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Energy of Phase Changes - AP Chemistry

  1. A sample of steam at 100°C condenses into liquid water at 100°C. Which statement best describes the energy change and molecular interactions during this process?
    Answer: The process is exothermic because energy is released as intermolecular hydrogen bonds are formed between water molecules. Condensation is an exothermic process (ΔH<0\Delta H < 0) because energy is released when gas particles come closer together and form intermolecular attractions. For water, these are strong hydrogen bonds. Answer A incorrectly states the process is endothermic. Answer C incorrectly states that the temperature decreases; temperature is constant during a phase change. Answer D incorrectly states that attractions are broken; they are formed during condensation.
  2. A 10.0 g sample of a solid melts at its melting point. The enthalpy of fusion is ΔHfus=250 J g−1\Delta H_{\text{fus}}=250\ \text{J g}^{-1}. How much energy is absorbed during melting?
    Answer: 2.50×10310^3 J This question tests the skill of calculating the energy absorbed during melting using the enthalpy of fusion. For the 10.0 g solid sample, the energy is 10.0 g × 250 J/g = 2.50×10310^3 J, reflecting the endothermic nature of melting. This heat is used to increase molecular disorder without altering temperature. Choice A matches this calculation correctly. Choice E, 2.50×10210^2 J, is a common distractor from the misconception of using 1.0 g instead of 10.0 g, perhaps due to a decimal error, leading to undercalculation. To handle such questions effectively, write out the formula q = m × ΔH_fus and plug in values carefully to avoid simple mistakes.
  3. The molar enthalpy of fusion (ΔHfus\Delta H_{\text{fus}}) of NaCl is 28 kJ/mol28 \text{ kJ/mol}, while that of solid methane (CH₄) is 0.94 kJ/mol0.94 \text{ kJ/mol}. Which statement best explains this large difference?
    Answer: Melting NaCl requires breaking strong ionic bonds, while melting CH₄ requires overcoming weak London dispersion forces. 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.
  4. Liquid benzene at its melting point is converted to solid benzene at the same temperature. If ΔHfus\Delta H_{fus} for benzene is 9.95 kJ/mol9.95\ \text{kJ/mol}, how much energy is released when 0.80 mol0.80\ \text{mol} of benzene freezes?
    Answer: 7.96 kJ 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.
  5. A sample of water at 0∘C0^\circ\text{C} freezes to form ice at 0∘C0^\circ\text{C}. If ΔHfus\Delta H_{fus} for water is 6.01 kJ/mol6.01\ \text{kJ/mol}, how much energy is released when 3.0 mol3.0\ \text{mol} of water freezes?
    Answer: 18.0 kJ 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 D: 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.
  6. Liquid ethanol at its boiling point is converted to ethanol vapor at the same temperature. If ΔHvap\Delta H_{vap} for ethanol is 38.6 kJ/mol38.6\ \text{kJ/mol}, how much energy is absorbed when 0.50 mol0.50\ \text{mol} of ethanol vaporizes?
    Answer: 19.3 kJ This question tests the ability to calculate energy changes during phase transitions using enthalpy of vaporization. The enthalpy of vaporization (ΔHvap) represents the energy required to convert one mole of liquid to vapor at constant temperature. To find the energy absorbed when 0.50 mol of ethanol vaporizes, we multiply: Energy = ΔHvap × moles = 38.6 kJ/mol × 0.50 mol = 19.3 kJ. A common misconception is to divide the enthalpy by the number of moles (choice E: 38.6 ÷ 8 ≈ 4.82), which incorrectly treats enthalpy as a total rather than a per-mole quantity. When solving phase change problems, always multiply the molar enthalpy value by the number of moles to find total energy.
  7. The molar enthalpy of fusion for iron (Fe) is 13.8 kJ/mol13.8 \text{ kJ/mol}. Calculate the heat, in Joules, required to melt a 11.2 g11.2 \text{ g} sample of solid iron at its melting point. (The molar mass of Fe is 55.85 g/mol55.85 \text{ g/mol}).
    Answer: 2770 J First, convert the mass of iron to moles: n=11.2 g55.85 g/mol≈0.2005 moln = \frac{11.2 \text{ g}}{55.85 \text{ g/mol}} \approx 0.2005 \text{ mol}. Next, calculate the heat in kJ: q=n⋅ΔHfus=(0.2005 mol)(13.8 kJ/mol)≈2.77 kJq = n \cdot \Delta H_{\text{fus}} = (0.2005 \text{ mol})(13.8 \text{ kJ/mol}) \approx 2.77 \text{ kJ}. Finally, convert kJ to J: 2.77 kJ×1000 J1 kJ=2770 J2.77 \text{ kJ} \times \frac{1000 \text{ J}}{1 \text{ kJ}} = 2770 \text{ J}. Distractor A fails to convert kJ to J. Distractor D results from incorrectly multiplying mass by molar enthalpy.
