Multistep Reaction Energy Profile - AP Chemistry

SECTION 1

Historical Context & Motivation

For much of the nineteenth century, chemists treated chemical reactions as single-step transformations—reactants converted to products in one concerted event. This view was sufficient for stoichiometric calculations but offered no insight into why some reactions were fast while others were agonizingly slow. The birth of chemical kinetics in the late 1800s, driven by systematic rate measurements, revealed that many apparently simple reactions actually proceed through multiple elementary steps. Understanding the energy changes at each step became essential for predicting reaction rates and designing catalysts.

1884 • Van 't Hoff's Études

Jacobus van 't Hoff published Études de dynamique chimique, introducing the concept that reaction rates depend on temperature and that reactions may proceed through multiple stages.

1889 • Arrhenius Equation

Svante Arrhenius proposed his famous equation relating the rate constant to an activation energy barrier, providing a quantitative link between energy profiles and observed kinetics.

1935 • Transition State Theory

Henry Eyring, Meredith Evans, and Michael Polanyi developed transition state theory, formalizing the energy saddle point through which reactants must pass—the foundation of modern energy profiles.

1953 • Lindemann–Hinshelwood Mechanism

Building on earlier work, Cyril Hinshelwood refined multistep mechanistic thinking for unimolecular reactions, demonstrating that even apparently first-order processes involve at least two elementary steps with distinct energy barriers.

1981 • Computational Energy Surfaces

Kenichi Fukui and Roald Hoffmann shared the Nobel Prize for applying molecular orbital theory to reaction pathways, enabling the computational mapping of multistep energy profiles that guide modern catalyst design.

These developments converge on a central question in kinetics: when a reaction mechanism involves two or more elementary steps, how does the energy landscape of each step determine the overall rate? The multistep reaction energy profile is the tool that answers this question, mapping potential energy against reaction coordinate to reveal intermediates, transition states, and the rate-determining step.

SECTION 2

Core Principles & Definitions

Before dissecting a multistep energy profile, you need a precise vocabulary. A reaction mechanism is the series of elementary steps whose sum gives the overall balanced equation. Each elementary step has its own activation energy, and between consecutive steps the system may pass through a reactive intermediate—a species that is produced in one step and consumed in a subsequent step. The energy profile encodes all of this information in a single diagram.

Transition State

The highest-energy configuration along a single elementary step, corresponding to a saddle point on the potential energy surface. It cannot be isolated because it exists for only the duration of a single molecular vibration.

Reactive Intermediate

A species that occupies a local energy minimum between two transition states. Although short-lived, intermediates are real molecules with finite lifetimes and can sometimes be detected spectroscopically.

Activation Energy (Eₐ)

The energy difference between a reactant or intermediate and the next transition state. Each elementary step has its own Eₐ. The step with the largest Eₐ is typically the rate-determining step.

Rate-Determining Step (RDS)

The slowest elementary step in the mechanism. On the energy profile, the RDS features the highest transition state relative to the starting material (i.e., the largest overall energy barrier the system must surmount).

Overall ΔE (or ΔH)

The net energy difference between products and reactants. A negative ΔE means the reaction is exothermic; a positive ΔE means endothermic. This value is independent of the pathway.

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KEY TAKEAWAY

Think of a multistep reaction as a hiking trail with several mountain passes. Intermediates are the valleys between passes where you can briefly rest, while transition states are the pass summits you must climb over. The highest pass relative to the trailhead determines how long the entire hike takes—that's your rate-determining step. A catalyst is like blasting a tunnel through the tallest peak: the trailhead and destination don't move, but the hardest climb gets much easier.

SECTION 3

Visual Explanation — The Two-Step Energy Profile

The diagram below illustrates a generic two-step exothermic reaction. The reaction coordinate (horizontal axis) tracks the progress of bond-breaking and bond-forming events, while the vertical axis plots potential energy. Two humps appear—one for each elementary step—separated by a valley that represents the reactive intermediate.

A two-step exothermic energy profile. TS₁ and TS₂ mark the transition states for steps 1 and 2, respectively. The valley between them corresponds to the reactive intermediate. Because TS₂ is the highest point on the entire profile, step 2 is the rate-determining step, even though it follows the intermediate. The overall ΔE is negative, confirming the reaction is exothermic.

