International A LevelChemistryOxford AQARevision NotesPhysical ChemistryEnergeticsBond Enthalpies

Bond Enthalpies (Oxford AQA International A Level Chemistry)

Revision Note

Alexandra Brennan

Written by: Alexandra Brennan

Reviewed by: Stewart Hird

Mean Bond Enthalpy Calculations

  • During a reaction, enthalpy changes take place because bonds are being broken and formed

  • Energy (in the form of heat) is needed to overcome attractive forces between atoms

    • Bond breaking is therefore endothermic

  • Energy is released from the reaction to the surroundings (in the form of heat) when new bonds are formed

    • Bond forming is therefore exothermic

Making and breaking bonds 

Chemical Energetics Bond Breaking and Forming,
Breaking bonds requires energy from the surroundings and making bonds releases energy to the surroundings

  • If more energy is required to break bonds than energy is released when new bonds are formed, the reaction is endothermic

  • If more energy is released when new bonds are formed than energy is required to break bonds, the reaction is exothermic

  • In reality, only some bonds in the reactants are broken and then new ones are formed

Bond Dissociation Enthalpy

  • The amount of energy required to break one mole of a specific covalent bond in the gas phase is called the bond dissociation enthalpy

  • Bond dissociation energy (E) is also known as exact bond energy or bond enthalpy

Mean bond enthalpy

  • Bond enthalpies are affected by other atoms in the molecule (the environment)

  • Therefore, an average of a number of the same type of bond but in different environments is calculated

  • This bond enthalpy is known as the mean bond enthalpy

  • Since bond enthalpies cannot be determined directly, enthalpy cycles are used to calculate the average bond enthalpy

How bond enthalpies are affected by other atoms in the molecule

Chemical Energetics Exact and Average Bond Energies, downloadable AS & A Level Chemistry revision notes

  • Bond enthalpies are used to find the ΔHꝋ of a reaction when this cannot be done experimentally

    • E.g. the Haber Process

  • The equation to calculate the standard enthalpy change of reaction using bond enthalpies is:

 ΔHθ = enthalpy change for bonds broken - enthalpy change for bonds formed

  • Enthalpy changes calculated using mean bond enthalpies differ from those determined using Hess's law

  • This is due to bond enthalpies being mean values, whereas a Hess's law cycle uses enthalpy data that has been taken from individual compounds where a bond has a specific value

Worked Example

Calculate the enthalpy change of reaction the Haber process reaction.

The relevant bond enthalpies are given in the table.

Bond

Mean bond enthalpy
kJ mol-1

Nidentical toN

945

H–H

436

N–H

391

Answer:

  • Step 1: The chemical equation for the Haber process is:

    • N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

  • Step 2: Set out the calculation as a balance sheet:

Bonds broken
kJ mol-1

Bonds formed
kJ mol-1

1 x Nidentical toN = 1 x 945 = 945

3 x H–H = 3 x 436 = 1308

6 x N–H = 6 x 391 = 2346

Total = 2253

Total = 2346

  • Step 3: Calculate the standard enthalpy of reaction:

    • ΔHθr = enthalpy change for bonds broken - enthalpy change for bonds formed

    • ΔHθr =2253 kJ mol-1 - 2346 kJ mol-1

    • ΔHθr = –93 kJ mol-1

  • A negative enthalpy change indicates the reaction is exothermic

Worked Example

The complete combustion of ethyne, C2H2, is shown in the equation below:

2H–Cidentical toC–H + 5O=O → 2H–O–H + 4O=C=O

Calculate, using the average bond enthalpies given in the table, the enthalpy of combustion of ethyne.

Bond

Mean bond enthalpy
kJ mol-1

C–H

414

Cidentical toC

839

O=O

498

C=O

804

O–H

463

O–C

358

Answer:

  • Step 1:

    • The enthalpy of combustion is the enthalpy change when one mole of a substance reacts in excess oxygen to produce water and carbon dioxide

    • The chemical reaction should be therefore simplified such that only one mole of ethyne reacts in excess oxygen:

H–Cidentical toC–H + 2 ½ O=O → H-O-H + 2O=C=O

  • Step 2: Set out the calculation as a balance sheet:

Bonds broken
kJ mol-1

Bonds formed
kJ mol-1

1 x Cidentical toC = 1 x 839 = 839

2 x C–H = 2 x 414 = 828

2½ x O=O = 2½ x 498 = 1245 

2 x O–H = 2 x 463 = 926

4 x C=O = 4 x 804 = 3216

Total = 2912

Total = 4142

  • Step 3: Calculate the standard enthalpy of reaction:

    • ΔHθr = enthalpy change for bonds broken - enthalpy change for bonds formed

    • ΔHθr = 2912 kJ mol-1– 4142 kJ mol-1

    • ΔHθr = –1230 kJ mol-1 

  • A negative enthalpy change indicates the reaction is exothermic

Examiner Tips and Tricks

When new bonds are formed the amount of energy released is equal to the amount of energy absorbed when the same bonds are broken

  • For example:

O2 (g) → 2O (g)   E (O=O) = +498 kJ mol-1

2O (g) → O2 (g)   E (O=O) –498 kJ mol-1

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    Alexandra Brennan

    Author: Alexandra Brennan

    Expertise: Chemistry

    Alex studied Biochemistry at Newcastle University before embarking upon a career in teaching. With nearly 10 years of teaching experience, Alex has had several roles including Chemistry/Science Teacher, Head of Science and Examiner for AQA and Edexcel. Alex’s passion for creating engaging content that enables students to succeed in exams drove her to pursue a career outside of the classroom at SME.

    Stewart Hird

    Author: Stewart Hird

    Expertise: Chemistry Lead

    Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Topic Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.