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AP Chemistry Unit 6: Thermochemistry

7 to 9% of the exam · 180 practice questions · 5 free-response questions

Every chemical and physical change moves energy. Unit 6 covers endothermic and exothermic processes, heat transfer, calorimetry, phase changes, and three ways to find the enthalpy of a reaction: bond enthalpies, enthalpies of formation, and Hess's law. Expect to calculate heat from calorimetry data and reason about which way energy flows between system and surroundings. A frequent error is miscounting the bonds broken and formed. Another is believing that breaking a bond releases energy. Breaking a bond always takes energy, and forming one releases it.

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White belt · 60 questionsRegular · 60 questionsBlack belt · 60 questionsFree response

White belt covers core terms and foundational ideas, Regular is set at exam level, and Black belt has harder multi-step questions.

Topics in this unit

Free-response questions

Sample questions

These are taken from the Regular set, with the answer and explanation shown.

Topic 6.3 Heat Transfer and Thermal Equilibrium

1. A hot iron bar is placed in cool water in an insulated container and the two reach thermal equilibrium. Which comparison is correct?

  1. AThe magnitude of q is larger for the iron, but the magnitudes of ΔT are equal.
  2. BThe magnitude of q is larger for the water, and the magnitude of ΔT is larger for the iron.
  3. CThe magnitudes of q and of ΔT are both equal for the iron and the water.
  4. DThe magnitudes of q for the iron and the water are equal, but the magnitudes of ΔT are different.

Answer: D. Energy is conserved, so q_iron = −q_water: equal sizes, opposite signs. Their temperature changes depend on each object's mass and specific heat, so they are generally different.

Common trap: Thinking equal energy means equal temperature change. ΔT = q/(mc), so objects with different mc values change by different amounts.

Topic 6.5 Energy of Phase Changes

2. At their boiling points, methanol (CH₃OH) has ΔHvap ≈ 35.2 kJ/mol and dimethyl ether (CH₃OCH₃) has ΔHvap ≈ 21.5 kJ/mol. Which best explains the difference?

  1. AMethanol has a larger molar mass, so its molecules have greater London dispersion forces.
  2. BDimethyl ether molecules form hydrogen bonds with each other, but methanol molecules cannot.
  3. CMethanol molecules have stronger covalent C–O bonds than dimethyl ether molecules have.
  4. DMethanol molecules form hydrogen bonds with each other, but dimethyl ether molecules cannot.

Answer: D. ΔHvap measures how much energy it takes to pull liquid molecules apart. Methanol has an O–H group, so its molecules hydrogen bond to each other, which takes more energy to overcome. Dimethyl ether has no H on its O.

Common trap: Choosing the molar-mass option. Methanol (32 g/mol) is actually lighter than dimethyl ether (46 g/mol), so dispersion forces cannot explain it.

Topic 6.8 Enthalpy of Formation

Standard enthalpies of formation, ΔH°f
SubstanceΔH°f (kJ/mol)
C₃H₈(g)−103.8
C₂H₅OH(l)−277.7
CO₂(g)−393.5
H₂O(l)−285.8
H₂O(g)−241.8
O₂(g)0

Standard enthalpies of formation at 25 °C and 1 atm, rounded to 0.1 kJ/mol. These are the standard reference values used in general chemistry courses.

3. Use the data in the table to calculate ΔH° for C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l).

  1. A−2,219.9 kJ/molrxn
  2. B−2,043.9 kJ/molrxn
  3. C−575.5 kJ/molrxn
  4. D+2,219.9 kJ/molrxn

Answer: A. Products: 3(−393.5) + 4(−285.8) = −2,323.7. Reactants: −103.8 + 5(0) = −103.8. ΔH° = −2,323.7 − (−103.8) = −2,219.9 kJ/molrxn.

Common trap: Using the H₂O(g) value instead of H₂O(l), which gives −2,043.9 kJ/molrxn. Read the phase labels carefully.

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