Several features deserve close attention. First, the number of peaks (humps) in the profile equals the number of elementary steps in the mechanism. A two-step mechanism has exactly two transition states and one intermediate; a three-step mechanism has three transition states and two intermediates. Second, the rate-determining step is identified by finding the transition state that lies highest in energy relative to the starting reactants—not just the tallest individual hump. In the diagram above, TS₂ is higher than TS₁ when measured from the reactant baseline, so step 2 controls the overall rate. Third, the overall thermodynamic favorability (ΔE or ΔH) is determined solely by the difference between products and reactants, completely independent of the pathway taken.

SECTION 4

Mathematical Framework — Connecting Eₐ to Rate

The energy profile is more than a qualitative picture—it connects directly to the Arrhenius equation and the rate law. Each elementary step has its own rate constant governed by its activation energy, and the slowest step dictates the form of the overall rate law observed experimentally.

ARRHENIUS EQUATION

k = A × e^(−Eₐ / RT)

k = rate constant; A = pre-exponential (frequency) factor; Eₐ = activation energy (J mol⁻¹); R = 8.314 J mol⁻¹ K⁻¹; T = absolute temperature (K). A larger Eₐ yields a smaller rate constant, meaning that step is slower.

TWO-POINT ARRHENIUS FORM

ln(k₂ / k₁) = (Eₐ / R) × (1/T₁ − 1/T₂)

This form allows you to calculate Eₐ from rate constants measured at two different temperatures, or to predict how a rate constant changes with temperature—both common AP Chemistry tasks.

RATE LAW FROM THE RDS

rate = k_RDS × [reactants in RDS]^(stoichiometric coefficients)

The overall rate law is written from the rate-determining step. If the RDS involves a reactive intermediate, the intermediate's concentration must be expressed in terms of observable reactant concentrations using the steady-state or pre-equilibrium approximation.

Consider a two-step mechanism where step 1 is a fast, reversible pre-equilibrium and step 2 is the slow, rate-determining step. If step 1 is A + B ⇌ C (fast) and step 2 is C + D → E (slow), the rate law from the RDS is rate = k₂[C][D]. Because C is an intermediate, we use the equilibrium expression from step 1: K₁ = [C] / ([A][B]), giving [C] = K₁[A][B]. Substituting yields rate = k₂K₁[A][B][D], which is the experimentally observable rate law. The energy profile encodes all of this: the shallow valley for C tells us the intermediate is relatively unstable, and the tall second hump tells us step 2 is rate-limiting.

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AP EXAM TIP

On the AP Chemistry exam, you may be asked to draw or interpret an energy profile, identify the RDS, count intermediates, or write a rate law consistent with a given mechanism. Remember: intermediates appear in valleys (local minima), transition states sit at peaks, and the number of peaks equals the number of elementary steps. The rate law is always derived from the RDS, never from the overall balanced equation.

SECTION 5

Catalyzed vs. Uncatalyzed Profiles

One of the most important applications of multistep energy profiles is visualizing how a catalyst accelerates a reaction. A catalyst provides an alternative mechanism—often with more elementary steps—such that every transition state along the catalyzed pathway lies lower in energy than the highest transition state of the uncatalyzed pathway. The catalyst is consumed in an early step and regenerated in a later step, so it does not appear in the overall balanced equation and does not alter the thermodynamic ΔE.

Comparison of uncatalyzed (red, single hump) and catalyzed (green, two humps) pathways for the same overall reaction. The catalyzed route introduces an additional intermediate (Int') and two lower transition states (TS₁', TS₂'). Crucially, the highest point on the catalyzed path (TS₂') is significantly lower than the uncatalyzed TS, yielding a smaller effective Eₐ and a faster overall rate. Both pathways share the same reactant and product energy levels, confirming that the catalyst does not change ΔE.

Notice that the catalyzed pathway has more elementary steps than the uncatalyzed pathway—yet it is faster. This seems counterintuitive until you remember that speed depends on the height of the tallest barrier, not the number of steps. Enzymes in biological systems exploit this principle spectacularly: many enzyme-catalyzed reactions proceed through five or six elementary steps, yet occur millions of times faster than the uncatalyzed reaction because every transition state is stabilized by specific interactions within the enzyme active site.

