Application of Hess's Law (i) ΔHθ1 A   B ΔHθ2      ΔHθ3 C Clearly there are two pathways from A to B direct and via C following the arrows direction so,  ΔHθ1  =  ΔHθ2  +  ΔHθ3 or    ΔHθ2  =  ΔHθ1  -  ΔHθ3 or    ΔHθ3  =  ΔHθ1  -  ΔHθ2	Application of Hess's Law (ii) ΔHθ1 A    B ΔHθ2      ΔHθ3 C In this case there are two pathways from A to C direct and via B following the arrows direction so,  ΔHθ2  =  ΔHθ1  +  ΔHθ3 or   ΔHθ1  =  ΔHθ2  -  ΔHθ3 or    ΔHθ3  =  ΔHθ2  -  ΔHθ1
Always take care with the ΔH signs which ever way you set up the triangle to apply Hess's Law	Although slightly more awkward, (ii) might be better suited to the way the experiment results are obtained or given data.

 

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2nd example of Hess's Law Cycle calculation

From the following thermochemical data

(1)  ½H_2(g) + ½Cl_2(g) ==> HCl_(g)   ΔH^θ_f(hydrogen chloride) = –92.3 kJmol^–1

(2)  2C_(s) + 3H_2(g) + ½O_2(g) ==> CH₃CH₂OH_(l)   ΔH^θ_f(ethanol) = –278.0 kJmol^–1

(3)  2C_(s) + 1½H_2(g) + ½O_2(g) + ½Cl_2(g) ==> CH₃COCl_(l)   ΔH^θ_f(ethanoyl chloride) = –275.0 kJmol^–1

(4)  4C_(s) + 4H_2(g) + O_2(g) ==> CH₃COOCH₂CH_3(l)   ΔH^θ_f(ethanol) = –481.0 kJmol^–1

Calculate the enthalpy change for the reaction

(5)  CH₃COCl_(l) + CH₃CH₂OH_(l) ==> CH₃COOCH₂CH_3(l) +  HCl_(g)  

ΔH^θ_r@(esterification) = ??? kJmol^–1

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3rd example of Hess's Law Cycle calculations

Problem solving from the following thermochemical data:

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1.2b(ii) Solving enthalpy problems using an 'algebraic–equation style' method

BUT, remember, the question might specify solving the problem via a Hess's Law cycle!

Given the following data

(1)  C_(s) + O_2(g) ==> CO_2(g)   ΔH^θ = –393 kJmol^–1

(2)  H_2(g) + ½O_2(g) ==> H₂O_(l)  ΔH^θ = –286 kJmol^–1

(3)  3C_(s) + 4H_2(g) ==> C₃H_8(g)  ΔH^θ = –104 kJmol^–1

Calculate the standard enthalpy of combustion of propane

(4)  C₃H_8(g) + 5O_2(g) ==> 3CO_2(g) + 4H₂O_(l)  ΔH^θ_c,298(propane) = ??? kJ mol^–1

What you do is rearrange, if necessary, the data equations and add up the results – both equation components and delta H values, cancelling out the equation components should leave you with the correct equation whose enthalpy value you require.

3C(s) + 3O2(g) ==> 3CO2(g)	(1)	add ΔHθ = 3 x –393 kJmol–1
4H2(g) + 2O2(g) ==> 4H2O(l)	(2)	plus ΔHθ = 4 x –286 kJmol–1
C3H8(g) ==> 3C(s) + 4H2(g)	(3)	plus ΔHθ = +104 kJmol–1  (equation and sign reversed)
C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)	(4)	= ΔHθc,298(propane) = –2219 kJ mol–1
adding up (1) + (2) + (3) gives (4), cancelling out unwanted equation components, the ==> treated as an = sign

Note that here you are using factors of 3 and 4 to make the cycle balance in molar terms

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2nd example of 'algebraic' calculation style

Given the following thermochemical data:

