AP Chemistry · Unit 5

Unit 5: Kinetics

How to find each order from initial-rate data and write a rate law with k and its units

How the collision model explains the effects of concentration, temperature and surface area

How to use integrated rate laws, pick the order from a straight-line plot and work with half-life

How a mechanism's slow step sets the predicted rate law, and how to spot intermediates and catalysts

How to read a reaction energy profile for activation energy, ΔH and the effect of a catalyst

On the exam

Roughly 9 percent of the multiple-choice section by our estimate, often in sets built on a rate table or a concentration-time graph

Free-response parts often ask for orders and a rate law from a data table, k with units, a half-life, or whether a mechanism fits the data

Key terms

20 terms. Say what each one means, then open the card to check.

  • Reaction rate

    How fast a reactant is used up or a product forms, as a change in concentration per unit time, usually in M/s.

    Example. If [O₂] rises from 0.010 M to 0.016 M in 30 s, oxygen forms at an average rate of 2.0 × 10⁻⁴ M/s.

  • Rate law

    An equation, found by experiment, that links the rate to reactant concentrations: rate = k[A]^m[B]^n. The exponents are found from data, not read from the balanced equation.

    Example. For 2N₂O₅ → 4NO₂ + O₂, experiments give rate = k[N₂O₅], so the reaction is first order even though the coefficient of N₂O₅ is 2.

  • Rate constant (k)

    The constant in a rate law. It is fixed at a given temperature, changes when temperature changes, and its units depend on the overall order.

    Example. A first-order reaction has k in s⁻¹, such as k = 0.015 s⁻¹; a second-order one has k in M⁻¹ s⁻¹.

  • Order of reaction

    The exponent of a reactant's concentration in the rate law. The overall order is the sum of the exponents.

    Example. In rate = k[NO]²[Br₂], the reaction is second order in NO, first order in Br₂ and third order overall.

  • Method of initial rates

    Finding each order by comparing trials in which only one starting concentration changes: the ratio of rates equals the concentration ratio raised to the order.

    Example. If tripling [C] with everything else fixed multiplies the rate by 3, the reaction is first order in C.

  • Collision model

    Particles react only when they collide with enough energy, at least the activation energy, and with a suitable orientation.

    Example. Raising the temperature makes more collisions energetic enough to react, so the rate goes up.

  • Integrated rate law

    An equation that gives concentration as a function of time. Each order has its own, and each makes a different plot a straight line.

    Example. For a first-order reaction, ln⁡[A]t=−kt+ln⁡[A]0\ln[\text{A}]_t = -kt + \ln[\text{A}]_0, so ln[A] against time is a straight line with slope −k.

  • Half-life

    The time for a reactant's concentration to fall to half its value. For a first-order reaction it is t1/2=0.693/kt_{1/2} = 0.693/k and does not depend on the starting concentration.

    Example. A first-order reaction with k = 0.0231 min⁻¹ has a half-life of 30.0 min, whether it starts at 1.0 M or 0.10 M.

  • First-order plot

    For a first-order reaction, a graph of ln[A] against time is a straight line with slope −k. Radioactive decay is a common first-order process.

    Example. If ln[A] falls by 0.050 every second in a straight line, the reaction is first order with k = 0.050 s⁻¹.

  • Second-order plot

    For a reaction that is second order in A, a graph of 1/[A] against time is a straight line with slope k, and each half-life is longer than the one before.

    Example. If 1/[A] rises from 4.0 to 7.0 M⁻¹ in 100 s in a straight line, k = 0.030 M⁻¹ s⁻¹.

  • Zero-order reaction

    A reaction whose rate does not depend on the reactant concentration. [A] against time is a straight line with slope −k.

    Example. A reaction with k = 1.5 × 10⁻³ M/s loses 0.090 M of reactant every minute until the reactant is gone.

  • Elementary step

    One collision or molecular event in a mechanism. Its rate law can be written from its own coefficients.

    Example. For the step 2NO₂ → N₂O₄, the rate law of that step is rate = k[NO₂]².

  • Molecularity

    The number of particles that collide in one elementary step: unimolecular (1), bimolecular (2) or, rarely, termolecular (3).

    Example. O₃ → O₂ + O is unimolecular; NO + O₃ → NO₂ + O₂ is bimolecular.

  • Reaction mechanism

    The series of elementary steps by which a reaction happens. The steps must add up to the overall equation and give a rate law that matches experiment.

    Example. Two steps, NO₂ + NO₂ → NO₃ + NO and NO₃ + CO → NO₂ + CO₂, add up to the overall reaction NO₂ + CO → NO + CO₂.

  • Rate-determining step

    The slowest step in a mechanism. It limits the overall rate, so the predicted rate law comes from this step.

    Example. If the slow step is NO₂ + F₂ → NO₂F + F, the predicted rate law is rate = k[NO₂][F₂].

  • Intermediate

    A species made in one step of a mechanism and used up in a later step. It does not appear in the overall equation or the final rate law.

    Example. In the steps NO + Br₂ → NOBr₂ and NOBr₂ + NO → 2NOBr, NOBr₂ is an intermediate.

  • Activation energy (Ea)

    The minimum energy colliding particles need to react. On an energy diagram it is the height of the peak above the reactants.

    Example. If reactants sit at 25 kJ/mol and the peak at 95 kJ/mol, the activation energy is 70 kJ/mol.

  • Transition state

    The highest-energy arrangement of atoms along a step, at the top of a peak on the energy diagram. It cannot be isolated.

    Example. In a one-step reaction, the single peak on the energy diagram is the transition state.

  • Catalyst

    A substance that speeds a reaction by giving it a path with a lower activation energy. It is used in one step and remade in a later one, and it does not change ΔH.

    Example. Platinum in a car's catalytic converter speeds the change of CO to CO₂ without being used up.

  • Reaction energy profile

    A graph of energy along the reaction path. Peaks are transition states, dips between peaks are intermediates, and products minus reactants gives ΔH.

    Example. A profile with two peaks and a dip between them shows a two-step mechanism with one intermediate.

In this unit

  1. Rate laws and the collision model
  2. Integrated rate laws and half-life
  3. Mechanisms, energy profiles and catalysts
  4. From rate data to a rate law
  5. Summary
  6. Mastery quiz