Gibbs Free Energy and Thermodynamic Favorability
Combine enthalpy and entropy, compare temperature sign cases and distinguish standard references from actual composition.
AP Chemistry Topic 9.3 · CED printed p. 168, 169 / PDF p. 175, 176. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Gibbs free energy combines the enthalpy and entropy contributions to a process at constant temperature and pressure. The superscript ° marks a standard-state comparison: pure condensed substances, solutes referenced to unit activity (often approximated by 1.0 M), and gases at a specified standard pressure, usually 1 bar in modern tables. AP problems may specify 1 atm instead; follow their stated data consistently. A standard state does not by itself mean 25 °C; temperature is separately specified. Negative ΔG° means the forward process is thermodynamically favored under that comparison. Positive ΔG° favors the reverse. “Favored” says nothing about whether the pathway is fast. For an actual nonstandard mixture, composition also matters, as equilibrium and cell-potential reasoning will show.
You can obtain ΔG° from standard formation free energies: ΔG°rxn = ΣνΔGf°products − ΣνΔGf°reactants. Use coefficients and phases just as in a Hess’s-law sum. An element in its standard reference form has ΔGf° = 0 by convention, but a different allotrope or state may not. Formation free energy is not the same quantity as absolute entropy. Reversing the equation changes the sign of ΔG°, and multiplying the equation multiplies its ΔG°.
The other route is . Use kelvin and matching energy units. An endothermic process can be favored if its positive entropy term outweighs the enthalpy cost, as some salt dissolutions do. An exothermic process can become unfavored at high temperature if its entropy change is negative. Freezing water illustrates the competition: it releases heat but reduces the system’s entropy, so temperature changes which contribution wins.
The four sign cases are useful: negative ΔH°/positive ΔS° is favored at all positive T in the constant-parameter model; positive ΔH°/negative ΔS° is favored at none. Both positive is favored at sufficiently high T; both negative at sufficiently low T. For competing signs in the expression, find the boundary from ΔG° = 0, T = ΔH°/ΔS°, after matching units. Treat it as an approximation if ΔH° or ΔS° changes substantially with temperature or phases change. Always connect the sign of the final free-energy value to the direction of the written equation.
Let's work through it
- For a process, ΔH° = −24.0 kJ mol−1 and ΔS° = −80.0 J mol−1 K−1 = −0.0800 kJ mol−1 K−1.
- At 250 K, ΔG° = −24.0 − 250(−0.0800) = −4.00 kJ mol−1. At 350 K it is +4.00 kJ mol−1.
- The boundary is (−24.0)/(−0.0800) = 300 K. The negative enthalpy favors the process, while the negative entropy opposes it increasingly as T rises.
- The process is favored below 300 K in this constant-parameter standard-state model. This conclusion predicts direction, not elapsed time.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Can positive ΔH° still give negative ΔG°?
See every recall question with its answer
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Retrieval 1
Can positive ΔH° still give negative ΔG°?
Worked answer and reason
Yes, if positive TΔS° is larger than the enthalpy cost.
The entropy contribution can outweigh endothermicity.
Retrieval 2
Does the superscript ° mean the temperature must be 25 °C?
Worked answer and reason
No. It indicates standard reference states at the stated temperature.
Temperature is an independent condition.
Catch the tempting mistake
- Thermodynamically favored means immediate and fast.
- A high activation barrier can make a favored reaction extremely slow.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 9 problem 3 · ap synthesis · symbolic
- Unit 9 problem 4 · ap application · data
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For the teacher
- Require an explicit J-to-kJ conversion and the written-reaction direction in the conclusion. If a student asserts an actual mixture must react from ΔG° alone, ask for its composition.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 9.3.A
- Explain whether a physical or chemical process is thermodynamically favored based on an evaluation of ΔG°.
- 9.3.A.1
- The Gibbs free energy change for a chemical process in which all the reactants and products are present in a standard state (as pure substances, as solutions of 1.0 M concentration, or as gases at a pressure of 1.0 atm (or 1.0 bar)) is given the symbol ΔG°.