(1)  ¹/₂H_2(g) + ¹/₂Cl_2(g) ==> HCl_(g) ΔH^θ = –92.3 kJmol^–1

(2)  2C_(s) + 3H_2(g) ==> CH₃CH_3(g) ΔH^θ = –84.7 kJmol^–1

(3)  2C_(s) + 2H_2(g) + Cl_2(g) ==> ClCH₂CH₂Cl_(l) ΔH^θ = –166.0 kJmol^–1

Calculate the enthalpy change for the reaction

(4)  2Cl_2(g) + CH₃CH_3(g) ==> ClCH₂CH₂Cl_(l) +  2HCl_(g) ΔH^θ = ??? kJmol^–1

H2(g) + Cl2(g) ==> 2HCl(g)	(1)	add ΔHθ = 2 x –92.3 kJmol–1
CH3CH3(g) ==> 2C(s) + 3H2(g)	(2)	plus ΔHθ = +84.7 kJmol–1  (equation and sign reversed)
2C(s) + 2H2(g) + Cl2(g) ==> ClCH2CH2Cl(l)	(3)	plus ΔHθ = –166 kJmol–1
2Cl2(g) + CH3CH3(g) ==> ClCH2CH2Cl(l) +  2HCl(g)	(4)	= ΔHθ298(chlorination reaction) = –265.9 kJ mol–1
adding up (1) + (2) + (3) gives (4), cancelling out unwanted equation components, ==> treated as an = sign

Note that you double the enthalpy of formation hydrogen chloride to make the cycle balance in molar terms

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1.2b(iii) A 3rd method using the summation of enthalpies of reactants and products

A very simple example of an enthalpy summation method

molecule	ethanoic acid	+   ethanol	==>	ethyl ethanoate	+    water
equation	CH3COOH(l)	+   CH3CH2OH(l)	==>	CH3COOCH2CH3(l)	+   H2O(l)
ΔHθf,298/kJmol–1	–487	–278		–481	–286

ΔH_reaction = ∑H_products – ∑H_reactants

_{(this is still a form of using Hess's Law, but it just doesn't look like it!)}

ΔH^θ_reaction,298 = ∑ΔH^θ_f,298(products) – ∑ΔH^θ_f,298(reactants)

ΔH^θ_{esterification} = {ΔH^θ_f(ethyl ethanoate) + ΔH^θ_f(water)} – {ΔH^θ_f(ethanoic acid) + ΔH^θ_f(ethanol)}

ΔH^θ_{esterification} = {–481 + –286} – {–487 + –278}

ΔH^θ(esterification reaction) = (–767) – (–765) = –2 kJmol^–1

What would be ΔH^θ(hydrolysis)? Answer! just reverse the sign! i.e. +2 kJmol^–1

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2nd example of enthalpy summation method

A more complicated example where you need to think more about mole ratios in the equation.

Given the following data from laboratory measurements

(1)  C_(s) + O_2(g) ==> CO_2(g)   ΔH^θ_f,298(carbon dioxide) = –393 kJmol^–1

(2)  H_2(g) + ¹/₂O_2(g) ==> H₂O_(l)  ΔH^θ_f,298(water) = –286 kJmol^–1

(3)  C₄H_10(g) + 6¹/₂O_2(g) ==> 4CO_2(g) + 5H₂O_(l)  ΔH^θ_c,298(butane) = –2877 kJ mol^–1

Calculate the standard enthalpy of formation of butane gas, which cannot be determined by experiment.

(4)  4C_(s) + 5H_2(g) ==> C₄H_10(g)  ΔH^θ = ??? kJmol^–1

To solve this you can use equation (3) and the data from equations (1) and (2) to obtain a value for equation (4)

ΔH_reaction = ∑H_products – ∑H_reactants

for equation (3) ΔH_combustion(butane) = ∑ΔH^θ_f(products) – ∑ΔH^θ_f(reactants)

ΔH^θ_c,298(butane) = {4 x ΔH^θ_f(carbon dioxide) + 5 x ΔH^θ_f,298(water)} – {ΔH^θ_f(butane) + ΔH^θ_f(oxygen)}

Since oxygen is an element, ΔH^θ_f(oxygen) = 0, therefore after rearranging we get

ΔH^θ_c,298(butane) = {4 x ΔH^θ_f(carbon dioxide) + 5 x ΔH^θ_f,298(water)} – {ΔH^θ_f(butane)}

–2877 = {(4 x –393) + (5 x –286)} – ΔH^θ_f(butane)

ΔH^θ_f(butane) = 2877 – 1572 – 1430 = –125 kJmol^–1

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Key revision points about using Hess's Law

Hess’s Law states that the enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.

