Why can dilution shift equilibrium even when every concentration falls?
You will be able to: Calculate the immediate quotient after dilution and distinguish amounts from concentrations.
Why can dilution shift equilibrium even when every concentration falls?
Adding water to a solution lowers all solute concentrations at once. If different powers appear above and below the division line in Q, those decreases do not cancel.
A useful starting point: How does comparing Q with K predict the next change? →
Words and symbols before equations
- Dilution factor d
- Final volume divided by initial volume before reaction responds.
- Immediate concentration
- Original concentration divided by d when solute amounts are unchanged.
- Amount vs concentration
- Moles versus moles per volume.
- Net dissociation
- Conversion into more independently dissolved species for the stated equation.
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. Dissolved A₂ ⇌ 2A initially [A₂]=0.10 M and [A]=0.20 M, Kc=0.40. Add solvent with additive volumes. No gas–solution partition equilibrium is modeled.
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.
- Volume factor=2. Immediate Q=0.2, K=0.40. Final [A]=0.12361 M; [A₂]=0.038197 M. In an initial 1 L sample, final moles A=0.24721 mol. Atom amount stays 0.40 mol.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
For dissolved A₂⇌2A, doubling volume halves both species concentrations. Q′=([A]/2)²/([A₂]/2)=Q/2.
At fixed T, K is unchanged. If the original solution was at equilibrium, Q′<K and net dissociation follows.
After adjustment there may be more moles of A even though its concentration is below its original value. A direction statement describes net conversion after the immediate dilution, not necessarily concentration relative to the original smaller volume.
Assume ideal dilute behavior, additive volumes and no solvent reaction. Dilution effects depend on the equation; for a one-to-one A⇌B expression, uniform dilution leaves Q unchanged.
A worked example, step by step
Initially A₂⇌2A has [A₂]=0.10 M and [A]=0.20 M, so Kc=0.40. Double the volume at fixed T and predict the immediate and subsequent changes.
- Immediately [A₂]=0.050 M and [A]=0.100 M.
- Q′=0.100²/0.050=0.20.
- Since Q′<0.40, net dissociation follows.
- Solving 0.40=(0.100+2x)²/(0.050−x) gives x≈0.01180 M: final A≈0.12361 M and A₂≈0.03820 M. A concentration rose after dilution but stays below the original 0.20 M.
A shift toward a species does not guarantee its final concentration exceeds its pre-dilution concentration.
What happens to Q for one-to-one A⇌B under uniform dilution?
Compare with an explanation
Its numerator and denominator fall by the same factor, so Q is unchanged in the ideal model.
Predict. Change one thing. Explain.
Increase volume by a factor from 1 to 4. Compare initial, immediately diluted and final concentrations; use volume to interpret moles.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Volume factor=2. Immediate Q=0.2, K=0.40. Final [A]=0.12361 M; [A₂]=0.038197 M. In an initial 1 L sample, final moles A=0.24721 mol. Atom amount stays 0.40 mol.
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. Dissolved A₂ ⇌ 2A initially [A₂]=0.10 M and [A]=0.20 M, Kc=0.40. Add solvent with additive volumes. No gas–solution partition equilibrium is modeled.
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 questionA dissolved equilibrium A+B⇌C is uniformly diluted to twice its volume at fixed T. Determine the factor by which Q changes and predict the net direction.
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Compare with the answer and four-point rubric
- 1 point: Q=[C]/([A][B]).
- 1 point: Each concentration halves immediately.
- 1 point: Q′=(C/2)/((A/2)(B/2))=2Q.
- 1 point: Initially Q=K, so Q′>K and net dissociation toward A+B follows.
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 1Does dilution change K at fixed T?
No in the stated model.
RECALL 2Why do powers matter?
They determine how common concentration factors cancel.
RECALL 3How do you obtain moles from concentration?
Multiply concentration by the current volume.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why can dilution shift equilibrium even when every concentration falls?
- Uniform concentration division by d gives Q′=Q/d^Δν for dissolved species.
- Separate immediate dilution from subsequent reaction.
Remember: A shift toward a species does not guarantee its final concentration exceeds its pre-dilution concentration.
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. Dissolved A₂ ⇌ 2A initially [A₂]=0.10 M and [A]=0.20 M, Kc=0.40. Add solvent with additive volumes. No gas–solution partition equilibrium is modeled.
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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