Enter stoichiometric coefficients and standard enthalpies of formation (ΔH°f, kJ/mol) for reactants and products into the Heat of Formation Calculator to calculate the standard enthalpy change of reaction (ΔH°rxn). You can also solve for one unknown ΔH°f or estimate ΔH° at a different temperature using Kirchhoff’s law.

Heat of Formation / Reaction Enthalpy Calculator

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Enter stoichiometric coefficients and ΔH°f values (kJ/mol) for each species. Positive coefficients only.

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Standard Reaction Enthalpy (ΔH°rxn) from ΔH°f Formula

The following example outlines the steps and information needed to calculate the standard enthalpy change of reaction (ΔH°rxn) using tabulated standard enthalpies of formation (ΔH°f).

\Delta H^\circ_{\mathrm{rxn}}=\sum \nu\,\Delta H^\circ_f(\text{products})-\sum \nu\,\Delta H^\circ_f(\text{reactants})

Variables:

  • ΔH°rxn is the standard enthalpy change of reaction (kJ/mol of reaction as written)
  • ν is the stoichiometric coefficient for each species (use positive values on each side)
  • ΔH°f is the standard enthalpy of formation of each species (kJ/mol), defined as the enthalpy change to form 1 mol of a compound from its elements in their standard states (typically at 298.15 K and 1 bar)

To calculate ΔH°rxn from ΔH°f values, sum νΔH°f for the products and subtract the sum νΔH°f for the reactants (elements in their standard states have ΔH°f = 0).

How to Calculate ΔH°rxn from ΔH°f?

The following steps outline how to calculate the standard enthalpy change of reaction.

  1. Write the balanced chemical equation and identify each species on the reactant and product sides.
  2. Find ΔH°f (kJ/mol) for each species (note: elements in their standard states have ΔH°f = 0).
  3. Compute Σ(νΔH°f) for the products and Σ(νΔH°f) for the reactants.
  4. Subtract: ΔH°rxn = Σ(νΔH°f)products − Σ(νΔH°f)reactants.
  5. Check your answer with the calculator above.

Example Problem: 

Compute ΔH°rxn for the formation of liquid water: 2 H2(g) + O2(g) → 2 H2O(l). Use ΔH°f[H2O(l)] = −285.83 kJ/mol and ΔH°f for H2(g) and O2(g) (standard states) = 0.

Σ(νΔH°f) products = 2 × (−285.83) = −571.66 kJ/mol

Σ(νΔH°f) reactants = 2 × 0 + 1 × 0 = 0 kJ/mol

ΔH°rxn = −571.66 − 0 = −571.66 kJ/mol (exothermic)