AP Chemistry · Handsworth Secondary 2026–27

Unit 5 · Kinetics

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 5, 7–9% of the AP exam, taught in 9 blocks. Thermodynamics told us whether a reaction can happen; kinetics tells us how fast. BC Chemistry 12 introduced reaction rate and the collision idea, here we go to AP depth: rate laws from initial-rates data, the three integrated rate laws and their straight-line graphs, half-life, energy profiles, multistep mechanisms with a rate-determining step, and catalysis.

Blocks:  1 → A  ·  2 → B  ·  3 → C  ·  4 → D  ·  5 → E–F  ·  6 → G  ·  7 → H–I  ·  8 → J  ·  9 → review + Rate-Law lab

Name:Block:Date:

A · Reaction Rate CED 5.1 · LO 5.1.A

Rate is how fast a concentration changes. Because reactants are consumed and products formed in a fixed ratio, the balanced equation locks all the rates together.

Core: rate = change in concentration ÷ time. For aA + bB → cC + dD the unique rate is −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = (1/d)Δ[D]/Δt. Reactant terms get a minus sign because their concentrations fall.
What changes a rate: reactant concentration, temperature, surface area, and catalysts.
Worked example
For N₂ + 3H₂ → 2NH₃, H₂ is disappearing at 0.090 mol·L⁻¹·s⁻¹. How fast is NH₃ forming, and what is the unique reaction rate?
Stoichiometry links them: −(1/3)Δ[H₂]/Δt = (1/2)Δ[NH₃]/Δt.
Δ[NH₃]/Δt = (2/3)(0.090) = 0.060 mol·L⁻¹·s⁻¹.
NH₃ forms at 0.060 M·s⁻¹; unique rate = (1/3)(0.090) = 0.030 M·s⁻¹.

Practice A

  1. For 2N₂O₅ → 4NO₂ + O₂, O₂ forms at 0.024 M·s⁻¹. Find the rate of disappearance of N₂O₅.
  2. In 2HI → H₂ + I₂, HI disappears at 0.050 M·s⁻¹. How fast does I₂ form?
  3. List three changes that would speed up a reaction, and tie each to particles colliding.

B · Rate Law & Method of Initial Rates CED 5.2 · LO 5.2.A

The rate law links rate to concentrations. Crucially, the exponents are experimentalyou read them off data, not off the balanced equation.

Core: rate = k[A]ᵐ[B]ⁿ. m and n are the orders in A and B; their sum is the overall order. k is the rate constant (temperature-dependent); its units depend on overall order.
Method of initial rates: compare two trials in which only one reactant's concentration changes. The factor the rate changes by = (concentration factor)^order, so solve for the order.
Worked example
For A + B → products:
Trial 1: [A]=0.10, [B]=0.10, rate = 2.0×10⁻³ M·s⁻¹
Trial 2: [A]=0.20, [B]=0.10, rate = 4.0×10⁻³ M·s⁻¹
Trial 3: [A]=0.10, [B]=0.20, rate = 8.0×10⁻³ M·s⁻¹
1→2: [A] doubles, rate doubles → 2 = 2ᵐ → m = 1 (first order in A).
1→3: [B] doubles, rate ×4 → 4 = 2ⁿ → n = 2 (second order in B).
k from Trial 1: k = rate/([A][B]²) = 2.0×10⁻³ / (0.10 × 0.10²) = 2.0.
rate = k[A][B]², overall order 3, k = 2.0 M⁻²·s⁻¹.

Practice B

  1. Doubling [X] quadruples the rate; tripling [Y] leaves the rate unchanged. Give the order in X, the order in Y, and the overall order.
  2. A reaction is rate = k[A]². What are the units of k? (rate in M·s⁻¹)
  3. If rate = k[A][B] and you triple both [A] and [B] at once, by what factor does the rate change?

C · Integrated Rate Laws, Straight-Line Graphs & Half-Life CED 5.3 · LO 5.3.A

Integrated rate laws tell you concentration as a function of time. The trick the AP loves: each order gives a different straight line, so the graph that comes out linear reveals the order.

The three forms (each is y = mx + b):  Zero order: [A]t = −kt + [A]₀ → plot [A] vs t (slope −k).  First order: ln[A]t = −kt + ln[A]₀ → plot ln[A] vs t (slope −k).  Second order: 1/[A]t = kt + 1/[A]₀ → plot 1/[A] vs t (slope +k).
Half-life (first order only is constant): t½ = 0.693 / k. Independent of starting concentration, which is why radioactive decay (first order) has a fixed half-life.
Worked example 1, first order
A first-order reaction has k = 0.025 s⁻¹ and [A]₀ = 0.50 M. Find [A] after 60 s and the half-life.
ln[A] = −kt + ln[A]₀ = −(0.025)(60) + ln(0.50) = −1.500 + (−0.693) = −2.193
[A] = e^(−2.193) = 0.112 M.
[A] ≈ 0.112 M;   t½ = 0.693/0.025 = 27.7 s.
Worked example 2, half-life counting
A first-order reactant starts at 0.80 M with t½ = 40 s. How much remains after 120 s?
120 s = 3 half-lives → halve three times: 0.80 → 0.40 → 0.20 → 0.10.
0.10 M remains.

