Exam Weight

7% – 9% of AP Exam

Topics Covered

11 Standard Topics (5.1 – 5.11)

Core Concepts

Rate Laws, Half-Life, Arrhenius, RDS

Launch Free Lab

Pillar 1: Empirical Rate Laws

Reaction Rates, Rate Laws, and Integrated Rate Laws (5.1–5.3)

The study of reaction rates links microscopic collisions to macroscopic observations of disappearing reactants and accumulating products over time.

Topic 5.1 Foundation

Reaction Rates

Reaction rate measures the change in concentration of a reactant or product per unit of time (mol / (L \cdot s) = M \cdot s^{-1}). Because rates are defined as positive values, reactant loss is expressed with a negative sign.

Stoichiometric Relationship: For reaction aA + bB \rightarrow cC + dD, the relative rate is:
Rate = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}.
Instantaneous vs Average Rate: Average rate spans a finite time interval, while instantaneous rate is the tangent slope of the concentration-versus-time graph at time t.
Initial Rate: The instantaneous rate at t = 0, evaluated before reverse reactions or byproduct interference occur.

AP Exam Watchdog: Coefficients matter! If 2 N_2O_5 \rightarrow 4 NO_2 + O_2, then NO_2 appears 4 times faster than O_2 appears, and twice as fast as N_2O_5 disappears.

Topic 5.2 Calculations & FRQ

Introduction to Rate Law

A differential rate law expresses rate as a function of reactant concentrations: Rate = k[A]^m[B]^n. Reaction orders m and n cannot be deduced from stoichiometric coefficients; they must be determined experimentally.

Method of Initial Rates: Compare two trials where only one reactant's concentration changes. If doubling [A] leaves rate unchanged, order is 0. If doubling [A] doubles rate, order is 1. If doubling [A] quadruples rate, order is 2.
Overall Reaction Order: The sum of individual exponents (m + n).
Units of Rate Constant k: Vary with overall order x:
Units = M^{1 - x} \cdot \text{time}^{-1}.

Units Alert: Omitting or writing incorrect units for k is one of the most common point-drops on AP Chemistry FRQs. Zero order: M \cdot s^{-1}; 1st order: s^{-1}; 2nd order: M^{-1} \cdot s^{-1}.

Topic 5.3 Heavyweight FRQ Topic

Concentration Changes Over Time & Integrated Rate Laws

Integrated rate laws relate reactant concentration directly to elapsed time t. Graphical analysis of experimental data determines reaction order by identifying which plot yields a straight line (y = mx + b).

Order	Differential Rate Law	Integrated Rate Law	Linear Plot (y vs x)	Slope (m)	Half-Life Expression (t_{1/2})
0	Rate = k	[A]_t = -kt + [A]_0	[A] vs t	-k	t_{1/2} = \frac{[A]_0}{2k} (depends on [A]_0)
1	Rate = k[A]	\ln[A]_t = -kt + \ln[A]_0	\ln[A] vs t	-k	t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k} (constant!)
2	Rate = k[A]^2	\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}	\frac{1}{[A]} vs t	+k	t_{1/2} = \frac{1}{k[A]_0} (increases as [A]_0 drops)

The First-Order Constant Half-Life Hallmark: For any first-order process (including radioactive nuclear decay), t_{1/2} is completely independent of initial concentration. Every successive half-life takes the exact same number of seconds or hours.
Slope Sign Note: The 2nd-order plot of 1/[A] vs t has a positive slope (+k), while 0th and 1st order plots have negative slopes (-k).

Graph Reading Skill: If the question provides three plots ([A] vs t, \ln[A] vs t, and 1/[A] vs t), find the only graph that is a straight line. If \ln[A] is linear, the reaction is strictly 1st order, and the rate constant is k = -\text{slope}.

Pillar 2: Collision Model & Energy Profiles

Elementary Steps, Collision Theory & Potential Energy (5.4–5.6)

Molecules must physically collide with proper three-dimensional orientation and sufficient kinetic energy to overcome the activation barrier.

Topic 5.4 Mechanistic Basis

Elementary Reactions & Molecularity

An elementary reaction occurs in a single collision event. Unlike overall reactions, the rate law of an elementary step is written directly from its stoichiometric coefficients.

