What is AP Chem unit 6?
Thermochemistry is the study of heat flow in chemical and physical processes. Every reaction either releases energy to the surroundings or absorbs energy from them, and the first law of thermodynamics guarantees that energy is conserved throughout. Unit 6 gives you the vocabulary, diagrams, and calculation methods to quantify those energy changes.
Unit 6 asks you to classify processes as endothermic or exothermic, calculate heat using q = mcΔT and q = nΔH, interpret energy diagrams, and find reaction enthalpy three ways: from bond energies, from standard enthalpies of formation, and by applying Hess's law.
Energy and temperature
A temperature increase in the surroundings signals an exothermic process (ΔH < 0); a temperature decrease signals an endothermic process (ΔH > 0). Topics 6.1-6.3 build this conceptual foundation using system-versus-surroundings language and molecular collision reasoning.
Quantitative heat calculations
Topics 6.4 and 6.5 introduce the two core equations: q = mcΔT for temperature changes and q = nΔH for phase changes. Calorimetry experiments use these equations together, and the first law requires that q_system + q_surroundings = 0.
Three methods for reaction enthalpy
Topics 6.6-6.9 give you three calculation routes: using q = nΔH_rxn directly from calorimetry data, estimating ΔH from average bond energies (bonds broken minus bonds formed), applying the formation equation ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants), and combining thermochemical equations with Hess's law.
Energy is conserved, and its direction of flow determines reaction characterThe first law of thermodynamics runs through every topic in Unit 6. Whether you are dissolving a salt, melting ice, or combusting methane, the total energy of the universe does not change. Thermochemistry gives you the tools to track where that energy goes and how much of it moves, which is the foundation for predicting reaction favorability in Unit 9.
Unit 6 review notes
6.1
Endothermic and Exothermic Processes
Every chemical or physical change either absorbs energy from the surroundings or releases energy to them. The sign of ΔH tells you which direction energy flows: positive ΔH means the system gains energy (endothermic), negative ΔH means the system loses energy (exothermic). Always define the system first before assigning a sign.
- System vs. surroundings: The system is the specific matter being studied; the surroundings are everything else. Heat flows between them, and their signs are always opposite.
- Exothermic process: Energy leaves the system, so ΔH < 0. The surroundings warm up. Examples include combustion, condensation, and freezing.
- Endothermic process: Energy enters the system, so ΔH > 0. The surroundings cool down. Examples include melting, vaporization, and many dissolving processes.
- Temperature as indicator: A measurable temperature change in the surroundings confirms that energy transfer occurred. No temperature change during a phase change does not mean no energy transfer.
- Enthalpy of solution: Dissolving can be exothermic or endothermic depending on whether the energy released forming solute-solvent interactions is greater or less than the energy required to separate solute and solvent particles.
If a reaction causes the solution in a coffee-cup calorimeter to cool down, is the reaction endothermic or exothermic? Explain using system-surroundings language.
| Feature | Exothermic | Endothermic |
|---|
| Sign of ΔH | Negative (ΔH < 0) | Positive (ΔH > 0) |
| Surroundings temperature | Increases | Decreases |
| Energy flow | System to surroundings | Surroundings to system |
| Example process | Combustion, condensation | Melting, many dissolving reactions |
6.2
Energy Diagrams
An energy diagram (reaction coordinate diagram) plots potential energy on the y-axis against reaction progress on the x-axis. The relative heights of reactants and products show whether the process is endothermic or exothermic, and the peak represents the transition state. For Unit 6, focus on reading ΔH from the diagram rather than activation energy, which is covered more in Unit 5.
- Exothermic diagram: Products sit lower than reactants on the y-axis. The energy difference is negative ΔH, and energy is released to the surroundings.
- Endothermic diagram: Products sit higher than reactants. The energy difference is positive ΔH, and energy is absorbed from the surroundings.
- Transition state: The highest-energy point on the diagram. It represents the activated complex and is not a stable intermediate.
- Activation energy (Ea): The energy difference between the reactants and the transition state. A catalyst lowers Ea without changing the overall ΔH.
- Phase change diagrams: Melting and vaporization are shown as uphill steps; freezing and condensation are downhill. The magnitude of the step equals the molar enthalpy of the phase change.