Note: on the AP, the three integrated equations and t½ = 0.693/k are on the equation sheet, you don't memorize them, you choose the right one and use it.

Practice C

  1. A plot of 1/[A] vs t is a straight line with slope 0.50 M⁻¹·s⁻¹. State the order and the value (and units) of k.
  2. Second order, k = 0.50 M⁻¹·s⁻¹, [A]₀ = 0.10 M. Find [A] after 120 s.
  3. A radioactive isotope has t½ = 8.0 days. What fraction remains after 24 days?

D · Elementary Reactions & Molecularity CED 5.4 · LO 5.4.A

An elementary reaction is a single collision event, one actual step. For these (and ONLY these) you may read the rate law straight from the coefficients.

Core: the order of an elementary step equals its molecularity (the number of particles colliding). Unimolecular A → P gives rate = k[A]; bimolecular A + B → P gives rate = k[A][B]; 2A → P gives rate = k[A]².
Reality check: three particles colliding at once (termolecular) is rare, so elementary steps are almost always uni- or bimolecular.
Worked example
Write the rate law for the elementary step NO + O₃ → NO₂ + O₂.
It is a single bimolecular collision of one NO with one O₃.
rate = k[NO][O₃] (first order in each, second order overall).

Watch out: this shortcut works only for a step you are told is elementary. For an overall reaction, orders must come from data (Section B).

Practice D

  1. Write the rate law for the elementary step 2NO₂ → NO₃ + NO.
  2. Write the rate law for the unimolecular elementary step O₃ → O₂ + O.
  3. Why can't you write the rate law for an overall (non-elementary) reaction from its coefficients?

E · Collision Model & Maxwell–Boltzmann CED 5.5 · LO 5.5.A

Why do concentration and temperature change rate? Because reactions happen one collision at a time, and only a fraction of collisions actually work.

A successful collision needs two things: enough energy (≥ the activation energy, Ea) AND the correct orientation so the right bonds can break and form. Most collisions fail one of these.
Maxwell–Boltzmann curve: shows the spread of particle energies at a temperature. The area to the right of Ea = the fraction of particles energetic enough to react. Raising T shifts the curve right and flattens it, so that area grows sharply, which is why heating speeds reactions far more than the small rise in collision frequency alone would.
Worked example
A reaction roughly doubles in rate when warmed from 25 °C to 35 °C. Explain at the particle level.
A 10 °C rise barely changes how often particles collide, so frequency alone can't double the rate.
The Maxwell–Boltzmann curve shifts right, so a much larger fraction of collisions now exceeds Ea, more effective collisions per second, so the rate roughly doubles.

Practice E

  1. Name the two requirements for an effective collision.
  2. On a Maxwell–Boltzmann diagram, what does the area to the right of the Ea line represent?
  3. Increasing concentration increases rate, explain in terms of collision frequency (not energy).

F · Reaction Energy Profile & the Arrhenius Idea CED 5.6 · LO 5.6.A

An energy profile graphs potential energy along the reaction coordinate, from reactants over a hill (the transition state) down to products.

Reading the profile: the height from reactants up to the peak = activation energy Ea (forward). The difference between reactant and product energies = ΔH (down = exothermic, up = endothermic). The peak itself is the transition state, highest energy, partial bonds, not isolable.
Arrhenius (qualitative): a higher temperature or a lower Ea means a larger fraction of collisions clears the barrier, so the rate constant k is larger. Same logic as the Maxwell–Boltzmann area.
Worked example
A profile shows reactants at 50 kJ, a peak at 200 kJ, and products at 90 kJ. Find Ea(forward), ΔH, and Ea(reverse).
Ea(forward) = 200 − 50 = 150 kJ.
ΔH = products − reactants = 90 − 50 = +40 kJ (endothermic).
Ea(reverse) = 200 − 90 = 110 kJ. Check: Ea(fwd) − Ea(rev) = 150 − 110 = 40 = ΔH ✓.

Not tested: calculations using the Arrhenius equation will not be assessed on the AP Exam (CED 5.6 Exclusion Statement). Know the relationship (higher T or lower Ea → faster) not the plug-in math.

Practice F

  1. A reaction is exothermic with Ea(forward) = 60 kJ and ΔH = −25 kJ. Find Ea(reverse).
  2. Sketch an energy profile for an endothermic reaction; label reactants, products, transition state, Ea, and ΔH.
  3. Two reactions are identical except reaction 2 has a higher Ea. Which has the larger k at the same temperature? Why?

G · Reaction Mechanisms & Intermediates CED 5.7 · LO 5.7.A

Most reactions happen as a sequence of elementary steps, a mechanism. The steps must add up to the overall equation.

Parts of a mechanism: reactants, products, intermediates (made in an early step, used up in a later step, never appear in the overall equation), and catalysts (present at the start, regenerated by the end). Add the steps and cancel anything that appears on both sides; what's left is the overall reaction.
Worked example
Step 1: NO₂ + NO₂ → NO₃ + NO   Step 2: NO₃ + CO → NO₂ + CO₂. Find the overall reaction and the intermediate.
Add the steps: 2NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂.
Cancel species on both sides: one NO₂ and the NO₃ cancel.
Overall: NO₂ + CO → NO + CO₂. Intermediate: NO₃ (made in step 1, consumed in step 2).

