An exothermic reaction is a chemical or physical process that releases energy to the surroundings as heat, meaning ΔH is negative and the products are lower in energy than the reactants on an energy diagram (AP Chem Topics 6.2 and 5.6).
An exothermic reaction releases energy, usually as heat, into the surroundings. In AP Chem terms, that means the enthalpy change is negative (ΔH < 0). On an energy diagram, the products end up lower than the reactants, and the gap between them is the energy released. The sign convention is the whole game here. Energy leaves the system, so the system's enthalpy goes down.
Why does energy get released? Bond energies. Forming bonds releases energy and breaking bonds costs energy. In an exothermic reaction, the new bonds formed in the products are stronger (lower energy) than the bonds broken in the reactants, so there's leftover energy that escapes as heat. One critical caution the CED hammers on (EK 9.3.A.2): exothermic does NOT automatically mean a reaction happens. Thermodynamic favorability depends on ΔG° = ΔH° − TΔS°, not on ΔH alone. A negative ΔH helps, but entropy and temperature get a vote too.
Exothermic is one of the few concepts that shows up in four different units. In Unit 6, LO 6.2.A asks you to draw or interpret energy diagrams that show whether a process is exothermic or endothermic. In Unit 5, LO 5.6.A has you read reaction energy profiles, where an exothermic reaction still needs activation energy to get over the transition state hump even though the products land lower. In Unit 7, LO 7.9.A uses Le Châtelier's principle to predict what happens when you heat an exothermic equilibrium (it shifts toward reactants, and per EK 7.10.A.2, K itself gets smaller). In Unit 9, LO 9.3.A makes you weigh ΔH° against TΔS° to decide if a process is thermodynamically favored. If you can classify a reaction as exothermic and explain what that does in each of these contexts, you've connected half the course.
Keep studying AP Chemistry Unit 7
Endothermic Reaction (Units 5-6)
The mirror image. Endothermic reactions absorb heat (ΔH > 0), so products sit higher than reactants on the energy diagram. The fastest exam check is the sign of ΔH or which side of the diagram is lower.
Activation Energy and Reaction Energy Profiles (Unit 5)
Even exothermic reactions have to climb to a transition state first. The activation energy is the gap from reactants up to the peak, and the exothermic ΔH is the drop from reactants down to products. A catalyst lowers the peak but never changes ΔH.
Le Châtelier's Principle (Unit 7)
Treat heat like a product in an exothermic equilibrium. Raise the temperature and the system shifts toward reactants to use up the added heat. This is the one stress that actually changes K, not just Q.
Gibbs Free Energy and ΔG° (Unit 9)
Exothermic (ΔH° < 0) is half the story. ΔG° = ΔH° − TΔS° decides favorability, so an exothermic reaction with a big negative ΔS° can still be unfavorable at high temperature. Don't equate exothermic with spontaneous.
Multiple-choice questions love energy diagrams. You'll be asked to pick the diagram showing an exothermic reaction (products lower than reactants), identify the activation energy on it, or choose the correct diagram for an exothermic reaction with a catalyst (same start and end heights, lower peak). The other big MCQ format is Le Châtelier with an observable change, like the classic Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ exothermic equilibrium, where heating shifts the system left and the red color fades. On FRQs, exothermic shows up through bond-energy reasoning. The 2025 long FRQ on white phosphorus combustion is the model, where you justify the sign of ΔH by comparing energy released forming product bonds to energy absorbed breaking reactant bonds. Always argue with energy, signs, and the diagram, not just the word 'exothermic.'
Exothermic means ΔH < 0. Thermodynamically favored means ΔG° < 0. They often go together but they are not the same thing. ΔG° = ΔH° − TΔS°, so an exothermic process can be unfavorable if entropy decreases a lot at high temperature, and an endothermic process can be favorable if entropy increases enough (ice melting above 0°C is the classic example). The CED explicitly warns against equating 'exothermic' with 'spontaneous,' and graders look for ΔG° reasoning, not just a ΔH sign.
An exothermic reaction releases heat to the surroundings, so ΔH is negative and products are lower in energy than reactants on an energy diagram.
Exothermic reactions release energy because the bonds formed in the products are stronger than the bonds broken in the reactants.
Exothermic reactions still require activation energy to reach the transition state; a catalyst lowers that barrier but does not change ΔH.
For an exothermic equilibrium, increasing temperature shifts the system toward reactants and decreases the value of K, since heat acts like a product.
Exothermic does not mean thermodynamically favored; you need ΔG° = ΔH° − TΔS° to be negative, so entropy and temperature matter too.
It's a process that releases heat to the surroundings, giving it a negative ΔH. On an energy diagram, the products sit lower than the reactants, and the difference is the energy released.
No. Favorability is decided by ΔG° = ΔH° − TΔS°, not ΔH alone. An exothermic reaction with a large entropy decrease can be unfavorable at high temperature, which is exactly the misconception EK 9.3.A.2 warns about.
Exothermic reactions release heat (ΔH < 0, products lower on the energy diagram), while endothermic reactions absorb heat (ΔH > 0, products higher). Check the sign of ΔH or compare reactant and product heights on the diagram.
The equilibrium shifts toward the reactants, and K decreases. Heat behaves like a product, so adding it pushes the reaction backward. In the Fe³⁺/SCN⁻ system, heating makes the red FeSCN²⁺ color fade.
Yes. Every elementary reaction must climb to a transition state first, even if the products end up lower in energy. That's why exothermic reactions like combustion still need a spark to start.