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AP Chemistry Unit 5 Progress Check MCQ Answers
AP Chemistry Unit 5 Progress Check MCQ Answers
Kinetics - Complete Guide with Full Explanations
By AP-Certified Chemistry Educators at TestPrepKart | Aligned with 2024–25 College Board AP Chemistry CED
Sub-Topics 5.1–5.11 · Exam Weight: 7–9% · Reviewed by AP Chemistry Teachers · Last Updated: 2025
βοΈ Kinetics Unit 5 π Sub-Topics 5.1–5.11 π Exam Weight: 7–9% β 12 Practice Questions π« No Calculator π 100% Free
Quick Answer: The AP Chemistry Unit 5 Progress Check MCQ covers Kinetics (sub-topics 5.1–5.11): rate laws, integrated rate laws (zero/first/second order), half-life, collision model, Arrhenius equation, reaction mechanisms, rate-determining step, and catalysis. This guide gives complete answers with full explanations — including why every wrong answer fails.
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Key Takeaways - Before You Start
π΄ Critical Rules to Know
- Rate laws must be determined EXPERIMENTALLY — never from balanced equations
- Rate-determining step (slowest) controls the overall rate law
- Catalysts lower Ea — they do NOT change ΔH or Keq
- Only first-order reactions have a CONSTANT half-life
- Temperature NEVER changes Ea — only catalysts can lower Ea
π Integrated Rate Law Plots
- [A] vs. t is linear → Zero order (slope = −k)
- ln[A] vs. t is linear → First order (slope = −k)
- 1/[A] vs. t is linear → Second order (slope = +k)
- Constant half-life → diagnostic for First order
11
Sub-Topics
7–9%
Exam Weight
12
Practice Q's
βββ
Hard Unit
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At a Glance - Unit 5 Progress Check MCQ
| Category | Details |
|---|---|
| Official Unit Name | Kinetics |
| Sub-Topics Covered | 5.1 – 5.11 (11 sub-topics) |
| AP Exam Weight | 7–9% of total AP Chemistry exam score (~4–5 MCQ questions on full exam) |
| Recommended Class Periods | 15–16 periods (College Board CED, 2024–25) |
| Progress Check MCQ Format | Multiple-choice | No calculator | Approximately 10–14 questions |
| Difficulty Level | Hard — one of the most difficult and data-intensive units in AP Chemistry |
| Hardest Sub-Topics | 5.5 (Collision Model), 5.7 (Reaction Mechanisms), 5.9 (Steady-State Approximation) |
| Most Tested MCQ Skills | Rate law from initial rate data, integrated rate law graphs, mechanism-to-rate-law derivation |
| Calculator Policy | No calculator permitted on the Progress Check MCQ |
| Prerequisite Knowledge | Unit 1 (Atomic Structure), Unit 4 (Chemical Reactions), Unit 3 (Ideal Gas Law for concentration) |
π§ͺ Where Can You Practice AP Chemistry Unit 5 Progress Check MCQ Answers?
| AP Chemistry Unit 5 Resource | What’s Included | Practice |
|---|---|---|
| Unit 5 Progress Check MCQ Answers - AP Chemistry | Official-style MCQs covering kinetics, rate laws, and reaction mechanisms with answers | Practice |
| Unit 5 MCQ Practice Questions | Exam-level multiple-choice questions with step-by-step explanations | Practice |
| Rate Law & Reaction Order Questions | Problems based on determining rate laws and reaction order from data | Practice |
| Integrated Rate Laws MCQs | Questions on first-order, second-order, and zero-order reactions | Practice |
| Half-Life & Kinetics Problems | Concept-based MCQs involving half-life and reaction rates | Practice |
| Collision Theory & Mechanisms | Questions on reaction mechanisms, intermediates, and rate-determining steps | Practice |
| Graph-Based Kinetics Questions | Interpretation of concentration vs time and rate graphs | Practice |
| Common Mistakes MCQs | Trap questions highlighting common kinetics mistakes | Practice |
| Mixed Concept MCQ Practice | Combined Unit 5 topics in exam-level format | Practice |
| Full Unit 5 MCQ Mock Test | Timed test with scoring and detailed explanations | Practice |
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All 11 CED Sub-Topics - AP Chemistry Unit 5 (Kinetics)
Every sub-topic from 5.1 to 5.11 is tested on the Unit 5 Progress Check MCQ. Use this table to map each question to the exact CED skill.
