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Enthalpy of Combustion or Heat of Combustion Chemistry Tutorial

Key Concepts

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Heat of Combustion (enthalpy of combustion) of Some Common Substances

The heat of combustion (enthalpy of combustion) of some common alkanes and alcohols used as fuels is tabulated below:

Substance
(fuel)
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.

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

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

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Measuring Heat of Combustion (Measuring Enthalpy of Combustion)

In the school laboratory it is possible to measure the heat of combustion (enthalpy of combustion) of a liquid fuel such as an alcohol using the procedure outlined below:1

  1. A known quantity of water is placed in a flask, beaker or can.
  2. A thermometer is positioned with bulb 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 final temperature recorded (Tf).
  7. The final quantity of fuel is measured and recorded.

Typical Results for an experiment where ethanol is used to heat 200g of water:

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 temperature
    = Tf - Ti
    = 75 - 20
    = 55°C
mass ethanol used
    = mi - mf
    = 37.25 - 35.50
    = 1.75 g

Calculation of Heat of Combustion of ethanol using these experimental results:

  1. 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
        = 1.75 ÷ 46.1
        = 0.0380 mol

  2. Calculate energy required to change temperature of water:

    energy = 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 = 4.184 J°C-1g-1 × 200 g × 55°C
        = 46024 J
    Convert energy in J to kJ by dividing by 1000:
        46024 J ÷ 1000 J/kJ
        = 46.024 kJ

  3. Calculate the heat of combustion of ethanol

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

    0.0380 mole ethanol produced 46.024 kJ of heat.

    Therefore 1 mole of ethanol would produce:
        = 46.024 kJ ÷ 0.0380 mol
        = 1211 kJ mol-1

    The heat of combustion of ethanol is 1211 kJ mol-1
    (Note the enthalpy change for the reaction is -1211 kJ mol-1, ΔH = - 1211 kJ mol-1

The experimentally determined value for the 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.

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Worked Example of Calculating Molar Enthalpy of Combustion

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 (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:
    = q(released) ÷ moles(CH3OH)
    = heat of combustion of methanol in J mol-1
    Divide by 1000 to get 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 ÷ 32.042
        = 0.03121 mol

    (b) Energy absorbed by the water, q(absorbed):
    q(absorbed) = m(H2O) × cg(H2O) × (Tf - Ti)
        = 100.0 × 4.184 × (55.0 - 25.0)
        = 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:
    = q(released) ÷ moles(CH3OH)
    = 12552 J ÷ 0.03121 mol
    = 402179 J mol-1
    = heat of combustion of ethanol in J mol-1
    Divide by 1000 to get heat of combustion in kJ mol-1
    = 402179 Jmol-1 ÷ 1000 J/kJ
    = 402 kJ mol-1

  5. Is your answer plausible?
    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 ethanol in kJ mol-1":

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

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Footnotes

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