When can temperature reverse thermodynamic favorability?
You will be able to: Use sign combinations and a crossover temperature to predict favorability.
When can temperature reverse thermodynamic favorability?
Ice can melt above its melting temperature and freeze below it. The competition between enthalpy and entropy changes with temperature even when the phase-change direction is held fixed.
A useful starting point: How do enthalpy and entropy compete in Gibbs free energy? →
Words and symbols before equations
- Crossover temperature
- Temperature where ΔG°=0 in a stated approximation.
- Enthalpy-favored
- Negative ΔH° contributes toward negative ΔG°.
- Entropy-favored
- Positive ΔS° makes −TΔS° negative.
- Constant-data approximation
- Treating ΔH° and ΔS° as unchanged over a limited T range.
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
If ΔH°<0 and ΔS°>0, both contributions favor the forward process for positive temperatures within the model. If ΔH°>0 and ΔS°<0, neither does.
If both are positive, high T makes the favorable entropy term large enough. If both are negative, increasing T enlarges the unfavorable −TΔS° term, so low T favors the process.
Set ΔG°=0 to find T*=ΔH°/ΔS° using matched energy units. A physically meaningful positive crossover occurs for same-sign nonzero values.
Real ΔH° and ΔS° can vary with temperature and phases can change. A constant-data graph is a stated approximation, not a guarantee across every possible temperature.
A worked example, step by step
A supplied process has ΔH°=−40.0 kJ/mol and ΔS°=−100 J/(mol·K). Find T* and the favored temperature side.
- Convert ΔS° to −0.100 kJ/(mol·K).
- T*=−40.0/(−0.100)=400 K.
- At 300 K, ΔG°=−40−300(−0.100)=−10 kJ/mol.
- The process is favored below 400 K and unfavored above it within the constant-data model.
For two negative values, low temperature favors the process. Do not use an absolute-value shortcut without examining the signs.
What happens when ΔS°=0 in this approximation?
Compare with an explanation
ΔG°=ΔH° and is independent of T; dividing by zero to find a crossover is invalid.
Predict. Change one thing. Explain.
Compare positive/positive and negative/negative choices with the same crossover. Explain the opposite graph slopes and the favored side of the zero line.
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 questionFor ΔH°=+60 kJ/mol and ΔS°=+150 J/(mol·K), find the crossover and compare 300 K with 500 K.
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Compare with the answer and four-point rubric
- 1 point: Convert ΔS° to +0.150 kJ/(mol·K).
- 1 point: T*=60/0.150=400 K.
- 1 point: At 300 K ΔG°=+15 kJ/mol, unfavored forward under standard conditions.
- 1 point: At 500 K ΔG°=−15 kJ/mol, favored; the constant-data assumption must be stated.
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 1Which sign pair favors only low T?
ΔH°<0 and ΔS°<0.
RECALL 2Which pair favors all positive T in the model?
ΔH°<0 and ΔS°>0.
RECALL 3When must crossover division be avoided?
When ΔS°=0.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When can temperature reverse thermodynamic favorability?
- Same-sign nonzero ΔH° and ΔS°: T*=ΔH°/ΔS°.
- Positive/positive favors high T; negative/negative favors low T.
Remember: For two negative values, low temperature favors the process. Do not use an absolute-value shortcut without examining the signs.
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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