AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/6/6-5)
Unit 6 · Topic 6.5
6.5 Energy of Phase Changes
Melting and boiling absorb energy because attractions between particles must be overcome, while freezing and condensing release exactly the same amount. During a phase change a pure substance's temperature stays constant, and the heat equals the moles times the molar enthalpy of the change.
Key terms
- enthalpy of fusion
- enthalpy of vaporization
- phase change
- heating curve
- condensation
Why phase changes involve energy
In a solid or liquid, particles are held together by intermolecular forces (or ionic or metallic attractions). Melting loosens those attractions, and boiling breaks them almost completely. Both take energy, so they are endothermic.
Going the other way, particles that come together and attract each other release energy. Freezing and condensing are exothermic. Sublimation (solid to gas) is endothermic and deposition (gas to solid) is exothermic.
Phase changes do not break covalent bonds inside molecules. When water boils, the bubbles are H₂O molecules, not H₂ and O₂.
Molar enthalpies of fusion and vaporization
The molar enthalpy of fusion, ΔH(fus), is the energy to melt one mole of solid at its melting point. The molar enthalpy of vaporization, ΔH(vap), is the energy to vaporize one mole of liquid. For water, ΔH(fus) = 6.01 kJ/mol at 0 °C and ΔH(vap) = 40.7 kJ/mol at 100 °C.
The reverse change has the same size and the opposite sign. Freezing water releases 6.01 kJ/mol, so ΔH(freezing) = −6.01 kJ/mol. Condensing steam releases 40.7 kJ/mol, so ΔH(condensation) = −40.7 kJ/mol.
To find the heat for a phase change, use q = n × ΔH, where n is moles. Convert grams to moles first.
ΔH(vap) is much larger than ΔH(fus). Melting only lets particles slide past each other, but vaporizing separates them completely. Substances with stronger intermolecular forces have larger values of both.
Reading a heating curve
A heating curve plots temperature (vertical axis) against heat added or time at steady heating (horizontal axis). Picture it for water starting as ice at −20 °C.
First a rising, sloped line: the ice warms. Then a flat segment at 0 °C: the ice melts. Then another sloped line: liquid water warms from 0 to 100 °C. Then a long flat segment at 100 °C: the water boils. Finally a sloped line as the steam warms.
On the flat segments, energy goes into separating particles (raising their potential energy), not into speeding them up, so the temperature, which reflects average kinetic energy, stays constant. The boiling plateau is much longer than the melting plateau because ΔH(vap) is much bigger than ΔH(fus).
On the sloped segments, q = mcΔT applies. A steeper slope means a smaller heat capacity for that phase. A cooling curve is the same shape read in reverse, with energy released.
Multistep problems
When a sample changes temperature and phase, split the process into pieces. Use q = mcΔT for each temperature change (with the specific heat of that phase) and q = nΔH for each phase change. Then add them, making sure all values are in the same unit.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Energy released by condensing steam
How much energy is released when 10.0 g of steam condenses to liquid water at 100 °C? ΔH(vap) of water = 40.7 kJ/mol; molar mass of water = 18.02 g/mol.
Show the solutionHide the solution
- Step 1: Moles of water: 10.0 g ÷ 18.02 g/mol = 0.555 mol.
- Step 2: Condensing is the reverse of vaporizing, so ΔH(condensation) = −40.7 kJ/mol.
- Step 3: q = nΔH = (0.555 mol)(−40.7 kJ/mol) = −22.6 kJ. The negative sign means energy leaves the water.
Answer: 22.6 kJ released (q = −22.6 kJ)
- Example 2Calculator allowed
Ice to warm water in three steps
How much heat is needed to turn 36.0 g of ice at −10.0 °C into liquid water at 25.0 °C? Use: specific heat of ice 2.09 J/(g·°C), specific heat of liquid water 4.18 J/(g·°C), ΔH(fus) = 6.01 kJ/mol, molar mass 18.02 g/mol.
Show the solutionHide the solution
- Step 1: Step 1, warm the ice from −10.0 °C to 0 °C: q₁ = (36.0 g)(2.09)(10.0 °C) = 752 J = 0.752 kJ.
- Step 2: Step 2, melt the ice at 0 °C: n = 36.0 ÷ 18.02 = 1.998 mol, so q₂ = (1.998 mol)(6.01 kJ/mol) = 12.01 kJ. The temperature doesn't change during this step, so q = mcΔT cannot be used here.
- Step 3: Step 3, warm the liquid from 0 °C to 25.0 °C: q₃ = (36.0 g)(4.18)(25.0 °C) = 3762 J = 3.762 kJ.
- Step 4: Add them in the same unit: 0.752 + 12.01 + 3.762 = 16.5 kJ.
- Step 5: Notice that melting takes far more energy than both warming steps combined. Forgetting to convert q₁ and q₃ from J to kJ is the most common way to get this wrong.
Answer: q ≈ 16.5 kJ absorbed
Common mistakes
- Using q = mcΔT across a phase change. During melting or boiling ΔT is zero, so you must use q = nΔH.
- Thinking temperature keeps rising while water boils. It stays at the boiling point until all the liquid is gone.
- Multiplying a molar enthalpy by grams instead of moles.
- Giving condensation or freezing a positive ΔH. The reverse of an endothermic change is exothermic, with the same size.
On the exam
- Heating-curve questions ask you to identify what is happening on a segment, to explain why temperature is constant on a plateau, or to compare plateau lengths using ΔH(fus) and ΔH(vap).
- In explanations, link the energy of a phase change to the strength of intermolecular forces being overcome.
Connected topics
Videos
Check yourself
5 questions on 6.5 Energy of Phase Changes. Pick an answer to see if you got it, and why.
The molar enthalpy of fusion of water is 6.01 kJ/mol. How much energy is needed to melt 36.0 g of ice at 0 °C?
A sample of pure water is heated at a constant rate. While the water is boiling at 100 °C, its temperature stays constant even though energy is still being added. Which of the following best explains this?
A burn from 10 g of steam at 100 °C is usually much worse than a burn from 10 g of liquid water at 100 °C. Which of the following best explains this?
A 1.00 mol sample of a pure solid is heated at a constant rate of 100 J/s, starting at −50 °C.
The temperature rises until 30 s, then stays at −20 °C until 80 s while the solid melts.
It then rises again until 210 s, and stays at 65 °C until 510 s while the liquid boils. After that, the temperature of the gas rises.
Hypothetical heating-curve data
What is the molar enthalpy of vaporization of the substance?
Which of the following best explains why the boiling plateau is much longer than the melting plateau?
0 of 5 answered