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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.
|Molar Heat of
||CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
||ΔH = -890
||C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
||ΔH = -1560
||C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
||ΔH = -2220
||C4H10(g) + 13/2O2(g) → 4CO2(g) + 5H2O(l)
||ΔH = -2874
||C8H18(g) + 25/2O2(g) → 8CO2(g) + 9H2O(l)
||ΔH = -5460
||CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l)
||ΔH = -726
||C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
||ΔH = -1368
||C3H7OH(l) + 9/2O2(g) → 3CO2(g) + 4H2O(l)
||ΔH = -2021
||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.
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
- A known quantity of water is placed in a flask, beaker or can.
- A thermometer is positioned with its bulb (reservoir) near the middle of the volume of water.
- A known quantity of fuel, such as an alcohol (alkanol), is placed in the spirit burner.
- The initial temperature of the water is measured and recorded (Ti).
- The wick on the spirit burner is lit, burning the fuel, and heating the water.
- 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).
- 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
|mass ethanol used in combustion reaction
= mi - mf
= 37.25 - 35.50
= 1.75 g
- 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:
- Calculate moles (n) of fuel used
moles = mass ÷ molar mass
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
g ÷ 46.1 g mol-1
= 0.0380 mol
- 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
- 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.