 # Molar Enthalpy of Combustion of Fuels or Molar Heat of Combustion Fuels Chemistry Tutorial

## Key Concepts

• Heat of combustion is also known as enthalpy of combustion.
• When a substance undergoes combustion it releases energy.
• Combustion is always exothermic, the enthalpy change for the reaction is negative, ΔH has a negative sign.
• Molar Heat of Combustion (molar enthalpy of combustion) of a substance is the heat liberated when 1 mole of the substance undergoes complete combustion with oxygen at constant pressure.
• By definition, the heat of combustion (enthalpy of combustion, ΔHc) is minus the enthalpy change for the combustion reaction, ie, -ΔH.
• So, by convention, the molar heat of combustion (molar enthalpy of combustion) is given in tables as a positive value.
• Tabulated values of molar enthalpy of combustion can be used to determine the amount of heat energy released (q) when a known amount of the substance (n) undergoes complete combustion:

q = n × molar enthalpy of combustion

• Molar Heat of Combustion (molar enthalpy of combustion) of a pure substance can be determined experimentally:

Step 1: Calculate moles of substance consumed in combustion reaction (n)

Step 2: Calculate the amount of energy released by the combustion reaction (q)

Step 3: Calculate energy released per mole of substance combusted = q ÷ n

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## Molar Heat of Combustion (molar enthalpy of combustion) of Some Common Substances Used as Fuels

Hydrocarbons, such as alkanes, and alcohols, such as alkanols, can be used as fuels.

When an alkane undergoes complete combustion in excess oxygen gas the products of the reaction are carbon dioxide (CO2(g)) and water (H2O(g) which will condense to H2O(l) at room temperature and pressure).

alkane + excess oxygen gas → carbon dioxide gas + water vapor

The molar heat of combustion of the alkane (molar enthalpy of combustion of the alkane) is the amount of heat energy released when 1 mole of the alkane combusts in excess oxygen gas.

When an alkanol undergoes complete combustion in excess oxygen gas the products of the reaction are carbon dioxide (CO2(g)) and water (H2O(g) which will condense to H2O(l) at room temperature and pressure).

alkanol + excess oxygen gas → carbon dioxide gas + water vapor

The molar heat of combustion of the alkanol (molar enthalpy of combustion of the alkanol) is the amount of heat energy released when 1 mole of the alkanol combusts in excess oxygen gas.

In order to determine the molar heat of combustion, we need to be able to determine how many moles of the substance were consumed in the combustion reaction so the substance must be a pure substance.1

The molar heat of combustion (molar enthalpy of combustion) of some common alkanes and alcohols used as fuels is tabulated below in units of kilojoules per mole (kJ mol-1)2.

Note that the chemical equations representing each of the combustion reactions is balanced so that 1 mole of the substance combusted, the fuel, is used.
The combustion reaction occurs in excess oxygen gas, excess O2(g), so it is quite OK to use fractions of O2(g) to balance the equation because we are really only interested in the energy released per mole of the fuel, not per mole of oxygen gas.

Substance
(fuel)
Molar Heat of
Combustion
(kJ mol-1)
Combustion Reaction ΔHreaction
(kJ mol-1)
methane 890 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -890
ethane 1560 C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1560
propane 2220 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH = -2220
butane 2874 C4H10(g) + 13/2O2(g) → 4CO2(g) + 5H2O(l) ΔH = -2874
octane 5460 C8H18(g) + 25/2O2(g) → 8CO2(g) + 9H2O(l) ΔH = -5460
methanol
(methyl alcohol)
726 CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l) ΔH = -726
ethanol
(ethyl alcohol)
1368 C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔH = -1368
propan-1-ol
(1-propanol)
2021 C3H7OH(l) + 9/2O2(g) → 3CO2(g) + 4H2O(l) ΔH = -2021
butan-1-ol
(1-butanol)
2671 C4H9OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l) ΔH = -2671

From the table we see that 1 mole of methane gas, CH4(g), undergoes complete combustion in excess oxygen gas releasing 890 kJ of heat.
The molar heat of combustion of methane gas is given in the table as a positive value, 890 kJ mol-1.
The enthalpy change for the combustion of methane gas is given in the table as a negative value, ΔH = -890 kJ mol-1, because the reaction produces energy (it is an exothermic reaction).
We could write a chemical equation to represent the combustion of 1 mole of methane gas as:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)     ΔH = -890 kJ mol-1

But how much energy is released if 2 moles of methane undergoes complete combustion?
When we write a chemical equation for this reaction we must multiply every term by two ( × 2)including the value of ΔH:

2 × CH4(g) + 2 × 2O2(g)2 × CO2(g) + 2 × 2H2O(g)     ΔH = 2 × -890 kJ mol-1

2CH4(g) + 4O2(g)2CO2(g) + 4H2O(g)     ΔH = -1780 kJ mol-1

2 moles of methane would combust completely to release 2 × 890 = 1780 kJ of heat.

