AP Chemistry

AP Chemistry Unit 6 review: Thermochemistry

Unit 6 tracks energy transferred as heat during reactions, temperature changes and phase changes. It is 7–9% of multiple choice. Expect to define the system, assign the sign of heat correctly and calculate enthalpy from calorimetry, bond energies or summed reactions.

Unit 6 on the May 2027 exam

AP Chemistry Unit 6 exam facts
2027 exam
Weight7–9% of the multiple-choice section
Exam dateThursday, May 6, 2027, afternoon session
Multiple choice60 questions in 90 minutes, worth 50% of the exam score
Free response7 questions in 105 minutes, worth 50%: three long questions (10 points each) and four short questions (4 points each).
CalculatorBoth sections allow scientific or approved graphing calculators. Four-function calculators are allowed but not recommended. Bluebook includes both Desmos scientific and graphing calculators.
Class timeAbout 10–11 class periods in College Board's pacing guide
2026 results15% of students scored a 5 on the whole exam, and about 76% scored 3 or higher

Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.

What Unit 6 covers

College Board's course description splits the unit into 9 topics. The second column is what you should be able to do for each one.

AP Chemistry Unit 6 topics
TopicWhat you should be able to do
6.1 Endothermic and Exothermic ProcessesIdentify the system and surroundings and the direction of heat transfer.
6.2 Energy DiagramsRepresent endothermic and exothermic processes with relative energy levels.
6.3 Heat Transfer and Thermal EquilibriumExplain thermal equilibrium and heat flow from warmer to cooler matter.
6.4 Heat Capacity and CalorimetryUse q = mcΔT or calorimeter heat capacity and relate measured heat to reaction enthalpy.
6.5 Energy of Phase ChangesCalculate heat for phase changes and distinguish them from temperature changes.
6.6 Introduction to Enthalpy of ReactionRelate a balanced equation and ΔH to the actual amount reacting.
6.7 Bond EnthalpiesEstimate reaction enthalpy from bonds broken minus bonds formed.
6.8 Enthalpy of FormationCalculate reaction enthalpy from standard enthalpies of formation.
6.9 Hess’s LawAdd, reverse and scale reactions and their enthalpy changes.

Select the heat calculation and keep its sign

Name what gains or loses heat before inserting numbers. All rows assume the stated energy model and units.

Select the heat calculation and keep its sign
EvidenceUseCommon sign or unit check
Temperature changeq = mc(Tfinal − Tinitial)A warming sample has positive q.
Calorimeter constantqcal = CcalΔTDo not multiply by mass again if Ccal is for the whole calorimeter.
Insulated reaction calorimeterqreaction = −(qsolution + qcal)A warming solution usually indicates an exothermic reaction.
Phase changeq = nΔHphaseVaporization and melting absorb heat; reverse changes release it.
Bond enthalpiesΔH ≈ ΣDbroken − ΣDformedCount every bond and every stoichiometric coefficient.
Formation enthalpiesΔH°rxn = ΣnΔH°f(products) − ΣnΔH°f(reactants)An element in its standard state has ΔH°f = 0.
Hess’s lawReverse: change sign; scale: multiply ΔHCancel species only after adjusting whole equations.

Common mistakes

Each of these loses points on questions you otherwise understand.

  • Counting molecules instead of individual bonds. Include the number of each bond per molecule and the equation coefficient. CED printed p. 109 identifies this error.
  • Giving the reaction the same heat sign as the warming solution. In the insulated model the reaction loses the heat the solution gains. CED topic 6.4.
  • Saying bond breaking releases energy. Breaking a bond requires energy; forming a bond releases energy. CED topic 6.7.
  • Using q = mcΔT during a constant-temperature phase change. Use the phase-change enthalpy for that interval. CED topic 6.5.

6 practice questions with answers

Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.

  1. Question 1 · 6.4

    A 75.0 g solution warms by 4.00 °C. Its specific heat is 4.00 J g⁻¹ °C⁻¹. How much heat does the solution absorb?

