Unit 13 — End of Topic Tests (reviewer copy: all keys and model answers visible)

End of Topic Test: Collisions and the energy barrier & Reaction rate and its graphs & Factors that change the rate & Catalysts and reading rate experiments (u13-t1 · 5 forms × 21 items · 105 total)

David's notes for James:
full notes file
# u13-t1 — notes/queries from David's side (U13-G1)

1. FRQ COMPOSITION - frq_per_form = 1 (your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - this gate spans Topics 1-4 (KEQ-001..020), the course's QUALITATIVE kinetics block: all 20 specs carry displayEquation null, no atomKind 'quant', KEQ-006's own specNotes exclude numeric rates and KEQ-004/019 exclude kJ values (HS-PS1-4 boundary). No formula is ever substituted, so a calculation FRQ has no production routine to assess and the 5-line calc scaffold is barred everywhere. The single FRQ is homed on the capstone KEQ-020 as a TYPED 3-part rate-experiment analysis, terminal at seq 21 on every form: (a) name the ONE changed condition, (b) which run was faster + the sooner-finished observation that shows it, (c) does the result match the taught direction of that factor, and why is the comparison fair (exactly one condition differed). Rubric 30/40/30 = 100; evidence = condition lists + timed observations (no figure read - graph reading is assessed at seq 19); nothing typed needs sub/superscripts. Changed condition rotates across forms: A concentration (descaling bath), B temperature (etching cream), C piece size (dolomite chips vs powder), D catalyst (generic powder), E temperature (bluing bath); all times fresh and one-cell.
2. ALLOCATION CONSEQUENCE OF THE 20-LESSON SCOPE (confirm): D5 scaling checked and NOT applied (rule scales down only; your target_size=20 concurs), so 20 objective rows over 20 scope lessons = EXACTLY ONE objective row per lesson - the 'calc/multi-case lessons 2-3 rows' guideline is unsatisfiable at full every-lesson coverage (every-lesson-per-form wins; U5-G1/U6-G1 oversubscribed-scope precedent). Multi-case lessons cover their case space ACROSS FORMS via case rotation instead of extra rows; the capstone KEQ-020 gets a second surface via the seq-21 FRQ (its seq-19 row and the FRQ are non-adjacent).
3. SEED USAGE - 51 of your 60 pool items placed (form A carries 20, B 16, C 15; D/E are fresh-authored parallels; 17 of the 51 freshened under the strict SEED RULING), 9 pool spares, 2 prod spares. Your demo_curated trio are byte-level duplicates of pool[57]/[58]/[59] - deduped at stage 1 with both refs cited (they hold the three KEQ-020 seq-19 placements). Your 2 prod_seeds duplicate KEQ-005's IN-LESSON production-draw fluency items - student-seen, so spares only per the standing rule; the cold list is ignored per standing answer #3.
4. FORMAT ADJUDICATIONS - two KEQ seeds spared on FORMAT grounds (confirm): the KEQ-002 figure-stimulus collision seed (p5) was spared and seq 5 runs the lesson's own VERBAL classify-the-collision format on all forms (format parity across forms); the KEQ-005 image-option seed (p12) and production-draw seed (p13) were spared because image options and drawing do not survive the typed test surface - seq 7 runs the lesson's own construction-judgment TEXT format (its gate2/p14 format, and your p14 IS placed verbatim at A#7). Not merit drops.
5. BARRED-NUMBER FRESHENS UNDER THE STRICT SEED RULING (confirm; the validator computes a barred set from the lessons' own article/fluency text - 1.5, 20, 25, 30, 40, 50, 90 - and every text-visible number outside 0..12 lives in exactly one (form,seq) cell): p48@A#4 '50 kg' -> 60 kg; p49@B#4 50/20/30 g -> 45/18/35 g; p51@A#16 '30 degC wash' -> 'a cool wash' (number dropped); p58@B#19 15/25/35 degC -> 16/27/38 degC; p59@C#19 40/95 s -> 44/96 s. All fields swept for the old values (QB 3.9).
6. INSTANCE FRESHENS UNDER THE STRICT SEED RULING (confirm; structure + misconception keys preserved, all fields swept): p3@A#5 'mixture of hydrogen gas...' recast instance-free ('the same reaction mixture' - both gas names are scanned tokens and the opener was a KEQ-001 rule-line gram); p29@C#11 zinc -> tin (zinc is KEQ-005/006/008-fenced and ledgered elsewhere); p30@A#12 marble -> seashell (marble ledgered to A#17); p31@B#12 steel wool + pure oxygen -> aluminum strip in concentrated acid (own-lesson tokens; all four option classes kept); p33@A#13 marble -> chalk powder; p36@A#14 zinc -> iron filings; p40@B#15 iron -> brass railings; p41@C#15 'with oxygen' dropped; p45@A#18 sugar-dust refinery explosion KEPT (its own home cell) with 'oxygen' -> 'the other reactant' in the key (the canonical MECH-SURFACE wording) and 'the reacting gas' in the richer-air distractor; p47@C#18 marble -> limestone; B#6 zinc+hydrochloric unnamed. Your equations, option structures, and misconception targets are untouched throughout.
7. 'lead' TOKEN ADAPTATION (content, not validator - confirm the wording): 'lead' is a scanned MATERIALS token that sits inside the taught clause 'lead to a reaction' (a verb, not the metal) in five lessons' own corpus fences (KEQ-002/003/011/016/018). Rather than weaken the scan, every canonical quote in this gate uses the KEQ-002 rule-line form 'leads to a reaction', and the token 'lead' appears nowhere in the gate.
8. COURSE-WIDE INSTANCE RULE - two token-level wording fixes to confirm: (1) your KEQ-017 gate item at A#4 read 'loads 50 kg of it into a reactor' - 'reactor' is on the course-wide banned-token list (literal, token-level; a chemical-plant vessel needs no nuclear context to trip it), and it was the bank's sole occurrence; now 'into a reaction vessel' (the 50->60 kg barred-number freshen rode the same edit). Our stage-1 claim that U13 seed text was banned-lexicon-clean was WRONG on this token - caught by the blind-packet pre-send guard. (2) The packaging re-sweep (the guard's own regex, which scans 'enrich' as a substring) found ONE further hit: A#18's richer-air distractor feedback said 'you enriched the surrounding air' (plain verb, zero nuclear context; blind reviewers never saw it - feedback fields are not rendered in packets) -> 'you made the surrounding air richer', all layers, meaning unchanged. Post-fix sweep over forms_final AND this export body: ZERO banned-token hits in student-facing content (uranium/plutonium/thorium/enrich*/reactor/weapon/warhead/fissile/fission/radioactiv*/fallout all clean; the only matches anywhere are provenance 'style enriched' notes in non-student source fields and these disclosure sentences).
9. 'None of the intervals' KEPT at seq 6 (confirm): it is KEQ-004's own canonical fourth option, present in all three of the lesson's fluency items - substantive taught content (the diagram may genuinely not mark the activation-energy interval), NOT a none-of-the-above dodge. V-T5 passed it outright. No NOTA/AOTA literals exist anywhere in your 60-item pool (zero replace_distractor flags) and none were authored.
10. AUTHORING-SIDE FRQ SCHEMA DEFECTS (disclosure, both fixed): (1) the five seq-21 FRQs were first authored with one-line pointer stems ('Answer all three parts...') while the full scenario + parts (a)-(c) lived only in frq.prompt - every prior gate ships stem == frq.prompt and the blind packet renders the stem, so blind round 1 was unanswerable on all five (recorded honestly as packet-defect verdicts). Stems now carry the full prompt; re-blind cold answers matched the model answers on all 5 forms x both reviewers. (2) At packaging, the shared exporter's rubric-splitter hard-failed on all 15 rubric criteria: they were authored opening 'Part (a) - ...' while the canon tag (every prior gate; exporter contract) is the lowercase parenthetical '(part a)'. Retagged 'Part (x) ' -> '(part x) ' on all 15 criteria across all three artifact layers - grading-machinery text only; no points, observables, or student-facing field changed; the fresh package validator run is byte-identical to the stage-7 final.
11. BLIND CATCHES FIXED (confirm the D#19 rework - it touches a fresh item, not your seeds): A#11/A#14 acid-coverage under-specification made a 'larger volume' distractor defensible -> stems now grant full coverage; C#16 share-vs-absolute-count ambiguity -> 'greater share of the collisions'; D#5 'high speed' did not rigorously grant the energy check -> explicit energy grant; D#19 was a stem-echo giveaway (the stem said the catalyst went into Run 2 AND the curves were drawn labeled Run 1/Run 2) -> reworked to the lost-labels pattern: curves P/Q, task = identify Run 2's curve from the catalyst's one trace (steeper climb, same final height; figure now load-bearing); B#15/D#15 the key was the only two-part option -> a distractor gained a wrong second arm (kills the option-logic solve).
12. SOL REALISM REWORKS (authoritative QC; confirm the system swaps, all on fresh forms): C#17 wax fire-starter -> pond-treatment block vs pellets (wax melts and coalesces - surface-area comparison was chemically unsound); D#13 disposable hand warmer -> antacid tablet in dilute acid (oxygen-transport and temperature confounds); D#14 dental curing lamp -> warmed gel pack (curing lamps work by photopolymerization, not thermal rate); C#20's referenced two-curve energy diagram was never attached -> the energy-diagram-two-paths stimulus is now authored (wording distinct from D#20/E#20). FRQ fair-test premises hardened: A#21 matched scale coatings, B#21 same-bench runs, E#21 identical cabinets. Two qualifier-dependent distractors (E#2, E#11) were re-keyed to categorical misconceptions after sol flagged their 'all/soon/only' hedges.
13. GATE_CONFIG - ZERO whitelist relief for this gate: entries are the unit-13 spec path pointer plus 13 extra_materials tokens (seashell, limestone, dolomite, charcoal, sherbet, thiosulfate, vanadium, limewater, resin, moss, grime, enzymes, ore) which make the freshness scan STRICTER (each token is ledgered to exactly one (form,seq) cell) - no extra idioms, labels, convention phrases, or figure relief. The shared validator ran UNMODIFIED to 12/12 PASS, exit 0; the 12 warn lines are the accepted other-lesson-material WARN class, each recorded with a memory-answerability rationale in seed_placement.json.
14. QC LEDGER (provenance, no action needed): blind round 1 (gpt-5.6-sol blind role + gemini, all 105 items) = 200/200 key picks, 9 hard fails (5 = the FRQ pointer-stem schema defect above; 4 = the r1 defensible-distractor catches above), 72 advisory -> triaged; re-blind rounds on every changed item (fresh tags r3-r8) = 44/44 key picks, 0 hard fails - blind totals 244/244. FRQ manual grading recorded per form with justifications (round-1 = packet-defect record; post-fix both reviewers match all 5 forms; post-premise-repair re-grades match with the confound flags gone). sol authoritative loops: 15 FAIL -> 2 -> 0; final ledger 105/105 pass with 3 recorded honest-survivor style minors (A#15 key states the activation-energy arm only with keyExplanation carrying both; C#15 weak-but-wrong D2; C#17 D2/D3 style notes - none affects key uniqueness).
15. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings anywhere - the topic is qualitative collision-theory kinetics on classroom-safe systems (descaling acid, antacid tablets, pond-treatment pellets, gel packs, sugar-dust explosion hazard exactly as the lesson teaches it); every scanned material token and every visible number outside 0..12 lives in exactly one (form,seq) cell gate-wide.

Form A

KEQ-001 · easy · recall · seq 1 · seed:U13-G1 pool[0] VERBATIM · mcq4
A student says, 'Once two gases are in the same flask, their particles react wherever they are — meeting is not required.' Which correction fixes the claim?
Particles can only react when they collide with enough energy and in the right orientation. KEY
The student is right for gases, because gas particles share their energy through the whole flask.
Meeting is required, and every meeting between reactant particles produces a reaction.
Meeting is required, but only between particles of the same substance.
KEQ-003 · easy · recall · seq 2 · seed:U13-G1 pool[6]; 'lead'->'leads' token adaptation recorded · mcq4
A student writes, 'Activation energy is the energy a reaction gives off as it happens.' Which correction fixes the definition?
Activation energy is the minimum energy colliding particles need before their collision leads to a reaction. KEY
The student is close — activation energy is the energy given off by the very first collision only.
Activation energy is the total energy of all the reactant particles added together.
Activation energy is the energy needed to keep a reaction going once it has started.
KEQ-006 · easy · recall · seq 3 · seed:U13-G1 pool[15] VERBATIM · mcq4
Two reactions each fill a jar with gas. Reaction 1 fills its jar in one minute; Reaction 2 takes a full day. A student says Reaction 2 has the higher rate because it ran for longer. Which correction is right?
Rate means how quickly the gas formed — Reaction 1 filled its jar in a minute, so its rate is higher. KEY
The student is right — the longer a reaction runs, the higher its rate.
The rates are equal, because both jars ended up holding the same amount of gas.
Neither reaction has a rate, because rate only applies to reactions that stay unfinished.
KEQ-017 · easy · recall · seq 4 · seed:U13-G1 pool[48]; 50 kg -> 60 kg (barred spec number 50) · mcq4
Vanadium(V) oxide speeds up a key reaction in sulfuric acid manufacture. A plant loads 60 kg of it into a reaction vessel. After a year of continuous running, roughly how much vanadium(V) oxide should remain, and why?
About 60 kg — a catalyst is not used up by the reaction it speeds up. KEY
Far less than 60 kg — a year of reactions gradually consumes the catalyst.
More than 60 kg — product builds up inside the catalyst and adds to its mass.
Almost none — a catalyst converts itself into product as it works.
KEQ-002 · medium · application · seq 5 · seed:U13-G1 pool[3]; named gases dropped (scanned tokens; 'mixture of hydrogen gas' is a KEQ-001 rule-line gram) · mcq4
Two collisions happen in the same reaction mixture. Collision 1: fast, the molecules lined up side by side — the right orientation for this reaction. Collision 2: equally fast, but the molecules hit end-to-end. Which classification is correct?
Collision 1 is successful; Collision 2 is unsuccessful. KEY
Both collisions are successful, because both are fast.
Both collisions are unsuccessful, because most collisions fail.
Collision 1 is unsuccessful; Collision 2 is successful.
KEQ-004 · medium · application · seq 6 · seed:U13-G1 pool[9] VERBATIM (style enriched to narrate interval endpoints) · mcq4
The figure shows the energy diagram for the decomposition of mercury(II) oxide, an endothermic reaction, with three marked intervals. Which interval shows the activation energy?
FIGURE (energy-diagram-barrier · stimulus) — reaction: "decomposition of mercury(II) oxide (endothermic)" · reactantLevel: "low" · productLevel: "higher than reactants" · intervals: [{"label": "J", "from": "bottom axis", "to": "top of the barrier"}, {"label": "K", "from": "reactants' level", "to": "top of the barrier"}, {"label": "L", "f… · style: "clean single-hump energy diagram with unnumbered axes (vertical axis 'energy', horizontal axis 'progress of the reaction'); the curve starts at a low reacta…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-barrier",
  "reaction": "decomposition of mercury(II) oxide (endothermic)",
  "reactantLevel": "low",
  "productLevel": "higher than reactants",
  "intervals": [
   {
    "label": "J",
    "from": "bottom axis",
    "to": "top of the barrier"
   },
   {
    "label": "K",
    "from": "reactants' level",
    "to": "top of the barrier"
   },
   {
    "label": "L",
    "from": "reactants' level",
    "to": "products' level"
   }
  ],
  "style": "clean single-hump energy diagram with unnumbered axes (vertical axis 'energy', horizontal axis 'progress of the reaction'); the curve starts at a low reactants' level, rises over one smooth hump, and ends at a products' level higher than the reactants' level; three labeled vertical interval arrows: J rises from the bottom axis to the top of the hump, K rises from the reactants' level to the top of the hump, L rises from the reactants' level to the products' level"
 }
}
Interval K KEY
Interval J
Interval L
None of the intervals
KEQ-005 · medium · application · seq 7 · seed:U13-G1 pool[14] VERBATIM · mcq4
On an endothermic reaction's energy diagram, how does the activation-energy arrow compare with the gap between the reactants' level and the products' level?
The arrow is taller — it climbs past the products' level to the top of the barrier. KEY
The arrow is exactly as tall as the gap, because the barrier top and the products' level coincide.
The arrow is shorter than the gap, because the barrier top sits between the two levels.
The two cannot be compared, because they sit on different parts of the diagram.
KEQ-007 · medium · application · seq 8 · seed:U13-G1 pool[18] VERBATIM (style enriched with both curves' directions) · mcq4
The figure shows the two curves, labeled 1 and 2, for a reaction with one reactant and one product. Which statement matches the graph?
FIGURE (concentration-time-graph · stimulus) — reaction: "a reaction with one reactant and one product, unnamed" · curves: [{"label": "1", "shape": "rising", "start": "zero", "end": "high"}, {"label": "2", "shape": "falling", "start": "high", "end": "near zero"}] · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only 1 and 2: curve 1 rises from zero to a high level, curve 2 falls fro…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "a reaction with one reactant and one product, unnamed",
  "curves": [
   {
    "label": "1",
    "shape": "rising",
    "start": "zero",
    "end": "high"
   },
   {
    "label": "2",
    "shape": "falling",
    "start": "high",
    "end": "near zero"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only 1 and 2: curve 1 rises from zero to a high level, curve 2 falls from a high start to near zero"
 }
}
Curve 2 shows the reactant and curve 1 shows the product. KEY
Curve 1 shows the reactant and curve 2 shows the product.
Curves 1 and 2 both show the reactant at different measuring times.
The graph cannot show which curve is which without the substances' names.
KEQ-008 · medium · application · seq 9 · seed:U13-G1 pool[21] VERBATIM (style enriched) · mcq4
The figure shows the reactant curves for two runs of the same solution reaction, labeled Run E and Run F. Which run is slower?
FIGURE (concentration-time-graph · stimulus) — reaction: "one solution reaction, two runs, reactant tracked" · curves: [{"label": "Run E", "shape": "falling", "steepness": "steep", "start": "high"}, {"label": "Run F", "shape": "falling", "steepness": "shallow", "start": "high… · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two falling curves from the same high start, labeled Run E and Run F: Run E falls steeply, …
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one solution reaction, two runs, reactant tracked",
  "curves": [
   {
    "label": "Run E",
    "shape": "falling",
    "steepness": "steep",
    "start": "high"
   },
   {
    "label": "Run F",
    "shape": "falling",
    "steepness": "shallow",
    "start": "high"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two falling curves from the same high start, labeled Run E and Run F: Run E falls steeply, Run F falls shallowly"
 }
}
Run F KEY
Run E
Both runs are equally slow.
The graph cannot show which run is slower.
KEQ-009 · medium · application · seq 10 · seed:U13-G1 pool[24] VERBATIM (style enriched) · mcq4
The figure shows the product's curve for one run. A student says, 'The reaction is fastest just before the curve turns flat, because the amount of product is largest there.' Which correction is right?
FIGURE (concentration-time-graph · stimulus) — reaction: "one run, product tracked" · curves: [{"label": "product", "shape": "rising", "profile": "steep at first, shallowing, flat at the end", "start": "zero", "end": "high"}] · curveCount: 1 · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': it climbs steeply at first, shallows through the middle…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one run, product tracked",
  "curves": [
   {
    "label": "product",
    "shape": "rising",
    "profile": "steep at first, shallowing, flat at the end",
    "start": "zero",
    "end": "high"
   }
  ],
  "curveCount": 1,
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': it climbs steeply at first, shallows through the middle, and runs exactly flat at the end at a high level"
 }
}
The rate is in the steepness — the curve is steepest at the start, so the reaction is fastest at the start. KEY
The student is right — the more product present, the faster the reaction is running.
The student picked the wrong end — the reaction is fastest exactly when the curve turns flat.
The reaction runs at one rate for the whole run, so no moment is fastest.
KEQ-010 · medium · application · seq 11 · seed:U13-G1 pool[27] VERBATIM · mcq4
A strip of magnesium ribbon lies fully covered by hydrochloric acid, with acid to spare. Which single change makes the reaction faster, with everything else unchanged?
Swapping the acid for a more concentrated acid. KEY
Swapping the acid for a more dilute acid.
Using a larger volume of the same acid.
Using half the volume of the same acid.
KEQ-011 · medium · application · seq 12 · seed:U13-G1 pool[30]; marble -> seashell (marble's one cell is (17,A)); blocking-layer distractor recast without the water token · mcq4
A seashell reacts more slowly in dilute acid than in concentrated acid. Explain the slower reaction, using collisions.
Fewer acid particles in the same space means collisions happen less often, so fewer succeed each second. KEY
In dilute acid, the collisions that happen are all unsuccessful.
In dilute acid, the acid particles move more slowly.
In dilute acid, the added liquid forms a layer that blocks the shell, so no acid particles can reach it at all.
KEQ-012 · medium · application · seq 13 · seed:U13-G1 pool[33]; marble chips -> a strip of chalk (marble's one cell is (17,A)); chalk home (13,A) · mcq4
The product curve for a strip of chalk reacting with acid turns flat after ten minutes and stays flat. In collision terms, why does no more product form?
A reactant has been used up, so collisions between the reactants' particles no longer happen. KEY
The product has filled all the space in the flask, leaving no room to react.
Collisions continue as often as before, but they stopped succeeding.
The reaction is resting and will speed up again later.
KEQ-013 · medium · application · seq 14 · seed:U13-G1 pool[36]; zinc -> iron nails (zinc's one cell is (6,B)); iron home (14,A) · mcq4
Iron nails sit completely under the surface of dilute acid in a flask, with plenty of depth to spare. Which single change makes the reaction faster WITHOUT changing the acid's concentration?
Warming the flask. KEY
Cooling the flask.
Using a larger volume of the same acid.
Sealing the flask so no gas escapes.
KEQ-014 · medium · application · seq 15 · seed:U13-G1 pool[39] VERBATIM · mcq4
A student explains why a warm mixture reacts quickly: 'Heating lowers the activation energy, so more collisions can clear it.' Which correction is right?
Heating does not change the activation energy — it speeds particles up, so more collisions clear the same barrier. KEY
The student's explanation is correct as written.
Heating actually raises the activation energy, but the particles speed up by even more than the barrier rises.
Heating removes the need for collisions altogether.
KEQ-018 · medium · application · seq 16 · seed:U13-G1 pool[51]; '30 °C wash' (x2) -> 'a cool wash' (30 is a barred spec number); 'wash water' -> 'wash liquid' (water-token ban) · mcq4
Enzymes in laundry detergent speed up the reactions that break down food stains, even in a cool wash. Why can those reactions run quickly at such a low temperature?
The enzymes provide paths with lower activation energies, so many collisions have enough energy to react. KEY
The enzymes warm the wash liquid around each stain until the break-down reactions run hot.
The enzymes make the stain particles collide far more often than they otherwise would.
The enzymes dissolve the stains directly, without any reaction taking place.
KEQ-015 · hard · transfer · seq 17 · seed:U13-G1 pool[43] VERBATIM · mcq4
Acid rain slowly attacks marble. Two marble pieces of equal mass stand outdoors: a plain solid sphere, and a statue carved with thin, lacy details. Predict which piece loses a gram of marble to the acid sooner.
The carved statue — its thin details give the same mass more surface area. KEY
The sphere — compact shapes react faster than thin ones.
Both lose marble equally fast, because their masses are equal.
Neither, because marble does not react with acid rain.
KEQ-016 · hard · transfer · seq 18 · seed:U13-G1 pool[45]; 'collisions with oxygen' -> the canonical 'collisions with the other reactant' and 'richer in oxygen' -> 'richer in the reacting gas' (oxygen is a KEQ-016 own-fence material) · mcq4
In 2008, sugar dust suspended in the air of a refinery exploded, though the same sugar burns only slowly as a solid pile. Which explanation is correct?
The dust's tiny pieces expose more sugar particles at surfaces, where collisions with the other reactant can happen. KEY
Floating dust is much hotter than piled sugar, and hotter particles collide with more energy, so the dust burns faster.
Grinding sugar into dust changes it into a new, more explosive substance.
The air between dust grains is richer in the reacting gas than the air above a sugar pile.
KEQ-020 · hard · transfer · seq 19 · seed:U13-G1 pool[57] VERBATIM (= demo_curated[0]; style enriched) · mcq4
Two runs of the same reaction differ only in acid concentration. On the graph, curve A is steeper than curve B, and both level off at the same height. A student concludes that run A made more product. What do the curves actually show?
FIGURE (concentration-time-two-runs · stimulus) — description: "concentration of product against time; curve A rises steeply and levels off early; curve B rises gently and levels off later; both flatten at exactly the sa… · curves: [{"label": "A", "shape": "rising", "steepness": "steep", "start": "zero", "end": "levels off early at the shared final height"}, {"label": "B", "shape": "ris… · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled A and B in-diagram: A climbs steeply and turns flat ea…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-two-runs",
  "description": "concentration of product against time; curve A rises steeply and levels off early; curve B rises gently and levels off later; both flatten at exactly the same final height",
  "curves": [
   {
    "label": "A",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero",
    "end": "levels off early at the shared final height"
   },
   {
    "label": "B",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero",
    "end": "levels off later at the same final height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled A and B in-diagram: A climbs steeply and turns flat early, B climbs gently and turns flat later, and both flat sections sit at exactly the same height"
 }
}
Both runs ended at the same final product concentration; run A simply got there faster. KEY
Run A made more product, because its curve climbed more steeply.
Run B made more product, because its curve was still rising after curve A had flattened.
Neither run finished, because the curves stop climbing before the graph ends.
KEQ-019 · hard · transfer · seq 20 · seed:U13-G1 pool[54] VERBATIM (style enriched) · mcq4
A student adds a 'with catalyst' curve to an energy diagram. The figure shows the student's sketch: the new curve has a lower peak, but it also ends at a lower product level than the original curve. What is wrong with the sketch?
FIGURE (energy-diagram-two-paths · stimulus) — description: "an energy diagram with an original tall curve from the reactants' level to the products' level, plus a student-drawn second curve with a lower peak that des… · style: "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); a solid curve rises from the reactants' level over a tall …
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-two-paths",
  "description": "an energy diagram with an original tall curve from the reactants' level to the products' level, plus a student-drawn second curve with a lower peak that descends past the products' level and ends visibly below it",
  "style": "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); a solid curve rises from the reactants' level over a tall peak and ends at the products' level; a dashed curve labeled 'with catalyst (student sketch)' rises from the same reactants' level over a lower peak, then descends past the products' level and ends visibly below it"
 }
}
The catalyzed curve must end at the original products' level — a catalyst lowers only the barrier. KEY
The catalyzed curve's peak should be higher than the original peak, not lower.
Nothing is wrong — a catalyst lowers every part of the diagram, products included.
The catalyzed curve should start below the reactants' level to show the saved energy.
KEQ-020 · medium · application · seq 21 · authored fresh (stage 2); concentration case (times 4/21 min, home (21,A)) · frq
A technician strips hard scale from two identical heating elements — each carrying the same amount of scale, spread the same way — in an acid bath. Run 1 — bath liquid: the descaling acid at full strength; element: standard size; bath warmth: room level; nothing else added. The scale was fully dissolved after 4 minutes. Run 2 — bath liquid: the same descaling acid mixed to half strength; element: standard size; bath warmth: room level; nothing else added. The scale was fully dissolved after 21 minutes.
A technician strips hard scale from two identical heating elements — each carrying the same amount of scale, spread the same way — in an acid bath. Run 1 — bath liquid: the descaling acid at full strength; element: standard size; bath warmth: room level; nothing else added. The scale was fully dissolved after 4 minutes. Run 2 — bath liquid: the same descaling acid mixed to half strength; element: standard size; bath warmth: room level; nothing else added. The scale was fully dissolved after 21 minutes. (a) Name the ONE condition that changed between the runs. (b) State which run was faster, and cite the recorded observation that shows it. (c) Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Part a. Name the ONE condition that changed between the runs.
Model: The concentration of the descaling acid — Run 1 used it at full strength, Run 2 at half strength. Everything else in the two runs matched.
Part b. State which run was faster, and cite the recorded observation that shows it.
Model: Run 1 was faster. Both runs finished the same job — dissolving the scale — and Run 1 finished it in 4 minutes while Run 2 needed 21 minutes. The sooner-finished run is the faster run.
Part c. Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Model: Yes. The taught direction is that a higher reactant concentration gives a faster reaction, and the full-strength run was the faster one. The comparison is fair because exactly one condition changed between the runs — the acid's concentration — while the elements, their matched scale coatings, the warmth, and everything else stayed the same.
Full model answer. (a) The concentration of the descaling acid — Run 1 used it at full strength, Run 2 at half strength. Everything else in the two runs matched. (b) Run 1 was faster. Both runs finished the same job — dissolving the scale — and Run 1 finished it in 4 minutes while Run 2 needed 21 minutes. The sooner-finished run is the faster run. (c) Yes. The taught direction is that a higher reactant concentration gives a faster reaction, and the full-strength run was the faster one. The comparison is fair because exactly one condition changed between the runs — the acid's concentration — while the elements, their matched scale coatings, the warmth, and everything else stayed the same.

