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AP02 · MIXED UNIT 5 PRACTICE

Connect kinetic models, measurements and evidence limits

Solve eight unfamiliar kinetics cases, compare competing models and distinguish calculation agreement from experimental support.

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Challenge a law that fits two anchors

A fictional sample has C₀=0.200 mol/L and C(5 s)=0.100 mol/L at fixed conditions. Zero-, first- and second-order laws are each calibrated on these two anchors. Which conclusion is supported? Predict each concentration at 10 s and identify why their k values cannot be ranked without units.

Quantity / representation	Packet value
Anchors	0.200 mol/L at 0 s; 0.100 at 5 s
Candidates	Constant amount loss, constant fractional loss, linear reciprocal concentration
Missing evidence	No later concentrations or uncertainty supplied

Your response

A first half-life always proves first-order kineticsAll three calibrated laws fit these anchors; a later matched observation can discriminate themThe largest numerical k must identify the fastest valid mechanism

Your reasoning for Challenge a law that fits two anchors

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Normalize a finite-window rate without double-counting a common blank

For 2 A→P at fixed volume, corrected concentration readings are 0.0800 and 0.0754 mol/L separated by 2 s. Each has a separately varying absolute reading bound ±0.0002mol/L, plus the same unknown constant blank offset. Enter the average reaction rate in mol L⁻¹s⁻¹. Explain its worst-case reading bound, blank cancellation and why this is not automatically the initial rate.

Quantity / representation	Packet value
Endpoints	0.0800 →0.0754mol/L over2s
Separate bounds	±0.0002mol/L per endpoint
Shared error	Same constant offset at both times; changing drift excluded by premise

Numerical response (units specified in the question) Your reasoning for Normalize a finite-window rate without double-counting a common blank

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Keep event repetition separate from molecularity

A stipulated route has slow D₂→2D followed by fast D+E→DE. Choose how to obtain D₂+2E→2DE and predict the slow-first-step law. Explain why even a correctly summed mechanism might fail a controlled rate record.

Quantity / representation	Packet value
Slow elementary event	D₂→2D
Fast elementary event	D+E→DE
Assumption	Rapid completion without appreciable intermediate accumulation

Your response

Repeat the fast event twice; the slow-event rate stays k[D₂]Combine the repeated fast events into one four-particle collision and square every concentration factorUse k[D₂][E]² because the net equation contains two E

Your reasoning for Keep event repetition separate from molecularity

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Distinguish a steady intermediate from pre-equilibrium

In the declared held-input model R⇌I,I+S→P, the drain/reversal ratio k₂[S]/kr is 0.25. At sufficiently late time, enter the pre-equilibrium rate overprediction as a percentage of the exact held-input rate. Explain whether waiting longer removes it and whether steady I means global equilibrium.

Quantity / representation	Packet value
Exact late rate	rexact=rpre/(1+ε)
Ratio	ε=0.25
Boundary	Reservoirs hold free R/S; product is removed

Numerical response (units specified in the question) Your reasoning for Distinguish a steady intermediate from pre-equilibrium

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Use the observed intersection rather than an assumed product

An original event ledger contains 50 collisions: 30 meet the energy rule, 20 meet the orientation rule and 7 meet both. Enter the accepted fraction for the rule requiring both. Explain the conditional orientation fraction and whether a marginal-product calculation is justified.

Quantity / representation	Packet value
Total / energetic / aligned	50 /30 /20
Joint count	7
Scope	Finite synthetic event packet, not a molecular rate constant

Numerical response (units specified in the question) Your reasoning for Use the observed intersection rather than an assumed product

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Measure a barrier from the adjacent minimum

A two-step schematic has reactants 0, first transition state 45, intermediate −25, second transition state 35 and products 15 kJ/mol. Enter the second local forward barrier. Explain its reverse barrier, the overall energy change and why the highest plotted peak alone need not identify the slowest overall step.

Quantity / representation	Packet value
Levels /kJ mol⁻¹	0 →45 →−25 →35 →15
Coordinate	Schematic reaction coordinate, not time
Missing kinetics	No populations, prefactors or reversibility data

Numerical response (units specified in the question) Your reasoning for Measure a barrier from the adjacent minimum

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Transfer fractional loss to a fresh starting concentration

An unfamiliar stipulated first-order sample starts at 0.240 mol/L with a 6 s half-life. Enter the concentration at 24 s in mol/L. Explain why subtracting the same amount each interval fails and whether observing just the first halving would prove the law.

Quantity / representation	Packet value
Initial concentration	0.240mol/L
Stipulated first-order half-life	6s
Elapsed time	24s

Numerical response (units specified in the question) Your reasoning for Transfer fractional loss to a fresh starting concentration

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Separate faster approach from demonstrated catalysis

Two fictional matched-temperature runs start with the same composition. With a proposed additive, time to product fraction 0.30 falls from 40 s to 10 s; both report a late plateau 0.60. Choose the supported conclusion and identify controls needed before asserting catalytic action or equilibrium.

Quantity / representation	Packet value
No additive / candidate added	Time to 0.30: 40 s /10 s
Late plateau	0.60 in both runs
Not supplied	Additive regeneration, independent equilibrium/reversibility or mechanism evidence

Your response

A shorter time proves the candidate changes the equilibrium constantMatching late plateaus proves global equilibrium and a unique catalyst mechanismThe packet is consistent with faster approach to the same reported plateau; regeneration, equilibrium and mechanism need additional evidence

Your reasoning for Separate faster approach from demonstrated catalysis

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Retain a snapshot for a human review

The report separates current editable explanations from retained first/latest answers. It includes comparison reasoning only for already exposed cases. Preparing a snapshot does not submit it or establish independent review.

