A-Level organic chemistry exam revision notes on alcohols

Organic Chemistry Part 4. The chemistry of ALCOHOLS - Enthalpy of combustion of alcohols

Part 4.4 The complete/incomplete combustion of alcohols - products, equations, enthalpies, use as fuels, octane numbers

(including a few comments on the combustion of isomeric ethers and accompanying equations)

Sub-index for this alcohol chemistry page 4.4 on enthalpies of combustion of aliphatic alcohols

4.4.0 Introduction to the combustion of alcohols

4.4.1 The combustion of alcohols and calorimetry

4.4.2 The trend in enthalpies of combustion of linear aliphatic alcohols

4.4.4 The manufacture and use of alcohols as fuels, 'knocking' effects and octane values

(brief general comments, links to relevant pages, but more detail on octane values of alcohols)

4.4.5 Comparison of the enthalpies of complete combustion of alcohols and isomeric ethers

4.4.6 Factors concerning the incomplete combustion

4.4.7 Practice exam questions on the combustion of alcohols and links to answers!

4.4.8 Learning objectives for the combustion of alcohols

For Use of Hess's Law cycle applied to combustion of alcohols, see

Advanced Introduction to enthalpy changes - enthalpies of reaction, formation, combustion

Thermochemistry - Hess's Law calculations, enthalpies of reaction, combustion, formation etc.

Bond Enthalpy Calculations (including combustion)

Enthalpy data patterns - combustion of alkanes linear aliphatic alcohols, bond enthalpies

See thermochemistry page how to calculate enthalpies of combustion using bond enthalpy calculations.

 All advanced A level organic chemistry notes

 All revision notes on the chemistry of alcohols

 Index of GCSE level: Oil - Useful Products Chemistry Revision Notes

[Author © Dr Phil Brown GRIC, PhD: Doc Brown's advanced level organic chemistry exam revision notes suitable for students of UK A level chemistry courses & US K12 grade 11, grade 12 and AP honors chemistry courses: The chemistry of alcohols - enthalpy of combustion [alcohols page 4. RE-EDIT]

4.4.0 Introduction to the combustion of alcohols

Simple enthalpy level diagrams for a very exothermic reaction like the combustion of alcohols.

Note the hump for the activation energy, which prohibits any 'spontaneous' combustion between an alcohol and oxygen containing air!

So, although the reaction is very exothermic, alcohols will not burn unless ignited by a spark (e.g. in car fuel formulation) or application of  naked flame.

In a good supply of air of oxygen, alcohols usually burn completely to carbon dioxide and water, but lack of oxygen can lead to the formation of carbon monoxide or soot (mainly carbon).

 

The standard Enthalpy of Combustion ΔH^θ_{c or combustion} of an alcohol

is the enthalpy change when 1 mole of an alcohol is burned completely in oxygen (or air containing oxygen) computed for standard conditions (298K/25^oC, 1 atm/101kPa).

You should ensure just 1 mole of fuel appears in the equation to accompany the delta H value which is always negative i.e. always exothermic.

CH₃CH₂OH_(l) + 3O_2(g) ===> 2CO_2(g) + 3H₂O_(l)

 

A note on the kinetics of alcohol combustion

Although the combustion is highly exothermic, alcohols do not spontaneously ignite at room temperature.

From the collision theory of chemical reactions you should appreciate that an activation energy is required.

Despite alcohols being thermodynamically unstable in the presence of oxygen, the C-C, C-H and C-O bond enthalpies in alcohols are very high and the activation energy is correspondingly very high to overcome these strong bonds.

However, once initiated, the combustion is rapid and complex via many free radical reactions.

This is an excellent case of thermodynamic feasibility versus inhibiting kinetics!

