Hess's Law Calculator
Calculate the enthalpy change of any reaction using Hess's law: combine known reactions with multipliers, or build the target reaction from standard formation enthalpies. Enter your values and get the result in kJ/mol and kcal/mol.
Find a Reaction's Enthalpy from Other Reactions or from Formation Enthalpies
Hess's Law Formula
Where:
- \(\Delta H^\circ_{\mathrm{rxn}}\) — standard enthalpy change of the target reaction (kJ/mol)
- \(n_i\) — multiplier applied to known reaction \(i\) (−1 reverses it, ½ takes half of it)
- \(\Delta H^\circ_i\) — enthalpy change of known reaction \(i\) (kJ/mol)
- \(\nu\) — stoichiometric coefficient of a substance in the balanced equation
- \(\Delta H^\circ_f\) — standard formation enthalpy of a substance (kJ/mol)
Enthalpy is a state function: it depends only on the initial and final states, not on the path taken. Both formulas express the same idea — the enthalpy change of a reaction is the same whether it happens in one step or in any number of steps.
Hess's Law – Calculation Example
Example: find the enthalpy of formation of carbon monoxide, \( \mathrm{C(graphite) + \tfrac{1}{2}O_2(g) \to CO(g)} \), from two measured reactions: the combustion of graphite \( \mathrm{C + O_2 \to CO_2} \) with \(\Delta H^\circ_1 = -393.5\) kJ/mol, and the combustion of carbon monoxide \( \mathrm{CO + \tfrac{1}{2}O_2 \to CO_2} \) with \(\Delta H^\circ_2 = -283.0\) kJ/mol.
The second reaction must be reversed so that CO appears as a product, so its multiplier is −1:
\[ \Delta H^\circ_{\mathrm{rxn}} = (1)\times(-393.5) + (-1)\times(-283.0) = -393.5 + 283.0 \]
Result: \(\Delta H^\circ_{\mathrm{rxn}} = -110.5\) kJ/mol, which equals −26.41 kcal/mol. These are exactly the calculator's default values, so you can reproduce the result immediately.
How to Use the Hess's Law Calculator
- Choose a Calculation Method: "Add known reactions (Hess cycle)" to combine measured reactions, or "From standard formation enthalpies (ΔH°f)" to build the reaction from tabulated values.
- Set the Energy Unit of Entered Values to kJ/mol or kcal/mol. The result is shown in both units regardless of your choice.
- For the Hess-cycle method, enter the ΔH of Each Known Reaction, separated by semicolons, spaces or new lines (e.g.
-393.5; -283.0). - Enter the Multiplier of Each Reaction in the same order. Use −1 to reverse a reaction, ½ for half of it, or leave the field empty to use 1 for every reaction.
- For the formation-enthalpy method, list the ΔH°f of Each Product and ΔH°f of Each Reactant, then give the matching stoichiometric coefficients. Use 0 for elements in their standard state, such as O₂(g) or C(graphite).
- Read the Reaction Enthalpy, its equivalent in kcal/mol, an interpretation (exothermic or endothermic) and a step-by-step calculation in the results panel.
What Is Hess's Law?
Hess's law states that the total enthalpy change of a chemical reaction is the same regardless of the route by which the reaction occurs. If a target reaction can be written as the sum (or difference) of other reactions, its enthalpy change is the corresponding sum of the individual enthalpy changes. This works because enthalpy is a state function: only the starting and ending states matter, not the path between them.
This principle is enormously practical. Some reactions are impossible to measure directly — carbon cannot be burned to pure CO without also producing some CO₂, so the formation enthalpy of CO must be obtained indirectly, exactly as in the example above. With Hess's law, any hard-to-measure reaction can be assembled from reactions that are easy to run in a calorimeter.
The second calculation method is a special case of the same idea. Every substance is assigned a standard formation enthalpy \(\Delta H^\circ_f\), the enthalpy change when one mole of it forms from its elements in their standard states. Since formation is a common "path" to both reactants and products, the reaction enthalpy is simply the sum of the products' formation enthalpies minus the sum of the reactants'. By definition, \(\Delta H^\circ_f = 0\) for any element already in its standard state.
To interpret the sign: a negative \(\Delta H^\circ_{\mathrm{rxn}}\) means the reaction is exothermic (heat is released, products lie lower in enthalpy than reactants), while a positive value means it is endothermic (heat is absorbed). The magnitude is the heat exchanged per mole of reaction as written, so doubling all coefficients doubles the enthalpy change.
Assumptions, Limitations and Tips
- Standard conditions: the calculator assumes all values refer to the same standard state (usually 298 K, 1 bar) and the same physical states of the substances. Mixing values for different states (e.g. H₂O(g) vs H₂O(l)) gives wrong results — check the state labels in your data table.
