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A bond-dissociation enthalpy is the molar enthalpy change for the specified homolysis at temperature $T$, with reactant and products in specified standard states. Its usual units are $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The superscript $\circ$ specifies standard states; it does not fix $T$ to room temperature.^[ref]

Write $\delta h_i(T)=H_{m,i}^{\circ }(T)-H_{m,i}^{\circ }(0)$ for each species' molar thermal enthalpy increment. With $D_0$ an energy per parent molecule,Here $N_{\mathrm{A}}$ is the Avogadro constant. The thermal increments include translational kinetic energy, molecular rotation, thermally excited molecular vibration and electronic states when populated, and the $pV$ contribution to enthalpy. They are additional to the zero-point energies already included in $D_0$. If values of standard enthalpy of formation at the same $T$ are available, Hess's law gives the same quantity as ${\Delta }_f{H}^{\circ }(A)+{\Delta }_f{H}^{\circ }(B)-{\Delta }_f{H}^{\circ }(AB)$.

Units alone do not identify the convention. Multiplying either $D_e$ or $D_0$ by the Avogadro constant gives molar energy, but does not add thermal enthalpy. One electronvolt per dissociation corresponds to $96.485332\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. A spectroscopic wavenumber $\tilde{\nu }$ represents energy $hc\tilde{\nu }$, using the Planck constant $h$ and speed of light $c$; $1\mathrm{c}{\mathrm{m}}^{-1}$ corresponds to $0.011962657\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

Homolysis splits a covalent bond with one bonding Electron assigned to each fragment. A neutral closed-shell molecule thereby produces fragments with unpaired Electrons, each a radical. The usual bond-dissociation enthalpy uses this product convention.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.