AP Chemistry Quiz: Energy Of Phase Changes

20 questions · exam conditions

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Energy Of Phase ChangesQuestion 1 of 20

The enthalpy of solidification for benzene (C₆H₆) is $-9.95 \text{ kJ/mol}$ . How many grams of liquid benzene can be frozen at its freezing point if $4.975 \text{ kJ}$ of heat is removed? (The molar mass of C₆H₆ is $78.11 \text{ g/mol}$ ).

19.6 g

39.1 g

78.1 g

156 g

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AP Chemistry Quiz

AP Chemistry Quiz: Energy Of Phase Changes

Practice Energy Of Phase Changes in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Energy Of Phase Changes, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

19.6 g
39.1 g (correct answer)
78.1 g
156 g

Explanation: First, calculate the moles of benzene that can be frozen. Since heat is removed, $q = -4.975 \text{ kJ}$ . Using $q = n \Delta H_{\text{solidification}}$ , we find $n = \frac{q}{\Delta H_{\text{solidification}}} = \frac{-4.975 \text{ kJ}}{-9.95 \text{ kJ/mol}} = 0.500 \text{ mol}$ . Then, convert moles to grams: $mass = 0.500 \text{ mol} \times 78.11 \text{ g/mol} = 39.1 \text{ g}$ .

Question 2

During the phase change of a pure substance at constant pressure, which of the following occurs?

The temperature of the substance changes while its potential energy remains constant.
Both the temperature and the potential energy of the substance remain constant.
The temperature of the substance remains constant while its potential energy changes. (correct answer)
Both the temperature and the average kinetic energy of the substance change while potential energy is constant.

Explanation: During a phase change, the added or removed energy alters the potential energy of the molecules by changing the distance between them and the strength of their intermolecular interactions. The temperature, which is a measure of the average kinetic energy of the molecules, remains constant until the phase change is complete.

Question 3

The molar enthalpy of fusion ( $\Delta H_{\text{fus}}$ ) of NaCl is $28 \text{ kJ/mol}$ , while that of solid methane (CH₄) is $0.94 \text{ kJ/mol}$ . Which statement best explains this large difference?

Melting NaCl requires breaking strong ionic bonds, while melting CH₄ requires overcoming weak London dispersion forces. (correct answer)
NaCl has a much higher molar mass than CH₄, which accounts for the significant energy difference.
The covalent bonds within the CH₄ molecule must be broken during melting, which is not the case for NaCl.
The change in entropy is much larger for melting NaCl than for melting CH₄, which requires a larger enthalpy input.

Explanation: The energy required for melting is determined by the strength of the forces holding the particles in the solid lattice. NaCl is an ionic solid with strong electrostatic attractions (ionic bonds) between ions. Methane is a molecular solid with only weak London dispersion forces between molecules. Overcoming the strong ionic bonds in NaCl requires far more energy than overcoming the weak dispersion forces in solid methane. Intramolecular covalent bonds are not broken during phase changes.

Question 4

For H₂O, the molar enthalpy of fusion is $6.02 \text{ kJ/mol}$ and the molar enthalpy of vaporization is $40.7 \text{ kJ/mol}$ . Which statement correctly explains the large difference between these values?

Vaporization requires significantly more energy because all intermolecular forces must be overcome, whereas in fusion they are only weakened. (correct answer)
Fusion requires more energy because breaking the rigid crystal lattice of ice is more difficult than separating liquid molecules.
Vaporization requires more energy because the process involves a much larger change in the kinetic energy of the molecules.
The energy difference is small when considering the change is for the same number of molecules in a one mole sample.

Explanation: Fusion (melting) involves weakening the intermolecular forces enough for molecules to move past each other, but significant attractions remain. Vaporization involves completely overcoming the intermolecular forces to separate molecules into the gas phase. This requires much more energy, hence $\Delta H_{\text{vap}}$ is much larger than $\Delta H_{\text{fus}}$ . Phase changes involve changes in potential energy, not kinetic energy, so C is incorrect.

