Calculate the heat of formation or reaction enthalpy from a balanced equation: enter coefficients and ΔH°f values, or solve for one unknown ΔH°f.

Use coefficients from a balanced equation. Leave both entries of an unused row blank; elements in their standard state have ΔH°f = 0.

Rows 3–4 and display precision

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).

Δ H^°ᵣₓₙ = Σ nu Δ H^°f(products) - Σ nu Δ H^°f(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 for formation of 1 mol of the substance from its elements in their reference states at the specified temperature and standard pressure (1 bar). Tables often use 298.15 K; standard state alone does not fix temperature

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 for each species in the correct phase at a common specified temperature. Elements in their reference states have ΔH°f = 0. Select the correct kJ/mol or thermochemical kcal/mol input unit for each row.
  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 per mol of reaction as written

Σ(νΔH°f) reactants = 2 × 0 + 1 × 0 = 0 kJ per mol of reaction as written

ΔH°rxn = −571.66 − 0 = −571.66 kJ per mol of reaction as written (exothermic). This doubled water-formation equation produces 2 mol of water; the molar formation enthalpy of H₂O(l) is −285.83 kJ/mol. Use the balanced equation basis consistently.