SECTION 6

Worked Example — Reading & Interpreting a Profile

The following worked example mirrors the type of problem you will encounter on the AP Chemistry exam. Suppose you are given a three-step mechanism with the energy values shown below, and asked to identify the rate-determining step, count intermediates, and determine whether the reaction is exothermic or endothermic.

Analyzing a Three-Step Energy Profile

1
Step 1 — Identify Given Information

A reaction proceeds via three elementary steps. The energy levels (in kJ mol⁻¹) are: Reactants = 0 (reference); TS₁ = 80; Intermediate 1 = 30; TS₂ = 120; Intermediate 2 = 50; TS₃ = 90; Products = −20.
2
Step 2 — Count Intermediates and Transition States

Three elementary steps produce three transition states (TS₁, TS₂, TS₃) and two reactive intermediates (Int 1 at 30 kJ mol⁻¹ and Int 2 at 50 kJ mol⁻¹). Intermediates sit in the valleys between consecutive peaks.

3 transition states, 2 intermediates
3
Step 3 — Calculate Eₐ for Each Step

Eₐ for each step is measured from the preceding minimum to the next transition state. Eₐ₁ = TS₁ − Reactants = 80 − 0 = 80 kJ mol⁻¹. Eₐ₂ = TS₂ − Int 1 = 120 − 30 = 90 kJ mol⁻¹. Eₐ₃ = TS₃ − Int 2 = 90 − 50 = 40 kJ mol⁻¹.

Eₐ₁ = 80, Eₐ₂ = 90, Eₐ₃ = 40 kJ mol⁻¹
4
Step 4 — Identify the Rate-Determining Step

The RDS is the step whose transition state is the highest point on the entire profile relative to the overall starting point. TS₂ at 120 kJ mol⁻¹ is the global maximum, so step 2 is the rate-determining step. Note that the individual Eₐ of step 2 (90 kJ mol⁻¹) is also the largest, but the definitive test is always the absolute height of the transition state measured from the reactant baseline.

Step 2 is the rate-determining step (TS₂ = 120 kJ mol⁻¹, highest point)
5
Step 5 — Determine Exothermic vs. Endothermic

The overall energy change is ΔE = E(products) − E(reactants) = −20 − 0 = −20 kJ mol⁻¹. Because ΔE is negative, the reaction is exothermic overall—the products are lower in energy than the reactants.

ΔE = −20 kJ mol⁻¹ → Exothermic

SECTION 7

Common Misconceptions & Comparisons

Energy profiles are a frequent source of errors on the AP exam, often because students confuse related but distinct concepts. The table below highlights common misconceptions alongside the correct interpretations.

Feature	Common Misconception	Correct Understanding
Transition state vs. intermediate	"They're the same thing—both are high-energy species."	Transition states are energy maxima (peaks) that cannot be isolated. Intermediates are energy minima (valleys) that have finite lifetimes.
Identifying the RDS	"The step with the tallest individual hump is always the RDS."	The RDS is the step whose transition state is the highest absolute point on the profile. A step starting from a high-energy intermediate may have a small individual Eₐ but its TS could still be the highest overall.
Effect of a catalyst	"A catalyst lowers ΔE and makes the reaction more exothermic."	A catalyst lowers Eₐ by providing an alternative pathway. It does not change ΔE, ΔH, or ΔG because it does not alter the energies of reactants or products.
Number of steps vs. speed	"More steps means a slower reaction."	Speed depends on Eₐ of the RDS, not the number of steps. A catalyzed path may have more steps yet be dramatically faster because all barriers are lower.
Rate law origin	"The rate law comes from the overall balanced equation."	The rate law is derived from the rate-determining step. Overall stoichiometric coefficients do not directly dictate reaction orders unless the reaction is a single elementary step.

Common misconceptions about multistep energy profiles

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KEY TAKEAWAY

When analyzing any energy profile, apply a systematic checklist: (1) count the peaks to determine the number of elementary steps; (2) identify valleys as intermediates; (3) measure each Eₐ from the preceding valley to the next peak; (4) find the highest absolute peak for the RDS; (5) compare reactant and product energy levels for overall thermodynamics. This five-step protocol prevents the most common AP exam errors.