4.4.1 The combustion of alcohols and calorimetry

• When burned, ethanol, like any alcohol, on complete combustion forms carbon dioxide and water
  • (i)  ethanol + oxygen ===> carbon dioxide + water
    • CH₃CH₂OH_(l) + 3O_2(g) ===> 2CO_2(g) + 3H₂O_(l)
    • As mentioned in section Fuels Survey, ethanol can be blended with petrol to fuel road vehicles.
  • Similarly, but the symbol equations may be more awkward to balance  ...
    • (ii)  methanol + oxygen ===> carbon dioxide + water
      • 2CH₃OH_(l) + 3O_2(g) ===> 2CO_2(g) + 4H₂O_(l)
    • (iii)  propan-1-ol + oxygen ===> carbon dioxide + water
      • 2CH₃CH₂CH₂OH_(l) + 9O_2(g) ===> 6CO_2(g) + 8H₂O_(l)
      • CH₃CH₂CH₂OH_(l) + 4.5O_2(g) ===> 3CO_2(g) + 4H₂O_(l)
    • (iv)  butan-1-ol + oxygen ===> carbon dioxide + water
      • CH₃CH₂CH₂CH₂OH_(l) + 6O_2(g) ===> 4CO_2(g) + 5H₂O_(l)
    • See also a Fuels Survey, ethanol can be blended with petrol to fuel road vehicles.
    • As already mentioned, ethanol is used in sprit burners where it burns much more cleanly with a blue flame - using a hydrocarbon in the same situation is more smelly and gives a more yellow smoky flame - less efficient combustion.
  • Measuring the enthalpy of combustion of alcohols
    • This can be determined using the simple copper calorimeter (diagram on the right).
    • You can compare the heat energy released by different alcohols e.g. methanol, ethanol, propanols and butanols.
    • The alcohol is poured into a little spirit burner which is then weighed.
    • The burner is placed under the copper calorimeter, which is filled with a known mass of water at a known start temperature.
    • After burning for e.g. 5 minutes, the flame is blown out and the final temperature noted.
    • The burner reweighed and the mass decrease is equal to the mass of alcohol burned.
    • From the mass of water, the heat capacity of water and the temperature change you can work out the heat energy released.
    • You can then work out the heat released per gram or per mole of alcohol.
    • Or more simply, you might just compare the mass of fuel burned to give the same temperature rise.
    • The results should show that the efficiency of an alcohol fuel combustion increases with increase in carbon chain length of the molecule i.e. in terms of J/g (J/fuel mass)
      • pentan-1-ol > butan-1-ol > propan-1-ol > ethanol > methanol
      • because the proportional 'hydrocarbon' content is increasing.
    • For lots more details on the method and calculations see
    • methods of measuring heat energy transfers in chemical reactions

  • Determining the enthalpy of combustion of an alcohol

    • 100 cm³ of water (100g) was measured into the calorimeter.

    • The spirit burner contained the fuel ethanol CH₃CH₂OH ('alcohol') and weighed 18.62g at the start.

    • After burning it weighed 17.14g and the temperature of the water rose from 18 to 89^oC.

    • The temperature rise = 89 - 18 = 71^oC (exothermic, heat energy given out).

    • Mass of fuel burned = 18.62-17.14 = 1.48g.

    • Heat given out to the water = mass of water x SHC_water x temperature change

      • = 100 x 4.18 x 71 = 29678 J (for 1.48g)

    • M_r(ethanol) = 46 (H=1, C=12, O=16)

    • Therefore 1.48g ethanol = 1.48/46 = 0.03217 mol

    • So, scaling up to 1 mole of ethanol burned gives 29678 x 1 / 0.03217 = 922536 J

    • Enthalpy of combustion of ethanol = ΔH_c(ethanol) = -923 kJmol^-1 

      • (only accurate to 3 sf at best)

    • for the reaction: CH₃CH₂OH_(l) + 3O_2(g) ===> 2CO_2(g) + 3H₂O_(l)

    • The data book value for the heat of combustion of ethanol is -1367 kJmol^-1, showing lots of heat loss in the experiment!

    • It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on combustion is known.

    • See thermodynamics page on

    • how to calculate enthalpies of combustion using bond enthalpy calculations.

    • and accurate enthalpy change values from a bomb calorimeter from which standardised international standard enthalpy values are derived.