Unimolecular: A single molecule rearranges or fragments (A \rightarrow \text{products}; Rate = k[A]).
Bimolecular: Two particles collide (A + B \rightarrow \text{products}; Rate = k[A][B], or 2A \rightarrow \text{products}; Rate = k[A]^2).
Termolecular: Three particles collide simultaneously. Extremely rare due to astronomical odds of three molecules meeting with proper orientation and energy at the same instant.

Golden Rule: Never write a rate law from balanced overall equation coefficients. You may only use coefficients if a step is explicitly stated to be elementary.

Topic 5.5 Conceptual Theory

Collision Model & Activation Energy

For a reaction to occur, colliding particles must possess: (1) kinetic energy equal to or exceeding activation energy (E_a), and (2) favorable steric orientation to form new bonds.

Temperature Influence: Higher temperature shifts the Maxwell-Boltzmann distribution toward higher kinetic energies, exponentially increasing the fraction of collisions with KE \ge E_a.
Arrhenius Equation: k = A e^{-E_a / RT}. Where A is the frequency factor (accounting for collision frequency and steric factor p), R = 8.314\text{ J/(mol}\cdot\text{K)}, and T is temperature in Kelvin.
Linear Arrhenius Form: \ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A. A plot of \ln k vs 1/T yields slope -E_a / R.

Why Rate Increases With Temp: Collision frequency increases slightly, but the dominant factor is the exponential increase in the fraction of molecules with E \ge E_a.

Topic 5.6 Visual & Diagnostic

Reaction Energy Profiles & The Transition State

A potential energy diagram graphs potential energy along the reaction coordinate (progress from reactants to products). It provides a direct visual representation of activation barriers and thermodynamic enthalpy changes.

Transition State / Activated Complex: The highest potential energy configuration along the path where old bonds are partially broken and new bonds are partially formed. Highly unstable, cannot be isolated.
Forward Activation Energy (E_{a,fwd}): Energy difference between reactants and the peak transition state (E_{a,fwd} = E_{TS} - E_{reactants}). Always positive!
Reverse Activation Energy (E_{a,rev}): Energy difference between products and the peak transition state (E_{a,rev} = E_{TS} - E_{products}).
Enthalpy Change (\Delta H): \Delta H = E_{products} - E_{reactants} = E_{a,fwd} - E_{a,rev}. Exothermic reactions feature \Delta H < 0 (products lower than reactants). Endothermic reactions feature \Delta H > 0 (products higher than reactants).

Thermodynamics vs Kinetics: \Delta H is a thermodynamic state function (depends only on initial and final states). E_a is a kinetic property (governs rate). A reaction can be highly thermodynamically favorable (\Delta G^\circ \ll 0) but imperceptibly slow if it has a high activation barrier (kinetic control).

Pillar 3: Mechanisms, Steady-State & Catalysis

Multistep Mechanisms, Pre-Equilibrium & Catalysis (5.7–5.11)

Chemical reactions typically proceed via a sequence of elementary steps. Validating a mechanism requires matching the overall stoichiometry and observed experimental rate law.

Topics 5.7 & 5.8 FRQ Mechanism Proofs

Reaction Mechanisms & The Rate-Determining Step

A proposed mechanism is plausible only if: (1) elementary steps sum to the overall balanced equation, and (2) the rate law of the rate-determining step matches the observed rate law.

Rate-Determining Step (RDS): The slowest elementary step in the mechanism. It has the highest activation barrier relative to its immediate reactant state and acts as the bottleneck for the entire reaction.
Intermediates vs Catalysts:
• Intermediate: Produced in an early step, then consumed in a later step. Does not appear in the overall balanced equation.
• Catalyst: Consumed in an early step, then regenerated in a later step. Present before and after reaction.
Slow First Step: If Step 1 is the slow step, the overall rate law is simply the rate law of Step 1.

Mechanism Check: An experimental rate law can disprove a mechanism, but can never prove one conclusively—it can only support it as consistent with data!

