Doc Brown's A Level Chemistry  Advanced Level Theoretical Physical Chemistry AS A2 Level Revision Notes Basic Thermodynamics

GCE Thermodynamicsthermochemistry subindex links below

Part 1 ΔH Enthalpy Changes The thermochemistry of enthalpies of reaction, formation, combustion and neutralisation

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

This page describes how to do enthalpy calculations involving bond enthalpies ('bond energies'). Bond enthalpy calculations using a Hess's Law cycle can be used to calculate unknown enthalpies but there are limitations to the results of these calculations since they are based on average bond enthalpies and only gaseous species.

Energetics index: GCSE Notes on the basics of chemical energy changes important to study and know before tackling any of the three Advanced Level Chemistry pages Parts 13 here * Part 1ab ΔH Enthalpy Changes 1.1 Advanced Introduction to enthalpy changes reaction, formation, combustion : 1.2a & 1.2b(i)(iii) Thermochemistry Hess's Law and Enthalpy Calculations reaction, combustion, formation etc. : 1.2b(iv) Bond Enthalpy Calculations  : 1.3ab Experimental methods for determining enthalpy changes and treatment of results : 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 BornHaber cycle : 2.1ac What happens when a salt dissolves in water and why? : 2.1de Enthalpy cycles involving a salt dissolving : 2.2ac The BornHaber Cycle *** Part 3 ΔS Entropy and ΔG Free Energy Changes : 3.1ag 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 : 3.4ad More on ΔG, Free energy changes, feasibility and applications : 3.5 Calculating Equilibrium Constants : 3.6 Kinetic stability versus thermodynamic feasibility * PLEASE note that delta H/S/G values vary slightly from source to source, so I apologise in advance for any inconsistencies that may arise as I've researched and developed each section.


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

Lots of simple 'starter' examples worked out on the GCSE energetics page in section 5

Unlike methods 1.2b (i, ii, iii) which all give precise and accurate enthalpy values, method 1.2b (iv) only gives an approximate value for reasons that will be explained later.

The bond enthalpy/energy is the energy required to break 1 mole of a specified bond for a gaseous species at 298K/25oC. (note gaseous species are specified)

i.e. AB(g) ==> A(g) + B(g)    ΔHBE(AB) = + ??? kJmol1

The diagram above shows, in terms of a dot and cross diagram, the breaking of a CCl bond by homolytic bond fission.

Note that bond breaking is always endothermic (+) and bond formation is always exothermic ().

How are bond enthalpies determined?

1. Just as spectroscopy can be used to determine ionisation energies (see hydrogen spectrum), spectroscopic techniques can be used to determine bond energies i.e. it is possible to estimate the frequency of radiation required to cause bond fission.

2. Electron impact methods in essence it is a method which measures the energy to fragment molecules.

3. Thermochemical calculations the method most appropriate to this pages level of study.

3. is illustrated by the diagram below to calculate the C=O bond enthalpy in carbon dioxide.

+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 mol1

therefore the C=O bond energy in CO2 = 1606/2 = 803 kJ mol1

A second example of using Hess's Law to calculate the average CH bond enthalpy in the methane molecule

ΔHf (methane) = 75 kJmol1
    C(s) + 2H2(g) (c) doc b CH4(g)

ΔHθsub(C(s)) + 2 x ΔHθBE(HH(g))

= (+715) + (2 x +436)

both endothermic

(c) doc b

C(g) +



4 x avΔHθBE(CH(g)) = X


formation of 4 CH bonds

The cycle for the standard enthalpy of formation of methane, (ΔHθf), based on the enthalpy of sublimation of carbon (graphite) and the HH and CH 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.

ΔHθf (methane) = {ΔHθsub(C(s)) + 2 x ΔHθBE(HH(g))} + {4 x avΔHθBE(CH(g))}  = 75 kJmol1

75 = +715 +872 X

X = (715 + 872 +75) = 1662

avΔHθBE(CH(g)) = 1662/4 = +415.5 kJmol1


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


methane + chlorine ==> chloromethane + hydrogen chloride

So, how can we theoretically calculate the energy change for this reaction using bond enthalpies?

CH4 + Cl2 ==> CH3Cl + HCl

Bond energies: The energy required to break or make 1 mole of a particular bond in kJ/mol

CH = 412, ClCl = 242 kJ/mol, CCl = 331 kJ/mol, HCl = 432

To appreciate all the bonds in the molecules its better to set out as follows ...

alkanes structure and naming (c) doc b + ClCl ===> (c) doc b + HCl

Then I'm using the diagram below to illustrate how you do the calculation and what its all about.

