AP Chemistry Unit 6 Review: Thermochemistry

Every Unit 6 topic, the formulas that actually show up, and the sign conventions that trip students up most.

Unit 6 is where energy stops being a side note and becomes the main variable. It's 7–9% of the multiple-choice section across 9 topics and 10–11 class periods, and nearly every calculation in it comes down to one idea: energy is conserved, and you're just tracking where it goes. Here's what's actually in it, topic by topic, verified against the current CED.

Unit 6 at a Glance

Unit 6 (Thermochemistry) shifts from Unit 5's question of how fast a reaction happens to how much energy it releases or absorbs. See the full AP Chemistry units breakdown for how it fits alongside the other 8 units, and the Course and Exam Description for the full framework.

Endothermic and Exothermic Processes

Every physical or chemical process either releases energy to the surroundings (exothermic, ΔH < 0) or absorbs energy from the surroundings (endothermic, ΔH > 0). The sign is always defined from the system's point of view: a negative ΔH means the system lost energy, a positive ΔH means it gained energy -- regardless of whether that energy shows up as heat, light, or another form.

Energy Diagrams

A Unit 6 energy diagram plots the total energy of a system before and after a process, with the vertical drop or rise between the two levels representing ΔH. This is a different diagram than Unit 5's reaction energy profile: Unit 5's version tracks the pathway through a transition state to explain reaction rate (activation energy), while Unit 6's version only cares about the net difference between the initial and final energy levels -- the pathway in between is irrelevant here.

Heat Transfer and Thermal Equilibrium

Heat flows spontaneously from a hotter object to a colder one until both reach the same temperature -- thermal equilibrium. In an isolated system (no energy escaping to the outside), this is just the first law of thermodynamics at work: energy is conserved, so whatever heat one substance loses, another substance in the same system gains. That single conservation statement, q_lost = q_gained, is the foundation every calorimetry calculation in this unit is built on.

Heat Capacity and Calorimetry

Two formulas cover almost every calorimetry problem: q = mcΔT for a temperature change (m = mass, c = specific heat capacity, ΔT = temperature change), and q = nΔH for a phase change or reaction (n = moles). Water's specific heat, c = 4.18 J/(g·°C), is the one value worth having memorized cold -- it shows up in the large majority of calorimetry problems.

Worked example: A 50.0 g piece of metal at 95.0°C is dropped into 100.0 g of water at 22.0°C. The final temperature is 24.5°C. Find the metal's specific heat.

Heat gained by water: q = mcΔT = (100.0 g)(4.18 J/g·°C)(24.5°C − 22.0°C) = 1045 J. By conservation of energy, the metal lost exactly this much heat: q_metal = −1045 J. Solving q = mcΔT for the metal's c: −1045 J = (50.0 g)(c)(24.5°C − 95.0°C), so c = −1045 / (50.0 × −70.5) = 0.296 J/(g·°C).

Sign convention: for a reaction measured in a coffee-cup calorimeter, the heat released or absorbed by the reaction is the negative of the heat measured by the surrounding solution: q_rxn = −q_solution. Getting this sign backwards is one of the most common calorimetry errors on the exam.

Energy of Phase Changes

A phase change absorbs or releases energy without changing temperature, using q = nΔH_fus (melting/freezing) or q = nΔH_vap (boiling/condensing). This is latent heat -- energy that goes into rearranging molecular arrangement, not into raising temperature. Compare that to sensible heat (q = mcΔT), where energy does raise temperature. On a heating curve, the sloped segments are sensible heat; the flat plateaus, where temperature holds steady while heat keeps flowing in, are phase changes -- latent heat.

Introduction to Enthalpy of Reaction

ΔH is a state function: it depends only on the initial and final states of a system, never on the path taken between them. Three consequences follow directly and matter for every later topic in this unit: ΔH scales proportionally if you multiply a reaction's coefficients, ΔH flips sign if you reverse a reaction, and ΔH for a multi-step path equals the sum of ΔH for each individual step -- which is exactly the idea Hess's Law formalizes below.

Bond Enthalpies

Breaking a bond always requires energy (endothermic, contributes positively); forming a bond always releases energy (exothermic, contributes negatively). That gives a reaction-enthalpy estimate: ΔH_rxn ≈ Σ(bonds broken) − Σ(bonds formed).