Form B

KEQ-001 · easy · recall · seq 1 · seed:U13-G1 pool[1] VERBATIM · mcq4
Two reactant molecules collide at high speed, but they hit lined up the wrong way. Does a reaction happen?
No — a reacting collision needs enough energy and the right orientation, and this one has the wrong orientation. KEY
Yes — a fast enough collision reacts however the molecules are lined up, because speed makes up for a poor line-up.
Yes — every collision between two reactant molecules leads to a reaction.
No — molecules can only react when they collide gently rather than at high speed.
KEQ-003 · easy · recall · seq 2 · seed:U13-G1 pool[8]; 'lead to' phrasing swapped for the safe token (recorded) · mcq4
In one collision, two reactant particles hit carrying less than the reaction's activation energy. In another, two identical particles hit carrying more than it. What is the difference in outcome?
Only the second collision can react; the first leaves its particles unchanged. KEY
Both collisions react, but the first one reacts more slowly than the second.
Neither collision reacts, because a reaction also needs a spark.
Both collisions react, and the second simply releases more energy.
KEQ-006 · easy · recall · seq 3 · seed:U13-G1 pool[16] VERBATIM · mcq4
Which observation is a statement about a reaction's rate?
The limewater turned cloudy within five seconds. KEY
The reaction warmed the flask as it ran.
The mixture turned from colorless to deep blue during the run.
The flask ended up holding two spoonfuls of white solid.
KEQ-017 · easy · recall · seq 4 · seed:U13-G1 pool[49]; masses 50/20/30 g -> 45/18/35 g (barred spec numbers) · mcq4
A reaction mixture is analyzed before and after the reaction. Substance P: 45 g before, 45 g after, and the reaction runs much faster when P is present. Substance Q: 18 g before, 0 g after. Substance R: 0 g before, 15 g after. Substance S: 35 g before, 35 g after, and the reaction runs at the same speed with or without S. Which substance is a catalyst?
Substance P KEY
Substance Q
Substance R
Substance S
KEQ-002 · medium · application · seq 5 · seed:U13-G1 pool[4] VERBATIM · mcq4
A student says, 'If a collision carries enough energy, it is always successful.' Which correction fixes the claim?
A successful collision needs enough energy AND the right orientation — an energetic collision in the wrong orientation is unsuccessful. KEY
The claim is right, because an energetic collision always knocks the molecules into the right orientation on impact.
The claim is backwards — orientation is the only thing that decides a collision's outcome.
The claim is right for very fast collisions — a hard enough hit reacts no matter how the molecules are lined up.
KEQ-004 · medium · application · seq 6 · seed:U13-G1 pool[11]; zinc + hydrochloric acid UNNAMED (the pair is KEQ-005's fluency-figure instance - spec-gram collision; the reaction plays no role in the read) · mcq4
The figure shows the energy diagram for an energy-releasing reaction. A student has labeled the drop from the top of the barrier to the products' level as 'activation energy'. Which correction is right?
FIGURE (energy-diagram-barrier · stimulus) — reaction: "an energy-releasing reaction (unnamed)" · reactantLevel: "high" · productLevel: "lower than reactants" · intervals: [{"label": "activation energy (student's label)", "from": "top of the barrier", "to": "products' level"}] · style: "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' leve…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-barrier",
  "reaction": "an energy-releasing reaction (unnamed)",
  "reactantLevel": "high",
  "productLevel": "lower than reactants",
  "intervals": [
   {
    "label": "activation energy (student's label)",
    "from": "top of the barrier",
    "to": "products' level"
   }
  ],
  "style": "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' level, rises over one smooth hump, then falls to a products' level below the reactants' level; one labeled vertical arrow sits on the downhill side, running from the top of the hump down to the products' level, carrying the student's label 'activation energy'"
 }
}
The activation energy is the climb from the reactants' level to the top of the barrier, not the drop on the far side. KEY
The label is right, because energy always runs downhill in an exothermic reaction.
The label belongs on the vertical gap between the reactants' level and the products' level, not on the drop.
The label belongs on the whole curve, from its start to its finish.
KEQ-005 · medium · application · seq 7 · authored fresh (stage 2); rule-line-0 start-point family · mcq4
A student is about to draw the activation-energy arrow on an energy diagram. At which height should the arrow start?
At the reactants' level. KEY
At the bottom axis.
At the products' level.
At the top of the barrier.
KEQ-007 · medium · application · seq 8 · seed:U13-G1 pool[19] VERBATIM · mcq4
During a run of a reaction, one substance's concentration rises steadily from zero. Nothing is added to the mixture after the start. What is that substance?
A product — the reaction is making it, so its concentration grows. KEY
A reactant — the reaction is using it, so its concentration grows.
A reactant that the experimenter keeps topping up.
It could be either — the direction a concentration changes says nothing about reactant or product.
KEQ-008 · medium · application · seq 9 · seed:U13-G1 pool[22] VERBATIM · mcq4
A student compares two runs on one graph and says, 'Run 2 is faster, because its curve is higher up the graph at the end.' Which correction is right?
Faster shows as steeper, not higher — compare how quickly the curves change, not where they end. KEY
The student is right — the higher a curve ends, the faster its run.
The student is almost right — the higher curve is faster, but only when both curves rise.
Neither height nor steepness can show which run is faster.
KEQ-009 · medium · application · seq 10 · seed:U13-G1 pool[25] VERBATIM (style enriched; Unicode subscripts) · mcq4
The figure marks three times on a reactant's curve: t₁, t₂, and t₃. At which marked time has the reaction already stopped?
FIGURE (concentration-time-graph · stimulus) — reaction: "one run, reactant tracked" · curves: [{"label": "reactant", "shape": "falling", "profile": "steep at first, shallowing, flat at the end", "start": "high", "end": "near zero"}] · markedTimes: [{"label": "t₁", "position": "steep early section"}, {"label": "t₂", "position": "shallowing middle section"}, {"label": "t₃", "position": "flat final sectio… · curveCount: 1 · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; one falling curve labeled 'reactant': steep fall at first, shallowing through the middle, e…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one run, reactant tracked",
  "curves": [
   {
    "label": "reactant",
    "shape": "falling",
    "profile": "steep at first, shallowing, flat at the end",
    "start": "high",
    "end": "near zero"
   }
  ],
  "markedTimes": [
   {
    "label": "t₁",
    "position": "steep early section"
   },
   {
    "label": "t₂",
    "position": "shallowing middle section"
   },
   {
    "label": "t₃",
    "position": "flat final section"
   }
  ],
  "curveCount": 1,
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; one falling curve labeled 'reactant': steep fall at first, shallowing through the middle, exactly flat at the end near zero; three vertical dashed markers on the time axis: t₁ in the steep early section, t₂ in the shallowing middle, t₃ in the flat final section"
 }
}
t₃ KEY
t₁
t₂
None of the marked times
KEQ-010 · medium · application · seq 11 · seed:U13-G1 pool[28] VERBATIM ('added water' -> 'added liquid': gate-wide water-token ban) · mcq4
A student says, 'Diluting the acid will not change the rate, because it is still the same acid.' What actually happens to the rate when the acid is diluted, everything else unchanged?
The reaction slows down — the reacting acid is now at a lower concentration. KEY
Nothing changes, just as the student says.
The reaction speeds up, because the added liquid helps the acid flow.
The reaction stops, because diluted acid cannot react.
KEQ-011 · medium · application · seq 12 · seed:U13-G1 pool[31]; steel wool + pure oxygen (own-lesson materials) -> aluminum strip in concentrated acid; all four option classes kept; aluminum home (12,B) · mcq4
An aluminum strip reacts faster in concentrated acid than in dilute acid. A student explains: 'Each collision in the concentrated acid carries more energy.' Which correction is right?
The collisions are not harder — there are simply more of them per second, so more succeed each second. KEY
The student is right — packing in more acid particles makes every collision more energetic.
The collisions are both harder and more frequent in the concentrated acid.
The concentrated acid lowers the reaction's activation energy.
KEQ-012 · medium · application · seq 13 · seed:U13-G1 pool[34] VERBATIM · mcq4
A student says, 'Reactions slow down over time because the particles get tired and lose their energy.' Which correction is right?
The particles' energy is unchanged — the reactant's concentration falls, so collisions happen less often each second. KEY
The student is right — colliding wears particles down until they stop reacting.
Reactions slow down because their activation energy grows as the run goes on.
Reactions slow down because the product's particles start colliding in place of the reactants'.
KEQ-013 · medium · application · seq 14 · seed:U13-G1 pool[37] VERBATIM ('cold air is drier' -> 'cold conditions are drier': air's one cell is (18,A)); oxygen home (14,B) · mcq4
Oil paint hardens by a chemical reaction with oxygen. A student says paint in an unheated winter shed stops hardening completely until spring. Which correction is right?
The hardening reaction keeps running in the cold — only more slowly. KEY
The student is right — below room temperature, reactions cannot run at all.
The cold actually speeds the hardening, because cold conditions are drier.
The cold has no effect, because the paint's ingredients are unchanged.
KEQ-014 · medium · application · seq 15 · seed:U13-G1 pool[40]; iron railings -> brass railings (iron's one cell is (14,A)); more-oxygen distractor recast (oxygen's one cell is (14,B)); brass home (15,B) · mcq4
Brass railings tarnish faster in a hot climate than in a cold one. Which explanation gives BOTH parts of the temperature effect?
The particles collide more often, and more of their collisions carry enough energy to react. KEY
The particles collide more often, but each collision carries less energy than it would in the cold.
More collisions carry enough energy — collision energy is the only change.
In a hot climate the railing simply meets more of the tarnishing gas, not faster particles.
KEQ-018 · medium · application · seq 16 · seed:U13-G1 pool[52] VERBATIM · mcq4
A student says: 'A catalyst works by giving the reacting particles extra energy, so their collisions are harder.' Which correction is right?
The particles' energies are unchanged — the catalyst provides a path that needs less energy, so more of the existing collisions succeed. KEY
The student is right that extra energy is handed out, but the catalyst gives it only to the very fastest particles.
The catalyst removes energy from the particles, and slower collisions are the ones that react.
The catalyst does not affect energy at all — it simply pushes the reactant particles closer together.
KEQ-015 · hard · transfer · seq 17 · seed:U13-G1 pool[44] VERBATIM · mcq4
Sherbet powder fizzes on the tongue as its ingredients react. A candy maker presses the same sherbet mixture into a hard candy. Predict how the pressed candy's fizzing compares with the loose powder's.
The pressed candy fizzes more slowly, because pressing joins the small pieces into one large piece. KEY
The pressed candy fizzes faster, because pressing packs the reacting mixture closer together.
The pressed candy fizzes at the same speed, because it contains the same ingredients.
The pressed candy cannot fizz at all, because pressing destroys the reactants.
KEQ-016 · hard · transfer · seq 18 · seed:U13-G1 pool[46] VERBATIM · mcq4
Crushing a solid reactant speeds up its reaction. Which statement gives the correct reason?
Crushing exposes more of the solid's particles at surfaces, where the other reactant can collide with them. KEY
Crushing creates brand-new particles of the solid, so there is simply more reactant available to react.
Crushing makes each particle smaller, so each one reacts more quickly.
Crushing heats the solid, and the leftover warmth is what speeds the reaction.
KEQ-020 · hard · transfer · seq 19 · seed:U13-G1 pool[58] (= demo_curated[1]); 15/25/35 °C -> 16/27/38 °C (25 is a barred spec number; triple refreshed for one-cell hygiene) · mcq4
Three runs of the same reaction differ only in temperature: 16 °C, 27 °C, and 38 °C. The graph shows the three product curves, labeled only X, Y, and Z — the temperature labels were lost. Which curve is the 16 °C run?
FIGURE (concentration-time-three-runs · stimulus) — description: "concentration of product against time; curve X rises steepest and levels off first; curve Y rises less steeply; curve Z rises most gently and levels off las… · curves: [{"label": "X", "shape": "rising", "steepness": "steep", "start": "zero", "end": "levels off first at the shared final height"}, {"label": "Y", "shape": "ris… · curveCount: 3 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; three rising curves labeled X, Y, Z in-diagram: X climbs steepest and flattens f…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-three-runs",
  "description": "concentration of product against time; curve X rises steepest and levels off first; curve Y rises less steeply; curve Z rises most gently and levels off last; all three flatten at the same final height",
  "curves": [
   {
    "label": "X",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero",
    "end": "levels off first at the shared final height"
   },
   {
    "label": "Y",
    "shape": "rising",
    "steepness": "moderate",
    "start": "zero",
    "end": "levels off next at the same final height"
   },
   {
    "label": "Z",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero",
    "end": "levels off last at the same final height"
   }
  ],
  "curveCount": 3,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; three rising curves labeled X, Y, Z in-diagram: X climbs steepest and flattens first, Y climbs at a moderate steepness, Z climbs most gently and flattens last; all three flat sections sit at the same height"
 }
}
Curve Z — the coldest run is the slowest, so it has the shallowest curve. KEY
Curve X — the coldest run reacts fastest.
Curve Y — the coldest run always sits in the middle.
The curves cannot be matched to temperatures without their labels.
KEQ-019 · hard · transfer · seq 20 · seed:U13-G1 pool[55] VERBATIM (style enriched) · mcq4
The diagram shows an energy diagram whose products' level sits ABOVE its reactants' level, with two barrier curves. Which statement about the catalyzed path is correct?
FIGURE (energy-diagram-two-paths · stimulus) — description: "an energy diagram with the products' level higher than the reactants' level; a taller curve labeled 'without catalyst' and a lower-peaked curve labeled 'wit… · style: "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the products' level sits higher than the reactants' level;…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-two-paths",
  "description": "an energy diagram with the products' level higher than the reactants' level; a taller curve labeled 'without catalyst' and a lower-peaked curve labeled 'with catalyst' both run from the reactants' level up over their peaks and down to the products' level",
  "style": "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the products' level sits higher than the reactants' level; two curves labeled in-diagram: 'without catalyst' rises over a tall peak, 'with catalyst' rises over a clearly lower peak; both start at the reactants' level and end at the products' level"
 }
}
Its activation energy is still the rise from the reactants' level to its own, lower peak. KEY
Its activation energy is the rise from the products' level to its peak, since the products sit higher here.
The catalyst turns this energy-absorbing reaction into one that gives energy out.
This diagram cannot show a catalyst, because catalyzed reactions must give out energy.
KEQ-020 · medium · application · seq 21 · authored fresh (stage 2); temperature case (times 8/26 min, home (21,B)) · frq
An artist frosts two identical glass panes with the same etching cream, which reacts with the glass surface. Run 1 — cream: one full tube; pane: standard; workspace: the workshop bench on a warm afternoon; nothing else added. The frosting was complete after 8 minutes. Run 2 — cream: one full tube of the same batch; pane: standard; workspace: the same workshop bench on a chilly morning, nothing else about the space changed; nothing else added. The frosting was complete after 26 minutes.
An artist frosts two identical glass panes with the same etching cream, which reacts with the glass surface. Run 1 — cream: one full tube; pane: standard; workspace: the workshop bench on a warm afternoon; nothing else added. The frosting was complete after 8 minutes. Run 2 — cream: one full tube of the same batch; pane: standard; workspace: the same workshop bench on a chilly morning, nothing else about the space changed; nothing else added. The frosting was complete after 26 minutes. (a) Name the ONE condition that changed between the runs. (b) State which run was faster, and cite the recorded observation that shows it. (c) Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Part a. Name the ONE condition that changed between the runs.
Model: The temperature — Run 1 ran on the workshop bench on a warm afternoon and Run 2 at the same bench on a chilly morning. Everything else about the two runs matched.
Part b. State which run was faster, and cite the recorded observation that shows it.
Model: Run 1 was faster. The same frosting job was complete after 8 minutes in the warm run, while the cold run needed 26 minutes. The sooner-finished run is the faster run.
Part c. Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Model: Yes. The taught direction is that raising the temperature makes a reaction faster, and the warm run was the faster one. The comparison is fair because exactly one condition changed — the temperature — while the cream, the panes, and everything else stayed the same.
Full model answer. (a) The temperature — Run 1 ran on the workshop bench on a warm afternoon and Run 2 at the same bench on a chilly morning. Everything else about the two runs matched. (b) Run 1 was faster. The same frosting job was complete after 8 minutes in the warm run, while the cold run needed 26 minutes. The sooner-finished run is the faster run. (c) Yes. The taught direction is that raising the temperature makes a reaction faster, and the warm run was the faster one. The comparison is fair because exactly one condition changed — the temperature — while the cream, the panes, and everything else stayed the same.

Form C

KEQ-001 · easy · recall · seq 1 · authored fresh (stage 2); rule-line-4 outcome family · mcq4
Two reactant particles bump into each other so gently that the collision does not carry enough energy. What happens to them?
Both particles stay unchanged — they bounce apart. KEY
They react partly, forming one new particle between them.
They stick together and finish reacting over the next few minutes.
The collision's energy is saved up and added to their next collision.
KEQ-003 · easy · recall · seq 2 · authored fresh (stage 2); rule-line-3 at-threshold family · mcq4
A collision between two reactant particles carries exactly the reaction's activation energy — no more, no less. What is the outcome?
The collision can cause a reaction — it meets the minimum. KEY
It cannot react — a collision must carry more than the minimum, not just meet it.
The particles react, but only partly.
The particles bounce apart, with the energy saved for their next collision.
KEQ-006 · easy · recall · seq 3 · authored fresh (stage 2); definition family (fluency stem's exact wording is corpus) · mcq4
In chemistry, what does it mean to say one reaction has a higher rate than another?
Its reactant is used up more quickly, or its product forms more quickly. KEY
It ends up making a larger total amount of product than the other reaction.
It gives off a larger amount of energy than the other reaction as it runs.
With the same amounts to react, it keeps running for a longer time before it finishes.
KEQ-017 · easy · recall · seq 4 · seed:U13-G1 pool[50] VERBATIM · mcq4
A lab manager says: 'We must buy fresh nickel catalyst for every batch, because the catalyst turns into product during the reaction.' Which correction is right?
The same nickel can be reused batch after batch — a catalyst is not used up and is chemically unchanged at the end. KEY
No correction is needed — a catalyst really is converted into product as it speeds the reaction.
Fresh catalyst is needed each batch, but only because the old catalyst turns into a reactant instead.
The nickel can be reused, but each batch permanently uses up half of it.
KEQ-002 · medium · application · seq 5 · authored fresh (stage 2); fails-both case, verbal (the spared p5's case in the lesson's verbal format) · mcq4
In a reaction mixture, two reactant molecules drift together slowly and meet end-to-end. For this reaction, only a side-by-side line-up can react. Classify the collision.
Unsuccessful — it lacks enough energy and the orientation is wrong. KEY
Unsuccessful — it lacks enough energy, though the orientation is right.
Unsuccessful — the orientation is wrong, though it carries enough energy.
Successful — it passes both checks.
KEQ-004 · medium · application · seq 6 · authored fresh (stage 2); interval-read family, unnamed exothermic, letters rotated · mcq4
The figure shows the energy diagram for an energy-releasing reaction, with three marked intervals. Which interval shows the activation energy?
FIGURE (energy-diagram-barrier · stimulus) — reaction: "an energy-releasing reaction (unnamed)" · reactantLevel: "high" · productLevel: "lower than reactants" · intervals: [{"label": "P", "from": "top of the barrier", "to": "products' level"}, {"label": "Q", "from": "reactants' level", "to": "products' level"}, {"label": "R", "… · style: "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' leve…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-barrier",
  "reaction": "an energy-releasing reaction (unnamed)",
  "reactantLevel": "high",
  "productLevel": "lower than reactants",
  "intervals": [
   {
    "label": "P",
    "from": "top of the barrier",
    "to": "products' level"
   },
   {
    "label": "Q",
    "from": "reactants' level",
    "to": "products' level"
   },
   {
    "label": "R",
    "from": "reactants' level",
    "to": "top of the barrier"
   }
  ],
  "style": "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' level, rises over one smooth hump, then falls to a products' level below the reactants' level; three labeled vertical interval arrows: P runs from the top of the hump down to the products' level, Q spans from the reactants' level to the products' level, R rises from the reactants' level to the top of the hump"
 }
}
Interval R KEY
Interval P
Interval Q
None of the intervals
KEQ-005 · medium · application · seq 7 · authored fresh (stage 2); start-end pair family (the three taught mistakes as distractors) · mcq4
To add the activation energy to any energy diagram, which start-and-end pair is correct for the arrow?
Start at the reactants' level; end at the very top of the barrier. KEY
Start at the bottom axis; end at the very top of the barrier.
Start at the reactants' level; end at the products' level.
Start at the products' level; end at the very top of the barrier.
KEQ-007 · medium · application · seq 8 · authored fresh (stage 2); which-curve read, letters rotated (canonical option family) · mcq4
The figure shows the concentration-time graph for a reaction with one reactant and one product, with curves labeled P and Q. Which curve shows the product?
FIGURE (concentration-time-graph · stimulus) — reaction: "a reaction with one reactant and one product, unnamed" · curves: [{"label": "P", "shape": "falling", "start": "high", "end": "near zero"}, {"label": "Q", "shape": "rising", "start": "zero", "end": "high"}] · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only P and Q: curve P falls from a high start to near zero, curve Q rise…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "a reaction with one reactant and one product, unnamed",
  "curves": [
   {
    "label": "P",
    "shape": "falling",
    "start": "high",
    "end": "near zero"
   },
   {
    "label": "Q",
    "shape": "rising",
    "start": "zero",
    "end": "high"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only P and Q: curve P falls from a high start to near zero, curve Q rises from zero to a high level"
 }
}
Curve Q KEY
Curve P
Both curves
Neither curve
KEQ-008 · medium · application · seq 9 · seed:U13-G1 pool[23] VERBATIM (style enriched) · mcq4
The figure shows the product curves for three runs of the same reaction, labeled Runs 1, 2, and 3. Which run is fastest?
FIGURE (concentration-time-graph · stimulus) — reaction: "one reaction, three runs, product tracked" · curves: [{"label": "Run 1", "shape": "rising", "steepness": "shallow", "start": "zero"}, {"label": "Run 2", "shape": "rising", "steepness": "moderate", "start": "zer… · curveCount: 3 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; three rising curves from the same zero start, labeled Run 1, Run 2, Run 3: Run 1 climbs sha…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one reaction, three runs, product tracked",
  "curves": [
   {
    "label": "Run 1",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero"
   },
   {
    "label": "Run 2",
    "shape": "rising",
    "steepness": "moderate",
    "start": "zero"
   },
   {
    "label": "Run 3",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero"
   }
  ],
  "curveCount": 3,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; three rising curves from the same zero start, labeled Run 1, Run 2, Run 3: Run 1 climbs shallowly, Run 2 at a moderate steepness, Run 3 steeply"
 }
}
Run 3 KEY
Run 1
Run 2
All three runs have the same rate.
KEQ-009 · medium · application · seq 10 · seed:U13-G1 pool[26] VERBATIM (style enriched) · mcq4
The figure shows a product's curve for one run. Between the start of the run and the flat end, what happens to the reaction's rate?
FIGURE (concentration-time-graph · stimulus) — reaction: "one run, product tracked" · curves: [{"label": "product", "shape": "rising", "profile": "steep at first, shallowing, flat at the end", "start": "zero", "end": "high"}] · curveCount: 1 · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': steep climb at first, shallowing through the middle, ex…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one run, product tracked",
  "curves": [
   {
    "label": "product",
    "shape": "rising",
    "profile": "steep at first, shallowing, flat at the end",
    "start": "zero",
    "end": "high"
   }
  ],
  "curveCount": 1,
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': steep climb at first, shallowing through the middle, exactly flat at the end at a high level"
 }
}
It keeps falling throughout the run, from fastest at the start to zero at the flat end. KEY
It rises steadily, from slowest at the start to fastest at the end.
It stays constant, then drops to zero all at once at the end.
It rises at first, then falls back to zero.
KEQ-010 · medium · application · seq 11 · seed:U13-G1 pool[29]; zinc -> tin (zinc is a KEQ-010 own-fence material); tin home (11,C) · mcq4
Two runs of the same tin–acid reaction differ only in the acid's concentration. On a concentration-time graph of the product, which run's curve climbs more steeply?
The run with the more concentrated acid. KEY
The run with the more dilute acid.
Both curves climb equally steeply.
Neither — the product curves fall rather than climb.
KEQ-011 · medium · application · seq 12 · seed:U13-G1 pool[32] VERBATIM · mcq4
Two runs of a solution reaction differ only in reactant concentration. Run H's reactant curve falls more steeply than Run L's. Which account of the two runs is correct?
Run H used the higher concentration — more particles in the same space collide more often, giving more successful collisions per second. KEY
Run L used the higher concentration — raising a reactant's concentration slows its reaction, so the slower run is the concentrated one.
The two runs used equal concentrations, and the steepness difference happened by chance.
Run H's curve tells you its particles collided harder, not more often.
KEQ-012 · medium · application · seq 13 · seed:U13-G1 pool[35] VERBATIM · mcq4
A solution reaction runs at a steady temperature, with nothing added after the start. At which point in the run are successful collisions happening most often per second?
At the very start of the run. KEY
Midway through the run.
Just before the curve turns flat.
Successful collisions happen equally often throughout the run.
KEQ-013 · medium · application · seq 14 · seed:U13-G1 pool[38] VERBATIM · mcq4
Two runs of the same reaction differ only in temperature. On a concentration-time graph of the product, which run's curve climbs more steeply?
The warmer run's. KEY
The cooler run's.
Both climb equally steeply.
The warmer run's curve ends higher but climbs at the same steepness.
KEQ-014 · medium · application · seq 15 · seed:U13-G1 pool[41]; 'reactions with oxygen' -> 'chemical reactions' and the more-oxygen distractor recast (oxygen's one cell is (14,B)) · mcq4
Car tires harden and crack through slow chemical reactions with gases around them. Tires stored in a hot shed harden sooner than tires in a cool garage, though the same outdoor breeze drifts through both buildings. Which particle-level account is right?
In the hot shed the particles move faster — they collide more often, and more collisions carry enough energy. KEY
The hot shed holds far more of the reacting gas than the cool garage does.
Heat melts the rubber, and melting is a physical change rather than a reaction.
In the cool garage the tires' particles do not react at all, so those tires never harden or crack.
KEQ-018 · medium · application · seq 16 · seed:U13-G1 pool[53] VERBATIM; copper home (16,C) · mcq4
A reaction can happen two ways: by its ordinary path, or by a path available only on a copper surface, which has a lower activation energy. For the same mixture at the same temperature, on which path does a greater share of the collisions carry enough energy to react?
The copper-surface path — its lower activation energy means weaker collisions can still succeed on it. KEY
The ordinary path — particles are free to collide anywhere, not just at the copper.
Both paths equally — the mixture and temperature are the same for both.
The copper-surface path — copper transfers extra energy into every collision that happens on it.
KEQ-015 · hard · transfer · seq 17 · authored fresh (stage 2); breaking-direction prediction; sol loop-2 system swap to pond-treatment pellets (previous instance melts, realism major); pellets home (17,C) · mcq4
A pond-treatment chemical reacts away as it cleans a garden pond. It is sold two ways: as one solid block, or as the same mass of small pellets. Predict which form is used up by its reaction sooner.
The pellets — small pieces give the same mass far more surface area, so they react away faster. KEY
The block — one large piece treats the pond faster than many small pieces.
Both together — the same mass of the same substance always reacts away in the same time.
Neither form reacts — pressing the chemical into a block or into pellets stops it from working.
KEQ-016 · hard · transfer · seq 18 · seed:U13-G1 pool[47]; marble -> limestone (marble's one cell is (17,A)); key's 'marble particles' -> 'of them' · mcq4
A single limestone sphere loses mass to acid more slowly than an equal mass of small limestone chips in the same acid. Which explanation is correct?
The sphere keeps more of its particles buried, so the acid has fewer of them to collide with. KEY
The sphere's particles are packed more tightly than the chips' particles, so the acid cannot squeeze between them.
The chips make the acid stronger by spreading through it.
The sphere reacts more slowly because its curved surface deflects the acid's particles.
KEQ-020 · hard · transfer · seq 19 · seed:U13-G1 pool[59] (= demo_curated[2]); 40/95 s -> 44/96 s (40 is a barred spec number); sodium home (19,C) · mcq4
Sodium thiosulfate solution slowly turns cloudy when acid is added, and a cross drawn under the flask disappears from view once enough cloudiness forms. Two runs were alike except for the thiosulfate solution used: run A's solution was more concentrated. Run A hid the cross in 44 seconds; run B took 96 seconds. Which statement do the results support?
The more concentrated solution reacted faster — it hid the cross sooner. KEY
Run B reacted faster, because its reaction kept going for more than twice as long.
The runs were equally fast, because both crosses eventually disappeared.
Concentration made no difference, because only temperature can change a reaction's rate.
KEQ-019 · hard · transfer · seq 20 · seed:U13-G1 pool[56] VERBATIM · mcq4
Two students read the same two-curve energy diagram. Ana says the catalyzed path's lower peak means the reaction gives out extra energy. Ben says the energy given out is unchanged. Who is right, and why?
FIGURE (energy-diagram-two-paths · stimulus) — description: "an energy diagram with two curves from the same reactants' level to the same products' level: a solid curve over a tall peak, and a broken-line curve marked… · style: "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); a solid curve rises from a high reactants' level over a ta…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-two-paths",
  "description": "an energy diagram with two curves from the same reactants' level to the same products' level: a solid curve over a tall peak, and a broken-line curve marked 'catalyst path' over a clearly lower peak; the reactants' level sits above the products' level and both curves end together",
  "style": "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); a solid curve rises from a high reactants' level over a tall peak and descends to a lower products' level; a second curve drawn as a broken line and marked 'catalyst path' leaves the same reactants' level, crosses a clearly lower peak, and lands on exactly the same products' level"
 }
}
Ben — the energy given out is read from the level gap, and neither level moves when a catalyst is added. KEY
Ana — a lower peak means less energy is swallowed by the barrier, so more is left over to give out.
Ana — the peak height is what sets how much energy a reaction gives out.
Ben — but only because the catalyst quietly gives out energy of its own to make up the difference.
KEQ-020 · medium · application · seq 21 · authored fresh (stage 2); piece-size case (dolomite, unscanned; times 6/19 min, home (21,C)) · frq
A student reacts dolomite with acid twice. Run 1 — solid: fist-sized dolomite chips; acid: one beaker at standard strength; bench temperature; nothing else added. The solid was gone after 19 minutes. Run 2 — solid: the same mass of dolomite ground to a powder; acid: a second beaker, same strength and amount; bench temperature; nothing else added. The solid was gone after 6 minutes.
A student reacts dolomite with acid twice. Run 1 — solid: fist-sized dolomite chips; acid: one beaker at standard strength; bench temperature; nothing else added. The solid was gone after 19 minutes. Run 2 — solid: the same mass of dolomite ground to a powder; acid: a second beaker, same strength and amount; bench temperature; nothing else added. The solid was gone after 6 minutes. (a) Name the ONE condition that changed between the runs. (b) State which run was faster, and cite the recorded observation that shows it. (c) Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Part a. Name the ONE condition that changed between the runs.
Model: The piece size of the solid — Run 1 used fist-sized chips and Run 2 used the same mass ground to a powder. Everything else about the two runs matched.
Part b. State which run was faster, and cite the recorded observation that shows it.
Model: Run 2 was faster. The same mass of solid was gone after 6 minutes in the powder run, while the chips needed 19 minutes. The sooner-finished run is the faster run.
Part c. Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Model: Yes. The taught direction is that breaking a solid reactant into smaller pieces makes its reaction faster, and the powder run was the faster one. The comparison is fair because exactly one condition changed — the piece size — while the mass, the acid, the temperature, and everything else stayed the same.
Full model answer. (a) The piece size of the solid — Run 1 used fist-sized chips and Run 2 used the same mass ground to a powder. Everything else about the two runs matched. (b) Run 2 was faster. The same mass of solid was gone after 6 minutes in the powder run, while the chips needed 19 minutes. The sooner-finished run is the faster run. (c) Yes. The taught direction is that breaking a solid reactant into smaller pieces makes its reaction faster, and the powder run was the faster one. The comparison is fair because exactly one condition changed — the piece size — while the mass, the acid, the temperature, and everything else stayed the same.

