AP Chemistry · Handsworth Secondary 2026–27

Unit 6 · Thermochemistry

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 6, 7–9% of the AP exam, taught in 8 blocks. You met energy changes and the law of conservation of energy in BC Chemistry 11 (exo/endothermic reactions, simple q = mcΔT); BC Chemistry 12 adds enthalpy and Hess's law. Here we push to AP depth: calorimetry math, enthalpy diagrams, Hess's law, standard enthalpies of formation, and bond-enthalpy estimates. Entropy and Gibbs free energy wait until Unit 9, this unit is all about heat and enthalpy.

Blocks:  1 → A  ·  2 → B–C  ·  3 → D  ·  4 → E  ·  5 → F  ·  6 → G  ·  7 → H  ·  8 → I + review

Name:Block:Date:

A · Endothermic & Exothermic Processes CED 6.1 · LO 6.1.A

Energy is conserved (first law of thermodynamics): it moves between the system (the reaction or substance we watch) and the surroundings (everything else). A temperature change is our visible signal that energy moved.

Core: Exothermic = system loses energy to the surroundings → surroundings warm up → ΔH < 0.  Endothermic = system gains energy from the surroundings → surroundings cool down → ΔH > 0.
Watch the sign of the surroundings: if the beaker feels hot, energy left the system (exo); if it feels cold, the system pulled energy in (endo). Dissolving a salt can go either way, depending on whether the energy to break apart the solid is more or less than the energy released when ions are surrounded by water.
Worked example
A student dissolves NH₄NO₃ in water and the beaker turns cold to the touch. Is the dissolution exo- or endothermic, and what is the sign of ΔH?
The surroundings (water + beaker) lost energy (their temperature dropped) so the system must have absorbed that energy.
Endothermic; ΔH > 0 (this is why NH₄NO₃ is used in instant cold packs).

Practice A

  1. A combustion reaction makes the surrounding air noticeably warmer. Exo- or endothermic? Sign of ΔH?
  2. When CaCl₂ dissolves, the solution warms up. Classify the process and give the sign of ΔH.
  3. Explain, in terms of system and surroundings, why your skin feels cold when rubbing alcohol evaporates off it.

B · Energy Diagrams CED 6.2 · LO 6.2.A

An energy diagram plots potential energy (y-axis) against the progress of the reaction (x-axis). It shows at a glance whether a process releases or absorbs energy.

How to read it: if the products sit lower than the reactants, energy was released → exothermic, ΔH < 0. If the products sit higher, energy was absorbed → endothermic, ΔH > 0. ΔH is the vertical gap: ΔH = E(products) − E(reactants).
Worked example
On a diagram, reactants sit at 150 kJ and products at 90 kJ. Find ΔH and classify the reaction.
ΔH = E(products) − E(reactants) = 90 − 150 = −60 kJ
ΔH = −60 kJ; products are lower → exothermic. Draw the energy arrow pointing down from reactants to products.

Practice B

  1. Sketch an energy diagram for an endothermic reaction; label reactants, products, and the ΔH arrow.
  2. A diagram shows reactants at 40 kJ and products at 175 kJ. Find ΔH and classify it.
  3. Two diagrams have the same reactant energy; one's products are at 30 kJ, the other's at 80 kJ. Which reaction is more exothermic?

C · Heat Transfer & Thermal Equilibrium CED 6.3 · LO 6.3.A

Temperature is a measure of the average kinetic energy of particles. Heat is energy in transit between objects at different temperatures.

Particle picture: particles in a warmer body move faster (greater average KE). When a warm and a cool body touch, collisions transfer energy from fast particles to slow ones. This continues until thermal equilibrium: both bodies reach the same temperature and the same average kinetic energy. Energy still flows both ways, but the net transfer is zero.
Worked example
A hot copper block is dropped into cool water in an insulated cup. Describe the energy flow until equilibrium.
Copper particles have higher average KE, so collisions at the boundary transfer energy to the water particles.
The copper cools and the water warms until both reach one shared temperature.
At equilibrium, average KE (and temperature) of copper and water are equal; net energy transfer stops.

Practice C

  1. Two blocks, one at 80 °C and one at 20 °C, are placed in contact. Which way does heat flow, and why, in terms of particle kinetic energy?
  2. At thermal equilibrium, are the two bodies' temperatures equal, or are their total energies equal? Explain the difference.
  3. Why does stirring speed up the approach to thermal equilibrium?

D · Heat Capacity & Calorimetry CED 6.4 · LO 6.4.A

Calorimetry measures heat by tracking a temperature change. The same energy raises the temperature of different substances by different amounts, that's heat capacity.

