# Energy of Phase Changes

> AP Chemistry · Unit 6: Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u6-energy-of-phase-changes/

This module covers molar enthalpies of fusion, vaporization, and sublimation, heating/cooling curve analysis, $q = n\Delta H$ calculations, sign conventions, and multi-step energy problems for AP Chemistry.

**Prerequisites:** [Enthalpy change and endo/exothermic sign conventions](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-basics/); [$q = mc\Delta T$ for temperature change calculations](https://www.owlsprep.com/study/ap-chemistry-calorimetry/); Basic definitions of solid/liquid/gas phase transitions

## Learning objectives

- Calculate heat energy for phase changes using $q = n\Delta H$
- Interpret heating/cooling curves and calculate total heat for multi-step processes
- Apply Hess's law to find unknown phase change enthalpies
- Correctly assign sign conventions for endo/exothermic phase changes

## Core Concepts of Phase Change Energy

Energy of phase changes (also called latent heat) refers to the heat energy absorbed or released when a pure substance undergoes a transition between solid, liquid, or gaseous phases, without changing temperature. Unlike heating that increases molecular kinetic energy (and thus temperature), phase change energy goes entirely into altering the potential energy of intermolecular forces between molecules.

**Isothermal Phase Change** — A phase transition that occurs at constant temperature, because all energy input or output changes intermolecular potential energy, not the kinetic energy that determines temperature.

All phase changes are classified as endothermic or exothermic, based on the direction of heat flow:

- **Endothermic**: Absorb energy from surroundings (melting, vaporization, sublimation)
- **Exothermic**: Release energy to surroundings (freezing, condensation, deposition)

**Check your understanding**

Check your basic understanding:

1. Which of the following phase changes is endothermic?

   - Freezing
   - Condensation
   - Melting
   - Deposition

   *Why:* Melting requires energy input to break the ordered solid structure, so it is endothermic. All other options are exothermic processes that release energy.

## Molar Enthalpies and Phase Change Calculations

**Molar Enthalpy of Phase Change** — The enthalpy change per mole of substance undergoing a specific phase transition. By convention, all values are reported as positive for the forward endothermic transition.

*Notation:* $\Delta H_{\text{phase}}$

The most commonly tested molar enthalpies are:

- $\Delta H_{\text{fus}}$: Molar enthalpy of fusion (solid → liquid melting)
- $\Delta H_{\text{vap}}$: Molar enthalpy of vaporization (liquid → gas vaporization)
- $\Delta H_{\text{sub}}$: Molar enthalpy of sublimation (solid → gas sublimation)

For the reverse transition (e.g., liquid → solid freezing), the enthalpy change is the negative of the forward value: $\Delta H_{\text{freezing}} = -\Delta H_{\text{fus}}$. The core formula for calculating total heat energy $q$ for a phase change is:

$$q = n \Delta H_{\text{phase}}$$

Where $n$ is moles of substance, and $\Delta H_{\text{phase}}$ is the molar enthalpy of the phase change occurring. If given mass instead of moles, convert to moles via $n = \frac{m}{M}$, where $m$ is mass in grams and $M$ is molar mass in g/mol. $\Delta H_{\text{vap}}$ is almost always much larger than $\Delta H_{\text{fus}}$ because vaporization requires breaking all intermolecular interactions, while melting only loosens them.

**Worked example:** Calculate the total heat absorbed when 75.0 g of ice at 0°C melts to liquid water at 0°C. The molar enthalpy of fusion of water is 6.02 kJ/mol, and the molar mass of water is 18.015 g/mol.

1. Convert mass of water to moles:

   $$n = \frac{m}{M} = \frac{75.0\ \text{g}}{18.015\ \text{g/mol}} = 4.163\ \text{mol}$$
2. Confirm the sign of $\Delta H$: melting is endothermic, so $q$ will be positive (heat absorbed by the water system).
3. Substitute into the core formula:

   $$q = n\Delta H_{\text{fus}} = (4.163\ \text{mol})(6.02\ \text{kJ/mol}) = 25.1\ \text{kJ}$$
4. Check units: moles cancel, leaving kJ, which is the correct unit for energy.

> **Exam tip:** Always confirm the direction of the phase change before assigning the sign of ΔH. AP exam questions often give you ΔHvap as a positive value and ask for the heat released during condensation, so you must add the negative sign explicitly to get the correct answer.

