Bond Enthalpies

Bond enthalpies provide another method for calculating enthalpy changes, particularly useful for reactions involving covalent molecules in the gas phase. This approach considers the energy required to break bonds in the reactants and the energy released when new bonds form in the products. This lesson covers mean bond enthalpies, the calculation method, limitations, and comparisons with Hess's law methods.

What Is a Bond Enthalpy?

A bond enthalpy (also called bond energy or bond dissociation enthalpy) is the energy required to break one mole of a specific covalent bond in the gaseous state, with all species in the gas phase.

For example, the O–H bond enthalpy in water refers to:

H–O–H(g) → H(g) + OH(g) ΔH = +463 kJ mol⁻¹

Notice that bond breaking is always endothermic (positive ΔH) — you must put energy in to overcome the attractive forces holding atoms together. Conversely, bond making is always exothermic (negative ΔH) — energy is released when atoms come together and form bonds.

Key rule: Breaking bonds = energy IN (endothermic). Making bonds = energy OUT (exothermic). This must be second nature for your exam.

Mean Bond Enthalpies

The exact energy needed to break a particular bond depends on the molecular environment. For example, the energy to break the first O–H bond in water is different from the energy to break the second O–H bond:

H₂O(g) → H(g) + OH(g) ΔH = +498 kJ mol⁻¹

OH(g) → O(g) + H(g) ΔH = +428 kJ mol⁻¹

These values are different because after the first hydrogen is removed, the electronic environment of the remaining O–H bond changes.

The mean bond enthalpy is the average value obtained from many different compounds containing that bond. For the O–H bond, the mean bond enthalpy is +463 kJ mol⁻¹, which is the average across many molecules containing O–H bonds (water, alcohols, carboxylic acids, etc.).

Data tables give mean bond enthalpies, which is why calculations using bond enthalpies give approximate rather than exact results.

Reference Table: Common Mean Bond Enthalpies

Bond	Mean bond enthalpy / kJ mol⁻¹	Bond	Mean bond enthalpy / kJ mol⁻¹
C–H	+413	O–H	+463
C–C	+347	O=O	+498
C=C	+614	N–H	+391
C≡C	+839	N≡N	+945
C–O	+358	H–H	+436
C=O	+805	Cl–Cl	+242
C–Cl	+346	H–Cl	+432
C–N	+286	H–Br	+366
C=N	+615	H–F	+568
C≡N	+887	Br–Br	+193

Calculating Enthalpy Changes Using Bond Enthalpies

The method follows this principle:

ΔH = Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed)

Or more concisely: Energy in (breaking) − Energy out (making)

Since breaking bonds requires energy (positive) and making bonds releases energy (negative), the calculation becomes:

If more energy is released in making bonds than is needed to break bonds → ΔH is negative → exothermic
If more energy is needed to break bonds than is released in making bonds → ΔH is positive → endothermic

Worked Example 1: Combustion of Methane

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Bonds broken (reactants):

4 × C–H = 4 × 413 = 1652 kJ
2 × O=O = 2 × 498 = 996 kJ
Total energy in = 2648 kJ

Bonds formed (products):

2 × C=O (in CO₂) = 2 × 805 = 1610 kJ
4 × O–H (in 2H₂O) = 4 × 463 = 1852 kJ
Total energy out = 3462 kJ

ΔH = 2648 − 3462 = −814 kJ mol⁻¹

The data book value (−890 kJ mol⁻¹) differs because we used mean bond enthalpies, not exact values for the specific bonds in these molecules. Also note the product is H₂O(g) in this calculation, not H₂O(l).

Worked Example 2: Hydrogenation of Ethene

C₂H₄(g) + H₂(g) → C₂H₆(g)

Bonds broken:

1 × C=C = 614 kJ
4 × C–H = 4 × 413 = 1652 kJ
1 × H–H = 436 kJ
Total = 2702 kJ

Bonds formed:

1 × C–C = 347 kJ
6 × C–H = 6 × 413 = 2478 kJ
Total = 2825 kJ

ΔH = 2702 − 2825 = −123 kJ mol⁻¹

The Hess's law value using combustion data was −137 kJ mol⁻¹. The difference is expected when using mean bond enthalpies.

Worked Example 3: Chlorination of Methane

CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)

Bonds broken:

1 × C–H = 413 kJ (only one C–H bond breaks, not all four)
1 × Cl–Cl = 242 kJ
Total = 655 kJ

Bonds formed:

1 × C–Cl = 346 kJ
1 × H–Cl = 432 kJ
Total = 778 kJ

ΔH = 655 − 778 = −123 kJ mol⁻¹

Common error: When a molecule like CH₄ reacts with Cl₂, only one C–H bond breaks in the first substitution step. Do NOT break all four C–H bonds. Read the equation carefully to identify which bonds break and form.

Why Bond Enthalpy Calculations Give Approximate Results

Several factors explain why bond enthalpy calculations differ from experimental values:

Mean values, not exact values. Mean bond enthalpies are averages across many compounds. The actual bond strength varies depending on the molecular environment.
Gas-phase only. Bond enthalpy calculations assume all species are gases. If the actual reaction involves liquids or solids, extra energy for phase changes is not accounted for. For example, combustion of methane produces H₂O(l) in reality, but bond enthalpy calculations assume H₂O(g).
Resonance and delocalisation. In molecules with delocalised electrons (e.g. benzene), the actual bond strengths differ from the tabulated values for simple single or double bonds.

