ΔH° (standard enthalpy change) is the heat a reaction absorbs or releases at constant pressure under standard conditions (1 atm, 298.15 K). A negative ΔH° means the reaction is exothermic (releases heat); a positive ΔH° means endothermic (absorbs heat).
ΔH° is the standard enthalpy change of a reaction. It measures how much heat flows in or out of a system at constant pressure when the reaction runs under standard conditions, defined as 1 atm of pressure and 298.15 K. The little degree symbol (°) is doing real work. It tells you the value was measured under those agreed-upon standard conditions, so chemists everywhere can compare numbers fairly.
The sign is the whole story. ΔH° < 0 means the reaction releases heat to the surroundings (exothermic), like the Haber process at -92 kJ/mol. ΔH° > 0 means the reaction absorbs heat from the surroundings (endothermic), like PCl₅ decomposition at +92.5 kJ/mol. A handy mental shortcut for equilibrium problems is to treat heat like a chemical species. If ΔH° is positive, write "heat" on the reactant side. If it's negative, write "heat" on the product side. Suddenly temperature changes behave exactly like adding or removing a reactant.
ΔH° shows up in two very different jobs across the course. In Unit 7, it's the input you need for learning objective 7.9.A, which asks you to predict how a system at equilibrium responds to stress using Le Châtelier's principle. Specifically, EK 7.9.A.1 lists temperature change as a stress, and you literally cannot predict the direction of the shift without knowing the sign of ΔH°. Raise the temperature on an endothermic reaction and equilibrium shifts toward products; raise it on an exothermic reaction and it shifts back toward reactants.
In Unit 3 (Topic 3.7, learning objective 3.7.A), enthalpy thinking sneaks into solutions and mixtures. Dissolving and dilution involve heat changes too, like the hydration of ions, so whether a dissolution process releases or absorbs heat connects the energetics of solutions to the same ΔH° logic. Across the whole exam, ΔH° is one of those quantities that links the macroscopic (a beaker getting warm) to the particulate (bonds breaking and forming).
Keep studying AP Chemistry Unit 7
Le Châtelier's Principle (Unit 7)
Temperature is the only stress where ΔH° decides the answer. Treat heat as a reactant (positive ΔH°) or a product (negative ΔH°), and a temperature increase becomes just another case of "adding a species," pushing equilibrium away from the side with the heat.
Exothermic and Endothermic Reactions (Unit 6)
These labels are literally just the sign of ΔH° in word form. Negative ΔH° is exothermic, positive is endothermic. ΔH° is the number; exo and endo are the vocabulary the exam uses to describe it.
Equilibrium Constant K and Temperature (Unit 7)
K only changes when temperature changes, and ΔH° tells you which direction. For an exothermic reaction, heating the system makes K smaller. Catalysts, concentration changes, and pressure changes never touch K, which is a classic MCQ trap.
Dilution and Hydration (Unit 3)
Dissolving a solute isn't energy-neutral. Hydration of ions releases energy while breaking up the solute costs energy, so dissolution has its own enthalpy sign. That's why some solutions get hot and others get cold when you mix them.
Multiple-choice questions almost never ask "what is ΔH°?" directly. Instead, they hand you a reaction with its ΔH° value and expect you to use the sign. Typical stems include the Haber process (N₂ + 3H₂ ⇌ 2NH₃, ΔH° = -92 kJ/mol) with a temperature increase, where you must recognize the exothermic reaction shifts toward reactants and K decreases. Another favorite is PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with ΔH° = +92.5 kJ/mol, where you combine stresses (heating plus expanding the volume) to maximize product. Watch for the catalyst trap. Adding a catalyst to a system at equilibrium changes nothing about K or the concentrations, no matter what ΔH° is.
On the free-response section, ΔH° appears in long FRQs (including 2017, 2019, and 2025) where you calculate or interpret reaction energies for real molecules like urea or P₄, then justify a claim using the sign. The verbs that score points are "predict the direction of the shift" and "justify using ΔH°," so always connect the sign to heat-as-a-species in your written explanation.
ΔH is the enthalpy change under whatever conditions the reaction actually runs. ΔH° is the same quantity pinned to standard conditions, 1 atm and 298.15 K. The degree symbol means "measured under standard conditions," which is what lets you compare tabulated values across reactions. For sign-based reasoning like Le Châtelier predictions, both work the same way.
ΔH° is the heat absorbed or released by a reaction at constant pressure under standard conditions (1 atm, 298.15 K).
A negative ΔH° means the reaction is exothermic and releases heat; a positive ΔH° means it is endothermic and absorbs heat.
For Le Châtelier problems, write heat as a reactant when ΔH° is positive and as a product when ΔH° is negative, then treat temperature changes like adding or removing that "heat" species.
Temperature is the only stress that changes the equilibrium constant K, and the sign of ΔH° tells you whether K goes up or down when you heat the system.
Adding a catalyst changes neither K nor the equilibrium concentrations, regardless of the value of ΔH°.
Dissolution and hydration in Unit 3 have enthalpy changes too, which is why mixing some solutions makes the beaker warm and others makes it cold.
ΔH° is the standard enthalpy change, the heat absorbed or released by a reaction at constant pressure under standard conditions (1 atm and 298.15 K). Its sign tells you whether the reaction is exothermic (negative) or endothermic (positive).
No. A positive ΔH° only means the reaction absorbs heat; it says nothing definitive about whether the reaction occurs. PCl₅ decomposition has ΔH° = +92.5 kJ/mol and still reaches equilibrium with real product. Spontaneity depends on more than enthalpy alone.
The degree symbol means standard conditions. ΔH° is the enthalpy change measured at 1 atm and 298.15 K, which makes values comparable across reactions, while plain ΔH applies to whatever conditions a reaction actually runs under.
No. A catalyst speeds up how fast equilibrium is reached but doesn't change ΔH°, the equilibrium constant K, or the final concentrations. This is one of the most common trap answers on equilibrium MCQs.
Use the sign to place heat in the equation. For N₂ + 3H₂ ⇌ 2NH₃ with ΔH° = -92 kJ/mol, heat is a product, so raising the temperature shifts equilibrium toward reactants and decreases K. For endothermic reactions, heating does the opposite and favors products.