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Thermodynamics questions list several named enthalpies, and students often treat them as separate facts. They are easier to handle as one family: each describes a start state and an end state per mole, and Hess's law links them. This post defines each one, shows when two are equal, and works examples with iodine, water and dry ice that differ from the ones in the NCERT text.
What does NCERT say, and which exams include it?
The NCERT Class 11 chapter Thermodynamics (Reprint 2026-27) defines:
Enthalpy of fusion, ΔfusH: melting one mole of solid; for water 6.00 kJ/mol at 273 K. All fusion enthalpies are positive.
Enthalpy of vaporisation, ΔvapH: vaporising one mole of liquid at constant temperature and 1 bar; for water 40.79 kJ/mol at its boiling point and 44.01 kJ/mol at 298 K.
Enthalpy of sublimation, ΔsubH: solid straight to vapour; dry ice sublimes at 195 K with 25.2 kJ/mol, and naphthalene has 73.0 kJ/mol.
Enthalpy of atomisation, ΔaH: breaking a substance completely into atoms in the gas phase, for example H₂(g) → 2H(g), 435.0 kJ/mol, and Na(s) → Na(g), 108.4 kJ/mol. For sodium, NCERT notes, atomisation is the same as sublimation.
The JEE (Main) 2026 syllabus lists enthalpies of "atomization, sublimation, phase transition" with bond dissociation, combustion and formation, and the NEET (UG) 2026 syllabus lists the same set. The JEE (Advanced) 2026 brochure names only fusion and vaporisation among these, alongside Hess's law. Check the syllabus for your exam year.
How are they related?
Write each as a start and end state per mole.
Fusion: solid → liquid.
Vaporisation: liquid → gas.
Sublimation: solid → gas. At one temperature, Hess's law gives ΔsubH = ΔfusH + ΔvapH.
Atomisation: substance → separate gaseous atoms.
When the gas is already made of single atoms, as for sodium vapour, atomisation and sublimation describe the same change. When the gas is made of molecules, atomisation also includes breaking the bonds.
Worked example 1: iodine, from solid to atoms
NIST data give the standard enthalpy of formation of I₂(g) as 62.42 kJ/mol at 298 K. Since solid I₂ is iodine's standard state, with enthalpy of formation zero, that number is the sublimation enthalpy. NIST gives 106.76 kJ/mol for the iodine atom in the gas phase.
Sublimation: I₂(s) → I₂(g), ΔH = +62.42 kJ/mol.
Atomisation: I₂(s) → 2I(g), ΔH = 2 × 106.76 = +213.52 kJ per mole of I₂, or 106.76 kJ per mole of iodine atoms.
Bond dissociation of gaseous I₂: I₂(g) → 2I(g), ΔH = 213.52 − 62.42 = +151.10 kJ/mol.
So turning solid iodine into atoms costs about 29% for separating the molecules and about 71% for breaking the I–I bonds.
Worked example 2: ice to steam
How much heat turns 36.0 g of ice at 0 °C into steam at 100 °C? Use NCERT's values: ΔfusH = 6.00 kJ/mol, specific heat of water 4.2 J g⁻¹ °C⁻¹, ΔvapH = 40.79 kJ/mol at 100 °C. Here 36.0 g is 2.00 mol.
Melt at 0 °C: 2.00 × 6.00 = 12.0 kJ.
Heat the liquid from 0 to 100 °C: 36.0 × 4.2 × 100 = 15,120 J = 15.1 kJ.
Vaporise at 100 °C: 2.00 × 40.79 = 81.6 kJ.
Total ≈ 108.7 kJ. Vaporisation is about three-quarters of it. ΔvapH is about 6.8 times ΔfusH for water, because melting only loosens the hydrogen-bonded network while boiling pulls the molecules fully apart.
Worked example 3: dry ice
How much heat does 1.00 kg of dry ice absorb as it sublimes at 195 K? Moles = 1000/44.01 = 22.7 mol, so the heat is 22.7 × 25.2 ≈ 573 kJ.
Which mistakes cost marks?
Mixing "per mole of substance" with "per mole of atoms". For I₂(s) → 2I(g), state which you mean.
Treating atomisation as sublimation for molecular solids. They match only when the vapour is atomic.
Adding enthalpies from different temperatures in a Hess cycle without saying it is an approximation.
Forgetting signs. Fusion, vaporisation, sublimation and atomisation are all positive; the reverse changes are negative.
Skipping the heating step in ice-to-steam problems.
These pieces feed the Born-Haber cycle, where sublimation and atomisation are separate steps; see our post on lattice enthalpy and the Born-Haber cycle. For building cycles in general, see Hess's law and enthalpy calculations.
A practice routine
Write the start and end state for each of the four enthalpies, with the sign.
Redo the iodine example for chlorine, using NCERT's 242 kJ/mol for Cl₂(g) → 2Cl(g), and give the answer per mole of atoms.
Repeat the ice-to-steam calculation for 9.0 g, then for 54.0 g.
Explain in two lines why ΔaH equals ΔsubH for sodium but not for iodine.
If these definitions keep blurring, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can review each enthalpy term on its own.
Today, draw the iodine cycle as three boxes and arrows, and check that the numbers add up.
