Energy of phase changes — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Molar enthalpies of fusion, vaporization, and sublimation; heating/cooling curve energy analysis; the $q=n\Delta H$ calculation formula; sign conventions for endo/exothermic phase changes; and multi-step energy change problems.

You should already know: Enthalpy change definition and sign conventions for endothermic/exothermic reactions. The $q=mc\Delta T$ formula for temperature change of a pure substance. Basic definitions of solid/liquid/gas phase transitions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Energy of phase changes?

Energy of phase changes 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. This topic is weighted approximately 7-9% of AP Chemistry Unit 6 (Thermodynamics) exam content, and it appears regularly in both multiple-choice (MCQ) sections (as multi-step calculations or heating curve interpretation) and free-response (FRQ) sections, where it is combined with other thermodynamics concepts like calorimetry or Hess's law. Synonyms you may see on the exam include enthalpy of phase change, heat of fusion/vaporization, and latent heat of phase transition. All phase changes are isothermal (constant temperature) because energy input/output only changes intermolecular potential energy, not the kinetic energy that determines temperature. Endothermic phase changes (melting, vaporization, sublimation) absorb energy from the surroundings, while exothermic phase changes (freezing, condensation, deposition) release energy to the surroundings.

2. Molar Enthalpies of Phase Change

Every pure substance has a characteristic molar enthalpy for each phase transition: the amount of enthalpy change per mole of substance undergoing the transition. The most commonly tested values are $\Delta H_{\text{fus}}$ (molar enthalpy of fusion, for solid → liquid melting), $\Delta H_{\text{vap}}$ (molar enthalpy of vaporization, for liquid → gas vaporization), and $\Delta H_{\text{sub}}$ (molar enthalpy of sublimation, for solid → gas sublimation). By convention, all three values are reported as positive numbers because the forward phase change they describe is endothermic. For the reverse transition (e.g., liquid → solid freezing), the enthalpy change is simply the negative of the forward value: $\Delta H_{\text{freezing}}=-\Delta H_{\text{fus}}$, $\Delta H_{\text{condensation}}=-\Delta H_{\text{vap}}$.

The core formula for calculating total heat energy $q$ for a complete 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 you are given mass instead of moles, convert mass to moles via $n=\frac{m}{M}$, where $m$ is mass in grams and $M$ is molar mass in g/mol. Intuitively, $\Delta H_{\text{vap}}$ is almost always much larger than $\Delta H_{\text{fus}}$ for the same substance: vaporization requires breaking all intermolecular interactions between molecules, while melting only loosens intermolecular forces to allow flow, so far less energy is needed.

Worked Example

Problem: 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 system, the water).
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.

3. 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:

1. Sloped regions: Only one phase is present, heat added changes the temperature of the substance. Use $q=mc\Delta T$ for these regions.
2. Flat (horizontal) regions: Temperature is constant, corresponding to a phase change. All added heat goes into breaking or forming intermolecular forces, so use $q=n\Delta H$ for these regions.

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

Problem: A 1 mol sample of a pure substance starts at -100°C. Its heating curve has a phase change (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: $\Delta T=(-78)-(-100)=22°C$. $q_1=n{c}_{\text{solid}}\Delta T=(1\text{mol})(0.05\text{kJ/molcdotp°C})(22°C)=1.1\text{kJ}$. Total heat used = 1.1 kJ, remaining heat = 45 - 1.1 = 43.9 kJ.
2. Calculate heat for complete sublimation: $q_2=n\Delta H_{\text{sub}}=(1\text{mol})(32\text{kJ/mol})=32\text{kJ}$. Total heat used = 1.1 + 32 = 33.1 kJ, remaining heat = 45 - 33.1 = 11.9 kJ.
3. Check if we reach the next phase change at 100°C: Heat required to warm gas from -78°C to 100°C is $\Delta T=178°C$, $q_3=(1\text{mol})(0.1\text{kJ/molcdotp°C})(178°C)=17.8\text{kJ}$. 11.9 kJ < 17.8 kJ, so we do not reach the next phase change.
4. Calculate final temperature: $\Delta T=\frac{q_{\text{remaining}}}{n{c}_{\text{gas}}}=\frac{11.9}{(1)(0.1)}=119°C$. Final T = $-78+119=41°C$.

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.

4. 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 $-\Delta H_{\text{sub}}=-\Delta H_{\text{fus}}-\Delta H_{\text{vap}}$. This works for any combination of phase changes, so you can always use Hess's law to find unknown enthalpies.

Worked Example

Problem: 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:
   1. $\text{Ice (s)}\to \text{Liquid water (l)}\Delta H_1=+6.01\text{kJ/mol}$
   2. $\text{Liquid water (l)}\to \text{Water vapor (g)}\Delta H_2=+40.7\text{kJ/mol}$
2. The target process is $\text{Ice (s)}\to \text{Water vapor (g)}$, 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: $\Delta H_{\text{sub}}=6.01+40.7=46.7\text{kJ/mol}$.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using mass in grams directly in $q=n\Delta H$ without converting to moles. Why: Students confuse $q=mc\Delta T$ (which uses mass, usually in grams) with $q=n\Delta H$ (which uses moles), so they plug mass directly in. Correct move: 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 move: Forgetting to add the heat from sloped (temperature change) regions when calculating total heat for a full heating process. Why: Students focus so much on the phase change regions that they skip the steps where the substance is heated between phase changes. Correct move: 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 move: Keeping ΔH positive for exothermic phase changes like condensation or freezing. Why: 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 move: 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 move: Using $q=mc\Delta T$ for a phase change region. Why: Students assume all energy change changes temperature, so they use the wrong formula for constant-temperature phase changes. Correct move: 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 move: Calculating ΔHsub as equal to only ΔHvap, forgetting to add ΔHfus. Why: Students think sublimation is directly solid to gas so it only requires vaporization energy, missing the fusion component. Correct move: 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.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

