1.4 Some enthalpy data patterns

1.4a The combustion of linear alkanes and linear aliphatic alcohols

The standard enthalpies of complete combustion (at 298K, 1 atm = 101kPa) are listed below (4 sf)

C. no.	alkane	formula	Mr	ΔHcomb kJ/mol	ΔHcomb kJ/g alkane	ΔHcomb  kJ/g CO2 produced	alcohol	formula	ΔHcomb kJ/mol
1	methane	CH4	16	–890	-55.6	-20.2	methanol	CH3OH	–726
2	ethane	C2H6	30	–1560	-52.0	-17.7	ethanol	CH3CH2OH	–1367
3	propane	C3H8	44	–2219	-50.4	-16.8	propan–1–ol	CH3(CH2)2OH	–2021
4	butane	C4H10	58	–2877	-49.6	-16.3	butan–1–ol	CH3(CH2)3OH	–2676
5	pentane	C5H12	72	–3509	-48.7	-16.0	pentan–1–ol	CH3(CH2)4OH	–3329
6	hexane	C6H14	86	–4163	-48.4	-15.8	hexan–1–ol	CH3(CH2)5OH	–3984
7	heptane	C7H16	100	–4817	-48.2	-15.6	heptan–1–ol	CH3(CH2)6OH	–4638
8	octane	C8H18	114	–5470	-48.0	-15.5	octan–1–ol	CH3(CH2)7OH	–5294
20	n-eicosane	C20H42	282.6	-13316	-47.1	-15.1

Note for reference:

For burning pure carbon the energy release figures are 32.75 kJ/g of carbon and 8.93 kJ/g of CO₂ emitted.

See note on global warming and carbon dioxide emissions

 

General formula of these homologous series: Alkanes C_nH_2n+2 and aliphatic alcohols H(CH₂)_nOH

and the general equations for complete combustion can be represented as ... (n = 1, 2, 3 etc.)

alkanes: C_nH_2n+2(g/l) + (1¹/₂n + ¹/₂)O_2(g) ===>  nCO_2(g) + (n + 1)H₂O_(l)

alcohols: H(CH₂)_nOH_(l) + 1¹/₂nO_2(g) ===> nCO_2(g) + (n + 1)H₂O_(l)

Graph interpretation and comments

The graph of ΔH_comb versus the number of carbon atoms shows an almost linear relationship as the combustion of each extra –CH₂– unit usually contributes an extra 632–670kJ to the molar enthalpy of combustion.

The first incremental rise in ΔH_c from C₁ to C₂ is slightly anomalous in both homologous series compared to the general trend.

I don't think this is particularly important, but it may due to the highest H/C ratio or the fact that the first molecule in each series doesn't have a C-C bond, whereas the rest have a carbon chain of >1 C atoms.

For the first 8 alkanes, this incremental rise ranges from 632 kJ to 670 kJ. For methane ==> ethane the incremental rise is 670 kJ. The increment for butane ==> pentane is 632 kJ and this lesser incremental rise corresponds to a the first change in state involved i.e. some of the energy released on burning pentane must be used to vapourise it and evaporation is an endothermic process. In fact ΔH_vap(C₅H₁₂) is +36 kJ mol^–1. This absorbed energy is not required by methane ==> butane which are already in the gaseous state. Apart from these two small anomalies all the other incremental rises are 653–658 kJ.

In the case of the first 8 alcohols, all liquids at 298K 101kPa, apart from the incremental rise of 641 kJ from methanol to ethanol, all the other incremental rises up this homologous series are 653–656 kJ and these are completely consistent with incremental rises for most alkane.

For the same carbon number (n) the values for alcohols are slightly smaller than those for alkanes because the alcohols are already partially oxidised i.e. the presence of a single oxygen atom in each alcohol molecule.

 

Note on global warming and carbon dioxide emissions

In the debate on fossil fuels its often quoted that natural gas (mainly methane) is 'greener' than heavier fuel oils, so I thought I'd put a data input to justify the statement or otherwise.

 

(a) In the table of alkane combustion data I've worked out the energy released per gram of pure alkane fuel.

This ranges from 55.6 kJ/g for methane down to 47.1 kJ/g for a long chain hydrocarbon (equates to a heating oil etc.).

Therefore a heavy fuel oil - central heating oil gives about 85% of the energy per unit mass compared to methane from natural gas.

So on this a basis there a case for methane being 'greener' on heat released per unit mass of fuel.

 

(b) Also, in the table of data, I've worked out the energy released per gram of carbon dioxide formed.

This ranges from 22.2 kJ/g CO₂ for methane down to 15.1 kJ/g CO₂ for a long chain hydrocarbon (equates to a heating oil etc.).

