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

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

Part 1.4 Enthalpy data patterns contd. (c) Enthalpy of neutralisation and enthalpy of oxide or hydride formation, (d) Enthalpy of hydrogenation and benzene structure

This pages discuses a selection of enthalpy of neutralisation and the enthalpy of formation of oxides and hydrides. The final section describes, with the aid of an enthalpy level diagram, how enthalpies of hydrogenation of unsaturated hydrocarbons (alkenes) support the theoretical model for the delocalised electron ring structure of the benzene molecule (the simplest 'aromatic ring').

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 1–3 here * Part 1a–b 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  : 1.3a–b 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 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 : 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.


1.4 Some enthalpy data patterns continued

1.4c(i) Enthalpies of Neutralisation and Enthalpies of Formation of oxides and hydrides

Neutralisation reaction: base + acid ==> salt + water ΔHneutralisation kJ mol–1
1. NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) –57.1
2. KOH(aq) + HCl(aq) ==> KCl(aq) + H2O(l) –57.2
3. NaOH(aq) + HNO3(aq) ==> NaNO3(aq) + H2O(l) –57.3
4. KOH(aq) + HNO3(aq) ==> KNO3(aq) + H2O(l) –57.3
5. Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l) –116.4 (–58.4 per H2O)
6. NaOH(aq) + HF(aq) ==> NaF(aq) + H2O(l) –68.6
7.  NH3(aq) + HCl(aq) ==> NH4Cl(aq) –52.2
8. NaOH(aq) + CH3COOH(aq) ==> CH3COONaaq) + H2O(l) –55.2
9. NH3(aq) + CH3COOH(aq) ==> CH3COONH4aq) –50.4
10. KOH(aq) + HCN(aq) ==> KCN(aq) + H2O(l) –11.7
11.  NH3(aq) + HCN(aq) ==> NH4CN(aq) –5.4

Notes on the above examples from the data table above.

Beware on comparing values!

(i) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)  ΔHθneutralisation = –57.1 kJ mol–1

(ii) Ba(OH)2(aq) + 2HNO3(aq) ==> Ba(NO3)2(aq) + 2H2O(l)  ΔHθneutralisation = –116.4 kJ mol–1

(iii) 1/2Ba(OH)2(aq) + HNO3(aq) ==> 1/2Ba(NO3)2(aq) + H2O(l)  ΔHθneutralisation = –58.2 kJ mol–1

Note! It looks as if the enthalpy of neutralisation of barium hydroxide is approximately double that of sodium hydroxide ie ~ twice as exothermic! Well yes it is! and no it isn't!

Yes – ~twice as much energy is released per mole of soluble base/alkali.

No – however, on the basis of heat released per mole of water formed, they are actually very similar.

In other words, which value you quote, depends on which point you want to make.

Yet another example of carefully qualifying enthalpy values with respect to the context.

Comparing Weak & strong acids or bases (soluble–alkali)

Neutralisation reactions 1., 2., 3., 4. and 5. are reactions between ~fully ionised strong acids and strong bases and the only real chemical change is formation of a water molecule from hydrogen/oxonium ions and hydroxide ions.

H3O+(aq) + OH(aq) ==> 2H2O(l)  or more simply H+(aq) + OH(aq) ==> H2O(l)  ΔH = ~–57 kJ mol–1

All the other ions e.g. Na+, K+, Ba2+, Cl, NO3, SO42– etc. are spectator ions and take no part in the reaction.

The resulting salt solution has a pH of ~7.

Wherever one of the acids or bases is 'weak' the energy change is less exothermic compared to a pair of 'strong' reactants.

The reason for this relates to the fact that if an acid or base is not fully ionised, then energy must be expended facilitate the ionisation. Weak acids or weak bases are on partially ionised (e.g. 2%) in aqueous solution. Or, you can argue theoretically another way in terms of salt hydrolysis i.e. the ions of the salt react with water to 'reform' some of the acid or the base. In other words, you don't really get complete neutralisation of the acid and base on a strictly stoichiometric molar basis. (for more on acid–base theory see Equilibria Part 5.)

Reaction 7. is the neutralisation of the weak base ammonia and the strong hydrochloric acid.

NH4+(aq) + H2O(l) (c) doc b H3O+(aq) + NH3(aq)

ammonium chloride solution has a pH of ~4

Reaction 8. is the neutralisation of the strong base sodium hydroxide and the weak organic carboxylic acid, ethanoic acid.

