Standard Enthalpies of Formation (ΔHf°) for Common Compounds
| Compound | Formula | State | ΔHf° (kJ/mol) | Category |
|---|---|---|---|---|
| Water (liquid) | H₂O | l | -285.8 | Inorganic |
| Water (gas) | H₂O | g | -241.8 | Inorganic |
| Carbon dioxide | CO₂ | g | -393.5 | Inorganic |
| Carbon monoxide | CO | g | -110.5 | Inorganic |
| Ammonia | NH₃ | g | -45.9 | Inorganic |
| Nitrogen dioxide | NO₂ | g | 33.2 | Inorganic |
| Nitric oxide | NO | g | 91.3 | Inorganic |
| Dinitrogen tetroxide | N₂O₄ | g | 9.2 | Inorganic |
| Nitrous oxide | N₂O | g | 81.6 | Inorganic |
| Sulfur dioxide | SO₂ | g | -296.8 | Inorganic |
| Sulfur trioxide | SO₃ | g | -395.7 | Inorganic |
| Hydrogen sulfide | H₂S | g | -20.6 | Inorganic |
| Hydrogen chloride | HCl | g | -92.3 | Inorganic |
| Hydrogen bromide | HBr | g | -36.3 | Inorganic |
| Hydrogen iodide | HI | g | 26.5 | Inorganic |
| Hydrogen fluoride | HF | g | -273.3 | Inorganic |
| Hydrogen peroxide (liquid) | H₂O₂ | l | -187.8 | Inorganic |
| Sodium chloride | NaCl | s | -411.2 | Salt |
| Sodium hydroxide | NaOH | s | -425.6 | Base |
| Potassium chloride | KCl | s | -436.5 | Salt |
| Calcium carbonate | CaCO₃ | s | -1206.9 | Salt |
| Calcium oxide | CaO | s | -634.9 | Oxide |
| Calcium hydroxide | Ca(OH)₂ | s | -985.2 | Base |
| Magnesium oxide | MgO | s | -601.6 | Oxide |
| Aluminum oxide | Al₂O₃ | s | -1675.7 | Oxide |
| Iron(III) oxide | Fe₂O₃ | s | -824.2 | Oxide |
| Iron(II) oxide | FeO | s | -272 | Oxide |
| Copper(II) oxide | CuO | s | -157.3 | Oxide |
| Silver chloride | AgCl | s | -127 | Salt |
| Barium sulfate | BaSO₄ | s | -1473.2 | Salt |
| Methane | CH₄ | g | -74.8 | Organic |
| Ethane | C₂H₆ | g | -84.7 | Organic |
| Propane | C₃H₈ | g | -103.8 | Organic |
| Ethylene (ethene) | C₂H₄ | g | 52.4 | Organic |
| Acetylene (ethyne) | C₂H₂ | g | 226.7 | Organic |
| Benzene | C₆H₆ | l | 49.1 | Organic |
| Ethanol | C₂H₅OH | l | -277.7 | Organic |
| Methanol | CH₃OH | l | -239.2 | Organic |
| Acetic acid | CH₃COOH | l | -484.5 | Organic |
| Glucose | C₆H₁₂O₆ | s | -1275 | Organic |
| Sucrose | C₁₂H₂₂O₁₁ | s | -2222 | Organic |
| Urea | CO(NH₂)₂ | s | -333.5 | Organic |
All values are at 298.15 K and 1 atm (1 bar in the strict modern convention; the difference is negligible for ΔHf°). Always match the physical state in your equation to the state in this table — the gap between H₂O(l) at −285.8 and H₂O(g) at −241.8 kJ/mol is exactly the enthalpy of vaporization (44.0 kJ/mol), and using the wrong phase silently throws off combustion calculations. Carbon's standard state is graphite, not diamond; ozone (O₃) is not the standard state of oxygen, so its ΔHf° is +142.7 kJ/mol, not zero. Sources: NIST Chemistry WebBook, CRC Handbook of Chemistry and Physics, Atkins' Physical Chemistry (12th ed.).
Frequently Asked Questions
How do you calculate the enthalpy of a reaction from enthalpies of formation?
Apply Hess's law: ΔH°rxn = Σ[n × ΔHf°(products)] − Σ[n × ΔHf°(reactants)], where n is the stoichiometric coefficient. For methane combustion, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): ΔH = [(−393.5) + 2(−285.8)] − [(−74.8) + 2(0)] = −890.3 kJ. The O₂ term is zero because O₂(g) is the standard state of oxygen. Watch the phases — using H₂O(g) instead of H₂O(l) here would change the answer by 88 kJ.
Why is the enthalpy of formation of O₂(g) zero?
It's a definitional convention, not a measured quantity. ΔHf° is defined as the enthalpy of forming a compound from its elements in their standard states — and since O₂(g) is itself the standard state of oxygen, the 'formation' reaction is just O₂ → O₂, which has zero enthalpy change. The same applies to N₂(g), H₂(g), C(graphite), Fe(s), and S₈(s). Allotropes that aren't the standard state get nonzero values: O₃(g) is +142.7 kJ/mol, and diamond is +1.9 kJ/mol relative to graphite.
Why does physical state matter for enthalpy of formation?
Phase transitions carry their own enthalpy. The 44.0 kJ/mol gap between H₂O(l) at −285.8 and H₂O(g) at −241.8 kJ/mol is the enthalpy of vaporization at 25 °C. If your reaction produces liquid water but you grab the gas value, every mole of water in the equation throws your answer off by 44 kJ. Check the (l), (g), (s), or (aq) labels in the balanced equation and pull the matching row. The hazard is highest for water, ethanol, ammonia, and any solvent that might be liquid or vapor depending on conditions.