Standard Enthalpy of Reaction from Formation Enthalpies

This is Hess's law with the bookkeeping already done for you. Every compound gets one tabulated number (the enthalpy of forming one mole of it from its elements in their standard states) and every element in its standard state is defined as exactly zero. Build any reaction out of "unmake the reactants back to elements, then make the products", and the whole route collapses to products minus reactants. Multiply each ΔH°f by its coefficient in the balanced equation before you add: the sums that go into this calculator are already coefficient-weighted.

The two classic slips are the minus sign and the physical state. Reactants are subtracted, so a strongly negative reactant enthalpy pushes ΔH°rxn up, not down. And ΔH°f depends on state: liquid water is −285.8 kJ/mol but water vapour is −241.8 kJ/mol, a 44 kJ/mol gap that is exactly the enthalpy of vaporisation and the entire difference between a fuel's higher and lower heating value.

Worked case: burning methane, CH₄ + 2O₂ → CO₂ + 2H₂O(l). Products: −393.5 + 2(−285.8) = −965.1 kJ/mol. Reactants: −74.6 + 2(0) = −74.6 kJ/mol, since O₂ is an element in its standard state. ΔH°rxn = −965.1 − (−74.6) = −890.5 kJ/mol, the number a gas utility is selling you, one mole at a time.