Topic 5.9 Advanced FRQ

Fast Equilibrium Preceding a Slow Step

When the rate-determining step is preceded by a fast reversible step, the rate law for the slow step contains an intermediate, which cannot appear in the final reported rate law.

Example Mechanism:
Step 1 (Fast Eq): \text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2 (k_1 forward, k_{-1} reverse)
Step 2 (Slow RDS): \text{NOBr}_2 + \text{NO} \rightarrow 2\text{NOBr} (k_2)
Deducing Rate Law:
1. RDS rate: Rate = k_2[\text{NOBr}_2][\text{NO}].
2. Set Step 1 forward = reverse: k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2].
3. Solve for intermediate: [\text{NOBr}_2] = \frac{k_1}{k_{-1}}[\text{NO}][\text{Br}_2].
4. Substitute into RDS: Rate = \left(\frac{k_2 k_1}{k_{-1}}\right)[\text{NO}]^2[\text{Br}_2] = k_{obs}[\text{NO}]^2[\text{Br}_2].

Intermediates Forbidden: Never leave an intermediate in the overall rate law expression on an AP FRQ! Always use pre-equilibrium algebra to express it in terms of actual reactants.

Topic 5.10 Diagrammatic Analysis

Multistep Reaction Energy Profiles

Multistep mechanisms produce energy diagrams with multiple peaks (transition states) and valleys (reaction intermediates).

Number of Steps: Number of peaks = number of elementary steps.
Number of Intermediates: Number of valleys between reactant and product = number of reaction intermediates.
Identifying the RDS: The transition state with the highest energy barrier relative to the preceding minimum represents the rate-determining step.

Key Distinction: Intermediates sit in local energy minima (valleys) and have finite (though short) lifetimes. Transition states sit at local energy maxima (peaks) and are transient configurations of breaking/forming bonds.

Topic 5.11 Core Conceptual

Catalysis: Mechanisms & Energetics

A catalyst accelerates reaction rates by providing an alternative reaction pathway with a lower activation energy (E_a). It does not alter the thermodynamic equilibrium constant (K) or \Delta H.

Types of Catalysts:
• Acid-Base Catalysis: Reactant gains or loses a proton, forming an active intermediate.
• Surface / Heterogeneous Catalysis: Reactants adsorb onto a metal surface, weakening bonds and aligning reactant orientation (e.g., catalytic converters).
• Enzyme / Biological Catalysis: Substrates bind to an active site via specific noncovalent interactions, stabilizing the transition state.
Equilibrium Invariance: Because a catalyst lowers E_{a,fwd} and E_{a,rev} by the exact same amount, it accelerates forward and reverse rates equally, leaving K_{eq} unchanged.

Zero Yield Change: A catalyst helps a system reach equilibrium faster, but it never increases the equilibrium yield or changes \Delta H^\circ, \Delta S^\circ, or \Delta G^\circ.

⚡ Kinetic Siege: The Reaction Rate Architect

Test your mastery of initial rates, integrated rate laws, and activation energy profiles. Free & interactive—no sign-up required.

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Reaction Trial A

Target: Determining Order with Respect to [A]

Experimental data analysis prompt...

Integrated Rate Law Diagnostic Arena

Examine the linear plot below and diagnose the reaction order, the slope identity, and the half-life behavior.

Select Diagnostic Mystery Reaction:

Activation Energy & Catalyst Simulator

Adjust reaction temperature and catalyst conditions to observe how the activation barrier and Maxwell-Boltzmann molecular fraction alter the rate constant k.

Temperature (T): 298 K (25 °C)

Uncatalyzed Activation Energy (E_a): 60 kJ/mol

Introduce Catalyst (Lowers E_a by 30 kJ/mol)

Continue Your AP Chemistry Mastery Route

Navigate directly to related foundational and advanced curriculum modules:

Unit 1: Atomic Structure Unit 2: Molecular & Ionic Bonding Unit 3: Intermolecular Forces Unit 4: Chemical Reactions Unit 5: Kinetics (Active) Unit 6: Thermodynamics Unit 7: Equilibrium

Master Kinetics: Reaction Rates, Differential & Integrated Rate Laws, Mechanisms, and Catalysis.