| Sub-Topic | Title | Key Skill | Frequency |
|---|---|---|---|
| 5.1 | Reaction Rates | Express rate as Δ[conc]/Δt; relate rates of reactants and products via stoichiometry | Very High |
| 5.2 | Introduction to Rate Law | Rate = k[A]α΅[B]βΏ; identify order from exponents; units of k | Very High |
| 5.3 | Concentration Changes Over Time | Integrated rate laws; zero, first, second order; half-life formulas | Very High |
| 5.4 | Elementary Reactions | Molecularity (unimolecular, bimolecular); rate law from elementary step | High |
| 5.5 | Collision Model | Effective collisions require proper energy AND orientation; temperature, concentration, surface area effects | High |
| 5.6 | Reaction Energy Profile | Activation energy Ea; transition state; catalyzed vs. uncatalyzed profiles | High |
| 5.7 | Introduction to Reaction Mechanisms | Multi-step mechanisms; elementary steps; overall equation; intermediates | High |
| 5.8 | Reaction Mechanism and Rate Law | Rate law from rate-determining step; identify intermediates and catalysts | Very High |
| 5.9 | Steady-State Approximation | Pre-equilibrium and steady-state approaches for deriving complex rate laws | Medium |
| 5.10 | Multistep Reaction Energy Profile | Energy diagram for multi-step reaction; identify Ea for each step; overall ΔH | Medium |
| 5.11 | Catalysis | Catalysts lower Ea; homogeneous vs. heterogeneous catalysis; enzymes | High |
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Complete Formula Reference for Unit 5 MCQ
β οΈ No formula sheet is provided on the AP Chemistry Progress Check MCQ. Every formula below must be memorized before exam day.
Rate Laws and Reaction Rates
Rate Law
Rate = k[A]α΅[B]βΏ
k = rate constant; m, n = orders (determined experimentally). Overall order = m + n.
β οΈ NEVER write rate law from balanced equation coefficients. Exponents come ONLY from experimental data.
Units of k by Order
Zero: M/s · First: 1/s · Second: 1/(M·s) · Third: 1/(M²·s)
For nth order reaction: units of k = M^(1−n)/s. Use this to verify your rate law on MCQs.
β οΈ Wrong units of k is a key elimination tool — use it before choosing between answer choices.
Rate from Stoichiometry
For aA → bB: −(1/a)Δ[A]/Δt = +(1/b)Δ[B]/Δt
Divide each species' rate by its stoichiometric coefficient to get the general reaction rate.
β οΈ Rate of A ≠ Rate of B unless stoichiometric coefficients are equal. Always use ratios.
Rate Law from Initial Rate Data
Rateβ/Rateβ = ([A]β/[A]β)α΅
Compare experiments where only one concentration changes. Solve for order m by inspection.
β οΈ If doubling [A] quadruples rate → m=2. If doubling [A] doubles rate → m=1. If no change → m=0.
Integrated Rate Laws - The Most Tested Table
| Order | Integrated Rate Law | Linear Plot | Slope | Half-Life |
|---|---|---|---|---|
| Zero | [A]t = [A]β − kt |
[A] vs. time | −k (negative) | t½ = [A]β/(2k) — depends on [A]β |
| First | ln[A]t = ln[A]β − kt |
ln[A] vs. time | −k (negative) | t½ = 0.693/k — CONSTANT β |
| Second | 1/[A]t = 1/[A]β + kt |
1/[A] vs. time | +k (positive!) | t½ = 1/(k[A]β) — depends on [A]β |
β οΈ Sign Trap: Zero and first order plots have slope = −k (negative). Second order 1/[A] plot has slope = +k (POSITIVE). This sign difference is tested almost every year.
