Mean Bond Enthalpies

SPEC 3.1.4.4 - ENERGETICS

Bond Enthalpies

Using mean bond enthalpies to calculate enthalpy changes for reactions in the gaseous phase

Reactions Involve Breaking and Making Bonds

Chemical reactions involve the breaking and forming of bonds, which result in enthalpy changes.

Key Principle

The overall enthalpy change (ΔH) of a reaction is the combined effect of these two processes:

1. Breaking bonds in the reactant molecules - An endothermic process (positive ΔH) that requires energy. The breaking of stronger bonds requires more energy.

2. Forming bonds to create the product molecules - An exothermic process (negative ΔH) that releases energy. The formation of stronger bonds releases more energy.

Overall ΔH Depends on Relative Amounts

If bond breaking requires more energy than is released during bond formation, the reaction has a positive ΔH (endothermic)

If bond breaking requires less energy than is released during bond formation, the reaction has a negative ΔH (exothermic)

Bond Breaking vs Bond Making

When a bond breaks, energy must be supplied. When a bond forms, energy is released.

ProcessEnergy ChangeSign of ΔH

Breaking bondsEnergy is absorbed (endothermic)Positive (+)

Making bondsEnergy is released (exothermic)Negative (−)

Definition

The amount of energy required to break one mole of a specific covalent bond in the gas phase is called the bond dissociation energy. This is usually just simplified to bond energy or bond enthalpy.

Mean bond enthalpy is an averaged value representing: the energy required to break one mole of the stated bond in a gaseous state, under standard conditions.

In symbols, the type of bond broken is written in brackets after E. For example, E(H−H) is the bond energy of a mole of single bonds between two hydrogen atoms.

Bond energies are affected by other atoms in the molecule (the environment). For example, it takes more energy to break an O−H bond in water than in methanol, because in methanol the oxygen is connected to a C-atom rather than another H-atom. Therefore, an average of a number of the same type of bond but in different environments is calculated. This is known as the average bond energy.

Important Points

They are always positive (endothermic) because breaking bonds requires energy.

They represent an average value across different molecular environments.

Since bond energies cannot be determined directly, enthalpy cycles are used to calculate the average bond energy.

Important

Because mean bond enthalpies are averages, calculations using them give approximate values of ΔH, not exact values. Average bond enthalpies are listed in data tables.

Calculating Reaction Enthalpy Changes

Bond energies are used to find the ΔH_r^⦵ of a reaction when this cannot be done experimentally. The enthalpy change is determined using average bond enthalpies for the bonds that are broken and those that are formed.

Formula

ΔH_r^⦵ = enthalpy change for bonds broken + enthalpy change for bonds formedNote: Values for bonds broken are positive (endothermic) and values for bonds formed are negative (exothermic) This can also be written as:ΔH_r = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed)where Σ = sum of

Step-by-Step Method

Identify and count the bonds broken and formed - set out the calculation as a balance sheet

Calculate total energy absorbed in breaking bonds (positive values)

Calculate total energy released in forming bonds (negative values)

Calculate ΔH by adding them together (or: bonds broken − bonds formed)

Critical: Gaseous State Only

Bond enthalpy calculations only apply to substances in the gaseous state. If a reactant or product is a liquid or solid, the calculation will not give an accurate answer because energy is also involved in overcoming intermolecular forces. You may need to include enthalpy of vaporisation or fusion to account for phase changes.

Energy Level Diagram

The reaction pathway goes from reactants up to gaseous atoms (breaking all bonds), then down to products (making all bonds).

Interpreting the Diagram

If the products are lower than reactants, ΔH is negative (exothermic). If products are higher, ΔH is positive (endothermic).

01Worked Example - Combustion of Methane

Calculate the enthalpy change (ΔH) for the combustion of methane:

CH₄ + 2O₂ → CO₂ + 2H₂O

Bond typeMean bond enthalpy (kJ mol⁻¹)

C−H+414

O=O+498

C=O+799

O−H+464

1

Set out the calculation as a balance sheet:

2

Calculate ΔH_r^⦵:

ΔH = enthalpy change for bonds broken + enthalpy change for bonds formed

= (+2652 kJ mol⁻¹) + (−3454 kJ mol⁻¹)

= −802 kJ mol⁻¹

The reaction is exothermic, releasing 802 kJ mol⁻¹.