It is crucial because many enthalpy changes cannot be measured directly, so Hess cycles and enthalpy data (formation, combustion, bond enthalpies) are used to calculate them.

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• Definition:

  • Hess’s Law: The total enthalpy change for a reaction is the same, no matter which route is taken, provided the starting and finishing conditions are the same.

  • Hess's Law is based on the law of conservation of energy.

• Mathematical Forms e.g.

  • ΔH_reaction = ∑H_products – ∑H_reactants

  • ΔH^θ_reaction,298 = ∑ΔH^θ_f,298(products) – ∑ΔH^θ_f,298(reactants)

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

• Energy Cycles:

  • Draw cycles with arrows showing alternative routes.

  • Label each arrow with the correct enthalpy change.

  • Use algebra to solve for the unknown enthalpy.

• Applications:

  • Calculating enthalpy of reaction from enthalpies of formation.

  • Calculating enthalpy of reaction from enthalpies of combustion.

  • Using bond enthalpies to estimate enthalpy changes.

  • Explaining why experimental values differ from theoretical (bond enthalpy averages, incomplete combustion, non-standard conditions).

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Common Misconceptions about Hess's Law

• Sign Errors:

  • Students often forget that combustion enthalpies are negative (exothermic).

  • Wrongly add instead of subtracting enthalpy values.

• Cycle Construction Mistakes:

  • Arrows pointing the wrong way in Hess cycles.

  • Forgetting to balance chemical equations before applying enthalpy data.

• Confusing Routes:

  • Mixing up formation versus combustion cycles.

  • Forgetting that bond enthalpy calculations are approximate (average values).

• Units:

  • Omitting “per mole” in kJ mol⁻¹.

  • Using kJ instead of kJ mol⁻¹ in exam answers.

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Exam Revision Tips when using Hess's Law

• Learn Definitions Word-for-Word: Examiners expect precise definitions of Hess’s Law.

• Practice Cycles: Draw neat, labelled Hess cycles for both formation and combustion routes.

• Check Signs: Always double-check whether enthalpy values are positive or negative.

• Balance Equations: Before using enthalpy data, ensure the chemical equation is balanced.

• Use Algebra Carefully: which ever method you use and watch the enthalpy signs related to direction of enthalpy change.

• Bond Enthalpy Calculations: Remember: breaking bonds = positive, making bonds = negative.

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

  • Definition of Hess’s Law.

  • Construction of Hess cycles.

  • Calculations using formation, combustion, or bond enthalpies.

  • Awareness of experimental vs theoretical discrepancies.

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Summary Table of typical errors when Hess's Law for calculations

Concept	Key Point	Common Error	Exam Tip
Hess’s Law	Enthalpy change independent of route	Forgetting conditions	Memorise definition
Formation Cycle	Use ΔHf values	Wrong states/elements	Balance equation first
Combustion Cycle	Use ΔHc values	Sign errors	Always negative values
Bond Enthalpy	Approximate via averages	Forget averages	State “approximate” in answers
Energy Cycles	Alternative routes	Wrong arrow direction	Draw arrows carefully

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Final thought: Hess’s Law is the bridge between theory and experiment.

It allows indirect calculation of enthalpy changes that are impossible to measure directly, making it one of the most powerful tools in energetics.

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for the next  problem!
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Enthalpy calculation problems with worked out answers – based on enthalpies of reaction, formation, combustion

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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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: Problem solving in thermochemistry using Hess's Law. 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