Critical rule: bond enthalpies used in this formula are average values taken across many different compounds containing that bond type -- not the exact bond strength in your specific molecule. A bond-enthalpy calculation is always an estimate of ΔH_rxn, not an exact value; a real experimental measurement (calorimetry) or a formation-enthalpy calculation is more precise.

Enthalpy of Formation

The standard enthalpy of formation, ΔH°_f, is the enthalpy change when 1 mole of a compound forms from its elements in their standard states. Because it's referenced to elements in their standard states, ΔH°_f of any element in its standard state is defined as zero. This gives a second, more precise way to find a reaction's enthalpy: ΔH°_rxn = ΣΔH°_f(products) − ΣΔH°_f(reactants).

Hess's Law

Because ΔH is a state function, thermochemical equations can be added like algebraic equations: if you reverse a given reaction, reverse the sign of its ΔH; if you multiply a reaction by a coefficient, multiply its ΔH by the same coefficient. Add the (possibly modified) given reactions together so everything except the target reaction's own reactants and products cancels out, and add their ΔH values to get the target reaction's ΔH.

Worked example: Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ, find ΔH for C(s) + ½O₂(g) → CO(g).

Keep the first equation as written. Reverse the second equation (flip its sign to +283.0 kJ) so CO₂ appears as a reactant and cancels: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Adding the two: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g). Cancel CO₂(g) from both sides and simplify O₂: C(s) + ½O₂(g) → CO(g). Sum the ΔH values: −393.5 + 283.0 = −110.5 kJ.

Common Mistakes in Unit 6

Treating q and ΔH as always interchangeable. They're only numerically equal at constant pressure (q_p = ΔH) -- which is exactly why an open coffee-cup calorimeter works, but it's a specific condition, not a general identity.
Forgetting to flip ΔH's sign when reversing a reaction in a Hess's Law problem. If you reverse a given equation to make it cancel correctly, its ΔH must flip sign too -- this is the single most common Hess's Law error.
Treating a bond-enthalpy calculation as an exact value. Average bond enthalpies give an estimate of ΔH_rxn, not the true experimental value.
Getting the calorimetry sign backwards. q_rxn = −q_solution -- mixing up which side is negative is a frequent source of a sign-flipped final answer.
Confusing sensible heat with latent heat on a heating curve. During a plateau (a phase change), temperature isn't changing, so q = mcΔT doesn't apply -- you need q = nΔH_fus or q = nΔH_vap instead.

How Unit 6 Connects to the Rest of the Course

Hess's Law-style scaling and combining of equations here requires the same stoichiometric mole-ratio skills Unit 4 introduced for balanced equations.
Enthalpy is one half of the free-energy equation ΔG = ΔH − TΔS that Unit 9 builds directly on top of this unit's ΔH.
Whether a reaction is exothermic or endothermic determines how it responds to a temperature change under Unit 7's Le Châtelier's principle -- heat effectively behaves like a reactant or product depending on that sign.

Related Resources

Frequently Asked Questions

What topics are in AP Chemistry Unit 6?

Endothermic and Exothermic Processes, Energy Diagrams, Heat Transfer and Thermal Equilibrium, Heat Capacity and Calorimetry, Energy of Phase Changes, Introduction to Enthalpy of Reaction, Bond Enthalpies, Enthalpy of Formation, and Hess's Law -- 9 topics in total.

How much is Unit 6 worth on the AP Chemistry exam?

Seven to nine percent of the multiple-choice section across roughly 10 to 11 class periods.

Are bond-enthalpy calculations of reaction enthalpy exact?

No. Bond enthalpies used in the calculation ΔH ≈ Σ(bonds broken) − Σ(bonds formed) are averages taken across many different compounds containing that bond type, not the exact bond strength in your specific molecule -- so the result is always an estimate, not a precise experimental value.

Is the first law of thermodynamics tested in Unit 6?

Yes. Energy conservation -- heat lost by one substance equals heat gained by another in an isolated system -- is the direct justification for every calorimetry calculation in this unit, from a simple metal-in-water problem to a full Hess's Law derivation.

What's the difference between q and ΔH?

q is the heat transferred during a process and is path-dependent in general. ΔH is a state function -- it only depends on initial and final states. The two become numerically equal only at constant pressure (qₚ = ΔH), which is exactly why an open, atmospheric-pressure coffee-cup calorimeter lets you read ΔH directly from a measured q.

Sourced from College Board's official AP Chemistry Course and Exam Description, Effective Fall 2024. This page describes the document's real content; it is not a copy of it and is not affiliated with or endorsed by College Board.