SCIENCE · INVESTIGATE AND EXPLAIN

Construct a composition that satisfies both equilibrium and conservation

Build independent A and B concentrations, distinguish reaction events from species changes, and reject a rate-balanced mixture with the wrong conserved inventory.

Learning outcome: Use unequal stoichiometry, opposing rates and a reaction quotient to justify an equilibrium composition for a specified closed batch.

Starting recap

For 2A ⇌ B, one forward event consumes two A units and forms one B. B contains two conserved units, so A + 2B is fixed at constant volume.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

A proposal has equal forward and reverse rates. Is that enough to identify the equilibrium of a batch containing a specified amount of material?

Open the experiment

Predict

Predict whether equal concentrations are necessary for equilibrium and whether equal rates can coexist with the wrong batch inventory.

Your prediction and reason

Experiment

Start with A = B = 0.50 in the first packet. Change A to zero and B to 0.50: this conserves the batch but does not balance rates. Repair both concentrations. Compare the second packet, where equal concentrations do work. Construct the transfer packet without first opening the solution. Capture one rejected proposal and a repaired composition.

Construct an equilibrium that also conserves material

Worked teaching model · synthetic values · Original hypothetical elementary 2A ⇌ B model at fixed temperature and volume. B contains two conserved A units. Forward event rate kf[A]² and reverse event rate kr[B] are stipulated; kf has units L/(mol s), kr has units s⁻¹. Ideal activities use c° = 1 mol/L. No measured species, competing reaction, time trajectory or general deduction of rate laws from net equations is claimed.

Closed-batch packet: Unequal equilibrium: M = 1.00 mol/L, K = 1 packet

Reset keeps your prediction, captures and practice history.

Build both concentrations independently

Reaction: 2A ⇌ B. Each B contains two conserved A units. This batch requires [A] + 2[B] = 1.00 mol/L. Changing the proposal does not add material to the target batch.

Proposed compositionProposed [A] (mol/L)Proposed [B] (mol/L)

Quantity	Proposal	Required check
[A]	0.50 mol/L	One conserved unit per A
[B]	0.50 mol/L	Two conserved units per B
[A] + 2[B]	1.50 mol/L	1.00 mol/L; fails

Assigned k_f = 0.5 L/(mol s); k_r = 0.5 s⁻¹. These are elementary event rates, normalized to one occurrence of the reaction as written.

Rate	Calculation	mol/(L s)
Forward event rate	0.5 × 0.50²	0.12500
Reverse event rate	0.5 × 0.50	0.25000
Net B change	forward − reverse	-0.12500
Net A change	−2 × (forward − reverse)	0.25000

Rate balance: fails. Batch inventory: fails. This proposal is not an equilibrium composition for the selected batch. A rate-balanced mixture with the wrong inventory could describe a different batch.

With c° = 1 mol/L, Q = ([B]/c°)/([A]/c°)² and K = k_fc°/k_r = 1. Q = 2.00000.

Check the equilibrium solution and model limits

For the selected inventory, substituting [B] = K[A]²/c° gives 2K[A]²/c° + [A] − M = 0. Its nonnegative solution is [A] = 0.50000 mol/L and [B] = 0.25000 mol/L. Equal concentrations are neither necessary nor sufficient in general. At equilibrium both opposing processes continue; their net species changes vanish.

The grid constructs candidate states; it does not simulate how a closed batch reaches them. Net A and B changes have different magnitudes because two A units participate per event. An actual non-elementary net reaction does not generally have these rate laws. All constants and batch quantities here are original teaching premises, not experimental estimates.

Check material conservation and rate balance separately: both are required.

Inventory residual 0.500 mol/L; forward 0.1250, reverse 0.2500 mol/(L s). Net species changes: A 0.2500, B -0.1250 mol/(L s).

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Try: two checks, not one

A closed batch can redistribute material, but cannot create it. For hypothetical 2A ⇌ B, each B contains two conserved units carried by A. At fixed volume, inventory is [A] + 2[B], not [A] + [B]. Square brackets mean concentration in mol/L.

Begin with the first packet and [A] = [B] = 0.50 mol/L. Required inventory is 1.00 mol/L. Inspect both ledgers: equal concentrations do not settle either question.

Predict: can one check pass alone?

Construct [A] = 0 and [B] = 0.50. Inventory is 1.00 mol/L, but forward rate is zero while reverse rate is 0.5 × 0.50 = 0.25 mol/(L s). B converts back to A. Predict species-change signs and relative magnitudes before inspecting them.