Arrhenius Equation
Arrhenius Equation — All Forms
Exponential form
k = A·e^(−Ea/RT)Linear form (plot)
ln k = ln A − Ea/RT → slope = −Ea/RTwo-temperature form
ln(kβ/kβ) = (Ea/R)(1/Tβ − 1/Tβ)Gas constant R
R = 8.314 J/mol·K (use J, not kJ, for Ea in J/mol)Key Rules: Higher T → larger k → faster rate. Lower Ea → larger k → faster rate. Temperature does NOT change Ea. Only a catalyst lowers Ea.
π§ͺ Rate Law Decision Tool (AP Chemistry Unit 5)
Enter how concentration changes affect rate to find the reaction order and rate law.
π Select values to determine the rate law.
πΊοΈ
Visual Concept Map - AP Chemistry Unit 5 Kinetics
See how all 11 sub-topics connect. Use this map to understand the big picture before diving into individual questions.
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MCQ Strategy - 8 Key Rules for AP Chemistry Unit 5
These strategies are compiled from College Board AP Chemistry Chief Reader Reports and AP educator analysis of the most common student errors on Unit 5 MCQs.
1
Always determine rate law experimentally from data
Never write rate law from a balanced equation alone. Rate law exponents come ONLY from experimental data or from the rate-determining step of a mechanism. This is the most tested trap in Unit 5.
2
For rate law from initial rate data: compare experiments
Find two experiments where one concentration is held constant. The change in rate tells you the order for the other reactant. Rate ratio = (concentration ratio)^order. Solve for order by inspection.
3
For integrated rate laws: look for which plot is linear
[A] vs. t straight → zero order. ln[A] vs. t straight → first order. 1/[A] vs. t straight → second order. Remember: second order slope is POSITIVE (+k), the others are negative (−k).
4
For half-life: check if it is constant
Constant half-life → first order ONLY. If t½ changes with initial concentration → zero or second order. First-order: t½ = 0.693/k (memorize this).
5
For mechanisms: find the slow step (rate-determining)
Write rate law from slow step only. If an intermediate appears, substitute it out using the fast pre-equilibrium expression. The final rate law must contain only initial reactant concentrations.
6
For catalysts: remember they lower Ea, NOT ΔH
Catalysts change the pathway (lower Ea for both forward and reverse equally) but do NOT change ΔH, Keq, or stoichiometry. The energy diagram shows a lower peak but same start and end heights.
7
For collision model: effective collisions need two conditions
Both sufficient energy (≥ Ea) AND proper molecular orientation are required. Not every collision leads to a reaction. Surface area increases collision frequency — NOT activation energy.
8
Unit check for k — use it as an elimination tool
Units of k depend on overall reaction order: Zero: M/s · First: 1/s · Second: 1/(M·s) · Third: 1/(M²·s). On MCQs, wrong units of k eliminates answer choices quickly — check units before solving.
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AP Chemistry Unit 5 Progress Check MCQ - 12 Practice Questions with Full Answers
Every question includes the correct answer with a complete explanation AND why each wrong answer choice is incorrect. Instructions: No calculator permitted. Allow 90 seconds per question. Attempt all questions before checking answers.
1
For the reaction 2A(g) + B(g) → 3C(g), if the rate of appearance of C is 0.06 M/s, what is the rate of disappearance of A?
A0.02 M/s
B0.06 M/s
C0.04 M/s β
D0.09 M/s
β Why C is Correct
Using stoichiometric ratios: Rate of disappearance of A = (coefficient of A / coefficient of C) × rate of C = (2/3) × 0.06 = 0.04 M/s. General formula: −(1/a)Δ[A]/Δt = +(1/c)Δ[C]/Δt.
