Why you can predict a reaction's heat without ever running it
Enthalpy is a state function. That single fact is the whole justification for this calculation: the enthalpy of a substance depends only on what it is and what state it is in, never on how it got there. So the heat of a reaction depends only on the difference between the products and the reactants, and any convenient route between them gives the same answer.
Chemists exploit that by choosing the most inconvenient imaginable route: take every reactant apart into its constituent elements, then assemble the products from those elements. The first leg costs the negative of the reactants' formation enthalpies; the second leg gains the products' formation enthalpies. Add them and you get ΔH° = Σ products − Σ reactants.
This is Hess's law in its most useful form. It means a single table of formation enthalpies — a few thousand numbers, each measured once in a calorimeter — lets you predict the heat of essentially any reaction those substances can undergo, including reactions that are too fast, too slow, too dangerous or too incomplete to measure directly.
The standard enthalpy of formation, ΔHf°, is defined as the enthalpy change when one mole of a compound forms from its elements in their most stable states at 298.15 K and 1 bar. Because an element forming from itself involves no change, ΔHf° of O₂(g), N₂(g), H₂(g), graphite and metallic iron are all exactly zero. That is a definition, not a measurement, and it is why the reactant column of a combustion calculation is usually so short.
Three things the equation demands and one it silently assumes
The equation must be balanced first. Every coefficient is a multiplier on a formation enthalpy, so an unbalanced equation gives an answer that is simply wrong rather than approximately wrong. For methane combustion, CH₄ + 2 O₂ → CO₂ + 2 H₂O, the coefficient 2 on water is not decoration — it doubles a −285.8 kJ/mol term into a −571.6 kJ contribution, which is nearly two-thirds of the total heat released.
The physical state must match the tabulated value. Liquid water has ΔHf° = −285.8 kJ/mol and gaseous water −241.8 kJ/mol; the 44.0 kJ/mol gap is the enthalpy of vaporisation. Methane burning to liquid water releases 890.5 kJ per mole, and burning to water vapour releases 802.5 kJ. Those are the higher and lower heating values a fuel engineer quotes, and confusing them is a 10% error in a boiler calculation.
The sign convention is fixed. Products minus reactants, always in that order. Getting it backwards flips exothermic to endothermic, which is the most conspicuous possible mistake and yet a common one, because the subtraction runs opposite to the way the equation reads.
The silent assumption is temperature. Tabulated formation enthalpies are for 298.15 K, so the ΔH° you compute is the enthalpy change at 25 °C. Reactions run at 800 °C in a furnace have a different ΔH, and correcting for it requires Kirchhoff's law — integrating the difference in heat capacities between products and reactants over the temperature range. For most purposes the correction is modest, a few percent over a few hundred degrees, but it is not zero.
Two further points about what ΔH° does and does not tell you. It is a constant-pressure quantity; a sealed bomb calorimeter measures ΔU instead, and the two differ by ΔnRT where Δn is the change in moles of gas — about 2.5 kJ per mole of gas at room temperature. And it says nothing about rate. A hugely exothermic reaction can sit unchanged for centuries if its activation barrier is high, which is why a mixture of hydrogen and oxygen is stable until you supply a spark. That side of the story belongs to the Arrhenius equation.
Worked example: the combustion of methane, CH₄ + 2 O₂ → CO₂ + 2 H₂O(l)
Natural gas burns to carbon dioxide and water. Take these standard formation enthalpies at 298.15 K: CH₄(g) = −74.6, O₂(g) = 0, CO₂(g) = −393.5, H₂O(l) = −285.8, all in kJ/mol.
- Confirm the balance. One carbon each side; four hydrogens each side; four oxygens on the left as 2 O₂ and four on the right as 2 in CO₂ plus 2 in the waters. Balanced.
- Weight each product. CO₂: 1 × (−393.5) = −393.5 kJ. H₂O: 2 × (−285.8) = −571.6 kJ. Sum of products = −965.1 kJ.
- Weight each reactant. CH₄: 1 × (−74.6) = −74.6 kJ. O₂: 2 × 0 = 0 kJ, because oxygen gas is an element in its standard state. Sum of reactants = −74.6 kJ.
- Subtract. ΔH° = −965.1 − (−74.6) = −890.5 kJ for the equation as written, which consumes one mole of methane.
- Express per gram. Methane's molar mass is 16.043 g/mol, so −890.5 ÷ 16.043 = −55.51 kJ/g. That is the highest energy density per gram of any common hydrocarbon fuel, because methane has the highest hydrogen-to-carbon ratio.
The published standard enthalpy of combustion of methane is −890.4 kJ/mol, so the calculation lands on the accepted value — as it must, since the accepted value was itself derived from these same tabulated formation enthalpies.
Now swap liquid water for water vapour. The product sum becomes −393.5 + 2 × (−241.8) = −877.1 kJ, and ΔH° = −802.5 kJ. The difference, 88.0 kJ, is exactly two moles of water times the 44.0 kJ/mol enthalpy of vaporisation. A condensing boiler recovers that 88 kJ; a conventional flue throws it away, which is the entire physical basis for the higher-efficiency rating of condensing appliances.