    • Improvements (with reference to the diagram of the calorimeter)

    • (i) You can male a few improvement to the method e.g. insulating cover and draught shields and the copper calorimeter can be surrounded with a layer on non-flammable insulating material.

    • (ii) Another approach is to calibrate a specific calorimeter.

    • If using a simple calorimeter that school students can use, like the one illustrated above, you burn an alcohol whose enthalpy of combustion is known.

    • You then do the calculation of its enthalpy of combustion.

    • Suppose you only get 80% of the data book value.

    • In other words your calorimeter is 80% efficient i.e. 20% of loss of heat energy.

    • If you repeat the experiment with an unknown value of the enthalpy of combustion of the alcohol, you can then multiply the experimental value by 100/80 to achieve a more accurate estimation of the enthalpy of combustion of the alcohol.

TOP OF PAGE and sub-index

4.4.2 Trend in enthalpies of combustion of linear aliphatic alcohols

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

Graph interpretation and comments

For alkanes and linear primary alcohols, the graph of ΔH^θ_comb versus the number of carbon atoms shows an almost linear relationship as the combustion of each extra -CH₂- unit in the carbon chain 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 i.e. at least one C-C bond.

For the graph above see also Enthalpy of combustion data for linear alkanes.

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 you see for alkanes.

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.

See thermodynamics page on how to calculate enthalpies of combustion using bond enthalpy calculations.

 

Explaining the enthalpy of combustion of alcohol trend in terms of a homologous series

Obviously from the graph the molar enthalpy of combustion becomes more exothermic as the alcohol chain length increases.

This trend is explained by the addition of an extra -CH₂- unit to the next alcohol as the chain length increases.

Although this means more C-C and C-H bonds have to be broken (endothermic), the formation of extra H-O and C=O bonds (exothermic) in water and carbon dioxide, more than compensates for the extra bond breaking enthalpies required and so as you go up the homologous series of alcohols, so the molar enthalpy of combustion increases.

This trend is quite general for any homologous as the carbon chain length increases by one carbon atom (as -CH₂-) at a time.

Alcohol isomers of molecular formula C₃H₈O

propan-1-ol, CH₃CH₂CH₂OH, ΔH^θ_{comb =} -2021 kJ mol^-1

propan-2-ol, CH₃CH(OH)CH₃, ΔH^θ_{comb =} -2006 kJ mol^-1

 

Alcohol isomers of C₄H₁₀O

butan-1-ol, CH₃CH₂CH₂CH₂OH, ΔH^θ_{comb =}  -2673 kJ mol^-1

butan-2-ol, CH₃CH₂CH(OH)CH₃, ΔH^θ_{comb =} -2660 kJ mol^-1

2-methylpropan-1-ol, (CH₃)₂CH₂CH₂OH, ΔH^θ_{comb =} -2665 kJ mol^-1

2-methylpropan-2-ol,(CH₃)₃COH , ΔH^θ_{comb =} -2643.8 kJ mol^-1

 

These alcohol enthalpies of combustion values suggest a general small decrease in value trend with increase in carbon chain branching.

And the -OH in butan-2-ol or propan-2-ol is effectively a sort of branching.

The differences in the enthalpy of combustion of isomeric alcohols will be due to small 'subtle' differences in C-C, C-H or C-O bond enthalpies of the alcohol molecules.

The results is it enthalpies of combustion seem be slightly less for the 2-ols (secondary alcohols) compared to 1-ols (primary alcohols)

If you are doing a practical assignment to experimentally determine the enthalpy of combustion of alcohols, school/college experiments are not accurate enough to distinguish between isomers i.e. those alcohols with the same molecular formula.

See thermodynamics page on

how to calculate enthalpies of combustion using bond enthalpy calculations.

TOP OF PAGE and sub-index

4.4.4 The manufacture and use of alcohols as fuels and 'knocking' effects and octane values of fuels e.g. alcohols versus alkanes

Important factors for comparing the use of alcohols as fuels

Energy density e.g. J/kg

Volatility, ease of vaporisation prior to combustion.

