Endothermic and exothermic processes — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definitions of endothermic and exothermic processes, sign conventions for enthalpy change, heat flow relative to the system, bond enthalpy calculations, and energy profile diagram classification for AP exam questions.

You should already know: System vs. surroundings thermodynamic definitions. The first law of thermodynamics. Enthalpy as a state function.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Endothermic and exothermic processes?

Endothermic and exothermic processes describe the direction of heat flow between a thermodynamic system and its surroundings during any physical or chemical change, and form the core foundational concept for AP Chemistry Unit 6 Thermodynamics, which accounts for 14-20% of the total AP exam weight. This topic appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a standalone classification question or as a required reasoning step for larger thermodynamics problems. The AP Chemistry CED uniformly uses the system (the reaction or change being studied) as the reference point for all energy measurements, a convention that is the source of most common student errors on the exam. Endothermic processes absorb heat from the surroundings into the system, while exothermic processes release heat from the system to the surroundings. This classification applies equally to chemical reactions (e.g., combustion, photosynthesis) and physical changes (e.g., melting, condensation).

2. Heat Flow and Enthalpy Sign Conventions

The most fundamental rule for working with endo- and exothermic processes relies on consistent sign convention, which AP Chemistry graders strictly enforce based on the system perspective. Recall that for processes at constant pressure (the condition for nearly all reactions run in open laboratory containers), the enthalpy change of a process $\Delta H$ equals the heat exchanged $q_p$, so $\Delta H=q_p$. The universal convention in chemistry is: heat gained by the system is positive, and heat lost by the system is negative. This means for an endothermic process, where heat flows from the surroundings into the system, $q>0$ and $\Delta H>0$ (positive enthalpy change). For an exothermic process, where heat flows from the system to the surroundings, $q<0$ and $\Delta H<0$ (negative enthalpy change). An intuitive check: if the reaction container feels warm to the touch, heat left the system to warm your hand (surroundings), so it is exothermic with negative $\Delta H$. If the container feels cold, heat left your hand (surroundings) to enter the system, so it is endothermic with positive $\Delta H$.

Worked Example

A student activates an instant cold pack by mixing solid ammonium nitrate with room temperature water. The temperature of the resulting solution drops from 21°C to 3°C. Identify the dissolution process as endothermic or exothermic, and state the sign of $\Delta H$.

1. Define the system: the dissolution of ammonium nitrate in water is the system; the beaker, air, and the student's hand are the surroundings.
2. Track heat flow: the temperature of the solution (part of the system) dropped because thermal energy left the surroundings and entered the system to drive the dissolution process.
3. By convention, enthalpy change is measured relative to the system: heat gained by the system gives a positive $\Delta H$.
4. A process that absorbs heat from the surroundings is classified as endothermic. Final answer: Process is endothermic, $\Delta H>0$.

Exam tip: If an FRQ asks you to justify your classification, always explicitly mention heat flow relative to the system, not just the temperature change. AP graders require this reasoning for full credit.

3. Enthalpy Change from Bond Enthalpies

Bond enthalpy (or average bond energy) is defined as the energy required to break one mole of a specific covalent bond in the gaseous state. Breaking any bond is always an endothermic process: energy must be added to the system to pull bonded atoms apart, so bond breaking has $\Delta H>0$. Forming any bond is always an exothermic: when atoms form a stable bond, energy is released from the system to the surroundings, so bond formation has $\Delta H<0$. To calculate the total enthalpy change for a gas-phase reaction, we add the enthalpy changes for all bond breaking and bond formation steps. Simplifying this sum gives the standard formula: $$\Delta H_{rxn}=\sum (\text{bond enthalpies of bonds broken})-\sum (\text{bond enthalpies of bonds formed})$$ After calculating $\Delta H_{rxn}$, a positive value means the overall reaction is endothermic, and a negative value means the overall reaction is exothermic.

Worked Example

Use the given average bond enthalpies to calculate $\Delta H_{rxn}$ for the reaction ${\text{H}}_2(g)+{\text{Br}}_2(g)\to 2\text{HBr}(g)$, then classify the reaction. Bond enthalpies (kJ/mol): H-H = 436, Br-Br = 193, H-Br = 366.

