---
title: "Enthalpy of Vaporization — AP Chem Definition & Exam Guide"
description: "Enthalpy of vaporization is the heat needed to turn one mole of liquid into gas at constant temperature. Master q = nΔHvap and heating curve problems for AP Chem Unit 6."
canonical: "https://fiveable.me/ap-chem/key-terms/enthalpy-of-vaporization"
type: "key-term"
subject: "AP Chemistry"
unit: "Unit 6"
---

# Enthalpy of Vaporization — AP Chem Definition & Exam Guide

## Definition

Enthalpy of vaporization (ΔHvap) is the energy required to convert one mole of a liquid into a gas at constant temperature and pressure. In AP Chem Topic 6.5, you use it in q = nΔHvap to find the heat absorbed during boiling, and its sign flips for condensation.

## What It Is

Enthalpy of vaporization (ΔHvap) is the amount of energy you have to add to convert one [mole](/ap-chem/key-terms/mole "fv-autolink") of a [liquid](/ap-chem/unit-3/solids-liquids-gases/study-guide/uNMmoAyjJQkL63ysUrJC "fv-autolink") into a gas at constant temperature and pressure. It's always positive, because vaporization is endothermic. The molecules in a liquid are held together by intermolecular forces, and turning that liquid into a gas means supplying enough energy to pull every molecule completely away from its neighbors.

Two details from the CED trip people up. First, the temperature of a [pure substance](/ap-chem/unit-1/elemental-composition-pure-substances/study-guide/EKrsYbL2G7PpzhUXDOSz "fv-autolink") does not change during a phase change (EK 6.5.A.1). All the energy you add at the boiling point goes into breaking intermolecular attractions, not into making molecules move faster. Second, vaporization and condensation are complementary processes (EK 6.5.A.2). Whatever energy a sample absorbs to vaporize, it releases the exact same amount when it condenses. So ΔHcond = -ΔHvap, same number, opposite sign.

## Why It Matters

This term lives in **Topic 6.5 (Energy of Phase Changes)** in **[Unit 6](/ap-chem/unit-6 "fv-autolink"): Thermochemistry**, directly supporting learning objective **6.5.A**, which asks you to explain the heat absorbed or released during a phase transition based on moles and the molar enthalpy of that transition. In practice, that means the calculation q = n × ΔHvap, where n is moles of substance. Notice the units. ΔHvap is given in kJ per *mole*, so if a problem hands you grams, converting with molar mass is step one. Enthalpy of vaporization is also the quantitative payoff of Unit 3's [intermolecular forces](/ap-chem/key-terms/intermolecular-forces "fv-autolink"). Stronger IMFs mean a larger ΔHvap, which is exactly the kind of cross-unit reasoning the exam loves.

## Connections

### [Heat of Fusion (Unit 6)](/ap-chem/key-terms/heat-of-fusion)

[Heat of fusion](/ap-chem/key-terms/heat-of-fusion "fv-autolink") is the same idea applied to melting instead of boiling. For water, ΔHvap (40.7 kJ/mol) is almost seven times larger than ΔHfus (6.02 kJ/mol), because vaporization fully separates molecules while melting only loosens them. Multi-step heating curve problems make you use both.

### [Condensation (Unit 6)](/ap-chem/key-terms/condensation)

[Condensation](/ap-chem/key-terms/condensation "fv-autolink") is vaporization run in reverse, so its enthalpy is exactly -ΔHvap. If 25.0 g of water absorbs a certain amount of heat to vaporize at 100°C, the same 25.0 g of steam releases that identical amount when it condenses. Energy in equals energy out.

### [Intermolecular Forces (Unit 3)](/ap-chem/key-terms/intermolecular-forces)

ΔHvap is basically a price tag on a liquid's intermolecular forces. Hydrogen-bonding liquids like HF cost more energy to vaporize than liquids with only dipole-dipole or dispersion forces. The 2023 free-response exam asked exactly this kind of comparison with HBr(l) and HF(l).