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

The energy absorbed equals that to break one CH bond (in methane molecule) plus energy to break one ClCl 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 CCl bond (in chloromethane molecule) is formed plus the energy released when one HCl 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.

Therefore ΔHreaction = 139 kJmol1



Further examples how you might solve them in exams!

Ex 1. Calculating the enthalpy of a reaction

Given the following bond enthalpies in kJ mol1: ClCl = 242, HH = 436, HCl = 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 ...

H2(g) + Cl2(g) (c) doc b 2HCl(g)

(c) doc b 2H(g) + 2Cl(g)

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

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

ΔHθreaction = {+436 +242} + {2 x 431}

ΔHθreaction = +(436 + 242 862) = 184 kJmol1

Note that ΔHθreaction /2 = 92 kJmol1 = ΔHθf(HCl(g))

Ex 2. Given the following bond dissociation enthalpies at 298K in kJmol1 ...
 Bond CC (single) CH C=O in CO2 OH O=O
Bond enthalpy +348 +412 +805 +463 +496

... calculate the enthalpy of combustion of butane. You construct a Hess's Law Cycle from the 'normal' combustion equation and 'atomise' the reactant molecules to give ALL of the 'theoretical' intermediate atoms. From this you can  theoretically calculate the enthalpy of the reaction:

ΔHcomb, 298 (butane) = ??? kJmol1
(g) + 61/2O=O(g) (c) doc b 4O=C=O(g) + 5HOH(g)

(3 x ΔHBE(CC))

+ (10 x ΔHBE(CH))

+ (6.5 x ΔHBE(O=O))

= (3 x +348) + (10 x +412) + (6.5 x +496)

(c) doc b



+ 10H(g)

+ 13O(g)



(8 x ΔHBE(C=O))

(10 x ΔHBE(OH))

= (8 x 743) (10 x 463)

ΔHcomb, 298 (butane) = 1044 + 4120 + 3224 6440 4630 = 2682 kJmol1

The true thermodynamic value from very accurate calorimetry experiments is 2877 kJmol1

So why the significant difference/error?

The difference is the theoretical value from the bond energy calculation is less exothermic by 195 kJ.

All enthalpy calculations done by this method will always be in error to some extent because the bond enthalpies quoted in data information are based on average values for that particular bond. Each 'AB' bond will differ slightly depending on the 'ambient' electronic situation e.g.

The CH bond in methane, alkanes structure and naming (c) doc b, will be slightly different than in ethane, alkanes structure and naming (c) doc b etc.

See also further discussion on methane's CH bonds at the end of section 1.4 namely 1.4b(iii) point (c)

There is a 2nd reason which for some reason many textbooks don't bother to mention. Since only gaseous species can be considered, if any reactant or product is a liquid or solid, then the enthalpy value for any state change is NOT taken into account. For example, in the above example, five molecules of water are formed which at 298K would condense to water and this is an exothermic process. ΔHvap for water is 41 kJmol1, so if water condensation takes place then 5 x 41 = 205 kJ extra would be released and this actually accounts for most of the error!


Ex 3. example of bond enthalpy calculation method

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

Hydrazine has been used as monopropellant in rocket engines because it can 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 kJmol1, given below, calculate the enthalpy changes for reactions (1) and (2)

 Bond NH Nalkene (c) doc bN NN HH OH OO
Bond enthalpy +388 +944 +163 +436 463 +146

For (1)  N2H4 ==> N2 + 2H2

H2NNH2 (c) doc b Nalkene (c) doc bN + 2HH


+ (4 x ΔHBE(NH))

= (+163) + (4 x +388)

(c) doc b


2N + 4H   


(ΔHBE(Nalkene (c) doc bN))

(2 x ΔHBE(HH))

= (+944) (2 x 436)

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

= +163 +1552 944 872 = 101 kJmol1

For (2)  N2H4 + 2H2O2 ==> N2 + 4H2O

H2NNH2 + 2H2O2 (c) doc b Nalkene (c) doc bN + 4HOH


+ (4 x ΔHBE(NH))

+ (2 x ΔHBE(OO))

+ (4 x ΔHBE(OH))  

= (+163) + (4 x +388) + (2 x +146) + (4 x +463)

(c) doc b


2N + 8H + 4O   


(ΔHBE(Nalkene (c) doc bN))

(8 x ΔHBE(OH))

= (+944) (8 x +463)

watch the signs!

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

ΔHreaction(combustion N2H4) = 789 kJmol1

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!

A set of enthalpy calculation problems with worked out answers based on enthalpies of reaction, formation, combustion and bond enthalpies


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