So a heavy fuel oil - central heating oil gives about 68% of the energy per unit mass of carbon dioxide emitted compared to methane from natural gas.

So on this a basis, there is an even greater case for methane being 'greener' on heat released per unit of carbon dioxide released into the environment.

 

I'm not using these figures to argue the case for continuing to use large quantities of fossil fuels for power generation or heating, but methane is definitely 'greener' than other hydrocarbon fuels.

Methane is also much more 'greener' than coal, judging from my calculations in (c) below!

 

(c) Burning theoretically 'pure' coal i.e. burning pure carbon

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

Energy released in terms of molar mass in grams:

393/12 = 32.8 kJ/g of carbon fuel burned, 49% energy released compared to methane.

393/44 = 8.93 kJ/g of carbon dioxide emitted, 44% energy released compared to methane.

Apart from the greater pollution from burning coal, it does not release nearly as much energy per mass of fuel or CO₂ emitted, compared to hydrocarbon fuels.

The difference is mainly due to the fact there is no hydrogen (and little in coal) to oxidise to water.

It should also be born in mind that coal is much less pure than processed natural gas, so more potentially more polluting and less energy releasing.

Burning coal doesn't produce water vapour, itself a greenhouse gas, but, unlike carbon dioxide, water can condense out until the equilibrium vapour pressure is achieved at the ambient temperature. On the other hand atmospheric carbon dioxide concentration can continue to build up and build up, and it is!

See Greenhouse effect, global warming, climate change, carbon footprint from fossil fuel burning

Energetics-Thermochemistry-Thermodynamics Notes INDEX

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1.4b Some patterns in bond enthalpies, bond length and bond order

1.4b(i) Examples of single versus multiple bond enthalpies

For the same pair of atoms (similar/dissimilar) the bond length shortens and the bond enthalpy increases in going from a single to double to triple bond (1, 2 and 3 electron pairs involved).

This is a rule in chemistry which is always true! The reason is quite simple. A covalent bond results from the sharing of electrons, which is actually the mutual attraction of two positive nuclei for the negative electrons between them.

The greater the number of electrons between the two nuclei the stronger the attraction between them and the two positive nuclei of the bonded atoms.

Therefore the nuclei are drawn together more closely to give a shorter bond length and more energy is required to 'pull them apart' i.e. a greater bond enthalpy (bond dissociation energy).

The number of bonding electrons for a particular bond divided by 2 is referred to as the bond order.

In other words bond order is the equivalent number of bond pairs involved between a bond between two atoms.

Bond order may not be needed for UK A level students but it is a useful and interesting concept.

e.g. look at the consistent pattern-trend for the five carbon-carbon bonds quoted.

Bond (must be based on a gaseous species)	Bond order	Bond enthalpy (kJmol–1)	Bond length (nm)	Examples (must be a gaseous species)
single bond C–C	1	+347	0.154	alkanes/diamond
CC	1.33	? not a gaseous species	0.142	graphite
CC	1.5	average +518	0.140	benzene compounds
double bond C=C	2	+612	0.134	alkenes
triple bond CC	3	+838	0.120	alkynes
single bond C–O	1	+358	0.143	alcohols/ethers
double C=O (not CO2)	2	+743	0.122	aldehydes/ketones
triple bond CO	3	+1077	0.113	carbon monoxide
single O–O	1	+146	0.148	peroxides
double O=O	2	+496	0.121	oxygen
single bond N–N	1	+163	0.146	hydrazines
double bond N=N	2	+409	0.120	azo compounds
triple NN	3	+944	0.110	nitrogen

Footnotes on bond order, bond enthalpy and bond length

(i) in CC the dots indicate the delocalised electrons (on average < 1) of the pi bond AND pm = nm x 1000.

(ii) In carbon dioxide, bond enthalpy of C=O is +805 kJ/mol, bond length is 0.116 nm

(iii) The triple bond of carbon monoxide comprises a double covalent bond plus a dative covalent bond.

(iv) Explaining the sequence for carbon-carbon bonds in increasing bond order for:

All arguments are based on carbon having four valency electrons, AND, all used in the various C-C bonding situations described below.

There is a consistent pattern-trend for the five carbon-carbon bonds quoted. 'On average', as the number of electrons each carbon atom contributes to the specified carbon-carbon bond increases, the bond length decreases and the bond strength (bond enthalpy) increases.

The more electrons between the two positive carbon nuclei, the greater the attractive force - the carbon atoms are more strongly attracted, shortening the bond length and more energy needed to 'pull them apart'.