Ease of ignition, can be related to volatility.

Carbon footprint, efficiency of production.

The practical use of alcohols in real engines, how 'smooth' does the alcohol fuel formulation burn - see notes on 'knocking' effects and the octane numbers of fuels like alcohols further down.

Notes on knocking effects in petrol engines and octane numbers of alcohol fuels

Knocking (also referred to as pinking pre-detonation) is premature auto-ignition of the petrol–air mixture in a car engine.

In a normal spark-ignition petrol engine, the mixture burns smoothly after the spark plug fires.

Under certain conditions (high temperature, high pressure, low-octane fuel), pockets of fuel mixture auto-ignite before the flame front reaches them.

These sudden pressure spikes create the characteristic “knock”.

Knocking causes loss of energy i.e. reduces useful energy conversion efficiency, increases engine component damage and produces more environmental pollution.

Modern engines use knock sensors to adjust timing and prevent it.

 

Measurement of the knocking effect of a fuels

The octane number of a fuel is a measure of a fuel's resistance to knocking and is based on a standard scale from data on specific test petrol engine.

The research octane number (RON) is defined relative to linear n‑heptane (0) and isooctane (2,2,4-trimethylpentane) (100).

Choice of alkane standards:

isooctane = 100; highly branched ==> very resistant to knocking.

heptane = 0; straight chain (linear) ==> auto‑ignites easily.

The higher the octane number of a molecule or fuel mixture, the more resistant to auto‑ignition (more resistant to knocking effects).

The octane number of an alcohol (or any fuel) is measured by running the fuel in a standardised test engine and comparing its knocking behaviour with pure or mixtures of the standard isooctane (100) and linear n-heptane (0).

 

Molecular structure factors affecting the relative octane number (RON)

1. The longer the carbon chain, the lower the octane number

You can see this in the steady trend RON table below

The longer the carbon chain, the lower the activation energy of C-C bond scission allowing the rapid, potentially uncontrolled, lots of branching free radical reactions, so less energy (and precise spark timing) to detonate.

The same rule applies to straight chain alcohols.

 

2. Branched alkanes have higher octane numbers than straight‑chain alkanes e.g. for the same molecular formula.

The more branching raises octane number compared to   straight chain alkanes with lower octane numbers (comparing isomers).

e.g. for isomers of C₈H₁₈, 2,4-trimethyl pentane is 100, straight chain octane is -25 (yes, minus 25).

Combustion is a series of rapid free radical reaction. The radicals formed by straight-chain (linear) alkanes are highly reactive, which leads to rapid, uncontrolled explosions from branched free radical chain reactions that cause premature ignition.

In contrast, highly branched alkanes such as isooctane, form more stable free radicals.  This stability slows the reaction rate and ensures the fuel burns more smoothly.

Also, long, unbranched chains have low activation energy for auto-ignition and form reactive peroxides more readily under compression - these readily split homolytically to give two free radicals.

The same rule applies to isomeric alcohols.

 

3. Arenes (aromatic hydrocarbons) have very high octane numbers

e.g. benzene C₆H₆ is ~104 (hexane is 25) and C₆H₅CH₃ methylbenzene ~117 (heptane is 0)

Aromatic hydrocarbons (like benzene and methylbenzene) have high octane numbers because their highly stable, aromatic ring molecular structures strongly resist premature self-ignition under high compression allowing for smooth, controlled burning.

When aromatic molecules are exposed to high heat and pressure, they form highly stable resonance-stabilized free radicals which are less reactive than those radicals produced by straight-chain alkanes.

 

4. Alcohols have higher octane numbers compared to alkanes due to the OH group

e.g. linear octane is -25 and octan-1-ol is 86 for C8 molecules

Alcohols resist knocking because the stronger O-H bond absorbs more heat energy and slows radical chain reactions because the high latent heat of vaporisation (due to H-bonding) cools the mixture.

Alcohols burn in a more controlled manner i.e. burn more smoothly, more cleanly (more completely) with fewer intermediate radicals.