Sketch an energy diagram for an endothermic reaction and label reactants, products, transition state, Ea, and ΔH.
| Diagram feature | Exothermic reaction | Endothermic reaction |
|---|
| Product energy vs. reactant energy | Products lower | Products higher |
| ΔH sign | Negative | Positive |
| Energy released or absorbed | Released | Absorbed |
6.3
Heat Transfer, Thermal Equilibrium, and Calorimetry
Heat transfer occurs at the particle level: warmer particles have greater average kinetic energy and transfer energy to cooler particles through collisions. This continues until thermal equilibrium is reached. Calorimetry quantifies that transfer using q = mcΔT, where q is heat in joules, m is mass in grams, c is specific heat capacity, and ΔT is the temperature change.
- Thermal equilibrium: Reached when two bodies in contact have the same average kinetic energy and temperature, so there is no net heat flow between them.
- q = mcΔT: The core calorimetry equation. Use it for any process where temperature changes but no phase change occurs. ΔT = T_final - T_initial, so cooling gives negative q.
- Specific heat capacity (c): The energy needed to raise 1 g of a substance by 1°C. Water's value is 4.184 J/g·°C, which is high compared to most metals.
- First law of thermodynamics: Energy is conserved. In a calorimetry experiment, q_system + q_surroundings = 0, so heat lost by one substance equals heat gained by the other.
- Coffee-cup vs. bomb calorimeter: A coffee-cup calorimeter operates at constant pressure and measures ΔH directly. A bomb calorimeter operates at constant volume and measures ΔE; corrections are needed to get ΔH.
A 50.0 g sample of metal at 95.0°C is placed in 100.0 g of water at 22.0°C. The final temperature is 27.5°C. Calculate the specific heat of the metal using q_metal = -q_water.
| Feature | Coffee-cup calorimeter | Bomb calorimeter |
|---|
| Pressure condition | Constant pressure | Constant volume |
| Quantity measured directly | ΔH | ΔE (internal energy) |
| Typical use | Dissolution, neutralization | Combustion reactions |
| Heat equation | q = mcΔT for solution | q = C_cal × ΔT |
6.5
Phase Change Energy and Enthalpy of Reaction
During a phase change, temperature stays constant while energy is absorbed or released. The calculation switches from q = mcΔT to q = nΔH, where n is moles and ΔH is the molar enthalpy of the phase transition. The same q = nΔH framework applies to chemical reactions, where ΔH_rxn gives the heat per mole of reaction at constant pressure.
- q = nΔH for phase changes: Multiply moles of substance by the molar enthalpy of fusion or vaporization. Melting and boiling are endothermic; freezing and condensation are exothermic with the same magnitude.
- Temperature plateau: On a heating or cooling curve, a flat region indicates a phase change. Temperature does not change because added energy breaks intermolecular forces rather than increasing kinetic energy.
- Enthalpy of vaporization vs. fusion: ΔH_vap is always larger than ΔH_fus for the same substance because more intermolecular forces must be overcome to convert liquid to gas than solid to liquid.
- ΔH_rxn sign convention: Negative ΔH_rxn means the reaction releases heat (exothermic). Positive ΔH_rxn means the reaction absorbs heat (endothermic). Units are kJ/mol of reaction as written.
- Scaling ΔH_rxn: If you double the moles of reactant, you double the heat released or absorbed. ΔH is an extensive property when tied to a specific balanced equation.
How much energy is required to vaporize 3.00 mol of water at 100°C if ΔH_vap = 40.7 kJ/mol? Is this process endothermic or exothermic?
| Process | Direction | ΔH sign | Equation |
|---|
| Melting (fusion) | Solid to liquid | Positive | q = nΔH_fus |
| Freezing | Liquid to solid | Negative | q = -nΔH_fus |
| Vaporization | Liquid to gas | Positive | q = nΔH_vap |
| Condensation | Gas to liquid | Negative | q = -nΔH_vap |
6.7
Bond Enthalpies
You can estimate ΔH_rxn by accounting for every bond broken in the reactants and every bond formed in the products. Breaking bonds always requires energy input (endothermic step); forming bonds always releases energy (exothermic step). The net ΔH equals the total energy of bonds broken minus the total energy of bonds formed.
- ΔH = Σ(bonds broken) - Σ(bonds formed): Add up average bond energies for all bonds broken in reactants, then subtract the sum for all bonds formed in products. A positive result means more energy was needed to break bonds than was released forming them.
- Average bond energy: A tabulated average value in kJ/mol for a specific bond type (e.g., C-H, O=O, N≡N). Values are averages across many compounds, so this method gives estimates, not exact values.