Reading the sign, the size, and what it does not predict
A negative ΔH° means the reaction releases heat under standard conditions. A positive value means it absorbs heat. Zero means the two sides sit at the same enthalpy, which happens for isomerisations and for any reaction written among elements alone.
Judge the magnitude by class. Combustion of an organic fuel runs from hundreds to thousands of kilojoules per mole, scaling roughly with the number of carbon and hydrogen atoms. Neutralisation of a strong acid by a strong base is about −57 kJ per mole of water formed. Most heats of solution fall inside ±50 kJ/mol. Decomposition of a stable mineral — limestone to lime — costs a couple of hundred kilojoules per mole, which is why cement kilns are among the most energy-intensive industrial processes.
Per-gram figures are the right comparison for fuels, because vehicles and boilers are limited by mass and volume rather than by moles. On this basis hydrogen leads at about 142 kJ/g for the higher heating value, methane follows near 55.5 kJ/g, and liquid hydrocarbons cluster around 45–48 kJ/g. Hydrogen's advantage evaporates on a volumetric basis, which is the central engineering problem of hydrogen fuel.
ΔH° does not predict whether a reaction happens. Spontaneity is governed by Gibbs free energy, ΔG = ΔH − TΔS, and plenty of endothermic reactions proceed readily because they increase entropy — dissolving ammonium nitrate, or the same limestone decomposition once the temperature is high enough. Feed your computed ΔH° into the Gibbs free energy calculator along with an entropy change to answer that question, and use the equilibrium constant calculator to see how far the reaction goes.
Standard enthalpies of formation at 298.15 K
| Substance | State | ΔHf° (kJ/mol) |
|---|---|---|
| H₂, O₂, N₂, Cl₂ | g | 0 |
| Carbon (graphite) | s | 0 |
| Water | l | −285.8 |
| Water | g | −241.8 |
| Carbon dioxide | g | −393.5 |
| Carbon monoxide | g | −110.5 |
| Methane | g | −74.6 |
| Ethyne (acetylene) | g | +227.4 |
| Ethene | g | +52.4 |
| Ethane | g | −84.0 |
| Propane | g | −103.8 |
| Methanol | l | −239.2 |
| Ethanol | l | −277.6 |
| Ammonia | g | −45.9 |
| Nitrogen monoxide | g | +91.3 |
| Nitrogen dioxide | g | +33.2 |
| Sulfur dioxide | g | −296.8 |
| Hydrogen chloride | g | −92.3 |
| Sodium chloride | s | −411.2 |
| Calcium carbonate (calcite) | s | −1207.6 |
| Calcium oxide | s | −634.9 |
| Iron(III) oxide | s | −824.2 |
| Aluminium oxide | s | −1675.7 |
Sources differ in the last significant figure — methane appears as −74.6 or −74.8 depending on the compilation. Use one table consistently rather than mixing values from several.
Mistakes that change the answer
- Subtracting in the wrong order. It is products minus reactants. Reversing it flips the sign and turns an exothermic reaction into an endothermic one.
- Forgetting the coefficients. Each formation enthalpy is multiplied by its stoichiometric coefficient. The two water molecules in methane combustion contribute −571.6 kJ, not −285.8.
- Using the wrong physical state. Liquid and gaseous water differ by 44.0 kJ/mol, which is 10% of a methane combustion. Match the state in your equation to the state in the table.
- Assigning a non-zero value to an element. O₂, N₂, H₂, graphite and every metal in its standard state are zero by definition. Entering a real number for them is double counting.
- Mixing values from different tables. Compilations disagree in the last digit and occasionally by more. Pick one source and stay in it.
- Quoting the answer per mole of the wrong species. ΔH° is for the equation as written. Divide by the coefficient of whichever species you want the answer per mole of, and say which one.
- Applying a 298 K value at furnace temperature. Kirchhoff's law corrects ΔH for temperature using the heat-capacity difference between products and reactants. Over several hundred degrees the correction becomes worth making.
- Confusing ΔH with ΔU. A bomb calorimeter measures constant-volume energy; converting to enthalpy adds ΔnRT for the change in moles of gas.
The alternatives to the formation-enthalpy route
Three other routes give the same ΔH, and each is preferable in some situations. Direct calorimetry measures it in one experiment — see the calorimetry calculator — and is the only option for a substance whose formation enthalpy has never been tabulated. Bond enthalpies estimate ΔH by counting bonds broken and formed, which is quick and requires no tables of compounds, but it is only an approximation because bond strengths depend on their molecular environment; expect errors of tens of kilojoules. Hess's law with a chain of known reactions works when you can construct your target equation as a sum of reactions with known ΔH, and it is how many formation enthalpies were determined in the first place.
Use formation enthalpies when the values exist, which is nearly always for common inorganic and organic substances. Use bond enthalpies for a rough gas-phase estimate. Use calorimetry when nothing is tabulated, or when you need to confirm the tables apply to your actual system.
Beyond enthalpy, the two quantities that complete the thermodynamic picture are entropy and free energy. Standard entropies are tabulated in the same way, they are combined by the same products-minus-reactants sum, and together with ΔH they give ΔG through the Gibbs free energy calculator. That is the number that decides direction, and from it the reaction quotient and equilibrium constant follow.