A level A level Thermochemistry Enthalpy of reaction Hess's Law cycle calculations KS5 GCE chemistry revision notes

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Doc Brown's A Level Chemistry - Advanced Level Theoretical Physical Chemistry – GCE AS A2 IB Level Revision Notes – Basic Thermodynamics thermochemistry

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

Parts 1.2a–b(i)–(iii) Thermochemistry – Hess's Law and Enthalpy Calculations

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Page introduction

This page describes Hess's Law of 'constant heat summation' and the use of thermochemical cycles to determine unknown enthalpy changes from known data e.g. involving the calculation of enthalpy of reaction, enthalpy of formation, enthalpy of combustion.
As well as Hess's Law calculations other methods of solving numerical enthalpy problems without using thermochemical cycles are also described.

1.2 Thermochemistry – Hess's Law and Enthalpy Calculations

1.2a Hess's Law and its importance

Hess's Law is a version of the general law of conservation of energy.
i.e. the first law of thermodynamics which can be stated as energy cannot be destroyed or created but merely changed in form or distributed in different ways.
Hess's Law states that the energy change from reactants A to products B is independent of the pathway taken no matter how many stages it involves.
This is shown in the diagram of enthalpy cycles on the right.

ΔH₁ = ΔH(A to B) Pathway 1 the most direct route with no intermediate stages,
or ΔH₁ = ΔH₂ + ΔH₃ pathway 2 involving one set of intermediates C, or
or ΔH₁ = ΔH₄ + ΔH₅ + ΔH₆ pathway 3 involving two sets of intermediates D and E
etc. etc. – there is no limit to the complexity of the Hess's Law Cycle as long as A and B are constant.

Many problems can be set out as a 'Hess's Law' triangle e.g.

Application of Hess's Law (i) ΔH^θ₁
A B
ΔH^θ₂				ΔH^θ₃
C
Clearly there are two pathways from A to B direct and via C following the arrows direction so, ΔH^θ₁ = ΔH^θ₂ + ΔH^θ₃ or ΔH^θ₂ = ΔH^θ₁ - ΔH^θ₃ or ΔH^θ₃ = ΔH^θ₁ - ΔH^θ₂

Application of Hess's Law (ii) ΔH^θ₁
A B
ΔH^θ₂				ΔH^θ₃
C
In this case there are two pathways from A to C direct and via B following the arrows direction so, ΔH^θ₂ = ΔH^θ₁ + ΔH^θ₃ or ΔH^θ₁ = ΔH^θ₂ - ΔH^θ₃ or ΔH^θ₃ = ΔH^θ₂ - ΔH^θ₁

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.

Note: Superscript ^θ means a standard enthalpy value (see further down the page)

and each ΔH^θ could be a composite of several ΔH^θ values - see enthalpy of combustion calculations below.

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1.2b Using Hess's Law to perform enthalpy calculations

1.2b(i) Using known enthalpies of reaction or combustion etc.

Using a Hess's Law Cycle you can calculate enthalpies of formation which you could not determine by laboratory experiment. However, using these calculated enthalpies of formation and experimentally determined enthalpies of combustion a huge variety of other enthalpies for other reactions can then be calculated.

The 1st example of calculating the enthalpy of formation of methane illustrates the principles of using Hess's Law, especially as this cannot be determined by direct laboratory experiment!

Given the following data below from text/data book calculate the enthalpy of formation of methane.

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

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

CH_4(g) + 2O_2(g) ==> CO_2(g) + 2H₂O_(l) ΔH^θ_c(CH_4(g)) = –890 kJmol^–1

All of the three enthalpies above can be very accurately determined by direct experiment in a calorimeter.

BUT ΔH^θ_f for C(s) + 2H₂(g) ===> CH₄(g) cannot be determined directly.