Experiment: repair the composition

Try [A] = 0.50 and [B] = 0.25. Inventory is 0.50 + 2 × 0.25 = 1.00 mol/L. Both event rates are 0.125 mol/(L s), although concentrations differ. Now keep the proposal and select the second packet: both constants and the target inventory are explicitly different. Its equilibrium is A = B = 0.50. Equal concentrations can work, but are not a general rule.

For a rate-balanced wrong-inventory counterexample, set A = 1.00 and B = 1.00 in the first packet. Both rates are 0.50, yet the proposal contains 3.00 mol/L of conserved units rather than 1.00. A correct quotient cannot repair missing or extra material. Capture the rejection and a repaired state.

Understand: events, species changes and the quotient

Assigned elementary forward rate is kf[A]²; reverse rate is kr[B]. kf has units L/(mol s), while kr has units s⁻¹. Their difference is net B formation rate. Net A change is minus twice that difference. The derivative of [A] + 2[B] therefore vanishes even away from equilibrium.

At equilibrium event rates are equal and nonzero for a nonempty batch. Concentrations need not be equal. Ideal activities are [A]/c° and [B]/c°, with c° = 1 mol/L. Dimensionless Q = ([B]/c°)/([A]/c°)², and K = kf c°/kr. When A is zero and B positive, Q diverges but the reverse direction remains well defined from rates. For an empty proposal Q is undefined. Zero opposing rates cannot establish equilibrium for a nonempty target batch.

Apply: different constants and inventory

In the transfer packet construct A = 1.00 and B = 0.50 mol/L. Verify inventory 2.00 mol/L and both rates 0.25 mol/(L s). Explain the factor of two in A change versus the normalized event rate. Complete the separate practice case, whose concentrations need not lie on the visual grid.

Continue: evidence and boundaries

Numbers, construction and diagrams are original teaching material. This hypothetical elementary reaction is not measured chemistry; a general rate law does not follow merely from an overall balanced equation. This activity evaluates states, not elapsed-time dynamics. Return to the HS dynamic-equilibrium trace for a distinct first-order time-course model.

The AP Chemistry framework, topics 7.1–7.3 motivates dynamic balance, direction and quotient representations. OpenStax Chemistry 2e, chemical equilibria explains elementary opposing processes. This does not establish experimental observations, full Unit 7 coverage or independent proficiency.

Transfer: a large K is not a universal particle-count ratio

For a separate idealized 2A ⇌ B system, let K = a(B)/a(A)² = 100, using concentration activities relative to 1 mol/L. Predict whether this necessarily means more B molecules than A molecules. It does not: at [A] = 0.0010 mol/L and [B] = 0.00010 mol/L, Q = 100 although there are ten times as many A molecules as B. Conserved A-unit inventory is 0.0012 mol/L; only one-sixth of those units reside in B.

A second equilibrium with the same K has [A] = 0.010 mol/L and [B] = 0.010 mol/L: equal molecule counts, but a different conserved inventory, 0.030 mol/L. Verify both quotients and ledgers. The powers in the expression and the available amount matter; “products favored” is shorthand about a specific equilibrium expression, not a universal statement about every species count or fraction converted.

Finally, doubling the written reaction to 4A ⇌ 2B squares K to 10,000 without changing either physical mixture. A larger number produced by rescaling an equation is not evidence that the same sample became more product-rich. These are synthetic comparison states outside the constructor’s fixed grid; explain the equations rather than forcing the sliders to represent them. Reference checked 30 October 2026: OpenStax equilibrium constants.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain the two independent acceptance checks, the factor of two in species rates, and why Q is undefined for an empty proposal. State why these elementary rate laws cannot be inferred for an arbitrary net reaction.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Apply molar phase energy only within the available phase inventory Next: Transform the equation and its equilibrium constant together

Save a science backup or import one

Science work is stored locally in this browser, separately from mathematics and AI. It does not sync automatically. Use this science-only backup; this checkout's older all-subject export does not include science.

A legacy file is available only when the complete snapshot fits the 36 MB limit. For larger snapshots, prepare a complete file set below; never discard notebooks to make a smaller file.

Choose a science backup (maximum 36 MB)

Matching imported notebook drafts replace their current drafts. Omitted lessons and known first attempts remain. Export your current work first if you want to retain both drafts.

Complete file sets and course backups

A file set can hold more work than one science file. Keep its manifest and every part together. Each file is limited to 36 MB. This format needs this version of LearnAI or a compatible newer version; older versions can use only the legacy single-file backup above.

Export scope

The set includes every saved science course and global continuation at this moment. Later edits need a new export. A conflicting tab exports its own visible work, which may differ from saved work in another tab.

Choose the manifest and all parts together

0 file(s) selected. File order and filenames do not determine identity. Missing, changed or conflicting parts block the entire restore; no good part is applied by itself.

Science storage: loading. No account or remote computation is required.