Colourful Solutions > Energy from fuels > Combustion

IB Chemistry Shop

Standard level

Combustion is probably the first chemical reaction that mankind knew and although it is the oldest, it is also one of the most interesting.

We live on the earth in an oxidising atmosphere that contains up to 20% oxygen. Oxygen is a very reactive gas which can oxidise even relatively unreactive substances.

Syllabus ref: R1.3.1

Reactivity 1.3.1 - Reactive metals, non-metals and organic compounds undergo combustion reactions when heated in oxygen.

  • Deduce equations for reactions of combustion, including hydrocarbons and alcohols.

Guidance

Tools and links

  • Reactivity 2.2 - Why is high activation energy often considered to be a useful property of a fuel?
  • Reactivity 3.2 - Which species are the oxidizing and reducing agents in a combustion reaction?

Combustion of metals

Reactive metals will burn readily in oxygen. The group 1 metals all burn easily in air.

Oxygen is able to remove electrons from the active metals leaving positive metal ions and creating oxides.

Sodium forms a simple oxide

4Na + O2 → 2Na2O

But in excess oxygen

Sodium forms a peroxide

2Na + O2 → Na2O2

Group 2 metals will also burn, but require a little more activation to start the reaction.

Calcium, strontium and magnesium form simple oxides. Most students are familiar with the blinding light produced when magnesium burns.

Magnesium reacts in air

2Mg + O2 → 2MgO

Aluminium, from group 13 will burn in air when powdered, and its a similar story for zinc and iron.

The oxides formed by group 1 metals are all highly basic, while those of group 2 metals are weakly basic. The peridic trend is clear.

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Combustion of non-metals

Some, but not all, non-metallic elements burn in oxygen. Those that do produce acidic oxides. The strength of acidity increases with increasing oxidation state and the periodic trend towards the right hand side.

Carbon burns in air to produce carbon dioxide if the supply of air is adequate, or carbon monoxide if the air supply is restricted.

Carbon forms carbon dioxide in excess air

C(s) + O2(g) → CO2(g)

Carbon forms carbon monoxide in limited air

2C(s) + O2(g) → 2CO(g)

Carbon dioxide is weakly acidic, while carbon monoxide is neutral.

Both phosphorus and sulfur burn readily in air forming acidic oxides.

Phosphorus forms phosphorus(V) oxide in excess air

4P(s) + 5O2(g) → P4O10(s)

Sulfur forms sulfur(IV) oxide in excess air

S(s) + O2(g) → SO2(s)

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Combustion of hydrocarbons

Since the 19th century hydrocarbons have been extracted from the earth to use as fuels for light and heat, and in the 20th century to power internal combustion engines.

These hydrocarbons, known as fossil fuels, burn easily in sufficient air, making carbon dioxide and water.

Generating combustion equations

In complete combustion, all of the carbon atoms in the fuel molecule ends up as carbon dioxide and all of the hydrogen atoms ends up in water molecules.

Combustion of propane, C3H8

There are 3 carbon atoms so 3 molecules of carbon dioxide must be made. There are 8 hydrogen atoms, enough to make 4 water molecules. So far we have:

C3H8 + oxygen → 3CO2(g) + 4H2O(l)

Now we must count up the oxygen atoms in the product to find how many molecules of oxygen are needed on the left hand side.

3CO2(g) + 4H2O(l) contains 10 oxygen atoms = 5O2

So the final equation is:

Complete combustion of propane

C3H8 + 5O2(g) → 3CO2(g) + 4H2O(l)

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Combustion of other organic compounds

Most organic compounds burn in air due to their high carbon and hydrogen content. Probably the most well known example of non-flammable organics are the chloro- and fluorocarbons. Tetrachloromethane finds use in fire extinguishers.

The alcohols are flammable and ethanol is used as a 50% fuel additive to petrol to make a mixture that is suitable for internal combustion, "gasahol".

Complete combustion of ethanol

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

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Worked examples

Q461-01 Which of the changes below occurs with the greatest increase in entropy?
  1. Na2O(s) + H2O(l) → 2Na+(aq) + 2OH-(aq)
  2. NH3(g) + HCl(g) → NH4Cl(s)
  3. H2(g) + I2(g) → 2HI(g)
  4. C(s) + CO2(g) → 2CO(g)
Answer

Entropy can be considered the degree of disorder of a chemical system. It is increased by the number of particles and their temperature. In this case it is important to examine the number of moles of free particles on both sides of the equation.

It may be seen that in equation D there are more moles of gas (maximum entropy) on the right hand side than on the left hand side. Thus the entropy increases from left to right. correct response

Although there are more free ions in A this is not as important in entropy terms as an increase in the number of moles of gas.

In equation B there is a large decrease in entropy (two gases make a solid) and in equation C the number of moles of gas on both sides is equal.

Q461-02 In which of the following reactions is the entropy change ( S) closest to zero
  1. SO2(g) + ½O2(g) → SO3(g)
  2. Br2(l) → Br2(g)
  3. H2(g) + I2(g) → 2HI(g)
  4. 3Ca(s) + N2 → Ca3N2(s)
Answer

Entropy can be considered the degree of disorder of a chemical system. It is increased by the number of particles and their temperature. In this case it is important to examine the number of moles of free particles, i.e. gas, on both sides of the equation.

Equation A the moles of gas decreases from reactants to products, ΔS is negative.

Equation B the moles of gas increases from 0 to 1, ΔS is positive.