This is why ethanol, propan-1-ol and butan-1-ol are attractive to use in biofuels mixtures.

 

Research octane number  RON Comparison Table for linear alkanes versus linear alcohols (primary alcohols 1-Alkanols from C1 to C10  (RON is essentially a relative octane number)

(PLEASE note data sources vary in quoted RON values)

Carbon atoms	Alkane	RON (approx.)	1-Alkanol	RON (approx.)
1	Methane	120+	Methanol	110
2	Ethane	105	Ethanol	109
3	Propane	110	Propan-1-ol	107
4	Butane	94	Butan-1-ol	99
5	Pentane	62	Pentan-1-ol	97
6	Hexane	25	Hexan-1-ol	90
7	Heptane	0 (definition)	Heptan-1-ol	88
8	Octane	-25	Octan-1-ol	86
9	Nonane	-40	Nonan-1-ol	84
10	Decane	-45	Decan-1-ol	82

Comments on the data table for linear alkanes and linear primary alcohols

RON values for alkanes drop sharply with increasing chain length because long straight chains auto-ignite more easily.

RON values for primary alcohols decrease only slowly with chain length because the OH group stabilises combustion and suppresses knock.

Negative RON values (e.g., for nonane, decane) mean the fuel knocks even more readily than pure heptane.

 

Comments on petrol pump formulations in the UK

In the UK, petrol formulation is a blend of hydrocarbons (mostly alkanes, cycloalkanes, and aromatics) plus regulated oxygenates including alcohols like ethanol.

Petrol formulation blend E5

Octane numbers 97 (premium) and 99 (super premium) with ≤5% v/v ethanol.

Despite having less higher octane number (109) ethanol than E10, the E5 formulation has a greater proportion of higher octane branched alkanes, which raises the octane number above that of 95 for the E10 formulation.

Petrol formulation blend E10

Octane number 95 with ≤10% v/v ethanol (quoted at 5.5-10.0%)

E5 contains more ethanol than the higher octane E5.

The source of ethanol is usually bioethanol from fermenting sugar or starch from plant based crops, but sometimes by hydration of ethene in the petrochemical industry itself.

For the industrial production of ethanol see

Alcohols - manufacture of ethanol (basic notes)

The laboratory synthesis and manufacture of alcohols (extra advanced level notes)

For a discussion on the use of ethanol as a biofuel from fermentation (biosynthetic route) or blending ethanol from the petrochemical industry with petrol see:

Biofuels & alternative fuels, hydrogen, biogas, biodiesel

and mention on 'uses of alcohols' page

and use of alcohols in fuel cells - the redox chemistry explained using an alcohol fuel

Key words and phrases: How to determine the enthalpy of combustion of linear alcohols (-1-ols), methanol, ethanol, propan-1-ol,  butan-1-ol,  pentan-1-ol,  hexan-1-ol,  heptan-1-ol,  octan-1-ol.  Data table of the enthalpy of combustion values for alcohols.  Description of a simple experiment to determine the enthalpy of combustion of alcohols. Balanced equations for the enthalpy of combustion of alcohols. How to calculate the enthalpy of combustion of alcohols from experimental data. A comparison of the enthalpy of combustion of alcohols with their isomeric ethers. Enthalpy of combustion data to compare alcohols and ethers. Graph showing the increase in the enthalpy of combustion with increase in chain length and a graph line comparing alcohols with linear alkanes. Explaining the trend in the enthalpy of combustion of linear alcohols. Explaining the importance of enthalpy of combustion of alcohols in organic chemistry, What you need to know about enthalpy of combustion of alcohols for organic chemistry, Explaining the use of enthalpy of combustion of alcohols knowledge in organic chemistry, Examples of enthalpy of combustion of alcohols explained when studying organic chemistry, What is the significance of enthalpy of combustion of alcohols in organic chemistry, What is the use of enthalpy of combustion of alcohols in organic chemistry  Describing and explaining the theory of enthalpy of combustion of alcohols when studying organic chemistry, exam revision notes for enthalpy of combustion of alcohols in exams, online help for enthalpy of combustion of alcohols, revision notes for enthalpy of combustion of alcohols, what do I need to learn for enthalpy of combustion of alcohols in exams? revision summary for enthalpy of combustion of alcohols, help in teaching enthalpy of combustion of alcohols, learning notes for enthalpy of combustion of alcohols, help to pass the enthalpy of combustion of alcohols exam, how to prepare for examination questions on enthalpy of combustion of alcohols?  Website content © Dr Phil Brown 2000+. All copyrights reserved on revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries & references to science course specifications are unofficial. Website content © Dr Phil Brown 2000+. All copyrights reserved on these organic chemistry exam revision notes on combustion equations for alcohols, these A level chemistry revision notes are suitable for use of pre-university students studying AQA advanced level organic chemistry revision notes on combustion equations for alcohols, Edexcel advanced level organic chemistry revision notes on combustion equations for alcohols, OCR advanced level organic chemistry revision notes on combustion equations for alcohols, IB advanced level organic chemistry revision notes on combustion equations for alcohols, WJEC (Eduqas) advanced level organic chemistry revision notes on combustion equations for alcohols, CIE Cambridge advanced level organic chemistry revision notes on combustion equations for alcohols, CCEA advanced level organic chemistry revision notes on combustion equations for alcohols, and useful for US grade 11 grade 12 AP honors organic chemistry courses involving combustion equations for alcohols