    • A0.300 kJ
    • B12.0 kJ
    • C3.00 kJ
    • D1.20 kJ
    Show the answer

    D. q = mcΔT = 75.0(4.00)(4.00) = 1,200 J = 1.20 kJ. The positive sign belongs to the warming solution.

  2. Question 2 · 6.4

    An insulated calorimeter’s solution absorbs 2.40 kJ while 0.0300 mol of reactant is consumed. Neglect calorimeter heat. What is reaction ΔH per mole consumed?

    • A+80.0 kJ mol⁻¹
    • B−80.0 kJ mol⁻¹
    • C−0.0720 kJ mol⁻¹
    • D+0.0720 kJ mol⁻¹
    Show the answer

    B. The reaction releases 2.40 kJ, so qreaction = −2.40 kJ. Dividing by 0.0300 mol gives −80.0 kJ mol⁻¹. The solution and reaction have opposite heat signs.

  3. Question 3 · 6.7

    For H₂ + Cl₂ → 2HCl, use bond enthalpies H–H = 436, Cl–Cl = 243 and H–Cl = 431 kJ mol⁻¹. What is estimated ΔH?

    • A−183 kJ mol⁻¹
    • B+183 kJ mol⁻¹
    • C+248 kJ mol⁻¹
    • D−679 kJ mol⁻¹
    Show the answer

    A. Break one H–H and one Cl–Cl bond: 436 + 243 = 679 kJ. Form two H–Cl bonds: 2(431) = 862 kJ released. ΔH ≈ 679 − 862 = −183 kJ per mole of reaction as written.

  4. Question 4 · 6.9

    For A → B, ΔH = +35 kJ mol⁻¹; for B → C, ΔH = −82 kJ mol⁻¹. What is ΔH for C → A?

    • A−117 kJ mol⁻¹
    • B−47 kJ mol⁻¹
    • C+47 kJ mol⁻¹
    • D+117 kJ mol⁻¹
    Show the answer

    C. Adding the original equations gives A → C with ΔH = 35 − 82 = −47 kJ mol⁻¹. Reversing to C → A changes the sign to +47 kJ mol⁻¹.

  5. Question 5 · 6.5

    At its boiling point, 0.250 mol of a liquid vaporizes with ΔHvap = 32.0 kJ mol⁻¹. What heat is absorbed?

    • A128 kJ
    • B32.0 kJ
    • C0.00781 kJ
    • D8.00 kJ
    Show the answer

    D. q = nΔHvap = 0.250(32.0) = 8.00 kJ. Temperature remains constant during the idealized phase change, but energy still enters the sample.

  6. Question 6 · 6.8

    For CO(g) + ½O₂(g) → CO₂(g), use ΔH°f(CO) = −111 and ΔH°f(CO₂) = −394 kJ mol⁻¹. What is ΔH°rxn?

    • A+283 kJ mol⁻¹
    • B−283 kJ mol⁻¹
    • C−505 kJ mol⁻¹
    • D+505 kJ mol⁻¹
    Show the answer

    B. ΔH°f for O₂ in its standard state is zero. Products minus reactants gives −394 − (−111) = −283 kJ mol⁻¹. Keep the negative sign on the reactant formation enthalpy inside parentheses.

How to study Unit 6

Draw a boundary around the reaction and arrows for heat entering or leaving. Calculate the measured sample’s heat first, then translate it to reaction heat and finally to enthalpy per mole. Keep joules and kilojoules separate until the unit conversion is explicit.

For a bond calculation, draw all reactant and product structures and tally bonds in a table. For Hess’s law, write the target equation first, scale and reverse source equations, and verify species cancellation before adding their enthalpies.

Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.

Questions

Still have a question? support@tryashore.com

It is 7–9% of the multiple-choice section, which is half the exam score. This is not a fixed share of the whole exam: free-response questions can combine units. Suggested pacing is 10–11 class periods.

Learn Chemistry one short lesson at a time

Free to start, no card. A short lesson for every topic in all 9 units.

Starting check

SAT Math · 1 of 5

Calculator allowed

A jacket's price is increased by 20% and then the new price is decreased by 20%. The final price is what percent of the original price?
  • A96%96\%
  • B100%100\%
  • C104%104\%
  • D80%80\%