02Worked Example - Synthesis of Ammonia

Calculate the enthalpy change (ΔH) for the synthesis of ammonia:

N₂ + 3H₂ → 2NH₃

Bond typeMean bond enthalpy (kJ mol⁻¹)

N≡N+945

H−H+436

N−H+391

1

Set out the calculation as a balance sheet:

2

Calculate ΔH_r^⦵:

ΔH = enthalpy change for bonds broken + enthalpy change for bonds formed

= (+2253 kJ mol⁻¹) + (−2346 kJ mol⁻¹)

= −93 kJ mol⁻¹

The reaction is exothermic, releasing 93 kJ mol⁻¹.

Hess Cycles Provide More Accurate Values

Calculating ΔH_r using mean bond enthalpies is less accurate than using Hess cycles because average bond enthalpies do not account for the unique molecular environment of each bond in a specific molecule.

Key Comparison

Hess cycles, in contrast, consider the actual enthalpy changes associated with the specific bonds in the reactants and products, providing a more precise estimation of ΔH_r.

MethodAccuracyWhy?

Mean bond enthalpiesApproximateUses average values that may not match actual bond strengths

Hess's lawMore accurateUses experimentally measured values for specific compounds

Example

For methane combustion, mean bond enthalpies give ΔH ≈ −802 kJ mol⁻¹, but the experimentally determined value is −890 kJ mol⁻¹.

Common Mistakes to Avoid

Mistake 1: Wrong Formula

Wrong: ΔH = bonds formed − bonds broken

Correct: ΔH = bonds broken − bonds formed

Mistake 2: Forgetting Stoichiometry

If the equation has coefficients (like 2H₂O), count ALL bonds. Two water molecules = 4 × O−H bonds.

Mistake 3: Missing Bonds

Draw structural formulas! CO₂ has TWO C=O bonds. O₂ has a double bond (O=O).

Mistake 4: Sign Errors

Bond enthalpies in data tables are positive. The minus sign only appears in your final ΔH if exothermic.

Practice

Question 1: Calculate the enthalpy change for:

H₂(g) + Cl₂(g) → 2HCl(g)

Bond enthalpies (kJ mol⁻¹): H−H = +436, Cl−Cl = +242, H−Cl = +431

1

Bonds broken: 1 × H−H + 1 × Cl−Cl = 436 + 242 = +678 kJ mol⁻¹

2

Bonds formed: 2 × H−Cl = 2 × 431 = +862 kJ mol⁻¹

3

ΔH = 678 − 862 = −184 kJ mol⁻¹

Practice

Question 2: The enthalpy change calculated using bond enthalpies was −285 kJ mol⁻¹. The experimental value is −296 kJ mol⁻¹.

(a) Explain why these values are different.

(b) Which value is more accurate?

(a)

The bond enthalpy calculation uses mean bond enthalpies, which are average values. The actual bond enthalpies in specific molecules may differ because they do not account for the unique molecular environment.

(b)

The experimentally determined value (−296 kJ mol⁻¹) is more accurate because it directly measures the enthalpy change for this specific reaction.

Practice

Question 3: For the decomposition of hydrogen peroxide:

2H₂O₂(g) → 2H₂O(g) + O₂(g)

(a) Calculate the enthalpy change for this reaction.

(b) Is this reaction endothermic or exothermic? Explain your answer.

Bond enthalpies (kJ mol⁻¹): O−H = +464, O−O = +146, O=O = +498

1

Bonds broken (in 2H₂O₂): 2 × O−O + 4 × O−H = (2 × 146) + (4 × 464) = 292 + 1856 = +2148 kJ

2

Bonds formed (in 2H₂O + O₂): 4 × O−H + 1 × O=O = (4 × 464) + (1 × 498) = 1856 + 498 = +2354 kJ

3

ΔH = 2148 − 2354 = −206 kJ mol⁻¹

(b)

The reaction is exothermic because ΔH is negative. More energy is released when forming the stronger O=O double bond and O−H bonds in products than is required to break the weaker O−O single bond in hydrogen peroxide.

Summary

Key Formulas and Concepts

ConceptDetails

Mean bond enthalpyEnergy to break 1 mol of a bond in gaseous state (average)

ΔH calculationΔH = Σ(bonds broken) − Σ(bonds formed)

Breaking bondsEndothermic (+ve) - requires energy

Making bondsExothermic (−ve) - releases energy

LimitationsMean values are approximate; Hess's law more accurate

Common Bond Enthalpies (AQA Data)

C−H = +414 | C−C = +347 | C=C = +612 | O=O = +498 | O−H = +464 | C=O = +799 | N≡N = +945 | N−H = +391 | H−H = +436 | Cl−Cl = +242 | H−Cl = +431