Form D

KEQ-001 · easy · recall · seq 1 · authored fresh (stage 2); requirement-statement family (flu0 class, fresh wording) · mcq4
Two reactant gases fill the same container. Before any new product can form, what must their particles do?
Collide with enough energy and in the right orientation. KEY
Simply drift near each other — closeness is all a reaction needs.
Touch each other — any contact between the two kinds makes product.
Exchange energy from a distance, without touching.
KEQ-003 · easy · recall · seq 2 · authored fresh (stage 2); rule-line-2 below-threshold family (fresh wording; the fluency stem's phrasing is corpus) · mcq4
During a run, two reactant particles collide, but the collision falls short of the reaction's activation energy. What is the outcome?
Both particles stay unchanged. KEY
The particles react, but more slowly than usual.
The particles form product that carries less energy than usual.
The shortfall is taken from nearby particles so the reaction can finish.
KEQ-006 · easy · recall · seq 3 · authored fresh (stage 2); time-vs-amount comparison family · mcq4
Two sealed flasks hold identical mixtures, and in both the reaction runs to the finish. Flask 1 finishes within an afternoon; Flask 2 needs a whole month. Which statement is right?
Flask 1's reaction had the higher rate — its reactant was used up more quickly. KEY
Flask 2's reaction had the higher rate, because it kept reacting for far longer.
The rates were equal, because both flasks made the same amount of product.
Neither run had a rate, because nothing was measured while the reactions ran.
KEQ-017 · easy · recall · seq 4 · authored fresh (stage 2); rule-line-3 mass read (generic powder) · mcq4
A powder added to a reaction acts as a catalyst. Its mass is measured before the reaction starts and again after the reaction ends. What does the second measurement show?
The same mass as before — a catalyst is not used up. KEY
About half the starting mass.
Almost nothing — the powder is gone once the reaction finishes.
More than before — the powder gains mass as product settles onto it.
KEQ-002 · medium · application · seq 5 · authored fresh (stage 2); passes-both case, verbal · mcq4
In a hot reaction mixture, two reactant molecules collide with more than enough energy for this reaction, lined up exactly the way this reaction needs. Classify the collision.
Successful — it carries enough energy and has the right orientation. KEY
Unsuccessful — it lacks enough energy.
Unsuccessful — the orientation is wrong.
Cannot be classified — whether a collision reacts is a matter of pure chance.
KEQ-004 · medium · application · seq 6 · authored fresh (stage 2); wrong-label family, gap variant (the lesson's fixed misread space) · mcq4
On the energy diagram for an energy-absorbing reaction, a student labels the vertical gap between the reactants' level and the products' level as 'activation energy'. The figure shows the labeled diagram. Which correction is right?
FIGURE (energy-diagram-barrier · stimulus) — reaction: "an energy-absorbing reaction (unnamed)" · reactantLevel: "low" · productLevel: "higher than reactants" · intervals: [{"label": "activation energy (student's label)", "from": "reactants' level", "to": "products' level"}] · style: "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a low reactants' level…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-barrier",
  "reaction": "an energy-absorbing reaction (unnamed)",
  "reactantLevel": "low",
  "productLevel": "higher than reactants",
  "intervals": [
   {
    "label": "activation energy (student's label)",
    "from": "reactants' level",
    "to": "products' level"
   }
  ],
  "style": "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a low reactants' level, rises over one smooth hump that peaks above both levels, and ends at a products' level higher than the reactants' level; one labeled vertical arrow spans from the reactants' level up to the products' level, carrying the student's label 'activation energy'"
 }
}
The activation energy is the climb from the reactants' level to the top of the barrier, not the gap between the two levels. KEY
The label is right — on an energy-absorbing diagram, the gap and the activation energy are the same interval.
The label belongs on the drop from the top of the barrier to the products' level.
The label belongs on the whole curve, from its start to its finish.
KEQ-005 · medium · application · seq 7 · authored fresh (stage 2); rule-line-3 endothermic-invariance family · mcq4
A reaction is endothermic, so its products' level sits above its reactants' level. How does that change where its activation-energy arrow is drawn?
It does not change it — the arrow still runs from the reactants' level up to the top of the barrier. KEY
The arrow now starts at the products' level instead, since that is the higher of the two levels.
The arrow now ends at the products' level instead of ending at the top of the barrier.
The arrow flips to point downward, because the reaction takes in energy rather than giving it out.
KEQ-007 · medium · application · seq 8 · authored fresh (stage 2); statement-match family, arrangement flipped from A, letters rotated · mcq4
A reaction has one reactant and one product, and the figure tracks both as curves M and N. Which statement matches the graph?
FIGURE (concentration-time-graph · stimulus) — reaction: "a reaction with one reactant and one product, unnamed" · curves: [{"label": "M", "shape": "rising", "start": "zero", "end": "high"}, {"label": "N", "shape": "falling", "start": "high", "end": "near zero"}] · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only M and N: curve M rises from zero to a high level, curve N falls fro…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "a reaction with one reactant and one product, unnamed",
  "curves": [
   {
    "label": "M",
    "shape": "rising",
    "start": "zero",
    "end": "high"
   },
   {
    "label": "N",
    "shape": "falling",
    "start": "high",
    "end": "near zero"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two curves labeled only M and N: curve M rises from zero to a high level, curve N falls from a high start to near zero"
 }
}
Curve M shows the product and curve N shows the reactant. KEY
Curve N shows the product and curve M shows the reactant.
Curves M and N both show products of the reaction.
The curves cannot be told apart without knowing the substances' names.
KEQ-008 · medium · application · seq 9 · authored fresh (stage 2); two-run product comparison, faster asked, letters rotated · mcq4
The figure shows the product curves for two runs of the same reaction, labeled Run J and Run K. Which run is faster?
FIGURE (concentration-time-graph · stimulus) — reaction: "one reaction, two runs, product tracked" · curves: [{"label": "Run J", "shape": "rising", "steepness": "shallow", "start": "zero"}, {"label": "Run K", "shape": "rising", "steepness": "steep", "start": "zero"}] · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two rising curves from the same zero start, labeled Run J and Run K: Run J climbs shallowly…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one reaction, two runs, product tracked",
  "curves": [
   {
    "label": "Run J",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero"
   },
   {
    "label": "Run K",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two rising curves from the same zero start, labeled Run J and Run K: Run J climbs shallowly, Run K climbs steeply"
 }
}
Run K KEY
Run J
Both runs have the same rate.
The graph cannot show which run is faster.
KEQ-009 · medium · application · seq 10 · authored fresh (stage 2); regions-marked fastest read (rule line 0) · mcq4
The figure shows a product's curve for one run, with time regions 1, 2, and 3 marked in order. During which region is the reaction fastest?
FIGURE (concentration-time-graph · stimulus) — reaction: "one run, product tracked" · curves: [{"label": "product", "shape": "rising", "profile": "steep at first, shallowing, flat at the end", "start": "zero", "end": "high"}] · markedRegions: [{"label": "1", "position": "steep early section"}, {"label": "2", "position": "shallowing middle section"}, {"label": "3", "position": "flat final section"}] · curveCount: 1 · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': steep climb at first, shallowing middle, exactly flat a…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one run, product tracked",
  "curves": [
   {
    "label": "product",
    "shape": "rising",
    "profile": "steep at first, shallowing, flat at the end",
    "start": "zero",
    "end": "high"
   }
  ],
  "markedRegions": [
   {
    "label": "1",
    "position": "steep early section"
   },
   {
    "label": "2",
    "position": "shallowing middle section"
   },
   {
    "label": "3",
    "position": "flat final section"
   }
  ],
  "curveCount": 1,
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; one rising curve labeled 'product': steep climb at first, shallowing middle, exactly flat at the end; the time axis is divided into three labeled regions — region 1 spans the steep early climb, region 2 the shallowing middle, region 3 the flat end"
 }
}
Region 1 KEY
Region 2
Region 3
It runs equally fast in all three regions.
KEQ-010 · medium · application · seq 11 · authored fresh (stage 2); dilution prediction, bare option set (canonical fluency options); scenario kept clear of the form-A FRQ's descaling bath (cross-form instance discipline) · mcq4
A patio-cleaning fluid is sold ready to use. A landscaper treats one flagstone with it straight from the bottle, and treats an identical flagstone with a half-strength mix of the same fluid. What happens to the rate of the cleaning reaction on the half-strength flagstone?
The reaction slows down. KEY
The reaction speeds up.
The rate stays the same.
The reaction stops completely.
KEQ-011 · medium · application · seq 12 · authored fresh (stage 2); frequency mechanism pick (harder-collisions distractor family) · mcq4
A tile-scrubbing fluid clears grime faster when used undiluted than when mixed half-and-half. Why, in collision terms?
Undiluted, more reacting particles share the same space, so collisions with the grime happen more often. KEY
Undiluted fluid's particles each carry more energy into their collisions.
Undiluted, every collision with the grime succeeds instead of only some.
Undiluted fluid reacts with itself as well as with the grime, doubling the reactions.
KEQ-012 · medium · application · seq 13 · authored fresh (stage 2); running-out slowdown; sol loop-2 system swap to antacid tablet in dilute acid (previous instance had oxygen/temperature confounds) · mcq4
An antacid tablet dropped into a glass of dilute acid bubbles hard at first, then more and more gently, and the bubbling dies away just as the last of the tablet disappears. In collision terms, why does the reaction slow and then stop?
The tablet is used up, so collisions between the reacting particles grow rarer and finally stop. KEY
The particles in the glass wear out and lose their energy after all the colliding.
The gas made by the reaction blocks every remaining collision from happening.
The reaction pauses to recharge and would restart if the glass were swirled.
KEQ-013 · medium · application · seq 14 · authored fresh (stage 2); warming prediction, bare option set (canonical fluency options) · mcq4
A dental filling resin hardens by a chemical reaction. The dentist holds a small warmed gel pack against the tooth while the resin sets, with everything else unchanged. What happens to the hardening reaction?
It speeds up. KEY
It slows down.
Its rate stays the same.
It stops until the tooth cools again.
KEQ-014 · medium · application · seq 15 · authored fresh (stage 2); two-arm account, fresh instance (silver home (15,D)) · mcq4
Silver bowls tarnish sooner in a hot dining hall than in a cool pantry. Which account gives BOTH parts of the temperature effect?
The particles collide more often, and more of those collisions carry enough energy to react. KEY
They collide more often, but each collision lands with less energy than in the cool pantry.
More of their collisions carry enough energy — how often they collide is unchanged.
The hot hall simply presses more tarnishing gas against the silver, not faster particles.
KEQ-018 · medium · application · seq 16 · authored fresh (stage 2); needed-energy comparison (rule lines 2/6) · mcq4
A reaction is run twice at the same temperature: once plainly, and once with a catalyst mixed in. Compare the energy a successful collision needs in the two runs.
Less is needed in the catalyzed run — the catalyst's path has a lower activation energy than the original path. KEY
More is needed in the catalyzed run — the added substance gets in the reactants' way.
The same is needed in both — a catalyst changes how often the particles collide, not the energy required.
No energy is needed in the catalyzed run — on a catalyzed path, every collision succeeds.
KEQ-015 · hard · transfer · seq 17 · authored fresh (stage 2); crumbling prediction (unscanned tokens) · mcq4
A bath fizzer reacts and fizzes in a full tub. One fizzer is dropped in whole; an identical fizzer is first crumbled into coarse crumbs. Predict which one is done fizzing first.
The crumbled one — the same mass in smaller pieces carries more surface area, so it reacts faster. KEY
The whole one — keeping the fizzer in one piece concentrates the fizzing in one spot.
Both finish together — equal masses always finish reacting at the same time.
Neither changes — crumbling alters how the fizz looks, not how fast the reaction runs.
KEQ-016 · hard · transfer · seq 18 · authored fresh (stage 2); MECH-SURFACE key verbatim (generic solid) · mcq4
A solid reactant can be added to a solution either as one large crystal or as the same mass of small grains. The grains react away far sooner. Why?
Breaking a solid into smaller pieces exposes more particles at the surface, where collisions with the other reactant can happen. KEY
The grains' particles each carry more energy than the crystal's particles, so their collisions hit harder.
Small grains are made of smaller, lighter particles that react more easily.
The solution passes through the middle of each grain, but cannot enter the crystal.
KEQ-020 · hard · transfer · seq 19 · authored fresh (stage 2); catalyst-condition read (generic catalyst powder; canonical option classes) · mcq4
One thing separated two otherwise matching runs of a fizzing reaction: a spoonful of a catalyst powder went into Run 2. The graph shows the two product curves, labeled only P and Q — the run numbers were left off. Which curve is Run 2's?
FIGURE (concentration-time-two-runs · stimulus) — description: "concentration of product against time; curve P rises gently and levels off later; curve Q rises steeply and levels off early; both flatten at exactly the sa… · curves: [{"label": "P", "shape": "rising", "steepness": "shallow", "start": "zero", "end": "levels off later at the shared final height"}, {"label": "Q", "shape": "r… · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled only P and Q in-diagram: Q climbs steeply and flattens…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-two-runs",
  "description": "concentration of product against time; curve P rises gently and levels off later; curve Q rises steeply and levels off early; both flatten at exactly the same final height",
  "curves": [
   {
    "label": "P",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero",
    "end": "levels off later at the shared final height"
   },
   {
    "label": "Q",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero",
    "end": "levels off early at the same final height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled only P and Q in-diagram: Q climbs steeply and flattens early, P climbs gently and flattens later; both flat sections sit at the same height"
 }
}
Curve Q — the catalyst made Run 2 faster, and Q climbs more steeply to the same final height. KEY
Curve P — it keeps rising for longer, showing the catalyst kept its run reacting.
Neither curve can be picked — both reach the same final height, so the two runs were identical.
It cannot be read from the graph, because a catalyst leaves no trace on a product curve.
KEQ-019 · hard · transfer · seq 20 · authored fresh (stage 2); which-feature read (canonical bare-feature option set, fresh stem) · mcq4
An energy diagram is redrawn to include a catalyst's path alongside the original path. The figure shows the result. Which feature differs between the two paths?
FIGURE (energy-diagram-two-paths · stimulus) — description: "an energy diagram with two barrier curves between the same reactant and product levels; the original curve peaks high, the catalyst's curve peaks lower" · style: "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the reactants' level sits higher than the products' level;…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-two-paths",
  "description": "an energy diagram with two barrier curves between the same reactant and product levels; the original curve peaks high, the catalyst's curve peaks lower",
  "style": "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the reactants' level sits higher than the products' level; two curves labeled in-diagram 'original path' and 'catalyst's path': both start at the reactants' level and end at the products' level, the original path over a tall peak, the catalyst's path over a clearly lower peak"
 }
}
The height of the barrier's peak. KEY
The reactants' energy level.
The products' energy level.
The energy gap between reactants and products.
KEQ-020 · medium · application · seq 21 · authored fresh (stage 2); catalyst case (generic powder; times 5/23 min, home (21,D)) · frq
A lab runs the same foam-producing reaction twice in identical flasks. Run 1 — mixture: the standard recipe; flask temperature: bench level; no additive. The foaming finished after 23 minutes. Run 2 — mixture: the standard recipe; flask temperature: bench level; additive: a spoonful of a catalyst powder, recovered unchanged afterward. The foaming finished after 5 minutes.
A lab runs the same foam-producing reaction twice in identical flasks. Run 1 — mixture: the standard recipe; flask temperature: bench level; no additive. The foaming finished after 23 minutes. Run 2 — mixture: the standard recipe; flask temperature: bench level; additive: a spoonful of a catalyst powder, recovered unchanged afterward. The foaming finished after 5 minutes. (a) Name the ONE condition that changed between the runs. (b) State which run was faster, and cite the recorded observation that shows it. (c) Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Part a. Name the ONE condition that changed between the runs.
Model: The presence of a catalyst — Run 2 had a spoonful of catalyst powder stirred in and Run 1 had none. Everything else about the two runs matched.
Part b. State which run was faster, and cite the recorded observation that shows it.
Model: Run 2 was faster. The same foaming job finished after 5 minutes with the catalyst, while the plain run needed 23 minutes. The sooner-finished run is the faster run.
Part c. Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Model: Yes. The taught direction is that a run with a catalyst present is the faster run, and the catalyst run was indeed the faster one. The comparison is fair because exactly one condition changed — the catalyst's presence — while the recipe, the flasks, the temperature, and everything else stayed the same.
Full model answer. (a) The presence of a catalyst — Run 2 had a spoonful of catalyst powder stirred in and Run 1 had none. Everything else about the two runs matched. (b) Run 2 was faster. The same foaming job finished after 5 minutes with the catalyst, while the plain run needed 23 minutes. The sooner-finished run is the faster run. (c) Yes. The taught direction is that a run with a catalyst present is the faster run, and the catalyst run was indeed the faster one. The comparison is fair because exactly one condition changed — the catalyst's presence — while the recipe, the flasks, the temperature, and everything else stayed the same.

Form E

KEQ-001 · easy · recall · seq 1 · authored fresh (stage 2); orientation-completion family · mcq4
A collision between two reactant molecules carries plenty of energy. What else must be true for the collision to make a reaction happen?
The molecules must be lined up in the right orientation. KEY
Nothing else — enough energy on its own guarantees a reaction.
The two molecules must be particles of the same substance.
The container must be crowded enough to press the molecules together.
KEQ-003 · easy · recall · seq 2 · authored fresh (stage 2); unlit-mixture family (rule line 4's class, generic gases) · mcq4
A fuel gas mixed with a second reactant gas sits unchanged in a sealed vessel for hours, even though their molecules collide constantly — yet the touch of a small flame would set the reaction off at once. Why do almost none of those collisions make product?
Almost none of them carry the reaction's activation energy — the minimum it needs. KEY
The molecules never actually meet inside the vessel.
The vessel holds far too little of the second gas for any reaction.
The two gases are simply unreactive — no reaction between them is possible.
KEQ-006 · easy · recall · seq 3 · authored fresh (stage 2); rate-statement identification family (parallel to B, fresh observation lines) · mcq4
A student watches a flask and wants to write down one line that records the reaction's rate. Which line should she write?
The froth rose quickly in the first minute, then far more slowly. KEY
The flask grew warm to the touch while the froth formed.
The mixture's smell changed from sharp to sweet during the run.
By the end, the froth filled half the flask.
KEQ-017 · easy · recall · seq 4 · authored fresh (stage 2); S-the-bystander isolate (the lesson's own gate design) · mcq4
A substance sits in a reaction mixture and is chemically unchanged when the reaction ends — but the reaction runs at the same speed with or without it. Is the substance a catalyst?
No — a catalyst must also speed the reaction up. KEY
Yes — being unchanged at the end is the whole definition of a catalyst.
Yes — any added substance that survives a reaction counts as its catalyst.
No — because a true catalyst must be used up as it works.
KEQ-002 · medium · application · seq 5 · authored fresh (stage 2); lacks-energy-only case, verbal · mcq4
Two reactant molecules meet lined up correctly for this reaction, but they are barely moving when they touch. Classify the collision.
Unsuccessful — it lacks enough energy. KEY
Successful — a correctly lined-up collision is successful at any speed.
Unsuccessful — the orientation is wrong.
Successful — a gentle touch gives the molecules more time to react.
KEQ-004 · medium · application · seq 6 · authored fresh (stage 2); interval-read family with the from-axis distractor drawn, letters rotated · mcq4
An energy-releasing reaction's energy diagram is shown, with three intervals marked on it. Which interval shows the activation energy?
FIGURE (energy-diagram-barrier · stimulus) — reaction: "an energy-releasing reaction (unnamed)" · reactantLevel: "high" · productLevel: "lower than reactants" · intervals: [{"label": "M", "from": "reactants' level", "to": "top of the barrier"}, {"label": "N", "from": "bottom axis", "to": "top of the barrier"}, {"label": "T", "f… · style: "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' leve…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-barrier",
  "reaction": "an energy-releasing reaction (unnamed)",
  "reactantLevel": "high",
  "productLevel": "lower than reactants",
  "intervals": [
   {
    "label": "M",
    "from": "reactants' level",
    "to": "top of the barrier"
   },
   {
    "label": "N",
    "from": "bottom axis",
    "to": "top of the barrier"
   },
   {
    "label": "T",
    "from": "top of the barrier",
    "to": "products' level"
   }
  ],
  "style": "clean single-hump energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of the reaction'); the curve starts at a high reactants' level, rises over one smooth hump, then falls to a products' level below the reactants' level; three labeled vertical interval arrows: M rises from the reactants' level to the top of the hump, N rises from the bottom axis to the top of the hump, T runs from the top of the hump down to the products' level"
 }
}
Interval M KEY
Interval N
Interval T
None of the intervals
KEQ-005 · medium · application · seq 7 · authored fresh (stage 2); fix-a-wrong-arrow family (verbal construction judgment, the lesson's mistake set) · mcq4
A student's activation-energy arrow runs from the bottom axis up to the top of the barrier. Which correction fixes the drawing?
Start the arrow at the reactants' level instead of the bottom axis. KEY
End the arrow at the products' level instead of the top of the barrier.
Stretch the arrow from the reactants' level all the way to the products' level.
Nothing needs fixing — activation-energy arrows always start at the axis.
KEQ-007 · medium · application · seq 8 · authored fresh (stage 2); falling-inference family (mirror of B, figure-less) · mcq4
Partway through a run, a chemist checks the records: one substance's concentration has fallen steadily since the start, and nothing has been removed from the mixture. What is that substance?
A reactant — the reaction is using it up, so its concentration falls. KEY
A product — a falling concentration marks what the reaction makes.
A product that keeps escaping from the mixture as it forms.
It could be either — direction alone cannot separate reactant from product.
KEQ-008 · medium · application · seq 9 · authored fresh (stage 2); two-run reactant comparison, faster asked, letters rotated · mcq4
Two runs of the same solution reaction are plotted as reactant curves, labeled Run R and Run S. Which run is faster?
FIGURE (concentration-time-graph · stimulus) — reaction: "one solution reaction, two runs, reactant tracked" · curves: [{"label": "Run R", "shape": "falling", "steepness": "shallow", "start": "high"}, {"label": "Run S", "shape": "falling", "steepness": "steep", "start": "high… · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; two falling curves from the same high start, labeled Run R and Run S: Run R falls shallowly…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one solution reaction, two runs, reactant tracked",
  "curves": [
   {
    "label": "Run R",
    "shape": "falling",
    "steepness": "shallow",
    "start": "high"
   },
   {
    "label": "Run S",
    "shape": "falling",
    "steepness": "steep",
    "start": "high"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; two falling curves from the same high start, labeled Run R and Run S: Run R falls shallowly, Run S falls steeply"
 }
}
Run S KEY
Run R
Both runs have the same rate.
Falling curves cannot show which run is faster.
KEQ-009 · medium · application · seq 10 · authored fresh (stage 2); description-match family (canonical option texts) · mcq4
A reaction in a beaker is tracked from its start to its finish, and the figure shows the reactant's curve. Which description fits what is drawn?
FIGURE (concentration-time-graph · stimulus) — reaction: "one run, reactant tracked" · curves: [{"label": "reactant", "shape": "falling", "profile": "steep at first, shallowing, flat at the end", "start": "high", "end": "near zero"}] · curveCount: 1 · style: "unnumbered axes, horizontal 'time' and vertical 'concentration'; one falling curve labeled 'reactant': steep fall at first, shallowing through the middle, e…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-graph",
  "reaction": "one run, reactant tracked",
  "curves": [
   {
    "label": "reactant",
    "shape": "falling",
    "profile": "steep at first, shallowing, flat at the end",
    "start": "high",
    "end": "near zero"
   }
  ],
  "curveCount": 1,
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration'; one falling curve labeled 'reactant': steep fall at first, shallowing through the middle, exactly flat at the end near zero"
 }
}
Steep at first, shallower later, flat at the end. KEY
Equally steep the whole way down.
Shallow at first, steepest at the end.
Flat at first, then steadily falling.
KEQ-010 · medium · application · seq 11 · authored fresh (stage 2); increase prediction, bare option set (canonical fluency options) · mcq4
A gardener mixes a moss-killing spray at double the recommended concentration, with everything else unchanged. What happens to the rate of the spray's reaction with the moss?
The reaction speeds up. KEY
The reaction slows down.
The rate stays the same.
The reaction finishes instantly — doubling the concentration makes it happen all at once.
KEQ-011 · medium · application · seq 12 · authored fresh (stage 2); chain-of-causes format (canonical chain option family, fresh scene) · mcq4
Doubling a dissolved reactant's concentration makes its reaction faster. Which cause-and-effect chain is right?
More particles in the same space → collisions happen more often → more successful collisions per second → faster reaction. KEY
More particles in the same space → each collision carries more energy → more successful collisions per second → faster reaction.
More particles in the same space → the activation energy falls → faster reaction.
More particles in the same space → more product can form in total → faster reaction.
KEQ-012 · medium · application · seq 13 · authored fresh (stage 2); fastest-at-start why (rule line 3), fresh instance-free scene · mcq4
Two dissolved chemicals react in a beaker at a steady temperature, with nothing added after the start. Why is the reaction fastest in its first moments?
The reactants' concentrations are highest at the start, so their particles collide most often then. KEY
The particles carry their greatest energy at the start and gradually tire.
The activation energy is lowest at the start of a run and rises as the run goes on.
Stirring at the start makes the first collisions stronger than the later ones.
KEQ-013 · medium · application · seq 14 · authored fresh (stage 2); cooling prediction with the never-stops boundary (rule line 3 class) · mcq4
Cooking oil slowly turns rancid through chemical reactions. A cook moves the bottle from a warm shelf into a chilly cellar. What happens to those reactions?
They slow down — but keep running. KEY
They stop completely in the cold.
They speed up, because the cellar is drier than the shelf.
They are unchanged — temperature only matters for fast reactions.
KEQ-014 · medium · application · seq 15 · authored fresh (stage 2); cooling-reverse two-arm item (rule line 5), instance-free · mcq4
A reaction mixture is moved into a chilled cabinet. Using both parts of the temperature effect, what happens at the particle level?
The particles move slower — colliding less often, with fewer collisions carrying enough energy. KEY
They collide less often, but each collision keeps the same chance of reacting.
They collide just as often, but each collision carries less energy.
They stop moving entirely, so the reaction stops.
KEQ-018 · medium · application · seq 16 · authored fresh (stage 2); same-temperature share comparison (canonical option family, fresh wording) · mcq4
Two identical mixtures sit at the same temperature, and a catalyst is stirred into one of them. In which mixture do more collisions succeed each second, and why?
The catalyzed one — its path needs less energy, so more of the same collisions are strong enough to react. KEY
The catalyzed one — the catalyst makes its particles move faster than the other mixture's.
Both equally — at the same temperature, the same share of collisions must succeed.
The plain one — the catalyst's particles get in the way of the reactants.
KEQ-015 · hard · transfer · seq 17 · authored fresh (stage 2); grinding prediction (charcoal unscanned; distinct task from KEQ-010's ember-in-oxygen instance) · mcq4
Grilling charcoal comes as solid briquettes. A camper grinds one briquette into a coarse powder of the same total mass. Predict how the powder's burning compares with a whole briquette's.
The powder burns away faster — grinding gives the same mass more surface area. KEY
The powder burns more slowly, because it settles into a dense heap.
Both burn at the same rate, because the mass of charcoal is unchanged.
The powder cannot burn — grinding a solid takes away its ability to react.
KEQ-016 · hard · transfer · seq 18 · authored fresh (stage 2); rule-line-2/3 buried-particles transfer (ore unscanned) · mcq4
At a processing plant, crushed ore reacts with a leaching solution far faster than the same mass of ore in fist-sized lumps. Which explanation is correct?
Crushing uncovers buried particles, so the solution has far more ore particles to collide with. KEY
Crushing heats the ore, and the leftover warmth speeds the leaching.
The solution becomes more concentrated around small pieces than around lumps.
Crushed ore is a different, more reactive substance than lump ore.
KEQ-020 · hard · transfer · seq 19 · authored fresh (stage 2); piece-size condition read (generic solid reactant) · mcq4
Two runs of a gas-making reaction matched in every condition but one: Run 1 used its solid reactant as coarse granules, Run 2 used the same mass ground to a powder. The graph shows the two product curves. Which conclusion is backed by the conditions and the curves together?
FIGURE (concentration-time-two-runs · stimulus) — description: "concentration of product against time; the Run 2 curve rises steeply and levels off early; the Run 1 curve rises gently and levels off later; both flatten a… · curves: [{"label": "Run 1", "shape": "rising", "steepness": "shallow", "start": "zero", "end": "levels off later at the shared final height"}, {"label": "Run 2", "sh… · curveCount: 2 · curveConstraint: "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height" · style: "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled Run 1 and Run 2 in-diagram: Run 2 climbs steeply and f…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-two-runs",
  "description": "concentration of product against time; the Run 2 curve rises steeply and levels off early; the Run 1 curve rises gently and levels off later; both flatten at the same final height",
  "curves": [
   {
    "label": "Run 1",
    "shape": "rising",
    "steepness": "shallow",
    "start": "zero",
    "end": "levels off later at the shared final height"
   },
   {
    "label": "Run 2",
    "shape": "rising",
    "steepness": "steep",
    "start": "zero",
    "end": "levels off early at the same final height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "curves are distinct, separately drawn paths; they may touch only at the shared start point and the shared final height",
  "style": "unnumbered axes, horizontal 'time' and vertical 'concentration of product'; two rising curves labeled Run 1 and Run 2 in-diagram: Run 2 climbs steeply and flattens early, Run 1 climbs gently and flattens later; both flat sections sit at the same height"
 }
}
Run 2 was faster — the powdered solid gave the steeper curve, matching the smaller-pieces direction. KEY
Run 1 was faster — coarse granules react faster than powder.
The runs made different amounts of product, because their curves rise at different speeds.
Piece size cannot change a reaction's rate, so some other condition must have differed.
KEQ-019 · hard · transfer · seq 20 · authored fresh (stage 2); statement-match family (canonical key/distractor texts, fresh stem) · mcq4
The figure shows one reaction drawn with two barrier curves, one for each path the reaction can take. Which statement is true of the drawn diagram?
FIGURE (energy-diagram-two-paths · stimulus) — description: "an energy diagram with two barrier curves between the same reactant and product levels, one labeled 'with catalyst' peaking lower than the other, labeled 'w… · style: "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the reactants' level sits higher than the products' level;…
raw spec
{
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "energy-diagram-two-paths",
  "description": "an energy diagram with two barrier curves between the same reactant and product levels, one labeled 'with catalyst' peaking lower than the other, labeled 'without catalyst'",
  "style": "clean energy diagram with unnumbered axes (vertical 'energy', horizontal 'progress of reaction'); the reactants' level sits higher than the products' level; two curves labeled in-diagram 'without catalyst' and 'with catalyst': both start at the reactants' level and end at the products' level, the 'without catalyst' curve over a tall peak, the 'with catalyst' curve over a clearly lower peak"
 }
}
Both paths start and end at the same levels; the catalyzed path clears a lower peak. KEY
The catalyzed path ends at a lower product level, so it gives out more energy.
The catalyzed path starts at a higher reactant level, giving the particles a head start.
The two paths differ only in how much time the reaction takes, which the horizontal axis shows.
KEQ-020 · medium · application · seq 21 · authored fresh (stage 2); temperature case, observations variant (times 17/55 min, home (21,E)) · frq
A quality-control team runs the same color-forming test reaction twice in identical sealed flasks. Run 1 — flask: placed in a heated cabinet; reagents: one standard measure each; nothing else added. The full color change was complete after 17 minutes. Run 2 — flask: placed in an identical cabinet beside the first with its heater off, running cold; reagents: one standard measure each of the same batch; nothing else added. The full color change was complete after 55 minutes.
A quality-control team runs the same color-forming test reaction twice in identical sealed flasks. Run 1 — flask: placed in a heated cabinet; reagents: one standard measure each; nothing else added. The full color change was complete after 17 minutes. Run 2 — flask: placed in an identical cabinet beside the first with its heater off, running cold; reagents: one standard measure each of the same batch; nothing else added. The full color change was complete after 55 minutes. (a) Name the ONE condition that changed between the runs. (b) State which run was faster, and cite the recorded observation that shows it. (c) Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Part a. Name the ONE condition that changed between the runs.
Model: The temperature — Run 1 sat in a heated cabinet and Run 2 in an identical unheated, cold cabinet beside it. Everything else about the two runs matched.
Part b. State which run was faster, and cite the recorded observation that shows it.
Model: Run 1 was faster. The same full color change was complete after 17 minutes in the heated run, while the cold run needed 55 minutes. The sooner-finished run is the faster run.
Part c. Say whether the result matches the taught direction of that rate factor, and explain why this comparison is a fair test.
Model: Yes. The taught direction is that raising the temperature makes a reaction faster, and the heated run was the faster one. The comparison is fair because exactly one condition changed — the temperature — while the reagents, the flasks, and everything else stayed the same.
Full model answer. (a) The temperature — Run 1 sat in a heated cabinet and Run 2 in an identical unheated, cold cabinet beside it. Everything else about the two runs matched. (b) Run 1 was faster. The same full color change was complete after 17 minutes in the heated run, while the cold run needed 55 minutes. The sooner-finished run is the faster run. (c) Yes. The taught direction is that raising the temperature makes a reaction faster, and the heated run was the faster one. The comparison is fair because exactly one condition changed — the temperature — while the reagents, the flasks, and everything else stayed the same.

