Syllabus Edition

First teaching 2023

First exams 2025

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Combustion Reactions (HL IB Chemistry)

Revision Note

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Chemistry

Combustion Reactions

  • Combustion is a relatively fast, thermochemical reaction that requires:
    • Fuel
      • These can include reactive metals, non-metals and organic compounds such as hydrocarbons and alcohols
    • Oxygen
      • This reaction with oxygen means that combustion can be categorised as oxidation
      • There are other reactions with oxygen that are oxidation but not combustion, e.g. the rusting of iron
    • A source of ignition / trigger
      • This does not automatically mean that a spark or flame is required
      • It can include the build up of heat in a volatile liquid or exposure to high levels of oxygen in certain types of coal, for example
  • It is more commonly referred to as burning
  • Combustion is accompanied by the generation of heat and light, in the form of flame
    • The heat released from combustion means that the reaction is exothermic
  • The two types of combustion to consider are:
    • Complete combustion
    • Incomplete combustion 

Combustion of metals

  • All metals can oxidise, but not all metals can combust
    • Some metals will only combust if they have a high surface area, e.g. they are finely divided as filings or a powder

Sparklers demonstrate the combustion of iron

Image of a sparkler

Photo by Jez Timms on Unsplash

Sparklers are coated with iron powder which gives the characteristic sparks when it undergoes combustion

  • Less reactive metals such as copper don't combust 
    • Copper does not combust because no flame is formed when it is directly heated in air
    • However, it does oxidise as the surface of the copper metal turns black as copper oxide is formed
  • More reactive metals such as those in the s-block will combust in air
    • The s-block metals form ionic oxides when they undergo combustion
    • The ionic oxides of these metals are basic, i.e. they will react with water to form solutions with a pH > 7
  • The standard example of a metal that combusts in air is magnesium, which burns with a bright white flame:

magnesium + oxygen →  magnesium oxide

2Mg (s) + O2 (g) → 2MgO (s)

  • So, the general word equation for the combustion of suitable metals is:

metal + oxygen →  metal oxide

Exam Tip

  • Careful: When some metals combust, they do not form the typical oxides:
    • Sodium forms sodium peroxide, Na2O2 
    • Iron forms iron(II, III) oxide, Fe3O4 
  • Since this knowledge is beyond the scope of the specification, you should achieve the marks in an exam for forming the typical oxides, e.g.
    • Sodium oxide, Na2O
    • Iron(III) oxide, Fe2O3 

Combustion of non-metals

  • Several non-metals show a variety of oxidation states in the different oxides that they form during combustion
  • p-block non-metals generally form covalent oxides when they undergo combustion
    • These covalent oxides are acidic, i.e. they will react with water to form solutions with a pH < 7
  • A common example of a non-metal that combusts in air is sulfur, which burns with a blue flame

sulfur + oxygen →  sulfur dioxide

S (s) + O2 (g) → SO2 (g) 

  • So, the general word equation for the combustion of non-metals is:

non-metal + oxygen →  non-metal oxide

Worked example

Combustion of metals and non-metals

  1. Potassium produces a lilac flame when it burns in air to form potassium oxide, K2O. Write a chemical equation for this reaction.
  2. Write a chemical equation for the combustion of white phosphorous, P4, to form phosphorous(V) oxide, P4O10.

 

Answer 1: 

  • The chemical symbol for potassium is K
  • The chemical formula for oxygen is O2 
  • The chemical formula of potassium oxide is given as K2O
  • So, the unbalanced chemical equation is:
    • K + O2 K2O
  • Doubling the K2O balances the oxygen atoms
  • Consequently, four potassium atoms are required on the reactant side to balance the equation
  • The balanced chemical equation is:
    • 4K + O2 → 2K2O

Answer 2: 