Not tested: collecting data to experimentally detect an intermediate will not be assessed (CED 5.7 Exclusion Statement). You only need to identify intermediates from a written mechanism.

Practice G

  1. Step 1: Cl + O₃ → ClO + O₂; Step 2: ClO + O → Cl + O₂. Give the overall reaction, the intermediate, and the catalyst.
  2. How do you tell an intermediate from a catalyst by where each appears in the steps?
  3. Why must an intermediate be absent from the overall balanced equation?

H · Mechanism & Rate Law: Slow First Step CED 5.8 · LO 5.8.A

A chain is only as fast as its slowest link. The slowest elementary step (the rate-determining step (RDS)) sets the rate law for the whole reaction.

Rule (when step 1 is the slow step): write the rate law from the molecularity of the slow step. Because step 1 uses only starting reactants, the overall rate law follows directly from its reactants.
Worked example
Step 1 (slow): NO₂ + NO₂ → NO₃ + NO   Step 2 (fast): NO₃ + CO → NO₂ + CO₂. Predict the overall rate law.
The slow step is the RDS; it is bimolecular in NO₂.
CO appears only in the fast step, so it is absent from the rate law.
rate = k[NO₂]², second order in NO₂, zero order in CO.
Validity test: a proposed mechanism is only acceptable if (1) its steps sum to the overall equation AND (2) the RDS gives a rate law that matches the experimental one.

Practice H

  1. Slow: A + A → C; Fast: C + B → D. Write the predicted rate law.
  2. A reaction's measured rate law is rate = k[H₂][I₂]. Could a slow step of H₂ + I₂ → 2HI be consistent? Explain.
  3. Why does a reactant that appears only after the rate-determining step not show up in the rate law?

I · Multistep Energy Profile CED 5.10 · LO 5.10.A

A multistep mechanism has one hill per elementary step, with a valley (the intermediate) between hills. Combine the steps' energetics into one profile.

Reading it: each peak is a transition state; each dip between peaks is an intermediate (a real, if short-lived, species). The tallest peak is the highest barrier, that step is the rate-determining step. Overall ΔH is still just (final products − initial reactants).
Worked example
A two-step profile: reactants at 30 kJ; first peak 120 kJ; intermediate valley 70 kJ; second peak 100 kJ; products 40 kJ. Which step is rate-determining, and what is the overall ΔH?
Step 1 barrier = 120 − 30 = 90 kJ. Step 2 barrier = 100 − 70 = 30 kJ.
Step 1 has the larger barrier (and the higher absolute peak), so it is slowest.
Step 1 is rate-determining; overall ΔH = 40 − 30 = +10 kJ (endothermic).

Practice I

  1. On a two-hump profile, how do you spot the intermediate and the rate-determining step?
  2. Reactants at 0 kJ, peaks at 80 and 50 kJ, intermediate at 20 kJ, products at −30 kJ. Identify the RDS and ΔH.
  3. Is an intermediate higher or lower in energy than the transition states on either side of it?

J · Catalysis CED 5.11 · LO 5.11.A

A catalyst speeds a reaction without being used up overall. It works by opening a new pathway with a lower activation energy.

Core: a catalyst provides an alternate mechanism with lower Ea (and/or increases the number of effective collisions), so more collisions clear the barrier. It is consumed in one step and regenerated in a later one, so its net concentration is unchanged. It does NOT change ΔH, only the hill height, not the endpoints.
Two flavours: homogeneous: catalyst in the same phase as reactants (e.g. aqueous, or Cl atoms in gas-phase ozone destruction). Heterogeneous (surface): catalyst a different phase; reactants adsorb and bond to its surface (e.g. metals in a catalytic converter).
Worked example
In ozone depletion: Step 1 Cl + O₃ → ClO + O₂; Step 2 ClO + O → Cl + O₂. Identify the catalyst, the intermediate, and the type of catalysis.
Cl is present at the start and regenerated in step 2 → catalyst. ClO is made then consumed → intermediate.
Both Cl and O₃ are gases (same phase) → homogeneous.
Catalyst = Cl; intermediate = ClO; homogeneous catalysis. Net reaction: O₃ + O → 2O₂.

Practice J

  1. Does a catalyst change Ea, ΔH, both, or neither? Explain what it does and does not affect.
  2. Classify each as homogeneous or heterogeneous: (a) solid Pt speeding a gas-phase reaction; (b) aqueous H⁺ speeding a reaction of dissolved molecules.
  3. On an energy profile, sketch how adding a catalyst changes the curve. Which features move and which stay fixed?
AP Chemistry · Unit 5, Kinetics · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 5.1–5.11) and BC Chemistry 11/12. Integrated rate-law equations and t½ = 0.693/k are provided on the AP equation sheet. The Rate-Law lab (fading of crystal violet via Beer's Law, CED Investigation 11) runs in Block 9.