Free Energy and Equilibrium
Later free-energy connection: distinguish standard-state preference from the current mixture’s Q.
Before this connection: Energy transfer and enthalpy; Entropy and standard free energy.
AP Chemistry Topic 9.5 · CED printed p. 171 / PDF p. 178. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- K magnitude
- Standard free energy
- Natural logarithms
The equilibrium constant and standard free energy describe the same preference in different languages: and . R = 8.314 J mol−1 K−1 when ΔG° is in joules; T is kelvin; ln is the natural logarithm. Negative ΔG° gives K > 1 and a product-favored standard-state comparison. Positive ΔG° gives K < 1. Zero ΔG° gives K = 1. Use the thermodynamic dimensionless K consistent with the reaction and standard-state convention.
The useful scale is RT, about 2.5 kJ mol−1 near room temperature. If |ΔG°| is much smaller than RT, K is close to one. If the magnitude is many times RT, exponentiation drives K far above or far below one. This is why even a moderately sized free-energy difference can produce a dramatic equilibrium preference. Write the dimensionless ratio ΔG°/(RT) before exponentiating, and keep kJ/J consistent.
The superscript matters. ΔG° is the reference-state quantity used to determine K. A mixture’s actual directional driving force also depends on its current Q. At equilibrium Q = K and the actual reaction free-energy change is zero, even when ΔG° is not zero. As a conceptual link, the actual change can be written ΔG = ΔG° + RT ln Q; it vanishes at Q = K. This explains why a product-rich starting mixture can react backward even for a reaction with negative ΔG°. Equilibrium composition and the route’s speed remain separate questions.
Let's work through it
- At 300 K, K = 10.0. Write ΔG° = −(8.314 J mol−1 K−1)(300 K)ln 10.0.
- ln 10.0 = 2.303, giving ΔG° = −5743 J mol−1 = −5.74 kJ mol−1.
- The negative sign agrees with K > 1. If K were 0.100 for the reversed equation, ΔG° would be +5.74 kJ mol−1.
- Neither value says the actual mixture is already at equilibrium. Measure or compute its Q and compare with its equation’s K.
Make the picture move
Interactive concept model
A potential is intensive; reaction energy scales with n.
Predict first. Compare the signs of E° and ΔG°. If a balanced equation is doubled, which of E°, n, ΔG°, and K changes?
ΔG° = −nFE°; ln K = nFE°/(RT)
- ΔG° with n = 2
- Labeled selected or reference state
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
E° = 0.500 V gives ΔG° = -96.48 kJ mol⁻¹ and K = 8.02 × 10^16 at 298.15 K.
Read the data and comparisons
| Reaction scaling | n | E° (V) | ΔG° (kJ mol⁻¹) | log₁₀ K |
|---|---|---|---|---|
| As written | 2 | 0.500 | -96.48 | 16.904 |
| Equation doubled | 4 | 0.500 | -193 | 33.809 |
Macroscopic / data. An ideal reversible cell potential links the chemical driving force to electrical work per charge.
Particle view. The balanced reaction fixes how many moles of electrons transfer per mole of reaction; electrons travel through the external circuit rather than the salt bridge.
Symbolic view. n = 2; T = 298.15 K; log₁₀ K = 16.904.
Why it works. F converts moles of electrons to charge; volts are joules per coulomb. The negative sign connects positive cell potential to negative Gibbs energy for the process as written. Doubling the balanced reaction doubles n and ΔG°, leaves E° unchanged, and squares K. Thermodynamic favorability does not establish a reaction rate.
Conditions and limits. Abstract balanced redox process with supplied E° and electron count, not a specific chemical cell. Changing temperature evaluates thermodynamic relationships using the supplied E° at that temperature; it does not assume a real cell has temperature-independent E°. F = 96485 C mol⁻¹ and R = 8.314 J mol⁻¹ K⁻¹. K and Q use dimensionless activities referenced to standard states; solids and pure liquids have unit activity. Standard states do not by themselves prescribe a temperature. Positive E applies to the reaction orientation as written.
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
What does ΔG° = 0 imply for K?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
What does ΔG° = 0 imply for K?
Worked answer and reason
K = 1.
The exponent −ΔG°/RT is zero.
Retrieval 2
At equilibrium must ΔG° be zero?
Worked answer and reason
No. Actual ΔG is zero when Q = K; ΔG° still specifies the reference-state preference.
Composition contributes to the actual driving force.
Catch the tempting mistake
- A product-favored K prevents every backward reaction.
- A current Q larger than K can drive net reverse reaction despite K > 1.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 9 problem 6 · ap synthesis · symbolic
- Unit 9 problem 7 · ap application · symbolic
Scientific calculator workspace
Use this workspace for arithmetic, logarithms and elementary functions. Keep intermediate precision and report final values with appropriate units and significant figures. Graphing and general equation solving need a suitable approved tool.
Scientific calculator
This calculator requires JavaScript. Use the physics formula sheets and show your working manually.
+ − * / ^ ! · pi · e · sqrt · sin, cos, tan · asin, acos, atan · atan2(y,x) · ln, log · abs · exp · hypot(x,y). Use * between factors. Keep units in your written working.
For the teacher
- Before calculator use, require a sign estimate and a comparison of |ΔG°| with RT. Reject answers using mismatched energy units.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 9.5.A
- Explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T.
- 9.5.A.1
- The phrase “thermodynamically favored” (ΔG° < 0) means that the products are favored at equilibrium (K > 1) under standard conditions.
- 9.5.A.2
- The equilibrium constant is related to free energy by the equations EQN: K = e−ΔG°/RT and EQN: ΔG° = −RT ln K.
- 9.5.A.3
- Connections between K and ΔG° can be made qualitatively through estimation. When ΔG° is near zero, the equilibrium constant will be close to 1. When ΔG° is much larger or much smaller than RT, the value of K deviates strongly from 1.
- 9.5.A.4
- Processes with ΔG° < 0 favor products (i.e., K > 1) and those with ΔG° > 0 favor reactants (i.e., K < 1).
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 · 5 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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Gibbs free energy
Gibbs free energy combines enthalpy and entropy contributions to a system at specified temperature and pressure. For a reaction at constant temperature and pressure, a negative actual reaction free-energy change favors its forward direction, while standard and actual values must be distinguished. For example, a favorable reaction can still be slow if its pathway has a large activation barrier.
Read more: Gibbs Free Energy and Thermodynamic Favorability →
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Equilibrium constant
An equilibrium constant is the value of the reaction quotient at equilibrium for a specified reaction and temperature. Its magnitude indicates equilibrium composition preference rather than reaction speed. For example, a large constant often indicates a product-favored equilibrium, but the starting composition still determines the immediate net direction.
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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.
Read more: Gibbs Free Energy and Thermodynamic Favorability →
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Chemical equilibrium
Chemical equilibrium is a dynamic state in which forward and reverse reaction rates are equal. Macroscopic composition remains constant in a closed system under fixed conditions, but reactant and product concentrations need not be equal. For example, a reversible reaction can reach equilibrium with much more reactant than product.
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Mixture
A mixture contains two or more substances whose proportions can vary. Its components retain their chemical identities until a chemical reaction changes them. For example, water and ethanol can form mixtures with many different compositions.