Standard Enthalpies of Formation (ΔH°f) for Common Compounds
| Substance | Formula | State | ΔH°f (kJ/mol) | ΔG°f (kJ/mol) | S° (J/(mol·K)) |
|---|---|---|---|---|---|
| Water (liquid) | H₂O(l) | liquid | -285.83 | -237.13 | 69.95 |
| Water (gas) | H₂O(g) | gas | -241.82 | -228.57 | 188.84 |
| Carbon dioxide | CO₂(g) | gas | -393.51 | -394.36 | 213.79 |
| Carbon monoxide | CO(g) | gas | -110.53 | -137.17 | 197.67 |
| Methane | CH₄(g) | gas | -74.87 | -50.72 | 186.26 |
| Ethane | C₂H₆(g) | gas | -84.68 | -32 | 229.6 |
| Ethylene (ethene) | C₂H₄(g) | gas | 52.47 | 68.15 | 219.56 |
| Acetylene (ethyne) | C₂H₂(g) | gas | 226.73 | 209.2 | 200.94 |
| Propane | C₃H₈(g) | gas | -103.85 | -23.4 | 270.2 |
| Benzene | C₆H₆(l) | liquid | 49.04 | 124.5 | 173.26 |
| Ethanol | C₂H₅OH(l) | liquid | -277.69 | -174.78 | 160.7 |
| Methanol | CH₃OH(l) | liquid | -238.4 | -166.27 | 126.8 |
| Acetic acid | CH₃COOH(l) | liquid | -484.5 | -389.9 | 159.8 |
| Glucose | C₆H₁₂O₆(s) | solid | -1273.3 | -910.4 | 212.1 |
| Sucrose | C₁₂H₂₂O₁₁(s) | solid | -2221.7 | -1544.6 | 360.2 |
| Ammonia | NH₃(g) | gas | -45.9 | -16.4 | 192.77 |
| Nitric oxide | NO(g) | gas | 91.3 | 87.6 | 210.76 |
| Nitrogen dioxide | NO₂(g) | gas | 33.1 | 51.3 | 240.06 |
| Dinitrogen tetroxide | N₂O₄(g) | gas | 9.16 | 99.8 | 304.38 |
| Nitrous oxide | N₂O(g) | gas | 81.6 | 103.7 | 220 |
| Hydrogen chloride | HCl(g) | gas | -92.31 | -95.3 | 186.9 |
| Hydrogen fluoride | HF(g) | gas | -273.3 | -275.4 | 173.78 |
| Hydrogen bromide | HBr(g) | gas | -36.29 | -53.43 | 198.7 |
| Hydrogen iodide | HI(g) | gas | 26.48 | 1.7 | 206.59 |
| Sulfur dioxide | SO₂(g) | gas | -296.83 | -300.13 | 248.22 |
| Sulfur trioxide | SO₃(g) | gas | -395.72 | -371.06 | 256.77 |
| Sulfuric acid | H₂SO₄(l) | liquid | -813.99 | -690 | 156.9 |
| Sodium chloride | NaCl(s) | solid | -411.15 | -384.14 | 72.11 |
| Sodium hydroxide | NaOH(s) | solid | -425.61 | -379.49 | 64.46 |
| Calcium carbonate | CaCO₃(s) | solid | -1206.9 | -1128.8 | 92.9 |
| Calcium oxide | CaO(s) | solid | -635.09 | -603.3 | 38.1 |
| Calcium hydroxide | Ca(OH)₂(s) | solid | -986.09 | -898.49 | 83.39 |
| Magnesium oxide | MgO(s) | solid | -601.6 | -569.3 | 27 |
| Aluminum oxide | Al₂O₃(s) | solid | -1675.7 | -1582.3 | 50.92 |
| Iron(III) oxide | Fe₂O₃(s) | solid | -824.2 | -742.2 | 87.4 |
| Iron(II) oxide | FeO(s) | solid | -272 | -255.2 | 60.75 |
| Copper(II) oxide | CuO(s) | solid | -157.3 | -129.7 | 42.63 |
| Copper(II) sulfate | CuSO₄(s) | solid | -771.4 | -662.2 | 109.2 |
| Silver chloride | AgCl(s) | solid | -127.01 | -109.79 | 96.25 |
| Hydrogen peroxide | H₂O₂(l) | liquid | -187.78 | -120.35 | 109.6 |
| Ozone | O₃(g) | gas | 142.7 | 163.2 | 238.93 |
| Phosphoric acid | H₃PO₄(l) | liquid | -1271.7 | -1123.6 | 150.8 |
| Urea | CO(NH₂)₂(s) | solid | -333.5 | -197.4 | 104.6 |
Conditions: 298.15 K, 1 atm (1 bar for gases under the modern convention; the difference is negligible here), and 1 M for aqueous species. ΔH°f and ΔG°f for any element in its standard state are zero by definition (O₂(g), N₂(g), C(graphite), Fe(s), H₂(g)) — but S° is not zero, because absolute entropy is referenced to 0 K via the third law. So you'll see N₂(g) at S° = 191.6 J/(mol·K) even though both formation values are zero. This catches students every year. Sources: NIST-JANAF Thermochemical Tables and the CRC Handbook of Chemistry and Physics.
Frequently Asked Questions
How do you calculate the enthalpy change of a reaction using standard enthalpies of formation?
Apply Hess's law: ΔH°rxn = Σ[n × ΔH°f(products)] − Σ[n × ΔH°f(reactants)], multiplying each ΔH°f by the stoichiometric coefficient n. For methane combustion, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): ΔH°rxn = [(−393.51) + 2(−285.83)] − [(−74.87) + 2(0)] = −890.30 kJ. The O₂ term drops out because elements in their standard states have ΔH°f = 0. Match the phase labels in your equation to the rows in the table — using H₂O(g) instead of H₂O(l) here would change the answer by 88 kJ.
Why is the standard enthalpy of formation of elements in their standard state zero?
It's a definitional zero. ΔH°f measures the enthalpy change for the reaction 'elements in their standard states → one mole of compound.' If the compound itself is already an element in its standard state, the reaction is the identity (e.g., O₂ → O₂) and the enthalpy change is exactly zero. The convention provides a consistent reference baseline. Note this only applies to ΔH°f and ΔG°f — absolute entropy (S°) is referenced to perfect crystals at 0 K via the third law, so elements in their standard states still have positive S° values.
What is the difference between ΔH°f and ΔG°f?
ΔH°f only tells you whether forming the compound is exothermic or endothermic. ΔG°f layers in entropy via ΔG° = ΔH° − TΔS°, and that's what determines spontaneity at a given temperature. A reaction can be exothermic (ΔH° < 0) but non-spontaneous if it produces a large entropy decrease, or endothermic but spontaneous if entropy rises enough. Use ΔG°f for spontaneity checks (ΔG°rxn < 0 ⇒ K > 1) and ΔH°f for heat-flow problems. The relationship ΔG° = −RT ln K also lets you extract equilibrium constants directly.