Similarly, if we have only half a mole of methane gas that undergoes complete combustion we must multiply every term in the chemical equation, including the value of ΔH, by ½ as shown in the chemical equations below:

½ × CH4(g) + ½ × 2O2(g)½ × CO2(g) + ½ × 2H2O(g)     ΔH = ½ × -890 kJ mol-1

½CH4(g) + O2(g)½CO2(g) + H2O(g)     ΔH = -445 kJ mol-1

½ mole of methane would combust to release ½ × 890 = 445 kJ of heat.

In general, the amount of heat energy released by the combustion of n moles of fuel is equal to the value of the molar heat of combustion of the fuel multiplied by the moles of fuel combusted

heat released (kJ) = n (mol) × molar enthalpy of combustion (kJ mol-1)

(See the Enthlapy Change Calculations for a Chemical Reaction Tutorial for more examples of these types of calculations)

In this section we looked at how to use tables of values for the molar enthalpy of combustion of pure substances to calculate how much heat energy would be released when known amounts of the substance were combusted in excess oxygen gas.
But where do these values come from?
Molar enthalpy of combustion values can be determined using laboratory experiments.
In the next section we will discuss an experiment you could do to determine the molar heat of combustion of an alcohol.

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## Measuring Molar Heat of Combustion of a Liquid Fuel (Measuring Molar Enthalpy of Combustion of a Liquid Fuel)

In the school laboratory it is possible to determine the molar heat of combustion (enthalpy of combustion) of a liquid fuel such as an alcohol using the procedure outlined below:3 1. A known quantity of water is placed in a flask, beaker or can.
2. A thermometer is positioned with its bulb (reservoir) near the middle of the volume of water.
3. A known quantity of fuel, such as an alcohol (alkanol), is placed in the spirit burner.
4. The initial temperature of the water is measured and recorded (Ti).
5. The wick on the spirit burner is lit, burning the fuel, and heating the water.
6. When the temperature of the water has risen an appreciable amount, the spirit burner is extinguished and the maximum temperature reached is recorded as the final temperature (Tf).
7. The final quantity of fuel is measured and recorded.

Typical results for an experiment where the energy released by the complete combustion of ethanol is used to heat 200 g of water are shown below:

 initial water temperature (Ti) = 20°C intial mass burner + ethanol (mi) = 37.25 g final water temperature (Tf)= 75°C final mass burner + ethanol (mf) = 35.50 g change in water temperature     = Tf - Ti     = 75 - 20     = 55°C mass ethanol used in combustion reaction     = mi - mf     = 37.25 - 35.50     = 1.75 g

Note:

• Temperature of the water increases because combustion of the fuel releases energy which heats the water.
• Mass of the fuel decreases because it is being consumed in the combustion reaction.

The results from this experiment can then be used to calculate the molar heat of combustion of ethanol (molar enthalpy of combustion of ethanol) as shown below:

Steps to calculate the molar enthalpy of combustion of ethanol using these experimental results:

1. Calculate moles (n) of fuel used

mass ethanol used = 1.75 g (from experiment)

molar mass (M) of ethanol (C2H5OH)

= (2 × 12.01) + (6 × 1.008) + 16.00 (from periodic table)

= 46.1 g mol-1

n(ethanol) = mass ÷ molar mass

= 1.75 g ÷ 46.1 g mol-1

= 0.0380 mol

2. Calculate energy required to change temperature of water:

energy absorbed by water = specific heat capacity of water × mass of water × change in water temperature

specific heat capacity of water = cg = 4.184 J°C-1g-1 (data sheet)

mass water = 200 g (from experiment)

change in water temperature = 55°C (from experiment)

energy absorbed by water = 4.184 J°C-1g-1 × 200 g × 55°C

= 46024 J

Convert energy in J to kJ by dividing by 1000:

energy = 46024 J ÷ 1000 J/kJ

= 46.024 kJ

3. Calculate the molar heat of combustion of ethanol (molar enthalpy of combustion of ethanol):

Assume all the heat produced from burning the ethanol has gone into heating the water, that is, no heat has been wasted.
Then we can say that

0.0380 mole ethanol produced 46.024 kJ of heat.

Therefore 1 mole of ethanol would produce:

heat energy produced per mole of ethanol = 46.024 kJ ÷ 0.0380 mol

= 1211 kJ mol-1

Therefore the molar heat of combustion of ethanol is 1211 kJ mol-1
(The molar enthalpy of combustion of ethanol is 1211 kJ mol-1).

(Note that the enthalpy change for the reaction is negative because the reaction is exothermic, so the enthalpy change for the reaction is -1211 kJ mol-1, ΔH = - 1211 kJ mol-1)

The experimentally determined value for the molar heat of combustion of ethanol is usually less than the accepted value of 1368 kJ mol-1 because some heat is always lost to the atmosphere and in heating the vessel.