CH3COO(aq) + H2O(l) (c) doc b CH3COOH(aq) + OH(aq)

sodium ethanoate solution (CH3COONa(aq)) solution has a pH of ~9

Reaction 9. is even less exothermic because this is the neutralisation of a weak base (ammonia) and a weak acid (ethanoic acid).

See the equations for reactions 7. and 8. above

Reaction 10. is the neutralisation of an extremely weak acid (HCN) and, despite KOH being a strong base, only about 1/5th of the energy of a strong acid – strong base neutralisation is released.

CN(aq) + H2O(l) (c) doc b HCN(aq) + OH(aq)

potassium cyanide solution (KCN(aq)) has a pH of over 10.

Reaction 11. is the neutralisation of an extremely weak acid (HCN) and a weak base (NH3) in which only a 1/10th of the maximum possible energy is released.

See equations for reaction 7 and 10.

Reactions 6 is unusually high despite the fact that hydrofluoric acid is a weak acid.


1.4c(ii) Examples of series of enthalpies of formation of similar compounds

Most of the time there is little point in comparing the enthalpies of formation of dissimilar compounds, but for a series of like compounds useful patterns emerge e.g.

Series and comments:

Members of the series:

Group VII Halogen Halides – the value becomes less exothermic from F ==> I, a reflection of the increasing size of the halogen atom, but emphasises the particularly high stability of hydrogen fluoride. HF(g) HCl(g) HBr(g) HI(g)
Enthalpy of formation ΔHf,298(HX(g)) for 1/2H2(g) + 1/2X2(g/l/s) ==> HX(g) in kJ mol–1 –271.1 –92.3 –36.2 +26.5
The organic homologous series of hydrocarbons the Alkanes – where the enthalpy of formation becomes progressively more exothermic as the series is ascended. CH4(g) C2H6(g) C3H8(g) C4H10(g)
Enthalpy of formation ΔHf,298(CnH2n+2) for nC(s) + (n+1)H2(g) ==> CnH2n+2(g) in kJ mol–1 –74.9 –84.7 –104.0 –125.0

Sometimes for a series of compounds different ways of viewing the enthalpy values help in comparing values e.g. the enthalpy of formation of the Period 3 oxides with the element in its highest oxidation state.

Period 3 oxide Na2O(s) MgO(s) Al2O3(s) SiO2(s) P4O10(s) SO3(g) Cl2O7(g)
Enthalpy of formation ΔHf,298(per mole of oxide) in kJ mol–1 –416 –601 –1676 –911 –2984 –395 +265
Enthalpy of formation ΔHf,298(per mole of oxygen) in kJ mol–1 –832 –1202 –1117 –911 –597 –264 +76
Oxidation state of the Period 3 element Na to Cl in its highest valency oxide +1 +2 +3 +4 +5 +6 +7
Type of structure ionic lattice ionic lattice ionic lattice giant covalent small molecule small molecule small molecule
  • By comparing the values per mole of oxygen you get a better comparison of the energetics of the reaction of the element with oxygen.

  • For example it looks as if phosphorus(V) oxide is the most energetically formed and seems way above the other oxides but when compared on a 'per mole of oxygen basis' it fits in with a more 'gentler' pattern of initially rising and then gradually falling enthalpy values of oxide formation.

  • As a very rough rule of thumb, very negative values suggest high thermodynamic stability e.g. the high energy needed via electrolysis to extract reactive metals like sodium, magnesium and aluminium or the very stable giant covalent structure of silicon dioxide.

  • The positive value of the enthalpy of formation of chlorine(VIII) oxide suggests it might be unstable and it is! It decomposes readily on heating and is a highly reactive oxidising agent.

  • Other variations in the data for a series of compounds can be due to differences in structure.

  • So, PLEASE do not always expect nice consistent patterns in enthalpy data that you see in enthalpies of combustion of alkanes/alcohols or enthalpies of formation of alkanes – here you are dealing with similar molecules with ~identical C–H, C–C or C–O covalent bonds and constant 'incremental' changes in the length of the molecule.


1.4d Enthalpies of Hydrogenation of unsaturated hydrocarbons and benzene structure

  • The enthalpy of hydrogenation can be defined as the energy released when one mole of an unsaturated hydrocarbon molecule becomes saturated on reaction with hydrogen.

    • For comparative purposes, the values of ΔHhydrogenation quoted here are for gaseous species only at 355K and 101.3 kPa.

