AP Chemistry · Handsworth Secondary 2026–27
Unit 5 · Study Guide
Dr. Ras Mulinta
Kinetics
Exam-focused review
A one-page map of what the Unit 5 test (and the AP exam) expects. This is a checklist, not a re-teach, if a line doesn't click, go back to that section of the notes package. The integrated rate laws and t½ = 0.693/k are on the AP equation sheet.
Must be able to do
- 5.1 Define rate; relate rates of reactants/products by stoichiometry: −(1/a)Δ[A]/Δt = (1/c)Δ[C]/Δt.
- 5.2 Write rate = k[A]ᵐ[B]ⁿ; find m and n by the method of initial rates; give overall order and the units of k.
- 5.3 Pick the integrated rate law by which plot is linear, [A] (zero), ln[A] (first), 1/[A] (second) vs t; use t½ = 0.693/k for first order.
- 5.4 For an elementary step only, write the rate law from molecularity (uni-/bimolecular).
- 5.5 Explain rate with collision theory (energy + orientation) and the Maxwell–Boltzmann curve vs temperature.
- 5.6 Read an energy profile: Ea(fwd), Ea(rev), ΔH, transition state; relate higher T / lower Ea → larger k.
- 5.7 Identify mechanism parts; spot intermediates (made then consumed) and catalysts (used then regenerated); sum steps to the overall equation.
- 5.8 Derive the rate law from a mechanism whose first step is rate-determining; test a mechanism against the measured rate law.
- 5.10 Build/read a multistep profile; tallest peak = rate-determining step; valleys = intermediates.
- 5.11 Explain how a catalyst lowers Ea via a new path (homogeneous vs heterogeneous) without changing ΔH.
The big idea that ties it together
Rate is the count of effective collisions per second. Everything in this unit is one of two levers on that count: how often particles collide (concentration, surface area) and what fraction succeedsneeding energy ≥ Ea and the right orientation (temperature, catalyst, Ea). Rate laws measure the first; the Maxwell–Boltzmann curve and energy profiles explain the second; mechanisms tell you which single slow collision (the RDS) controls the whole thing.
Don't waste time on (excluded by the CED)
Arrhenius-equation calculations (CED 5.6), know that higher T or lower Ea → faster, but no plug-in math. Experimentally detecting/collecting data on intermediates (CED 5.7, 5.8), you only identify intermediates from a written mechanism. Pre-equilibrium (5.9) derivations stay light: focus on the slow-first-step case (5.8).
Quick self-check (answer in your head, then verify)
- For 2N₂O₅ → 4NO₂ + O₂, O₂ forms at 0.020 M·s⁻¹. How fast does N₂O₅ disappear?
- Doubling [A] doubles the rate; doubling [B] quadruples it. Give the rate law and overall order.
- A plot of ln[A] vs t is linear with slope −0.030 s⁻¹. State the order, k, and the half-life.
- First order, k = 0.010 s⁻¹, [A]₀ = 0.40 M. Find [A] after two half-lives.
- Step 1 (slow): 2NO₂ → NO₃ + NO; Step 2 (fast): NO₃ + CO → NO₂ + CO₂. Give the overall rate law and the intermediate.
- An energy profile: reactants 60 kJ, peak 180 kJ, products 100 kJ. Find Ea(fwd), ΔH, Ea(rev).
- Does adding a catalyst change Ea, ΔH, or both?
Check yourself: 1) N₂O₅ disappears at 2×0.020 = 0.040 M·s⁻¹ (ratio 2:1). 2) rate = k[A][B]², overall order 3. 3) first order; k = 0.030 s⁻¹; t½ = 0.693/0.030 = 23 s. 4) two half-lives → 0.40 → 0.20 → 0.10 M. 5) rate = k[NO₂]² (slow step, bimolecular in NO₂; CO is after the RDS); intermediate = NO₃. 6) Ea(fwd) = 180−60 = 120 kJ; ΔH = 100−60 = +40 kJ (endothermic); Ea(rev) = 180−100 = 80 kJ. 7) Ea only (lowers it); ΔH is unchanged.