The design of the experiment can be improved by trying to minimise the heat lost to the surroundings, for example, by surrounding the whole experimental set-up with metal walls.
By far the best way of minimising heat loss to the surroundings is to use a bomb calorimeter.

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## Worked Example of Calculating Molar Enthalpy of Combustion Using Experimental Results

Question: A spirit burner used 1.00 g of methanol to raise the temperature of 100.0 g of water in a metal can from 25.0°C to 55.0°C. Assuming the heat capacity of water is 4.184 J°C-1g-1, calculate the molar enthalpy of combustion of methanol in kJ mol-1.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate the molar enthalpy of combustion of methanol in kJ mol-1

2. What data (information) have you been given in the question?

Extract the data from the question:

fuel name: methanol (condensed structural formula: CH3OH)
m(CH3OH) = mass methanol used = 1.00 g
m(H2O) = mass water heated = 100.0 g
cg = heat capacity of water = 4.184 J°C-1g-1
Ti = temperature of water before methanol combustion = 25.0°C
Tf = temperature of water after methanol combustion = 55.0°C
3. What is the relationship between what you know and what you need to find out?
(a) Calculate moles of methanol, n(CH3OH), used:

n(CH3OH) = m(CH3OH) ÷ molar mass(CH3OH)

(b) Energy absorbed by the water, q(absorbed):

q(absorbed) = m(H2O) × cg(H2O) × (Tf - Ti)

Assuming there is no heat lost, then all the energy released by combustion of methanol, q(released), is used to heat the water, q(absorbed):

q(released) = q(absorbed)

(c) Energy released per mole of methanol:

energy released = q(released) ÷ moles(CH3OH)

energy released = molar heat of combustion of methanol in J mol-1

Divide molar heat of combustion of methanol in J mol-1 by 1000 J/kJ to get molar heat of combustion in kJ mol-1

4. Perform the calculations
(a) Calculate moles of methanol, n(CH3OH), used:

n(CH3OH) = m(CH3OH) ÷ Mr(CH3OH)

Mr(CH3OH) = molar mass(CH3OH)

= 12.01 + (4 × 1.008) + 16.00

= 32.042 g mol-1

n(CH3OH) = 1.00 g ÷ 32.042 g mol-1

= 0.03121 mol

(b) Energy absorbed by the water, q(absorbed):

q(absorbed) = m(H2O) × cg(H2O) × (Tf - Ti)

= 100.0g × 4.184 Jg-1°C-1 × (55.0 - 25.0)°C

= 100.0 × 4.184 × 30.0

= 12552 J

Assuming there is no heat lost, then all the energy released by combustion of methanol, q(released), is used to heat the water, q(absorbed):

q(released) = q(absorbed) = 12552 J

(c) Energy released per mole of methanol:

energy released per mole methanol = q(released) ÷ moles(CH3OH)

= 12552 J ÷ 0.03121 mol

= 402179 J mol-1

= heat of combustion of methanol in J mol-1

Divide by 1000 J/kJ to get heat of combustion in kJ mol-1

heat of combustion of methanol = 402179 Jmol-1 ÷ 1000 J/kJ

= 402 kJ mol-1

Work backwards using your calculated value for heat of combustion to calculate the expected change in water temperature:

heat of combustion ≈ 400 kJ mol-1
n(CH3OH) ≈ 1.00 g ÷ 32 ≈ 0.03 mol
heat released ≈ 400 kJ mol-1 × 0.03 mol = 12 kJ = 12,000 J
heat absorbed by water ≈ 12,000 J = m × cg × ΔT
12,000 = 100 × 4 × ΔT
ΔT = 30°C

Since this value for the temperature change expected is the same as that given in the question (55°C - 25°C = 30°C) we are reasonably confident that our answer for the heat of combustion is plausible.

6. State your solution to the problem "molar enthalpy of combustion of methanol in kJ mol-1":

molar enthalpy of combustion of methanol is 402 kJ mol-1

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Footnotes

1. If the substance to be combusted is not a pure substance, you can still determine how much energy is released when it combusts, BUT, the units will not be kilojoules per mole.
If you know the mass of mixture combusted you could determine the energy released in units of kilojoules per gram for instance, or kilojoules per kilogram, etc.
If you know the volume of a liquid fuel, you could determine the energy released in units of kilojoules per milliltre or kilojoules per litre etc.
For example, biodiesel and vegetable oils are both mixtures of substances so their heats of combustion are usually given in units of J g-1

2. The Ninth International Conference on Weights and Measures (1948) recommended the use of the joule (volt coulomb) as the unit of heat.
The joule is a derived SI unit for the measurement of energy.
The SI base unit for the measurement of energy is kg.m2 s-2
1 J = 1 kg.m2 s-2

The calorie is also a unit of heat. 1 cal = 4.1840 J
If you need to convert between joules (J) and calories (cal) go to Energy Conversions Tutorial

3. An alternative method for determining heat of combustion (enthalpy of combustion) using a bomb calorimeter is outlined in the calorimetry tutorial.