Core equation: q = mcΔT, where m = mass (g), c = specific heat capacity (J·g⁻¹·°C⁻¹), ΔT = T_final − T_initial. A positive q means the substance absorbed heat; negative means it released heat. Water's c = 4.18 J·g⁻¹·°C⁻¹.
First law in a calorimeter: energy is conserved, so in an insulated cup q_lost = −q_gainedthe heat released by the hot object equals the heat absorbed by the cool one.
Worked example
Part 1, straight q = mcΔT
How much heat is needed to warm 50.0 g of water from 25.0 °C to 75.0 °C?
q = (50.0 g)(4.18 J·g⁻¹·°C⁻¹)(75.0 − 25.0 °C) = (50.0)(4.18)(50.0)
q = 10450 J
Part 2, finding an unknown c
A 25.0 g metal at 100.0 °C is dropped into 100.0 g of water at 22.0 °C; the mixture settles at 24.6 °C. Find the metal's specific heat.
q_water = (100.0)(4.18)(24.6 − 22.0) = +1086.8 J (water absorbed this)
q_metal = −q_water = −1086.8 J; ΔT_metal = 24.6 − 100.0 = −75.4 °C
c = q_metal / (m·ΔT) = (−1086.8) / [(25.0)(−75.4)]
q = 10450 J ≈ 1.05 × 10⁴ J (10.5 kJ);   c_metal = 0.577 J·g⁻¹·°C⁻¹

Practice D

  1. How much heat is released when 120. g of water cools from 65.0 °C to 30.0 °C? (c = 4.18 J·g⁻¹·°C⁻¹)
  2. A 45.0 g sample of aluminum (c = 0.897 J·g⁻¹·°C⁻¹) absorbs 1.50 kJ. What is its temperature change?
  3. A 35.0 g hot iron piece (c = 0.449) at 95.0 °C is dropped into 80.0 g of water at 20.0 °C. Set up the q_lost = −q_gained equation for the final temperature (you do not have to solve it).

E · Energy of Phase Changes CED 6.5 · LO 6.5.A

During a phase change, added energy goes into rearranging particles rather than speeding them up, so the temperature stays constant while the substance melts or boils.

Core: q = n·ΔH_phase, where n = moles and ΔH_phase is the molar enthalpy of the transition (kJ/mol). Melting and boiling absorb energy (endothermic, +); freezing and condensing release the same magnitude (exothermic, −). So ΔH_condensation = −ΔH_vaporization.
Worked example
How much energy is needed to melt 2.50 mol of ice? (ΔH_fusion of water = +6.01 kJ/mol)
q = n·ΔH_fusion = (2.50 mol)(+6.01 kJ/mol)
q = +15.0 kJ absorbed (endothermic, the energy breaks attractions, not raises temperature).

Practice E

  1. How much energy is released when 1.80 mol of steam condenses? (ΔH_vap of water = +40.7 kJ/mol)
  2. Melting 36.0 g of ice requires how much energy? (M_water = 18.02 g/mol; ΔH_fus = +6.01 kJ/mol)
  3. Why does the temperature of boiling water stay at 100 °C even as you keep adding heat?

F · Introduction to Enthalpy of Reaction CED 6.6 · LO 6.6.A

The enthalpy change ΔH of a reaction is the heat released or absorbed at constant pressure. It scales with the amount of substance reacting, just like phase changes scale with moles.

Core: q = n·ΔH_rxn. A negative ΔH releases heat (exothermic); positive absorbs it (endothermic). ΔH stated for a balanced equation refers to the molar amounts in that equation, so always relate your moles back to the coefficients.

Not tested: the technical distinction between enthalpy and internal energy. At the AP level reactions are at constant pressure, where ΔH simply equals the heat of reaction, treat "enthalpy change" and "heat of reaction" as the same thing here.

Worked example
Combustion of methane releases 445 kJ when 0.500 mol of CH₄ burns. Find ΔH per mole of CH₄.
"Releases" → q = −445 kJ for 0.500 mol.
ΔH = q / n = (−445 kJ) / (0.500 mol)
ΔH = −890 kJ/mol CH₄ (exothermic).

Practice F

  1. A reaction has ΔH = −286 kJ/mol. How much heat is released when 3.00 mol react?
  2. Burning 0.250 mol of propane releases 555 kJ. Find ΔH per mole of propane.
  3. For 2 H₂ + O₂ → 2 H₂O, ΔH = −572 kJ. How much heat is released per mole of H₂O formed?

G · Bond Enthalpies CED 6.7 · LO 6.7.A

Breaking bonds costs energy (endothermic); forming bonds releases energy (exothermic). Average bond enthalpies let us estimate a reaction's ΔH from the bonds that change.

