How does free energy tell you which side equilibrium favors?
You will be able to: Connect ΔG°, K and temperature with correct logarithms and units.
How does free energy tell you which side equilibrium favors?
An equilibrium mixture can contain much more product than reactant even though forward and reverse rates are equal. The size of K describes that balance, and ΔG° gives another way to express it.
A useful starting point: Why can a favored reaction appear not to happen? →
Words and symbols before equations
- K
- Dimensionless equilibrium constant using activities; dilute models use normalized concentrations.
- ln
- Natural logarithm, with base e rather than base ten.
- R
- Gas constant, 8.314 J mol⁻¹ K⁻¹ in these calculations.
- Standard free energy
- ΔG° for the balanced reaction at the stated temperature.
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. K is dimensionless, using activities. The supplied K and T are independently chosen states, not a prediction of how K varies with T for one physical reaction. Natural logarithm is used in ΔG°=−RT ln K.
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.
- K=10, T=300 K and ΔG°=-5.743 kJ/mol. Products are favored in the equilibrium expression. Exact conversion is not determined by K alone.
- 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°=−RT ln K. If K>1, ln K is positive and ΔG° is negative. If K<1, ΔG° is positive. If K=1, ΔG°=0.
The magnitude matters relative to RT: a large negative ΔG° compared with RT gives a very large K. Merely saying a value is close to zero requires a relevant energy scale.
To find K, rearrange to K=exp(−ΔG°/RT). Convert kJ to J when using R=8.314, and use kelvin.
K is not generally a direct percent conversion. Its relation to composition depends on the balanced expression, starting mixture and other constraints.
A worked example, step by step
At 300 K, K=10.0 for a supplied reaction. Calculate ΔG° using R=8.314 J mol⁻¹ K⁻¹.
- ln(10.0)=2.303.
- RT=8.314×300=2494.2 J/mol.
- ΔG°=−2494.2×2.303≈−5743 J/mol=−5.74 kJ/mol.
- K>1 and negative ΔG° consistently indicate products are favored at equilibrium.
Use ln, not log₁₀, with ΔG°=−RT ln K. Do not confuse standard ΔG° with actual ΔG at equilibrium.
Does equilibrium always require K=1?
Compare with an explanation
No. Equilibrium requires Q=K; K can be very large or small.
Predict. Change one thing. Explain.
At fixed 300 K, compare K=0.1, 1 and 10 using the log₁₀K control. Predict the ΔG° signs and explain why K=1 is a special case.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
K=10, T=300 K and ΔG°=-5.743 kJ/mol. Products are favored in the equilibrium expression. Exact conversion is not determined by K alone.
Original teaching model with supplied rounded data. Numerical states, units and assumptions are specified below; no measured reaction rate is implied. K is dimensionless, using activities. The supplied K and T are independently chosen states, not a prediction of how K varies with T for one physical reaction. Natural logarithm is used in ΔG°=−RT ln K.
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 300 K, a supplied reaction has ΔG°=+5.74 kJ/mol. Estimate K and describe what it favors.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Convert to about +5740 J/mol.
- 1 point: Exponent=−5740/(8.314×300)≈−2.30.
- 1 point: K≈e⁻²⋅³≈0.10.
- 1 point: Reactants are favored at equilibrium; the exact composition still depends on the reaction expression and constraints.
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 sign pair matches K>1?
ΔG°<0.
RECALL 2Which logarithm appears with R?
The natural logarithm.
RECALL 3Does K equal percent yield?
No.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does free energy tell you which side equilibrium favors?
- ΔG°=−RT ln K.
- K=exp(−ΔG°/RT); match J or kJ units with R.
Remember: Use ln, not log₁₀, with ΔG°=−RT ln K. Do not confuse standard ΔG° with actual ΔG at equilibrium.
Conditions: Original teaching model with supplied rounded data. Numerical states, units and assumptions are specified below; no measured reaction rate is implied. K is dimensionless, using activities. The supplied K and T are independently chosen states, not a prediction of how K varies with T for one physical reaction. Natural logarithm is used in ΔG°=−RT ln K.
Refresh Kid · AP Chemistry Unit 9 · Objectives 9.5.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 9.5, objective 9.5.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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