SECTION 8

Connection to Advanced Theory

The energy profiles encountered in AP Chemistry are two-dimensional cross-sections of a much richer landscape. In advanced physical chemistry and computational chemistry courses, you will encounter the full potential energy surface (PES), a multidimensional surface where each axis represents a different internal coordinate (bond length, bond angle, dihedral angle). Transition states correspond to first-order saddle points on this surface, and the minimum energy path connecting reactants to products through these saddle points is the intrinsic reaction coordinate (IRC)—the theoretical foundation for the simple reaction coordinate axis you draw on the AP exam.

AP Chemistry Concept	Advanced / College Extension
Reaction coordinate (1-D plot)	Intrinsic reaction coordinate on a multidimensional PES
Eₐ from Arrhenius equation	Gibbs free energy of activation (ΔG‡) from Eyring–Polanyi equation: k = ( $k_B$ T / h) × e^(−ΔG‡ / RT)
Pre-equilibrium approximation	Steady-state approximation; Michaelis–Menten kinetics for enzyme mechanisms
Qualitative profile sketch	Density functional theory (DFT) calculations to map precise energy profiles with quantitative barrier heights
Catalyst lowers Eₐ	Transition state stabilization via orbital interactions, electrostatic effects, and strain release (Woodward–Hoffmann rules, Hammond's postulate)

AP-level concepts mapped to their advanced counterparts

One particularly useful bridge concept is Hammond's postulate: for an exothermic elementary step, the transition state resembles the reactants in structure and energy; for an endothermic step, it resembles the products. This principle connects the shape of the energy profile to the molecular geometry of the transition state—a concept that becomes central in organic chemistry when predicting regioselectivity and stereoselectivity.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A reaction energy profile shows three peaks and two valleys between the reactant and product energy levels. Which of the following statements is correct?

A. The mechanism has three elementary steps, three transition states, and three intermediates.
B. The mechanism has three elementary steps, three transition states, and two intermediates.
C. The mechanism has two elementary steps, three transition states, and two intermediates.
D. The mechanism has three elementary steps, two transition states, and two intermediates.

Show answer (B)

The correct answer is B. Each peak on the energy profile corresponds to one transition state and one elementary step, so three peaks mean three elementary steps and three transition states. The valleys between consecutive peaks correspond to reactive intermediates. With three peaks, there are exactly two valleys, hence two intermediates. Option A is wrong because it claims three intermediates—there is always one fewer intermediate than the number of steps. Option C incorrectly states only two elementary steps despite three peaks. Option D incorrectly states only two transition states.

Question 2: BASIC CALCULATION

A two-step reaction has the following energy values (kJ mol⁻¹): Reactants = 0, TS₁ = 65, Intermediate = 20, TS₂ = 95, Products = −10. What is the activation energy of the rate-determining step?

A. 65 kJ mol⁻¹
B. 75 kJ mol⁻¹
C. 95 kJ mol⁻¹
D. 30 kJ mol⁻¹

Show answer (B)

The correct answer is B. The RDS is identified by the highest transition state on the profile: TS₂ at 95 kJ mol⁻¹ is higher than TS₁ at 65 kJ mol⁻¹, so step 2 is the RDS. The activation energy for step 2 is measured from the intermediate (its starting point) to TS₂: Eₐ₂ = 95 − 20 = 75 kJ mol⁻¹. Option A (65 kJ mol⁻¹) is the Eₐ for step 1 (65 − 0), which is not the RDS. Option C (95 kJ mol⁻¹) confuses the absolute energy of TS₂ with the activation energy—Eₐ is the difference from the preceding minimum, not from zero. Option D (30 kJ mol⁻¹) appears to be 95 − 65, confusing TS₂ − TS₁, which is meaningless.

Question 3: INTERMEDIATE

Consider the following two-step mechanism for the reaction of NO₂ with CO: Step 1 (slow): NO₂(g) + NO₂(g) → NO₃(g) + NO(g) Step 2 (fast): NO₃(g) + CO(g) → NO₂(g) + CO₂(g) Overall: NO₂(g) + CO(g) → NO(g) + CO₂(g) Which of the following correctly identifies the intermediate and the rate law?