Prepare Unit 5 text snapshotCurrent Unit 5 report textscience:ap-chemistry-unit5-review@1 PUBLIC PRACTICE REPORT — not an independent assessment or course-completion certificate. First/latest numerical or choice responses are retained; explanation text is the CURRENT EDITABLE DRAFT, not a locked first explanation. Keys are public. No automatic writing grade is assigned. No unrecorded key or reasoning is included below. unit5-two-anchor-ambiguity — Challenge a law that fits two anchors A fictional sample has C₀=0.200 mol/L and C(5 s)=0.100 mol/L at fixed conditions. Zero-, first- and second-order laws are each calibrated on these two anchors. Which conclusion is supported? Predict each concentration at 10 s and identify why their k values cannot be ranked without units. Anchors: 0.200 mol/L at 0 s; 0.100 at 5 s Candidates: Constant amount loss, constant fractional loss, linear reciprocal concentration Missing evidence: No later concentrations or uncertainty supplied Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-window-and-common-blank — Normalize a finite-window rate without double-counting a common blank For 2 A→P at fixed volume, corrected concentration readings are 0.0800 and 0.0754 mol/L separated by 2 s. Each has a separately varying absolute reading bound ±0.0002mol/L, plus the same unknown constant blank offset. Enter the average reaction rate in mol L⁻¹s⁻¹. Explain its worst-case reading bound, blank cancellation and why this is not automatically the initial rate. Endpoints: 0.0800 →0.0754mol/L over2s Separate bounds: ±0.0002mol/L per endpoint Shared error: Same constant offset at both times; changing drift excluded by premise Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-repeated-elementary-event — Keep event repetition separate from molecularity A stipulated route has slow D₂→2D followed by fast D+E→DE. Choose how to obtain D₂+2E→2DE and predict the slow-first-step law. Explain why even a correctly summed mechanism might fail a controlled rate record. Slow elementary event: D₂→2D Fast elementary event: D+E→DE Assumption: Rapid completion without appreciable intermediate accumulation Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-pre-equilibrium-error — Distinguish a steady intermediate from pre-equilibrium In the declared held-input model R⇌I,I+S→P, the drain/reversal ratio k₂[S]/kr is 0.25. At sufficiently late time, enter the pre-equilibrium rate overprediction as a percentage of the exact held-input rate. Explain whether waiting longer removes it and whether steady I means global equilibrium. Exact late rate: rexact=rpre/(1+ε) Ratio: ε=0.25 Boundary: Reservoirs hold free R/S; product is removed Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-joint-collision — Use the observed intersection rather than an assumed product An original event ledger contains 50 collisions: 30 meet the energy rule, 20 meet the orientation rule and 7 meet both. Enter the accepted fraction for the rule requiring both. Explain the conditional orientation fraction and whether a marginal-product calculation is justified. Total / energetic / aligned: 50 /30 /20 Joint count: 7 Scope: Finite synthetic event packet, not a molecular rate constant Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-local-energy-barrier — Measure a barrier from the adjacent minimum A two-step schematic has reactants 0, first transition state 45, intermediate −25, second transition state 35 and products 15 kJ/mol. Enter the second local forward barrier. Explain its reverse barrier, the overall energy change and why the highest plotted peak alone need not identify the slowest overall step. Levels /kJ mol⁻¹: 0 →45 →−25 →35 →15 Coordinate: Schematic reaction coordinate, not time Missing kinetics: No populations, prefactors or reversibility data Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-half-life-transfer — Transfer fractional loss to a fresh starting concentration An unfamiliar stipulated first-order sample starts at 0.240 mol/L with a 6 s half-life. Enter the concentration at 24 s in mol/L. Explain why subtracting the same amount each interval fails and whether observing just the first halving would prove the law. Initial concentration: 0.240mol/L Stipulated first-order half-life: 6s Elapsed time: 24s Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: unit5-catalyst-evidence-limit — Separate faster approach from demonstrated catalysis Two fictional matched-temperature runs start with the same composition. With a proposed additive, time to product fraction 0.30 falls from 40 s to 10 s; both report a late plateau 0.60. Choose the supported conclusion and identify controls needed before asserting catalytic action or equilibrium. No additive / candidate added: Time to 0.30: 40 s /10 s Late plateau: 0.60 in both runs Not supplied: Additive regeneration, independent equilibrium/reversibility or mechanism evidence Attempts: 0/3; known feedback exposure: no First response: No response Latest response: No response CURRENT EDITABLE EXPLANATION: Reviewer follow-up: request a new case, inspect assumptions and representations, and assess the explanation separately. This text is an exported snapshot; later edits do not update it.

Source and model boundaries

All eight packets are original public teaching records. The College Board CED, Unit 5 supplies the capability framework. The time-course, pre-equilibrium, collision and energy-profile activities explain assumptions and source boundaries. No restricted exam material or authentic experimental dataset is claimed.

A human reviewer should request a fresh case, inspect units and assumptions, and assess practical evidence separately. Matching a key or submitting prose does not establish full-unit understanding. Actual kinetics, broader catalytic mechanisms and independent practical review remain open.

Return to the separate Unit 4 reaction review.

Continue to the separate Unit 6 thermochemistry review.

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