End of Topic Test: Reversible reactions and dynamic equilibrium (u13-t2 · 5 forms × 21 items · 105 total)

David's notes for James:
full notes file
# u13-t2 — notes/queries from David's side (U13-G2)

1. FRQ COMPOSITION - frq_per_form = 1 (your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - Topic 5 (KEQ-021..027) is purely qualitative: all seven scope atoms are fact/explanation/discrimination/interpretation, no lesson carries a displayEquation, and zero of your 21 pool seeds are numeric. The single FRQ is the QB section-4 explain-why baseline, homed on KEQ-025 (the topic's explanation atom, MECH-DYNAMIC-EQ) and terminal at seq 21 on every form: (a) state the rate condition, (b) explain WHY the amount holds constant in the taught wording, (c) rebut a rotated wrong claim (stopped-reactions / container-credit / surplus-tops-it-up / one-direction-only / restart-later). Rubric 25/45/30 = 100; model answers are word-only (nothing typed requires sub/superscripts). D5 scaling evaluated and NOT triggered: 7 scope lessons support the full 20-objective allocation (every lesson 2-3 rows; capstone KEQ-027 and contrast atom KEQ-026 get 3 each). Total 20 mcq + 1 FRQ = 21 items/form.
2. SEED USAGE - 20 of your 21 pool items placed, all on form A (forms B-E are fresh-authored parallels): your demo_curated trio are byte-level duplicates of pool[18]/[19]/[20] modulo option order - deduped at stage 1 with both refs cited on the placements (they hold the three KEQ-027 figure rows at seqs 20/15/18). prod_seeds = [] for this topic and the cold list is ignored by standing answer #3. The ONE spare: pool[14], the water-tank-with-pump analogy (KEQ-025) - 'water' is a base MATERIALS-lexicon word sitting inside KEQ-025's own sink-analogy rule line, so the item cannot be placed without discarding the analogy instance itself; a fresh flow analog (stadium crowd, constant headcount with people still entering and leaving) was authored in its D#8-family slot instead. Not a merit drop.
3. SEED FRESHENINGS UNDER THE STRICT SEED RULING (no freshness exemption; the shared validator was never weakened) - 4 dispositions to confirm, every one preserving your structure and misconception keys with all fields swept (QB 3.9): (a) pool[1]@A#1 stem 'Methane burns in oxygen...' -> 'The complete combustion of CH₄ is written...' ('oxygen' lives in KEQ-021's own fluency corpus; the word 'methane' dropped with it), distractor 'extra oxygen' -> 'extra O₂', keyExplanation closing clause recast to formulas - your equation and keys untouched. (b) pool[4]@A#2 opener 'Hydrogen and carbon dioxide react reversibly' -> 'Two gases react reversibly' ('hydrogen'/'carbon' are KEQ-022 own-corpus words); your equation and options verbatim. (c) pool[9]@A#4 stem opener reworded to 'Once a sealed reversible mixture arrives at equilibrium...' (the original four-word run was verbatim KEQ-024 setup text). (d) pool[20]@A#18 product flattening time 6 -> 4 minutes in stem + one distractor + figureSpec ('flat from 6 minutes' is verbatim in KEQ-027's worked-example line); all fields swept for the old value.
4. COURSE-WIDE INSTANCE RULE - one wording fix to confirm: D#7's stem said 'A sealed reactor's readout' - 'reactor' is on the course-wide banned-token list (literal, no nuclear context needed), and it was the bank's sole occurrence; reworded to 'A sealed reaction vessel's readout' with zero other changes. Re-swept at packaging over forms_final AND this export body: zero banned-token hits anywhere (uranium/plutonium/enrichment/reactor/weapon/fissile/meltdown/dose all clean).
5. SEQ-18 TRANSFER FAMILY AMENDED AT BLIND QC (confirm - touches your seed A#18 and its demo_curated twin): the blind round's sol reviewer correctly observed that EXACT one-curve flatness while the other curve still changes is stoichiometrically impossible for a single sealed reversible reaction, which made B#18's faulty-graph distractor genuinely defensible. The whole seq-18 family (A/B/D/E; C already had it) now frames the early flat curve as APPARENT - 'has looked level ... at the graph's scale' - and B#18's faulty-graph distractor was replaced with a mismatched-settling-times-preclude-equilibrium misconception. Your taught rule (equilibrium needs EVERY curve flat) is unchanged and remains each item's key discrimination; A#18's figure and correct answer are untouched beyond the framing sentence.
6. OTHER SEED-ITEM EDITS AT QC (confirm, all on form A): A#9 'changing as it is measured' -> 'still drifting downward on its own' (killed a measurement-disturbance misread); A#10 the stem's own 'at equilibrium' declaration removed - it made the key circular; steadiness evidence now carries the inference your item intended; A#16 'each hold a gas mixture' -> 'each hold the same reversible reacting gas mixture' (kills a nonreacting-mixture defense of one distractor). Structure, keys, and misconception targets all yours.
7. B#14/E#14 CONSTRUCT FIX AT AUTHORITATIVE QC: sol objected that picking the correct statement-PAIR required a fact the options did not carry; the keys on the fresh forms now instantiate KEQ-024's OWN rule sentence ('the amounts stop changing even though the reaction has not stopped'), options re-parallelized. Your seed pair item (A#14, water-gas) needed no change. RELATED KEPT-AS-IS: A#2/A#14 reuse the water-gas system within form A - that is YOUR bank's own pairing, the rows are non-adjacent, and no blind reviewer raised telegraphing across three rounds.
8. E#15 REBUILT AT AUTHORITATIVE QC (fresh item, disclosure): the draft measured the KEQ-025 mechanism while its graph carried no load; the shipped item is a completion-misread whose load-bearing evidence is the reactant curve's visible nonzero flat height (KEQ-027 graph reading; the ran-to-completion contrast is KEQ-021 course content).
9. D#13 DISTRACTOR DISPOSITION (course-local authority, recorded): a blind reviewer wanted steady-state nuance on the leaky-container option, but KEQ-023's own rule line IS 'if a substance leaks away, the balance is never struck' - the taught rule wins over beyond-course nuance, so the key stands verbatim (same class as U10-G1's idealized-tester ruling). The distractor that DID hinge on a disallowed 'only because' qualifier was replaced with a flexing-walls misconception.
10. GATE_CONFIG - ZERO whitelist relief for this gate: the only gate_config entry is the unit-13 spec path override. No extra_materials, no task idioms, no convention phrases, no A6 figure relief. The shared validator ran UNMODIFIED to 12/12 PASS, exit 0, with ZERO warning lines - no adjudication wrapper needed (first-class result; the qualitative topic plus formula-notation carry made full instance exclusivity satisfiable).
11. NOTA/AOTA: zero literal none-of-the-above/all-of-the-above hits across your 21-seed pool (V-T5 regex clone at stage 1); zero replace_distractor flags needed; V-T5 PASS on all forms.
12. QC LEDGER (provenance, no action needed): blind round 1 (gpt-5.6-sol blind role + gemini, all 105 items) = 200/200 key picks, 2 hard fails (the B#18 and E#9 defensible-distractor catches above - E#9's fix: two point-readings a night apart cannot exclude a very slow approach, so the evidence is now a continuous recorder trace and the slow-approach distractor became the constant-not-equal trap); re-blind rounds on every changed item (fresh tags r3-r6) = 32/32 key picks, 0 hard fails. FRQ cold answers: both reviewers hit all three rubric observables on every form; B#21 re-graded after its loop-2 rebuttal fix (both match). sol authoritative loops: 10 FAIL -> 0; final ledger 105/105 pass with 10 recorded honest-survivor style minors (none affects key uniqueness). Blind-packet disclosure: the shared packet builder rendered the two graph-figure families style-only, fixed per-gate by an augmenter that renders the drawn content structurally (generic _shared addition recommended; U10-G1 precedent class).
13. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings anywhere - the topic is qualitative equilibrium chemistry on classroom-safe systems (kiln limestone, desiccant-card color change, ester equilibria, dimerizations); every named system is a real equilibrium and appears at exactly one seq (the water-gas pairing above is your own bank design).

Form A

KEQ-021 · easy · recall · seq 1 · seed:U13-G2/pool[1] (FRESHENED) · mcq4
The complete combustion of CH₄ is written CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), with a single arrow. What does the single arrow tell you, in contrast to ⇌?
The reaction is written as one-way: the products are not shown re-forming the reactants. KEY
The reaction is faster than any reaction written with ⇌.
The reaction only works when extra O₂ is supplied.
The equation counts as unbalanced until the single arrow is replaced by the symbol ⇌.
KEQ-022 · easy · recall · seq 2 · seed:U13-G2/pool[4] (FRESHENED) · mcq4
Two gases react reversibly: CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g). Which change is the reverse reaction?
CO and H₂O reacting to re-form CO₂ and H₂. KEY
CO₂ and H₂ reacting to form CO and H₂O.
CO₂ reacting with CO to form H₂ and H₂O.
H₂O condensing out of the mixture as a liquid.
KEQ-023 · easy · recall · seq 3 · seed:U13-G2/pool[7] · mcq4
Why is the equilibrium reached by a sealed reversible reaction mixture called DYNAMIC equilibrium?
Because both reactions keep running even though the mixture is balanced. KEY
Because the mixture keeps swinging between mostly reactants and mostly products.
Because the equilibrium arrives suddenly, with a burst of reaction.
Because the container must be shaken for the balance to hold.
KEQ-024 · easy · recall · seq 4 · seed:U13-G2/pool[9] (FRESHENED) · mcq4
A student says: 'Once a sealed reversible mixture arrives at equilibrium, the reactant concentrations must fall to zero.' Which correction is right?
At equilibrium every concentration — reactants' included — holds constant at a value above zero. KEY
The student is right — equilibrium is the point where the reactants are finally used up.
Only the concentration of the product falls to zero at equilibrium.
The reactant concentrations keep falling forever, just more and more slowly.
KEQ-021 · easy · application · seq 5 · seed:U13-G2/pool[0] · mcq4
Limestone decomposes in a kiln: CaCO₃(s) ⇌ CaO(s) + CO₂(g). Which statement about this reaction is correct?
CaO and CO₂ can react together to re-form CaCO₃. KEY
Once CaCO₃ has decomposed, the change can never be undone.
The ⇌ shows that the CaCO₃ is only half real and half decomposed.
The equation is wrongly written, because decomposition reactions cannot carry ⇌.
KEQ-022 · easy · application · seq 6 · seed:U13-G2/pool[3] · mcq4
A student describes a sealed reversible reaction mixture: 'The forward reaction must use up the reactants completely before the reverse reaction can begin.' Which correction is right?
The two reactions run at the same time — products start re-forming reactants while reactants are still forming products. KEY
The student has the order backward: the reverse reaction runs first, and the forward reaction follows it.
The student is right, except that the switch happens when half of the reactants are used up.
The student is right for gases, but liquids run both directions at the same time.
KEQ-023 · medium · application · seq 7 · seed:U13-G2/pool[6] · mcq4
A sealed tube holds the reversible reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at dynamic equilibrium. Which statement about the SO₃ is correct?
SO₃ is being formed exactly as fast as it is breaking apart. KEY
SO₃ is no longer being formed, because the reaction has finished.
SO₃ is being formed faster than it breaks apart, so the tube slowly fills with SO₃.
SO₃ breaks apart only when the tube is opened.
KEQ-025 · medium · application · seq 8 · seed:U13-G2/pool[12] · mcq4
A student explains a sealed equilibrium mixture this way: 'The amounts stay constant because both reactions have stopped.' Which correction is right?
The forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up. KEY
The student is nearly right — only the reverse reaction has fully stopped, and the forward one runs far too slowly to matter.
The amounts stay constant because the sealed container blocks any change.
The student is right, and the amounts will start changing again once the reactions restart.
KEQ-024 · medium · application · seq 9 · seed:U13-G2/pool[10] · mcq4
A sealed flask of the reversible reaction PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) is checked twice, 30 minutes apart, with no disturbance in between. The first check finds the Cl₂ concentration still drifting downward on its own. What do the readings show about the first check?
The mixture had not yet reached equilibrium when the first check was made. KEY
The mixture was at equilibrium, because some change is always allowed there.
The flask must have been leaking during the first check.
The Cl₂ reading must be an error, because concentrations can never change in a sealed flask.
KEQ-026 · medium · application · seq 10 · seed:U13-G2/pool[15] · mcq4
A sealed flask of CO(g) + 2H₂(g) ⇌ CH₃OH(g) is analyzed: it holds a great deal of CO and H₂ and only a little CH₃OH, and all three amounts have been steady for hours. Which statement is correct?
The flask is at equilibrium — a mostly-reactant mixture with steady amounts qualifies fully. KEY
The flask is not at equilibrium, because most of the reactants have not yet been converted.
The flask is only halfway to equilibrium, and the CH₃OH amount will keep climbing until it matches the CO.
The analysis must be wrong, because equilibrium mixtures contain mostly product.
KEQ-021 · medium · application · seq 11 · seed:U13-G2/pool[2] · mcq4
A student reads the equation N₂(g) + O₂(g) ⇌ 2NO(g) and says: 'The ⇌ means the equation is balanced.' Which correction is right?
The ⇌ means the reaction is reversible; balance is a separate fact, checked from the atom counts. KEY
The student is right — ⇌ is the balancing check mark chemists add to finished equations.
The ⇌ means the reaction is unbalanced, and a single arrow would mean balanced.
The ⇌ means the two sides hold equal masses of gas at all times.
KEQ-022 · medium · application · seq 12 · seed:U13-G2/pool[5] · mcq4
In a sealed flask of the reversible reaction H₂(g) + Br₂(g) ⇌ 2HBr(g), a chemist observes at one instant that HBr molecules are breaking apart into H₂ and Br₂. What must also be true at that same instant?
Somewhere in the flask, H₂ and Br₂ are reacting to form HBr. KEY
Every HBr molecule in the flask is breaking apart at that instant.
No HBr can be forming, because forming and breaking apart cannot share a flask.
The flask must be about to run out of HBr completely.
KEQ-023 · medium · application · seq 13 · seed:U13-G2/pool[8] · mcq4
Two flasks hold the same reversible gas reaction at the same temperature. Flask 1 is sealed. Flask 2 is left open, and one gaseous product steadily escapes. Which flask can reach dynamic equilibrium, and why?
Flask 1 — equilibrium needs a closed container, where nothing escapes. KEY
Flask 2 — removing product makes room for the balance to form.
Both flasks — once the rates are equal anywhere, containment stops mattering.
Neither flask — a gas reaction cannot reach equilibrium at all.
KEQ-024 · medium · application · seq 14 · seed:U13-G2/pool[11] · mcq4
A sealed mixture of H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g) is at dynamic equilibrium. Which pair of statements is correct together?
The CO concentration is constant, and CO is still being made and used. KEY
The CO concentration is constant, and no CO is being made or used anymore.
The CO concentration is rising, and CO is still being made and used.
The CO concentration is falling, and no CO is being made anymore.
KEQ-027 · medium · application · seq 15 · seed:U13-G2/pool[19] · mcq4
The graph shows a sealed run of CO(g) + 2H₂(g) ⇌ CH₃OH(g) with both curves flat from 10 minutes. At 14 minutes, a certain amount of CH₃OH breaks apart each second. Which statement about 14 minutes is correct?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 16 minutes; the CO curve falls then runs flat from 10 minutes; the CH₃OH curve rises then runs flat from 10 minutes at … · style: "unit-standard concentration-time graph, curves labeled CO and CH₃OH in-diagram" · curves: [{"label": "CO", "shape": "falling-then-flat", "start": "high", "flatFrom": "10 minutes"}, {"label": "CH₃OH", "shape": "rising-then-flat", "start": "zero", "… · curveCount: 2 · curveConstraint: "two distinct curve paths, one falling and one rising to different flat heights; both flat from 10 minutes"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 16 minutes; the CO curve falls then runs flat from 10 minutes; the CH₃OH curve rises then runs flat from 10 minutes at a different height",
  "style": "unit-standard concentration-time graph, curves labeled CO and CH₃OH in-diagram",
  "curves": [
   {
    "label": "CO",
    "shape": "falling-then-flat",
    "start": "high",
    "flatFrom": "10 minutes"
   },
   {
    "label": "CH₃OH",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "10 minutes",
    "note": "flat at a different height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths, one falling and one rising to different flat heights; both flat from 10 minutes"
 }
}
CH₃OH is also being formed at that same rate, which is why its curve stays flat. KEY
The flat curve proves that no CH₃OH is breaking apart at 14 minutes.
The CH₃OH curve must actually be falling slightly at 14 minutes.
The breaking apart will drag the mixture out of equilibrium within minutes.
KEQ-026 · medium · application · seq 16 · seed:U13-G2/pool[16] · mcq4
Two sealed flasks each hold the same reversible reacting gas mixture. Flask A: the two substances' amounts are exactly equal at this moment, but both amounts are still changing. Flask B: the two substances' amounts are very different, and both have held steady all day. Which flask is at equilibrium?
Flask B — steady amounts mark equilibrium, however unequal they are. KEY
Flask A — equal amounts are the definition of equilibrium.
Both flasks — each satisfies one half of the definition of equilibrium.
Neither flask — equilibrium needs the amounts to be both equal and steady.
KEQ-025 · medium · application · seq 17 · seed:U13-G2/pool[13] · mcq4
A sealed flask of CO(g) + 2H₂(g) ⇌ CH₃OH(g) is at dynamic equilibrium. In one minute, a certain number of CH₃OH molecules break apart. What must be true of that same minute?
About the same number of CH₃OH molecules were made in that minute. KEY
No CH₃OH molecules were made in that minute, because the amounts are constant.
Far more CH₃OH molecules were made than broke apart, to keep the flask stocked.
The broken-apart molecules were the last ones — the flask is running out of CH₃OH.
KEQ-027 · hard · transfer · seq 18 · seed:U13-G2/pool[20] (FRESHENED) · mcq4
A graph of a sealed reversible run shows the reactant curve still falling gently at 9 minutes while the product curve looks flat from 4 minutes. Has the mixture reached equilibrium by 9 minutes?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the product curve rises and looks flat from 4 minutes at the graph's scale; the reactant curve falls steepl… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "product", "shape": "rising-then-flat", "start": "zero", "flatFrom": "4 minutes"}, {"label": "reactant", "shape": "falling", "start": "high", "not… · curveCount: 2 · curveConstraint: "two distinct curve paths; the product curve is flat from 4 minutes while the reactant curve is still visibly falling through 9 minutes"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 12 minutes; the product curve rises and looks flat from 4 minutes at the graph's scale; the reactant curve falls steeply at first and is still falling gently through 9 minutes, flattening only near 11 minutes",
  "style": "unit-standard concentration-time graph, curves labeled reactant and product in-diagram",
  "curves": [
   {
    "label": "product",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "4 minutes"
   },
   {
    "label": "reactant",
    "shape": "falling",
    "start": "high",
    "note": "falls steeply at first, still falling gently through 9 minutes, flattening only near 11 minutes"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths; the product curve is flat from 4 minutes while the reactant curve is still visibly falling through 9 minutes"
 }
}
No — equilibrium needs BOTH curves flat, and the reactant curve is still falling at 9 minutes. KEY
Yes — one flat curve is enough to mark equilibrium.
Yes — the product curve flattened at 4 minutes, and the product is what equilibrium is about.
No — and the mixture never will reach equilibrium, because its curves flattened at different times.
KEQ-026 · hard · transfer · seq 19 · seed:U13-G2/pool[17] · mcq4
The figure reproduces a textbook diagram of an equilibrium mixture's two concentration bars across three drawings at later and later times. A student looks at it and says the artist has drawn equilibrium wrongly. What should the student's teacher say?
FIGURE (qualitative-bars-over-time · stimulus) — series: [{"label": "reactant", "height": "tall", "note": "same tall height at every time"}, {"label": "product", "height": "short", "note": "same short height at eve… · timePoints: ["earlier", "later", "later still"] · style: "no numbers on any axis; the article's bar-triplet template with the sides flipped (reactant tall, product short)"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "qualitative-bars-over-time",
  "series": [
   {
    "label": "reactant",
    "height": "tall",
    "note": "same tall height at every time"
   },
   {
    "label": "product",
    "height": "short",
    "note": "same short height at every time"
   }
  ],
  "timePoints": [
   "earlier",
   "later",
   "later still"
  ],
  "style": "no numbers on any axis; the article's bar-triplet template with the sides flipped (reactant tall, product short)"
 }
}
The drawing is correct — each bar holds its own height over time, and the heights need not match. KEY
The student is right — the bars must be drawn at the same height at equilibrium.
The student is right — the taller bar should shrink a little in each later drawing.
The drawing is wrong, but only because the product bar should be the taller one.
KEQ-027 · hard · transfer · seq 20 · seed:U13-G2/pool[18] · mcq4
The graph shows a sealed reversible run in which the falling reactant curve CROSSES the rising product curve at 3 minutes, and both curves become flat from 8 minutes. A student says equilibrium was reached at 3 minutes, where the curves meet. What does the graph actually show?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls from high and the product curve rises from zero; the two curves cross at 3 minutes… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "8 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "z… · curveCount: 2 · curveConstraint: "two distinct curve paths crossing exactly once, at 3 minutes, while both are still visibly changing; both flat from 8 minutes at different heights"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 12 minutes; the reactant curve falls from high and the product curve rises from zero; the two curves cross at 3 minutes while both are still visibly changing; both curves bend and run flat from 8 minutes at different heights",
  "style": "unit-standard concentration-time graph, curves labeled reactant and product in-diagram",
  "curves": [
   {
    "label": "reactant",
    "shape": "falling-then-flat",
    "start": "high",
    "flatFrom": "8 minutes"
   },
   {
    "label": "product",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "8 minutes",
    "note": "flat at a different height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths crossing exactly once, at 3 minutes, while both are still visibly changing; both flat from 8 minutes at different heights"
 }
}
Equilibrium arrived at 8 minutes — the crossing at 3 minutes is just the moment the two concentrations happened to match. KEY
The student is right — equal concentrations are what equilibrium means.
Equilibrium arrived at 3 minutes, but only for the two substances that crossed.
The graph shows no equilibrium at all, because curves that have crossed once can never settle into balance.
KEQ-025 · medium · application · seq 21 · fresh-authored (U13-G2 stage-2; QB sec.4 explain-why FRQ, KEQ-025 home) · frq
A sealed vessel holds two colorless gases that react reversibly to form a single product gas. A monitor shows the product's amount has not changed for the past three hours, yet a tracer test proves the product is still reacting away continuously.
A sealed vessel holds two colorless gases that react reversibly to form a single product gas. A monitor shows the product's amount has not changed for the past three hours, yet a tracer test proves the product is still reacting away continuously. (a) State what must be true of the rates of the forward and reverse reactions in this vessel. (b) Explain WHY the product's amount stays constant even though the product keeps reacting away. Use the wording the lesson taught. (c) A classmate concludes: 'Three hours with no change means both reactions have stopped.' In one or two sentences, explain what is wrong with that conclusion and what is really happening.
Part a. State what must be true of the rates of the forward and reverse reactions in this vessel.
Model: The rate of the forward reaction equals the rate of the reverse reaction, and neither rate is zero — both reactions are still running.
Part b. Explain WHY the product's amount stays constant even though the product keeps reacting away. Use the wording the lesson taught.
Model: The product's amount stays constant because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Making and using cancel exactly, so the measured amount never moves even though the product keeps reacting.
Part c. A classmate concludes: 'Three hours with no change means both reactions have stopped.' In one or two sentences, explain what is wrong with that conclusion and what is really happening.
Model: The classmate has read steady readings as stopped chemistry. The amounts stop changing even though the reactions have not stopped — the tracer test itself shows the product still reacting, and the forward reaction replaces every loss at the same rate.
Full model answer. (a) The rate of the forward reaction equals the rate of the reverse reaction, and neither rate is zero — both reactions are still running. (b) The product's amount stays constant because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Making and using cancel exactly, so the measured amount never moves even though the product keeps reacting. (c) The classmate has read steady readings as stopped chemistry. The amounts stop changing even though the reactions have not stopped — the tracer test itself shows the product still reacting, and the forward reaction replaces every loss at the same rate.

Form B

KEQ-021 · easy · recall · seq 1 · fresh-authored (U13-G2 stage-2) · mcq4
A chemist writes a gas reaction with the two-way symbol ⇌ between reactants and products. What is the ⇌ saying about the reaction?
It runs in both directions — the products can react to re-form the reactants. KEY
It has reached its end, and no further change is possible.
It gives out energy in whichever direction it runs.
It forms its products only while the mixture is kept heated.
KEQ-022 · easy · recall · seq 2 · fresh-authored (U13-G2 stage-2) · mcq4
The gas COCl₂ decomposes reversibly: COCl₂(g) ⇌ CO(g) + Cl₂(g). Which change is the forward reaction?
COCl₂ breaking apart into CO and Cl₂. KEY
CO and Cl₂ combining to re-form COCl₂.
COCl₂ reacting with CO to make Cl₂.
Cl₂ molecules splitting into separate Cl atoms.
KEQ-023 · easy · recall · seq 3 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible mixture has settled at dynamic equilibrium. What are the forward and reverse reactions doing from that point on?
Both keep going, each exactly as fast as the other. KEY
Both have shut down now that the mixture is balanced.
Only the forward reaction is still going, very slowly.
They alternate — one runs while the other waits its turn.
KEQ-024 · easy · recall · seq 4 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible gas mixture spent the past hour at dynamic equilibrium, undisturbed. Over that hour, what happened to the concentration of each substance in it?
Each one held exactly the value it had when the hour began. KEY
Each one drifted slowly downward toward zero.
Each one crept upward as the reactions kept running.
Each one swung above and below an average value.
KEQ-021 · easy · application · seq 5 · fresh-authored (U13-G2 stage-2) · mcq4
In hot AlCl₃ vapor, pairs of molecules join up reversibly: 2AlCl₃(g) ⇌ Al₂Cl₆(g). Which statement about this change is correct?
Al₂Cl₆ can break apart again to give back AlCl₃. KEY
Once an Al₂Cl₆ molecule has formed, it can never return to AlCl₃.
The ⇌ shows the pairing-up is stuck at the halfway point.
The 2 in front of AlCl₃ makes the ⇌ unnecessary.
KEQ-022 · easy · application · seq 6 · fresh-authored (U13-G2 stage-2) · mcq4
A lab notebook entry reads: 'In our sealed reversible mixture, the reverse reaction will stay switched off until something disturbs the flask.' Which correction is right?
No switch is needed — both directions run together on their own, at every moment. KEY
The entry is right — an undisturbed mixture runs its reactions one direction at a time.
The entry is nearly right — the reverse reaction starts by itself, but only once half the reactants have reacted.
The entry is backward — it is the forward reaction that waits for a disturbance.
KEQ-023 · medium · application · seq 7 · fresh-authored (U13-G2 stage-2) · mcq4
In a hot sealed tube, cyclohexane(g) ⇌ methylcyclopentane(g) sits at dynamic equilibrium. Which statement about the methylcyclopentane is correct?
It is being formed exactly as fast as it is converting back. KEY
It is no longer being formed — the conversion has run its course.
It is being formed slightly faster than it converts back, so the tube gradually fills with it.
It converts back only if the tube is cooled first.
KEQ-025 · medium · application · seq 8 · fresh-authored (U13-G2 stage-2) · mcq4
A probe shows one product's amount in a sealed reversible mixture unchanged for hours, while checks confirm that reactions are still running inside. WHY does the amount hold still?
Because the forward and reverse reactions keep happening at the same rate, so the substance is made exactly as fast as it is used up. KEY
Because the sealed walls physically prevent the amounts inside from changing.
Because the product molecules have become too stable to take part in any reaction.
Because equilibrium locks every particle of the mixture rigidly in place from the moment the balance first arrives.
KEQ-024 · medium · application · seq 9 · fresh-authored (U13-G2 stage-2) · mcq4
A probe left in a sealed reversible gas mixture shows one reactant's concentration still creeping downward. What does that reading say about the mixture?
It has not reached equilibrium yet. KEY
It is at equilibrium — a slow creep is small enough to count as steady.
It must have sprung a leak, since nothing else could move a reading.
The probe must be broken, because sealed mixtures keep every reading fixed.
KEQ-026 · medium · application · seq 10 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed tube is analyzed: it holds far more product than reactant, and every amount has held its value since early morning. A skeptical student insists something must be wrong with the analysis. Which reply is right?
Nothing is wrong — unchanging amounts mean equilibrium, no matter how far ahead the product sits. KEY
The student is right — at equilibrium the reactant and product amounts have to match.
The student is right — a reactant that scarce would have been used up completely by now.
The tube is still mid-journey, and the product's lead will shrink as the balance forms.
KEQ-021 · medium · application · seq 11 · fresh-authored (U13-G2 stage-2) · mcq4
Reading an equation written with ⇌, a student concludes: 'This reaction must be unusually fast, in both directions.' Which correction is right?
⇌ marks reversibility only — it says nothing about how fast either direction runs. KEY
The student is right — the double arrow doubles the speed.
The student has it inverted — ⇌ actually marks the slowest reactions.
Speed is shown by the coefficients, not the arrow, so the student looked in the wrong place.
KEQ-022 · medium · application · seq 12 · fresh-authored (U13-G2 stage-2) · mcq4
A particle-level simulation of a sealed reversible mixture is paused on a single frame. The frame clearly shows several product particles in the act of re-forming reactant particles. What else must a close look at that same frame find?
Reactant particles elsewhere in the frame reacting to form product. KEY
Every product particle in the frame caught mid-breakup.
No product being formed anywhere, since a mixture runs one direction per moment.
A product supply on the verge of running out.
KEQ-023 · medium · application · seq 13 · fresh-authored (U13-G2 stage-2) · mcq4
The same reversible gas reaction is run twice: once under a tight stopper, once under a loose lid that lets one gaseous product seep away. Why does only the stoppered run settle into dynamic equilibrium?
Because equilibrium needs everything kept in — with a substance seeping away, the two rates never come to match. KEY
Because the stopper presses the particles closer together until they balance.
It doesn't — the loose-lidded run balances more easily, having less product to manage.
Both runs settle the same way in the end — lids and stoppers make no difference to how a reaction behaves.
KEQ-024 · medium · application · seq 14 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible gas mixture sits at dynamic equilibrium. For any one REACTANT in it, which two claims are both true?
Its concentration has stopped changing, even though its particles have not stopped reacting. KEY
Its concentration has stopped changing, because its particles are done reacting.
Its concentration is still slowly falling, even though its particles keep reacting.
Its concentration is still slowly falling, because its particles are done reacting.
KEQ-027 · medium · application · seq 15 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run whose two curves have been level since 7 minutes. Which claim about the mixture at 10 minutes does the graph support?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls steeply then runs flat from 7 minutes; the product curve rises from zero then runs… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "7 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "z… · curveCount: 2 · curveConstraint: "two distinct curve paths, one falling and one rising to different flat heights; both flat from 7 minutes"
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Both reactions are running at the same, nonzero rate. KEY
Both reactions ended at 7 minutes, when the curves leveled out.
Only the forward reaction survives past 7 minutes.
The product's level line is really a very slow decline, too gentle to draw.
KEQ-026 · medium · application · seq 16 · fresh-authored (U13-G2 stage-2) · mcq4
Watching a sealed run's live readings, a student sees the reactant and product amounts pass through the same value — while both are still visibly changing — and announces that equilibrium has arrived. Which correction is right?
Momentary equality is not the test — equilibrium is amounts that have stopped changing. KEY
No correction is needed — the moment the values match, equilibrium exists.
The student stopped one match too soon — equilibrium arrives the second time the values line up.
The student watched the wrong pair — matching only counts between two products.
KEQ-025 · medium · application · seq 17 · fresh-authored (U13-G2 stage-2) · mcq4
Over one hour, a particle counter watching a product in a sealed mixture at dynamic equilibrium tallies a large number of that product's molecules being used up. What did the hour's tally of NEWLY made molecules of that product look like?
About the same large number — it was made exactly as fast as it was used up. KEY
Zero — nothing new is made once the amounts have settled.
A far larger number, building a reserve against future losses.
A shrinking number each minute, as the making wound down.
KEQ-027 · hard · transfer · seq 18 · fresh-authored (U13-G2 stage-2) · mcq4
On the graph of a sealed reversible run, the product curve has looked level since 3 minutes, but the reactant curve is still sliding downward at 6 minutes. A lab report written at that 6-minute point claims the mixture is at equilibrium. Is the claim right?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 10 minutes; the product curve rises quickly and looks level from 3 minutes at the graph's scale; the reactant curve fal… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "product", "shape": "rising-then-flat", "start": "zero", "flatFrom": "3 minutes"}, {"label": "reactant", "shape": "falling", "start": "high", "not… · curveCount: 2 · curveConstraint: "two distinct curve paths; the product curve looks level from 3 minutes while the reactant curve is still visibly falling through 6 minutes"
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No — a curve still sliding means the mixture is still approaching, whatever the other curve does. KEY
Yes — the product settled at 3 minutes, and that is when equilibrium begins.
Yes — a sliding reactant is normal at equilibrium, since reactants are being used.
No — and it never will be: curves that settle at different times rule equilibrium out for good.
KEQ-026 · hard · transfer · seq 19 · fresh-authored (U13-G2 stage-2) · mcq4
The figure shows a student's poster of an equilibrium mixture: a tall product bar and a short reactant bar, drawn identically at three successive times. A classmate insists the bars must be redrawn to equal heights before the poster can go up. Who is right?
FIGURE (qualitative-bars-over-time · stimulus) — series: [{"label": "product", "height": "tall", "note": "same tall height at every time"}, {"label": "reactant", "height": "short", "note": "same short height at eve… · timePoints: ["first check", "second check", "third check"] · style: "no numbers on any axis; poster-style bar pairs repeated identically at three checks (product tall, reactant short)"
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The poster's artist — each bar keeps its own height, and equal heights were never required. KEY
The classmate — equilibrium is only shown properly when the two bars match.
The classmate — but the fix is to make the reactant bar the taller one.
Neither — bars are the wrong representation, since equilibrium amounts cannot be drawn as fixed heights.
KEQ-027 · hard · transfer · seq 20 · fresh-authored (U13-G2 stage-2) · mcq4
On the graph of a sealed reversible run, the two curves meet once at 5 minutes and both level out from 9 minutes. A revision guide's caption reads: 'Equilibrium is found where the lines meet.' What should the caption say instead?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls from high and the product curve rises from zero; the curves cross once at 5 minute… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "9 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "z… · curveCount: 2 · curveConstraint: "two distinct curve paths crossing exactly once, at 5 minutes, while both are still visibly changing; both level from 9 minutes at different heights"
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Equilibrium is found where both curves become level — here 9 minutes, not the meeting point. KEY
The caption is fine as printed — meeting lines are the graphical sign of equilibrium.
The caption should say equilibrium lasts only from the meeting point to the leveling.
The caption should say this run has no equilibrium, since its lines met on the way.
KEQ-025 · medium · application · seq 21 · fresh-authored (U13-G2 stage-2; QB sec.4 explain-why FRQ, KEQ-025 home) · frq
In a teaching lab, a reversible gas reaction runs inside a sealed chamber. The wall display labels the product 'no net change' and has done so all afternoon, although the chamber log confirms reactions are still proceeding inside.
In a teaching lab, a reversible gas reaction runs inside a sealed chamber. The wall display labels the product 'no net change' and has done so all afternoon, although the chamber log confirms reactions are still proceeding inside. (a) State the condition on the two reaction rates that the chamber must have reached. (b) Explain WHY 'no net change' can be true of the product while reactions keep proceeding. Use the wording the lesson taught. (c) A visitor suggests: 'It is the chamber's thick sealed walls that hold the product's amount fixed.' In one or two sentences, explain why the walls cannot be the reason and give the real one.
Part a. State the condition on the two reaction rates that the chamber must have reached.
Model: The chamber has reached the condition where the rate of the forward reaction equals the rate of the reverse reaction — and both are still running, not zero.
Part b. Explain WHY 'no net change' can be true of the product while reactions keep proceeding. Use the wording the lesson taught.
Model: 'No net change' holds because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Every loss is replaced in step, so the display reads steady while the chemistry continues.
Part c. A visitor suggests: 'It is the chamber's thick sealed walls that hold the product's amount fixed.' In one or two sentences, explain why the walls cannot be the reason and give the real one.
Model: The walls cannot be the reason: sealing only keeps substances from escaping, and any sealed mixture still shows changing amounts while it is approaching equilibrium. What fixes the product's amount is the matched rates: it is made exactly as fast as it is used up.
Full model answer. (a) The chamber has reached the condition where the rate of the forward reaction equals the rate of the reverse reaction — and both are still running, not zero. (b) 'No net change' holds because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Every loss is replaced in step, so the display reads steady while the chemistry continues. (c) The walls cannot be the reason: sealing only keeps substances from escaping, and any sealed mixture still shows changing amounts while it is approaching equilibrium. What fixes the product's amount is the matched rates: it is made exactly as fast as it is used up.

