How does comparing Q with K predict the next change?
You will be able to: Predict net direction using a correctly matched Q and K.
How does comparing Q with K predict the next change?
A mixture can contain both reactants and products without being at equilibrium. Its present composition must be compared with the equilibrium relationship for that exact reaction and temperature.
A useful starting point: How do warming and a catalyst affect equilibrium differently? →
Words and symbols before equations
- Q
- Current composition expression.
- K
- Target equilibrium value of the same expression at stated T.
- Net forward / reverse
- Overall conversion toward written products / reactants.
- Redistribution
- Changes in species amounts that bring Q toward K.
What this picture assumes
Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. A₂ ⇌ 2A with current [A₂]=0.10 M and Kc=0.80. Compare independent compositions; the control does not conserve one closed inventory.
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.
- Qc=0.4, Kc=0.80: net forward. At fixed temperature, composition changes to bring Qc toward Kc.
- 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 Q<K, the product terms are too small relative to the reactant terms for equilibrium. Net forward change increases Q toward K.
If Q>K, net reverse change reduces Q. If Q=K, there is no net change although both directions continue.
Do not compare raw product and reactant totals instead of evaluating the full expression. Powers and concentration or pressure conventions matter.
The comparison predicts direction, not rate, completion time or the exact final composition. Those require additional constraints such as conserved material and, for time, kinetic data.
| Comparison | Net change | Meaning |
|---|---|---|
| Q < K | Toward products | Q increases toward K |
| Q = K | No net change | Forward and reverse continue |
| Q > K | Toward reactants | Q decreases toward K |
A worked example, step by step
A₂⇌2A has Kc=0.80, [A]=0.20 M and [A₂]=0.10 M. Predict the net direction.
- Use the matching concentration expression Qc=[A]²/[A₂].
- Qc=0.20²/0.10=0.40.
- 0.40<0.80, so net dissociation produces more A.
- This raises the numerator and lowers the denominator until Qc=Kc; it does not determine how fast that occurs.
Q<K does not mean that K will fall to meet Q. At fixed T, the composition adjusts toward the same K.
If Q=K, are the reactions stopped?
Compare with an explanation
No. Their opposing rates are equal.
Predict. Change one thing. Explain.
Hold [A₂]=0.10 M and Kc=0.80 fixed. Change [A] across the value that makes Qc=Kc. Predict direction before reading the comparison.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Qc=0.4, Kc=0.80: net forward. At fixed temperature, composition changes to bring Qc toward Kc.
Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. A₂ ⇌ 2A with current [A₂]=0.10 M and Kc=0.80. Compare independent compositions; the control does not conserve one closed inventory.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using relative rates, particle conservation, the Q/K comparison or the stated dissolution equilibrium. 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 A₂⇌2A, Kc=0.50 and the mixture has [A₂]=0.20 M, [A]=0.50 M. Calculate Q, predict direction, and state what remains fixed and what the comparison cannot tell you.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Q=0.50²/0.20=1.25.
- 1 point: Q>K, so net association forms A₂.
- 1 point: Kc remains 0.50 at fixed T; concentrations change.
- 1 point: Q/K alone does not provide a rate or time to equilibrium.
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 if Q>K?
Net change toward reactants.
RECALL 2Does K follow a concentration disturbance?
No, at fixed temperature K stays fixed.
RECALL 3What additional information predicts the endpoint?
A conservation relation or initial inventory together with K.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How does comparing Q with K predict the next change?
- Q<K: net forward. Q>K: net reverse. Q=K: no net change.
- Compare the same equation, temperature and concentration/pressure convention.
Remember: Q<K does not mean that K will fall to meet Q. At fixed T, the composition adjusts toward the same K.
Conditions: Ideal dilute concentrations in mol/L (M); fixed temperature, fixed volume except when explicitly changed, and no side reactions. Supplied K values use the stated AP concentration convention. A₂ ⇌ 2A with current [A₂]=0.10 M and Kc=0.80. Compare independent compositions; the control does not conserve one closed inventory.
Refresh Kid · AP Chemistry Unit 7 · Objectives 7.10.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.10, objective 7.10.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 7: Equilibrium, Topics 7.1–7.12. 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. Converting between Kc and Kp and calculations for a dissolved species in equilibrium with its gas phase are excluded from assessed Unit 7 scope. Concentrations use mol/L and gas partial pressures use the stated pressure convention. Supplied constants are teaching data at fixed temperature unless otherwise specified. Ideal dilute-solution and ideal-gas approximations are stated. 3D views show inventories, not molecular trajectories, measured structures or proof of equilibrium from a single snapshot. Approximation checks are explicit; a small K alone does not justify neglecting every change.
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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