Energy from the Combustion of Fuels (VCE SSCE Chemistry): Revision Notes

Energy from the Combustion of Fuels

Introduction to combustion

Combustion is a chemical process that releases energy in the form of heat and light. Bushfires are a dramatic example of uncontrolled combustion, capable of destroying homes, habitats, and vast areas of land. Understanding the chemistry of combustion helps us control fire and use it safely as an energy source.

Combustion reactions are exothermic reactions in which a reactant combines with oxygen to produce oxides. This type of reaction is also called an oxidation reaction because oxygen is added to form oxides. When hydrocarbons undergo combustion with sufficient oxygen, the products are carbon dioxide and water.

Complete combustion

Complete combustion occurs when there is a plentiful supply of oxygen. Under these conditions, the fuel burns efficiently to produce only two products: carbon dioxide ($\text{CO}_2$) and water ($\text{H}_2\text{O}$).

For example, the complete combustion of methane can be represented by the equation:

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$

This equation shows that one mole of methane gas reacts with two moles of oxygen gas to produce one mole of carbon dioxide gas and two moles of liquid water. The products are completely oxidised, meaning no carbon monoxide or carbon particles are formed.

Incomplete combustion

When the oxygen supply is limited, incomplete combustion occurs. Under these conditions, not all the carbon in the fuel can be converted into carbon dioxide. Instead, the products include carbon monoxide ($\text{CO}$) and/or solid carbon ($\text{C}$), along with water.

Incomplete combustion is easily identified by a yellow, smoky, or sooty flame. The yellow colour and smoke result from glowing carbon particles that have not been fully oxidised.

The equation for the incomplete combustion of methane to form carbon monoxide is:

$2\text{CH}_4\text{(g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO(g)} + 4\text{H}_2\text{O(l)}$

The danger of carbon monoxide

The effects of carbon monoxide depend on its concentration:

• At concentrations as low as 10 parts per million (ppm), carbon monoxide can cause drowsiness, dizziness, and headaches
• At approximately 200 ppm, carbon monoxide can be lethal
• In large cities, the average concentration is around 7 ppm, mainly from incomplete combustion of fuels in vehicles
• At busy intersections during heavy traffic, concentrations can reach as high as 120 ppm

Car exhaust gases often contain high levels of carbon monoxide due to incomplete combustion of petrol and diesel fuels. This makes proper ventilation crucial when running engines in enclosed spaces.

Writing equations for complete combustion

Balancing combustion equations for hydrocarbons

Writing balanced equations for the complete combustion of hydrocarbons follows a systematic process. Since the products are always carbon dioxide and water, the main challenge is determining the correct coefficients.

This systematic approach ensures that the equation is properly balanced and follows chemical conventions. The state symbols indicate that the combustion occurs with gaseous reactants producing gaseous carbon dioxide and liquid water.

Balancing combustion equations for alcohols

Alcohols are carbon-based fuels that contain oxygen in their molecular structure. The process for balancing their combustion equations is similar to that for hydrocarbons, with one additional consideration: we must account for the oxygen already present in the alcohol molecule.

Writing equations for incomplete combustion

Incomplete combustion equations can also be written systematically. When oxygen supply is insufficient, the carbon products are carbon monoxide and/or solid carbon, along with water.

Heat of combustion

The heat of combustion is the heat energy released when a specified amount of a substance burns completely in oxygen. It is expressed as a positive value and measured under standard laboratory conditions of 298 K and 100 kPa. Under these conditions, water produced is in the liquid state.

Heat of combustion can be expressed in different units:

• kJ mol⁻¹ for pure substances
• kJ g⁻¹ or kJ L⁻¹ for fuel mixtures like wood, coal, or kerosene that don't have specific chemical formulas

Heat of combustion values for common fuels

Substance	Heat of combustion (kJ mol⁻¹)
methane	890
ethane	1560
propane	2220
butane	2880
octane	5460
methanol	725
ethanol	1360
hydrogen	282
carbon (graphite)	394

These values show that larger hydrocarbon molecules generally release more energy per mole when burned. This is because they contain more carbon-hydrogen bonds that can be broken and reformed into carbon dioxide and water, releasing energy in the process.