Form C

KEQ-021 · easy · recall · seq 1 · fresh-authored (U13-G2 stage-2) · mcq4
In a data book, some equations carry the single arrow → and others carry ⇌. Which kind of reaction earns the ⇌?
One whose products can react to re-form its reactants. KEY
One that runs faster than most other reactions.
One that has been checked and is correctly balanced.
One that runs only when a catalyst is present.
KEQ-022 · easy · recall · seq 2 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed bulb holds the reversible reaction SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g). Which change is the reverse reaction?
SbCl₃ and Cl₂ reacting to re-form SbCl₅. KEY
SbCl₅ breaking apart into SbCl₃ and Cl₂.
SbCl₅ reacting with SbCl₃ to release Cl₂.
Cl₂ escaping from the bulb as the pressure rises.
KEQ-023 · easy · recall · seq 3 · fresh-authored (U13-G2 stage-2) · mcq4
A textbook glossary is being written. What should the entry say the word 'dynamic' records about a mixture at dynamic equilibrium?
That the balanced mixture is never still — both reactions continue inside it. KEY
That the mixture's makeup keeps drifting back and forth around a middle value.
That the balance appeared suddenly rather than building up gradually.
That the mixture must be kept moving for the balance to survive.
KEQ-024 · easy · recall · seq 4 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible gas mixture is at dynamic equilibrium. Which of its substances keep a constant concentration?
Every reactant and every product alike. KEY
Only the products.
Only the reactants.
None of them, since the reactions are still running.
KEQ-021 · easy · application · seq 5 · fresh-authored (U13-G2 stage-2) · mcq4
At low temperatures, two gases combine reversibly: NO(g) + NO₂(g) ⇌ N₂O₃(g). Which statement about this change is correct?
N₂O₃ can break apart to re-form NO and NO₂. KEY
After N₂O₃ forms, the change cannot run the other way.
The ⇌ means the flask holds matching amounts of the three gases.
The equation must be miswritten, because two gases joining into one cannot be reversible.
KEQ-022 · easy · application · seq 6 · fresh-authored (U13-G2 stage-2) · mcq4
A revision card claims: 'A reversible reaction runs forward for a stretch, then switches over and runs in reverse for a stretch.' Which correction is right?
There is no switching — the two directions run at the same time in the same mixture. KEY
The card is right, but the switching is too quick to notice.
The card is right once equilibrium arrives — that is when the switching begins.
The card is right for small containers, where there is only room for one direction at a time.
KEQ-023 · medium · application · seq 7 · fresh-authored (U13-G2 stage-2) · mcq4
A stoppered flask holds the reversible ester reaction CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) at dynamic equilibrium. Which statement about the ester, CH₃COOC₂H₅, is correct?
It is being made and being broken apart at matching rates. KEY
It stopped being made when the CH₃COOH began to run low.
None of it is still being produced — the flask simply stores what formed earlier.
It breaks apart only while the flask is being heated.
KEQ-025 · medium · application · seq 8 · fresh-authored (U13-G2 stage-2) · mcq4
A revision poster reads: 'Amounts hold constant at equilibrium because the sealed flask stops anything inside from changing.' Which correction is right?
Sealing only keeps substances in — the amounts hold because each substance is made exactly as fast as it is used up. KEY
The poster is right — a good seal is what fixes the amounts in place.
The poster names the wrong cause — the amounts hold because both reactions have shut down.
The poster names the wrong cause — the amounts hold because only the forward reaction is left running.
KEQ-024 · medium · application · seq 9 · fresh-authored (U13-G2 stage-2) · mcq4
Two careful measurements of a product's concentration in a sealed, undisturbed reversible mixture — one taken early in the morning, one taken later the same morning — give different values. What follows?
The mixture was still approaching equilibrium between the two measurements. KEY
The mixture was at equilibrium, and its amounts simply wander a little over a morning.
Someone must have opened the flask between the measurements.
The later measurement has to be the wrong one, since the first fixed the true value.
KEQ-026 · medium · application · seq 10 · fresh-authored (U13-G2 stage-2) · mcq4
An analysis of a sealed reversible mixture whose measured composition has held the same values all day finds only a trace of product alongside plenty of unreacted starting material. Can this mixture really be at equilibrium?
Yes — steady amounts mean equilibrium, however small the product's share is. KEY
No — too little of the reactants has been converted for the mixture to count.
No — equilibrium requires the two sides to hold at least comparable amounts.
Yes — but only because the reaction never actually started in this flask.
KEQ-021 · medium · application · seq 11 · fresh-authored (U13-G2 stage-2) · mcq4
A tutee tells you: 'When an equation carries ⇌, it means the reaction gave up partway and left a half-finished mixture.' Which correction is right?
⇌ says the reaction runs in both directions — it is not a mark of an unfinished change. KEY
The tutee is right — ⇌ flags reactions that quit before completing.
The tutee is nearly right — ⇌ means the products broke down again the instant they formed.
The tutee is right, but only for reactions between gases.
KEQ-022 · medium · application · seq 12 · fresh-authored (U13-G2 stage-2) · mcq4
A monitoring system flags a sealed reversible mixture with the note: 'reverse direction active at this timestamp'. What must have been true of the forward direction at that same timestamp?
It was active too, in the same mixture. KEY
It was paused while the reverse direction had its turn.
It was due to start a moment later, once the reverse direction eased off.
Nothing — the note itself must be a sensor fault, since one direction cannot run alone.
KEQ-023 · medium · application · seq 13 · fresh-authored (U13-G2 stage-2) · mcq4
A student plans to reach dynamic equilibrium by running a reversible gas reaction in an open beaker on the bench. What is the flaw in the plan?
Gaseous substances will drift out of the open beaker, so the matched-rate balance can never be struck. KEY
Nothing is flawed — the mixture will simply take longer to balance in the open.
The beaker is merely too small — a larger open vessel would reach the balance.
The mixture needs stirring, not sealing — motionless mixtures cannot balance.
KEQ-024 · medium · application · seq 14 · fresh-authored (U13-G2 stage-2) · mcq4
Which two facts about a PRODUCT in a sealed mixture at dynamic equilibrium sit together correctly?
A steady amount, with its making and unmaking still going on. KEY
A steady amount, with its making and unmaking finished.
A growing amount, with its making and unmaking still going on.
A shrinking amount, with its making finished and only its unmaking left.
KEQ-027 · medium · application · seq 15 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run, level in both curves since the 4-minute mark. A student announces that by 8 minutes the reverse reaction has shut down for good. What does the graph actually support?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 9 minutes; the reactant curve falls then runs flat from 4 minutes; the product curve rises from zero then runs flat fro… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "4 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "z… · curveCount: 2 · curveConstraint: "two distinct curve paths, one falling and one rising to different flat heights; both flat from 4 minutes"
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Both directions are still running at 8 minutes, at matched rates — level curves are not stopped chemistry. KEY
The student's announcement — a level curve leaves the reverse reaction nothing to do.
A stronger claim — both directions shut down at 4 minutes.
The opposite claim — only the reverse reaction is running at 8 minutes.
KEQ-026 · medium · application · seq 16 · fresh-authored (U13-G2 stage-2) · mcq4
Two sealed mixtures were each measured at 10 a.m. and again at 11 a.m. Mixture 1: its two substances matched each other at the first measurement but not at the second. Mixture 2: its two substances differed from each other, but each gave the same value at both measurements. Which mixture was at equilibrium through that hour?
Mixture 2 — repeated identical readings for each substance are the mark of equilibrium. KEY
Mixture 1 — substances that match each other are the mark of equilibrium.
Both mixtures — each shows one kind of matching, and either kind will do.
Neither mixture — a single hour is too short for equilibrium to mean anything.
KEQ-025 · medium · application · seq 17 · fresh-authored (U13-G2 stage-2) · mcq4
A tracer study of a sealed mixture at dynamic equilibrium finds that every second, a small fraction of the product's molecules convert back into reactants. Why doesn't the product's amount drift downward over time?
Because the forward reaction re-makes it at that same rate, second after second. KEY
Because the converted fraction is too small to affect the total.
Because the sealed container returns the converted molecules to the product.
Because the converting stops the moment the amount would start to fall.
KEQ-027 · hard · transfer · seq 18 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run in which the reactant curve looks level from 5 minutes, while the product curve is still creeping upward through 8 minutes. A student certifies the mixture 'at equilibrium since 5 minutes'. What does the graph actually show?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls and looks level from 5 minutes; the product curve rises steeply then keeps creepin… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "5 minutes"}, {"label": "product", "shape": "rising", "start": "zero", "not… · curveCount: 2 · curveConstraint: "two distinct curve paths; the reactant curve is level from 5 minutes while the product curve is still visibly rising through 8 minutes"
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The mixture was still approaching at 8 minutes — the creeping product curve rules out equilibrium until every curve is level. KEY
The certification stands — the reactant is the substance that decides equilibrium.
The certification is early but harmless — on any real graph, a gently creeping curve like this one counts as level.
The graph shows the run failed — a product that creeps this long never settles.
KEQ-026 · hard · transfer · seq 19 · fresh-authored (U13-G2 stage-2) · mcq4
The figure shows a notebook sketch meant to depict a mixture AT equilibrium: two concentration bars drawn at three successive times, with the taller bar drawn slightly shorter at each later time. What is wrong with the sketch?
FIGURE (qualitative-bars-over-time · stimulus) — series: [{"label": "substance 1", "height": "tall", "note": "drawn slightly shorter at each later time"}, {"label": "substance 2", "height": "short", "note": "same s… · timePoints: ["start", "soon after", "later"] · style: "no numbers on any axis; notebook-sketch bar pairs in which the tall bar visibly shrinks across the three times while the short bar repeats its height"
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A bar may not shrink — at equilibrium each bar keeps exactly the same height at every time. KEY
Nothing is wrong with the sketch — a slow shrink in the taller bar is how equilibrium is supposed to look.
The real flaw is that the two bars start at different heights.
The real flaw is that the shrinking bar should be growing instead.
KEQ-027 · hard · transfer · seq 20 · fresh-authored (U13-G2 stage-2) · mcq4
The graph of a sealed reversible run shows its two curves crossing at 2 minutes and running level from 6 minutes. According to the graph, what actually happened at 2 minutes?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 10 minutes; the reactant curve falls from high and the product curve rises from zero; the curves cross once at 2 minute… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "6 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "z… · curveCount: 2 · curveConstraint: "two distinct curve paths crossing exactly once, at 2 minutes, while both are still visibly changing; both level from 6 minutes at different heights"
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The two concentrations passed through the same value while both were still changing — nothing to do with equilibrium. KEY
Equilibrium began there, at the moment of equality — matching concentrations are the arrival the run was working toward.
The reaction switched direction there, from forward to reverse.
The two substances traded roles there, the reactant becoming the product.
KEQ-025 · medium · application · seq 21 · fresh-authored (U13-G2 stage-2; QB sec.4 explain-why FRQ, KEQ-025 home) · frq
A sealed reaction tube holds a reversible gas reaction that settled long ago: every measured amount, product included, has repeated its value at each check this week, while instruments confirm ongoing reaction in both directions.
A sealed reaction tube holds a reversible gas reaction that settled long ago: every measured amount, product included, has repeated its value at each check this week, while instruments confirm ongoing reaction in both directions. (a) State the relationship between the forward and reverse reaction rates in the tube this week. (b) Explain WHY the product's amount repeats its value at every check even though it is continuously being used up. Use the wording the lesson taught. (c) A lab partner proposes: 'The making must run a tiny bit ahead of the using — that small surplus is what keeps the amount from slipping.' In one or two sentences, explain why the proposal fails and state the real relationship.
Part a. State the relationship between the forward and reverse reaction rates in the tube this week.
Model: The rate of the forward reaction equals the rate of the reverse reaction, with both directions still running.
Part b. Explain WHY the product's amount repeats its value at every check even though it is continuously being used up. Use the wording the lesson taught.
Model: The amount repeats its value because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. The making and the using cancel exactly, so each check finds the same value.
Part c. A lab partner proposes: 'The making must run a tiny bit ahead of the using — that small surplus is what keeps the amount from slipping.' In one or two sentences, explain why the proposal fails and state the real relationship.
Model: A surplus would not hold the amount — it would grow it: making that runs ahead of using adds product week after week. The readings hold because the two run at exactly the same rate, every loss matched by a gain.
Full model answer. (a) The rate of the forward reaction equals the rate of the reverse reaction, with both directions still running. (b) The amount repeats its value because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. The making and the using cancel exactly, so each check finds the same value. (c) A surplus would not hold the amount — it would grow it: making that runs ahead of using adds product week after week. The readings hold because the two run at exactly the same rate, every loss matched by a gain.

Form D

KEQ-021 · easy · recall · seq 1 · fresh-authored (U13-G2 stage-2) · mcq4
Four changes are described below. Which one would correctly be written with the symbol ⇌?
A change whose products react together to give back the starting substances. KEY
A change that happens very slowly at room temperature.
A change that needs constant stirring to keep going.
A change that gives more than one product.
KEQ-022 · easy · recall · seq 2 · fresh-authored (U13-G2 stage-2) · mcq4
But-2-ene exists in two forms that convert into each other reversibly: cis-but-2-ene(g) ⇌ trans-but-2-ene(g). Which change is the forward reaction?
The cis form converting into the trans form. KEY
The trans form converting back into the cis form.
The two forms exchanging atoms with each other.
A but-2-ene molecule splitting into two smaller molecules.
KEQ-023 · easy · recall · seq 3 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible mixture settles into dynamic equilibrium. Which condition defines that state?
The rate of the forward reaction equals the rate of the reverse reaction. KEY
The amounts of reactant and product have become the same as each other.
Every reactant particle has been converted into product.
Both reactions have come to a complete stop.
KEQ-024 · easy · recall · seq 4 · fresh-authored (U13-G2 stage-2) · mcq4
A lab partner claims: 'Even at equilibrium, a sealed mixture shows small, steady rises in its product concentration.' Which correction is right?
From equilibrium on, no concentration rises or falls at all. KEY
The claim is right — small changes are always allowed at equilibrium.
The claim is wrong only about the product — reactant concentrations are the ones that keep rising.
The claim is right if the small rises are later canceled by equal small falls, averaging out steady.
KEQ-021 · easy · application · seq 5 · fresh-authored (U13-G2 stage-2) · mcq4
The gas N₂F₄ splits reversibly into two identical pieces: N₂F₄(g) ⇌ 2NF₂(g). Which statement about this change is correct?
NF₂ molecules can join together to re-form N₂F₄. KEY
The splitting cannot be undone once it has happened.
The ⇌ means exactly half of the N₂F₄ has split so far.
A second, separate arrow must be added before the change can run backward.
KEQ-022 · easy · application · seq 6 · fresh-authored (U13-G2 stage-2) · mcq4
A classmate says: 'While product particles are re-forming reactants, the forward reaction has to pause and wait its turn.' Which correction is right?
Neither direction pauses — reactants keep forming products at the same moment. KEY
The classmate is right — brief pauses let each direction have the mixture to itself.
The classmate is right, but only in mixtures that hold very little product.
The forward reaction does more than pause — it stops for good once the reverse reaction begins.
KEQ-023 · medium · application · seq 7 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reaction vessel's readout marks one product of a reversible gas reaction as 'holding steady', and the mixture is known to be at dynamic equilibrium. At the particle level, what is happening to that product?
It is still being made and still being used up, at equal rates. KEY
Nothing happens to it anymore — its particles are finished reacting.
It is made in short bursts, with quiet stretches in between.
It is used up slightly faster than it is made, too slowly for the readout to notice.
KEQ-025 · medium · application · seq 8 · fresh-authored (U13-G2 stage-2) · mcq4
At dynamic equilibrium, each substance's particles keep reacting, yet its measured amount never moves. What squares the two facts?
The forward and reverse reactions keep happening at the same rate, so each substance is made exactly as fast as it is used up. KEY
Every substance in the mixture has settled at one shared concentration, so no exchange between them can shift anything.
The forward reaction finished first, and the leftover reverse reaction is now too slow to show on any readout.
The sealed container absorbs whatever change the reactions produce.
KEQ-024 · medium · application · seq 9 · fresh-authored (U13-G2 stage-2) · mcq4
A lab notebook for a sealed reversible run contains the entry: 'Product concentration: still rising.' Had the mixture reached equilibrium at the time of that entry?
No — at equilibrium the product's concentration would have stopped changing. KEY
Yes — a rising product is allowed at equilibrium, as long as the rise is smooth.
Yes, provided the reactant concentrations had already leveled off.
The entry must be mistaken, because concentrations cannot rise inside sealed apparatus.
KEQ-026 · medium · application · seq 10 · fresh-authored (U13-G2 stage-2) · mcq4
In a sealed run, the product's amount climbed for a while, then settled at a level clearly below the reactant's amount and has stayed there ever since. What state has the mixture reached?
Equilibrium — the amounts are steady, and they never needed to match. KEY
A stalled state partway to equilibrium, from which the climb will eventually resume.
A false balance — a true equilibrium would put the product level above the reactant's.
A finished reaction, with the leftover reactant no longer able to react.
KEQ-021 · medium · application · seq 11 · fresh-authored (U13-G2 stage-2) · mcq4
Seeing ⇌ in an equation, a student decides the equation 'may only be read from right to left'. Which correction is right?
The symbol licenses both readings — the reaction runs left to right and right to left. KEY
The student is right — the reverse reading is the official one for ⇌ equations.
The student picked the wrong side — ⇌ equations may only be read left to right.
The correct reading direction depends on which side of the equation holds more substance.
KEQ-022 · medium · application · seq 12 · fresh-authored (U13-G2 stage-2) · mcq4
Partway through a sealed reversible run — well before equilibrium — instruments confirm that product is being made at this very moment. What is the reverse direction doing right now?
Running as well — product particles elsewhere are re-forming reactants. KEY
Waiting until the forward direction finishes its work.
Waiting until the mixture reaches equilibrium.
Nothing — the two directions cannot share the mixture at one moment.
KEQ-023 · medium · application · seq 13 · fresh-authored (U13-G2 stage-2) · mcq4
A reversible gas reaction runs inside a balloon with a pinhole, and product leaks out steadily. Can the mixture in this leaking balloon still strike the matched-rate balance?
No — while a substance keeps leaving, the two rates never get to match. KEY
Yes — the leak only delays the balance, which forms in the end.
Yes — a balloon counts as a closed container even with a pinhole.
No — a stretchy balloon could never host the balance anyway, because its walls flex.
KEQ-024 · medium · application · seq 14 · fresh-authored (U13-G2 stage-2) · mcq4
About a sealed mixture at dynamic equilibrium, a student writes: sentence 1 — 'Every concentration in it is constant'; sentence 2 — 'So both reactions must have come to a stop.' Which judgment of the two sentences is right?
Sentence 1 is correct; sentence 2 does not follow — the reactions keep running. KEY
Both sentences are correct — constant readings can only come from stopped reactions.
Both sentences are wrong — concentrations keep changing at equilibrium.
Sentence 1 is wrong and sentence 2 is correct — the reactions stop while the amounts drift.
KEQ-027 · medium · application · seq 15 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run whose two curves settled at clearly different heights from 9 minutes onward. Which reading of the mixture at 11 minutes is correct?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls then runs flat from 9 minutes at a mid height; the product curve rises from zero t… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "9 minutes", "note": "flat at a mid height"}, {"label": "product", "shape":… · curveCount: 2 · curveConstraint: "two distinct curve paths, one falling and one rising to clearly different flat heights; both flat from 9 minutes"
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Each concentration is holding its own constant value, with both reactions still running. KEY
The two settled heights will slowly move together until they meet.
The different heights show the mixture failed to reach equilibrium.
The reactions ended at 9 minutes, leaving the heights stranded where they were.
KEQ-026 · medium · application · seq 16 · fresh-authored (U13-G2 stage-2) · mcq4
To decide whether a sealed reversible mixture is at equilibrium, which single question should you ask of its measurements?
Does each amount give the same value whenever it is measured? KEY
Are the reactant and product amounts the same as each other?
Is there more product than reactant in the mixture?
Has the mixture been reacting for long enough?
KEQ-025 · medium · application · seq 17 · fresh-authored (U13-G2 stage-2; replaces spared pool[14] water-tank analogy with a fresh flow analog) · mcq4
A stadium holds a steady crowd while people stream in through the gates and out through the exits. A sealed mixture at dynamic equilibrium holds a steady amount of product while two reactions run. What makes the two situations alike?
In both, the level holds because what comes in runs exactly as fast as what goes out. KEY
In both, the level holds because the flows in and out have been closed off.
In both, the walls are what fix the level in place.
They are not alike — a crowd can hold steady, but a reacting substance's amount cannot.
KEQ-027 · hard · transfer · seq 18 · fresh-authored (U13-G2 stage-2) · mcq4
The graph tracks THREE substances in one sealed reversible run: two reactant curves that look level from 4 minutes, and one product curve still rising through 7 minutes. Was the mixture at equilibrium at 7 minutes?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 10 minutes; two reactant curves fall from different starting heights and both look level from 4 minutes at the graph's … · style: "unit-standard concentration-time graph, curves labeled reactant 1, reactant 2, and product in-diagram" · curves: [{"label": "reactant 1", "shape": "falling-then-flat", "start": "high", "flatFrom": "4 minutes"}, {"label": "reactant 2", "shape": "falling-then-flat", "star… · curveCount: 3 · curveConstraint: "three pairwise-distinct curve paths; the two reactant curves are level from 4 minutes at different heights while the product curve is still visibly rising t…
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No — every curve must be level, and the product curve was still rising. KEY
Yes — two level curves out of three is a clear majority.
Yes — the reactants settled at 4 minutes, and reactants are what equilibrium is judged on.
No — and with three curves this run will never reach any equilibrium.
KEQ-026 · hard · transfer · seq 19 · fresh-authored (U13-G2 stage-2) · mcq4
The figure shows an exam sketch of an equilibrium mixture: its two concentration bars happen to be EQUAL in height, and both repeat that height at all three times. A student says the equal bars are what prove the artist understood equilibrium. What is the right verdict on the sketch?
FIGURE (qualitative-bars-over-time · stimulus) — series: [{"label": "reactant", "height": "medium", "note": "same medium height at every time"}, {"label": "product", "height": "medium", "note": "same medium height … · timePoints: ["earlier", "later", "later still"] · style: "no numbers on any axis; exam-sketch bar pairs with the two bars equal in height and identical across all three times"
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It is correct because each bar holds its height — the equal heights are a coincidence, not the requirement. KEY
The student is right — equal heights are exactly what equilibrium demands.
The sketch is wrong — equilibrium amounts must always differ, so equal bars are impossible.
The sketch is wrong — equal bars can only depict a mixture whose reactions have stopped.
KEQ-027 · hard · transfer · seq 20 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run whose two curves level out from 7 minutes without ever having crossed. A student objects: 'These curves never met, so this run never truly reached equilibrium.' What does the graph actually show?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 10 minutes; the reactant curve falls from high and levels out from 7 minutes while still clearly above the product curv… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "7 minutes", "note": "levels out clearly above the product curve"}, {"label… · curveCount: 2 · curveConstraint: "two distinct curve paths that never touch or cross; both level from 7 minutes at clearly different heights"
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The run reached equilibrium at 7 minutes — curves never need to meet for that. KEY
The student is right — a crossing is the required signature of a completed approach.
The student is right — without a crossing the two rates can never have matched.
The run reached a partial equilibrium that a crossing would have completed.
KEQ-025 · medium · application · seq 21 · fresh-authored (U13-G2 stage-2; QB sec.4 explain-why FRQ, KEQ-025 home) · frq
A demonstration cell contains a reversible gas reaction, sealed and left running. Its product readout has been perfectly steady since noon, and a molecular sensor confirms product particles are still being consumed all the time.
A demonstration cell contains a reversible gas reaction, sealed and left running. Its product readout has been perfectly steady since noon, and a molecular sensor confirms product particles are still being consumed all the time. (a) State what the steadiness since noon says about the rates of the forward and reverse reactions. (b) Explain WHY the readout can stay perfectly steady while product particles are consumed all the time. Use the wording the lesson taught. (c) A student argues: 'Only the forward reaction is still going, just very slowly — that is why nothing seems to change.' In one or two sentences, explain why that cannot be right and describe what the two reactions are actually doing.
Part a. State what the steadiness since noon says about the rates of the forward and reverse reactions.
Model: The steadiness says the rate of the forward reaction equals the rate of the reverse reaction — and neither is zero, since the sensor still sees product being consumed.
Part b. Explain WHY the readout can stay perfectly steady while product particles are consumed all the time. Use the wording the lesson taught.
Model: The readout stays steady because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Consumption never shows on the readout because every consumed particle is replaced in step.
Part c. A student argues: 'Only the forward reaction is still going, just very slowly — that is why nothing seems to change.' In one or two sentences, explain why that cannot be right and describe what the two reactions are actually doing.
Model: If only the forward reaction were running, product would be made with nothing using it up, and the readout would climb — slowly, but it would climb. In fact both reactions are running together, each exactly as fast as the other.
Full model answer. (a) The steadiness says the rate of the forward reaction equals the rate of the reverse reaction — and neither is zero, since the sensor still sees product being consumed. (b) The readout stays steady because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Consumption never shows on the readout because every consumed particle is replaced in step. (c) If only the forward reaction were running, product would be made with nothing using it up, and the readout would climb — slowly, but it would climb. In fact both reactions are running together, each exactly as fast as the other.