1. Count bonds broken: 1 mol of H-H bonds and 1 mol of Br-Br bonds are broken. Sum of bonds broken = $436+193=629$ kJ.
2. Count bonds formed: 2 mol of H-Br bonds are formed. Sum of bonds formed = $2\times 366=732$ kJ.
3. Calculate $\Delta H_{rxn}$ using the formula: $\Delta H_{rxn}=629-732=-103$ kJ.
4. Classify: $\Delta H<0$, so the reaction is exothermic overall. Final answer: $\Delta H=-103$ kJ, reaction is exothermic.

Exam tip: Always confirm the reaction is fully in the gas phase before using bond enthalpies. Bond enthalpy calculations for liquid or solid phase reactions omit phase change enthalpy terms, leading to incorrect results.

4. Energy Profile Diagrams and Process Classification

Energy profile (or reaction coordinate) diagrams plot the total enthalpy of the system versus reaction progress (the reaction coordinate) for a chemical process. These diagrams are regularly tested on the AP exam to assess understanding of the relationship between enthalpy change and process classification. The key feature for classification is the relative enthalpy of reactants versus products, which follows directly from the definition of enthalpy change as a state function: $\Delta H=H_{final}-H_{initial}=H_{products}-H_{reactants}$. For an exothermic process: products have lower enthalpy than reactants, so $\Delta H<0$, and the diagram slopes downward from reactants to products. For an endothermic process: products have higher enthalpy than reactants, so $\Delta H>0$, and the diagram slopes upward from reactants to products. Note that activation energy (the energy barrier from reactants to the activated complex) does not affect the overall classification of the reaction.

Worked Example

A reaction coordinate diagram shows reactants at an enthalpy of 120 kJ/mol, an activated complex at 300 kJ/mol, and products at 75 kJ/mol. Calculate $\Delta H_{rxn}$, classify the reaction, and confirm your result.

1. Recall the definition of enthalpy change for a reaction: $\Delta H_{rxn}=H_{products}-H_{reactants}$.
2. Substitute the given values: $\Delta H=75\text{ kJ/mol}-120\text{ kJ/mol}=-45\text{ kJ/mol}$.
3. A negative $\Delta H$ means the system lost enthalpy overall, which corresponds to an exothermic reaction.
4. Confirm: products (75 kJ/mol) have lower enthalpy than reactants (120 kJ/mol), which matches the definition of an exothermic process. Final answer: $\Delta H=-45$ kJ/mol, reaction is exothermic.

Exam tip: Do not confuse activation energy (the height of the reaction barrier) with the overall enthalpy change of the reaction. An exothermic reaction can still have a high activation energy (e.g., gasoline combustion requires a spark to start).

5. Common Pitfalls (and how to avoid them)

• Wrong move: Classifying a cold pack reaction as exothermic because the solution temperature drops, concluding heat left the system. Why: Students confuse the temperature change of the system with the direction of heat flow. If the system's temperature drops, it absorbed heat from the surroundings. Correct move: Always state "heat gained by system = endo, positive ΔH; heat lost by system = exo, negative ΔH" before classifying.
• Wrong move: Reversing the bond enthalpy formula, writing $\Delta H=\sum B{E}_{formed}-\sum B{E}_{broken}$. Why: Students mix up that breaking bonds requires energy, so it adds a positive contribution to ΔH. Correct move: Use the mnemonic "Broken First Minus Made" (B-F-M) to remember the formula order.
• Wrong move: Assigning a negative sign to the bond enthalpy of bonds broken. Why: Students incorrectly apply a negative sign to the endothermic bond breaking step, instead of using the formula that already accounts for sign. Correct move: Bond enthalpy values are always positive (they are the energy required to break bonds), so plug all BE values into the formula as positive.
• Wrong move: Classifying processes relative to the surroundings, getting the opposite sign: for example, saying an exothermic reaction has positive ΔH because the surroundings gain heat. Why: Students forget the AP Chem convention always uses the system as the reference point. Correct move: For any sign question, first write "ΔH is relative to the system" in your working before assigning a sign.
• Wrong move: Calculating ΔH for an energy diagram as $H_{reactants}-H_{products}$ to force a negative sign for exothermic reactions. Why: Students rearrange the formula to get the sign they expect instead of following the state function rule. Correct move: Always use $\Delta H=H_{final}-H_{initial}$, where final = products and initial = reactants, no exceptions.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

