The enthalpy calculator applies Hess's law to find the standard reaction enthalpy (ΔH°rxn) from standard enthalpies of formation. Select reactants and products from a 40+ compound database or enter custom ΔHf° values, set stoichiometric coefficients, and instantly see whether the reaction is exothermic or endothermic with a full step-by-step breakdown.
Quick examples:
R Reactants
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Results
Hess's Law Formula
Calculation Steps
How to Use the Enthalpy Calculator
This enthalpy calculator uses Hess's law to find the standard enthalpy change (ΔH°rxn) for chemical reactions. By combining the standard enthalpies of formation of all reactants and products, you can determine whether a reaction releases or absorbs heat — and how much.
Step 1: Enter Reactants
In the Reactants panel, select each compound from the dropdown (which includes 40+ common substances) and enter its stoichiometric coefficient. For example, the combustion of methane has CH4(g) and 2 O2(g) as reactants — set coefficient to 1 for methane and 2 for oxygen.
Step 2: Enter Products
In the Products panel, add each product with its coefficient. For methane combustion: 1 CO2(g) and 2 H2O(l). Use the "Custom" option to enter ΔHf° values for compounds not in the database.
Step 3: Calculate
Click "Calculate ΔH°rxn" to apply Hess's law: ΔH°rxn = Σ(n × ΔHf° products) − Σ(n × ΔHf° reactants). The result shows in kJ/mol with an exothermic/endothermic classification.
Worked Example: Combustion of Methane
The reaction CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l):
- Products: ΔHf°(CO2) = −393.5, 2×ΔHf°(H2O,l) = 2×(−285.8) = −571.6
- Reactants: ΔHf°(CH4) = −74.8, 2×ΔHf°(O2) = 2×0 = 0
- ΔH°rxn = (−393.5 − 571.6) − (−74.8 + 0) = −965.1 − (−74.8) = −890.3 kJ/mol
The negative value confirms this is an exothermic reaction — burning methane releases 890.3 kJ per mole of methane burned.
About Standard Enthalpies of Formation
Standard enthalpies of formation (ΔHf°) are measured at 298.15 K (25°C) and 1 bar pressure. Elements in their standard states (H2(g), O2(g), graphite for carbon) have ΔHf° = 0 by definition. The more negative a ΔHf° value, the more stable the compound relative to its elements. Water has ΔHf° = −285.8 kJ/mol (liquid) or −241.8 kJ/mol (gas), reflecting the significant energy released when hydrogen and oxygen combine.
FAQ
What is Hess's law?
Hess's law states that the total enthalpy change of a reaction is the same regardless of the path taken. This means ΔH°rxn = Σ(n × ΔHf° of products) − Σ(n × ΔHf° of reactants), where n is the stoichiometric coefficient and ΔHf° is the standard enthalpy of formation.
Is this calculator free?
Yes, completely free with no signup or account required. All calculations run in your browser.
Is my data private?
Yes. All calculations run locally in your browser. No data is sent to any server.
What is the standard enthalpy of formation?
The standard enthalpy of formation (ΔHf°) is the heat change when one mole of a compound forms from its elements in their standard states at 298.15 K and 1 atm. By definition, ΔHf° = 0 for all elements in their standard states (H2(g), O2(g), C(graphite), etc.).
What does exothermic vs endothermic mean?
An exothermic reaction has ΔH° < 0 — it releases heat to the surroundings (burning, rusting). An endothermic reaction has ΔH° > 0 — it absorbs heat from the surroundings (dissolving ammonium nitrate, photosynthesis).
Can I enter custom enthalpy values?
Yes. If a compound is not in the built-in database, select 'Custom' and enter the ΔHf° value in kJ/mol directly. This is useful for exotic compounds or when using values from your textbook that differ from standard references.
What is the combustion enthalpy of methane?
The combustion of methane (CH4 + 2O2 → CO2 + 2H2O) has ΔH°rxn = [ΔHf°(CO2) + 2×ΔHf°(H2O(l))] − [ΔHf°(CH4) + 2×0] = [−393.5 + 2×(−285.8)] − [−74.8] = −890.3 kJ/mol. This is a highly exothermic reaction.