Form E

KEQ-021 · easy · recall · seq 1 · fresh-authored (U13-G2 stage-2) · mcq4
The same reaction could be written once with → and once with ⇌. What would the two versions claim differently?
The ⇌ version shows the reaction running in both directions; the → version shows only one. KEY
The ⇌ version claims a faster reaction than the → version.
The → version is the balanced form, and the ⇌ version is the unbalanced draft.
The ⇌ version claims the two sides are present in matching amounts.
KEQ-022 · easy · recall · seq 2 · fresh-authored (U13-G2 stage-2) · mcq4
The gas NOCl decomposes reversibly: 2NOCl(g) ⇌ 2NO(g) + Cl₂(g). Which change is the reverse reaction?
NO and Cl₂ reacting to re-form NOCl. KEY
NOCl breaking apart into NO and Cl₂.
NOCl reacting with NO to give off Cl₂.
NO molecules pairing up with each other.
KEQ-023 · easy · recall · seq 3 · fresh-authored (U13-G2 stage-2) · mcq4
Chemists report that a sealed reversible mixture 'reached dynamic equilibrium overnight'. What exactly became equal?
The rates of the two opposite reactions. KEY
The concentrations of the reactants and the products.
The number of molecules on the left and right sides of the equation.
The lengths of time each direction spends running.
KEQ-024 · easy · recall · seq 4 · fresh-authored (U13-G2 stage-2) · mcq4
A data logger tracks one product's concentration in a sealed reversible mixture that settled into equilibrium at midday and was then left alone. What does the afternoon's trace look like?
A level line holding one value. KEY
A gentle downward slope as the product is used up.
A staircase that climbs each time the forward reaction runs.
A repeating wave as the two reactions trade places.
KEQ-021 · easy · application · seq 5 · fresh-authored (U13-G2 stage-2) · mcq4
The blue solid on a moisture-indicator card is CoCl₂. As the card takes up moisture it turns pink, and gentle warming drives the moisture out and turns it blue again. What kind of change is happening on the card?
A reversible reaction — the pink product can turn back into the blue starting substance. KEY
A one-way change — the pink form is a new substance that cannot change back.
Two unrelated changes that happen to share the same card.
A physical change only, since the solid never becomes anything new.
KEQ-022 · easy · application · seq 6 · fresh-authored (U13-G2 stage-2) · mcq4
A study-group summary states: 'The reverse reaction cannot begin until the mixture holds more product than reactant.' Which correction is right?
The reverse reaction runs as soon as any product exists, whatever the amounts. KEY
The summary is right — a product majority is what unlocks the reverse direction.
The summary sets the bar too high — the reverse reaction begins once the amounts are level.
The summary is right about the start, but the reverse reaction then keeps going even after the product runs low.
KEQ-023 · medium · application · seq 7 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed vessel reached dynamic equilibrium yesterday and has sat untouched since. Which statement describes what its product has been doing overnight?
Being formed and being converted back, continuously and at the same rate. KEY
Sitting unchanged — the reactions wound down once the balance arrived.
Being formed all night, with the reverse direction resting until morning.
Slowly drifting downward in amount as the night wore on.
KEQ-025 · medium · application · seq 8 · fresh-authored (U13-G2 stage-2) · mcq4
In a sealed reversible gas mixture, the product's amount has held steady for hours. A classmate proposes: 'The forward reaction runs a little faster than the reverse, and that small surplus keeps the product topped up and steady.' Which correction is right?
A surplus would make the amount rise — it is steady because making and using run at exactly the same rate. KEY
The proposal is right — a small surplus is exactly what steady amounts require.
The proposal has it backward — the reverse reaction runs slightly faster, and the small deficit keeps the amount trimmed and steady.
The proposal is right, and the surplus leaves through the container walls, which is why nothing builds up.
KEQ-024 · medium · application · seq 9 · fresh-authored (U13-G2 stage-2) · mcq4
A sealed reversible mixture is left overnight with a recorder logging every substance continuously, and nothing touches the flask. Every trace holds perfectly level from evening to morning, though the traces sit at different heights. What is the best reading of the situation?
The mixture is at equilibrium — concentrations holding constant are its signature. KEY
The reactions must have stopped overnight, leaving the readings frozen.
It cannot be at equilibrium yet — at equilibrium every substance's trace would settle at one shared height.
The instruments must have stuck, since a reacting mixture cannot give the same reading twice.
KEQ-026 · medium · application · seq 10 · fresh-authored (U13-G2 stage-2) · mcq4
Two chemists study a sealed mixture whose readings have been flat for a week. One argues its wildly unequal amounts rule out equilibrium; the other says the flat week settles it. Who is right, and why?
The second chemist — steadiness over time, not equality between substances, is the test. KEY
The first chemist — amounts that unequal cannot belong to a balanced mixture.
Both, in part — unequal amounts and flat readings each satisfy half of the test.
Neither — a week of readings is not enough evidence either way.
KEQ-021 · medium · application · seq 11 · fresh-authored (U13-G2 stage-2) · mcq4
A study guide's margin note claims: 'A ⇌ in an equation is a warning that the products are too unstable to keep.' Which correction is right?
⇌ carries one message — the products can react to re-form the reactants; it is no warning label. KEY
The note is right — chemists reserve ⇌ for products that fall apart on storage.
The note points at the wrong side — ⇌ warns about unstable reactants instead.
The note is right whenever the mixture is heated, since heat makes the products fall apart.
KEQ-022 · medium · application · seq 12 · fresh-authored (U13-G2 stage-2) · mcq4
A molecular-scale movie of a sealed reversible mixture shows, within one single moment, some particles joining into product while others split back into reactants. A student calls the movie badly edited. Which response is correct?
The movie is faithful — both directions of a reversible reaction really do run at the same moment. KEY
The student is right — a real mixture would show one direction per scene.
The student is right — the footage must splice together two different containers.
The student is half right — such two-way scenes only occur while the mixture is being heated.
KEQ-023 · medium · application · seq 13 · fresh-authored (U13-G2 stage-2) · mcq4
A technician must choose apparatus in which a reversible gas reaction can settle into dynamic equilibrium. Which choice does the job?
A sealed vessel that keeps every substance inside. KEY
A vessel with a vent that releases product as it forms.
A wide dish, open to the room, giving the gases space to settle.
Any container at all — the apparatus has no bearing on equilibrium.
KEQ-024 · medium · application · seq 14 · fresh-authored (U13-G2 stage-2) · mcq4
For a sealed reversible mixture at dynamic equilibrium, what does a concentration readout show, and what is going on underneath it?
A flat readout — the amount has stopped changing even though the reaction has not stopped. KEY
A flat readout — the chemistry inside has stopped for good.
A gently climbing readout — the running chemistry keeps nudging the value upward.
A flickering readout — the value wobbles as the reactions start and stop in turns.
KEQ-027 · medium · application · seq 15 · fresh-authored (U13-G2 stage-2) · mcq4
The graph shows a sealed reversible run, level in both curves from 5 minutes, with the reactant curve settled well above zero. A classmate reads the flat lines as proof the reaction 'ran to completion'. What does the graph actually show?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 10 minutes; the reactant curve falls then runs flat from 5 minutes at a height well above zero; the product curve rises… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "5 minutes", "note": "flat height well above zero"}, {"label": "product", "… · curveCount: 2 · curveConstraint: "two distinct curve paths, one falling and one rising to different flat heights; both flat from 5 minutes"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 10 minutes; the reactant curve falls then runs flat from 5 minutes at a height well above zero; the product curve rises from zero then runs flat from 5 minutes at a different height",
  "style": "unit-standard concentration-time graph, curves labeled reactant and product in-diagram",
  "curves": [
   {
    "label": "reactant",
    "shape": "falling-then-flat",
    "start": "high",
    "flatFrom": "5 minutes",
    "note": "flat height well above zero"
   },
   {
    "label": "product",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "5 minutes",
    "note": "flat at a different height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths, one falling and one rising to different flat heights; both flat from 5 minutes"
 }
}
An equilibrium with reactant left over — the reactant's flat line sits well above zero. KEY
The reading is correct — curves only go flat when the reactants are completely used up.
The reactant line is only pausing — left longer, it will still fall the rest of the way to zero.
The run did complete, and the leftover reactant line belongs to a second, slower reaction.
KEQ-026 · medium · application · seq 16 · fresh-authored (U13-G2 stage-2) · mcq4
A flash card reads: 'Equilibrium = the reactant and product amounts have become equal.' How should the card be corrected?
Equilibrium = each amount constant over time, whatever the amounts happen to be. KEY
Keep the card, but add that the amounts must also be constant — both conditions are required.
Change it to: the amounts must be equal at least once during the run.
Change it to: only the product's amount needs to hold constant.
KEQ-025 · medium · application · seq 17 · fresh-authored (U13-G2 stage-2) · mcq4
Take any stretch of time you like in a sealed mixture at dynamic equilibrium. Molecule for molecule, how does the making of one substance over that stretch compare with its using-up?
They match — every loss is balanced by a gain at the same rate. KEY
The making runs slightly ahead, so the mixture never runs short.
The using-up runs slightly ahead, which is why mixtures eventually fade.
Both are zero over any stretch — that is what being at equilibrium means.
KEQ-027 · hard · transfer · seq 18 · fresh-authored (U13-G2 stage-2) · mcq4
On the graph of a sealed reversible run, the product curve looks level from 2 minutes, while the reactant curve keeps easing downward and only levels out at 9 minutes. From the graph, at what time was equilibrium reached?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the product curve rises quickly from zero and looks level from 2 minutes at the graph's scale; the reactant… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "product", "shape": "rising-then-flat", "start": "zero", "flatFrom": "2 minutes"}, {"label": "reactant", "shape": "falling-then-flat", "start": "h… · curveCount: 2 · curveConstraint: "two distinct curve paths; the product curve is level from 2 minutes while the reactant curve keeps falling and levels only at 9 minutes"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 12 minutes; the product curve rises quickly from zero and looks level from 2 minutes at the graph's scale; the reactant curve falls steeply then eases downward, leveling only at 9 minutes at a different height",
  "style": "unit-standard concentration-time graph, curves labeled reactant and product in-diagram",
  "curves": [
   {
    "label": "product",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "2 minutes"
   },
   {
    "label": "reactant",
    "shape": "falling-then-flat",
    "start": "high",
    "flatFrom": "9 minutes",
    "note": "eases downward gradually before leveling"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths; the product curve is level from 2 minutes while the reactant curve keeps falling and levels only at 9 minutes"
 }
}
At 9 minutes — the time from which every curve stays flat. KEY
At 2 minutes — the first leveling marks the arrival.
Midway between the two levelings — the arrival splits the difference.
Never — curves that level out at different times never produce an equilibrium.
KEQ-026 · hard · transfer · seq 19 · fresh-authored (U13-G2 stage-2) · mcq4
The figure shows a printed diagram of a mixture said to be at equilibrium across two times: at the first time one bar is tall and the other short, and at the second time the two bars have traded heights. A reviewer says the printer must have swapped the panels; the illustrator insists a normal equilibrium can do this. Who is right?
FIGURE (qualitative-bars-over-time · stimulus) — series: [{"label": "substance 1", "height": "tall then short", "note": "tall at the first time, short at the second"}, {"label": "substance 2", "height": "short then… · timePoints: ["first time", "second time"] · style: "no numbers on any axis; two printed panels in which the two bars visibly trade heights between the times"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "qualitative-bars-over-time",
  "series": [
   {
    "label": "substance 1",
    "height": "tall then short",
    "note": "tall at the first time, short at the second"
   },
   {
    "label": "substance 2",
    "height": "short then tall",
    "note": "short at the first time, tall at the second"
   }
  ],
  "timePoints": [
   "first time",
   "second time"
  ],
  "style": "no numbers on any axis; two printed panels in which the two bars visibly trade heights between the times"
 }
}
The reviewer — at equilibrium each amount keeps its own steady value, so the heights cannot trade places. KEY
The illustrator — equilibrium means the two substances take turns holding the larger amount.
The illustrator — provided the two heights always sum to the same total, any redistribution is allowed.
Neither — two time points are too few to judge a drawing of equilibrium.
KEQ-027 · hard · transfer · seq 20 · fresh-authored (U13-G2 stage-2) · mcq4
On the graph of a sealed reversible run, the curves cross at 4 minutes and both become level from 10 minutes. A study group decides that 'equilibrium happened gradually across the stretch from 4 to 10 minutes'. Which correction is right?
FIGURE (concentration-time-approach · stimulus) — description: "concentration against time from 0 to 12 minutes; the reactant curve falls from high and the product curve rises from zero; the curves cross once at 4 minute… · style: "unit-standard concentration-time graph, curves labeled reactant and product in-diagram" · curves: [{"label": "reactant", "shape": "falling-then-flat", "start": "high", "flatFrom": "10 minutes"}, {"label": "product", "shape": "rising-then-flat", "start": "… · curveCount: 2 · curveConstraint: "two distinct curve paths crossing exactly once, at 4 minutes, while both are still visibly changing; both level from 10 minutes at different heights"
raw spec
{
 "figureRequired": true,
 "figureShape": "stimulus",
 "figureSpec": {
  "figureType": "concentration-time-approach",
  "description": "concentration against time from 0 to 12 minutes; the reactant curve falls from high and the product curve rises from zero; the curves cross once at 4 minutes while both are still visibly changing; both become level from 10 minutes at different heights",
  "style": "unit-standard concentration-time graph, curves labeled reactant and product in-diagram",
  "curves": [
   {
    "label": "reactant",
    "shape": "falling-then-flat",
    "start": "high",
    "flatFrom": "10 minutes"
   },
   {
    "label": "product",
    "shape": "rising-then-flat",
    "start": "zero",
    "flatFrom": "10 minutes",
    "note": "level at a different height"
   }
  ],
  "curveCount": 2,
  "curveConstraint": "two distinct curve paths crossing exactly once, at 4 minutes, while both are still visibly changing; both level from 10 minutes at different heights"
 }
}
The stretch was the approach — equilibrium holds from 10 minutes on, once both curves are level. KEY
No correction needed — equilibrium is a stretch, opened by the crossing and closed by the leveling.
The group has the stretch backward — equilibrium ran from the start until the crossing ended it.
The group should average the two marks and date equilibrium at 7 minutes.
KEQ-025 · medium · application · seq 21 · fresh-authored (U13-G2 stage-2; QB sec.4 explain-why FRQ, KEQ-025 home) · frq
An undisturbed, sealed flask of a reversible gas reaction has shown the same product reading at every check since it settled. A monitoring strip confirms that molecules of the product are still converting back to reactants around the clock.
An undisturbed, sealed flask of a reversible gas reaction has shown the same product reading at every check since it settled. A monitoring strip confirms that molecules of the product are still converting back to reactants around the clock. (a) State the condition relating the forward and reverse reaction rates in the settled flask. (b) Explain WHY the product reading never moves even though its molecules keep converting back. Use the wording the lesson taught. (c) A classmate writes: 'The reactions are resting at equilibrium; if the flask were warmed they would start up again.' In one or two sentences, explain what is wrong with the 'resting' picture and what the reactions are doing right now.
Part a. State the condition relating the forward and reverse reaction rates in the settled flask.
Model: The rate of the forward reaction equals the rate of the reverse reaction, and both reactions are running — nothing is paused.
Part b. Explain WHY the product reading never moves even though its molecules keep converting back. Use the wording the lesson taught.
Model: The reading never moves because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Every molecule that converts back is matched by a newly made one, so the measured value holds.
Part c. A classmate writes: 'The reactions are resting at equilibrium; if the flask were warmed they would start up again.' In one or two sentences, explain what is wrong with the 'resting' picture and what the reactions are doing right now.
Model: The reactions are not resting — the monitoring strip catches product converting back around the clock, and the forward reaction is matching it at this very moment. There is nothing to restart: both directions have been running continuously the whole time.
Full model answer. (a) The rate of the forward reaction equals the rate of the reverse reaction, and both reactions are running — nothing is paused. (b) The reading never moves because the forward and reverse reactions keep happening at the same rate, so the product is made exactly as fast as it is used up. Every molecule that converts back is matched by a newly made one, so the measured value holds. (c) The reactions are not resting — the monitoring strip catches product converting back around the clock, and the forward reaction is matching it at this very moment. There is nothing to restart: both directions have been running continuously the whole time.

End of Topic Test: Shifting an equilibrium: adding and removing substances & Temperature, pressure, and choosing conditions (u13-t3 · 5 forms × 21 items · 105 total)

David's notes for James:
full notes file
# u13-t3 — notes/queries from David's side (U13-G3)

1. FRQ COMPOSITION - frq_per_form = 1 (your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - Unit 13 is the qualitative kinetics/equilibrium unit: QB-v2 section 4's calculation-heavy list (stoichiometry, gas laws, molarity/dilution, calorimetry, mole conversions) names nothing in this scope, zero of the 13 scope atoms are quant (all fact/discrimination/explanation/representation, no displayEquation anywhere), and the only numeric production is the KEQ-036 gas-mole COUNT. The single FRQ = the KEQ-040 condition-choosing capstone (its spec's own D5 exception slot), terminal at seq 21 on every form: (a) which substance to add and which to remove, with the counteract logic; (b) raise or lower the temperature, justified from the written heat; (c) count the gas moles on each side, call the pressure move, and state what a catalyst would and would not do. Rubric 30/35/35 = 100; model answers word-only (nothing typed requires sub/superscripts). D5 scaling: min(20, 6 x 13) = 20 -> full 20 objective rows (every lesson >=1 row; 7 lessons carry an extra row per the blueprint rationale; KEQ-040 = 2 objective rows + the FRQ). KEQ-036's count is a live numeric ENTRY row (seq 11 on every form) per the HS numeric-entry rule - 5 numeric slots total, V-T6 fully armed and PASSING.
2. SEED USAGE - 37 of your 39 pool items placed (form A fully seeded with 20, B carries 12, C carries 5; D and E are fresh-authored parallels). demo_curated[0..5] are KEQ-020/KEQ-027 items from EARLIER gates' scopes -> recorded out_of_scope_curated (U6-G4 precedent, not placeable under V-T11 one-lesson-per-item); demo_curated[6..8] dedup to pool[36..38] (both refs cited on the placements). prod_seeds = [] for this topic pair; the cold list is ignored by standing answer #3. The TWO spares (not merit drops): pool[23] (KEQ-035 hard cooling rate-vs-amount mirror - KEQ-035 holds one medium row and the rate-vs-amount transfer construct is carried at seq 17 under KEQ-034, whose row owns that surface; the PCl5 system is also lesson corpus) and pool[31] (KEQ-038 does-NOT-shift multi-equation transfer - the construct is homed at seq 18 under KEQ-037 on every form; V-T11 bars the cross-lesson placement).
3. SEED FRESHENING UNDER THE STRICT SEED RULING (no freshness exemption; the shared validator was never weakened): the scope lessons consume the unit's whole taught-equilibria stock as corpus, so test instances draw on a fresh REAL-equilibria palette PARTITIONED BY FORM (each named system lives on exactly one form; all real chemistry with true exo/endo signs, classifications supplied in every stem per assetPolicy). 27 must_freshen_instance placements were executed at stage 2 - construct, option logic, and misconception keys preserved, system swapped to the placement form's palette, ALL fields swept for the old instance (QB 3.9). 10 verbatim-candidate seeds were field-mapped only. One numeric remint: pool[25]'s answer was a bare power of ten (V-T6 bars it) - reminted with the freshened system; your written-heat-not-counted twist is preserved at A#11 (answer 2, the heat term never counts as a gas mole).
4. STAGE-3 DE-BROMINATION AMENDMENT (confirm - recorded blueprint amendment): the V-T7 gram scan tokenizes formulas, so the non-lexicon token 'br' made every Br2-bearing equation collide across forms (B NOBr / C HBr / E IBr). Bromine is confined to FORM C, protecting your verbatim HBr seed p0: form B's NOBr seats moved to SbCl5(g) = SbCl3(g) + Cl2(g) (real, endothermic, 1-vs-2 gas moles; raise-T + decrease-P FRQ polarity preserved) and form E's IBr seats moved to I2(g) + Cl2(g) = 2ICl(g) (real, exothermic, 2-vs-2). Two stray Br2 leftovers in non-C explanation fields were caught at QC and swept (E#11 gradingNote/keyExplanation, E#20 keyExplanation); a final Br sweep over non-C forms finds provenance notes only.
5. A#1/D#1 CLASS - YOUR SEED p1 WAS VERBATIM AND THE CLASS MAY APPLY UPSTREAM: blind review hard-failed the amount-read framing 'after a disturbance, the mixture holds less X - which way did it shift?' because a bare 'disturbance' could BE a direct addition/removal of X itself, making the direction indeterminate (removing X directly leaves less X with no reactant-ward shift). Both items now pin the disturbance to a temperature-only change ('After its temperature is changed...'), which makes the less-product -> shifted-toward-reactants read airtight. You may want the same pin on the lesson-bank original.
6. C#16 CLASS (KEQ-040 condition-set MCQs on equal-gas-moles systems): a distractor that differs from the key ONLY by a null pressure move is outcome-identical and genuinely defensible (blind hard fail). The pressure-myth distractor must carry a real cost - C#16's now keeps the HBr in the vessel for its pressure push (losing the removal benefit), the same pattern as A#16. Worth checking any condition-set items you author on 2-vs-2 systems.
7. SEALED-REGISTER RULING (sol authoritative, recorded): 'sealed' + a ONE-SHOT add/remove event = the lessons' own register and is KEPT (incl. the KEQ-032 'drawn off from the sealed...' idiom on the one-shot withdrawal items C#8/E#8); 'sealed' + CONTINUOUS operation (remove-as-it-forms production runs) is internally inconsistent and 'sealed' was dropped from every continuous-feed/withdrawal item (seq 16/20/21 + D#5/E#5), including items sol passed (deterministic sweep).
8. KEQ-034 vs KEQ-035 SKILL SPLIT (construct discipline): heating and cooling are SEPARATE lesson skills in this scope. A blind de-duplication wave briefly flipped C#9/E#9 to cooling; sol correctly failed them for lesson misalignment (KEQ-034's atom is raising temperature) and they were reverted to warming - the duplication was instead fixed by SYSTEM (different equilibrium than their #17 twins), not by direction.
9. C#21 RUBRIC ALTERNATE (chemistry correctness): on CO2(g) + 3H2(g) = CH3OH(g) + H2O(g) + heat, the rubric and model answer license removing the methanol OR the water as it forms - both are products, and removing either shifts the equilibrium toward the products. The graded observable is the counteract logic, not which product is named.
10. WITHIN-FORM PALETTE CUEING (recorded survivor class, deliberate): each form reuses its own small stock of real systems across rows - the KEQ-038 lesson's own specNotes call this 'deliberate cross-lesson stock reuse' and 20-item single-topic forms cannot avoid it. Mitigations executed: the seq-11 numeric systems were moved OFF each form's seq-18 KEY system and off exact-duplicate mole-count pairs; the pressure/temperature MCQs partially foreshadow the FRQ dials on the same palette, mitigated by the FRQ rubric requiring per-dial justification (stating a direction earns nothing without the written-heat / mole-count logic).
11. COURSE-WIDE INSTANCE RULE (binding, wave 6) - one lesson-canon exemption to know about: the authored bank is 'reactor'-free (a pre-blind guard caught 7 stem occurrences of industrial-register 'reactor'; all reworded to vessel/flask/plant apparatus words before any review), and a packaging re-sweep over forms_final AND this export body shows ZERO banned-token hits in student-facing content (uranium/plutonium/enrichment/reactor/weapon/fissile/meltdown/dose all clean). The one scoped exemption: KEQ-039's OWN Haber-dials rule line reads '...the reactor delivers its maximum NH3' - lesson canon we cannot rewrite - so the blind-packet conventions block (which quotes rule lines verbatim) carries a whitelist for exactly that sentence. Nothing student-facing in the test carries the word.
12. GATE_CONFIG ADJUDICATIONS (scoped whitelists; the shared validator ran UNMODIFIED to 12/12 PASS, exit 0): [1] 11 extra_task_idioms = the scope lessons' own FIXED TASK PHRASINGS that sit below the A2 auto-derivation threshold (setup-clause or single-item phrasings): the KEQ-037/KEQ-038 piston and container-volume pressure framings, 'is compressed into a smaller volume', 'is allowed to expand (into a larger volume)', the KEQ-033 written-heat display phrasings ('is shown with heat written in', '(g) + heat is shown'), the KEQ-032 draw-off register ('is drawn off from the sealed'), and KEQ-031 F3's full what-ends-the-shift task clause - all instance-free (no material words, no numbers; instances still scan via the material/number lanes). [2] extra_convention_phrases '(l)' and '(aq)' = state-symbol convention tokens (A4 class): '(g)' already enters the canonical lexicon via the rule lines' gas equations, but this scope's rule lines contain no liquid/aqueous species, so the two symbols were missing and every H2O(l)/(aq) formula fragment scanned as an instance gram. Substance tokens keep their single-form scan discipline; materials and numbers still scan in full.
13. VALIDATOR WARNS ROUTED TO BLIND REVIEW (11 warn lines, all expected classes, zero telegraphing hard fails across six blind rounds): A#5 'water' = a material taught elsewhere in scope sitting inside your verbatim ester seed (distilling the water out IS the item's chemistry); 10 form-C non-adjacent 'butane' pair lines across seqs 3/6/9/15/18 - form C is the butane-isomer form by palette design (n-butane = isobutane isomerization; the formula fragments carry no scan weight, only the word does).
14. QC LEDGER (provenance, no action needed): pre-blind banned-token guard -> 8 apparatus-word edits; blind round 1 (r1 = gpt-5.6-sol blind role, r2 = gemini; all 105 items) = 190/190 key picks, 3 hard fails (A#1/D#1 amount-read class, C#16 null-pressure-move distractor), 73 advisory flags triaged into two class-sweep patch waves (31 + 8 edits); re-blind rounds on every changed item (fresh tags r3-r10 + gemini g2/g3) = 96/96 key picks total, 1 intermediate hard fail (A#12 adiabatic-compression ambiguity -> isothermal + settled-state pins, class-swept) then clean. FRQ/numeric manual grading: 10 keyless slots (5 mole-counts + 5 FRQs), r1/r2 both match on all 10; changed items re-graded on every subsequent round - all match, incl. form B's raise-T + decrease-P polarity and the C#21 either-product alternate. sol authoritative loops: 13 FAIL -> 3 -> 0; final ledger 105/105 pass, coverage audited item-by-item (no stale verdicts). Deterministic validator: 12/12 PASS exit 0 at stage 3, re-proven at finalize and again fresh at packaging (byte-identical output).
15. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings anywhere - the gate is qualitative equilibrium chemistry on classroom-safe real systems (industrial-adjacent but benign: methanation, reverse water-gas shift, ester equilibria, dimerizations, isomerizations, halogen interconversions); every named system appears on exactly one form and every exo/endo classification is chemically true and SUPPLIED in the stem.