A. Intermediate = NO₃; Rate = k[NO₂]²
B. Intermediate = NO₃; Rate = k[NO₂][CO]
C. Intermediate = NO; Rate = k[NO₂]²
D. Intermediate = CO₂; Rate = k[NO₂][CO]

Show answer (A)

The correct answer is A. NO₃ is produced in step 1 and consumed in step 2, fitting the definition of a reactive intermediate—it appears in the mechanism but not in the overall equation. The rate law is derived from the rate-determining step (step 1): rate = k₁[NO₂][NO₂] = k₁[NO₂]². CO does not appear in the rate law because it participates only in the fast step after the RDS. Option B is wrong because it includes [CO], which is in the fast step. Option C incorrectly identifies NO as the intermediate—NO is a product that appears in the overall balanced equation. Option D incorrectly identifies CO₂ as the intermediate—CO₂ is a final product.

Question 4: APPLIED

5 pts

An industrial chemist is studying the decomposition of ozone: 2 O₃(g) → 3 O₂(g). The proposed mechanism is: Step 1 (fast equilibrium): O₃ ⇌ O₂ + O (K₁ = k₁/k₋₁) Step 2 (slow): O₃ + O → 2 O₂ (k₂) The uncatalyzed reaction has an effective Eₐ of 105 kJ mol⁻¹. A chlorine-catalyzed pathway has an effective Eₐ of 2.1 kJ mol⁻¹. (a) Write the overall rate law predicted by this mechanism. (b) Identify the intermediate. (c) On an energy profile, explain how the catalyzed pathway differs from the uncatalyzed pathway. (d) Using the Arrhenius equation, estimate the ratio of the catalyzed to uncatalyzed rate constants at 250 K. (R = 8.314 J mol⁻¹ K⁻¹)

Show answer

Rubric (5 points): (a) [1 point] Rate law derivation: The rate from the slow step is rate = k₂[O₃][O]. Since O is an intermediate, use the fast equilibrium: K₁ = [O₂][O]/[O₃], so [O] = K₁[O₃]/[O₂]. Substituting: rate = k₂K₁[O₃]²/[O₂]. Award 1 point for correct substitution yielding rate = k_obs[O₃]²[O₂]⁻¹. (b) [1 point] Identify intermediate: The reactive intermediate is atomic oxygen, O. It is produced in step 1 and consumed in step 2, and does not appear in the overall balanced equation. (c) [1 point] Energy profile comparison: The catalyzed pathway has lower transition states than the uncatalyzed pathway while sharing the same reactant and product energy levels. Both ΔE and ΔH are unchanged. The catalyzed path may have more elementary steps (additional peaks and valleys), but the highest peak is significantly lower, yielding a reduced effective Eₐ. Award 1 point for clearly stating that Eₐ is lowered without changing ΔE/ΔH. (d) [2 points] Arrhenius ratio calculation: k_cat/k_uncat =

e^{(Eₐ,uncat − Eₐ,cat)/(RT)}

. ΔEₐ = 105,000 − 2,100 = 102,900 J mol⁻¹. RT = 8.314 × 250 = 2078.5 J mol⁻¹. Exponent = 102,900 / 2078.5 ≈ 49.5. k_cat/k_uncat =

e^{49.5}

≈ 3.5 × 10²¹. Award 1 point for correct setup and 1 point for correct numerical result (within an order of magnitude). This enormous ratio illustrates why catalytic destruction of ozone by chlorine radicals is so environmentally significant.

Question 5: CRITICAL THINKING

6 pts

A student performs an experiment to study the reaction 2A → D and collects the following data at 300 K: | Experiment | [A]₀ (M) | Initial Rate (M s⁻¹) | |---|---|---| | 1 | 0.10 | 2.0 × 10⁻³ | | 2 | 0.20 | 8.0 × 10⁻³ | | 3 | 0.30 | 1.8 × 10⁻² | The student proposes the following mechanism: Step 1 (fast equilibrium): A ⇌ B K₁ Step 2 (slow): B + A → D k₂ (a) Using the data, determine the overall order of the reaction. (b) Write the predicted rate law from the proposed mechanism and show that it is consistent with the data. (c) Identify all intermediates in the mechanism. (d) Sketch a qualitative energy profile for this mechanism, labeling all transition states, intermediates, reactants, and products. Indicate which step is the RDS. (e) If a catalyst were added that lowered the Eₐ of step 2 by 15 kJ mol⁻¹, predict the qualitative effect on the energy profile and the reaction rate.