 

Alkanes (C-C bond order 1)

Saturated hydrocarbons with only single C-C bonds

or or for the alkane ethane

and its no different for propane etc.

The C-C bond length is similar in diamond (an allotrope of the element carbon).

Each carbon atom, contributes one electron to each carbon-carbon bond.

 

Graphite (allotrope of carbon) (bond order 1.33, can you see why?)

In graphite, three C-C bonds emanate from each carbon atom using three of carbon's four valency electrons. However, the fourth delocalised electron is shared between three C-C bonds, so the bond order is actually 1.33.

I have seen this quoted as 1.5 on the internet, but this is incorrect, and graphite's electronic structure is being confused with that of benzene (see below).

Each carbon atom, 'on average', contributes 1.33 (1 + 1/3) electrons to each carbon-carbon bond.

You can see why the bond order can't be 1 or 2, but intermediate, by drawing a sort of Kekule style structure (above right)  for a layer of graphite/graphene, so it looks a bit like conjoined benzene rings.

The more usual simplified structure (above left) seems to indicate three single bonds to each carbon atom. Three bond yes, but strictly speaking, not single bonds, since there is pi bonding involved due to delocalised electrons.

Unfortunately, for 'necessary simplicity', the 4th delocalised electron doesn't show up in the 'usual' diagrams for graphite or graphene. Such diagrams indicate, 'incorrectly', that a carbon atom forms three single bonds with three other carbon atoms.

See also

Covalent Bonding – macromolecules and giant covalent structures

From fullerenes & bucky balls to carbon nanotubes

Graphene, graphene oxide and fluorographene

 

Benzene ( bond order 1.5)

The original Kekule style of displaying benzene was

i.e. showing at as cyclotriene, this ignores the delocalised ring of pi electrons, but, it does show clearly the average bond order is 1.5 as each carbon atom contributes, on average, 1.5 electrons to each bond.

Benzene molecular formula C₆H₆, skeletal formula , structural/displayed formula

Each carbon atom of the bonds 'on average' contributes 1.5 electrons (1 + 1/2) to each carbon-carbon bond.

The confusion on bond order between graphite/graphene and benzene arises from the fact that in the former, each carbon atom is linked to three other carbon atoms, but in benzene, each carbon atom is only linked to two other carbon atoms. In other words the fourth (delocalised) electron is distributed in bonding between 3 or 2 other carbon atoms - and this makes a subtle difference in electron sharing!

These ideas are important when considering the 'lower than expected' enthalpy of hydrogenation of benzene.

 

Alkenes (>C=C< bond order 2)

All alkenes have at least one carbon-carbon double bond

or or  

similarly for other alkenes like propene etc.

Each carbon atom of the double bond contribute two electrons to each carbon-carbon bond.

One electron contributes to the sigma bond and the other to the pi bond.

 

Alkynes (-C≡C- bond order 3)

and alkynes e.g. ethyne and propyne have a triple bond between two adjacent carbon atoms in the carbon atom chain.

Each carbon atom of the tripe bond contribute three electrons to each carbon-carbon bond.

One electron contributes to the sigma bond and the other two to the 'double' pi bond.

 

(iv) Reactivity patterns linked to bond order:

Since reactions usually involve collision and initiated by bond fission you might think that single bonds would automatically be more reactive i.e. have a lower activation energy due to a smaller bond enthalpy.

BUT, particularly in organic chemistry, the nature of the 'attacking' reagent is a major factor in the feasibility of a reaction.

For example, unsaturated alkenes (>C=C< functional group) and alkynes (–CC– functional group) are much more reactive than saturated alkanes with only single C–C bonds.

Despite being a 'double bond', the pi electron clouds of the unsaturated hydrocarbons are very susceptible to attack by electrophilic (electron pair seeking) reagents like bromine Br₂, hydrogen bromide HBr etc.

The polarised carbonyl group (>C^δ+=O^δ–) in aldehydes and ketones is susceptible to attack at the δ+ carbon by nucleophilic electron pair donors and much more so than the similarly polarised C^δ+–O^δ– bond in alcohols (C^δ+–O^δ––H) or ethers (C^δ+–O^δ––C).

However in the more inorganic situations the expected pattern is observed.

Nitrogen, with its triple bond is extremely stable, hence the need for a catalyst and high temperature to make it combine with hydrogen in the Haber Synthesis of ammonia.