Form A

KEQ-028 · easy · recall · seq 1 · seed:U13-G3.pool[1] (KEQ-028 gateItems, James) - VERBATIM (field mapping only; system corpus-free) · mcq4
After its temperature is changed, a sealed H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g) mixture settles into a new equilibrium holding less CO than before. Which way did the equilibrium shift?
Toward the reactants — the new mixture holds less product. KEY
Toward the products — CO is one of the products, so any change in it is a product shift.
Toward the reactants — but only the CO moved, while the other amounts stayed put.
The mixture cannot have shifted, because CO was still present at the end.
KEQ-029 · easy · recall · seq 2 · seed:U13-G3.pool[3] (KEQ-029 gateItems, James) - VERBATIM (field mapping only; instance-free) · mcq4
A student claims: 'When a change is made to an equilibrium mixture, the mixture shifts so as to undo the change completely, returning everything to exactly how it was.' Which correction fixes the claim?
The shift only partly undoes the change — it counteracts it, but never cancels it completely. KEY
No correction is needed, because a counteracting shift always restores the mixture exactly.
The shift actually reinforces the change, pushing the mixture further in the same direction.
The mixture does not shift at all, because equilibrium mixtures ignore changes in their conditions.
KEQ-033 · easy · recall · seq 3 · seed:U13-G3.pool[15] (KEQ-033 gateItems, James) - FRESHENED (water-gas CO+H₂O⇌CO₂+H₂ is scope corpus -> form-A ethene hydration, a real exothermic equilibrium; construct and options preserved) · mcq4
The reaction C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) is exothermic. On which side is heat written?
On the product side. KEY
On the reactant side.
On both sides.
Heat is not written into equations.
KEQ-030 · medium · application · seq 4 · seed:U13-G3.pool[6] (KEQ-030 gateItems, James) - FRESHENED (2NO+O₂⇌2NO₂ is scope corpus -> form-A SO₂Cl₂⇌SO₂+Cl₂; add-reactant construct and option logic preserved) · mcq4
The sulfuryl chloride mixture SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) is at equilibrium in a sealed container when more SO₂Cl₂ is pumped in. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
It shifts until the two sides hold equal amounts of gas.
KEQ-032 · medium · application · seq 5 · seed:U13-G3.pool[13] (KEQ-032 gateItems, James) - VERBATIM (field mapping only; ester system corpus-free, form-A home) · mcq4
An ester factory runs the liquid equilibrium CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) — ethanoic acid plus ethanol making ethyl ethanoate and water — and continuously distills the water out. Why does this raise the amount of ester made?
Each removal of water makes the equilibrium shift toward the products again, replacing part of what was taken out. KEY
Removing the water leaves more room in the flask, so the remaining particles collide more often and react faster in both directions.
Removing the water makes the reverse reaction impossible, so the forward reaction converts everything completely.
Distilling heats the mixture, and hotter mixtures always end up holding more product.
KEQ-030 · medium · application · seq 6 · seed:U13-G3.pool[7] (KEQ-030 gateItems, James) - FRESHENED (2SO₂+O₂⇌2SO₃ is scope corpus -> form-A dry reforming CH₄+CO₂⇌2CO+2H₂; add-product amount-consequence construct preserved) · mcq4
More CO is added to the sealed CH₄(g) + CO₂(g) ⇌ 2CO(g) + 2H₂(g) equilibrium. What happens to the amount of CH₄?
It increases. KEY
It decreases.
It stays exactly the same.
It falls to zero.
KEQ-031 · medium · application · seq 7 · seed:U13-G3.pool[9] (KEQ-031 gateItems, James) - FRESHENED (CH₄+H₂O steam reforming is scope corpus -> form-A SO₂Cl₂; sequence-of-events construct and option logic preserved) · mcq4
More SO₂Cl₂ is injected into the sealed SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) equilibrium. Which sequence of events follows?
The forward rate jumps, product builds up while the gap between the rates closes, and a new equilibrium forms with more Cl₂. KEY
The reverse rate jumps, reactant builds up while the gap between the rates closes, and a new equilibrium forms with less Cl₂ than before.
Both rates stay equal throughout, so the amounts in the mixture never change.
The forward rate jumps and then stays permanently higher than the reverse rate.
KEQ-032 · medium · application · seq 8 · seed:U13-G3.pool[12] (KEQ-032 gateItems, James) - FRESHENED (H₂+I₂⇌2HI is scope corpus -> form-A ethene hydration; rates-at-the-moment construct and option logic preserved) · mcq4
C₂H₅OH is drawn out of the sealed C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) equilibrium. What happens to the two rates at the moment of removal?
The reverse rate falls, and the forward rate is unchanged at first. KEY
The forward rate falls, and the reverse rate is unchanged at first.
Both rates fall together by the same amount, so the mixture stays settled.
The forward rate rises the moment C₂H₅OH is removed, because the remaining particles speed up.
KEQ-034 · medium · application · seq 9 · seed:U13-G3.pool[18] (KEQ-034 gateItems, James) - FRESHENED (PCl₃+Cl₂⇌PCl₅ is scope corpus -> form-A CO+NO₂⇌CO₂+NO + heat, real exothermic; hot-oven construct and options preserved) · mcq4
The equilibrium CO(g) + NO₂(g) ⇌ CO₂(g) + NO(g) + heat is moved from a cool room into a hot oven. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
It shifts toward the reactants and then back to the products.
KEQ-035 · medium · application · seq 10 · seed:U13-G3.pool[21] (KEQ-034/035 family seed; C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-A ester-1 + heat; cold-store construct and options preserved) · mcq4
The liquid equilibrium CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) + heat is moved from a warm lab into a cold store. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
It shifts toward the products and then back to the reactants.
KEQ-036 · medium · application · seq 11 · seed:U13-G3.pool[25] (KEQ-036 gateItems, James) - FRESHENED (CO+2H₂⇌CH₃OH is scope corpus -> form-A heat + dry reforming) + REMINTED (seed answer 1 = bare power of ten, V-T6 -> asked side counts 2); written-heat-not-counted twist preserved · numeric
The reversible reaction heat + CH₄(g) + CO₂(g) ⇌ 2CO(g) + 2H₂(g) is shown with heat written in. How many moles of gas are on the reactant side?
Answer: 2 moles of gas KEY
also accepted: 2 moles of gas, 2 moles, two
wrongValueDiagnostics
[
 {
  "value": 3,
  "misconception": "you counted the written heat as a mole of gas — heat is energy, not a substance, so it never joins the count",
  "feedback": "This answer is wrong — the written heat is a label for energy flow, not a substance. Count only CH₄ (1) and CO₂ (1): 2 moles of gas."
 },
 {
  "value": 4,
  "misconception": "you counted the product side — the question asks for the reactant side, which holds 1 + 1 = 2",
  "feedback": "This answer is wrong — 2CO + 2H₂ is the PRODUCT side. The reactant side holds CH₄ (1) + CO₂ (1) = 2 moles of gas."
 },
 {
  "value": 8,
  "misconception": "you counted atoms instead of coefficients — CH₄ and CO₂ contain 8 atoms between them, but moles of gas come from the coefficients alone",
  "feedback": "This answer is wrong — subscripts count atoms inside each molecule. Moles of gas come from the coefficients: 1 + 1 = 2."
 }
]
feedback
{
 "keyExplanation": "Add the coefficients of the substances marked (g) on the reactant side: CH₄ counts as 1 and CO₂ counts as 1, giving 2 moles of gas. The written heat is energy, not a substance, so it never joins the count.",
 "anticipatedWrong": [
  {
   "value": 3,
   "misconception": "you counted the written heat as a mole of gas — heat is energy, not a substance, so it never joins the count",
   "feedback": "This answer is wrong — the written heat is a label for energy flow, not a substance. Count only CH₄ (1) and CO₂ (1): 2 moles of gas."
  },
  {
   "value": 4,
   "misconception": "you counted the product side — the question asks for the reactant side, which holds 1 + 1 = 2",
   "feedback": "This answer is wrong — 2CO + 2H₂ is the PRODUCT side. The reactant side holds CH₄ (1) + CO₂ (1) = 2 moles of gas."
  },
  {
   "value": 8,
   "misconception": "you counted atoms instead of coefficients — CH₄ and CO₂ contain 8 atoms between them, but moles of gas come from the coefficients alone",
   "feedback": "This answer is wrong — subscripts count atoms inside each molecule. Moles of gas come from the coefficients: 1 + 1 = 2."
  }
 ]
}
KEQ-037 · medium · application · seq 12 · seed:U13-G3.pool[27] (KEQ-037 gateItems, James) - FRESHENED (C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-A ethene hydration; compressed amount-consequence construct preserved) · mcq4
A piston compresses the sealed C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) equilibrium to half its volume, with the temperature held steady. What happens to the amount of C₂H₅OH by the time the mixture settles at its new equilibrium?
It increases. KEY
It decreases.
It stays exactly the same.
It depends on how fast the piston was pushed in.
KEQ-036 · medium · application · seq 13 · seed:U13-G3.pool[24] (KEQ-036 gateItems, James) - FRESHENED (C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-A SO₂Cl₂⇌SO₂+Cl₂; which-side-more construct with molecule-size bait preserved, polarity flipped to a 1v2 decomposition) · mcq4
Which side of SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g) holds more moles of gas?
The product side. KEY
The reactant side.
The two sides hold equal moles of gas.
It cannot be decided from the equation alone.
KEQ-038 · medium · application · seq 14 · seed:U13-G3.pool[30] (KEQ-038 gateItems, James) - FRESHENED (2SO₂+O₂⇌2SO₃ is scope corpus -> form-A dry reforming; expand amount-consequence construct preserved) · mcq4
The sealed CH₄(g) + CO₂(g) ⇌ 2CO(g) + 2H₂(g) equilibrium is allowed to expand to twice its volume while the temperature is kept constant. What happens to the amount of CH₄ by the time the mixture settles at its new equilibrium?
It decreases. KEY
It increases.
It stays exactly the same.
It depends on how quickly the expansion happened.
KEQ-039 · medium · application · seq 15 · seed:U13-G3.pool[33] (KEQ-039 gateItems, James) - FRESHENED ('iron oxide' is an own-lesson material word and water-gas is scope corpus -> form-A CO+NO₂ + platinum; ends-with compare construct preserved) · mcq4
A platinum catalyst is used with the sealed CO(g) + NO₂(g) ⇌ CO₂(g) + NO(g) equilibrium reaction. Compared with the same mixture left uncatalyzed, the catalyzed mixture ends with —
the same amount of NO, reached sooner. KEY
more NO, reached sooner.
more NO, reached after the same time.
less NO, reached later, because the catalyst gets in the reaction's way.
KEQ-040 · medium · application · seq 16 · seed:U13-G3.pool[37]=demo_curated[7] (KEQ-040 gateItems, James demo-gate) - FRESHENED (heat+N₂+O₂⇌2NO is scope corpus -> form-A reverse water-gas, real endothermic 2v2; equal-moles twist and keep-product-in bait preserved) · mcq4
A plant runs the equilibrium heat + H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g) to make CO. Which set of conditions gives the most CO?
Add H₂, remove CO as it forms, and raise the temperature — pressure changes give no shift here. KEY
Add H₂, keep all the CO in the vessel so the pressure builds, and raise the temperature — the pressure push shifts the mixture toward CO.
Add H₂, remove CO as it forms, and lower the temperature.
Remove H₂, add extra CO, and raise the temperature.
KEQ-034 · hard · transfer · seq 17 · seed:U13-G3.pool[20] (KEQ-034 gateItems, James) - FRESHENED (C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-A ester-1 + heat; rate-vs-amount claim construct and options preserved) · mcq4
A student says: 'Heating always gives more product, because hot reactions run faster.' The liquid equilibrium CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) + heat is then heated. What actually happens to the amount of the ester, CH₃COOC₂H₅?
It decreases, because the equilibrium shifts away from the side where heat is written. KEY
It increases, because the faster a reaction runs, the more product it always ends up making.
It stays the same, because temperature only changes how fast the mixture reaches equilibrium.
It falls to zero, because heating destroys the ester completely.
KEQ-037 · hard · transfer · seq 18 · seed:U13-G3.pool[28] (KEQ-037 gateItems, James) - key equation KEPT (corpus-free); distractor equations FRESHENED (NOCl = form-B palette home, steam reforming + N₂O₄ = scope corpus -> form-A palette unequal systems), feedback counts rewritten · mcq4
Which of these equilibria does NOT shift when the pressure on it is increased?
CO(g) + NO₂(g) ⇌ CO₂(g) + NO(g) KEY
SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)
CH₄(g) + CO₂(g) ⇌ 2CO(g) + 2H₂(g)
C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)
KEQ-039 · hard · transfer · seq 19 · seed:U13-G3.pool[35] (KEQ-039 gateItems, James) - FRESHENED (heat+N₂O₄⇌2NO₂ is scope corpus -> form-A reverse water-gas with written heat; odd-one-out construct and option family preserved) · mcq4
Which change to the sealed equilibrium heat + H₂(g) + CO₂(g) ⇌ H₂O(g) + CO(g) leaves the settled amounts of all four gases unchanged?
Adding a catalyst. KEY
Raising the temperature.
Adding more H₂.
Removing some CO.
KEQ-040 · hard · transfer · seq 20 · seed:U13-G3.pool[36]=demo_curated[6] (KEQ-040 gateItems, James demo-gate) - FRESHENED (2NO+O₂⇌2NO₂+heat is scope corpus -> form-A ethene hydration + heat; five-setting audit trimmed to four options, against-setting preserved) · mcq4
A design proposal for a production vessel running C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) + heat lists four settings: add C₂H₄, remove C₂H₅OH as it forms, raise the temperature, increase the pressure. Which listed setting works AGAINST making more C₂H₅OH?
Raising the temperature. KEY
Adding C₂H₄.
Removing C₂H₅OH as it forms.
Increasing the pressure.
KEQ-040 · hard · transfer · seq 21 · fresh (stage 2, FINAL wave; FRQ-reserved form-A system, real exothermic ethyne hydrogenation) · frq
A factory operates the equilibrium C₂H₂(g) + 2H₂(g) ⇌ C₂H₆(g) + heat and wants to make as much ethane as possible.
A factory operates the equilibrium C₂H₂(g) + 2H₂(g) ⇌ C₂H₆(g) + heat and wants to make as much ethane as possible. (a) Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products. (b) Should the running temperature be raised or lowered? Justify the choice from the written heat. (c) Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Part a. Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products.
Model: Add more ethyne (or more of the other reactant gas): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the ethane as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again.
Part b. Should the running temperature be raised or lowered? Justify the choice from the written heat.
Model: Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more ethane.
Part c. Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Model: The reactant side holds 3 moles of gas (1 plus 2) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Full model answer. (a) Add more ethyne (or more of the other reactant gas): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the ethane as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again. (b) Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more ethane. (c) The reactant side holds 3 moles of gas (1 plus 2) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.

Form B

KEQ-028 · easy · recall · seq 1 · seed:U13-G3.pool[2] (KEQ-028 gateItems, James) - VERBATIM (field mapping only; instance-free) · mcq4
A sealed reversible mixture at equilibrium is disturbed, and for a while the forward reaction runs faster than the reverse. Describe the mixture's amounts from the disturbance until things settle.
Product builds up while the rates differ, and the amounts stop changing once the two rates are equal again. KEY
The amounts never change, because the mixture was at equilibrium when the disturbance arrived.
Product builds up forever, because the forward reaction stays ahead once it gets ahead.
Reactant builds up while the rates differ, then the mixture returns to its original amounts.
KEQ-029 · easy · recall · seq 2 · seed:U13-G3.pool[4] (KEQ-029 gateItems, James) - FRESHENED (2SO₂+O₂⇌2SO₃ is scope corpus -> form-B palette SO₂+NO₂⇌SO₃+NO; removal-kind construct and option logic preserved) · mcq4
Some NO is removed from the sealed SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g) equilibrium. What kind of response does Le Chatelier's principle predict?
A shift in the direction that counteracts the removal. KEY
A shift in the direction that reinforces the removal.
No response, because the principle covers only substances that are added.
A permanent stop of the forward and reverse reactions.
KEQ-033 · easy · recall · seq 3 · seed:U13-G3.pool[16] (KEQ-033 gateItems, James) - VERBATIM (field mapping only; NOCl corpus-free, form-B home) · mcq4
The decomposition 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) absorbs heat in the forward direction. Which equation writes heat correctly?
heat + 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) KEY
2NOCl(g) ⇌ 2NO(g) + Cl₂(g) + heat
heat + 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) + heat
heat + 2NO(g) + Cl₂(g) ⇌ 2NOCl(g)
KEQ-030 · medium · application · seq 4 · seed:U13-G3.pool[8] (KEQ-030 gateItems, James) - FRESHENED (C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-B SO₂+NO₂⇌SO₃+NO; claim + add-reactant construct preserved) · mcq4
A student says: 'Adding more of a reactant cannot change an equilibrium mixture, because the mixture is already settled.' More SO₂ is then added to the sealed SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g) equilibrium. What actually happens?
The equilibrium shifts toward the products, and the new mixture holds more SO₃. KEY
The equilibrium shifts toward the reactants, and the new mixture holds more NO₂ instead.
Nothing happens, because a settled equilibrium cannot respond to any change.
Both reactions stop, and the amounts freeze at their old values.
KEQ-032 · medium · application · seq 5 · seed:U13-G3.pool[14] (KEQ-032 gateItems, James) - FRESHENED (2NO+O₂⇌2NO₂ is scope corpus -> form-B SbCl₅⇌SbCl₃+Cl₂, real antimony-pentachloride equilibrium; two-flask add/remove contrast preserved with product-side polarity; stage-3 de-bromination replaced the earlier NOBr pick — bromine tokens are form C's) · mcq4
Two sealed flasks hold identical SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g) equilibrium mixtures. Into flask 1, extra Cl₂ is forced in. From flask 2, some Cl₂ is drawn out. Which pair of shifts follows?
Flask 1 shifts toward the reactants; flask 2 shifts toward the products. KEY
Flask 1 shifts toward the products; flask 2 shifts toward the reactants.
Both flasks shift toward the reactants.
Both flasks shift toward the products.
KEQ-030 · medium · application · seq 6 · fresh (stage 2, FINAL wave; form-B palette methanation CO₂+4H₂⇌CH₄+2H₂O; add-reactant amount read) · mcq4
More CO₂ is added to the sealed CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g) equilibrium. What happens to the amount of CH₄?
It increases. KEY
It decreases.
It stays exactly the same.
It first increases and then falls back to its old value.
KEQ-031 · medium · application · seq 7 · seed:U13-G3.pool[10] (KEQ-031 gateItems, James) - FRESHENED (Fe³⁺/SCN⁻ is scope corpus -> form-B CoCl₄ complex, real pink/blue observable; MECH-EQ-SHIFT-RATES verbatim key preserved) · mcq4
More Cl⁻ ions are added to the pink Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) ⇌ CoCl₄²⁻(aq) + 6H₂O(l) equilibrium, and the solution turns bluer as CoCl₄²⁻ builds up. A student says this happens 'because the equilibrium likes to restore balance.' What is the correct mechanism?
The change makes one direction temporarily faster than the other, and the mixture keeps changing until the two rates are equal again. KEY
The mixture counts the particles on each side of the equation and moves particles to the emptier side until both sides hold equal amounts.
The added Cl⁻ ions attract the other ions toward themselves without any change in the rate of either reaction.
The reverse reaction shuts down while the extra Cl⁻ is present, so only the forward reaction runs.
KEQ-032 · medium · application · seq 8 · fresh (stage 2, FINAL wave; form-B palette SO₂+NO₂⇌SO₃+NO; remove-reactant mirror of the rate-moment read) · mcq4
Some SO₂ is drawn out of the sealed SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g) equilibrium. What happens to the two rates at the moment of removal?
The forward rate falls, and the reverse rate is unchanged at first. KEY
The reverse rate falls, and the forward rate is unchanged at first.
Both rates fall together by the same amount, so the mixture stays settled.
Both rates rise, because the mixture rushes to replace what was lost.
KEQ-034 · medium · application · seq 9 · seed:U13-G3.pool[19] (KEQ-034 gateItems, James) - VERBATIM (field mapping only; NOCl corpus-free, form-B home) · mcq4
The reaction 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) is endothermic in the forward direction. The sealed mixture is heated. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift, because no heat appears in the equation as written.
The forward and reverse reactions both stop.
KEQ-035 · medium · application · seq 10 · fresh (stage 2, FINAL wave; form-B palette heat+SbCl₅⇌SbCl₃+Cl₂, real endothermic decomposition; cooled direction read; stage-3 swap from the CoCl₄ ice-bath draft — 'ice' is a KEQ-035 own-lesson corpus material word) · mcq4
The equilibrium heat + SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g) is cooled. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
The forward and reverse reactions both stop.
KEQ-036 · medium · application · seq 11 · fresh (stage 2, FINAL wave; form-B palette 2NOCl⇌2NO+Cl₂; product-side count) · numeric
How many moles of gas are on the product side of 2NOCl(g) ⇌ 2NO(g) + Cl₂(g)?
Answer: 3 moles of gas KEY
also accepted: 3 moles of gas, 3 moles, three
wrongValueDiagnostics
[
 {
  "value": 2,
  "misconception": "you counted the reactant side — the question asks for the product side, which holds 2 + 1 = 3",
  "feedback": "This answer is wrong — 2NOCl is the reactant side. The product side holds 2NO (2) + Cl₂ (1) = 3 moles of gas."
 },
 {
  "value": 5,
  "misconception": "you added the two sides together — count one side at a time; the product side alone holds 2 + 1 = 3",
  "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
 },
 {
  "value": 6,
  "misconception": "you counted atoms instead of coefficients — the product side holds 6 atoms, but moles of gas come from the coefficients alone",
  "feedback": "This answer is wrong — you counted atoms; atoms live inside the molecules. Moles of gas come from the coefficients: 2 + 1 = 3."
 }
]
feedback
{
 "keyExplanation": "Add the coefficients of the substances marked (g) on the product side: 2NO counts as 2 and Cl₂ counts as 1, giving 3 moles of gas.",
 "anticipatedWrong": [
  {
   "value": 2,
   "misconception": "you counted the reactant side — the question asks for the product side, which holds 2 + 1 = 3",
   "feedback": "This answer is wrong — 2NOCl is the reactant side. The product side holds 2NO (2) + Cl₂ (1) = 3 moles of gas."
  },
  {
   "value": 5,
   "misconception": "you added the two sides together — count one side at a time; the product side alone holds 2 + 1 = 3",
   "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
  },
  {
   "value": 6,
   "misconception": "you counted atoms instead of coefficients — the product side holds 6 atoms, but moles of gas come from the coefficients alone",
   "feedback": "This answer is wrong — you counted atoms; atoms live inside the molecules. Moles of gas come from the coefficients: 2 + 1 = 3."
  }
 ]
}
KEQ-037 · medium · application · seq 12 · seed:U13-G3.pool[29] (KEQ-037 gateItems, James) - FRESHENED (2NO+O₂⇌2NO₂ is scope corpus -> form-B SbCl₅⇌SbCl₃+Cl₂, real; why-mechanism construct with sides-as-regions distractor preserved, polarity mirrored to the reactant side; stage-3 de-bromination replaced the earlier NOBr pick) · mcq4
The pressure on the sealed SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g) equilibrium is increased by shrinking its container. Why does the mixture end up with more SbCl₅?
The shift toward the reactants cuts the gas from 2 moles to 1, which counteracts the higher pressure. KEY
The squeeze pushes the particles physically toward the reactant side of the container, where they pile up as SbCl₅.
Higher pressure always favors the reactants, in every gas-phase equilibrium.
The higher pressure raises the temperature, and that heat drives the reverse reaction.
KEQ-036 · medium · application · seq 13 · seed:U13-G3.pool[26] (KEQ-036 gateItems, James) - FRESHENED (H₂+I₂⇌2HI is scope corpus AND its equal-read would cue no-shift rows -> form-B methanation 5v3; statement-pick FORM preserved, answer moves to the unequal read) · mcq4
Which statement about the moles of gas in CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g) is correct?
The reactant side holds more moles of gas — 5 against 3. KEY
The product side holds more moles of gas.
Both sides hold the same number of moles of gas, because the equation is balanced.
The equation does not give enough information to count.
KEQ-038 · medium · application · seq 14 · seed:U13-G3.pool[32] (KEQ-038 gateItems, James) - VERBATIM (field mapping only; NOCl corpus-free, form-B home) · mcq4
The pressure on the 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) equilibrium is decreased by enlarging its container. Why does the mixture end up with more NO?
The shift toward the products raises the gas from 2 moles to 3, which counteracts the lower pressure. KEY
Lower pressure always favors the products, in every gas-phase equilibrium.
The extra space stretches the NOCl molecules until they snap into NO and Cl₂, without any collisions involved.
The lower pressure cools the mixture, and the cold drives the forward reaction.
KEQ-039 · medium · application · seq 15 · seed:U13-G3.pool[34] (KEQ-039 gateItems, James) - FRESHENED (CO+2H₂⇌CH₃OH methanol converter is scope corpus -> form-B methanation; right-about/missing option architecture preserved; catalyst left unnamed) · mcq4
A student designing a methanation unit for CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g) says: 'Skip the expensive catalyst — it would not give us any more CH₄ anyway.' What is the student right about, and what is missing?
Right that the final amount is unchanged; missing that the catalyst brings the mixture to that amount much sooner. KEY
Wrong on both counts — a catalyst raises the final amount of product as well as the speed with which the mixture gets there.
Right on both counts — a catalyst changes neither the final amount nor the time the mixture takes.
Wrong about the amount — a catalyst raises the final amount of product but leaves the time unchanged.
KEQ-040 · medium · application · seq 16 · fresh (stage 2, FINAL wave; form-B palette heat+2NOCl⇌2NO+Cl₂; endo-more-moles set-pick) · mcq4
A plant runs the equilibrium heat + 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) to make Cl₂. Which set of conditions gives the most Cl₂?
Add NOCl, remove Cl₂ as it forms, raise the temperature, and decrease the pressure. KEY
Add NOCl, remove Cl₂ as it forms, lower the temperature, and increase the pressure.
Remove NOCl, add extra Cl₂, raise the temperature, and decrease the pressure.
Add NOCl, remove Cl₂ as it forms, raise the temperature, and increase the pressure.
KEQ-034 · hard · transfer · seq 17 · fresh (stage 2, FINAL wave; form-B palette methanation + heat; faster-means-more claim variant) · mcq4
A student says: 'The faster a reaction runs, the more product it ends up making.' The equilibrium CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g) + heat is then heated. What actually happens to the amount of CH₄?
It decreases — both reactions do run faster, but the shift runs away from the written heat. KEY
It increases, because the hotter mixture reacts faster and faster reactions build more product.
It stays the same, because the two directions speed up together and cancel out.
It rises at first and then falls to zero.
KEQ-037 · hard · transfer · seq 18 · fresh (stage 2, FINAL wave; form-B palette SO₂+NO₂⇌SO₃+NO; single-system equal-moles no-shift read) · mcq4
A piston squeezes the sealed SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g) equilibrium into half its volume. Which way does the equilibrium shift?
It does not shift. KEY
Toward the products.
Toward the reactants.
It cannot be predicted without knowing the temperature.
KEQ-039 · hard · transfer · seq 19 · fresh (stage 2, FINAL wave; form-B palette CoCl₄ complex with written heat; odd-one-out on the aqueous system) · mcq4
Which change to the equilibrium heat + Co(H₂O)₆²⁺(aq) + 4Cl⁻(aq) ⇌ CoCl₄²⁻(aq) + 6H₂O(l) leaves the settled amounts unchanged?
Adding a catalyst. KEY
Warming the solution.
Adding more Cl⁻ ions.
Removing some Cl⁻ ions.
KEQ-040 · hard · transfer · seq 20 · seed:U13-G3.pool[38]=demo_curated[8] (KEQ-040 gateItems, James demo-gate) - FRESHENED (C₂H₄+H₂⇌C₂H₆+heat is scope corpus and 'nickel' is a corpus catalyst word -> form-B methanation + heat, catalyst unnamed; engineer-contrast construct preserved) · mcq4
A manager asks two engineers how to get more CH₄ from the production vessel running CO₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂O(g) + heat. Engineer A says: cool the vessel and compress the mixture. Engineer B says: just add more of the catalyst. Whose plan raises the settled amount of CH₄?
Engineer A's — cooling and compressing both shift the equilibrium toward the products; the catalyst changes speed, not amounts. KEY
Engineer B's — a catalyst speeds up the forward reaction, and a faster forward reaction always builds up a larger amount of product in the end.
Both plans — every change made to an equilibrium mixture raises the amount of product it finally holds.
Neither plan — the final amounts of an equilibrium mixture are fixed and cannot be moved by any change of conditions.
KEQ-040 · hard · transfer · seq 21 · fresh (stage 2, FINAL wave; FRQ form-B system heat+SbCl₅⇌SbCl₃+Cl₂, real endothermic decomposition — the one raise-T + decrease-P FRQ; stage-3 de-bromination replaced the earlier NOBr pick) · frq
A factory operates the equilibrium heat + SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g) and wants to make as much Cl₂ as possible.
A factory operates the equilibrium heat + SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g) and wants to make as much Cl₂ as possible. (a) Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products. (b) Should the running temperature be raised or lowered? Justify the choice from the written heat. (c) Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Part a. Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products.
Model: Add more antimony pentachloride: the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the chlorine as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again.
Part b. Should the running temperature be raised or lowered? Justify the choice from the written heat.
Model: Raise the temperature. Heat is written on the reactant side, and raising the temperature shifts an equilibrium away from the side where heat is written — toward the products, so the settled mixture holds more chlorine.
Part c. Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Model: The reactant side holds 1 mole of gas and the product side holds 2 (1 plus 1). Decrease the pressure: a pressure drop is counteracted by making more gas particles, so the equilibrium shifts toward the side with more moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Full model answer. (a) Add more antimony pentachloride: the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the chlorine as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again. (b) Raise the temperature. Heat is written on the reactant side, and raising the temperature shifts an equilibrium away from the side where heat is written — toward the products, so the settled mixture holds more chlorine. (c) The reactant side holds 1 mole of gas and the product side holds 2 (1 plus 1). Decrease the pressure: a pressure drop is counteracted by making more gas particles, so the equilibrium shifts toward the side with more moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.

Form C

KEQ-028 · easy · recall · seq 1 · seed:U13-G3.pool[0] (KEQ-028 gateItems, James) - VERBATIM (field mapping only; HBr corpus-free, form-C home) · mcq4
A student says: 'When the H₂(g) + Br₂(g) ⇌ 2HBr(g) equilibrium shifts toward the products, the reverse reaction must stop running.' Which correction is right?
Both reactions keep running throughout — the forward one is temporarily faster, until the rates match again. KEY
The student is right — a shift is only possible while the reverse reaction is switched off.
The student has it backward — it is the forward reaction that stops during this shift.
Both reactions stop briefly, and the mixture jumps straight to its new amounts.
KEQ-029 · easy · recall · seq 2 · seed:U13-G3.pool[5] (KEQ-029 gateItems, James) - VERBATIM (field mapping only; instance-free) · mcq4
A chemist wants one sentence that predicts which way an equilibrium mixture shifts when a change is made to its conditions. Which sentence does the job?
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change. KEY
In a mixture at dynamic equilibrium, the forward and the reverse reactions always run at the same rate, whatever happens to the conditions.
An equilibrium mixture shifts toward whichever side of the equation holds the smaller amounts, evening the two sides out.
An equilibrium mixture always shifts in whichever direction releases heat.
KEQ-033 · easy · recall · seq 3 · seed:U13-G3.pool[17] (KEQ-033 gateItems, James) - FRESHENED (C₂H₄+H₂⇌C₂H₆ is scope corpus -> form-C n-butane⇌isobutane + heat, a real exothermic isomerization; construct and options preserved) · mcq4
The equation n-butane(g) ⇌ isobutane(g) + heat is shown. What is true of the REVERSE direction?
It absorbs heat. KEY
It releases heat.
It neither releases nor absorbs heat.
It destroys the heat that the forward direction released.
KEQ-030 · medium · application · seq 4 · fresh (stage 2, FINAL wave; form-C palette 2H₂S⇌2H₂+S₂; add-product mirror of the taught rule) · mcq4
More H₂ is pumped into the sealed 2H₂S(g) ⇌ 2H₂(g) + S₂(g) equilibrium. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift, because H₂ was already in the mixture.
It shifts, but the direction cannot be predicted from the equation.
KEQ-032 · medium · application · seq 5 · fresh (stage 2, FINAL wave; form-C palette H₂+Br₂⇌2HBr; remove-reactant direction production, stated leak) · mcq4
A valve fault lets some Br₂ leak out of the sealed H₂(g) + Br₂(g) ⇌ 2HBr(g) equilibrium mixture. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift, because only additions disturb an equilibrium.
The forward and reverse reactions both stop.
KEQ-030 · medium · application · seq 6 · fresh (stage 2, FINAL wave; form-C palette n-butane⇌isobutane; add-product amount read) · mcq4
More isobutane is added to the sealed n-butane(g) ⇌ isobutane(g) equilibrium. What happens to the amount of n-butane?
It increases. KEY
It decreases.
It stays exactly the same.
It falls to zero.
KEQ-031 · medium · application · seq 7 · seed:U13-G3.pool[11] (KEQ-031 gateItems, James) - FRESHENED (N₂O₄⇌2NO₂ is scope corpus -> form-C CO₂+3H₂⇌CH₃OH+H₂O; collision-mechanism construct and option logic preserved) · mcq4
The moment more H₂ is added to the sealed CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g) equilibrium, the forward rate rises. Why?
The reactant's concentration rises, so collisions happen more often — and more frequent or more energetic collisions mean more successful collisions per second. KEY
The added particles raise the mixture's temperature, and hotter particles crash together more often and more energetically, which speeds up both directions at once.
Each added H₂ particle reacts instantly on arrival without needing to collide with anything.
The activation energy of the forward reaction falls when more reactant is present.
KEQ-032 · medium · application · seq 8 · fresh (stage 2, FINAL wave; form-C palette H₂+Br₂⇌2HBr; which-direction-outruns read) · mcq4
HBr is drawn off from the sealed H₂(g) + Br₂(g) ⇌ 2HBr(g) equilibrium. Immediately afterward, which direction is running faster, and why?
The forward direction — removing HBr starved the reverse reaction, and the forward rate has not changed yet. KEY
The reverse direction — the mixture speeds up whichever reaction replaces what was lost.
Neither — both rates fall together by the same amount, so the mixture stays settled.
The forward direction — removing HBr makes the remaining particles move faster.
KEQ-034 · medium · application · seq 9 · fresh (stage 2, FINAL wave; exo-word chain of KEQ-033; blind-wave1 flipped to cooling to break the C#17 system+direction duplicate; sol-loop2: cooling is KEQ-035's skill, so REVERTED to warming and moved onto n-butane⇌isobutane — different system from C#17's ester keeps the duplication fix) · mcq4
The isomerization n-butane(g) ⇌ isobutane(g) is exothermic in the forward direction. The sealed mixture is strongly warmed. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift, because no heat appears in the equation as written.
It shifts toward the reactants at first, then drifts back to where it started.
KEQ-035 · medium · application · seq 10 · seed:U13-G3.pool[22] (KEQ-035 gateItems, James) - FRESHENED (NOCl is form-B's single-form palette home -> form-C 2H₂S⇌2H₂+S₂, real endothermic decomposition; endo-word cooled construct preserved) · mcq4
The reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) is endothermic in the forward direction. The sealed mixture is cooled. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift, because no heat appears in the equation as written.
The forward and reverse reactions both stop.
KEQ-036 · medium · application · seq 11 · fresh (stage 2, FINAL wave; blind-wave1: moved off CO₂+3H₂⇌CH₃OH+H₂O — exact-duplicate gas-mole task with C#13 and C#21(c) — onto form-C palette 2H₂S⇌2H₂+S₂ product side) · numeric
How many moles of gas are on the product side of 2H₂S(g) ⇌ 2H₂(g) + S₂(g)?
Answer: 3 moles of gas KEY
also accepted: 3 moles of gas, 3 moles, three
wrongValueDiagnostics
[
 {
  "value": 2,
  "misconception": "you counted the reactant side — the question asks for the product side, which holds 2 + 1 = 3",
  "feedback": "This answer is wrong — 2H₂S is the reactant side. The product side holds 2H₂ (2) + S₂ (1) = 3 moles of gas."
 },
 {
  "value": 5,
  "misconception": "you added the two sides together — count one side at a time; the product side alone holds 2 + 1 = 3",
  "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
 },
 {
  "value": 6,
  "misconception": "you counted atoms instead of coefficients — subscripts count atoms inside each molecule; moles of gas come from the coefficients alone",
  "feedback": "This answer is wrong — you counted the atoms inside the molecules instead of the coefficients. Moles of gas come from the coefficients: 2 + 1 = 3."
 }
]
feedback
{
 "keyExplanation": "Add the coefficients of the substances marked (g) on the product side: 2H₂ counts as 2 and S₂ counts as 1, giving 2 + 1 = 3 moles of gas.",
 "anticipatedWrong": [
  {
   "value": 2,
   "misconception": "you counted the reactant side — the question asks for the product side, which holds 2 + 1 = 3",
   "feedback": "This answer is wrong — 2H₂S is the reactant side. The product side holds 2H₂ (2) + S₂ (1) = 3 moles of gas."
  },
  {
   "value": 5,
   "misconception": "you added the two sides together — count one side at a time; the product side alone holds 2 + 1 = 3",
   "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
  },
  {
   "value": 6,
   "misconception": "you counted atoms instead of coefficients — subscripts count atoms inside each molecule; moles of gas come from the coefficients alone",
   "feedback": "This answer is wrong — you counted the atoms inside the molecules instead of the coefficients. Moles of gas come from the coefficients: 2 + 1 = 3."
  }
 ]
}
KEQ-037 · medium · application · seq 12 · fresh (stage 2, FINAL wave; form-C palette 2H₂S⇌2H₂+S₂; squeeze direction production) · mcq4
A piston is pushed in, increasing the pressure on the sealed 2H₂S(g) ⇌ 2H₂(g) + S₂(g) equilibrium. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
It shifts toward the reactants at first, then drifts back to where it began as the piston holds.
KEQ-036 · medium · application · seq 13 · fresh (stage 2, FINAL wave; form-C palette CO₂+3H₂⇌CH₃OH+H₂O; which-side-more comparison) · mcq4
Which side of CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g) holds more moles of gas?
The reactant side. KEY
The product side.
The two sides hold equal moles of gas.
It cannot be decided from the equation alone.
KEQ-038 · medium · application · seq 14 · fresh (stage 2, FINAL wave; form-C palette 2H₂S⇌2H₂+S₂; expansion direction production — deliberate mirror of C seq 12's squeeze, matching the KEQ-037/038 mirror-lesson pair) · mcq4
A piston is pulled out, decreasing the pressure on the sealed 2H₂S(g) ⇌ 2H₂(g) + S₂(g) equilibrium. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
The equilibrium breaks, and the mixture stops being reversible.
KEQ-039 · medium · application · seq 15 · fresh (stage 2, FINAL wave; form-C palette n-butane⇌isobutane; catalyst-at-equilibrium amount read) · mcq4
A catalyst is added to the sealed n-butane(g) ⇌ isobutane(g) mixture, which is already at equilibrium. How does the amount of isobutane respond?
It stays the same. KEY
It increases.
It decreases.
It rises quickly at first and then falls back below its old value.
KEQ-040 · medium · application · seq 16 · fresh (stage 2, FINAL wave; form-C palette H₂+Br₂⇌2HBr + heat; exo-equal set-pick with the pressure-no-shift twist) · mcq4
A plant runs the equilibrium H₂(g) + Br₂(g) ⇌ 2HBr(g) + heat to make HBr. Which set of conditions gives the most HBr?
Add Br₂, remove HBr as it forms, and lower the temperature — pressure changes give no shift here. KEY
Add Br₂, keep the HBr in the vessel so the pressure builds, and lower the temperature — the pressure push shifts the mixture toward the HBr.
Add Br₂, remove HBr as it forms, and raise the temperature.
Remove Br₂, add extra HBr, and lower the temperature.
KEQ-034 · hard · transfer · seq 17 · fresh (stage 2, FINAL wave; form-C palette ester-3 + heat; speed-only claim variant — distinct construct from C seq 9's plain direction read, recorded) · mcq4
A student says: 'Temperature only changes how fast a mixture reaches equilibrium — it never changes the settled amounts.' The liquid equilibrium HCOOH(l) + C₂H₅OH(l) ⇌ HCOOC₂H₅(l) + H₂O(l) + heat is then heated. What actually happens to the amount of the ester, HCOOC₂H₅?
It decreases — a temperature change moves the settled amounts too, away from the written heat. KEY
It stays exactly the same, just as the student says — the mixture only settles faster.
It increases, because hot mixtures always end up holding more product.
It falls to zero, because heating undoes the reaction completely.
KEQ-037 · hard · transfer · seq 18 · fresh (stage 2, FINAL wave; form-C palette n-butane⇌isobutane; single-system equal-moles no-shift read on a 1v1 isomer pair) · mcq4
The sealed n-butane(g) ⇌ isobutane(g) equilibrium is squeezed into a smaller container. Which way does the equilibrium shift?
It does not shift. KEY
Toward the products.
Toward the reactants.
It cannot be predicted without knowing the temperature.
KEQ-039 · hard · transfer · seq 19 · fresh (stage 2, FINAL wave; form-C palette ester-3 with written heat; odd-one-out on the liquid system) · mcq4
Which change to the liquid equilibrium HCOOH(l) + C₂H₅OH(l) ⇌ HCOOC₂H₅(l) + H₂O(l) + heat leaves the settled amounts unchanged?
Adding a catalyst. KEY
Cooling the mixture.
Adding more HCOOH.
Distilling out the ester as it forms.
KEQ-040 · hard · transfer · seq 20 · fresh (stage 2, FINAL wave; form-C palette heat+2H₂S⇌2H₂+S₂; proposal audit with the pressure dial as the against-setting) · mcq4
A proposal for making more S₂ from the equilibrium heat + 2H₂S(g) ⇌ 2H₂(g) + S₂(g) lists four settings: add H₂S, remove S₂ as it forms, raise the temperature, increase the pressure. Which listed setting works AGAINST making more S₂?
Increasing the pressure. KEY
Adding H₂S.
Removing S₂ as it forms.
Raising the temperature.
KEQ-040 · hard · transfer · seq 21 · fresh (stage 2, FINAL wave; FRQ form-C system CO₂+3H₂⇌CH₃OH+H₂O+heat, real green-methanol route) · frq
A green-methanol factory operates the equilibrium CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g) + heat and wants to make as much methanol as possible.
A green-methanol factory operates the equilibrium CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g) + heat and wants to make as much methanol as possible. (a) Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products. (b) Should the running temperature be raised or lowered? Justify the choice from the written heat. (c) Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Part a. Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products.
Model: Add more of either reactant gas: the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the methanol (or the water) as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again.
Part b. Should the running temperature be raised or lowered? Justify the choice from the written heat.
Model: Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more methanol.
Part c. Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Model: The reactant side holds 4 moles of gas (1 plus 3) and the product side holds 2 (1 plus 1). Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Full model answer. (a) Add more of either reactant gas: the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the methanol (or the water) as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again. (b) Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more methanol. (c) The reactant side holds 4 moles of gas (1 plus 3) and the product side holds 2 (1 plus 1). Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.