### Heat Capacity and Calorimetry (Unit 6)

Heating curve problems stitch q = mcΔT segments ([temperature](/ap-chem/unit-5/reaction-rates/study-guide/4V94d3BwjoPaOOyQtDKQ "fv-autolink") rising within one phase) together with q = nΔH segments (flat plateaus during phase changes). Knowing which formula applies to which part of the curve is the whole game in Topic 6.5.

## On the AP Exam

Enthalpy of vaporization shows up in two main ways. The first is straight calculation. A typical multiple-choice or short FRQ part gives you a mass, asks you to convert to moles, and apply q = nΔHvap. For example, vaporizing 45.0 g of ethanol (ΔHvap = 38.6 kJ/mol) means dividing by ethanol's molar mass first, then multiplying. Harder versions chain steps together, like taking 18.0 g of ice at 0°C all the way to steam at 100°C, which requires ΔHfus, then q = mcΔT for heating the liquid, then ΔHvap. The second way is conceptual. You might be asked why measured enthalpies of vaporization and condensation should be equal and opposite, or why temperature stays constant during boiling. The 2023 short-response FRQ (Q6) asked about HBr(l) and HF(l), connecting vaporization energy to the strength of intermolecular forces. Watch your signs and your units. Vaporization is positive, condensation is negative, and ΔHvap is per mole, not per gram.

## enthalpy of vaporization vs Heat of Fusion

Both are molar enthalpies of phase transitions, but heat of fusion is for melting (solid to liquid) while enthalpy of vaporization is for boiling (liquid to gas). ΔHvap is always much larger for the same substance, because boiling has to completely separate the molecules from each other while melting just lets them slide past one another. For water, that's 40.7 kJ/mol versus 6.02 kJ/mol. On a heating curve, fusion is the first (shorter) plateau and vaporization is the second (longer) one.

## Key Takeaways

- Enthalpy of vaporization (ΔHvap) is the energy needed to convert one mole of liquid to gas at constant temperature and pressure, and it is always positive because vaporization is endothermic.
- Use q = n × ΔHvap, and remember that ΔHvap is per mole, so convert grams to moles with molar mass before plugging in.
- Condensation releases exactly as much energy as vaporization absorbs, so the enthalpy of condensation equals -ΔHvap.
- The temperature of a pure substance stays constant while it boils, because all added energy goes into overcoming intermolecular forces rather than raising kinetic energy.
- ΔHvap is larger than ΔHfus for the same substance because vaporization fully separates molecules while melting only partially disrupts their attractions.
- Substances with stronger intermolecular forces, like hydrogen-bonding liquids, have higher enthalpies of vaporization.

## FAQs

### What is enthalpy of vaporization in AP Chem?

It's the energy required to turn one mole of a liquid into a gas at constant temperature and pressure. In Topic 6.5, you use it in the equation q = nΔHvap, where n is moles. For water, ΔHvap = 40.7 kJ/mol.

### Does temperature increase while a liquid is vaporizing?

No. The temperature of a pure substance stays constant during a phase change. Every joule you add at the boiling point goes into breaking intermolecular attractions, not into speeding up the molecules. That's why heating curves have flat plateaus.

### What's the difference between enthalpy of vaporization and heat of fusion?

Heat of fusion is the energy to melt a solid into a liquid, while enthalpy of vaporization is the energy to boil a liquid into a gas. ΔHvap is always larger because boiling completely separates the molecules. For water, it's 40.7 kJ/mol versus 6.02 kJ/mol.

### Is enthalpy of vaporization positive or negative?

Vaporization is always positive (endothermic) because the system absorbs energy. The reverse process, condensation, is negative with the exact same magnitude, so ΔHcond = -ΔHvap.

### Why does water have such a high enthalpy of vaporization?

Water molecules form hydrogen bonds, which are strong intermolecular attractions that take a lot of energy to overcome. This Unit 3 connection is fair game on the exam, like the 2023 FRQ comparing HBr(l) and HF(l).

## Related Study Guides

- [6.5 Energy of Phase Changes](/ap-chem/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh)

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