Standard Enthalpies of Formation and Combustion
The standard enthalpy of formation (ΔHf°) is the enthalpy change when 1 mol of a compound forms from its elements in their standard states, and the standard enthalpy of combustion (ΔHc°) is the enthalpy change when 1 mol of a substance burns completely in oxygen. In brief: any reaction enthalpy can be found with ΔH°rxn = ΣnΔHf°(products) − ΣnΔHf°(reactants); for propane combustion this gives −2220 kJ/mol.
Tables of standard enthalpies make it possible to calculate the heat of thousands of reactions that were never measured directly. The same data are used to compare fuels and to estimate the energy content of foods. This article defines both quantities, gives data tables, explains the calculation and shows three typical problems.
What is a standard state?
The standard state of a substance is its most stable form at a pressure of 1 bar (very close to 1 atm) and a stated temperature, usually 298.15 K (25 °C). For solutions the standard state is a concentration of 1 mol/L. The degree symbol (°) shows that the value refers to standard conditions. Examples of standard states are O2(g), H2(g), N2(g), C(graphite), Br2(l) and Hg(l).
What is the enthalpy of formation?
The formation reaction produces exactly 1 mol of the compound from elements in their standard states. For water: H2(g) + ½O2(g) → H2O(l), ΔHf° = −285.8 kJ/mol. Fractional coefficients are allowed. By definition, the enthalpy of formation of an element in its standard state is zero: ΔHf°(O2) = 0, ΔHf°(C, graphite) = 0. A negative value means that the compound is more stable than its elements. A positive value, as for NO, means that it is less stable in terms of enthalpy.
Standard enthalpies of formation
| Substance | ΔHf° | Substance | ΔHf° |
|---|---|---|---|
| H2O(l) | −285.8 | CH4(g) | −74.6 |
| H2O(g) | −241.8 | C3H8(g) | −103.8 |
| CO2(g) | −393.5 | C2H5OH(l) | −277.6 |
| CO(g) | −110.5 | C6H12O6(s) | −1273 |
| NH3(g) | −45.9 | C2H4(g) | +52.4 |
| NO(g) | +90.3 | NaCl(s) | −411.2 |
Values differ slightly between data sources; the table gives commonly used textbook values.
How do you calculate the enthalpy of a reaction?
Because enthalpy is a state function, a reaction can be thought of as the decomposition of the reactants into elements, followed by the formation of the products from the elements. This leads to:
ΔH°rxn = Σ n ΔHf°(products) − Σ m ΔHf°(reactants)
where n and m are the stoichiometric coefficients. The calculation steps are: balance the equation, look up each ΔHf° (zero for elements), multiply by the coefficients, add the products, add the reactants and subtract.
What is the enthalpy of combustion?
The standard enthalpy of combustion is the heat released when 1 mol of a substance burns completely in oxygen to give stable products, normally CO2(g) and H2O(l) (and N2 for nitrogen compounds). Combustion enthalpies are always negative. They are relatively easy to measure in a bomb calorimeter, and they are used to find enthalpies of formation of organic compounds that cannot be made directly from the elements.
Enthalpies of combustion of common fuels
| Fuel | Formula | ΔHc° (kJ/mol) | Molar mass (g/mol) | Energy per gram (kJ/g) |
|---|---|---|---|---|
| Hydrogen | H2 | −285.8 | 2.016 | 141.8 |
| Methane | CH4 | −890 | 16.04 | 55.5 |
| Propane | C3H8 | −2220 | 44.10 | 50.3 |
| Octane | C8H18 | −5471 | 114.23 | 47.9 |
| Carbon (graphite) | C | −393.5 | 12.011 | 32.8 |
| Ethanol | C2H5OH | −1367 | 46.07 | 29.7 |
| Glucose | C6H12O6 | −2803 | 180.16 | 15.6 |
Hydrocarbons release more energy per gram than alcohols or sugars, because they contain no oxygen already bound in the molecule. Hydrogen has the highest energy per gram, but it is a gas of very low density and needs to be compressed or liquefied.
Worked examples
Example 1: enthalpy of combustion of methane
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
- Products: 1(−393.5) + 2(−285.8) = −393.5 − 571.6 = −965.1 kJ.
- Reactants: 1(−74.6) + 2(0) = −74.6 kJ.
- ΔH°rxn = −965.1 − (−74.6) = −890.5 kJ.
Example 2: enthalpy of combustion of propane
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
Products: 3(−393.5) + 4(−285.8) = −1180.5 − 1143.2 = −2323.7 kJ. Reactants: −103.8 kJ. ΔH°rxn = −2323.7 + 103.8 = −2219.9 kJ, about −2220 kJ per mole of propane. A 1.00 kg cylinder (22.68 mol) releases 22.68 × 2219.9 = 50,350 kJ, or about 5.0 × 104 kJ.
Example 3: finding ΔHf° from a combustion enthalpy
The enthalpy of combustion of ethanol, C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), is −1366.8 kJ/mol. Find ΔHf°(ethanol).
- −1366.8 = [2(−393.5) + 3(−285.8)] − ΔHf°(ethanol).
- The bracket is −787.0 − 857.4 = −1644.4.
- ΔHf°(ethanol) = −1644.4 + 1366.8 = −277.6 kJ/mol.
How were the data collected?
Systematic measurements of heats of reaction began in the 19th century with the work of Julius Thomsen in Denmark and Marcellin Berthelot in France, who measured the heats of hundreds of reactions with calorimeters. In the 20th century the data were collected into critical tables by organizations such as the National Bureau of Standards (now NIST). The use of a table of formation enthalpies for the calculation of any reaction heat follows from Hess’s law, which was formulated in 1840.
Which mistakes are common?
- Subtracting in the wrong order: it is products minus reactants.
- Forgetting the coefficients: ΔHf° must be multiplied by the number of moles in the equation.
- Using the wrong state: H2O(l) and H2O(g) have different values.
- Giving elements a non-zero value: only for the standard state; for example, O3, C(diamond) and Br2(g) are not zero.
- Mixing up combustion and formation: the formation of CO2 equals the combustion of graphite, but this is the exception.