- 9.3.A.2
- The standard Gibbs free energy change for a chemical or physical process is a measure of thermodynamic favorability. Historically, the term “spontaneous” has been used to describe processes for which ΔG° < 0. The phrase “thermodynamically favored” is preferred instead so that common misunderstandings (equating “spontaneous” with “suddenly” or “without cause”) can be avoided. When ΔG° < 0 for the process, it is said to be thermodynamically favored.
- 9.3.A.3
- The standard Gibbs free energy change for a physical or chemical process may also be determined from the standard Gibbs free energy of formation of the reactants and products. EQN: ΔG° reaction = ΣΔGf ° products − ΣΔGf ° reactants
- 9.3.A.4
- In some cases, it is necessary to consider both enthalpy and entropy to determine if a process will be thermodynamically favored. The freezing of water and the dissolution of sodium nitrate are examples of such phenomena.
- 9.3.A.5
- Knowing the values of ΔH° and ΔS° for a process at a given temperature allows ΔG° to be calculated directly. EQN: ΔG° = ΔH° − T ΔS°
- 9.3.A.6
- In general, the temperature conditions for a process to be thermodynamically favored (ΔG° < 0) can be predicted from the signs of ΔH° and ΔS° as shown in the table below: ΔH° < 0, ΔS° > 0, symbols < >, ΔG° < 0 favored at all T; ΔH° > 0, ΔS° < 0, symbols > <, ΔG° < 0 favored at no T; ΔH° > 0, ΔS° > 0, symbols > >, ΔG° < 0 favored at high T; ΔH° < 0, ΔS° < 0, symbols < <, ΔG° < 0 favored at low T. In cases where ΔH° < 0 and ΔS° > 0, no calculation of ΔG° is necessary to determine that the process is thermodynamically favored (ΔG° < 0). In cases where ΔH° > 0 and ΔS° < 0, no calculation of ΔG° is necessary to determine that the process is thermodynamically unfavored (ΔG° > 0).
Science references for this teaching
- AP Chemistry Course and Exam Description, effective Fall 2024 — Required topic-numbered learning objectives and essential knowledge, all continuations and AP exclusions; supplied PDF SHA256 b5dfe8677ef3d88c613865d2e2a3e8d6125d652e2b24c71ef1e8ce4e011094f0.
- OpenStax Chemistry 2e §16.4, Free Energy — Enthalpy–entropy competition and ΔG°/equilibrium connection.
Glossary and deeper reading · 7 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Enthalpy
Enthalpy is a state function defined from internal energy, pressure, and volume. At constant pressure with only pressure–volume work, an enthalpy change equals the heat transferred to the system. For example, a negative reaction enthalpy indicates heat release for the written reaction under those conditions.
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Entropy
Entropy is a state function related to the number of microscopic arrangements compatible with a macroscopic state. Changes depend on both matter and energy distribution, so visual messiness alone is not a reliable measure. For example, an ideal gas expanding at fixed temperature has more accessible positional arrangements and increased entropy.
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Standard state
A standard state is a defined reference for a substance's thermodynamic properties. It specifies activities or reference forms and does not by itself fix the temperature at 25 °C. For example, gases commonly use a one-bar reference pressure while the table states temperature separately.
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Solute
A solute is a component dispersed in a solvent to form a solution. It need not be a solid, and a solution need not be liquid. For example, carbon dioxide is a gaseous solute in carbonated water.
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Standard molar entropy
Standard molar entropy is the molar entropy of a substance in its specified standard state at a stated temperature. It is an absolute-property value rather than an entropy of formation. For example, an element can have a positive standard molar entropy even though its standard formation enthalpy is zero.
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Endothermic process
An endothermic process absorbs heat into the chosen system. Under constant-pressure conditions with only pressure–volume work, its enthalpy change is positive. For example, melting ice absorbs heat while the phase change occurs.
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Exothermic process
An exothermic process transfers heat from the chosen system to its surroundings. Under constant-pressure conditions with only pressure–volume work, its enthalpy change is negative. For example, freezing water releases heat to the surroundings.