ΔH^Ø_f (methane) = ??? kJmol^–1
C_(s) + 2H_2(g) CH_4(g)
ΔH^θ_c(C_(s)) + 2 x ΔH^θ_c(H_2(g)) = (–393) + (2 x –286) C => CO₂ and 2H₂ => 2H₂O	+2O_2(g)	+2O_2(g)	ΔH^θ_c(CH_4(g)) = –890 OR change the arrow round and change the signs as below, remember, change direction, you change the sign BUT not the numerical energy value!
OR ΔH^θ_f(CO_2(g)) + 2 x ΔH^θ_f(H₂O_(l)) same numbers due to coincidence of enthalpy names	+2O_2(g)	–2O_2(g)	= –ΔH^θ_c(CH_4(g)) = –(–890) = +890
CO_2(g) + 2H₂O_(l) Note that you double the enthalpy of combustion of hydrogen to make the cycle balance in molar terms
The cycle for the standard enthalpy of formation of methane, ΔH^θ_f 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. ΔH^θ_f (methane) = (–393) + (2 x –286) + (+890) = –75 kJmol^–1

2nd example of Hess's Law Cycle calculation

From the following thermochemical data

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

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

(3) 2C_(s) + 1¹/₂H_2(g) + ¹/₂O_2(g) + ¹/₂Cl_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

ΔH^Ø_reaction(esterification) = ??? kJmol^–1
CH₃COCl_(l) + CH₃CH₂OH_(l) CH₃COOCH₂CH_3(l) + HCl_(g)
–ΔH^θ_f(CH₃COCl_(l)) + {–ΔH^θ_f(CH₃CH₂OH_(l))} = –(–275.0) + {–(–278.0)}				ΔH^θ_f(HCl_(g)) + ΔH^θ_f(CH₃COOCH₂CH_3(l)) = (–92.3) + (–481.0)
4C_(s) + 4¹/₂H_2(g) + O_2(g) + ¹/₂Cl_2(g) from {2C_(s) + 1¹/₂H_2(g) + ¹/₂O_2(g) + ¹/₂Cl_2(g)} + {2C_(s) + 3H_2(g) + ¹/₂O_2(g)} Note that you didn't have to double or treble etc any enthalpy value because the molar ratio is 1 : 1 ==> 1 : 1
Adding up all the ΔH^θ's in the lower route of the cycle ΔH^θ₂₉₈(esterification) = (+275.0) + (+278.0) + (–92.3) + (–481.0) = –20.3 kJmol^–1 Note that the sign of enthalpy of formation of ethanoyl chloride and ethanol is reversed to fit in with the direction of change.

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3rd example of Hess's Law Cycle calculation

Problem solving from the following thermochemical data:

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) + ¹/₂O_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) + 3O_2(g) ==> 3CO_2(g) (1) add ΔH^θ = 3 x –393 kJmol^–1

4H_2(g) + 2O_2(g) ==> 4H₂O_(l) (2) plus ΔH^θ = 4 x –286 kJmol^–1

C₃H_8(g) ==> 3C_(s) + 4H_2(g) (3) plus ΔH^θ = +104 kJmol^–1 (equation and sign reversed)

C₃H_8(g) + 5O_2(g) ==> 3CO_2(g) + 4H₂O_(l)
(4) = ΔH^θ_c,298(propane) = –2219 kJ mol^–1

adding up (1) + (2) + (3) gives (4), cancelling out unwanted equation components, ==> treated as an = sign

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

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

H_2(g) + Cl_2(g) ==> 2HCl_(g)	(1)	add ΔH^θ = 2 x –92.3 kJmol^–1
CH₃CH_3(g) ==> 2C_(s) + 3H_2(g)	(2)	plus ΔH^θ = +84.7 kJmol^–1 (equation and sign reversed)
2C_(s) + 2H_2(g) + Cl_2(g) ==> ClCH₂CH₂Cl_(l)	(3)	plus ΔH^θ = –166 kJmol^–1
2Cl_2(g) + CH₃CH_3(g) ==> ClCH₂CH₂Cl_(l) + 2HCl_(g)	(4)	= ΔH^θ₂₉₈(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

1.2b(iii) A 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)	==>	CH₃COOCH₂CH_3(l)	+ H₂O_(l)
ΔH^θ_f,298/kJmol^–1	–487	–278		–481	–286

ΔH_reaction = ∑H_products – ∑H_reactants

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

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

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 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?

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