1.2b(iv) Bond Enthalpy (bond dissociation energy) calculations for Enthalpy of Reaction

Using ΔH^θ_reaction = ∑ΔH^θ_{(product bonds formed)} – ∑ΔH^θ_{(reactant bonds broken)}

_{Watch the enthalpy sign (+ve or -ve) related to the direction of change!}

+498 is the bond enthalpy of the oxygen molecule (O=O) or twice the enthalpy of atomisation of oxygen.

+715 is the enthalpy of atomisation of carbon.

–393 is the enthalpy of combustion of carbon or the enthalpy of formation of carbon dioxide.

This becomes +393 in the Hess's Law cycle, arrow and sign reversed.

x is equal to twice the bond enthalpy of a C=O bond in a carbon dioxide molecule.

x is the only one of the four delta H values that cannot be determined by experiment.

However using Hess's Law

x = +393 +715 +498 = +1606 kJ mol^–1

therefore the C=O bond energy in CO₂ = 1606/2 = 803 kJ mol^–1

Using a Hess's Law thermochemical cycle to perform bond enthalpy calculations to determine the theoretical enthalpy of reaction - in this case to determine the C-H bond enthalpy in methane

ΔHØf (methane) = –75 kJmol–1
C(s) + 2H2(g) CH4(g)
ΔHθsub(C(s)) + 2 x ΔHθBE(H–H(g)) = (+715) + (2 x +436) both endothermic	C(g) +	4H(g)	–4 x avΔHθBE(C–H(g)) = –X exothermic formation of 4 C–H bonds
The cycle for the standard enthalpy of formation of methane, (ΔHθf), based on the enthalpy of sublimation of carbon (graphite) and the H–H and C–H bond enthalpies. From Hess's Law, add up the sequence of enthalpy changes via the lower 'staged route' to get the overall enthalpy change for the enthalpy of formation of methane from its elements in their normal stable states. Using ΔHθreaction = ∑ΔHθ(product bonds formed) – ∑ΔHθ(reactant bonds broken) ΔHθf (methane) = {ΔHθsub(C(s)) + 2 x ΔHθBE(H–H(g))} + {–4 x avΔHθBE(C–H(g))}  = –75 kJmol–1 –75 = +715 +872 –X X = (715 + 872 +75) = 1662 avΔHθBE(C–H(g)) = 1662/4 = +415.5 kJmol–1

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Bond enthalpy calculations using average bond enthalpies

• Bond enthalpy calculations using a Hess's Law cycle can be used to calculate unknown enthalpy changes for a reaction.

• BUT there are limitations to the results of these calculations since they are based on ..

  • (i) average bond enthalpies (i.e. typical vales for a particular bond) and

  • (ii) only valid gaseous reactants AND products (quite restrictive, this because bond enthalpies are defined, measured and based on gaseous species only).

Introductory example to illustrate the method of calculating the enthalpy of a reaction using bond enthalpies

The diagram illustrates

Using ΔH^θ_reaction = ∑ΔH^θ_{(product bonds formed)} – ∑ΔH^θ_{(reactant bonds broken)}

First, imagine which bonds must be broken to enable the reaction to proceed.

The energy absorbed equals that to break one C–H bond (in methane molecule) plus energy to break one Cl–Cl bond (in chlorine molecule), both endothermic changes – 'bond breaking'.

Theoretically imagine you've got these atomic or molecular fragments, put them together to form the products, in doing so, work out which bonds must be formed to give the products.

The energy released is that given out when C–Cl bond (in chloromethane molecule) is formed plus the energy released when one H–Cl bond (in hydrogen chloride molecule) is formed, both exothermic changes – 'bond making'.

Calculating the difference in the two sums gives the numerical energy change and since more heat energy is given out to the surroundings in forming the bonds than that absorbed in breaking bonds, the reaction must be exothermic.

Energy absorbed in bond breaking = 412 + 242 = +654 kJ/mol

Energy released in bond formation = -331 + -432 = -763 kJ/mol

Enthalpy of the chlorination reaction = (energy released) - (energy absorbed)

= -763 - (+654), watching the signs!

Therefore ΔH_reaction = –109 kJmol^–1

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Further examples – how you might solve them in exams!

Ex 1. Calculating the enthalpy of a reaction

Given the following bond enthalpies in kJ mol^–1: Cl–Cl = 242, H–H = 436, H–Cl = 431

Calculate the enthalpy of the reaction of forming 2 moles of hydrogen chloride from its elements in their standard states ... the Hess's Law cycle for this is ...

H_2(g) + Cl_2(g) 2HCl_(g)

2H_(g) + 2Cl_(g)

For simple Q's like this, unless asked for, you can solve easily without drawing a cycle, either way ...