Core: ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Add up the bond energies of all bonds broken in the reactants, subtract all bonds formed in the products. If more energy is released forming bonds than was spent breaking them, ΔH is negative (exothermic).
Worked example
Estimate ΔH for CH₄ + 2 O₂ → CO₂ + 2 H₂O(g). Bond energies (kJ/mol): C–H 414, O=O 498, C=O 799, O–H 463.
Bonds broken: 4 (C–H) + 2 (O=O) = 4(414) + 2(498) = 1656 + 996 = 2652 kJ
Bonds formed: 2 (C=O) + 4 (O–H) = 2(799) + 4(463) = 1598 + 1852 = 3450 kJ
ΔH ≈ 2652 − 3450
ΔH ≈ −798 kJ (exothermic). Note: this is an estimate from average bond energies, close to but not identical to the −890 kJ from formation data.

Practice G

  1. Estimate ΔH for H₂ + Cl₂ → 2 HCl. Bond energies: H–H 436, Cl–Cl 242, H–Cl 431 kJ/mol.
  2. For N₂ + 3 H₂ → 2 NH₃, identify every bond broken and every bond formed (count them, this is the step students miss).
  3. If breaking the reactant bonds needs 600 kJ and forming the product bonds releases 550 kJ, is the reaction endo- or exothermic? Give ΔH.

H · Enthalpy of Formation CED 6.8 · LO 6.8.A

The standard enthalpy of formation ΔH°f is the enthalpy change to make one mole of a compound from its elements in their standard states. Tables of ΔH°f let us find any reaction's ΔH.

Core: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), each weighted by its coefficient. The ΔH°f of an element in its standard state is 0 (e.g. O₂, N₂, C(graphite)).
Worked example
Find ΔH°rxn for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l). ΔH°f (kJ/mol): CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8, O₂ 0.
Products: ΔH°f(CO₂) + 2·ΔH°f(H₂O) = (−393.5) + 2(−285.8) = −393.5 − 571.6 = −965.1 kJ
Reactants: ΔH°f(CH₄) + 2·ΔH°f(O₂) = (−74.8) + 2(0) = −74.8 kJ
ΔH°rxn = (−965.1) − (−74.8)
ΔH°rxn = −890.3 kJ (exothermic).

Practice H

  1. Find ΔH°rxn for 2 H₂O₂(l) → 2 H₂O(l) + O₂(g). ΔH°f: H₂O₂(l) −187.8, H₂O(l) −285.8, O₂ 0 kJ/mol.
  2. Why is ΔH°f of O₂(g) equal to zero, while ΔH°f of O₃(g) is not?
  3. For C₂H₄(g) + 3 O₂ → 2 CO₂ + 2 H₂O(l), set up (don't solve) the ΣΔH°f expression with coefficients.

I · Hess's Law CED 6.9 · LO 6.9.A, LO 6.9.B

Because energy is conserved, the enthalpy of an overall reaction equals the sum of the enthalpies of any steps that add up to it, no matter what path you take.

Three rules: (i) reverse a reaction → flip the sign of ΔH; (ii) multiply a reaction by a factor c → multiply ΔH by c; (iii) add reactions → add their ΔH values. Arrange the given steps so the unwanted species cancel and you're left with the target equation.

Not tested: the formal concept of "state functions." You apply the three rules above; you are not asked to define or invoke state functions by name.

Worked example
Find ΔH for C(s) + ½ O₂(g) → CO(g) using:  (1) C(s) + O₂ → CO₂, ΔH₁ = −393.5 kJ   (2) CO + ½ O₂ → CO₂, ΔH₂ = −283.0 kJ.
Keep (1) as written: C + O₂ → CO₂, ΔH = −393.5
Reverse (2) so CO ends up as a product: CO₂ → CO + ½ O₂, ΔH = +283.0
Add: C + O₂ + CO₂ → CO₂ + CO + ½ O₂. Cancel one CO₂ and ½ O₂ from each side → C + ½ O₂ → CO
ΔH = (−393.5) + (+283.0)
ΔH = −110.5 kJ.

Practice I

  1. Given N₂ + O₂ → 2 NO, ΔH = +180 kJ and 2 NO + O₂ → 2 NO₂, ΔH = −112 kJ, find ΔH for N₂ + 2 O₂ → 2 NO₂.
  2. If a target reaction is the reverse of a given one with ΔH = −250 kJ, what is the target's ΔH?
  3. You must double a given reaction to match the target. The given ΔH is −146 kJ. What ΔH do you use?
AP Chemistry · Unit 6, Thermochemistry · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 6.1–6.9) and BC Chemistry 11/12 energy changes. Entropy & Gibbs free energy are deferred to Unit 9. Specific heats, ΔH°f, and bond energies from standard reference tables; atomic masses from the IUPAC periodic table.