AP Chemistry · Learn by Concept

AP Chemistry Help: Energy Of Phase Changes

Review real example questions for Energy Of Phase Changes in AP Chemistry.

Question 1 / 10

0 of 10 answered

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 $78.11 \text{ g/mol}$ ).

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.