Enthalpy of Reaction from Formation Enthalpies Calculator

Enter Calculation

Σ ΔHf° of Products (kJ/mol)

Σ ΔHf° of Reactants (kJ/mol)

Enter the stoichiometrically-weighted sum of standard formation enthalpies for each side. Multiply each species' ΔHf° by its coefficient before summing. Elements in their standard state have ΔHf° = 0.

Results

Standard Enthalpy of Reaction (ΔH°rxn)

-393.5

kJ/mol

Reaction nature	Exothermic (releases heat)
Σ ΔHf° products	-393.5 kJ/mol
Σ ΔHf° reactants	0 kJ/mol

What This Calculator Does

This tool computes the standard enthalpy of reaction (\(\Delta H^{\circ}_{rxn}\)) using Hess's law of constant heat summation. By subtracting the total standard enthalpy of formation of the reactants from that of the products, you obtain the net heat change of a chemical reaction under standard conditions (1 bar, 298.15 K). A negative result means the reaction is exothermic; a positive result means it is endothermic.

How to Use It

For each side of the balanced equation, multiply every species' standard formation enthalpy (\(\Delta H_f^{\circ}\), in kJ/mol) by its stoichiometric coefficient, then add them up. Remember that pure elements in their standard reference state (such as O₂, N₂, or solid C as graphite) have \(\Delta H_f^{\circ} = 0\). Enter the product sum in the first field and the reactant sum in the second, then read off \(\Delta H^{\circ}_{rxn}\).

The Formula Explained

The governing equation is $$\Delta H^{\circ}_{rxn} = \sum \Delta H_f^{\circ}\,\text{Products} - \sum \Delta H_f^{\circ}\,\text{Reactants}$$ Because enthalpy is a state function, the path taken between reactants and products does not matter — only the initial and final states. This lets us treat tabulated formation enthalpies as building blocks and combine them algebraically.

The reaction enthalpy is the difference between the summed formation enthalpies of products and reactants.

Worked Example

Consider combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O(l). Products: \(\Delta H_f^{\circ}(\text{CO}_2) = -393.5\) plus $$2 \times (-285.8) = -393.5 - 571.6 = -965.1\ \text{kJ/mol}$$ Reactants: \(\Delta H_f^{\circ}(\text{CH}_4) = -74.8\) plus \(2 \times 0\) (O₂) \(= -74.8\ \text{kJ/mol}\). So $$\Delta H^{\circ}_{rxn} = -965.1 - (-74.8) = -890.3\ \text{kJ/mol}$$ a strongly exothermic reaction.

Hess's law: both reactants and products are referenced to the same elemental standard states.

FAQ

Why is the formation enthalpy of O₂ zero? By definition, the standard formation enthalpy of any element in its most stable form at standard conditions is zero, serving as the reference point.

What does a positive \(\Delta H^{\circ}_{rxn}\) mean? The reaction absorbs heat from the surroundings (endothermic).

Must I include stoichiometric coefficients? Yes — multiply each species' \(\Delta H_f^{\circ}\) by its coefficient before summing each side, or your result will be incorrect.