ANSWERS to advanced A-level Chemistry Questions on the Combustion of Alcohols

• Determining the energy change for the combustion of the alcohol butan-1-ol

• 100 cm³ of water (100g) was measured into a simple copper calorimeter.

• The spirit burner contained the fuel butan-1-ol CH₃CH₂CH₂CH₂OH and weighed 20.52 g at the start.

• The initial temperature of the water is taken.

• After burning some time, the flame is extinguished, the water stirred gently and the final water temperature is taken to get the temperature rise.

• The burner and fuel are then reweighed to see how much fuel had been burned.

• After burning it weighed 19.92 g and the temperature of the water rose from 20 to 66^oC.

• The specific heat capacity of water is 4.2 Jg^-1K^-1

• Calculate the thermal energy change of combustion in J/g butan-1-ol

• The temperature rise = 66 – 20 = 46^oC (exothermic, heat energy given out).

• Mass of fuel burned = 20.52 - 19.92 = 0.60 g.

• Heat absorbed by the water = mass of water x SHC_water x temperature rise

  • = 100 x 4.2 x 46 = 19320 J (for 1.48g)

  • heat energy released per g = energy supplied in J / mass of fuel burned in g

  • heat energy released on burning = 19320 / 0.60 = 32200 J/g of butan-1-ol

• Calculate the enthalpy change of combustion in kJmol^-1 of butan-1-ol

  • Relative atomic masses A_r: C = 12.01, H = 1.01, O = 16

  • M_r(C₄H₉OH) = (4 x 12.01) + (1.01 x 10) + 16 = 74.14, so 1 mole = 72.12 g

  • Heat energy released (given out) by 1 mole of C₄H₉OH = 74.14 x 32200

  •  = 2387308 J/mole or 2387 kJ/mol

  • ~-2390 kJmol^-1 (3 sf)

  • Enthalpy of combustion of butan=1-ol = ΔH_combustion (butan-1-ol) = ~2390 kJmol^–1

  • This means ~2390 kJ of heat energy is released on burning 74g of butan-1-ol

  • The data book value for the heat of combustion of butan-1-ol is –2676 kJmol^–1, showing lots of heat loss in the experiment!

  • It is possible to get more accurate values by calibrating the calorimeter with a substance whose energy release on combustion is known, but still not very good, a bomb calorimeter is the answer!

