Chemistry · Thermodynamics · NEET
It is the amount of heat you must add to break 1 mole of a substance into separate, free atoms, and those atoms must end up in the gas (vapour) state. For example, breaking 1 mole of H2 gas gives 2 moles of free H atoms. Because you are pulling atoms apart, you always have to supply energy, so the value is positive. The symbol is Delta_a H and the standard value is written Delta_a H (with a small circle for standard conditions).
Yes. To separate atoms you must break bonds (or break the forces holding a solid or liquid together). Breaking any bond needs energy, so heat is absorbed and Delta_a H is always positive (endothermic). If you ever see a negative sign in a question for atomization, it is a trap or a printing error. For NEET, remember: atomization is always positive.
For a diatomic molecule (a molecule with only two atoms, like H2, Cl2, O2), they are exactly equal. Breaking H2 into 2 H atoms breaks one H-H bond, so Delta_a H = bond dissociation enthalpy. But for molecules with more than two atoms (like CH4), they are NOT the same. In CH4, atomization breaks all four C-H bonds at once, so Delta_a H = sum of all four bonds = 1665 kJ/mol, while one bond dissociation enthalpy is only about one bond. See the next concept, bond dissociation enthalpy vs mean bond enthalpy, for the full comparison.
Only for metals and other solids that turn directly into single gaseous atoms. Example: solid sodium turning into Na atoms in the gas phase. Here atomization = sublimation, because sublimation already produces free atoms. But for a molecule like H2 or CH4, sublimation would give gaseous molecules, not atoms, so atomization is a bigger, separate step. So they match only for atomic (monatomic vapour) solids like Na, Fe, C(graphite).
There is no single algebra formula; it is defined per reaction. You write the atomization reaction (substance turning into free gaseous atoms) and Delta_a H is the enthalpy change for making 1 mole of the substance break into atoms. Example: H2(g) -> 2H(g), Delta_a H = 435.0 kJ/mol. The unit is kJ/mol (kilojoule per mole).
CH4 has four C-H bonds. Atomization of CH4 breaks all four bonds: CH4(g) -> C(g) + 4H(g), and this needs 1665 kJ/mol total. If you divide by 4 you get the MEAN (average) C-H bond enthalpy = 416 kJ/mol. So atomization enthalpy is the total of all bonds, while the average bond energy is that total divided by the number of bonds. They are only equal for diatomic molecules that have just one bond.
The bond dissociation energies of X2, Y2 and XY are in the ratio 1 : 0.5 : 1. Delta H for the formation of XY is -200 kJ/mol. The bond dissociation energy of X2 will be:
For the reaction 2Cl(g) -> Cl2(g), the correct option is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It links directly to bond enthalpy questions, Hess's law calculations, and formation enthalpy problems, which are common in the Thermodynamics chapter. Knowing that atomization is always positive and equals the sum of all bond enthalpies helps you solve numericals quickly.
For H2(g) -> 2H(g), Delta_a H = 435.0 kJ/mol. Because H2 is diatomic, this is also the H-H bond dissociation enthalpy. They are the same value.
No. Atomization always breaks bonds or breaks the forces in a solid or liquid to make free atoms, and that always absorbs energy. So the value is always positive (endothermic).
For metals the atoms are already single atoms in the solid, so atomization just means turning the solid into gaseous atoms. This equals the enthalpy of sublimation. Example: Na(s) -> Na(g), Delta_a H = Delta_sub H.
kJ/mol (kilojoule per mole), because it is the energy per one mole of the substance being broken into atoms.