Show answer

Rubric (6 points): (a) [1 point] Determining reaction order: Comparing Experiments 1 and 2: [A] doubles (0.10 → 0.20) and rate quadruples (2.0 × 10⁻³ → 8.0 × 10⁻³). Rate ∝ [A]ⁿ, so (0.20/0.10)ⁿ = (8.0 × 10⁻³)/(2.0 × 10⁻³) → 2ⁿ = 4 → n = 2. Confirming with Experiments 2 and 3: (0.30/0.20)ⁿ = (1.8 × 10⁻²)/(8.0 × 10⁻³) → 1.5ⁿ = 2.25 → n = 2. The overall reaction order is 2. Award 1 point for correctly determining second-order kinetics from the data. (b) [2 points] Rate law from mechanism: The rate-determining step (step 2, slow) gives: rate = k₂[B][A]. B is an intermediate produced in step 1. From the fast equilibrium in step 1: K₁ = [B]/[A], so [B] = K₁[A]. Substituting: rate = k₂(K₁[A])[A] = k₂K₁[A]² = k_obs[A]². This predicts second-order kinetics, consistent with the experimental data. Verification: Exp 1→2: (0.20/0.10)² = 4.0, and (8.0 × 10⁻³)/(2.0 × 10⁻³) = 4.0 ✓ Exp 2→3: (0.30/0.20)² = 2.25, and (1.8 × 10⁻²)/(8.0 × 10⁻³) = 2.25 ✓ Award 1 point for correctly deriving rate = k₂K₁[A]² using the pre-equilibrium method to eliminate the intermediate B. Award 1 point for demonstrating consistency with the experimental data. Mechanism check — summing the two steps: Step 1: A ⇌ B Step 2: B + A → D Overall: 2A → D ✓ (B cancels as an intermediate) (c) [1 point] Identifying intermediates: B is the only intermediate. It is produced in step 1 (A ⇌ B) and consumed in step 2 (B + A → D). It does not appear in the overall reaction (2A → D). Award 1 point for correctly identifying B as the intermediate. (d) [1 point] Energy profile sketch: The profile should show two energy peaks (two transition states) separated by one valley (intermediate B). The second peak (TS₂) should be higher than the first peak (TS₁) because step 2 is the rate-determining step (slow step). Reactants (2A) should be labeled at the start, the valley labeled as intermediate B, and products (D) at the end. The overall energy change (ΔE) is the difference between reactant and product energy levels. Award 1 point for a correctly structured two-step energy profile with TS₁ < TS₂, the intermediate valley labeled as B, and proper labeling of reactants and products. (e) [1 point] Effect of catalyst: A catalyst that lowers the Eₐ of step 2 by 15 kJ mol⁻¹ would lower the second peak (TS₂) on the energy profile while leaving the first peak (TS₁), the intermediate valley (B), and the reactant/product energy levels unchanged. Since step 2 is the RDS, lowering its activation energy directly increases the overall reaction rate. The overall thermodynamics (ΔE for the reaction) are unchanged because a catalyst does not alter the energies of reactants or products. Note: if the new TS₂ drops below TS₁, then step 1 could become the new rate-determining step. Award 1 point for explaining that the second peak is lowered, the overall rate increases, and ΔE remains unchanged.

SUMMARY

Lesson Summary

Summary

A multistep reaction energy profile plots potential energy against the reaction coordinate for a mechanism containing two or more elementary steps. Each peak represents a transition state (an energy maximum that cannot be isolated), while each valley between peaks represents a reactive intermediate (a real, short-lived species occupying a local energy minimum). The number of peaks equals the number of elementary steps; the number of valleys between peaks equals the number of intermediates.

The rate-determining step (RDS) is the elementary step whose transition state is the highest point on the entire profile, measured from the reactant baseline. The activation energy (Eₐ) for each step is measured from the preceding local minimum to the next peak and is related to the rate constant through the Arrhenius equation. A catalyst provides an alternative pathway with a lower maximum transition state energy, thereby lowering the effective Eₐ and increasing the rate—without altering the overall ΔE of the reaction. Mastering these profiles is essential for connecting reaction mechanisms to experimental rate laws on the AP Chemistry exam.