How do enthalpy and entropy compete in Gibbs free energy?
You will be able to: Calculate ΔG° with matched units and explain standard-state favorability.
How do enthalpy and entropy compete in Gibbs free energy?
Some dissolving processes cool their surroundings and still occur. An unfavorable enthalpy contribution can be outweighed by a favorable entropy contribution at the stated temperature.
A useful starting point: How do coefficients enter an entropy calculation? →
Words and symbols before equations
- Gibbs free energy change, ΔG
- Criterion for thermodynamic direction at constant T and pressure.
- ΔG°
- Free energy change when species are in their specified standard states.
- ΔH°
- Standard reaction enthalpy, often in kJ per mole of reaction.
- ΔS°
- Standard reaction entropy, often in J per mole of reaction per kelvin.
- Kelvin
- Absolute temperature scale, T(K)=T(°C)+273.15.
What this picture assumes
Original teaching model with supplied rounded data. Numerical states, units and assumptions are specified below; no measured reaction rate is implied. Constant ΔH° and ΔS° approximation across this temperature range, with no phase changes. The graph describes standard-state favorability, not reaction rate or an arbitrary nonstandard mixture.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- ΔG°=-10 kJ/mol: forward standard-state favored. Constant-data crossover at 200 K. No reaction speed is predicted.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Use ΔG°=ΔH°−TΔS°. A negative ΔH° favors a negative ΔG°, while a positive ΔS° also favors a negative ΔG° because the entropy term is subtracted.
Match units before subtracting. If ΔH° is kJ/mol and ΔS° is J/(mol·K), divide ΔS° by 1000 or convert ΔH° to joules. Temperature must be kelvin.
ΔG°<0 indicates forward standard-state favorability; ΔG°>0 favors the reverse under standard conditions. Actual mixtures need composition information as well, and a negative value does not predict speed.
The standard-state symbol does not automatically mean 298 K. Each value has a specified temperature. A fixed ΔH°/ΔS° explorer approximates them as constant over its chosen range.
A worked example, step by step
At 300 K a supplied reaction has ΔH°=+20.0 kJ/mol reaction and ΔS°=+100 J mol-reaction⁻¹ K⁻¹. Find ΔG°.
- Convert entropy: +100 J/(mol·K)=+0.100 kJ/(mol·K).
- TΔS°=300×0.100=30.0 kJ/mol.
- ΔG°=20.0−30.0=−10.0 kJ/mol.
- The positive entropy contribution outweighs endothermic enthalpy, so the forward reaction is favored under standard conditions at 300 K.
Do not subtract joules from kilojoules or use Celsius in TΔS. Favorability is not a promise of a fast reaction.
Can an endothermic process have negative ΔG°?
Compare with an explanation
Yes, if the positive TΔS° term is large enough to outweigh positive ΔH°.
Predict. Change one thing. Explain.
Set ΔH°=20 kJ/mol and ΔS°=100 J/(mol·K). Compare 100, 200 and 300 K. Explain which term changes and where the sign changes.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
ΔG°=-10 kJ/mol: forward standard-state favored. Constant-data crossover at 200 K. No reaction speed is predicted.
Original teaching model with supplied rounded data. Numerical states, units and assumptions are specified below; no measured reaction rate is implied. Constant ΔH° and ΔS° approximation across this temperature range, with no phase changes. The graph describes standard-state favorability, not reaction rate or an arbitrary nonstandard mixture.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using energy and entropy contributions, electron and ion bookkeeping, or the stated cell reaction. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionAt 400 K, ΔH°=+30 kJ/mol and ΔS°=+100 J/(mol·K). Calculate ΔG° and explain the competing contributions.
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Compare with the answer and four-point rubric
- 1 point: Convert ΔS° to 0.100 kJ/(mol·K).
- 1 point: TΔS°=40 kJ/mol.
- 1 point: ΔG°=30−40=−10 kJ/mol.
- 1 point: The favorable entropy contribution exceeds the unfavorable enthalpy contribution at this temperature.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What does a standard-state negative ΔG° indicate?
Forward thermodynamic favorability under standard conditions.
RECALL 2Which entropy sign favors negative ΔG°?
Positive ΔS°.
RECALL 3Does ° specify 298 K?
No; temperature is stated separately.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do enthalpy and entropy compete in Gibbs free energy?
- ΔG°=ΔH°−TΔS° at specified T.
- Use kelvin and consistent energy units.
Remember: Do not subtract joules from kilojoules or use Celsius in TΔS. Favorability is not a promise of a fast reaction.
Conditions: Original teaching model with supplied rounded data. Numerical states, units and assumptions are specified below; no measured reaction rate is implied. Constant ΔH° and ΔS° approximation across this temperature range, with no phase changes. The graph describes standard-state favorability, not reaction rate or an arbitrary nonstandard mixture.
Refresh Kid · AP Chemistry Unit 9 · Objectives 9.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.3, objective 9.3.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 9: Thermodynamics and Electrochemistry, Topics 9.1–9.11. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Numerical thermodynamic examples state standard conditions, temperature, reaction scaling and unit conventions. Supplied data and schematic geometry are teaching models. Standard ΔG° describes standard-state favorability and relates to K; actual direction depends on composition. Thermodynamic favorability does not predict rate. Nonstandard cell potential is taught through Q, distance from equilibrium and qualitative Nernst reasoning; algorithmic substitution alone does not demonstrate the assessed understanding. Electrode positive/negative labeling is excluded from assessed scope. Oxidation at the anode and reduction at the cathode remain essential. Faraday calculations assume the stated current efficiency and electron stoichiometry. Rotatable particle models are schematic inventories, not measured molecular trajectories. Virtual models do not replace required supervised laboratory work.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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