Enthalpy change (ΔH) is the heat energy a system absorbs or releases during a process at constant pressure. A negative ΔH means the reaction is exothermic (releases heat); a positive ΔH means endothermic (absorbs heat). On the AP exam, ΔH plugs into ΔG = ΔH − TΔS to judge thermodynamic favorability.
Enthalpy change, written ΔH, measures the heat that flows into or out of a chemical system at constant pressure. That "constant pressure" detail matters because almost every reaction you run in a beaker happens under constant atmospheric pressure, so the heat you measure with a thermometer is the enthalpy change. If ΔH is negative, the reaction releases heat to the surroundings (exothermic, like combustion). If ΔH is positive, the reaction pulls heat in (endothermic, like melting ice).
Here's the intuition that makes it click. ΔH tracks what happens to bond energy. Breaking bonds costs energy and forming bonds releases it, so a reaction that forms stronger bonds than it breaks ends up exothermic. In Unit 9, ΔH becomes one half of the favorability story. The Gibbs free energy equation, ΔG = ΔH − TΔS, weighs the enthalpy term against the entropy term to decide whether a process is thermodynamically favored. A negative ΔH pushes toward favorability, but it never gets the final say alone.
Enthalpy lives in Unit 9 (Thermodynamics and Electrochemistry) and supports Topic 9.5, Free Energy and Equilibrium. Learning objective 9.5.A asks you to explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T. You can't do that without ΔH, because ΔG° = ΔH° − TΔS° is how you get ΔG° in the first place, and then ΔG° = −RT ln K connects favorability to the equilibrium constant. The sign of ΔH also determines whether temperature helps or hurts favorability. An exothermic reaction with negative ΔS° is favored only at low T, while an endothermic reaction with positive ΔS° needs high T. That sign analysis shows up constantly on the exam, and ΔH is always one of the two levers.
Keep studying AP® Chemistry Unit 9
Gibbs Free Energy and Spontaneity (Unit 9)
ΔG = ΔH − TΔS is the equation that turns enthalpy into a verdict. A negative ΔH votes for favorability, but entropy and temperature get votes too. This is the single tightest connection on the exam, since LO 9.5.A questions almost always start from ΔH° and ΔS° values.
Hess's Law (Unit 6)
Because enthalpy is a state function, ΔH depends only on the start and end points, not the path. Hess's Law cashes that in by letting you add up the ΔH values of steps to get the ΔH of the overall reaction. It's the main calculation tool for finding an enthalpy you can't measure directly.
Standard Enthalpy Change (Unit 6)
ΔH° is just ΔH measured under standard conditions (1 atm, usually 25°C). The degree symbol matters in Unit 9, because the equations linking free energy to K (like ΔG° = −RT ln K) only work with standard-state values.
Heat Capacity and Calorimetry (Unit 6)
Calorimetry is how you measure ΔH in a lab. The heat absorbed by the water (q = mcΔT) equals the heat released by the reaction at constant pressure, so a temperature change in a coffee-cup calorimeter is ΔH wearing a lab coat.
Enthalpy shows up two ways. In multiple choice, expect to interpret the sign of ΔH (exothermic vs. endothermic), use Hess's Law or bond enthalpies to compute it, or combine ΔH° and ΔS° signs to decide at what temperatures a process is favored. In free-response thermodynamics questions, you're typically given or asked to calculate ΔH°, then expected to plug it into ΔG° = ΔH° − TΔS° and justify whether the process is thermodynamically favored, sometimes connecting that ΔG° to K with ΔG° = −RT ln K. The most common point lost is sign logic, so practice stating clearly why a negative ΔH° plus a positive ΔS° guarantees ΔG° < 0 at all temperatures.
ΔH tells you about heat flow; ΔG tells you about favorability. They are not the same thing, and assuming "exothermic = spontaneous" is one of the most graded-against mistakes in AP Chem. Ice melting above 0°C is endothermic (ΔH > 0) yet thermodynamically favored (ΔG < 0) because the TΔS term wins. ΔH is one input to ΔG = ΔH − TΔS, not the answer itself.
Enthalpy change (ΔH) is the heat absorbed or released by a process at constant pressure, with negative ΔH meaning exothermic and positive ΔH meaning endothermic.
ΔH is a state function, which is why Hess's Law lets you add the enthalpies of individual steps to find the enthalpy of an overall reaction.
A negative ΔH does not guarantee a reaction is thermodynamically favored; you have to check ΔG = ΔH − TΔS, because entropy and temperature can flip the outcome.
When ΔH° and ΔS° have the same sign, temperature decides favorability. Exothermic with negative ΔS° favors low T, and endothermic with positive ΔS° favors high T.
On Topic 9.5 problems, ΔH° feeds into ΔG° = ΔH° − TΔS°, and then ΔG° = −RT ln K links favorability to whether K is greater or less than 1.
Enthalpy change (ΔH) is the heat a system absorbs or releases during a process at constant pressure. Negative ΔH means exothermic (heat released, like combustion) and positive ΔH means endothermic (heat absorbed, like melting).
No. Favorability depends on ΔG = ΔH − TΔS, not ΔH alone. Ice melts above 0°C even though ΔH is positive, because the positive TΔS term makes ΔG negative. Exothermic helps, but it's never the whole answer.
ΔH measures heat flow (energy), while ΔS measures the change in dispersal of matter and energy (disorder, roughly). They're separate quantities that combine in ΔG = ΔH − TΔS to determine whether a process is thermodynamically favored.
They're equal only at constant pressure. q is heat transferred under any conditions, while ΔH is specifically the heat transferred at constant pressure. Since coffee-cup calorimetry runs at constant atmospheric pressure, the q you measure there equals ΔH.
The degree symbol means standard conditions, which is 1 atm pressure with solutions at 1 M, usually at 25°C. You need standard-state values like ΔH° and ΔS° to compute ΔG° and connect it to K through ΔG° = −RT ln K in Topic 9.5.
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