Question 5

Liquid benzene at its melting point is converted to solid benzene at the same temperature. If $\Delta H_{fus}$ for benzene is $9.95\ \text{kJ/mol}$ , how much energy is released when $0.80\ \text{mol}$ of benzene freezes?

7.96 kJ (correct answer)
12.4 kJ
9.95 kJ
19.9 kJ
3.98 kJ

Explanation: This question tests the calculation of energy released during freezing (liquid to solid transition). When benzene freezes, it releases energy equal to its enthalpy of fusion multiplied by the number of moles. The energy released = ΔHfus × moles = 9.95 kJ/mol × 0.80 mol = 7.96 kJ. A common error is to use the enthalpy value directly without accounting for the partial mole (choice C: 9.95 kJ), treating the molar enthalpy as if it were the total energy for any amount. For phase transitions, always multiply the per-mole enthalpy value by the actual number of moles present.

Question 6

A 27.0 g sample of ice at $0^\circ\text{C}$ melts completely at $0^\circ\text{C}$ . The enthalpy of fusion of water is $\Delta H_{fus}=6.01\ \text{kJ/mol}$ . How much energy is absorbed? (Molar mass of water $=18.0\ \text{g/mol}$ .)

6.01 kJ
1.50 kJ
0.667 kJ
9.02 kJ (correct answer)
3.01 kJ

Explanation: This question tests the ability to calculate the energy absorbed during melting using the enthalpy of fusion and the mass of the substance. Convert 27.0 g of ice to moles with 18.0 g/mol, yielding 1.50 moles. Multiply by 6.01 kJ/mol to get 9.02 kJ absorbed, reflecting the endothermic nature of breaking hydrogen bonds in ice. This matches choice B, the energy for complete melting at 0°C. Choice C, 6.01 kJ, tempts those who omit the mole conversion, confusing the per-mole value with the total energy. A transferable strategy is to identify whether the phase change is endothermic or exothermic by considering if it's increasing or decreasing molecular disorder.

Question 7

A 10.0 g sample of iodine, $\text{I}_2(s)$ , sublimes at its sublimation point. The enthalpy of sublimation is $\Delta H_{sub}=62.4\ \text{kJ/mol}$ . How much energy is absorbed during sublimation? (Molar mass of $\text{I}_2$ $=254\ \text{g/mol}$ .)

2.46 kJ (correct answer)
0.246 kJ
62.4 kJ
6.24 kJ
24.6 kJ

Explanation: This question tests the ability to calculate the energy absorbed during sublimation using the enthalpy of sublimation and the mass of the substance. Convert 10.0 g of iodine to moles using 254 g/mol, resulting in about 0.0394 moles. Multiply by 62.4 kJ/mol to find approximately 2.46 kJ absorbed, as sublimation is endothermic, directly transitioning solid to gas and requiring energy to break bonds. This corresponds to choice B, the energy input for the phase change. Choice D, 62.4 kJ, attracts those who skip the mole calculation and use the enthalpy value alone, mistakenly treating it as per gram instead of per mole. A transferable strategy is to remember that enthalpies of phase changes are molar values, so always scale by the number of moles involved.

Question 8

A 36.0 g sample of water at $0^\circ\text{C}$ melts completely at $0^\circ\text{C}$ . The enthalpy of fusion of water is $\Delta H_{fus}=6.01\ \text{kJ/mol}$ . How much energy is absorbed during the melting process? (Molar mass of water $=18.0\ \text{g/mol}$ .)

12.0 kJ (correct answer)
24.0 kJ
2.00 kJ
0.334 kJ
6.01 kJ

Explanation: This question tests the ability to calculate the energy absorbed during melting using the enthalpy of fusion and the mass of the substance. Start by converting the 36.0 g of water to moles using its molar mass of 18.0 g/mol, resulting in exactly 2.00 moles. Multiply this by the enthalpy of fusion, 6.01 kJ/mol, to find 12.0 kJ absorbed, as melting is endothermic and energy is needed to overcome lattice forces in the solid. This corresponds to choice A, indicating the total energy for the phase transition from solid to liquid at constant temperature. Choice E, 24.0 kJ, is a common distractor from doubling the correct value, perhaps from mistakenly using twice the moles or confusing fusion with vaporization enthalpies. A transferable strategy is to ensure the sign of energy reflects whether the process is endothermic (absorbed) or exothermic (released) based on the phase change direction.