A0.02 M/s — divides 0.06 by 3 only (0.06/3=0.02), forgetting to multiply by coefficient of A (2). Must use ratio a/c = 2/3.
B0.06 M/s — copies the rate of C without applying stoichiometric ratios. Different species have rates related by their stoichiometric coefficients.
D0.09 M/s — inverts the ratio: 0.06 × (3/2) = 0.09. Rate of A = (2/3)(rate of C), not (3/2)(rate of C).
β AP Exam Tip: Divide ALL rates by their stoichiometric coefficients to get the general reaction rate. Species with larger coefficients have proportionally larger rates of change.
2
In experiments studying A + B → products:
Exp 1: [A]=0.10M, [B]=0.10M, Rate=0.020 M/s
Exp 2: [A]=0.20M, [B]=0.10M, Rate=0.080 M/s
Exp 3: [A]=0.10M, [B]=0.20M, Rate=0.020 M/s
What is the experimental rate law?
Exp 1: [A]=0.10M, [B]=0.10M, Rate=0.020 M/s
Exp 2: [A]=0.20M, [B]=0.10M, Rate=0.080 M/s
Exp 3: [A]=0.10M, [B]=0.20M, Rate=0.020 M/s
What is the experimental rate law?
ARate = k[A][B]
BRate = k[A]² β
CRate = k[A]²[B]
DRate = k[A][B]²
β Why B is Correct
Order of A (Exp 1 vs 2, B constant): Rate ratio = 0.080/0.020 = 4. Conc ratio of A = 0.20/0.10 = 2. Since 2α΅ = 4 → m = 2. A is 2nd order.
Order of B (Exp 1 vs 3, A constant): Rate ratio = 0.020/0.020 = 1. Conc ratio of B = 2. Since 2βΏ = 1 → n = 0. B is zero order. Rate law: Rate = k[A]²
ARate = k[A][B] — assumes both first order. Data shows B has zero effect (Exp 1 vs 3: doubling [B] has no effect on rate).
CRate = k[A]²[B] — correctly identifies A as 2nd order but incorrectly includes B as 1st order. B is zero order — doubling [B] does not change rate.
DRate = k[A][B]² — both orders wrong. Data shows A is 2nd order (not 1st) and B is zero order (not 2nd).
β AP Exam Tip: Step-by-step method: (1) Find two experiments where only ONE concentration changes. (2) Rate ratio = (conc ratio)^order. (3) Repeat for each reactant. Never guess from balanced equation.
3
A reaction has the rate law: Rate = k[X]²[Y]. What are the overall reaction order and units of the rate constant k?
A3rd order; L²/mol²·s β
B2nd order; L/mol·s
C3rd order; L/mol·s
D2nd order; L²/mol²·s
β Why A is Correct
Overall order = 2 + 1 = 3rd order. Units of k: Rate = k[X]²[Y] → M/s = k × M³ → k = M/s ÷ M³ = Mβ»²/s = L²/mol²·s. Pattern: for nth order, units of k = M^(1−n)/s.
B2nd order; L/mol·s — wrong order (2nd not 3rd) and units match 2nd order only. This reaction is 3rd order (exponents 2+1=3).
C3rd order; L/mol·s — correct order but wrong units. L/mol·s = 1/(M·s) = units for 2nd order. For 3rd order: L²/mol²·s.
D2nd order; L²/mol²·s — wrong order stated (2nd). Units L²/mol²·s are correct for 3rd order, but the stated order is wrong.
β Units Shortcut: Zero: M/s · First: 1/s · Second: 1/(M·s) · Third: 1/(M²·s). Memorize this pattern for quick elimination.
4
A student plots three graphs: (i) [A] vs. time, (ii) ln[A] vs. time, (iii) 1/[A] vs. time. Only graph (iii) gives a straight line. What is the reaction order and what is the significance of the slope?