Q4.2(a) Write balanced equations, including state symbols, for the combustion of propan-1-ol for:

(a) complete combustion of the alcohol

(b) formation of carbon monoxide

(c) formation of carbon (soot)

Answers

(a) CH₃CH₂CH₂OH_(l) +  4.5O_2(g) ===> 3CO_2(g) +  4H₂O_(l)

(b) CH₃CH₂CH₂OH(l)  +  3O₂(g) ==>  3CO(g)  +  4H₂O(l)

(c) CH₃CH₂CH₂OH(l)  +  1.5O₂(g) ==>  3C(s)  +  4H₂O(l)

Note that hydrogen in the alcohol molecule is preferentially oxidised.

A The ΔH^θ_comb of alcohols increases ~ linearly with increase in carbon atom number of the molecule.

B The ΔH^θ_comb of alcohols is less than that for alkanes of the same carbon atom number.

C The incomplete combustion of alcohols can produce C(s) and CO(g), but not H₂(g)

D The ΔH^θ_comb of isomeric ethers and alcohols are significantly different.

E Generally speaking, the octane number of a linear alcohol is higher than that of a linear alkane molecule with the same number of carbon atoms.

ANSWER D is FALSE because there is only one significant bond difference between any pair of isomers, only alcohols have an O-H bond and only ethers have an extra C-O bond. (This results in ~ 100 kJ/mol difference in the ΔH^θ_comb values and apart from methane/methanol <10 % difference and significant decreasing % difference with increase in C number).

Q4.3(a) Given the standard enthalpies of formation at 298K and 1 atm/101 kPa pressure in kJmol^-1, use a Hess's Law cycle to calculate the enthalpy of combustion of propan-1-ol

ΔH^θ_f (propan-1-ol) = -302.7; ΔH^θ_f (carbon dioxide) = -393.5  and  ΔH^θ_f (water) = -285.8;

Construction of the Hess's Law Cycle to theoretically calculate the enthalpy of combustion of propan-1-ol
CH3CH2CH2OH(l) + 4.5O2(g)  == ΔHcf (propan-1-ol)  ==> 3CO2(g) + 4H2O(l)
-ΔHθf (propan-1-ol)				3 x ΔHθf (carbon dioxide) + 4 x ΔHθf (water)
3C(s)  +  4H2(g) + 5O2(g)

Watch out for the minus sign on the left arrow ΔH.

Therefore using the principle of a Hess's Law cycle

ΔH^θ_c (propan-1-ol)  =  -ΔH^θ_f (propan-1-ol)  +  3 x ΔH^θ_f (carbon dioxide)  + 4 x ΔH^θ_f (water)

ΔH^θ_c (propan-1-ol)  =  -302.7  +  3 x -393.5  +  4 x -285.8

ΔH^θ_c (propan-1-ol)  =  -302.7  +  1180.5  +  1143.2  = 2021 kJmol^-1

It is OK here to quote the answer to 5 sig. figs. since ALL data quoted to 4 sig. figs.

Q4.4 Using the average bond enthalpies listed below,

(a) Calculate the theoretical enthalpy change for the complete combustion of propan-1-ol.

(b) Explain why this value differs from the standard enthalpy of combustion computed in Q4.2 (and in data tables).

• C-H = 413 kJ mol⁻¹
• C-C = 348 kJ mol⁻¹
• C-O = 358 kJ mol⁻¹
• O-H = 463 kJ mol⁻¹
• O=O = 498 kJ mol⁻¹
• C=O (in CO₂) = 805 kJ mol⁻¹
• O–H (in H₂O) = 463 kJ mol⁻¹

For (a) CH₃CH₂CH₂OH +  4.5O₂ ===> 3CO +  4H₂O _{(all assumed to be in a gaseous state)}

+ 4.5 O=O  ==> 3 O=C=O  +  4 H-O-H

(i) Bonds broken = (7 x C-H)  + (1 x C-O)  +  (1 x O-H)  +  (3.5 x O=O)

= (2 x C-C)  +  (7 x C-H)  +  (1 x C-O)  +  (1 x O-H)  +  (3.5 x O=O)

= (2 x 348)  +  (7 x 413)  + (358)  +  (463)  +  (4.5 x 498)

= (696)  +  (2891)  + (358)  +  (463)  +  (1743)

endothermic change = +6649 kJ

(ii) Bonds made = (6 x C=O)  +  (8 x O-H)

= (6 x C=O)  +  (8 x O-H)

= (4830)  +  (3704)

exothermic change = -8534 kJ

Since (ii) > (i) the reaction is overall exothermic and the ? = -1885

(b) Reasons for difference in enthalpy of combustion values

(i) The calculation in (a) is based on all species being in a gaseous state, whereas for standard enthalpy of combustion values, the alcohol and water would be in the liquid state.