Calculating energy from combustion

When we know the heat of combustion of a fuel, we can calculate the total energy released when a given mass of the fuel burns. The calculation involves two steps:

1. Calculate the number of moles: $n = \frac{m}{M}$
2. Calculate energy released: $\text{Energy} = n \times \Delta H$

Energy content per gram

For practical comparisons, especially for transport fuels, energy content is often expressed per gram. For pure substances, this can be calculated by dividing the heat of combustion per mole by the molar mass.

For example, for ethanol: $\text{Heat of combustion per gram} = \frac{1360}{46.0} = 29.6 \text{ kJ g}^{-1}$

Substance	State	Heat of combustion (kJ g⁻¹) (approx.)
kerosene	liquid	18
wood	solid	21
diesel	liquid	25
black coal	solid	25
natural gas	gas	30

These values help us compare different fuels. Natural gas has the highest energy content per gram, making it highly efficient. However, other factors such as storage, transport, and availability also influence fuel choice.

Thermochemical equations

Thermochemical equations provide comprehensive information about chemical reactions by including both the balanced equation and the enthalpy change ($\Delta H$). These equations are powerful tools for understanding energy changes in combustion reactions.

For example, the thermochemical equation for complete combustion of propane is:

$\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \quad \Delta H = -2220 \text{ kJ}$

Relationship between mole ratios and enthalpy changes

The $\Delta H$ value in a thermochemical equation corresponds directly to the mole amounts specified by the coefficients. If you change the coefficients, the $\Delta H$ value changes proportionally.

For methanol combustion: $\text{CH}_3\text{OH(l)} + \frac{3}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \quad \Delta H = -726 \text{ kJ}$

This equation shows that 726 kJ is released when 1 mole of methanol reacts. If we double all coefficients, the energy released also doubles:

$2\text{CH}_3\text{OH(l)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \quad \Delta H = -1452 \text{ kJ}$

Similarly, tripling the coefficients triples the $\Delta H$ value:

$3\text{CH}_3\text{OH(l)} + \frac{9}{2}\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)} \quad \Delta H = -2178 \text{ kJ}$

This proportional relationship allows us to calculate energy changes for any amount of fuel burned.

Reversing chemical equations

When a chemical equation is reversed, the sign of $\Delta H$ changes but the magnitude remains the same. This reflects the principle that the energy released by a reaction equals the energy required to reverse it.

Forward reaction (combustion of methane): $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \quad \Delta H = -890 \text{ kJ}$

Reverse reaction: $\text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)} \rightarrow \text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \quad \Delta H = +890 \text{ kJ}$

The forward reaction is exothermic (releases energy), whilst the reverse reaction is endothermic (requires energy input).

The importance of state symbols

For example, boiling water is an endothermic process: $\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)} \quad \Delta H = +40.7 \text{ kJ}$

Energy must be supplied to convert liquid water into steam.

For propane combustion, the $\Delta H$ value depends on whether water is produced as a gas or liquid:

With water as gas: $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(g)} \quad \Delta H = -2057 \text{ kJ}$

With water as liquid: $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} \rightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \quad \Delta H = -2220 \text{ kJ}$

Case study: Explosives and chemical energy

Explosives demonstrate the rapid release of chemical energy through combustion reactions. Humans have used explosives since 919 BCE, when the Chinese first mixed saltpetre (potassium nitrate), sulfur, and charcoal to create gunpowder. Today, explosives are essential tools for mining, road construction, tunnelling, and controlled demolition.

Whilst fuels like petrol and natural gas release thermal energy through combustion, their reaction rate is limited by oxygen availability. Explosives are fundamentally different because they contain sufficient oxygen within their molecular structure for complete (or nearly complete) reaction to occur almost instantaneously.

Chemical explosives such as ammonium nitrate, trinitrotoluene (TNT), and nitroglycerine decompose rapidly, releasing large quantities of energy and gaseous products. The thermochemical equation for nitroglycerine decomposition is:

$4\text{C}_3\text{H}_5\text{N}_3\text{O}_9\text{(l)} \rightarrow 12\text{CO}_2\text{(g)} + 10\text{H}_2\text{O(g)} + 6\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \quad \Delta H = -1456 \text{ kJ}$

The ratio of products to reactants is particularly significant. In the nitroglycerine equation, 4 moles of liquid reactant produce 29 moles of gaseous products. This dramatic increase in moles, combined with the phase change from liquid to gas, causes the explosive expansion.

Explosives require an activation trigger (such as a detonator or spark) to initiate the reaction. Once started, the reaction proceeds extremely rapidly because it is highly exothermic and generates its own heat to sustain the process.