AFirst order; slope = −k
BZero order; slope = −k
CSecond order; slope = −k
DSecond order; slope = +k β
β Why D is Correct
If 1/[A] vs. time is linear → second order. Integrated 2nd-order law: 1/[A]t = 1/[A]β + kt. This is y = mx + b where slope = +k (positive). The slope is POSITIVE for the second-order plot.
AFirst order — first order produces linear ln[A] vs. t (graph ii), not the 1/[A] plot.
BZero order — zero order produces linear [A] vs. t (graph i).
CCorrectly identifies second order but slope is WRONG. The 1/[A] plot has slope = +k, not −k. This sign trap is tested almost every year.
β Sign Trap: Zero order [A] plot: slope = −k. First order ln[A] plot: slope = −k. Second order 1/[A] plot: slope = +k (positive!) The AP exam specifically tests this sign difference.
5
A radioactive isotope undergoes first-order decay. At t=0, the concentration is 80 μg. After 20 years, the concentration is 40 μg. After 40 years, what will the concentration be?
A0 μg
B20 μg β
C10 μg
D30 μg
β Why B is Correct
Half-life = 20 years (constant for first-order). After t=20: 80→40 μg (one half-life). After t=40: 40→20 μg (two half-lives). Formula: [A] = [A]β × (1/2)βΏ = 80 × (1/2)² = 80 × 0.25 = 20 μg.
A0 μg — First-order decay never reaches zero mathematically. The concentration approaches but never equals zero.
C10 μg — This would be after THREE half-lives (80→40→20→10). Only two half-lives elapsed at t=40 years.
D30 μg — Results from subtracting 10 μg per half-life instead of halving. Half-life means concentration is MULTIPLIED by 0.5, not reduced by a fixed amount.
β AP Exam Tip: [A] after n half-lives = [A]β × (1/2)βΏ. Constant half-life is the diagnostic feature of first-order reactions.
6
A 20 g sample of solid iron reacts slowly when exposed to oxygen gas. When the same 20 g of iron is ground into fine powder and exposed to oxygen, the reaction is explosive. Which correctly explains why the powdered iron reacts faster?
APowdered iron has higher kinetic energy than solid iron
BPowdered iron has a lower activation energy than solid iron
CPowdered iron has greater surface area, allowing more collision frequency β
DPowdered iron contains more iron atoms than the solid sample
β Why C is Correct
Grinding iron dramatically increases surface area exposed to oxygen. More surface area → more iron atoms in contact with Oβ → more frequent collisions → much faster reaction rate. Total iron and activation energy are unchanged.
AKinetic energy is determined by temperature, not physical form. Both samples are at the same temperature, so same average KE. Grinding does not increase kinetic energy.
BActivation energy is a property of the chemical reaction — it does NOT change based on physical form of iron. Ea can only be lowered by a catalyst.
DBoth samples have the same mass (20 g) → same number of iron atoms. Grinding changes surface area, not total atom count.
β Classic AP Trap: "Iron nail vs. iron filings" — same mass, same atoms, same Ea, same ΔH. ONLY surface area changes → more collision frequency → faster rate.