  8. Solid carbon dioxide at its sublimation temperature is converted to carbon dioxide gas at the same temperature. If ΔHsub\Delta H_{sub} for CO2\text{CO}_2 is 25.2 kJ/mol25.2\ \text{kJ/mol}, how much energy is absorbed when 44 g44\ \text{g} of CO2(s)\text{CO}_2(s) sublimates? (Molar mass of CO2\text{CO}_2 = 44 g/mol44\ \text{g/mol}.)
    Answer: 25.2 kJ 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.
  9. A 40.0 g sample of methanol is vaporized at its boiling point. The enthalpy of vaporization of methanol is ΔHvap=35.3 kJ/mol\Delta H_{vap}=35.3\ \text{kJ/mol}. How much energy is absorbed? (Molar mass of methanol =32.0 g/mol=32.0\ \text{g/mol}.)
    Answer: 44.1 kJ 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.
  10. A 12.0 g sample of a liquid freezes at its freezing point. The enthalpy of fusion is ΔHfus=400 J g−1\Delta H_{\text{fus}}=400\ \text{J g}^{-1}. How much energy is released during freezing?
    Answer: 4.80×10310^3 J 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 = 4.80×10310^3 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, 1.20×10310^3 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.
  11. A block of dry ice (solid CO₂) is left on a countertop at room temperature and slowly disappears without forming a liquid. Which statement correctly describes the thermodynamics of this process?
    Answer: The sublimation of CO₂ is endothermic (ΔH>0\Delta H > 0), as energy is absorbed from the surroundings to overcome intermolecular forces. The process described is sublimation, the direct conversion of a solid to a gas. This process requires energy input from the surroundings to overcome the intermolecular forces holding the CO₂ molecules in a solid lattice. Therefore, the process is endothermic, and the enthalpy change (ΔHsub\Delta H_{\text{sub}}) is positive.
  12. The enthalpy of solidification for benzene (C₆H₆) is −9.95 kJ/mol-9.95 \text{ kJ/mol}. How many grams of liquid benzene can be frozen at its freezing point if 4.975 kJ4.975 \text{ kJ} of heat is removed? (The molar mass of C₆H₆ is 78.11 g/mol78.11 \text{ g/mol}).
    Answer: 39.1 g First, calculate the moles of benzene that can be frozen. Since heat is removed, q=−4.975 kJq = -4.975 \text{ kJ}. Using q=nΔHsolidificationq = n \Delta H_{\text{solidification}}, we find n=qΔHsolidification=−4.975 kJ−9.95 kJ/mol=0.500 moln = \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 mol×78.11 g/mol=39.1 gmass = 0.500 \text{ mol} \times 78.11 \text{ g/mol} = 39.1 \text{ g}.
  13. A 25.0 g sample of ethanol(l) is vaporized to ethanol(g) at its boiling point. The enthalpy of vaporization of ethanol is ΔHvap=840 J g−1\Delta H_{\text{vap}}=840\ \text{J g}^{-1}. How much energy is absorbed during the phase change?
    Answer: 2.10×10410^4 J This question tests the skill of calculating the energy absorbed during vaporization using the enthalpy of vaporization. For the 25.0 g sample of ethanol, the energy absorbed is 25.0 g × 840 J/g = 2.10×10410^4 J, corresponding to the heat needed to convert the liquid to gas at the boiling point. This process is endothermic, absorbing energy to overcome strong intermolecular attractions. The correct choice is A, aligning with this calculation. Choice B, 2.50×10310^3 J, is a common distractor resulting from the misconception of using 100 g instead of 25.0 g or misplacing the decimal, underestimating the energy. A transferable strategy is to double-check the mass value and perform the multiplication step-by-step to avoid arithmetic errors in phase change energy calculations.
  14. A 40.0 g sample of a liquid is vaporized at its boiling point. The enthalpy of vaporization is ΔHvap=500 J g−1\Delta H_{\text{vap}}=500\ \text{J g}^{-1}. How much energy is absorbed during vaporization?
