bond enthalpy averages the energies needed for complete atomisation over the number of bonds.

Successive methane step	Enthalpy / kJ mol−1
$C H_{4} (g) \to C H_{3} (g) + H (g)$	427
$C H_{3} (g) \to C H_{2} (g) + H (g)$	439
$C H_{2} (g) \to CH (g) + H (g)$	452
$CH (g) \to C (g) + H (g)$	347

Example — Averaging methane’s dissociations

Problem
Find the mean C–H bond enthalpy from the four steps in the table.

1.Add all stages, cancelling intermediate fragments: the net process is $C H_{4} (g) \to C (g) + 4H (g)$ .
2.Its enthalpy is $427 + 439 + 452 + 347 = 1665 kJ mo l^{- 1}$ .
3.Divide by the four moles of C–H bonds broken per mole of methane: $1665/4 = 416.25 kJ mo l^{- 1}$ , or about 416 kJ/mol.
4.This mean is not the energy of every successive step. C–H mean values also vary somewhat between different compounds, which limits their precision in reaction estimates.

Selected single bond	Mean enthalpy / kJ mol−1	Selected multiple bond	Mean enthalpy / kJ mol−1
H–H	435.8	C=C	611
C–H	414	C≡C	837
C–C	347	N=N	418
N–H	389	N≡N	946
O–H	464	C=N	615
H–Cl	431	C≡N	891
Cl–Cl	243	C=O	741
O–O	138	C≡O	1070
F–F	155	O=O	498

The table illustrates how bond type and bond order affect the energy needed. A triple bond is generally stronger than a corresponding double bond, which is generally stronger than a single bond, though the numerical values depend on the chemical environment. For example, the O=O dissociation value used here is about 498 kJ/mol; dissociation values belong to specified bonds, not merely to an element’s name.

Breaking first, forming second

A Hess route can take gaseous reactants apart into atoms and then build gaseous products. Atomising the reactants costs energy. Building product bonds releases the negative of their dissociation sum. Subtracting these contributions gives an approximate reaction enthalpy when average bond values are used.

A bond-energy route through separated atoms— The uphill atomisation contribution is positive; making the product bonds contributes a negative enthalpy change.

Bond-enthalpy estimateLaTeX

Δ_{r} H^{\circ} \approx \sum D (bonds broken) - \sum D (bonds formed)

D values are positive dissociation enthalpies. Count each bond with its multiplicity. The direct formula assumes gaseous reactants and products; phase changes need additional corrections.

Example — Hydrogen chloride from bond data

Problem
Estimate $Δ H$ for $H_{2} (g) + C l_{2} (g) \to 2HCl (g)$ using H–H = 435.8, Cl–Cl = 243, and H–Cl = 431 kJ/mol.

1.Break one mole of H–H and one mole of Cl–Cl bonds: required energy is $435.8 + 243 = 678.8 kJ$ .
2.Form two moles of H–Cl bonds: the released energy has magnitude $2 (431) = 862 kJ$ .
3.Subtract formed-bond dissociation sums: $Δ H \approx 678.8 - 862 = - 183.2 kJ mo l^{- 1}$ per written reaction.
4.Bond making releases more energy than bond breaking requires, so the overall reaction is exothermic.

Common mistake

Breaking bonds absorbs energy; it does not release the energy of combustion by itself. The overall release occurs because forming product bonds gives out more energy than reactant bond breaking requires. Also, mean bond values provide estimates and cannot be mixed directly with liquid or solid endpoints without corrections.

Quick check

Why must water’s product state be specified in a combustion equation?

Quick check

What is the sign of the enthalpy change when a covalent bond is broken into the specified gaseous fragments?

Quick check

Why do the successive C–H dissociations of methane have different values?

Quick check

If complete atomisation of one mole of a molecule breaks four equivalent-type bonds and needs 1200 kJ, what is the mean bond enthalpy?

Quick check

Which expression estimates a gas-phase reaction enthalpy from positive dissociation enthalpies?

Quick check

Using the benzene combustion dataset in this lesson, what is its standard formation-enthalpy sign?

Practice Problems

Write the balanced combustion equation for liquid ethanol producing carbon dioxide gas and liquid water. Explain why its enthalpy value must be tied to those states.
Use methane, graphite, and hydrogen combustion enthalpies −890.3, −393.5, and −285.8 kJ/mol to infer methane’s formation enthalpy.
Calculate the heat released by forming 35.2 g of carbon dioxide from graphite and oxygen, using carbon combustion enthalpy −393.5 kJ/mol and carbon dioxide molar mass 44.0 g/mol.
Explain why atomisation of sodium metal and sublimation of sodium have the same written endpoint, while vaporising water is not atomisation.
Estimate the enthalpy for $N_{2} (g) + 3 H_{2} (g) \to 2N H_{3} (g)$ using N≡N = 946, H–H = 435.8, and N–H = 389 kJ/mol.
Methanol combustion to carbon dioxide gas and liquid water has enthalpy −726 kJ/mol. Using −393 and −286 kJ/mol for the product formation values, find liquid methanol’s formation enthalpy.

Key Takeaways

Combustion enthalpy depends on the written amount and the phases of the products. Atomisation ends with gaseous atoms; its amount basis must be explicit. Successive bond dissociations differ because the fragments change; mean values are averages. A gas-phase estimate subtracts energy released by forming bonds from energy required to break bonds.

Previous · Lesson 5

Formation Enthalpies and Hess’s Law

Next · Lesson 7

Lattice, Solution, and Dilution Enthalpies

Energy released by combustion

Making separated gaseous atoms

Individual bonds and mean values

Breaking first, forming second

Quiz

Practice Problems

Key Takeaways