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1.4b(ii) Some Group VII (Group 17) Halogens trends

Halogen X	fluorine		chlorine		bromine		iodine
molecule or bond	bond length nm	bond enthalpy kJmol–1	bond length nm	bond enthalpy kJmol–1	bond length nm	bond enthalpy kJmol–1	bond length nm	bond enthalpy kJmol–1
X–X, X2	0.142	+158	0.199	+242	0.228	+193	0.267	+151
H–X, HX	0.092	+562	0.128	+431	0.141	+366	0.160	+299
C–X, R–X	0.138	+484	0.177	+338	0.193	+276	0.214	+238

Some general observations, most of which relate to smaller radii giving shorter stronger bonds:

Halogen molecules X₂

From fluorine to iodine the bond length increases and, except for fluorine, the bond enthalpy decreases as the radius of the halogen atom increases with increasing number of filled inner electron shells.

Fluorine is distinctly anomalous with a much lower than expected bond dissociation energy, though the bond length fits the general trend.

This is explained by the close proximity of the small fluorine atoms causing repulsion between them due to the closeness of the outer electron orbitals.

Hydrogen halides HX

From hydrogen fluoride HF_(g) to hydrogen iodide HI_(g), there is clear trend in increasing bond length and decreasing bond enthalpy.

One result is the increasing ease of aqueous ionisation from hydrofluoric acid to hydriodic acid so that the HX_(aq) acids become stronger down the group.

In fact, hydrofluoric acid HF_(aq) is a relatively weak acid but hydrochloric, hydrobromic and hydriodic acids are all very strong.

The latter three are so strong in aqueous media you don't really see the difference e.g. from pH readings, but in non–aqueous media the differences can be clearly measured.

Halogenoalkanes R₃C–X

Based on polarisation of the bond (C^δ+–X^δ–), you might expect the reactivity order with respect to nucleophiles (electron pair donors) attacking the δ+ carbon bond to be R–F > R–Cl > R–Br > R–I as the electronegativity difference decreases from C–F to C–I.

However, it is the decreasing bond enthalpies from C–F to C–I that override this polarisation trend giving the reactivity trend R–I > R–Br > R–Cl > R–F.

See Nucleophilic substitution in halogenoalkanes

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1.4b(iii) The effect of bond polarity, electronic situations and problems in using average bond enthalpies

If the electronegativity difference between two atoms of a covalent bond increases then the polarity of the bond increases but does the bond enthalpy increases with this increased 'ionic' character?

Bond	Atomic covalent  radius nm	Electronegativity difference	Bond enthalpy	bond length
C–H	C = 0.077 nm	0.4	+413	0.109 nm
N–H	N = 0.074 nm	0.9	+388	0.101 nm
O–H	O = 0.074 nm	1.4	+463	0.096 nm
F–H	F = 0.072 nm	1.9	+562	0.092 nm

From N-H to F-H there is an increase in bond energy as the bond polarity increases with an increasing difference in electronegativity, BUT ...

(a) This does also coincide with a decreasing covalent atomic radius across Period 2 which would contribute to a decrease in bond length and the increase in bond enthalpy of X–H from left to right - which is a general expected trend.

(b) For the ~non–polar C–H bond, the bond enthalpy is +413 which doesn't quite fit in with the trend.

(c) If the polarity of the bond is 'shared out' the bond energy decreases e.g.

(i) P–Cl bond energy in gaseous PCl₃ is +319, but the P–Cl bond energy in gaseous PCl₅ is only +258 kJmol^–1.

Although both are covalent molecules in the gaseous state there is a significant electronic structure difference which results in quite different bond enthalpy values.

(ii) The three titanium chlorides show a similar pattern

The Ti–Cl bond enthalpy values are +502 in TiCl₂, +456 in TiCl₃ and + 427 kJmol^–1 in TiCl₄.

 

(d) These examples also illustrate the difficulties of using average bond enthalpies in theoretical calculations

Like it or not, the actual bond enthalpy of an 'A–B' bond  is quite dependent on the particular 'electronic' situation even for a particular pair of covalently bonded atoms A and B.

This point can further be emphasised by considering the stepwise deprotonation of methane in which the enthalpy of each step corresponds to the particular C–H bond enthalpy of the homolytic fission of each individual C–H bond.

CH_4(g) ==> CH_3(g) + H_(g)  ΔH^θ₂₉₈(C–H bond) = +425 kJmol^–1

CH_3(g) ==> CH_2(g) + H_(g)  ΔH^θ₂₉₈(C–H bond) = +470 kJmol^–1

CH_2(g) ==> CH_(g) + H_(g)   ΔH^θ₂₉₈(C–H bond) = +416 kJmol^–1

CH_(g) ==> C_(g) + H_(g)       ΔH^θ₂₉₈(C–H bond) = +335 kJmol^–1

The average of these values is 411.5, but look at the variation!, one need say no more!