    Answer: 2.00×10410^4 J This question tests the skill of calculating the energy absorbed during vaporization using the given enthalpy value. For the 40.0 g sample, the energy is 40.0 g × 500 J/g = 2.00×10410^4 J, representing the endothermic heat required to vaporize the liquid. This process involves breaking intermolecular bonds at the boiling point without a temperature change. Choice A is correct based on this multiplication. A common distractor is choice C, 2.00×10310^3 J, which occurs from the misconception of using 4.0 g instead of 40.0 g, perhaps by dropping a zero, resulting in a lower value. To approach similar questions, verify the phase change type and apply q = m × ΔH consistently while checking significant figures.
  15. The process of liquid water evaporating from skin feels cool. Which statement provides the correct thermodynamic explanation for this phenomenon?
    Answer: The phase change from liquid to gas is endothermic, absorbing thermal energy from the skin. Evaporation (vaporization) is an endothermic process, meaning it requires an input of energy to overcome intermolecular forces. The water molecules on the skin absorb this energy in the form of heat from the skin, causing the skin's temperature to drop and creating a cooling sensation.
  16. A 15 g15\,\text{g} sample of Substance Y condenses at its boiling point. The enthalpy of vaporization is ΔHvap=500 J/g\Delta H_{\text{vap}}=500\,\text{J/g}. How much energy is released during condensation?
    Answer: 7,500 J This question tests the skill of calculating energy released during condensation using enthalpy of vaporization. Energy released during condensation equals mass times enthalpy of vaporization: Energy = 15 g × 500 J/g = 7,500 J. Condensation is the reverse of vaporization, releasing energy as gas molecules come together to form liquid and intermolecular forces are established. The same amount of energy absorbed during vaporization is released during condensation at the same temperature. Answer B (500 J) reflects the misconception of using only the enthalpy value without accounting for the sample mass, forgetting that ΔHvap is defined per gram. When working with phase changes, always identify whether energy is absorbed (solid→liquid→gas) or released (gas→liquid→solid) to ensure proper interpretation.
  17. A 36 g36\ \text{g} sample of ice at 0∘C0^\circ\text{C} melts to liquid water at 0∘C0^\circ\text{C}. If ΔHfus\Delta H_{fus} for water is 6.01 kJ/mol6.01\ \text{kJ/mol}, how much energy is absorbed during melting? (Molar mass of water = 18 g/mol18\ \text{g/mol}.)
    Answer: 12.0 kJ This question tests the calculation of energy absorbed during melting when given mass. First convert 36 g of ice to moles: 36 g ÷ 18 g/mol = 2.0 mol. Then calculate energy absorbed: ΔHfus × moles = 6.01 kJ/mol × 2.0 mol = 12.0 kJ. A common error is to use only the enthalpy value without considering the amount of substance (choice A: 6.01 kJ), which would be correct for only 1 mole. When solving phase change problems with mass, always convert to moles first, then multiply by the molar enthalpy to get total energy.
  18. A 10.0 g sample of iodine, I2(s)\text{I}_2(s), sublimes at its sublimation point. The enthalpy of sublimation is ΔHsub=62.4 kJ/mol\Delta H_{sub}=62.4\ \text{kJ/mol}. How much energy is absorbed during sublimation? (Molar mass of I2\text{I}_2 =254 g/mol=254\ \text{g/mol}.)
    Answer: 2.46 kJ 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.
  19. Solid naphthalene at its melting point melts to liquid naphthalene at the same temperature. If ΔHfus\Delta H_{fus} for naphthalene is 19.0 kJ/mol19.0\ \text{kJ/mol}, how much energy is absorbed when 0.30 mol0.30\ \text{mol} of naphthalene melts?
    Answer: 5.70 kJ This question tests the calculation of energy absorbed during melting using enthalpy of fusion. The energy required to melt solid naphthalene equals ΔHfus × moles = 19.0 kJ/mol × 0.30 mol = 5.70 kJ. A common mistake is to use the enthalpy value directly without accounting for the fractional mole amount (choice B: 19.0 kJ), which would only be correct for exactly 1.0 mol. For accurate phase change calculations, always multiply the per-mole enthalpy by the actual number of moles, even when working with fractional amounts.
  20. A 15.0 g sample of benzene freezes at its melting point. The enthalpy of fusion of benzene is ΔHfus=9.95 kJ/mol\Delta H_{fus}=9.95\ \text{kJ/mol}. How much energy is released during freezing? (Molar mass of benzene =78.0 g/mol=78.0\ \text{g/mol}.)
    Answer: 1.91 kJ 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.

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