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Score Interpretation Guide
| Score | Performance Level | Recommended Next Step |
|---|---|---|
| 11–12 correct | β Excellent - AP Exam Ready | Unit 5 mastered. Move to Unit 6 (Thermochemistry). Review mechanism-to-rate-law derivation for FRQ. |
| 8–10 correct | π΅ Good - Minor Gaps | Re-read full explanations for wrong answers. Focus on integrated rate law plots and mechanism rate law derivation. |
| 5–7 correct | π‘ Developing - Review Needed | Re-study sub-topics for each wrong answer. Focus: rate law from data (5.2), integrated rate laws (5.3), mechanisms (5.7–5.8). |
| 0–4 correct | π΄ Needs Full Re-Study | Restart Unit 5 from sub-topic 5.1. Watch AP Classroom video lessons before retesting. Kinetics is foundational for Units 6–9. |
β οΈ
Top 7 Common Mistakes on AP Chemistry Unit 5 MCQ
| Mistake | Why It Loses Points | How to Fix It |
|---|---|---|
| Writing rate law from balanced equation | Rate law exponents are NOT the same as stoichiometric coefficients in the overall equation | Rate law must come from experimental data or from the rate-determining step — never from coefficients |
| Wrong sign for slope in integrated rate law | Saying slope of 1/[A] vs. t plot is −k instead of +k | Memorize: zero and first order slopes = −k. Second order slope = +k (positive). Different signs! |
| Thinking catalysts change ΔH | Writing that catalyst makes reaction more exothermic or changes energy of products | Catalysts ONLY lower Ea. ΔH, Keq, and stoichiometry are ALL unchanged by a catalyst |
| Confusing intermediates with catalysts | Labeling an intermediate as a catalyst or vice versa | Intermediate: produced then consumed (absent from overall equation). Catalyst: consumed then regenerated (present in overall equation) |
| Leaving intermediate in rate law | Writing Rate = k[intermediate][reactant] without substituting out the intermediate | Always substitute intermediates out using equilibrium expressions from fast pre-equilibrium steps |
| Thinking temperature changes Ea | Writing that higher temperature lowers activation energy | Temperature NEVER changes Ea. Higher temperature shifts Maxwell-Boltzmann distribution so MORE molecules have KE ≥ Ea. Only a catalyst changes Ea. |
| Confusing reaction order with molecularity | Applying molecularity rules to overall reactions | Molecularity (unimolecular, bimolecular) only applies to ELEMENTARY steps. Rate law for overall reaction comes from experimental data. |
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AP Chemistry Unit 5 Study Plan - 5-Phase Priority Guide
| Phase | Sub-Topics | Focus and Actions |
|---|---|---|
| Phase 1 — Rate Fundamentals | 5.1, 5.2 | Master rate expressions, stoichiometric rate relationships, and experimental rate law determination from initial rate data tables. Do 10 rate law problems from data daily. |
| Phase 2 — Integrated Rate Laws | 5.3 | Memorize all three integrated rate law equations and their linear plots with correct slope signs. Practice graph identification: given a linear plot, identify order and k. |
| Phase 3 — Collision Model + Energy | 5.5, 5.6 | Connect collision frequency (surface area, temperature, concentration) to reaction rate. Practice reading energy diagrams for Ea and ΔH. Memorize all 5 catalyst rules. |
| Phase 4 — Mechanisms | 5.7, 5.8 | Master the rate-determining step method. Practice deriving overall rate laws from mechanisms with pre-equilibrium steps. Identify intermediates vs. catalysts. |
| Phase 5 — Arrhenius and Catalysis | 5.9, 5.10, 5.11 | Practice Arrhenius equation problems. Master multistep energy profiles. Distinguish homogeneous from heterogeneous catalysis. Review enzyme catalysis. |
| Final 3 Days | All 5.1–5.11 | Complete all 12 practice questions timed (90 sec each). Review every wrong answer. Complete 2–3 official AP Chemistry past FRQ kinetics questions from AP Central with scoring rubrics. |
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11. Self-Check Quiz - Test Your Unit 5 Kinetics Knowledge
8 rapid-fire questions covering all major Unit 5 concepts. Select your answer and click Check for instant feedback. Try to score 8/8!
Progress0 of 8 answered
1
Which plot gives a straight line for a FIRST-ORDER reaction?
2
What is the half-life formula for a FIRST-ORDER reaction?
3
Rate = k[A]²[B]. What are the units of k for this 3rd order reaction?
4
Which correctly describes the effect of a catalyst on a reaction?
5
A species is PRODUCED in Step 1 and CONSUMED in Step 3 of a mechanism. What is it?
6
What does the Arrhenius equation say about temperature and rate?
7
A solid metal catalyst speeds up a reaction between two gases. This is an example of:
8
Doubling [A] quadruples the rate while doubling [B] has no effect. What is the rate law?