A Hess's law cycle thermochemical calculation gives an average C–H bond enthalpy of +415.5 kJmol^–1 for methane.

 

Other examples of electronically different situations for the same bond:

For the O–H bond in water (HO–H) is +494 kJmol^–1, but for the O–H bond in the OH radical itself the bond enthalpy is +430 kJmol^–1.

The C=O bond enthalpy in carbon dioxide (OC=O) is +531kJmol^–1, but, for the C=O bond in carbon monoxide itself, the bond enthalpy is +1075 kJmol^–1. This is due to the C-O bond in carbon monoxide being a triple bond (it involves a dative covalent bond as well as the expected double bond with oxygen.

 

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Enthalpy calculation problems with worked out answers – based on enthalpies of reaction, formation, combustion

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Key revision points for this page on (a) enthalpies of linear alkanes and (b) linear alcohols and bond enthalpy versus bond length

Students should understand that

(i) the enthalpy of combustion of alcohols and alkanes becomes more exothermic with increasing chain length due to more C–C and C–H bonds available for oxidation

(ii) bond enthalpy trends correlate with bond length (shorter bonds = stronger bonds).

Exams test both the quantitative calculations and the conceptual reasoning behind these patterns, alongside common pitfalls like incomplete combustion, misapplied sign conventions, and confusion between bond enthalpy and enthalpy change of reaction.

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(1.4a) Trends in Enthalpy of Combustion of Linear Alcohols and Alkanes

• General Trend:

  • As chain length increases, enthalpy of combustion becomes more negative (more exothermic) because more bonds are broken and formed during combustion.

  • Example: Methanol < Ethanol < Propanol in magnitude of ΔHc.

• Alcohols versus Alkanes:

  • Alcohols already contain an oxygen atom, so their combustion enthalpy per mole is slightly less exothermic than alkanes of similar chain length.

  • Alkanes release more energy per mole because they require full oxidation of only C and H atoms.

• Experimental versus Literature Values:

  • Practical calorimetry often gives lower values due to heat loss, incomplete combustion, evaporation of alcohols.

• Exam Board Focus:

  • All require practicals on combustion of alcohols, error analysis.

  • BUT, but not neglect an emphasis on thermodynamic reasoning and linking combustion enthalpies to energy density of fuels.

Misconceptions about trends in enthalpy of combustion

• Assuming enthalpy of combustion increases linearly without considering per mole of CH₂ group.

• Forgetting to use negative signs for exothermic values.

• Confusing enthalpy per mole of fuel with enthalpy per mole of carbon atoms.

• Ignoring incomplete combustion (formation of CO or C).

Exam Tips for questions involving enthalpy of reaction calculations

• Always state assumptions (complete combustion, no heat loss).

• Compare experimental versus literature values and explain discrepancies.

• Use graphs of ΔHc versus number of carbons to show linear trend.

• Be precise with units: kJ mol⁻¹.

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(1.4b) Bond enthalpy and bond length patterns

• Bond Enthalpy Definition: Energy required to break one mole of bonds in the gaseous state.

• Trend with Bond Length:

  • Shorter bonds = stronger bonds = higher bond enthalpy.

  • Example: C≡C (triple bond) > C=C (double bond) > C–C (single bond).

  • Similarly, H–F > H–Cl > H–Br > H–I due to decreasing bond strength with increasing atomic radius.

• Mean Bond Enthalpies: Used because bond strength varies depending on molecular environment (e.g., CH bonds in methane differ slightly).

• Exam Board Emphasis:

  • Calculations using average bond enthalpies to estimate ΔH.

  • BUT, don't neglect trends across the periodic table and correlation with atomic size.

Misconceptions about bond length and bond enthalpy

• Thinking bond enthalpy is the same as enthalpy change of reaction (it’s only an average measure).

• Forgetting bond enthalpies are always positive values (energy required to break bonds).

• Assuming bond enthalpy values are exact rather than averages.

• Confusing bond enthalpy trends with electronegativity trends.

Exam Tips for questions involving bond length and bond enthalpy

• Use Hess cycles or bond enthalpy calculations to estimate ΔH.

• Always apply ΔH = Σ(bonds formed -ve) – Σ(bonds broken +ve)

  • and watch the signs

• Remember bond enthalpy values are averages, so calculated ΔH values are approximate.

• Link bond length to atomic radius and number of shared electrons.

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Final thought:

Across all boards, examiners expect students to connect macroscopic trends (enthalpy of combustion) with microscopic reasoning (bond enthalpy and bond length).

Practising past papers will highlight recurring themes: calculation accuracy, clear sign conventions, and thoughtful evaluation of experimental limitations.

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