Enthalpy of Reaction Calculator
Enter the standard enthalpy of formation ΔH_f for each reactant and product and their stoichiometric coefficients; using ΔH° = ΣΔH_f(products) − ΣΔH_f(reactants) the tool instantly computes the standard reaction enthalpy.
Input Data
Results
At a glance:The enthalpy of reaction (standard reaction enthalpy, ΔH°_rxn) is the heat absorbed or released at constant pressure when a reaction proceeds as written with all substances in their standard states (1 bar, usually 298 K). It is the most central quantity in thermochemistry — combustion enthalpy, neutralisation enthalpy and formation enthalpy are all reaction enthalpies under specific stoichiometries. The key to computing it is Hess's law: enthalpy is a state function, so the enthalpy change of a reaction depends only on the initial and final states, not on the path. Consequently the reaction enthalpy can be obtained as 'sum of enthalpies of formation of products minus sum of enthalpies of formation of reactants', each multiplied by its stoichiometric coefficient: ΔH°_rxn = Σ (ν × ΔH_f)products − Σ (ν × ΔH_f)reactants, where ν is the stoichiometric coefficient (positive integers in the balanced equation). The 'standard enthalpy of formation' ΔH_f° is the enthalpy change when 1 mol of a compound is formed from its elements in their standard states; by definition the ΔH_f° of an element in its standard state (O₂ gas, N₂ gas, C graphite, etc.) is 0, which is why equations with elementary gases often include zero terms. Thus for a reaction with reactants A, B and products C, D: ΔH° = [ν_C·ΔH_f(C) + ν_D·ΔH_f(D)] − [ν_A·ΔH_f(A) + ν_B·ΔH_f(B)]. The sign indicates the heat direction: ΔH° < 0 exothermic (releases heat, surroundings warm), ΔH° > 0 endothermic (absorbs heat, surroundings cool). Using this tool's default (combustion of methane CH₄ + 2O₂ → CO₂ + 2H₂O): ΔH_f of CH₄ = −74.8, O₂ = 0, CO₂ = −393.5, H₂O(g) = −241.8 kJ/mol; ΔH° = [1·(−393.5) + 2·(−241.8)] − [1·(−74.8) + 2·0] = −802.3 kJ/mol, a strongly exothermic combustion. Notes: ΔH_f values are conventionally per mole of substance; the stoichiometric coefficient turns them into 'per mole of reaction' for summation; all ΔH_f should come from the same table and same phase/state (especially H₂O(l) vs H₂O(g) differ by ~44 kJ/mol) and same temperature (mostly 298 K); if you lack product/reactant ΔH_f you can get ΔH° from bond enthalpies or via Hess's law from other known reactions. In short, the enthalpy-of-reaction calculator turns 'tables of formation enthalpies plus a balanced equation' into a single line ΔH° — an essential tool for DSE Chemistry thermochemistry (combustion, neutralisation, formation) and energy-budget estimates.
Formula
Standard reaction enthalpy: ΔH°_rxn = Σ ν·ΔH_f(products) − Σ ν·ΔH_f(reactants).
Hess's law: enthalpy is a state function, path-independent.
Sign: ΔH° < 0 exothermic (releases heat); ΔH° > 0 endothermic (absorbs heat).
By definition ΔH_f° of an element in its standard state = 0.
$$\Delta H^\circ_{\text{rxn}} = \sum \nu \Delta H_f^\circ(\text{products}) - \sum \nu \Delta H_f^\circ(\text{reactants})$$How to Use
- Enter each reactant's stoichiometric coefficient and ΔH_f (kJ/mol); the tool multiplies them and sums.
- Enter each product's stoichiometric coefficient and ΔH_f (kJ/mol).
- The right panel instantly shows ΔH° and whether the reaction is exothermic or endothermic.
Standard enthalpies of formation ΔH_f° of common substances (298 K, kJ/mol)
| Substance | State | ΔH_f° (kJ/mol) |
|---|---|---|
| CO₂ | g | −393.5 |
| H₂O | l | −285.8 |
| H₂O | g | −241.8 |
| CH₄ | g | −74.8 |
| O₂ | g | 0 |
| C (graphite) | s | 0 |
ΔH_f° of an element in its standard state is 0; ΔH°_rxn = Σν·ΔH_f(products) − Σν·ΔH_f(reactants).
Case Studies
Combustion of methane
CH₄ + 2O₂ → CO₂ + 2H₂O(g); ΔH_f: CH₄=−74.8, O₂=0, CO₂=−393.5, H₂O(g)=−241.8 kJ/mol.
ΔH° = [1×(−393.5) + 2×(−241.8)] − [1×(−74.8) + 2×0] = −802.3 kJ/mol.
Strongly exothermic, the heat released by burning natural gas for cooking and heating in Hong Kong.
Formation of water
H₂ + ½O₂ → H₂O(l); ΔH_f: H₂=0, O₂=0, H₂O(l)=−285.8 kJ/mol.
ΔH° = 1×(−285.8) − [1×0 + 0.5×0] = −285.8 kJ/mol, the standard enthalpy of formation of liquid water.
If H₂O(g) is used instead (−241.8), the result differs by the latent heat of vaporisation, so state must be watched.
FAQ
What is Hess's law and why does the formula work?
Hess's law: enthalpy is a state function, so the enthalpy change of a reaction depends only on the initial and final states, not on the path. Therefore the reaction enthalpy equals the total formation enthalpy of the products minus that of the reactants, each multiplied by its stoichiometric coefficient.
Why is some ΔH_f taken as 0?
By definition the standard enthalpy of formation of an element in its standard state (O₂ gas, N₂ gas, C graphite, Fe solid, etc.) is 0, because forming it from itself requires no enthalpy change. So elementary substances in a balanced equation contribute a zero term.
What does a positive or negative ΔH° mean?
ΔH° < 0 means exothermic (the reaction releases heat, temperature of surroundings rises); ΔH° > 0 means endothermic (absorbs heat, surroundings cool). The magnitude is the heat per mole of reaction.
Why must the state be specified?
ΔH_f is state-dependent; the same formula H₂O as liquid (−285.8) and gas (−241.8) differ by ~44 kJ/mol of latent heat. Always use the state actually in the equation and keep all ΔH_f from the same table at the same temperature (usually 298 K).
What if I only have bond enthalpies?
You can estimate ΔH° ≈ Σ bond enthalpies broken (reactants) − Σ bond enthalpies formed (products). Bond enthalpies are average values, so the result is approximate and less accurate than using tabulated ΔH_f.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.