Revision notes on thermochemistry-energetics - solving enthalpy problems using Hess's Law - for Advanced Level Theoretical-Physical Chemistry:

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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)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 Cdirect 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.

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

and each ΔHθ could be a composite of several ΔHθ 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) + O2(g) ==> CO2(g)   ΔHθc(C(s)) = ΔHθf(CO2(g)) = –393 kJmol–1

H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθc(H2(g)) = ΔHθf(H2O(l)) = –286 kJmol–1

CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(l)  ΔHθc(CH4(g)) = –890 kJmol–1

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

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

 ΔHØf (methane) = ??? kJmol–1 C(s) + 2H2(g) CH4(g) ΔHθc(C(s)) + 2 x ΔHθc(H2(g)) = (–393) + (2 x –286) C => CO2 and 2H2 => 2H2O +2O2(g) +2O2(g) ΔHθc(CH4(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 ΔHθf(CO2(g)) + 2 x ΔHθf(H2O(l)) same numbers due to coincidence of enthalpy names +2O2(g) –2O2(g) = –ΔHθc(CH4(g)) = –(–890) = +890 CO2(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, ΔHθf 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) = (–393) + (2 x –286) + (+890) = –75 kJmol–1

2nd example of Hess's Law Cycle calculation

From the following thermochemical data

(1)  1/2H2(g) + 1/2Cl2(g) ==> HCl(g)   ΔHθf(hydrogen chloride) = –92.3 kJmol–1

(2)  2C(s) + 3H2(g) + 1/2O2(g) ==> CH3CH2OH(l)   ΔHθf(ethanol) = –278.0 kJmol–1

(3)  2C(s) + 11/2H2(g) + 1/2O2(g) + 1/2Cl2(g) ==> CH3COCl(l)   ΔHθf(ethanoyl chloride) = –275.0 kJmol–1

(4)  4C(s) + 4H2(g) + O2(g) ==> CH3COOCH2CH3(l)   ΔHθf(ethanol) = –481.0 kJmol–1

Calculate the enthalpy change for the reaction

(5)  CH3COCl(l) + CH3CH2OH(l) ==> CH3COOCH2CH3(l) +  HCl(g)   ΔHθr@(esterification) = ??? kJmol–1

 ΔHØreaction(esterification) = ??? kJmol–1 CH3COCl(l) + CH3CH2OH(l) CH3COOCH2CH3(l) + HCl(g) –ΔHθf(CH3COCl(l)) + {–ΔHθf(CH3CH2OH(l))} = –(–275.0) + {–(–278.0)}  ΔHθf(HCl(g)) + ΔHθf(CH3COOCH2CH3(l)) = (–92.3) + (–481.0) 4C(s) + 41/2H2(g) + O2(g) + 1/2Cl2(g) from {2C(s) + 11/2H2(g) + 1/2O2(g) + 1/2Cl2(g)} + {2C(s) + 3H2(g) + 1/2O2(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  ΔHθ's in the lower route of the cycle ΔHθ298(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) + O2(g) ==> CO2(g)   ΔHθ = –393 kJmol–1

(2)  H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθ = –286 kJmol–1

(3)  3C(s) + 4H2(g) ==> C3H8(g)  ΔHθ = –104 kJmol–1

Calculate the standard enthalpy of combustion of propane

(4)  C3H8(g) + 5O2(g) ==> 3CO2(g) + 4H2O(l)  ΔHθc,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, ==> 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)  1/2H2(g) + 1/2Cl2(g) ==> HCl(g) ΔHθ = –92.3 kJmol–1

(2)  2C(s) + 3H2(g) ==> CH3CH3(g) ΔHθ = –84.7 kJmol–1

(3)  2C(s) + 2H2(g) + Cl2(g) ==> ClCH2CH2Cl(l) ΔHθ = –166.0 kJmol–1

Calculate the enthalpy change for the reaction

(4)  2Cl2(g) + CH3CH3(g) ==> ClCH2CH2Cl(l) +  2HCl(g) ΔHθ = ??? 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

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) ==> CH3COOCH2CH3(l) + H2O(l) ΔHθf,298/kJmol–1 –487 –278 –481 –286

ΔHreaction = ∑Hproducts – ∑Hreactants

ΔHθreaction,298 = ∑ΔHθf,298(products) – ∑ΔHθf,298(reactants)

ΔHθesterification = {ΔHθf(ethyl ethanoate) + ΔHθf(water)} – {ΔHθf(ethanoic acid) + ΔHθf(ethanol)}

ΔHθesterification = {–481 + –286} – {–487 + –278}

ΔHθ(esterification reaction) = (–767) – (–765) = –2 kJmol–1

What would be ΔHθ(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) + O2(g) ==> CO2(g)   ΔHθf,298(carbon dioxide) = –393 kJmol–1

(2)  H2(g) + 1/2O2(g) ==> H2O(l)  ΔHθf,298(water) = –286 kJmol–1

(3)  C4H10(g) + 61/2O2(g) ==> 4CO2(g) + 5H2O(l)  ΔHθc,298(butane) = –2877 kJ mol–1

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

(4)  4C(s) + 5H2(g) ==> C4H10(g)  ΔHθ = ??? 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)

ΔHreaction = ∑Hproducts – ∑Hreactants

for equation (3) ΔHcombustion(butane) = ∑ΔHθf(products) – ∑ΔHθf(reactants)

ΔHθc,298(butane) = {4 x ΔHθf(carbon dioxide) + 5 x ΔHθf,298(water)} – {ΔHθf(butane) + ΔHθf(oxygen)}

Since oxygen is an element, ΔHθf(oxygen) = 0, therefore after rearranging we get

ΔHθc,298(butane) = {4 x ΔHθf(carbon dioxide) + 5 x ΔHθf,298(water)} – {ΔHθf(butane)}

–2877 = {(4 x –393) + (5 x –286)} – ΔHθf(butane)

ΔHθf(butane) = 2877 – 1572 – 1430 = –125 kJmol–1

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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) : 1.2b(iv) : 1.3a–b 1.4 : 1.4a : 1.4b : 1.4c : 1.4d : Extra Q page ** Part 2 : 2.1a–c 2.1d–e : 2.2a–c *** Part 3 : 3.1a–g : 3.2 * 3.3a : 3.3b : 3.4a–d : 3.5 Doc Brown's Chemistry Website content © Dr Phil Brown 2000+.  All copyrights reserved on revision notes, images, quizzes, worksheets. Copying of website material is NOT permitted. Exam revision summaries and references to science course specifications are all unofficial but all based on the official specifications.

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