Question 9

A 40.0 g sample of methanol is vaporized at its boiling point. The enthalpy of vaporization of methanol is $\Delta H_{vap}=35.3\ \text{kJ/mol}$ . How much energy is absorbed? (Molar mass of methanol $=32.0\ \text{g/mol}$ .)

44.1 kJ (correct answer)
1.41 kJ
22.1 kJ
35.3 kJ
11.0 kJ

Explanation: This question tests the ability to calculate the energy absorbed during vaporization using the enthalpy of vaporization and the mass of the substance. Divide 40.0 g of methanol by 32.0 g/mol to obtain 1.25 moles. Multiply by 35.3 kJ/mol, resulting in 44.1 kJ absorbed, as vaporization demands energy to separate liquid molecules into gas. This is choice A, quantifying the endothermic process at the boiling point. Choice D, 35.3 kJ, is a distractor from forgetting to multiply by moles and using the raw enthalpy, underestimating the energy for the given mass. A transferable strategy is to use dimensional analysis to confirm that units cancel correctly to kJ.

Question 10

A 15.0 g sample of benzene freezes at its melting point. The enthalpy of fusion of benzene is $\Delta H_{fus}=9.95\ \text{kJ/mol}$ . How much energy is released during freezing? (Molar mass of benzene $=78.0\ \text{g/mol}$ .)

9.95 kJ
0.191 kJ
1.91 kJ (correct answer)
0.383 kJ
3.83 kJ

Explanation: This question tests the ability to calculate the energy released during freezing using the enthalpy of fusion and the mass of the substance. Divide the 15.0 g of benzene by its molar mass of 78.0 g/mol to get approximately 0.192 moles. Multiply by the enthalpy of fusion, 9.95 kJ/mol, yielding about 1.91 kJ released, since freezing is exothermic as molecules form a more ordered solid structure. This matches choice B, the energy liberated in the liquid-to-solid phase change. Choice C, 9.95 kJ, is a distractor for those who neglect the mole conversion and apply the enthalpy to the mass directly, confusing molar quantities with mass-based ones. A transferable strategy is to double-check unit consistency, ensuring mass is converted to moles when using molar enthalpies.

Question 11

A 22.4 g sample of a substance condenses at its boiling point. The enthalpy of vaporization for the substance is $\Delta H_{vap}=28.0\ \text{kJ/mol}$ , and its molar mass is $56.0\ \text{g/mol}$ . How much energy is released during condensation?

2.80 kJ
28.0 kJ
11.2 kJ (correct answer)
1.12 kJ
5.60 kJ

Explanation: This question tests the ability to calculate the energy released during condensation using the enthalpy of vaporization and the mass of the substance. Convert the 22.4 g to moles by dividing by 56.0 g/mol, giving 0.400 moles. Since condensation is the reverse of vaporization, the energy released is the same magnitude, so multiply 0.400 moles by 28.0 kJ/mol to get 11.2 kJ released, as it's exothermic and strengthens intermolecular forces. This aligns with choice A, representing the heat given off during the gas-to-liquid transition. Choice B, 28.0 kJ, tempts those who forget to convert to moles and use the enthalpy directly, misunderstanding that enthalpy is per mole, not per gram. A transferable strategy is to recognize that for reverse processes like condensation or freezing, the energy magnitude is identical to the forward process but opposite in sign.

Question 12

A 12.0 g sample of a liquid freezes at its freezing point. The enthalpy of fusion is $\Delta H_{\text{fus}}=400\ \text{J g}^{-1}$ . How much energy is released during freezing?