Form D

KEQ-028 · easy · recall · seq 1 · fresh (stage 2, FINAL wave; form-D palette system 2O₃⇌3O₂) · mcq4
A sealed 2O₃(g) ⇌ 3O₂(g) mixture at equilibrium has its temperature changed. When it settles into its new equilibrium, it holds more O₂ than before. Which way did the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It did not shift — a disturbed mixture always returns to its original amounts.
It shifted toward both sides at once.
KEQ-029 · easy · recall · seq 2 · fresh (stage 2, FINAL wave; instance-free statement pick — key is the taught rule-line verbatim, A3 lane) · mcq4
On a quiz, a student is asked to write down Le Chatelier's principle. Which sentence should the student write?
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that counteracts the change. KEY
When a change is made to a system at dynamic equilibrium, the equilibrium shifts in the direction that reinforces the change.
When a change is made to a system at dynamic equilibrium, the equilibrium always shifts toward the products, no matter what the change is.
When a change is made to a system at dynamic equilibrium, the forward and reverse reactions both stop until the change is undone.
KEQ-033 · easy · recall · seq 3 · fresh (stage 2, FINAL wave; form-D palette 2CO₂⇌2CO+O₂, real endothermic high-temperature decomposition; parallel to the lesson's endo write-it pattern) · mcq4
At very high temperatures the decomposition 2CO₂(g) ⇌ 2CO(g) + O₂(g) runs as an equilibrium, absorbing heat in the forward direction. Which equation writes heat correctly?
heat + 2CO₂(g) ⇌ 2CO(g) + O₂(g) KEY
2CO₂(g) ⇌ 2CO(g) + O₂(g) + heat
heat + 2CO₂(g) ⇌ 2CO(g) + O₂(g) + heat
heat + 2CO(g) + O₂(g) ⇌ 2CO₂(g)
KEQ-030 · medium · application · seq 4 · fresh (stage 2, FINAL wave; form-D palette H₂+Cl₂⇌2HCl; add-reactant direction production) · mcq4
More Cl₂ is added to the sealed H₂(g) + Cl₂(g) ⇌ 2HCl(g) equilibrium. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
The forward and reverse reactions both stop.
KEQ-032 · medium · application · seq 5 · fresh (stage 2, FINAL wave; form-D palette 2CO₂⇌2CO+O₂; industrial draw-off read parallel to the lesson's factory rule line) · mcq4
A 2CO₂(g) ⇌ 2CO(g) + O₂(g) mixture runs at very high temperature, and the O₂ is removed from the vessel as it forms. Why does this raise the total amount of CO made?
Each removal of O₂ makes the equilibrium shift toward the products again, replacing part of what was taken out. KEY
Removing the O₂ speeds up the forward reaction itself, so the CO₂ decomposes faster at the same temperature.
Removing the O₂ leaves more room in the flask, so the remaining particles collide more often, reacting faster in both directions.
Drawing gas off chills the vessel, and that temperature drop is what pushes the decomposition forward.
KEQ-030 · medium · application · seq 6 · fresh (stage 2, FINAL wave; form-D palette NO+NO₂⇌N₂O₃; the other-reactant-falls subtle read) · mcq4
More NO is added to the sealed NO(g) + NO₂(g) ⇌ N₂O₃(g) equilibrium. What happens to the amount of NO₂?
It decreases. KEY
It increases.
It stays exactly the same.
It falls to zero.
KEQ-031 · medium · application · seq 7 · fresh (stage 2, FINAL wave; form-D palette 2O₃⇌3O₂; which-rate-changes-first read, canonical option set) · mcq4
More O₃ is added to the sealed 2O₃(g) ⇌ 3O₂(g) equilibrium. What happens to the two rates at the moment of the addition?
The forward rate rises, and the reverse rate is unchanged at first. KEY
The reverse rate rises, and the forward rate is unchanged at first.
Both rates rise instantly by exactly the same amount, so the mixture stays settled.
Both rates fall, because the mixture has been disturbed.
KEQ-032 · medium · application · seq 8 · fresh (stage 2, FINAL wave; form-D palette H₂+Cl₂⇌2HCl; rate-moment read, product removal) · mcq4
HCl is drawn out of the sealed H₂(g) + Cl₂(g) ⇌ 2HCl(g) equilibrium. What happens to the two rates at the moment of removal?
The reverse rate drops, while the forward rate is unchanged at first. KEY
The forward rate drops, while the reverse rate is unchanged at first.
Both rates drop together by the same amount, so the mixture stays settled.
The reverse rate rises, because the mixture works to replace the removed HCl.
KEQ-034 · medium · application · seq 9 · fresh (stage 2, FINAL wave; form-D palette 2O₃⇌3O₂ + heat; heated amount-consequence read) · mcq4
The equilibrium 2O₃(g) ⇌ 3O₂(g) + heat is heated. What happens to the amount of O₂?
It decreases. KEY
It increases.
It stays exactly the same.
It falls to zero.
KEQ-035 · medium · application · seq 10 · fresh (stage 2, FINAL wave; form-D palette heat+2CO₂⇌2CO+O₂; cooled direction read) · mcq4
The high-temperature equilibrium heat + 2CO₂(g) ⇌ 2CO(g) + O₂(g) is cooled. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
The direction cannot be known without measuring how much CO₂ the mixture holds.
KEQ-036 · medium · application · seq 11 · fresh (stage 2, FINAL wave; blind-wave1: moved off H₂+Cl₂⇌2HCl — it rehearsed the mole count of D#18's KEY system — onto form-D palette 2O₃⇌3O₂ product side) · numeric
How many moles of gas are on the product side of 2O₃(g) ⇌ 3O₂(g)?
Answer: 3 moles of gas KEY
also accepted: 3 moles of gas, 3 moles, three
wrongValueDiagnostics
[
 {
  "value": 1,
  "misconception": "you counted 3O₂ as a single substance worth one mole — the coefficient 3 counts three moles of it",
  "feedback": "This answer is wrong — you counted 3O₂ as a single mole because it is one substance. Its coefficient is 3, so the product side holds 3 moles of gas."
 },
 {
  "value": 5,
  "misconception": "you added the two sides together — count one side at a time; the product side alone holds 3",
  "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
 },
 {
  "value": 6,
  "misconception": "you counted atoms — 3O₂ holds 6 atoms, but moles of gas come from the coefficient alone",
  "feedback": "This answer is wrong — you counted the six O atoms in 3O₂ instead of its coefficient. Moles of gas come from the coefficient: 3O₂ counts as 3 moles of gas."
 }
]
feedback
{
 "keyExplanation": "Add the coefficients of the substances marked (g) on the product side. The only product is 3O₂, which counts as 3 moles of gas.",
 "anticipatedWrong": [
  {
   "value": 1,
   "misconception": "you counted 3O₂ as a single substance worth one mole — the coefficient 3 counts three moles of it",
   "feedback": "This answer is wrong — you counted 3O₂ as a single mole because it is one substance. Its coefficient is 3, so the product side holds 3 moles of gas."
  },
  {
   "value": 5,
   "misconception": "you added the two sides together — count one side at a time; the product side alone holds 3",
   "feedback": "This answer is wrong — 5 is both sides combined. The product side alone holds 3 moles of gas."
  },
  {
   "value": 6,
   "misconception": "you counted atoms — 3O₂ holds 6 atoms, but moles of gas come from the coefficient alone",
   "feedback": "This answer is wrong — you counted the six O atoms in 3O₂ instead of its coefficient. Moles of gas come from the coefficient: 3O₂ counts as 3 moles of gas."
  }
 ]
}
KEQ-037 · medium · application · seq 12 · fresh (stage 2, FINAL wave; form-D palette 2O₃⇌3O₂; squeeze direction production) · mcq4
The sealed 2O₃(g) ⇌ 3O₂(g) equilibrium is compressed into a smaller volume. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
It shifts toward the reactants and then back to the products.
KEQ-036 · medium · application · seq 13 · fresh (stage 2, FINAL wave; form-D palette 2CO₂⇌2CO+O₂; which-side-more comparison) · mcq4
Which side of 2CO₂(g) ⇌ 2CO(g) + O₂(g) holds more moles of gas?
The product side. KEY
The reactant side.
The two sides hold equal moles of gas.
It cannot be decided from the equation alone.
KEQ-038 · medium · application · seq 14 · fresh (stage 2, FINAL wave; form-D palette NO+NO₂⇌N₂O₃; expansion direction production) · mcq4
The sealed NO(g) + NO₂(g) ⇌ N₂O₃(g) equilibrium is allowed to expand into a larger volume. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
It shifts toward the reactants and then back to the products.
KEQ-039 · medium · application · seq 15 · fresh (stage 2, FINAL wave; instance-light on-the-way read; stem reworded off the lesson's own fluency phrasing) · mcq4
A catalyst is dropped into a sealed reversible gas mixture that is still approaching its equilibrium. What does the catalyst change?
How soon the mixture reaches equilibrium. KEY
Which direction the equilibrium finally favors.
The final amounts of reactants and products in the mixture.
The side of the equation on which heat is written.
KEQ-040 · medium · application · seq 16 · fresh (stage 2, FINAL wave; form-D palette 2O₃⇌3O₂ + heat; exo-more-moles set-pick) · mcq4
A researcher runs the 2O₃(g) ⇌ 3O₂(g) + heat equilibrium to make O₂. Which set of conditions gives the most O₂?
Add O₃, remove O₂ as it forms, lower the temperature, and decrease the pressure. KEY
Add O₃, remove O₂ as it forms, raise the temperature, and increase the pressure.
Remove O₃, add extra O₂, lower the temperature, and decrease the pressure.
Add O₃, remove O₂ as it forms, lower the temperature, and increase the pressure.
KEQ-034 · hard · transfer · seq 17 · fresh (stage 2, FINAL wave; form-D palette heat+2CO₂⇌2CO+O₂; right-for-wrong-reason transfer variant) · mcq4
A student says: 'Heating always gives more product, because hot reactions run faster.' The high-temperature equilibrium heat + 2CO₂(g) ⇌ 2CO(g) + O₂(g) is then heated further. What actually happens, and why?
The amount of CO rises — but because the shift runs away from the written heat, not because faster reactions make more product. KEY
The amount of CO rises, exactly as the student reasons — faster reactions always end up making more product.
The amount of CO falls, because heating always costs product.
Nothing changes, because temperature only affects how fast equilibrium is reached.
KEQ-037 · hard · transfer · seq 18 · fresh (stage 2, FINAL wave; four-equation validity-condition pick built from the form-D palette) · mcq4
Which ONE of these equilibria does NOT shift when its pressure is increased?
H₂(g) + Cl₂(g) ⇌ 2HCl(g) KEY
2CO₂(g) ⇌ 2CO(g) + O₂(g)
NO(g) + NO₂(g) ⇌ N₂O₃(g)
2O₃(g) ⇌ 3O₂(g)
KEQ-039 · hard · transfer · seq 19 · fresh (stage 2, FINAL wave; form-D palette H₂+Cl₂⇌2HCl + heat; odd-one-out — pressure options deliberately excluded on this 2v2 system so the catalyst stays the ONLY non-shifting change) · mcq4
Which change to the sealed equilibrium H₂(g) + Cl₂(g) ⇌ 2HCl(g) + heat leaves the settled amounts unchanged?
Adding a catalyst. KEY
Raising the temperature.
Adding more Cl₂.
Removing some HCl.
KEQ-040 · hard · transfer · seq 20 · fresh (stage 2, FINAL wave; form-D palette heat+2CO₂⇌2CO+O₂; proposal audit with the temperature dial as the against-setting) · mcq4
A proposal for making more CO from the high-temperature equilibrium heat + 2CO₂(g) ⇌ 2CO(g) + O₂(g) lists four settings: add CO₂, remove O₂ as it forms, lower the temperature, decrease the pressure. Which listed setting works AGAINST making more CO?
Lowering the temperature. KEY
Adding CO₂.
Removing O₂ as it forms.
Decreasing the pressure.
KEQ-040 · hard · transfer · seq 21 · fresh (stage 2, FINAL wave; FRQ form-D system NO+NO₂⇌N₂O₃+heat, real exothermic N₂O₃ formation) · frq
A laboratory operates the equilibrium NO(g) + NO₂(g) ⇌ N₂O₃(g) + heat and wants to make as much dinitrogen trioxide as possible.
A laboratory operates the equilibrium NO(g) + NO₂(g) ⇌ N₂O₃(g) + heat and wants to make as much dinitrogen trioxide as possible. (a) Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products. (b) Should the running temperature be raised or lowered? Justify the choice from the written heat. (c) Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Part a. Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products.
Model: Add more of either nitrogen oxide reactant (more NO, say): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the dinitrogen trioxide as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again.
Part b. Should the running temperature be raised or lowered? Justify the choice from the written heat.
Model: Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more dinitrogen trioxide.
Part c. Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Model: The reactant side holds 2 moles of gas (1 plus 1) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Full model answer. (a) Add more of either nitrogen oxide reactant (more NO, say): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the dinitrogen trioxide as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again. (b) Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more dinitrogen trioxide. (c) The reactant side holds 2 moles of gas (1 plus 1) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.

Form E

KEQ-028 · easy · recall · seq 1 · fresh (stage 2, FINAL wave; form-E palette system cis/trans-2-butene) · mcq4
In a sealed cis-2-butene(g) ⇌ trans-2-butene(g) mixture at equilibrium, a change makes the reverse reaction temporarily faster than the forward one. What does the mixture settle into?
A new equilibrium holding more cis-2-butene and less trans-2-butene than before. KEY
A new equilibrium holding more trans-2-butene and less cis-2-butene than before.
Its original amounts, restored once the two rates are equal again.
A mixture holding only cis-2-butene, the trans form having been used up completely.
KEQ-029 · easy · recall · seq 2 · fresh (stage 2, FINAL wave; form-E palette CS₂ process; kind-of-response construct at principle level) · mcq4
More CH₄ is added to the sealed CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g) equilibrium. What kind of response does Le Chatelier's principle predict?
A shift in the direction that counteracts the addition. KEY
A shift in the direction that reinforces the addition.
No response of any kind, because a settled equilibrium mixture cannot change.
A complete stop of both the forward and the reverse reactions.
KEQ-033 · easy · recall · seq 3 · fresh (stage 2, FINAL wave; form-E palette I₂+Cl₂⇌2ICl, real mildly exothermic interhalogen formation; read-back construct per the lesson's F3 pattern; stage-3 de-bromination: bromine is form C's single-form token, so E's interhalogen is ICl) · mcq4
The equation I₂(g) + Cl₂(g) ⇌ 2ICl(g) + heat is shown with heat written in. What does the written heat say about the reaction?
The forward reaction releases heat — the reaction is exothermic. KEY
The forward reaction absorbs heat — the reaction is endothermic.
The mixture is at a high temperature.
Heat is one of the chemical substances that the reaction makes.
KEQ-030 · medium · application · seq 4 · fresh (stage 2, FINAL wave; form-E palette C₂H₄+HCl⇌C₂H₅Cl, real ethyl-chloride synthesis; add-product direction production) · mcq4
More C₂H₅Cl is added to the sealed C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g) equilibrium. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift.
It shifts toward the reactants and then back to the products.
KEQ-032 · medium · application · seq 5 · fresh (stage 2, FINAL wave; form-E palette cis/trans-2-butene; continuous draw-off read) · mcq4
In a production vessel, trans-2-butene is continuously drawn off from the cis-2-butene(g) ⇌ trans-2-butene(g) equilibrium as it forms. Which way does the equilibrium keep shifting?
Toward the products. KEY
Toward the reactants.
It does not shift, because drawing off a product is not a change to the mixture.
It shifts toward the products once, then never again.
KEQ-030 · medium · application · seq 6 · fresh (stage 2, FINAL wave; form-E palette CS₂ process; add-product, other-product-falls read) · mcq4
More H₂ is forced into the sealed CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g) equilibrium. What happens to the amount of CS₂?
It decreases. KEY
It increases.
It stays exactly the same.
It falls to zero.
KEQ-031 · medium · application · seq 7 · fresh (stage 2, FINAL wave; form-E palette I₂+Cl₂⇌2ICl; what-ends-the-shift read, canonical option set) · mcq4
More I₂ is added to the sealed I₂(g) + Cl₂(g) ⇌ 2ICl(g) equilibrium, and the mixture starts shifting toward the products. What brings the shift to an end?
Reactants are used up and products build up, until the forward and reverse rates are equal again. KEY
The added I₂ runs out completely, which stops the forward reaction.
The temperature of the mixture falls back to its starting value, which slows the forward reaction down again.
The forward reaction stops for good once enough ICl has formed.
KEQ-032 · medium · application · seq 8 · fresh (stage 2, FINAL wave; form-E palette CS₂ process; rate-moment read, product removal) · mcq4
CS₂ is drawn off from the sealed CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g) equilibrium. What happens to the two rates at that moment?
The reverse rate falls first; the forward rate is untouched at that moment. KEY
The forward rate falls first; the reverse rate is untouched at that moment.
Both rates fall by the same amount at that moment, so the amounts never change.
The forward rate jumps at that moment, because removing CS₂ adds energy to the mixture.
KEQ-034 · medium · application · seq 9 · fresh (stage 2, FINAL wave; exo-word chain; blind-wave1 flipped to cooling to break the E#17 duplicate; sol-loop2: cooling is KEQ-035's skill, so REVERTED to warming and moved onto C₂H₄+HCl⇌C₂H₅Cl — different system from E#17's butenes keeps the duplication fix) · mcq4
The reaction C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g) is exothermic in the forward direction. The sealed mixture is strongly warmed. Which way does the equilibrium shift?
Toward the reactants. KEY
Toward the products.
It does not shift, because no heat appears in the equation as written.
It shifts toward the reactants briefly, then settles back at the amounts it held before the warming.
KEQ-035 · medium · application · seq 10 · fresh (stage 2, FINAL wave; form-E palette C₂H₄+HCl⇌C₂H₅Cl + heat; cooled direction read) · mcq4
The equilibrium C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g) + heat is cooled. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
Both directions slow down by the same amount, so the amounts stay the same.
KEQ-036 · medium · application · seq 11 · fresh (stage 2, FINAL wave; blind-wave1: moved off I₂+Cl₂⇌2ICl — E#18's KEY system — onto the form-E CS₂-process reactant side; also removes the stage-3 'Br₂' leftover in gradingNote/keyExplanation/feedback) · numeric
How many moles of gas are on the reactant side of CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)?
Answer: 3 moles of gas KEY
also accepted: 3 moles of gas, 3 moles, three
wrongValueDiagnostics
[
 {
  "value": 5,
  "misconception": "you counted the product side — the question asks for the reactant side, which holds 1 + 2 = 3",
  "feedback": "This answer is wrong — CS₂ + 4H₂ is the product side. The reactant side holds CH₄ (1) + 2H₂S (2) = 3 moles of gas."
 },
 {
  "value": 8,
  "misconception": "you added the two sides together — count one side at a time; the reactant side alone holds 1 + 2 = 3",
  "feedback": "This answer is wrong — 8 is both sides combined. The reactant side alone holds 3 moles of gas."
 },
 {
  "value": 2,
  "misconception": "you counted the substances instead of their coefficients — 2H₂S counts as 2 moles on its own, so the side holds 1 + 2 = 3",
  "feedback": "This answer is wrong — count coefficients, not substances: CH₄ (1) + 2H₂S (2) = 3 moles of gas."
 }
]
feedback
{
 "keyExplanation": "Add the coefficients of the substances marked (g) on the reactant side: CH₄ has no written coefficient, so it counts as 1; 2H₂S counts as 2. 1 + 2 = 3 moles of gas.",
 "anticipatedWrong": [
  {
   "value": 5,
   "misconception": "you counted the product side — the question asks for the reactant side, which holds 1 + 2 = 3",
   "feedback": "This answer is wrong — CS₂ + 4H₂ is the product side. The reactant side holds CH₄ (1) + 2H₂S (2) = 3 moles of gas."
  },
  {
   "value": 8,
   "misconception": "you added the two sides together — count one side at a time; the reactant side alone holds 1 + 2 = 3",
   "feedback": "This answer is wrong — 8 is both sides combined. The reactant side alone holds 3 moles of gas."
  },
  {
   "value": 2,
   "misconception": "you counted the substances instead of their coefficients — 2H₂S counts as 2 moles on its own, so the side holds 1 + 2 = 3",
   "feedback": "This answer is wrong — count coefficients, not substances: CH₄ (1) + 2H₂S (2) = 3 moles of gas."
  }
 ]
}
KEQ-037 · medium · application · seq 12 · fresh (stage 2, FINAL wave; form-E palette C₂H₄+HCl⇌C₂H₅Cl; squeeze amount-consequence read) · mcq4
The pressure on the sealed C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g) equilibrium is increased by shrinking its container. What happens to the amount of C₂H₅Cl?
It increases. KEY
It decreases.
It stays exactly the same.
It falls to zero.
KEQ-036 · medium · application · seq 13 · fresh (stage 2, FINAL wave; form-E palette cis/trans-2-butene; the equal-counts read on a 1v1 isomer pair) · mcq4
Which statement about the moles of gas in cis-2-butene(g) ⇌ trans-2-butene(g) is correct?
Both sides hold 1 mole of gas. KEY
The product side holds more moles of gas.
The reactant side holds more moles of gas.
The equation does not give enough information to count.
KEQ-038 · medium · application · seq 14 · fresh (stage 2, FINAL wave; form-E palette CS₂ process; expansion direction production) · mcq4
The pressure on the sealed CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g) equilibrium is decreased by enlarging its container. Which way does the equilibrium shift?
Toward the products. KEY
Toward the reactants.
It does not shift.
The forward and reverse reactions both stop.
KEQ-039 · medium · application · seq 15 · fresh (stage 2, FINAL wave; MECH-CATALYST-BOTH verbatim key = A3 rule-line/fluency-key lane; stem reworded off the lesson's F3 phrasing) · mcq4
A catalyst is added to a sealed gas mixture at equilibrium, and careful measurement shows the settled amounts do not move at all. What explains this?
A catalyst speeds up the forward and reverse reactions equally, so it never moves the settled amounts — its only effect is how soon a mixture settles. KEY
A catalyst speeds up only the forward reaction, and the reverse reaction then speeds up on its own until the two rates match again, canceling the catalyst's effect on the amounts.
A catalyst does not take part in the reaction at all, so it cannot change anything about the mixture.
The catalyst is used up before it can shift the equilibrium.
KEQ-040 · medium · application · seq 16 · fresh (stage 2, FINAL wave; form-E palette heat+CH₄+2H₂S⇌CS₂+4H₂, the real CS₂ manufacturing route; endo-more-moles set-pick) · mcq4
A plant runs the equilibrium heat + CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g) to make CS₂. Which set of conditions gives the most CS₂?
Add CH₄, remove CS₂ as it forms, raise the temperature, and decrease the pressure. KEY
Add CH₄, remove CS₂ as it forms, lower the temperature, and increase the pressure.
Remove CH₄, add extra CS₂, raise the temperature, and decrease the pressure.
Add CH₄, remove CS₂ as it forms, raise the temperature, and increase the pressure.
KEQ-034 · hard · transfer · seq 17 · fresh (stage 2, FINAL wave; form-E palette cis/trans-2-butene + heat; hotter-holds-more claim variant) · mcq4
A student says: 'A hotter mixture always ends up holding more product.' The equilibrium cis-2-butene(g) ⇌ trans-2-butene(g) + heat is then heated. What actually happens to the amount of trans-2-butene?
It decreases, because the equilibrium shifts away from the side where heat is written. KEY
It increases, because heat drives every reaction toward its products.
It stays the same, because temperature only changes how fast equilibrium is reached.
It rises without limit as long as the heating continues.
KEQ-037 · hard · transfer · seq 18 · fresh (stage 2, FINAL wave; form-E palette I₂+Cl₂⇌2ICl; single-system equal-moles no-shift read) · mcq4
The sealed I₂(g) + Cl₂(g) ⇌ 2ICl(g) equilibrium is squeezed into a smaller container; its temperature does not change. Which way does the equilibrium shift?
It does not shift. KEY
Toward the products.
Toward the reactants.
It cannot be predicted without knowing the temperature.
KEQ-039 · hard · transfer · seq 19 · fresh (stage 2, FINAL wave; form-E palette C₂H₄+HCl⇌C₂H₅Cl + heat; odd-one-out with the lower-T option family) · mcq4
Which change to the sealed equilibrium C₂H₄(g) + HCl(g) ⇌ C₂H₅Cl(g) + heat leaves the settled amounts unchanged?
Adding a catalyst. KEY
Lowering the temperature.
Adding more HCl.
Removing some C₂H₅Cl.
KEQ-040 · hard · transfer · seq 20 · fresh (stage 2, FINAL wave; form-E palette I₂+Cl₂⇌2ICl + heat; proposal audit with the remove-a-reactant against-setting; no pressure setting listed on this equal-moles system) · mcq4
A proposal for making more ICl from the equilibrium I₂(g) + Cl₂(g) ⇌ 2ICl(g) + heat lists four settings: add I₂, remove ICl as it forms, lower the temperature, and bleed off some of the Cl₂ to keep the vessel from crowding. Which listed setting works AGAINST making more ICl?
Bleeding off some of the Cl₂. KEY
Adding I₂.
Removing ICl as it forms.
Lowering the temperature.
KEQ-040 · hard · transfer · seq 21 · fresh (stage 2, FINAL wave; FRQ form-E system propene hydrogenation C₃H₆+H₂⇌C₃H₈+heat) · frq
A factory operates the equilibrium C₃H₆(g) + H₂(g) ⇌ C₃H₈(g) + heat and wants to make as much propane as possible.
A factory operates the equilibrium C₃H₆(g) + H₂(g) ⇌ C₃H₈(g) + heat and wants to make as much propane as possible. (a) Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products. (b) Should the running temperature be raised or lowered? Justify the choice from the written heat. (c) Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Part a. Name one substance to add during the run and one substance to remove as it forms, and say why each move shifts the equilibrium toward the products.
Model: Add more propene (or more of the other reactant gas): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the propane as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again.
Part b. Should the running temperature be raised or lowered? Justify the choice from the written heat.
Model: Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more propane.
Part c. Count the moles of gas on each side, state whether the pressure should be increased or decreased and why, and say what adding a catalyst would and would not do.
Model: The reactant side holds 2 moles of gas (1 plus 1) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.
Full model answer. (a) Add more propene (or more of the other reactant gas): the equilibrium shifts in the direction that uses up the added substance, so the shift runs toward the products. Remove the propane as it forms: the equilibrium shifts in the direction that replaces the removed substance, so each draw-off pushes the equilibrium toward the products again. (b) Lower the temperature. Heat is written on the product side, and lowering the temperature shifts an equilibrium toward the side where heat is written — toward the products, so the settled mixture holds more propane. (c) The reactant side holds 2 moles of gas (1 plus 1) and the product side holds 1. Increase the pressure: a squeeze is counteracted by making fewer gas particles, so the equilibrium shifts toward the side with fewer moles of gas — the product side. Adding a catalyst would bring the mixture to equilibrium sooner, but it would not change the settled amounts: a catalyst speeds up the forward and reverse reactions equally, so the mixture reaches equilibrium sooner but with the same final amounts.