- Bond order and strength: Triple bonds (N≡N, ~945 kJ/mol) are stronger and require more energy to break than double bonds, which require more than single bonds. Shorter bonds are stronger.
- Counting bonds with coefficients: Multiply the number of each bond type by the stoichiometric coefficient of that molecule in the balanced equation before summing bond energies.
- Limitation of bond enthalpies: Bond energy calculations apply to gas-phase species. Results differ from ΔH°f-based calculations because average bond energies do not account for specific molecular environments.
For the reaction H2(g) + Cl2(g) → 2 HCl(g), use bond energies (H-H: 436, Cl-Cl: 243, H-Cl: 432 kJ/mol) to calculate ΔH and determine whether the reaction is endothermic or exothermic.
6.8
Enthalpy of Formation
The standard enthalpy of formation (ΔH°f) is the enthalpy change when exactly 1 mole of a compound forms from its elements in their most stable standard states at 1 bar and 298 K. By definition, ΔH°f = 0 for any element in its standard state. Tabulated ΔH°f values let you calculate ΔH°rxn precisely using the products-minus-reactants formula.
- ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants): Multiply each ΔH°f value by its stoichiometric coefficient, sum the products side, sum the reactants side, and subtract. This gives the standard enthalpy of reaction.
- Standard state: The most stable physical form of an element or compound at 1 bar and 298 K. For carbon, graphite is the standard state (ΔH°f = 0), not diamond.
- ΔH°f = 0 for elements: Any element in its standard state has ΔH°f = 0 by definition. Examples: O2(g), H2(g), C(graphite), Na(s).
- State symbols matter: ΔH°f values differ for different physical states. For example, ΔH°f for H2O(l) and H2O(g) are not the same; always match the state in the balanced equation.
- Stoichiometric coefficients in summation: Each ΔH°f value must be multiplied by the coefficient of that species in the balanced equation before summing. Forgetting this is a common calculation error.
Using ΔH°f values, calculate ΔH°rxn for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l). Identify which species have ΔH°f = 0.
6.9
Hess's Law
Hess's law states that the enthalpy change of an overall reaction equals the sum of the enthalpy changes of any series of steps that add up to that reaction. Because enthalpy is a state function, the path does not matter, only the initial and final states. You manipulate thermochemical equations by reversing them (flip the sign of ΔH) or scaling them (multiply ΔH by the same factor) until they add up to the target reaction.
- State function property: Enthalpy depends only on the initial and final states, not the pathway. This is why you can combine any set of steps that give the correct overall equation.
- Reversing a reaction: If you reverse a thermochemical equation, the magnitude of ΔH stays the same but the sign changes. An exothermic forward reaction becomes an endothermic reverse reaction.
- Scaling a reaction: If you multiply all coefficients by a factor c, multiply ΔH by the same factor c. Halving a reaction halves its ΔH.
- Canceling intermediates: Species that appear on both sides of the combined equations cancel out, just like in algebraic addition. The remaining species should match the target equation exactly.
- Connection to ΔH°f: The ΔH°f formula is a specific application of Hess's law: you are summing formation reactions for products and subtracting formation reactions for reactants to get the overall reaction enthalpy.
Given: (1) C(s) + O2(g) → CO2(g), ΔH = -393.5 kJ; (2) H2(g) + 1/2 O2(g) → H2O(l), ΔH = -285.8 kJ; (3) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), ΔH = -890.3 kJ. Use Hess's law to find ΔH for C(s) + 2 H2(g) → CH4(g).
| Method | What you need | Best used when |
|---|
| Bond enthalpies | Average bond energy table, Lewis structures | Gas-phase reactions, quick estimates |
| ΔH°f formula | Table of standard formation enthalpies | Standard conditions, precise values needed |
| Hess's law | Set of related thermochemical equations | Target reaction not directly measurable |
Practice AP Chem unit 6 questions
Try stimulus-based AP practice questions and written prompts after you review the notes.
A student is analyzing the thermochemistry of nitrogen oxides. The diagram represents an enthalpy cycle connecting nitrogen and oxygen gas to nitrogen dioxide and dinitrogen tetroxide, with arrows indicating specific reaction pathways.
QuestionTo calculate the value of ΔH2 using Hess's Law, which quantities from the diagram must be known?
The values of ΔH3 and ΔH1 only, because ΔH2 equals ΔH3 minus ΔH1.