9.60×10^3 J
1.20×10^3 J
4.80×10^4 J
4.00×10^2 J
4.80×10^3 J (correct answer)

Explanation: This question tests the skill of calculating the energy released during freezing using the enthalpy of fusion. The energy released by the 12.0 g liquid sample is 12.0 g × 400 J/g = 4.80×10^3 J, as freezing is exothermic. This represents the heat given off as the substance solidifies and forms a more ordered structure. Choice A is the correct answer. Choice B, 1.20×10^3 J, could tempt due to the misconception of using one-tenth of the mass or dividing instead of multiplying, yielding a smaller value. A transferable strategy for phase change energy problems is to identify the process as endothermic or exothermic and use the absolute value of ΔH for magnitude calculations.

Question 13

Which statement accurately describes the energy of a system of pure H₂O molecules as it undergoes freezing from a liquid to a solid at 0°C?

The potential energy of the system decreases as intermolecular forces become stronger, while the average kinetic energy remains constant. (correct answer)
The average kinetic energy of the system decreases as molecules slow down, while the potential energy remains constant.
Both the potential energy and the average kinetic energy of the system decrease during freezing.
The potential energy of the system increases as molecules become more ordered, while the average kinetic energy remains constant.

Explanation: Freezing is an exothermic process where liquid turns into a more ordered solid. Energy is released as stronger, more stable intermolecular forces (hydrogen bonds in an ice lattice) are formed. This means the potential energy of the system decreases. Because the phase change occurs at a constant temperature (0°C), the average kinetic energy of the molecules does not change.

Question 14

A $24\,\text{g}$ sample of ice melts at $0^\circ\text{C}$ . For water, $\Delta H_{\text{fus}}=334\,\text{J/g}$ . How much energy is absorbed during melting?

8,016 J (correct answer)
801.6 J
7,200 J
334 J
5,344 J

Explanation: This question tests the skill of calculating energy absorbed during ice melting using enthalpy of fusion. Energy absorbed is calculated as mass times enthalpy of fusion: Energy = 24 g × 334 J/g = 8,016 J. When ice melts, energy is absorbed to break the hydrogen bonds holding water molecules in the rigid ice structure, allowing them to move more freely as liquid. The temperature remains at 0°C throughout melting as all energy goes into the phase change. Answer D (334 J) shows the common mistake of using only the enthalpy value without considering the 24 g mass, forgetting that ΔHfus represents energy per gram not total energy. To solve phase change problems systematically, always identify the phase change direction, determine if energy is absorbed or released, then multiply the given mass by the appropriate enthalpy value.

Question 15

A student melts a $25\,\text{g}$ sample of Substance X at its melting point. The enthalpy of fusion of X is $\Delta H_{\text{fus}}=200\,\text{J/g}$ . How much energy is absorbed by the sample during the melting process?

5,000 J (correct answer)
200 J
225 J
1,000 J
4,000 J

Explanation: This question tests the skill of calculating energy changes during phase transitions using enthalpy of fusion. To find the energy absorbed during melting, we multiply the mass of the sample by the enthalpy of fusion: Energy = mass × ΔHfus = 25 g × 200 J/g = 5,000 J. During melting, energy is absorbed to break intermolecular forces while the temperature remains constant at the melting point. A common misconception leading to answer B (200 J) is using only the enthalpy value without multiplying by mass, forgetting that enthalpy of fusion is given per gram. When solving phase change problems, always multiply the given enthalpy per gram by the total mass of the sample to find the total energy change.

Question 16

A $12\,\text{g}$ sample of a substance boils at its boiling point. The enthalpy of vaporization is $\Delta H_{\text{vap}}=900\,\text{J/g}$ . How much energy is absorbed during boiling?

10,800 J (correct answer)
1,080 J
9,000 J
8,100 J
900 J

Explanation: This question tests the skill of calculating energy absorbed during boiling using enthalpy of vaporization. The energy absorbed is found by multiplying mass by enthalpy of vaporization: Energy = 12 g × 900 J/g = 10,800 J. Boiling is vaporization occurring throughout the liquid at the boiling point, requiring energy to overcome intermolecular forces and convert liquid to gas. The temperature remains constant during boiling as all added energy goes into the phase change rather than increasing kinetic energy. Answer E (900 J) shows the misconception of using the enthalpy value directly without considering the mass, treating ΔHvap as total energy rather than energy per gram. Always set up phase change calculations with clear unit analysis to ensure proper multiplication of mass and specific enthalpy.

