The standard enthalpy of reaction $\Delta_r H^\circ$ is the enthalpy change when reactants in their standard states (pure form at 1 bar and a stated temperature, usually 298 K) convert completely to products in their standard states. It is calculated from standard enthalpies of formation using
$\Delta_r H^\circ=\sum \nu_p\,\Delta_f H^\circ(\text{products})-\sum \nu_r\,\Delta_f H^\circ(\text{reactants})$
where $\nu$ are stoichiometric coefficients. By convention the standard enthalpy of formation of an element in its most stable form (e.g. $\text{O}_2(g)$, graphite) is zero.
Several named enthalpy changes appear repeatedly. The enthalpy of formation $\Delta_f H^\circ$ is for forming one mole of a compound from its elements. The enthalpy of combustion $\Delta_c H^\circ$ is for complete combustion of one mole in oxygen (always negative). The enthalpy of neutralisation is the heat released when one mole of water forms from acid–base reaction; for strong acid + strong base it is a near-constant $-57.1$ kJ/mol because only $\text{H}^++\text{OH}^-\rightarrow\text{H}_2\text{O}$ occurs.
Other key terms: enthalpy of atomisation $\Delta_a H^\circ$ (breaking one mole of substance into gaseous atoms), enthalpy of sublimation $\Delta_{sub} H^\circ$ (solid to gas directly) and ionisation enthalpy (removing an electron from a gaseous atom). These building blocks let us assemble cycles for changes that cannot be measured directly.
The cornerstone is Hess’s law of constant heat summation: the total enthalpy change for a reaction is the same whether it occurs in one step or in several steps. This follows directly because $H$ is a state function — the change depends only on initial and final states, not the path. Practically, we add up the enthalpies of a sequence of steps (reversing a step changes the sign; multiplying a step multiplies its $\Delta H$) to obtain the enthalpy of an overall reaction.
Bond enthalpy is the energy needed to break one mole of a particular bond in the gaseous state. Because breaking bonds absorbs energy and forming bonds releases it, an estimate of reaction enthalpy is
$\Delta_r H=\sum(\text{bond enthalpies of bonds broken})-\sum(\text{bond enthalpies of bonds formed})$
The lattice enthalpy $\Delta_{lattice} H^\circ$ of an ionic solid is the enthalpy change when one mole of the solid separates into gaseous ions. It cannot be measured directly, so we obtain it from a Born–Haber cycle — a Hess’s-law cycle that links sublimation, dissociation, ionisation, electron gain and lattice formation back to the standard enthalpy of formation, allowing us to solve for the one unknown lattice enthalpy.