0/8
Quiz Complete!
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12. FAQ - AP Chemistry Unit 5 Progress Check MCQ
What topics are on the AP Chemistry Unit 5 Progress Check MCQ?
The Unit 5 Progress Check MCQ covers Kinetics (sub-topics 5.1–5.11): reaction rates and stoichiometric rate relationships, experimental rate law determination from initial rate data, integrated rate laws (zero, first, second order), half-life, collision model, reaction energy profiles, elementary reactions, reaction mechanisms and rate-determining step, steady-state approximation, multistep energy profiles, and catalysis (homogeneous and heterogeneous).
How do you determine the rate law from experimental data?
(1) Find two experiments where only ONE concentration changes. (2) Calculate rate ratio: Rateβ/Rateβ. (3) Calculate concentration ratio: [A]β/[A]β. (4) Solve: rate ratio = (conc ratio)^order. (5) Repeat for each reactant. (6) Write: Rate = k[A]^m[B]^n. You can NEVER determine a rate law from stoichiometric coefficients of the balanced equation alone.
What is the difference between an intermediate and a catalyst in a mechanism?
An intermediate is PRODUCED in one elementary step and CONSUMED in a subsequent step — it does not appear in the overall balanced equation. A catalyst is CONSUMED in an early step and REGENERATED in a later step — it appears in the overall equation but is ultimately unchanged. On energy diagrams: an intermediate appears as a valley (local energy minimum) between peaks. A catalyst is not consumed overall.
How does temperature affect reaction rate?
Temperature increases reaction rate by increasing kinetic energy of molecules, which shifts the Maxwell-Boltzmann distribution so a greater fraction of molecules have KE ≥ Ea (activation energy). More effective collisions per unit time → higher k (Arrhenius: k = Ae^(−Ea/RT)). Temperature does NOT change the activation energy Ea — only a catalyst can lower Ea. Temperature does not change ΔH or Keq.
How do you derive the rate law from a reaction mechanism?
Step 1: Identify the slow step (rate-determining step). Step 2: Write the rate law from that slow step using the species in it as written. Step 3: Check if any intermediates appear in the rate law. Step 4: If an intermediate appears, use the fast equilibrium step before it to write the intermediate concentration in terms of reactant concentrations. Step 5: Substitute and simplify. The final rate law must be expressed only in terms of initial reactants.
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Conclusion
The AP Chemistry Unit 5 Progress Check MCQ answers are only valuable when you understand the reasoning behind each correct answer AND why each wrong answer fails. The most important skills to master before exam day: determining rate law from experimental data (never from balanced equations), reading integrated rate law graphs with correct slope signs, deriving rate laws from mechanisms by eliminating intermediates, and knowing what catalysts do and do not change (Ea yes, ΔH no, Keq no).
Your Action Plan
- Practice 15–20 Unit 5 kinetics problems per day
- Draw and read all three integrated rate law plots until automatic
- For mechanisms: always write slow step rate law first, then substitute out intermediates
- Complete 2–3 AP Chemistry past FRQ kinetics questions from AP Central with scoring rubrics
- Review every wrong answer — understand the reasoning, not just the correct letter
Official Resources
- AP Chemistry Unit 5 Progress Check — AP Classroom (collegeboard.org/ap-classroom)
- AP Chemistry CED (Unit 5, 2024–25) — apcentral.collegeboard.org
- AP Chemistry Unit 5 AP Daily videos — AP Classroom
- AP Chemistry past FRQ kinetics questions — AP Central
- AP Chemistry Chief Reader Reports — AP Central (released after each May exam)
Final Insight: Kinetics skills appear in AP Chemistry FRQs almost every year. Every concept in this guide — rate law from data, integrated rate law graphs, mechanism analysis, Arrhenius equation, and catalysis — transfers directly to the full AP Chemistry exam. Use this guide alongside official AP Classroom Unit 5 materials.