Equation C the moles of gas stays the same from reactants to products, ΔS = 0. correct response

Equation D the moles of gas decreases from 1 to 0, ΔS is negative.
Q461-03 Estimate, without doing a calculation, the magnitude of the entropy change for the following reaction.
Fe2O3(s) + 2Al(s) 2Fe(s) + Al2O3(s)
Answer

Examination of the equation reveals that the compounds on both sides of the equation are in the solid state. As solids have very low entropy it is safe to estimate that the entropy difference between reactants and products is negligible. Hence ΔS = 0.
Q461-04 Consider the following reaction:
N2(g) + 3H2(g) → 2NH3(g)

The absolute entropy values, S, at 300K for N2(g), H2(g) and NH3(g) are 193, 131 and 192 JK-1 mol-1 respectively. Calculate ΔSo for the reaction and explain the sign of So.

Answer

On the left hand side there is one mole of nitrogen and three moles of hydrogen. Their entropy = 193 + (3 x 131) = 586 JK-1

On the right hand side there are two moles of ammonia. Entropy = (2 x 192) = 384 JK-1

The entropy change, ΔSo, is 384 - 586 = -202 JK-1

The negative sign indicates that the entropy has decreased from reactants to products.
Q461-05 Which reaction has the greatest positive entropy change?
  1. CH4(g) + 1½O2(g) → CO(g) + 2H2O(g)
  2. CH4(g) + 1½O2(g) → CO(g) + 2H2O(l)
  3. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
  4. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Answer

A positive entropy change means that the products have more entropy than the reactants. Gases have the largest entropy values, therefore we are looking for the reaction that produces the greatest positive change in moles of gas.

reaction 1 2½ moles gas → 3 moles of gas. An increase by ½ mole gas correct response

reaction 2 2½ moles gas → 1 mole of gas. A decrease of 1½ mole gas

reaction 3 3 moles gas → 3 moles of gas. No change in moles

reaction 4 3 moles gas → 1 mole of gas. A decrease of 2 moles of gas
Q461-06 Which reaction occurs with the largest increase in entropy?
  1. Pb(NO3)2(s) + 2KI(s) → PbI2(s) + 2KNO3(s)
  2. CaCO3(s) → CaO(s) + CO2(g)
  3. 3H2(g) + N2(g) → 2NH3(g)
  4. H2(g) + I2(g) → 2HI(g)
Answer

An increase in entropy change means that the products have more entropy than the reactants. Gases have the largest entropy values, therefore we are looking for the reaction that produces the greatest positive change in moles of gas.

reaction 1 0 moles gas → 0 moles of gas. No change in moles of gas

reaction 2 0 moles gas → 1 mole of gas. A increase of 1 mole of gas correct response

reaction 3 4 moles gas → 2 moles of gas. A decrease of 2 moles of gas

reaction 4 2 moles gas → 2 mole of gas. No change in moles of gas
Q461-07 Some chlorine gas is placed in a flask of fixed volume at room temperature. What change will cause a decrease in entropy?
  1. Adding a small amount of hydrogen
  2. Adding a small amount of chlorine
  3. Cooling the flask
  4. Exposing the flask to sunlight
Answer

Anything that increases the disorder, such as mixing two gases, or increasing the temperature, increases the entropy. The reverse is also tru. Hence decreasing the temperature decreases the entropy, e.g. Cooling the flask
Q461-08 Which reaction has the largest positive value of ΔSo?
  1. CO2(g) + 3H2(g) → CH3OH(g) + H2O(g)
  2. 2Al(s) + 3S(s) → Al2S3(s)
  3. CH4(g) + H2O(g) → 3H2(g) + CO(g)
  4. 2S(s) + 3O2(g) → 2SO3(g)
Answer

An increase in entropy change means that the products have more entropy than the reactants. Gases have the largest entropy values, therefore we are looking for the reaction that produces the greatest positive change in moles of gas.

reaction 1 4 moles gas → 2 moles of gas. A decrease of 2 moles of gas, ΔSo = negative

reaction 2 0 moles gas → 0 mole of gas. No change in moles of gas, ΔSo = 0 (approx)

reaction 3 2 moles gas → 4 moles of gas. An increase of 2 moles of gas, ΔSo = positive correct response

reaction 4 3 moles gas → 2 mole of gas. A decrease of 1 mole of gas, ΔSo = negative

Q461-09 Which equation represents a change with a negative value for ΔS?
  1. 2H2(g) + O2(g) → 2H2O(g)
  2. H2O(s) → H2O(g)
  3. H2(g) + Cl2(g) → 2HCl(g)
  4. 2NH3(g) → N2(g) + 3H2(g)
Answer

A negative value for ΔS means that the products have less entropy than the reactants. There are fewer moles of gas in the products than in the reactants.

reaction 1 3 moles gas → 2 moles of gas. A decrease of 1 moles of gas, ΔS = negative correct response

reaction 2 0 moles gas → 1 mole of gas. An increase by 1 mole of gas, ΔS = positive

reaction 3 2 moles gas → 2 moles of gas. No change in moles of gas , ΔS = 0 (approx)

reaction 4 2 moles gas → 4 mole of gas. An increase by 2 moles of gas, ΔS = positive

Q461-10 Which change does not lead to an increase in entropy?
  1. Mixing nitrogen and oxygen gases at room temperature
  2. Cooling steam so that it condenses to water
  3. Heating hexane to its boiling point
  4. Dissolving sugar in water
Answer

Entropy is increased by:

  • Temperature increase
  • Increased number of free particles
  • Mixing

From the choices given, only cooling steam reduces the entropy of the system

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