Question 17

A $30\,\text{g}$ sample of Substance Q solidifies at its melting point. The enthalpy of fusion is $\Delta H_{\text{fus}}=120\,\text{J/g}$ . How much energy is released when the sample solidifies?

3,600 J (correct answer)
120 J
2,400 J
3,000 J
4,800 J

Explanation: This question tests the skill of calculating energy released during solidification using enthalpy of fusion. Energy released when a liquid solidifies equals mass times enthalpy of fusion: Energy = 30 g × 120 J/g = 3,600 J. Solidification (freezing) is the reverse of melting, releasing energy as particles arrange into a more ordered solid structure and intermolecular forces strengthen. The same amount of energy absorbed during melting is released during solidification at the same temperature. Answer B (120 J) represents the error of using only the per-gram enthalpy value without scaling by the total mass of the sample. To master phase change calculations, remember that enthalpy values are intensive properties (per gram) that must be converted to extensive properties (total energy) by multiplying by mass.

Question 18

A $60\,\text{g}$ sample of water freezes at $0^\circ\text{C}$ . The enthalpy of fusion of water is $\Delta H_{\text{fus}}=334\,\text{J/g}$ . How much energy is released when the water freezes?

20,040 J (correct answer)
2,004 J
19,000 J
334 J
6,000 J

Explanation: This question tests the skill of calculating energy released during freezing using enthalpy of fusion. When water freezes, it releases energy equal to mass times enthalpy of fusion: Energy = 60 g × 334 J/g = 20,040 J. Freezing is the reverse of melting, so while melting absorbs energy, freezing releases the same amount of energy as intermolecular forces form. The energy is released to the surroundings as the water molecules arrange into the ordered ice structure. Answer D (334 J) represents the common error of using the enthalpy value alone without considering the mass, treating ΔHfus as the total energy rather than energy per gram. Remember that phase changes are reversible processes where the magnitude of energy is the same but the direction (absorbed vs. released) is opposite.

Question 19

Solid carbon dioxide at its sublimation temperature is converted to carbon dioxide gas at the same temperature. If $\Delta H_{sub}$ for $\text{CO}_2$ is $25.2\ \text{kJ/mol}$ , how much energy is absorbed when $44\ \text{g}$ of $\text{CO}_2(s)$ sublimates? (Molar mass of $\text{CO}_2$ = $44\ \text{g/mol}$ .)

12.6 kJ
25.2 kJ (correct answer)
50.4 kJ
2.52 kJ
100.8 kJ

Explanation: This question tests the calculation of energy absorbed during sublimation when given mass rather than moles. First convert mass to moles: 44 g CO₂ ÷ 44 g/mol = 1.0 mol. Then calculate energy absorbed: ΔHsub × moles = 25.2 kJ/mol × 1.0 mol = 25.2 kJ. A common error is to multiply the enthalpy by the mass value (choice E: 25.2 × 4 ≈ 100.8), treating grams as if they were moles. For phase change calculations involving mass, always use the two-step process: convert mass to moles using molar mass, then multiply by the molar enthalpy value.

Question 20

A sample of water at $0^\circ\text{C}$ freezes to form ice at $0^\circ\text{C}$ . If $\Delta H_{fus}$ for water is $6.01\ \text{kJ/mol}$ , how much energy is released when $3.0\ \text{mol}$ of water freezes?

3.00 kJ
12.0 kJ
36.1 kJ
6.01 kJ
18.0 kJ (correct answer)

Explanation: This question tests the ability to calculate energy changes during freezing using enthalpy of fusion. When water freezes, it releases energy equal to the enthalpy of fusion, making this an exothermic process. The energy released equals ΔHfus × moles = 6.01 kJ/mol × 3.0 mol = 18.0 kJ (note that energy is released, not absorbed, during freezing). A common error is using only the enthalpy value without considering the number of moles (choice C: 6.01 kJ), which ignores the stoichiometric relationship. Remember that freezing releases the same amount of energy that melting absorbs, so multiply the molar enthalpy by the number of moles.