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Enthalpy Change Calculator

Method

-74.6

0

-393.5

-285.8

Pick from the list for built-in values (state matters: H2O(l) vs H2O(g)), or type any species and enter its ΔHf°. Elements in their standard state are 0.

Try an example

Enthalpy change

ΔH = -890.5 kJ

Exothermic: the reaction releases heat. Per the reaction as written (per mole of reaction).

ΔH∘=∑n ΔHf∘(products)−∑n ΔHf∘(reactants)=−965.1−(−74.6)=−890.5 kJ\Delta H^\circ = \sum n\,\Delta H_f^\circ(\text{products}) - \sum n\,\Delta H_f^\circ(\text{reactants}) = -965.1 - (-74.6) = -890.5\ \text{kJ}

Calculate the enthalpy change of a reaction with Hess's law from standard enthalpies of formation, with built-in values for common substances, or estimate it from average bond enthalpies.

Formula

ΔH°=∑nΔHf°(products)−∑nΔHf°(reactants)\Delta \text{H}° = \sum \text{n} \Delta \text{Hf}°(\text{products}) - \sum \text{n} \Delta \text{Hf}°(\text{reactants})
from bonds:ΔH≈∑bonds broken−∑bonds formed\text{from bonds}: \Delta \text{H} \approx \sum \text{bonds broken} - \sum \text{bonds formed}

Worked Example

ReactionCH₄ + 2O₂ → CO₂ + 2H₂O(l)
ΔHf° valuesCH₄ −74.6, O₂ 0, CO₂ −393.5, H₂O(l) −285.8 kJ/mol
Result: ΔH° = −890.5 kJ (exothermic)

Products: −393.5 + 2 × (−285.8) = −965.1 kJ. Reactants: −74.6 + 0 = −74.6 kJ. ΔH° = −965.1 − (−74.6) = −890.5 kJ per mole of methane burned.

Frequently Asked Questions

What does a negative enthalpy change mean?

The reaction releases heat to its surroundings (exothermic), as in combustion. A positive ΔH means it absorbs heat (endothermic), as when limestone is decomposed into lime and carbon dioxide.

Why is the ΔHf° of O₂ zero?

By definition, the standard enthalpy of formation of an element in its most stable form at 25 °C and 1 bar is zero: O₂ gas, graphite, H₂ gas and so on. Ozone (O₃) or diamond, which are not the standard forms, have non-zero values.

Why does the bond-enthalpy method give a different answer?

Bond enthalpies are averages over many molecules and refer to gases, so they ignore the energy of condensing water and the exact bonding in each compound. Estimates from bond energies are often off by a few percent; values from enthalpies of formation are the accurate ones.