When liquid water freezes to form solid ice at constant pressure, which of the following correctly describes the process and the sign of $\Delta H$? A) Endothermic, $\Delta H>0$ B) Endothermic, $\Delta H<0$ C) Exothermic, $\Delta H>0$ D) Exothermic, $\Delta H<0$

Worked Solution: Freezing is the conversion of liquid water to solid water. To form solid ice from liquid, water molecules must lose energy to the surroundings. The system here is the water being converted. Heat is lost by the system, so by AP convention, $\Delta H<0$. A process that releases heat from the system is exothermic. This matches option D. Correct answer: D.

Question 2 (Free Response)

The combustion of gaseous methane is given by the equation: ${\text{CH}}_4(g)+2{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(g)$. (a) Using the bond enthalpy table below, calculate $\Delta H_{rxn}$ for this reaction. Show all work. Bond enthalpies (kJ/mol): C-H = 413, O=O = 498, C=O = 799, O-H = 464. (b) Is this reaction endothermic or exothermic? Justify your answer. (c) For an energy profile diagram of this reaction, state whether products have higher or lower enthalpy than reactants, and explain your reasoning.

Worked Solution: (a) Count bonds from Lewis structures: Reactants: 4 mol C-H bonds, 2 mol O=O bonds. Sum of bonds broken: $(4\times 413)+(2\times 498)=1652+996=2648$ kJ. Products: 2 mol C=O bonds, 4 mol O-H bonds. Sum of bonds formed: $(2\times 799)+(4\times 464)=1598+1856=3454$ kJ. $\Delta H_{rxn}=\sum B{E}_{broken}-\sum B{E}_{formed}=2648-3454=-806$ kJ.

(b) The reaction is exothermic. $\Delta H_{rxn}=-806$ kJ, which is less than 0, meaning the system releases heat to the surroundings overall.

(c) Products have lower enthalpy than reactants. For an exothermic reaction, $\Delta H=H_{products}-H_{reactants}<0$, which rearranges to $H_{products}<H_{reactants}$.

Question 3 (Application / Real-World Style)

Portable camping stoves use the combustion of butane fuel to release heat for cooking. The reaction for complete combustion of butane is $2{\text{C}}_4{\text{H}}_{10}(g)+13{\text{O}}_2(g)\to 8{\text{CO}}_2(g)+10{\text{H}}_2\text{O}(l)$ $\Delta H=-5754$ kJ/mol of reaction. A small canister contains 100.0 g of butane ($C_4{H}_{10}$, molar mass = 58.12 g/mol). How much total heat is released when all the butane is combusted, and is the process endothermic or exothermic?

Worked Solution:

1. Calculate moles of butane: $n(C_4{H}_{10})=\frac{100.0\text{ g}}{58.12\text{ g/mol}}=1.721$ mol.
2. The given $\Delta H$ is for 2 moles of butane reacted, so heat released per mole of butane is $\frac{5754\text{ kJ}}{2\text{ mol}}=2877$ kJ/mol.
3. Total heat released = $1.721\text{ mol}\times 2877\text{ kJ/mol}=4951$ kJ.
4. $\Delta H$ is negative, so the process is exothermic. In context, this means ~4950 kJ of heat is released to the surroundings to cook food, which matches the intended function of the camping stove.

7. Quick Reference Cheatsheet

8. What's Next

Mastering endothermic and exothermic process classification and sign conventions is the non-negotiable foundation for all remaining thermodynamics topics in AP Chemistry Unit 6. Next you will apply these rules to calorimetry calculations, where you will measure heat flow to determine enthalpy changes of reaction, then extend the concept to Hess's law for calculating ΔH for multi-step reactions. Without correctly identifying the sign of ΔH and classifying heat flow, every subsequent calculation in thermodynamics will give the wrong answer, even if your arithmetic is correct. This topic also feeds into the larger concepts of enthalpy of formation, entropy, and Gibbs free energy, where the sign of ΔH is critical for predicting reaction spontaneity.

Calorimetry and heat capacity Hess's law Enthalpy of formation Gibbs free energy and spontaneity