Energy would be absorbed to evaporate the alcohol, but energy would be released on the condensation of water.

(ii) The bond energies from data tables are based on average values, but accurate bond enthalpies for a specific bond vary slightly for different molecular context e.g. the C-O bond in propan-1-ol might be slightly different than in isomeric propan-2-ol or the methoxypropanes.

Q4.4(c) Which of the following statements about the combustion of alcohols is FALSE and why?

A The ΔH^θ_comb of alcohols increases ~ linearly with increase in carbon atom number of the molecule.

B The ΔH^θ_comb of alcohols is less than that for alkanes of the same carbon atom number.

C The incomplete combustion of alcohols can produce C(s) and CO(g), but not H₂(g)

D The ΔH^θ_comb of isomeric ethers and alcohols are quite similar.

E Generally speaking, the octane number of a linear alcohol is lower than that of a linear alkane molecule with the same number of carbon atoms.

ANSWER E is FALSE because alcohols are partially oxygenated and burn more in a more controlled manor and less prone to pre-ignition 'knocking' effects (so have higher octane number). Combustion is very complex free radical chemistry and apparently alcohols form fewer free radicals on compression - 'splintering' alkanes produce more diverse free radical pathways.

Q4.5(a) Which of the following 'fuel' molecules has the lowest octane number and which the highest octane number?

A  CH₃CH₂CH₂CH₂CH₂CH₂OH

C6 molecule, hexan-1-ol, linear primary alcohol, octane number 90, has a longer C chain than B or C  which decreases octane number and no branching.

B  (CH₃)₃CCH₂OH

C5 molecule, 2,2-dimethylpropan-1-ol, a highly branched primary alcohol, octane number 102, both branching lower C number both factors increase the octane number. B has the highest octane number

C  CH₃CH₂CH₂CH₂CH₂OH

C5 molecule, pentan-1-ol, linear primary alcohol, octane number 97, lower C number than A or D but no branching.

D  CH₃CH₂CH₂CH₂CH₂CH₂CH₂OH

C7 molecule, heptan-1-ol, linear primary alcohol, octane number 88, has longest C chain and no branching both factors that decrease the octane number of a fuel molecule. D has the lowest octane number

Q4.5(b) Which of the following statements about the combustion of alcohols is FALSE?

A The ΔH^θ_comb of alcohols increases ~ linearly with increase in carbon atom number of the molecule.

B The ΔH^θ_comb of alcohols is less than that for alkanes of the same carbon atom number.

C The incomplete combustion of alcohols can produce CO and H₂ gases.

D The ΔH^θ_comb of isomeric ethers and alcohols are quite similar.

E Generally speaking, the octane number of a linear alcohol is higher than that of a linear alkane molecule with the same number of carbon atoms.

ANSWER C is FALSE because the incomplete combustion of alcohols produce C(s) and CO(g), but not H₂(g) as hydrogen atoms are preferentially oxidised compared to carbon atoms or carbon monoxide molecules.

Q4.5(c) How many moles of oxygen are required to completely oxidise (burn)

(i) 1 mole of 3-methylbutan-1-ol?

(ii) 1 mole of octan-3-ol?

(i) 1 mole of 3-methylbutan-1-ol?

+ 7.5O₂(g) ==> 5CO₂(g)  + 6H₂O(l)  and don't forget 1 O atom already in the alcohol molecule

(ii) 1 mole of octan-3-ol?

+ 12.0O₂(g) ==> 8CO₂(g)  + 9H₂O(l)  again, don't forget 1 O atom already in the alcohol molecule