    • Since ΔHθf(hydrogen) = 0 by definition for an element, at 298K and 101.3 kPa

    • ΔHθhydrogenation,298(compound) = ΔHθf(saturated compound) = ΔHθf(unsaturated compound)

  • Examples:

    1. CH2=CH2 + H2 ==> CH3CH3  ΔHhydrogenation(ethene) = –157 kJ mol–1

    2. CH3CH=CH2 + H2 ==> CH3CH2CH3  ΔHhydrogenation(propene) = –126 kJ mol–1

    3. CH3CH2CH=CH2 + H2 ==> CH3CH2CH2CH3  ΔHhydrogenation(but–1–ene) = –127 kJ mol–1

    4. CH3CH=CHCH3 + H2 ==> CH3CH2CH2CH3  ΔHhydrogenation(but–2–ene) = –120/116 kJ mol–1

      • The two values are for the geometrical isomers, cis(Z)/trans(E))

    5. CH2=CH–CH=CH3 + 2H2 ==> CH3CH2CH2CH3  ΔHhydrogenation(but–1,3–diene/1,3–butadiene) = –239 kJ mol–1

    6. + H2 ==> alkanes structure and naming (c) doc b or alkenes structure and naming (c) doc b + H2 ==> alkanes structure and naming (c) doc b

      • ΔHhydrogenation(cyclohexene) = –120 kJ mol–1

    7. + 2H2 ==> alkanes structure and naming (c) doc b or + 2H2 ==> alkanes structure and naming (c) doc b

      • ΔHhydrogenation(cyclohexa–1,3–diene/1,3–cyclohexadiene) = –232 kJ mol–1

    8. C6H6 + 3H2 ==> C6H12 or  (c) doc b + 3H2 ==> alkanes structure and naming (c) doc b

      • ΔHhydrogenation(benzene) = –208 kJ mol–1

    9. C6H5CH3 + 3H2 ==> C6H11CH3 or  (c) doc b + 3H2 ==>

      • ΔHhydrogenation(methylbenzene) = –205 kJ mol–1

  • Discussion of the above enthalpy of hydrogenation values:

    • Apart from the first and simplest alkene, ethene, most enthalpies of hydrogenation are about –120±6 kJ mol–1 (for 1 mole H2 per mole alkene) and in the case of dienes about double that for two moles of hydrogen per diene, which is what you might reasonably expect.

    • The special case of benzene – the aromatic ring structure of arenes – aromaticity

    • However, on the basis of these trends, the expected value for benzene and other aromatic compounds with a single benzene ring would be around 3 x –120 = –360 kJ mol–1, but not so! In fact the energy released on hydrogenating benzene (208 kJ/mole) is even less than hydrogenating a diene!

    • So, something must be different about benzene but it can be explained with the enthalpy level diagram shown below and an examination of possible molecular structures.

    • The 'top' molecule in the diagram shows the theoretical structure of a triene with the same molecular formula of benzene (C6H6) and, if it existed in this form it would be called cyclohexa–1,3,5–triene or 1,3,5–cyclhexatriene.

    • This molecular structure assumes there are simple alternate single (C–C) and double (C=C) carbon carbon bonds.

    • BUT, according to the actual thermochemical data calculated and derived from e.g. enthalpies of combustion, benzene is already more stable by ~152kJ mol–1, so, whatever its structure, it cannot have this 'triene' structure.

    • (c) doc b(c) doc bWhat happens in reality, is that the equivalent of 3 double bonds (C=C) and 3 single bonds (C–C) 'merge' to form a six equal bonds each involve an average of 3 shared electrons. Two of the electrons for each bond are concentrated between the two carbon atoms equivalent to a single bond known as a sigma (σ) bond. The 3rd electron per atom is located in a ring orbital, of two sections, above and below the plane of the hexagonal ring of carbon atoms. These are known as pi (Π) orbitals and each contains 3 pi (Π) electrons which are delocalised around the ring. Whenever charge is delocalised or 'spread out' the potential energy of the system is lowered and in the case of benzene about 160 kJ per mole. This is indicated by the ring in the centre of the ring either in skeletal formula or structural formula

    • Apart from the thermochemical evidence argued above, actual bond length measurements from X–ray crystallography back up the aromatic ring theory and show that there are six carbon carbon bonds all of equal length and intermediate between single and double bonds – this gives the 'benzene or aromatic ring' a symmetrical hexagonal shape.

      • Typical bond lengths: single bond C–C is 0.154 nm (bond order 1) e.g. in alkanes, aromatic ring bond is 0.139 nm (bond order 1.5) e.g. in benzene or methylbenzene etc. and a double bond C=C is 0.134 nm (bond order 2)  e.g. in alkenes. Incidentally the triple bond in alkynes has a typical length of 0.120 nm (bond order 3). There term bond order refers to the 'electronsworth' of the bond i.e. the average number of shared electrons involved in that particular bond.

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


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