*Calculator:* allowed

## Heating and Cooling Curve Analysis

Heating (or cooling) curves plot the temperature of a substance versus the total heat added to the substance as it is heated from solid to gas (or cooled from gas to solid for cooling curves). The curve has two distinct region types:

- **Sloped regions**: Only one phase is present, heat added changes temperature. Use $q = mc\Delta T$.
- **Flat (horizontal) regions**: Temperature is constant, corresponds to a phase change. Use $q = n\Delta H$.

For a heating curve starting from a low-temperature solid, the order of regions is: (1) heat solid to melting point, (2) melt solid to liquid, (3) heat liquid to boiling point, (4) vaporize liquid to gas, (5) heat gas to final temperature. A common AP exam question asks you to calculate the total heat required to heat a substance from an initial cold temperature to a final hot temperature, which requires adding the $q$ from every region in sequence.

**Worked example:** A 1 mol sample of a pure substance starts at -100°C. It undergoes sublimation at -78°C, and the next phase change would occur at 100°C. Given: specific heat of solid = 0.05 kJ/mol·°C, $\Delta H_{\text{sub}} = 32$ kJ/mol, specific heat of gas = 0.1 kJ/mol·°C. If 45 kJ of total heat is added, what is the final state and temperature of the sample?

1. Calculate heat to warm solid from -100°C to -78°C:

   $$\Delta T = (-78) - (-100) = 22^\circ\text{C} \\ q_1 = n c_{\text{solid}} \Delta T = (1\ \text{mol})(0.05\ \text{kJ/mol·°C})(22°C) = 1.1\ \text{kJ}$$
2. Total heat used = 1.1 kJ, remaining heat = 45 - 1.1 = 43.9 kJ. Calculate heat for complete sublimation:

   $$q_2 = n \Delta H_{\text{sub}} = (1\ \text{mol})(32\ \text{kJ/mol}) = 32\ \text{kJ}$$
3. Total heat used = 1.1 + 32 = 33.1 kJ, remaining heat = 45 - 33.1 = 11.9 kJ. Check if we reach the next phase change at 100°C:

   $$q_{\text{required to warm gas}} = (1\ \text{mol})(0.1\ \text{kJ/mol·°C})(178°C) = 17.8\ \text{kJ}$$
4. 11.9 kJ < 17.8 kJ, so we do not reach the next phase change. Calculate final temperature:

   $$\Delta T = \frac{q_{\text{remaining}}}{n c_{\text{gas}}} = \frac{11.9}{(1)(0.1)} = 119^\circ\text{C} \\ \text{Final } T = -78 + 119 = 41^\circ\text{C}$$
5. Final result: The sample is pure gas at 41°C.

> **Exam tip:** When calculating total heat for a multi-step heating process, always add the q from every single region in order—never skip a region even if it seems small. AP exam questions often award partial credit for correctly calculating each region's q, so write out every term separately.

*Calculator:* allowed

## Hess's Law for Phase Change Enthalpy

Hess's law (the total enthalpy change for a process is independent of the path taken) applies to phase changes just as it does to chemical reactions. For example, sublimation (solid → gas) can occur either directly, or via an indirect path: solid → liquid (fusion) then liquid → gas (vaporization). The enthalpy of sublimation is therefore the sum of the enthalpies of fusion and vaporization at the same temperature and pressure:

$$\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$$

This relationship is often tested when you are given two of the three values and asked to calculate the third, or when you need to find the enthalpy of a reverse transition like deposition (gas → solid), which equals $-\Delta H_{\text{sub}}$.

**Worked example:** At 1 atm, the molar enthalpy of fusion of ice is 6.01 kJ/mol, and the molar enthalpy of vaporization of liquid water is 40.7 kJ/mol. Estimate the molar enthalpy of sublimation of ice, assuming enthalpies do not change significantly with temperature.

1. Write the given processes and their enthalpies:

   $$\text{Ice (s)} \rightarrow \text{Liquid water (l)} \quad \Delta H_1 = +6.01\ \text{kJ/mol} \\ \text{Liquid water (l)} \rightarrow \text{Water vapor (g)} \quad \Delta H_2 = +40.7\ \text{kJ/mol}$$
2. The target process is $\text{Ice (s)} \rightarrow \text{Water vapor (g)}$, which is the sum of the two given processes, with liquid water canceling out when adding.
3. Apply Hess's law to add the enthalpies:

   $$\Delta H_{\text{sub}} = \Delta H_1 + \Delta H_2$$
4. Calculate the final result:

   $$\Delta H_{\text{sub}} = 6.01 + 40.7 = 46.7\ \text{kJ/mol}$$

**Check your understanding**

Test your calculation skills with this AP-style multiple choice question:

1. How much heat is released when 125 g of ethanol ($\text{C}_2\text{H}_5\text{OH}$, molar mass 46.07 g/mol) condenses from gaseous ethanol to liquid ethanol at its boiling point? The molar enthalpy of vaporization of ethanol is 38.6 kJ/mol.