The values of ΔH3 and ΔH1 only, because ΔH2 equals ΔH3 plus ΔH1.
The value of ΔH1 only, because ΔH2 is the exact reverse of the ΔH1 step.
The value of ΔH3 only, because ΔH2 is exactly equal to the ΔH3 pathway.
For the phase change H2O(l)→H2O(g), standard enthalpies of formation for H2O(l) and H2O(g) are provided.
QuestionWhich calculation justifies the claim that vaporization requires energy?
ΔH∘=−242−(−286)=+44 kJ/mol because vaporization is endothermic.
ΔH∘=−242−(−286)=−44 kJ/mol because vaporization is exothermic.
ΔH∘=−242+(−286)=−528 kJ/mol because vaporization is endothermic.
ΔH∘=−242−(−286)=+44 kJ/mol because vaporization releases energy.
7. A scientist investigates the thermodynamics of the combustion of liquid ethanol, C2H5OH(l). The reaction proceeds according to the balanced equation below.
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)
The scientist generates an energy diagram for the reaction, which is shown in Figure 1.
Figure 1. Potential energy diagram for the combustion of liquid ethanol, showing the reactant energy level at 0 kJ/mol, the transition-state maximum at 150 kJ/mol, and the product energy level at −1370 kJ/mol.
Table 1. Standard Enthalpies of Formation
Substance | Standard Enthalpy of Formation, ΔHf∘ (kJ/mol) |
|---|
C2H5OH(l) | -277.7 |
CO2(g) | -393.5 |
H2O(l) | -285.8 |
i. Calculate the standard enthalpy change, ΔHrxn∘, for the combustion of ethanol using the data in Table 1.
ii. Using your answer to part B(i), calculate the amount of heat released, in kJ, when a 5.00 g sample of C2H5OH(l) (molar mass 46.07 g/mol) is burned completely.
6. A student conducts an experiment to determine the molar enthalpy of combustion of 1-propanol, C3H7OH(l). The student uses the calorimetry setup shown in Figure 1 and records the data shown in Table 1 during the experiment.
Table 1. Experimental data for the combustion of 1-propanol
Measurement | Value |
|---|
Mass of water in beaker | 150.0 g |
Initial temperature of water | 22.0 °C |
Final temperature of water | 44.5 °C |
Initial mass of burner and fuel | 185.45 g |
Final mass of burner and fuel | 184.82 g |
Figure 1. Calorimetry experimental setup for determining the molar enthalpy of combustion of 1-propanol
Table 2. Thermochemical data for selected fuels
Fuel Name | Formula | Molar Mass (g/mol) | Standard Enthalpy of Combustion (ΔHcomb∘) (kJ/mol) |
|---|
Methanol | CH3OH | 32.04 | -726 |
Ethanol | C2H5OH | 46.07 | -1368 |
Octane | C8H18 | 114.23 | -5470 |
2. Answer the following questions about hydrazine, N₂H₄.
Hydrazine, N₂H₄(l), is a liquid fuel used in rocket propulsion. It reacts with oxygen gas, O₂(g), to produce nitrogen gas and water. A student investigates the enthalpy of combustion of hydrazine using the experimental setup shown in Figure 1.
N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(l)
i. Calculate the amount of heat, q, in Joules, absorbed by the water.
ii. Calculate the experimental molar enthalpy of combustion, ΔH_comb, of hydrazine in kJ/mol_rxn.
The student considers the gas-phase reaction represented by the equation below. Lewis electron-dot diagrams for the species involved are shown in Figure 2, and average bond enthalpies are provided in Table 1.
N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g)
Figure 2. Lewis diagrams for reactants and products in hydrazine combustion
Table 1. Average Bond Enthalpies
Bond | Bond Enthalpy (kJ/mol) |
|---|
N–H | 391 |
N–N | 163 |
N≡N | 941 |
O=O | 495 |
O–H | 463 |
i. Using the Lewis diagrams in Figure 2 and the bond enthalpies in Table 1, calculate the approximate enthalpy change, ΔH, for the gas-phase reaction N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g).
The standard enthalpy of formation, ΔH°f, for H₂O(l) is -286 kJ/mol. The standard enthalpy of formation for N₂(g) is 0 kJ/mol.
An energy diagram for the combustion of hydrazine is shown in Figure 3.
Figure 3. Potential energy diagram for the combustion reaction of hydrazine
Figure 1. Calorimetry setup used to measure heat released from combustion of hydrazine