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Standard Enthalpy of Vaporization: Meaning, Sign and Value

Chemistry · Thermodynamics · NEET

Standard enthalpy of vaporization (Δvap H°) is the heat needed to turn one mole of a liquid into gas at its boiling point and standard pressure (1 bar). It is always positive because you must add energy to pull the liquid molecules apart. For water, Δvap H° = +40.79 kJ/mol. Memory hook: "Vapor eats heat" — boiling always takes energy IN, so the sign is +.
Standard Enthalpy of VaporizationLiquidH2O(l)Gas / VapourH2O(g)heat absorbed (endothermic)Δvap H° = +40.79 kJ/molAt boiling point (373 K) and 1 bar. Sign is always positive (+).
Vaporization turns liquid into gas by absorbing heat. Because energy is added to pull molecules apart, Δvap H° is always positive. For water it is +40.79 kJ/mol at the boiling point and 1 bar.

Your doubts, answered

Is enthalpy of vaporization positive or negative?

It is always positive. Vaporization means liquid turns into gas, and this needs heat to be added. You are breaking the attractions between liquid molecules, which costs energy. So Δvap H° is always greater than zero. For NEET, remember any phase change that INCREASES disorder (liquid to gas) absorbs heat, so the sign is +.

Why is enthalpy of vaporization always positive?

In a liquid the molecules attract each other and stay close. To make them fly apart as gas, you must supply energy to break these attractions. Adding heat = positive enthalpy change. The system absorbs heat, so vaporization is endothermic and Δvap H° > 0 for every substance.

What is the standard enthalpy of vaporization of water?

For water, Δvap H° = +40.79 kJ/mol (NCERT value). This is the heat needed to boil one mole of water at its boiling point (373 K, 100°C) under 1 bar pressure. The equation is: H2O(l) → H2O(g), Δvap H° = +40.79 kJ/mol. This is a common NEET number to remember.

What does 'standard' and 'molar' mean in enthalpy of vaporization?

'Standard' means it is measured at standard pressure of 1 bar. 'Molar' means it is for exactly one mole of the liquid. The '°' (or ⊖) symbol on Δvap H° tells you it is the standard value. So Δvap H° = heat to vaporize 1 mole of liquid at 1 bar, at the boiling point.

How is enthalpy of vaporization different from enthalpy of fusion?

Fusion (Δfus H°) is solid → liquid (melting). Vaporization (Δvap H°) is liquid → gas (boiling). Both are positive because both absorb heat. But vaporization needs MORE energy than fusion for the same substance, because in boiling you must fully separate the molecules, while melting only loosens them. Example: water fusion is +6.0 kJ/mol but vaporization is +40.79 kJ/mol.

What is the sign of enthalpy of condensation?

Condensation is the reverse of vaporization (gas → liquid), so its enthalpy is NEGATIVE and equal in size but opposite in sign to vaporization. If Δvap H° = +40.79 kJ/mol for water, then condensation releases −40.79 kJ/mol. NEET can trap you by reversing the direction, so always check which way the arrow points.

⚠️ The NEET trap
✗ Enthalpy of vaporization can be negative because energy is released when a gas forms.
✓ Enthalpy of vaporization is ALWAYS positive. Liquid → gas absorbs heat (endothermic). Only the reverse, condensation (gas → liquid), releases heat and is negative.
🧠 Arrow decides sign: liquid→gas = + (absorb), gas→liquid = − (release). Read the arrow before choosing.

Real NEET questions

NEET 2016 Phase 1

Consider the following liquid–vapour equilibrium: Liquid ⇌ Vapour. Which of the following relations is correct?

A · d(ln G)/dT² = ΔHv/RT²
B · d(ln P)/dT = −ΔHv/RT
C · d(ln P)/dT² = −ΔHv/T²
D · d(ln P)/dT = ΔHv/RT² ✓
Solution: This is the Clausius–Clapeyron relation for liquid–vapour equilibrium, where ΔHv is the enthalpy of vaporization. Starting from P = K·e^(−ΔHv/RT), take ln on both sides: ln P = ln K − ΔHv/RT. Differentiate with respect to T (ΔHv constant): d(ln P)/dT = +ΔHv/RT². So option D is correct. The positive sign matches the fact that Δvap H is positive, so vapour pressure rises as temperature increases.

Solved Thermodynamics NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 28 Thermodynamics NEET PYQs ›
Next concept: Standard Enthalpy of Sublimation (ΔsubH°) →Keep learning — 2 minFeeling ready? Solve the Thermodynamics NEET PYQs ›Or practice on your phone — get the free MedicNEET app ›

Frequently asked

Is vaporization endothermic or exothermic?

Endothermic. It absorbs heat, so Δvap H° is positive for every substance.

At what temperature is standard enthalpy of vaporization measured?

At the boiling point of the liquid (for water, 373 K / 100°C) and under standard pressure of 1 bar.

Why does vaporization need more heat than fusion?

Because boiling must fully separate the molecules into gas, while melting only loosens them slightly from a rigid solid. So Δvap H° > Δfus H° for the same substance.

What is the unit of enthalpy of vaporization?

kilojoules per mole (kJ/mol), because it is a molar quantity (per one mole of liquid).

How are vaporization, fusion and sublimation related?

For the same substance, Δsub H° = Δfus H° + Δvap H°. Sublimation (solid → gas) equals melting plus boiling because it does both steps in one go.

Related concepts:
Standard Enthalpy of Fusion (Melting)Standard Enthalpy of SublimationWhy Phase-Change Enthalpies Are PositiveStandard Enthalpy of Reaction and Standard StateEntropy Order: Solid < Liquid < Gas
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Thermodynamics PYQs