When to Use Bond Enthalpies vs. Hess's Law

Method	Best used when	Precision
Hess's law with ΔH°f	Formation data is available for all species	Exact (within data precision)
Hess's law with ΔH°c	Combustion data is available for all species	Exact (within data precision)
Bond enthalpies	Only bond enthalpy data is available, or the question specifically asks for it	Approximate

If both methods are possible, Hess's law with formation or combustion data gives more accurate results.

Bond Enthalpy and Bond Strength

Bond enthalpy is a direct measure of bond strength. Trends in bond enthalpy reveal important chemistry:

Triple bonds > Double bonds > Single bonds: C≡C (839) > C=C (614) > C–C (347). More shared electrons means stronger bonds.
Bond length decreases as bond strength increases: Shorter bonds are stronger.
Electronegativity differences affect bond strength: H–F (568 kJ mol⁻¹) is stronger than H–I (298 kJ mol⁻¹) partly because of the greater polarity of the H–F bond.

Real-World Application: Fuel Efficiency

The energy released by burning a fuel depends on the bonds broken and formed. Fuels with many C–H bonds per molecule (like octane, C₈H₁₈, with 18 C–H bonds) release large amounts of energy. Hydrogen fuel (H₂) has a very high energy per gram because H–H bonds are light and the product (H₂O) involves very strong O–H bonds. This is why hydrogen is being explored as a clean fuel alternative.

Summary

Bond enthalpy = energy to break one mole of bonds in the gas phase.
ΔH = Σ(bonds broken) − Σ(bonds formed).
Mean bond enthalpies give approximate answers because they are averages.
Always identify which bonds break and form from the balanced equation — do not break all bonds in a molecule if only some react.
Hess's law methods are more precise when formation or combustion data is available.

A-Level Deep Dive: Bond Enthalpies

Spec mapping

Edexcel 9CH0 specification Topic 8.3 requires the use of mean (average) bond enthalpies to estimate the enthalpy change of a gaseous reaction via the relationship ΔH°_r ≈ Σ(bonds broken) − Σ(bonds formed) (refer to the official specification document for exact wording). Candidates must explain why bond enthalpy estimates differ from values derived from ΔH°_f data: mean bond enthalpies are averages across many molecular environments, while ΔH°_f values are specific to one compound. Bond-enthalpy questions appear on Paper 1 (estimation problems) and Paper 3 (synoptic comparison with experimental values). The Edexcel data booklet provides a table of mean bond enthalpies; questions may also supply specific values within the question stem. The treatment links forward to lattice enthalpy in Topic 13 and back to bonding theory in Topic 2.

Worked example with full mark scheme

Question (8 marks):

Use mean bond enthalpies to estimate the standard enthalpy of combustion of methane:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

Mean bond enthalpies (kJ mol⁻¹): C–H = +413; O=O = +498; C=O = +805; O–H = +464.

(a) Calculate the energy required to break all bonds in the reactants. (2) (b) Calculate the energy released forming all bonds in the products. (2) (c) Calculate the estimated ΔH°_c. (2) (d) Explain why this estimate differs from the data-book ΔH°_c[CH₄(g)] = −890 kJ mol⁻¹. (2)

Solution with mark scheme:

(a) Bonds broken: 4 × C–H + 2 × O=O = 4(413) + 2(498) = 1652 + 996 = +2648 kJ.

M1 — correct identification of bond types and counts. A1 — correct sum.

(b) Bonds formed: 2 × C=O + 4 × O–H = 2(805) + 4(464) = 1610 + 1856 = +3466 kJ (released).

M1 — correct identification (4 O–H, not 2 — water has two O–H bonds per molecule, and there are two water molecules). A1 — correct sum.

M1 — bonds broken minus bonds formed (correct sign convention). A1 — final answer with sign.

(d) The estimate (−818 kJ mol⁻¹) is less exothermic than the data-book value (−890 kJ mol⁻¹) by 72 kJ mol⁻¹. B1 — bond enthalpies in the table are mean values averaged over many molecular environments, not specific to CH₄ or H₂O. B1 — additionally, the equation as written produces H₂O(g), but standard ΔH°_c is defined for H₂O(l); condensing 2 mol of water releases ~2 × 41 = 82 kJ, accounting for almost all of the discrepancy.

Total: 8 marks (M3 A3 B2).

Specimen question modelled on the Edexcel 9CH0 Paper 1 format

Question (6 marks):

Hydrogenation of ethene: CH₂=CH₂(g) + H₂(g) → CH₃CH₃(g).

Mean bond enthalpies (kJ mol⁻¹): C=C = +612; C–C = +347; C–H = +413; H–H = +436.

(a) Calculate ΔH°_r for the hydrogenation. (4) (b) Comment on the sign of ΔH and the industrial implication. (2)

Mark scheme decomposition by AO: