Enthalpy change (ΔH) is the heat energy a system absorbs or releases at constant pressure during a chemical reaction or phase change. A negative ΔH means the reaction is exothermic (releases heat); a positive ΔH means it is endothermic (absorbs heat).
Enthalpy change (ΔH) measures how much heat flows in or out of a system at constant pressure. The sign does all the talking. If ΔH is negative, the reaction is exothermic and dumps thermal energy into the surroundings (that's why your coffee cup calorimeter warms up). If ΔH is positive, the reaction is endothermic and pulls thermal energy from the surroundings, so the surroundings cool down. This is exactly how the CED frames it in 6.6.A.1.
The useful trick is that ΔH is a molar quantity, usually reported in kJ/mol of reaction. That means it scales with amount. If burning 1 mole of methane releases 890 kJ, burning 2 moles releases 1780 kJ. You convert between q (the actual heat in joules for your specific sample) and ΔH (the per-mole value) using moles, which is the calculation skill LO 6.6.A asks for. ΔH also describes physical processes, not just reactions. Melting ice has a positive ΔH (heat of fusion goes in), and freezing has the equal-but-opposite negative ΔH.
ΔH is the backbone of Unit 6 (Thermochemistry) and comes back in Unit 9 (Thermodynamics and Electrochemistry). LO 6.6.A has you calculate q from moles and the molar enthalpy of reaction. LO 6.5.A applies the same logic to phase transitions, where temperature stays flat while ΔH does the work. LOs 6.9.A and 6.9.B build Hess's Law on top of it, because enthalpy changes of steps add up to the enthalpy change of the whole process (that's just conservation of energy at constant pressure). Then in Topic 9.3, ΔH° becomes half of ΔG° = ΔH° − TΔS°, so it directly feeds into deciding whether a process is thermodynamically favored. If you can read and use a ΔH value, you've unlocked a huge fraction of the thermo questions on the exam.
Keep studying AP Chemistry Unit 9
Enthalpy of Reaction and Hess's Law (Unit 6)
Hess's Law says ΔH for an overall process equals the sum of the ΔH values of its steps. It works because enthalpy is a state function, so the path doesn't matter, only where you start and end. This is also why enthalpies of formation let you compute ΔH°rxn as products minus reactants.
Heat of Fusion and Phase Changes (Unit 6)
Phase changes have their own ΔH values, like ΔH of fusion for melting. During a phase change the temperature of a pure substance stays constant, so all the energy goes into changing potential energy between particles, not kinetic energy. Melting absorbs exactly the heat that freezing releases.
Gibbs Free Energy (Unit 9)
ΔH° is one of the two inputs in ΔG° = ΔH° − TΔS°. An exothermic ΔH pushes a process toward being thermodynamically favored, but it doesn't decide alone. Entropy and temperature can flip the verdict, which is why exothermic does not automatically mean favored.
Specific Heat Capacity and Calorimetry (Unit 6)
Calorimetry is how you measure ΔH in a lab. The surroundings (usually water) change temperature, you compute q = mcΔT, flip the sign to get the system's heat, then divide by moles to get ΔH. Specific heat capacity is the bridge between a thermometer reading and an enthalpy value.
Expect ΔH in both MCQs and FRQs, usually as a calculation or a sign interpretation. The classic setup is a coffee cup calorimeter, like mixing 50.0 mL of 1.0 M HCl with 50.0 mL of 1.0 M NaOH and watching the temperature rise. You compute q = mcΔT, recognize the temperature increase means the reaction is exothermic (ΔH < 0), and convert to kJ/mol. The 2018 long FRQ asked exactly this kind of experimental determination of ΔHrxn. Other common stems give you standard enthalpies of formation and ask for ΔH°rxn (products minus reactants), or give you ΔH per mole of reaction and ask you to scale it for a different number of moles. Watch two traps. First, the sign flip: heat absorbed by the calorimeter water was released by the reaction. Second, the stoichiometry: if ΔH = −572 kJ for 2 mol of H₂, then 4 mol of H₂ releases 1144 kJ.
Heat (q) is the actual energy transferred in your specific experiment, measured in joules or kilojoules. Enthalpy change (ΔH) is the per-mole, constant-pressure version, in kJ/mol of reaction. At constant pressure they're connected by q = n × ΔH, so q depends on how much stuff reacted while ΔH is a fixed property of the reaction. On calorimetry problems you measure q first, then divide by moles to report ΔH.
ΔH is the heat absorbed or released at constant pressure, with negative ΔH meaning exothermic and positive ΔH meaning endothermic.
In an exothermic reaction, thermal energy flows from the system to the surroundings, which is why the calorimeter temperature goes up.
ΔH is reported per mole of reaction, so you scale it with stoichiometry: q = n × ΔH.
Because enthalpy is a state function, Hess's Law lets you add the ΔH values of individual steps to get the ΔH of the overall process.
Phase changes have enthalpy changes too, and the temperature of a pure substance stays constant while the phase change happens.
ΔH° plugs into ΔG° = ΔH° − TΔS° in Unit 9, so exothermic helps favorability but entropy and temperature can override it.
Enthalpy change (ΔH) is the amount of heat a system absorbs or releases at constant pressure during a reaction or phase change. Negative ΔH means exothermic, positive ΔH means endothermic, and it's usually reported in kJ/mol.
No. Negative ΔH means the reaction is exothermic, but thermodynamic favorability depends on ΔG° = ΔH° − TΔS°. An exothermic reaction with a large entropy decrease can be unfavored at high temperature, and an endothermic one can be favored if entropy increases enough.
q is the heat transferred in your actual experiment (in J or kJ), while ΔH is the heat per mole of reaction at constant pressure (in kJ/mol). They're related by q = n × ΔH, so calorimetry problems have you measure q with q = mcΔT and then divide by moles.
During melting or boiling, the added energy goes into breaking intermolecular attractions (raising potential energy) rather than speeding up molecules (kinetic energy). That's why ice at 0°C absorbs its heat of fusion without warming up until it's fully melted.
A temperature increase means the reaction released heat, so ΔH is negative. Calculate q = mcΔT for the solution, flip the sign for the reaction, then divide by the moles of limiting reactant. That's exactly what the 2018 FRQ asked with a thiosulfate redox reaction in a calorimeter.