- Multipliers scale the reaction: when you multiply a reaction by \(n\), its enthalpy change is also multiplied by \(n\), and when you reverse a reaction the sign of \(\Delta H\) flips. The calculator applies this automatically.
- Per mole of reaction: the result refers to the reaction exactly as you balanced it. Always confirm the stoichiometric coefficients before comparing with literature values.
- Hess cycles must cancel cleanly: intermediate species must appear on both sides and cancel out. The calculator checks the arithmetic; verifying that the cycle is chemically valid is up to you.
- Unit conversion: the calculator converts at 1 kcal = 4.184 kJ (exact) and reports the answer in both units.
Common Standard Formation Enthalpies (kJ/mol, 298 K)
| Substance | \(\Delta H^\circ_f\) (kJ/mol) | Substance | \(\Delta H^\circ_f\) (kJ/mol) |
|---|---|---|---|
| CO₂(g) | −393.5 | CH₄(g) | −74.8 |
| H₂O(l) | −285.8 | C₂H₆(g) | −84.7 |
| H₂O(g) | −241.8 | C₃H₈(g) | −103.8 |
| CO(g) | −110.5 | NH₃(g) | −46.1 |
| NO(g) | +90.3 | SO₂(g) | −296.8 |
| HCl(g) | −92.3 | Elements in standard state | 0 (by definition) |
Values may vary slightly between reference tables; use one consistent source for all entries. To measure enthalpy changes experimentally, see the Calorimetry Calculator. Once you know \(\Delta H^\circ\), you can connect it to equilibrium behaviour with the Equilibrium Constant Calculator or study its temperature dependence with the Van 't Hoff Equation Calculator.
Frequently Asked Questions
How do I solve a Hess's law problem step by step by hand?
First, write the target reaction and the known reactions underneath. Second, reverse any known reaction so that its products/reactants point the way you need, flipping the sign of its \(\Delta H\). Third, multiply reactions so intermediate species cancel exactly when you add everything up. Finally, sum the modified enthalpy changes — the total is the target reaction's \(\Delta H\). If an intermediate species does not cancel, your multipliers are wrong.
Does Hess's law also apply to entropy and Gibbs free energy?
Yes. Hess's law works for any state function, so \(\Delta S^\circ\) and \(\Delta G^\circ\) can be combined across reactions exactly like \(\Delta H^\circ\): reverse a reaction and the sign flips, double the coefficients and the value doubles. The same is not true for temperature-dependent quantities like equilibrium constants, which must be converted to \(\Delta G\) first via \(\Delta G^\circ = -RT\ln K\).
Why is the enthalpy of formation of an element zero?
Because formation enthalpy is defined relative to the element in its own standard state. Forming a substance from itself means no change, so \(\Delta H^\circ_f = 0\) by definition. Note that "standard state" matters: oxygen gas O₂ is zero, but atomic O(g) is not, and carbon is zero only as graphite, not as diamond. Entering 1–2 kJ/mol instead of 0 for these is a common source of small errors.
How do I convert kJ/mol to kcal/mol myself?
Divide the value in kJ/mol by 4.184, which is the exact definition of the thermochemical calorie: \(1\ \text{kcal} = 4.184\ \text{kJ}\). For example, \(-110.5 \div 4.184 \approx -26.41\) kcal/mol. To go the other way, multiply by 4.184. A quick mental approximation is dividing by 4, but use the exact factor for anything you need to report.
Does a negative ΔH mean the reaction will happen on its own?
Not necessarily. A negative \(\Delta H\) only says the reaction releases heat; spontaneity depends on the Gibbs free energy, \(\Delta G = \Delta H - T\Delta S\). An exothermic reaction with a large entropy decrease can be non-spontaneous at high temperature, and many endothermic reactions proceed spontaneously because \(\Delta S\) is favorable. Use the enthalpy result together with entropy data before predicting whether a process occurs.
Can I use average bond enthalpies instead of formation enthalpies?
You can, but expect less accuracy. The bond-enthalpy version of Hess's law is \(\Delta H \approx \sum \) (bonds broken) \(- \sum\) (bonds formed), using tabulated average bond energies in kJ/mol. It works well for simple gas-phase reactions but can deviate by tens of kJ/mol because real bond strengths depend on the molecule. Standard formation enthalpies are the more precise route whenever they are available.
What are the most common mistakes in Hess's law calculations?
The top errors are: forgetting to flip the sign when reversing a reaction; forgetting to multiply \(\Delta H\) by the same factor as the coefficients; mixing physical states (H₂O(g) vs H₂O(l) differ by 44 kJ/mol); comparing your answer to a literature value for a differently balanced equation; and using \(\Delta H^\circ_f\) values from inconsistent reference tables. Always double-check the state labels and the coefficients before trusting the result.