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

Worked Solution: First, convert mass of ethanol to moles: $n=\frac{125\text{g}}{46.07\text{g/mol}}=2.71\text{mol}$. Condensation is the reverse of vaporization, so $\Delta H_{\text{condensation}}=-\Delta H_{\text{vap}}=-38.6$ kJ/mol. Calculate q: $q=n\Delta H=(2.71\text{mol})(-38.6\text{kJ/mol})=-105\text{kJ}$. The negative sign indicates heat is released by the system, which matches the question prompt. The correct answer is C.

Question 2 (Free Response)

A 36.0 g sample of ice (${\text{H}}_2\text{O}$, $M=18.0$ g/mol) starts at -20.0°C and is heated until it becomes water vapor at 120.0°C at 1 atm. Use the following data to answer the questions below: specific heat of ice = 2.09 J/g·°C, $\Delta H_{\text{fus}}=6.02$ kJ/mol, specific heat of liquid water = 4.18 J/g·°C, $\Delta H_{\text{vap}}=40.7$ kJ/mol, specific heat of water vapor = 2.01 J/g·°C. (a) Calculate the total heat required (in kJ) for the entire process. Show your work for each step. (b) Explain why $\Delta H_{\text{vap}}$ is much larger than $\Delta H_{\text{fus}}$ for water, in terms of intermolecular forces. (c) What is the sign of the total enthalpy change for this process, and why?

Worked Solution: (a) There are 5 regions to calculate:

1. Heat ice from -20.0°C to 0°C: $q_1=mc\Delta T=(36.0\text{g})(2.09\text{J/gcdotp°C})(20°C)=1505\text{J}=1.51\text{kJ}$.
2. Melt ice to liquid at 0°C: $n=\frac{36.0\text{g}}{18.0\text{g/mol}}=2.00\text{mol}$, $q_2=n\Delta H_{\text{fus}}=(2.00\text{mol})(6.02\text{kJ/mol})=12.04\text{kJ}$.
3. Heat liquid from 0°C to 100°C: $q_3=(36.0\text{g})(4.18\text{J/gcdotp°C})(100°C)=15048\text{J}=15.05\text{kJ}$.
4. Vaporize liquid to gas at 100°C: $q_4=n\Delta H_{\text{vap}}=(2.00\text{mol})(40.7\text{kJ/mol})=81.4\text{kJ}$.
5. Heat gas from 100°C to 120°C: $q_5=(36.0\text{g})(2.01\text{J/gcdotp°C})(20°C)=1447\text{J}=1.45\text{kJ}$. Total $q=1.51+12.04+15.05+81.4+1.45=111.45$ kJ ≈ 111 kJ.

(b) Fusion (melting) only requires enough energy to partially overcome intermolecular forces between water molecules, allowing the solid to flow as a liquid while most intermolecular interactions remain intact. Vaporization requires breaking all intermolecular interactions between water molecules to separate them into the gas phase, which requires much more energy, so ΔHvap is much larger than ΔHfus.

(c) The total enthalpy change is positive. The entire process involves adding heat to the system (the water sample) to go from solid at low temperature to gas at high temperature, and every step is either endothermic heating or an endothermic phase change, so total enthalpy change is positive.

Question 3 (Application / Real-World Style)

Cold packs used for sports injuries use the absorption of heat from the injury to reduce swelling. One design uses melting ice at 0°C to absorb heat. How many grams of ice (at 0°C) must melt to absorb 150 kJ of heat from an injury? $\Delta H_{\text{fus}}$ of ice is 6.02 kJ/mol, and molar mass of water is 18.02 g/mol.

Worked Solution: We need $q=+150$ kJ (heat absorbed by melting ice from the injury). Rearrange $q=n\Delta H_{\text{fus}}$ to solve for n: $n=\frac{q}{\Delta H_{\text{fus}}}=\frac{150\text{kJ}}{6.02\text{kJ/mol}}=24.9\text{mol}$. Convert moles to mass: $m=n\times M=24.9\text{mol}\times 18.02\text{g/mol}=449$ g ≈ 450 g. In context, this means approximately 450 grams (1 pound) of ice is needed to absorb 150 kJ of heat from an injured area, which matches the typical size of a disposable ice pack.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for understanding calorimetry of chemical reactions, where phase changes of the surroundings or products can alter the total measured heat change. It also connects directly to enthalpy of reaction calculations via Hess's law, a core skill for Unit 6 and Unit 9 (thermodynamics of reactions). Next, you will apply the energy calculation skills you learned here to constant-pressure calorimetry and bomb calorimetry problems, where you will measure enthalpy changes for chemical reactions instead of just phase changes. Without mastering the sign conventions and calculation methods for phase change energy, you will struggle to correctly calculate total heat change in calorimetry problems that involve phase changes of the solvent or products. This topic also feeds into the study of vapor pressure and phase diagrams in intermolecular forces, where enthalpy of vaporization is used to calculate vapor pressure at different temperatures.

Calorimetry and enthalpy of reaction Hess's law of enthalpy addition Phase diagrams and intermolecular forces Clausius-Clapeyron equation