   - A) 38.6 kJ
   - B) 105 kJ
   - C) -105 kJ
   - D) -4820 kJ

   *Why:* Convert mass to moles: $n = 125 / 46.07 = 2.71$ mol. Condensation is the reverse of vaporization, so $\Delta H = -38.6$ kJ/mol. $q = (2.71)(-38.6) = -105$ kJ, where the negative sign correctly indicates heat released by the system.

> **Exam tip:** Always check that your phase change equations add up correctly, canceling out any intermediate phases, just like you do for chemical reactions in Hess's law problems. A common mistake is reversing one of the enthalpies when it is not needed.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using mass in grams directly in $q = n\Delta H$ without converting to moles.
  - Why it fails: Students confuse $q = mc\Delta T$ (which uses mass) with $q = n\Delta H$ (which uses moles), so they plug mass directly in.
  - Correct: When given mass, always convert to moles via $n = m/M$ first before plugging into $q = n\Delta H$. Write the conversion step explicitly on your paper to avoid forgetting.
- **Wrong:** Forgetting to add the heat from sloped (temperature change) regions when calculating total heat for a full heating process.
  - Why it fails: Students focus so much on the phase change regions that they skip the steps where the substance is heated between phase changes.
  - Correct: Always draw the heating curve and label every region from initial T to final T, calculate q for each region separately, then sum all q values.
- **Wrong:** Keeping ΔH positive for exothermic phase changes like condensation or freezing.
  - Why it fails: Reference tables always report ΔHfus, ΔHvap as positive values for the forward endothermic process, so students forget to reverse the sign for the reverse process.
  - Correct: For any problem, write down the direction of the phase change first, assign the sign: + for endothermic (melting, vaporization, sublimation), - for exothermic (freezing, condensation, deposition) before calculating q.
- **Wrong:** Using $q = mc\Delta T$ for a phase change region.
  - Why it fails: Students assume all energy change changes temperature, so they use the wrong formula for constant-temperature phase changes.
  - Correct: Only use $q = mc\Delta T$ for single-phase regions where temperature is changing. Use $q = n\Delta H$ for constant-temperature phase change regions.
- **Wrong:** Calculating ΔHsub as equal to only ΔHvap, forgetting to add ΔHfus.
  - Why it fails: Students think sublimation is directly solid to gas so it only requires vaporization energy, missing the fusion component.
  - Correct: Always remember that $\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$, so add both enthalpies when calculating sublimation enthalpy from fusion and vaporization.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Heat of phase change | $q = n \Delta H_{\text{phase}}$ | n = moles; ΔH positive for endo, negative for exo. Only for constant-temperature phase changes. |
| Heat of temperature change | $q = mc\Delta T$ | m = mass (g), c = specific heat (J/g·°C). Only for single-phase changing temperature. |
| Enthalpy of sublimation | $\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$ | Applies at same T/P; ΔHsub is always positive. |
| Reverse phase change enthalpy | $\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}$ | e.g., $\Delta H_{\text{freezing}} = -\Delta H_{\text{fus}}$, $\Delta H_{\text{condensation}} = -\Delta H_{\text{vap}}$. |
| Convert mass to moles | $n = \frac{m}{M}$ | m = mass (g), M = molar mass (g/mol). Required for all $q = n\Delta H$ calculations with given mass. |
| Total multi-step heat | $q_{\text{total}} = \sum q_i$ | Add q from all sloped (temperature) and flat (phase change) regions in order. |

## What's next

Understanding energy of phase changes is a critical foundation for more advanced thermodynamics topics in AP Chemistry Unit 6, including calorimetry for chemical reactions, Hess's law for reaction enthalpies, and enthalpy of formation calculations. This topic also connects to intermolecular forces and properties of solids, liquids, and gases, where you will explore how intermolecular strength affects the magnitude of phase change enthalpies. Mastery of multi-step energy calculations here will prepare you for combined FRQ questions that blend phase changes with other thermodynamics concepts on the AP exam.

- [Unit 6 Thermodynamics Overview](https://www.owlsprep.com/study/ap-chemistry-u6-overview/)
- [Introduction to Enthalpy of Reaction](https://www.owlsprep.com/study/ap-chemistry-u6-introduction-to-enthalpy-of-reaction/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-chemistry-u6-energy-of-phase-changes/