Bonds broken, bonds formed.

An energy staircase: reactant bonds are broken up to free gaseous atoms, then product bonds form down to products; ΔrH = broken − formed is the net drop or rise.

Break and form every bond for H2 + Cl2 → 2HCl and land on ΔrH = −185 kJ (435 + 242 − 2×431).

Using products minus reactants with bond enthalpies.

• Standard enthalpy of combustion ΔcH° is the enthalpy change per mole of a substance burnt completely in oxygen, all species in standard states; combustion is always exothermic.
• Enthalpy of atomisation is the enthalpy change when all the bonds in one mole of a substance are broken to give separate gaseous atoms; for H₂ it is 435.0 kJ mol⁻¹, and for CH₄ it is 1665 kJ mol⁻¹.
• Bond dissociation enthalpy applies to one specific bond in a diatomic or particular molecule; for polyatomic molecules a mean bond enthalpy is used (C–H in CH₄: 1665/4 = 416 kJ mol⁻¹).
• Estimating reaction enthalpy for gas-phase reactions: ΔᵣH° = Σ bond enthalpies of reactants − Σ bond enthalpies of products (bonds broken minus bonds formed).
• Mean bond enthalpies give only approximate reaction enthalpies because a given bond's strength varies from molecule to molecule.
• Lattice enthalpy is the enthalpy needed to pull one mole of an ionic solid fully apart into its gaseous ions; for NaCl it is +788 kJ mol⁻¹.
• Lattice enthalpy cannot be measured directly; it is found from a Born-Haber cycle, which applies Hess's law to steps such as sublimation, ionisation, dissociation and electron gain.
• Enthalpy of solution ΔsolH is the enthalpy change for dissolving one mole of a substance in a stated amount of solvent; for an ionic solid, ΔsolH = ΔlatticeH + ΔhydH. For NaCl, +788 − 784 = +4 kJ mol⁻¹, so dissolving it barely changes the temperature.
• ΔsolH is positive for most ionic salts, so their solubility in water rises with temperature; a very high lattice enthalpy can stop a salt dissolving at all.
• Enthalpy of dilution is the enthalpy change when extra solvent is added to a solution; it depends on the starting concentration and on how much solvent is added. For HCl, going from HCl·25 aq to HCl·40 aq gives −72.79 − (−72.03) = −0.76 kJ mol⁻¹.

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