  • The chemical formula for white phosphorous is given as P4
  • The chemical formula for oxygen is O2 
  • The chemical formula of phosphorus(V) oxide is given as P4O10 
  • So, the unbalanced chemical equation is:
    • P4 + O2  P4O10 
  • The phosphorous atoms are balanced on both sides of the equation
  • There are 10 atoms of oxygen on the products side of the equation, which means that 5O2 are required on the reactant side to balance the equation
  • The balanced chemical equation is:
    • P4 + 5O2 → P4O10 

Complete combustion of organic compounds

  • Many organic compounds are used as fuels because they release relatively large amounts of energy when combusted
  • They do not usually undergo spontaneous combustion because their combustion reactions have a high activation energy
    • This makes fuels easy and safe to transport and store
    • For more information about activation energy, see our revision note on activation energy
  • The organic compounds that are commonly used as fuels include:
    • Hydrocarbons - particularly alkanes
    • Alcohols

What is complete combustion?

  • When fuels such as hydrocarbons and alcohols are burnt in excess (plenty of) oxygen, complete combustion takes place
    • This means that all carbon and hydrogen will be oxidised
    • Therefore, the products of complete combustion are carbon dioxide and water 
  • The word equation for complete combustion is:

fuel + oxygen → carbon dioxide + water

Combustion of hydrocarbons

  • For example, the word and chemical equations for the complete combustion of methane are:
    • Complete combustion of methane word equation:

methane + oxygen → carbon dioxide + water

    • Complete combustion of methane chemical equation:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

Combustion of Alcohols

  • Alcohols react with oxygen in the air when ignited and undergo complete combustion to form carbon dioxide and water

alcohol + oxygen → carbon dioxide + water

  • For example, the word and chemical equations for the complete combustion of ethanol are:
    • Complete combustion of ethanol word equation:

ethanol + oxygen → carbon dioxide + water

    • Complete combustion of ethanol chemical equation:

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)

Exam Tip

Be careful when balancing equations for the combustion of alcohol, as students often forget to count the oxygen in the alcohol

  • Lower alcohols burn with an almost invisible flame and make good fuels
  • Ethanol can be produced sustainably as a fuel by the fermentation of sugars
  • However, the energy density (the amount of energy in kJ per kg of fuel) is lower than gasoline so cars that run on ethanol must either have a larger fuel tank or fill up more often
  • Blending ethanol with gasoline or diesel increases the energy density and makes it safer in case of fires as it is easier to see the flames compared to pure ethanol burning
  • However, there are socio-economic concerns about using large quantities of farmland to produce crops for fermentation, which could be better used for food production

Worked example

Complete combustion of hydrocarbons and alcohols

Write chemical equations for the complete combustion of:

  1. Propane, C3H8 
  2. Propan-1-ol, C3H7OH

 

Answer 1: 

  • Since this is complete combustion, the products will be carbon dioxide and water
  • So, the unbalanced chemical equation is:
    • C3H8 + O2 CO2 + H2O
  • The 3 carbons in propane will form 3CO2 
  • The 8 hydrogens in propane will form 4H2O
  • This updates the unbalanced chemical equation to:
    • C3H8 + O2 3CO2 + 4H2O
  • There are now 10 oxygens in total on the product's side, which means that 5O2 are required on the reactant side to balance the equation
  • The balanced chemical equation is:
    • C3H8 + 5O2 → 3CO2 + 4H2O

Answer 2: 

  • Since this is complete combustion, the products will be carbon dioxide and water
  • So, the unbalanced chemical equation is:
    • C3H7OH + O2 CO2 + H2O
  • The 3 carbons in propan-1-ol will form 3CO2 
  • The 8 hydrogens in propan-1-ol will form 4H2O
  • This updates the unbalanced chemical equation to:
    • C3H7OH + O2 3CO2 + 4H2O
  • There are now 10 oxygens in total on the product's side AND one oxygen on the reactants side
    • This means that 4½O2 are required on the reactant side to balance the equation
  • The balanced chemical equation is:
    • C3H7OH + 4½O2 → 3CO2 + 4H2O
      OR
      2C3H7OH + 9O2 → 6CO2 + 8H2O, giving whole number coefficients

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Richard

Author: Richard

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.