When is it safe to neglect a small equilibrium change?
You will be able to: Check a small-x approximation against the concentration it replaces.
When is it safe to neglect a small equilibrium change?
Subtracting one cent from a hundred dollars hardly changes the total; subtracting it from two cents changes a lot. Neglecting a chemical change also depends on the size of the starting quantity.
A useful starting point: How do coefficients change an equilibrium calculation? →
Words and symbols before equations
- Approximation
- A deliberate simplification whose effect must be checked.
- Relative change
- Change divided by the original amount.
- Five-percent check
- A common classroom tolerance, not a universal law.
- Exact solution
- Solution of the stated model without neglecting the change term.
- Base-10 logarithm, log₁₀
- The exponent on 10: log₁₀(0.001)=−3 because 0.001=10⁻³. For a concentration slider, first divide by 1 M so the logarithm uses a numerical ratio.
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, Kc=10⁻⁴, initial A=0. Compare the exact solution with neglecting x only in C−x. Five percent is a chosen classroom screening tolerance.
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.
- Exact x=0.0015687 M; Qc=1×10⁻⁴=Kc. Atom concentration [A]+2[A₂]=0.2 M. Approximate x=0.0015811 M, 1.5811% of initial C: passes the chosen 5% screen. Passing is not exact equality.
- 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 A₂⇌2A, K=4x²/(C−x). If x is small compared with C, replace only the denominator C−x by C, giving x≈√(KC/4).
Do not erase x everywhere: it is still the source of product concentration. The approximation concerns subtraction from a much larger initial quantity.
Check 100xapprox/C. If it exceeds the chosen tolerance, solve the full equation. Even below 5%, a question may require a tighter accuracy target.
A small K alone is insufficient. At a small initial C the fractional dissociation can be substantial, so the same K can require an exact calculation.
A worked example, step by step
Use K=1.0×10⁻⁴ and C=0.100 M. Estimate x, apply a 5% check and compare with the full quadratic.
- Assume C−x≈C: xapprox=√(1.0×10⁻⁴×0.100/4).
- xapprox≈0.001581 M.
- Relative change≈1.58%, below the stated 5% screening tolerance.
- Exact x≈0.001569 M. The approximation is close for these conditions; report its assumption rather than calling it exact.
Small K does not by itself prove that x is small relative to every starting concentration.
What happens at C=0.001 M with the same K?
Compare with an explanation
xapprox/C≈15.8%, so the 5% screening check fails.
Predict. Change one thing. Explain.
Hold K=10⁻⁴ fixed and lower C from 0.10 to 0.001 M. Predict when the 5% check fails and compare approximate versus exact x.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Exact x=0.0015687 M; Qc=1×10⁻⁴=Kc. Atom concentration [A]+2[A₂]=0.2 M. Approximate x=0.0015811 M, 1.5811% of initial C: passes the chosen 5% screen. Passing is not exact equality.
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, Kc=10⁻⁴, initial A=0. Compare the exact solution with neglecting x only in C−x. Five percent is a chosen classroom screening tolerance.
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 student obtains xapprox=0.006 M for C=0.080 M using a small-change assumption. Evaluate a 5% check, explain what must happen next and distinguish the tolerance from a physical law.
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Compare with the answer and four-point rubric
- 1 point: Relative change is 0.006/0.080=0.075.
- 1 point: That is 7.5%, above 5%.
- 1 point: Solve the unapproximated equation and verify physical amounts and K.
- 1 point: Five percent is a chosen classroom accuracy screen, not a condition defining 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 1Which x is neglected?
Only the small subtraction relative to a larger starting quantity.
RECALL 2Can the same K pass and fail at different C?
Yes; the fractional change varies.
RECALL 3What if the check fails?
Use the full equation or a justified more accurate method.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
When is it safe to neglect a small equilibrium change?
- For this dissociation model: xapprox=√(KC/4).
- Check xapprox/C against the required tolerance and substitute back if needed.
Remember: Small K does not by itself prove that x is small relative to every starting 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. A₂ ⇌ 2A, Kc=10⁻⁴, initial A=0. Compare the exact solution with neglecting x only in C−x. Five percent is a chosen classroom screening tolerance.
Refresh Kid · AP Chemistry Unit 7 · Objectives 7.7.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.7, objective 7.7.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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