Chemistry · Thermodynamics · NEET
It is the enthalpy change when one mole of a solid turns directly into vapour (gas) at a constant temperature and under standard pressure of 1 bar. The word 'standard' just means the change happens at 1 bar. 'Sublimation' means the solid skips the liquid stage and becomes gas in one step. Example from NCERT: dry ice (solid CO2) sublimes with ΔsubH° = 25.2 kJ/mol, and naphthalene with ΔsubH° = 73.0 kJ/mol. This matters for NEET because it appears in Hess's law problems and Born-Haber cycles.
It is always positive. To pull the tightly packed particles of a solid apart into free-moving gas particles, the substance must absorb heat from the surroundings. Absorbing heat means ΔH is positive (endothermic). You will never see a negative ΔsubH° for a normal sublimation. This is the same reason ΔfusH (melting) and ΔvapH (boiling) are also positive.
Enthalpy is a state function, so the total heat depends only on start and end states, not the path. Going from solid to gas directly (sublimation) has the same enthalpy change as going solid to liquid (fusion) and then liquid to gas (vaporisation). So you just add the two: ΔsubH = ΔfusH + ΔvapH. This is Hess's law. Remember: the values must be taken at the same temperature for the addition to be exact.
Fusion (ΔfusH) is solid to liquid (melting). Vaporisation (ΔvapH) is liquid to gas (boiling). Sublimation (ΔsubH) is solid straight to gas. All three are positive because heat is absorbed. Sublimation is the biggest of the three for the same substance, because it equals fusion plus vaporisation combined.
Yes. Going from an ordered solid to a spread-out gas increases disorder (randomness), so ΔS is positive. NEET has directly asked this: sublimation of a solid to gas has ΔS greater than 0. Do not confuse this with a reaction like 2H(g) → H2(g), where gas particles combine and entropy decreases.
The size of the enthalpy change depends on how strongly the particles attract each other in the solid. Naphthalene molecules have stronger intermolecular forces than solid CO2 molecules, so more heat (73.0 kJ/mol vs 25.2 kJ/mol) is needed to separate them into gas. Stronger forces means larger ΔsubH.
In which case is the change in entropy negative?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
ΔsubH = ΔfusH + ΔvapH, taken at the same temperature. This comes from Hess's law because enthalpy is a state function.
Endothermic. The solid absorbs heat to break apart into gas, so ΔsubH° is positive.
NCERT gives ΔsubH° = 25.2 kJ/mol for solid CO2 (dry ice), which sublimes at 195 K.
It means the process happens at the standard pressure of 1 bar. The value is reported per one mole of the solid.
In Hess's law calculations, in finding ΔfusH or ΔvapH when the other two are known, and as the first step (metal solid → metal gas) in Born-Haber cycles.
