ΔG° = ΔH° − TΔS° is the Gibbs free energy equation in AP Chem Topic 9.3 that combines a reaction's enthalpy change (ΔH°), entropy change (ΔS°), and absolute temperature (T, in Kelvin) into one number: if ΔG° < 0, the process is thermodynamically favored.
ΔG° = ΔH° − TΔS° is the equation that decides whether a chemical or physical process is thermodynamically favored. It takes the two driving forces of chemistry, the push toward lower energy (ΔH°, enthalpy) and the push toward more dispersal of matter and energy (ΔS°, entropy), and weighs them against each other at a given temperature. The little degree symbols mean standard state: pure substances, 1.0 M solutions, gases at 1.0 atm.
The verdict comes from the sign of ΔG°. If ΔG° < 0, the process is thermodynamically favored (the CED deliberately prefers this phrase over the older word "spontaneous," because spontaneous sounds like "happens instantly," and ΔG° says nothing about speed). Notice that temperature is a multiplier on the entropy term only. That's the whole game. At low T, the ΔH° term dominates; at high T, the TΔS° term dominates. That single idea explains why ice melts above 0°C but not below it, and why endothermic decompositions become favored when you crank up the heat.
This equation lives in Topic 9.3 (Gibbs Free Energy and Thermodynamic Favorability) in Unit 9 and directly supports learning objective 9.3.A: explain whether a process is thermodynamically favored based on an evaluation of ΔG°. It's the payoff of the whole thermodynamics arc. Unit 6 taught you ΔH°, Topics 9.1-9.2 taught you ΔS°, and this equation is where they finally combine into a single yes-or-no answer about favorability. It's also one of the most calculation-friendly equations on the exam, which makes it a favorite for both multiple choice and FRQ math. If you can analyze the four sign combinations of ΔH° and ΔS°, you can answer an entire family of AP questions without touching a calculator.
Keep studying AP® Chemistry Unit 9
Enthalpy Change (Units 6 & 9)
ΔH° is half of the Gibbs equation. Everything you learned in Unit 6 about exothermic vs endothermic reactions feeds directly into ΔG°. An exothermic reaction (ΔH° < 0) gets a head start toward being favored, but it's not the whole story, because a negative ΔS° can flip the verdict at high temperature.
ΔG°reaction = ΣΔG°f products − ΣΔG°f reactants (Unit 9)
This is the other route to the same destination. If a problem hands you ΔH° and ΔS° data, use ΔG° = ΔH° − TΔS°. If it hands you a table of standard free energies of formation, use the products-minus-reactants sum instead. Same ΔG°, different starting data.
Entropy and Particulate Models (Unit 9)
AP questions love showing a particle diagram, like a solid lattice breaking into gas molecules, and asking you to infer the sign of ΔS° from it. Solid to gas means ΔS° > 0, so even an endothermic decomposition becomes favored once T is large enough to make TΔS° outweigh ΔH°.
Kinetics vs Thermodynamics (Unit 5)
ΔG° tells you whether a reaction is favored, not how fast it goes. A reaction with very negative ΔG° can still sit there for years if its activation energy is huge (think diamond turning to graphite). This favored-but-slow distinction is exactly why the CED retired the word "spontaneous."
Multiple choice questions hit this equation from several angles. One classic stem gives you ΔH° > 0 and ΔS° > 0 and asks about temperature dependence (answer: favored only at high T, where TΔS° beats ΔH°). Another shows a graph of ΔG° vs T and asks for the signs of ΔH° and ΔS°. The trick is recognizing the equation as a line, y = mx + b, where the slope is −ΔS° and the y-intercept is ΔH°. A negative slope with a positive intercept means ΔS° > 0 and ΔH° > 0. Particulate-model questions are also common: you're shown ions forming an ordered lattice or a solid decomposing into gases, and you have to deduce the signs of ΔS° and ΔH° and argue favorability from them. On FRQs, expect to actually calculate ΔG° from given ΔH° and ΔS° values, then justify whether the process is thermodynamically favored. Watch the two classic point-killers: forgetting to use Kelvin, and forgetting that ΔH° is usually given in kJ while ΔS° is in J/(mol·K). Convert before you subtract.
Both equations calculate the same quantity, ΔG°, but from different data. Use ΔG° = ΔH° − TΔS° when you have enthalpy and entropy values (or need to analyze how temperature changes favorability). Use the formation-sum equation when you're given a table of ΔG°f values, which only works at the temperature the table was measured at (usually 298 K). If a question asks how favorability changes with temperature, only ΔH° − TΔS° can answer it, because T appears explicitly in the equation.
ΔG° = ΔH° − TΔS° combines enthalpy, entropy, and temperature into one number, and ΔG° < 0 means the process is thermodynamically favored.
Temperature multiplies only the entropy term, so reactions with ΔH° > 0 and ΔS° > 0 become favored at high temperatures, while reactions with ΔH° < 0 and ΔS° < 0 are favored only at low temperatures.
If both signs push the same way, temperature doesn't matter: ΔH° < 0 with ΔS° > 0 is favored at all temperatures, and ΔH° > 0 with ΔS° < 0 is never favored.
T must be in Kelvin, and you almost always have to convert ΔS° from J/(mol·K) to kJ/(mol·K) before plugging in, since ΔH° is given in kJ.
On a graph of ΔG° vs T, the slope is −ΔS° and the y-intercept is ΔH°, so a negative slope means positive entropy change.
Thermodynamically favored does not mean fast; ΔG° says nothing about reaction rate, which is why the CED prefers 'favored' over 'spontaneous.'
It calculates the standard Gibbs free energy change from enthalpy (ΔH°), entropy (ΔS°), and absolute temperature. If the result is negative, the process is thermodynamically favored. It's the central equation of Topic 9.3 in Unit 9.
No. ΔG° < 0 means the reaction is thermodynamically favored, but it says nothing about speed. A favored reaction can be extremely slow if its activation energy is high, which is kinetics (Unit 5), not thermodynamics. This is exactly why the AP exam uses 'thermodynamically favored' instead of 'spontaneous.'
Use ΔH° − TΔS° when you're given enthalpy and entropy data, or when the question involves changing temperature. Use ΣΔG°f(products) − ΣΔG°f(reactants) when you're given a table of formation values, which only applies at the table's temperature (usually 298 K).
Yes, always. T is absolute temperature, so 25°C becomes 298 K. Also watch units: ΔH° is usually in kJ/mol while ΔS° is in J/(mol·K), so divide ΔS° by 1000 before multiplying. Unit mismatches are one of the most common errors on this calculation.
Set ΔG° = 0 and solve: T = ΔH°/ΔS°. Below or above that crossover temperature, whichever term dominates wins. This only matters when ΔH° and ΔS° have the same sign; if they have opposite signs, the verdict is the same at every temperature.
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