ΔH^θ_reaction = {endothermic ΔH for H₂ and Cl₂ bonds broken} + {exothermic ΔH for HCl bonds formed}

ΔH^θ_reaction = {+436 +242} + {2 x –431}

ΔH^θ_reaction = +(436 + 242 –862) = –184 kJmol^–1

Note that ΔH^θ_reaction /2 = –92 kJmol^–1 = ΔH^θ_f(HCl_(g))

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Ex 2. Given the following bond dissociation enthalpies at 298K in kJmol^–1 ...

Bond	C–C (single)	C–H	C=O in CO2	O–H	O=O
Bond enthalpy	+348	+412	+805	+463	+496

... calculate the enthalpy of combustion of butane.

Using a Hess's Law thermochemical cycle to perform bond enthalpy calculations to determine the theoretical enthalpy of reaction - in this case to determine the enthalpy of complete combustion of butane with oxygen.

You construct a Hess's Law Cycle from the 'normal' combustion equation (but of gaseous species only) for butane gas and 'atomise' the reactant molecule to give ALL of the 'theoretical' intermediate atoms. From this you can  theoretically calculate the enthalpy of the reaction:

Ex 3. Another example of a bond enthalpy calculation method

Using a Hess's Law thermochemical cycle to perform bond enthalpy calculations to determine the theoretical enthalpy of reaction - in this case to determine the enthalpy of complete combustion of hydrazine with oxygen or hydrogen peroxide.

Liquid Hydrazine N₂H₄ and hydrogen peroxide H₂O₂ have both been used in rocket fuels.

Hydrazine has been used as monopropellant in rocket engines because it can be catalysed to decompose into nitrogen and hydrogen gas very rapidly and exothermically.

(1)  N₂H₄ ==> N₂ + 2H₂

Hydrazine can also be used to power rockets in combination with hydrogen peroxide (a bipropellant fuel).

(2)  N₂H₄ + 2H₂O₂ ==> N₂ + 4H₂O

From the list of bond enthalpies in kJmol^–1, given below, calculate the enthalpy changes for reactions (1) and (2)

Assume all substances are in the gaseous state i.e. (g).

Bond	N–H	NN	N–N	H–H	O–H	O–O
Bond enthalpy	+388	+944	+163	+436	463	+146

For (1)  N₂H_4(g) ==> N_2(g) + 2H_2(g)

ΔH_reaction(decomp N₂H₄) = +163 + (4 x 388) –944 – (2 x 436)

= +163 +1552 –944 –872 = –101 kJmol^–1

 

For (2)  N₂H_4(g)  +  2H₂O_2(g)  ===>  N_2(g)  +  4H₂O(g)

ΔH_reaction(combustion N₂H₄) = +163 +1552 +292 +1852 –944 –3704

ΔH_reaction(combustion N₂H₄) = –789 kJmol^–1

which is considerably more exothermic than the catalysed thermal decomposition of hydrazine into nitrogen and hydrogen,

but the mixture is more costly and more dangerous to handle!

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Ex 4. Given the following bond dissociation enthalpies at 298K in kJmol^–1 ...

Bond	C–C (single)	C–H	C-O	C=O in CO2	O–H	O=O
Bond enthalpy	+348	+412	+358	+805	+463	+496

... calculate the enthalpy of combustion of butan-1-ol

Using a Hess's Law thermochemical cycle to perform bond enthalpy calculations to determine the theoretical enthalpy of reaction - in this case to determine the enthalpy of complete combustion of butan-1-ol (1-butanol) with oxygen.

You construct a Hess's Law Cycle from the 'normal' combustion equation (but of gaseous species only) for butan-1-ol,  and 'atomise' the reactant molecule to give ALL of the 'theoretical' intermediate atoms. From this you can  theoretically calculate the enthalpy of the reaction:

Enthalpy calculation problems with worked out answers – based on enthalpies of reaction, formation, combustion

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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Key revision points about bond enthalpy calculations

Bond enthalpy is the energy required to break one mole of a specific covalent bond in the gaseous state.

ΔH^θ_reaction,298 = ∑ΔH^θ₂₉₈(bonds formed) – ∑ΔH^θ₂₉₈(bonds broken)

They are important for estimating reaction enthalpies but limited because they rely on average values and apply only to gaseous molecules, making them less accurate than Hess’s Law cycles.

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• Definition:

  • Bond enthalpy (bond dissociation energy) = energy required to break one mole of a covalent bond in the gas phase.

  • Units: kJ mol⁻¹.

  • Mean bond enthalpy = average value across different molecules (since bond strength varies with environment).

• Breaking versus Making Bonds:

  • Breaking bonds = endothermic (+ΔH).

  • Making bonds = exothermic (–ΔH).

  • Reaction enthalpy:
     

  • ΔH^θ_reaction = ∑ΔH^θ_{(product bonds formed)} – ∑ΔH^θ_{(reactant bonds broken)}

• Importance:

  • Provides a way to estimate enthalpy changes when experimental data (formation/combustion enthalpies) are unavailable.

  • Useful for comparing bond strengths and predicting stability/reactivity.

  • Helps explain trends in bond strength (e.g., C–C versus C=C versus C≡C).

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Common Misconceptions about bond enthalpy calculations

• Sign Confusion:

  • Students often forget breaking bonds is endothermic (+), making bonds is exothermic (–).

• Exactness Assumed:

  • Bond enthalpy values are averages, not exact for a specific molecule. Calculations give approximate values.

• States Ignored:

  • Bond enthalpies apply to gaseous molecules only. Using them for liquids/solids introduces error.

• Equation Errors:

  • Forgetting to balance chemical equations before applying bond enthalpy calculations.

  • Mixing up “bonds broken – bonds formed” with the reverse.

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Exam Revision Tips for bond enthalpy calculations

• Memorise Definition: Word-perfect definition of mean bond enthalpy is expected.

• Balance Equations First: Always balance before identifying bonds broken/formed.

• Check Signs: Write “+” for bonds broken, “–” for bonds formed.

• Units: Always quote kJ mol⁻¹.

• State Approximation: In exam answers, note that bond enthalpy calculations are approximate.

• Compare Methods: Be ready to explain why Hess’s Law cycles are more accurate than bond enthalpy calculations.

• Cross-board Alignment: All exam boards (AQA, Edexcel, OCR, Salters, WJEC, CCEA, CIE, IB, AP) require:

  • Definition of bond enthalpy.

  • Calculations using bond enthalpy data.

  • Understanding of limitations (averages, gaseous state).

  • Comparison with experimental enthalpy values.

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Summary Table

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Final thoughts about bond enthalpy calculations:

Bond enthalpy calculations are a first approximation tool.

Examiners expect you to highlight their limitations and contrast them with more accurate Hess’s Law or experimental data.

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Learning objectives for this page

Be able to construct a Hess's Law cycle to determine an unknown bond enthalpy.

Be able to construct a Hess's Law cycle to determine a theoretical unknown enthalpy change for a given reaction expressed in terms of gaseous species.

Know that the calculated enthalpy change involves assumptions and is therefore subject to errors.

Know that more accurate values can be obtained using enthalpies of evaporation or condensation to correct the initial theoretical calculated enthalpies of reaction.

QUICK INDEX for Energetics: GCSE Notes on the basics of chemical energy changes – important to study and know before tackling any of the three Advanced Level Chemistry pages INDEX of ALL advanced level pages on thermochemistry and thermodynamics Parts 1–3 here * Part 1a–b ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes of reaction, formation, combustion etc. : 1.2a & 1.2b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations – reaction, combustion, formation etc. : 1.2b(iv) Enthalpy of reaction from bond enthalpy calculations  : 1.3a–b Experimental methods for determining enthalpy changes and treatment of results and calculations : 1.4 Some enthalpy data patterns : 1.4a The combustion of linear alkanes and linear aliphatic alcohols : 1.4b Some patterns in Bond Enthalpies and Bond Length : 1.4c Enthalpies of Neutralisation : 1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and evidence of aromatic ring structure in benzene : Extra Q page A set of practice enthalpy calculations with worked out answers ** Part 2 ΔH Enthalpies of ion hydration, solution, atomisation, lattice energy, electron affinity and the Born–Haber cycle : 2.1a–c What happens when a salt dissolves in water and why? : 2.1d–e Enthalpy cycles involving a salt dissolving : 2.2a–c The Born–Haber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1a–g Introduction to Entropy : 3.2 Examples of entropy values and comments * 3.3a ΔS, Entropy and change of state : 3.3b ΔS, Entropy changes and the feasibility of a chemical change : 3.4a–d More on ΔG, free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants from ΔG the free energy change : 3.6 Kinetic stability versus thermodynamic feasibility - can a chemical reaction happen? and will it happen?

Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's Chemistry revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Advanced level pre-university/college chemistry revision notes and key points on energetics, thermochemistry, thermodynamics including enthalpy, entropy and free energy changes: Using bond enthalpies to theoretically calculate the enthalpy change of a reaction. All suitable for chemistry students taking the WJEC advanced A level chemistry, CCEA advanced A level chemistry, Cambridge CIE advanced A level chemistry, AQA advanced A level chemistry, Edexcel advanced A level chemistry, OCR advanced A level chemistry, Salters advanced A level chemistry, IB advanced level chemistry, US grade 11-12 K12 AP Honors chemistry courses INDEX of all my notes on thermochemistry and thermodynamics