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Unit 5 · 7–9% of the exam

Kinetics

Unit 5 asks how fast reactions go and why. You will find rate laws from experimental data, use integrated rate laws and half-lives, and connect rates to collisions, activation energy and step-by-step mechanisms. Catalysts, which speed up reactions without being used up, tie the ideas together.

Collision and rate simulator

Raise the temperature or add a catalyst and watch the fraction of collisions that clear the activation energy.

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Topics in this unit

  1. 5.1

    Reaction Rates

    Reaction rate is the change in concentration per unit time; it depends on concentration, temperature, surface area and catalysts.

  2. 5.2

    Introduction to Rate Law

    A rate law, rate = k[A]ᵐ[B]ⁿ, has orders that must be determined experimentally, for example by the method of initial rates; they do not come from the balanced equation.

  3. 5.3

    Concentration Changes Over Time

    Integrated rate laws give a straight line for ln[A] vs t (first order), 1/[A] vs t (second order) or [A] vs t (zero order); first-order half-life is constant.

  4. 5.4

    Elementary Reactions

    For an elementary step, the rate law follows directly from its molecularity (the number of particles colliding in that step).

  5. 5.5

    Collision Model

    Particles react only when they collide with enough energy (at least Eₐ) and the correct orientation.

  6. 5.6

    Reaction Energy Profile

    An energy profile shows the activation energy barrier and the transition state; higher temperature increases the fraction of collisions that can get over the barrier.

  7. 5.7

    Introduction to Reaction Mechanisms

    A mechanism is a sequence of elementary steps that adds up to the overall equation; intermediates are formed in one step and consumed in a later one.

  8. 5.8

    Reaction Mechanism and Rate Law

    The slowest (rate-determining) step controls the overall rate law, which must be consistent with the experimental one.

  9. 5.9

    Pre-Equilibrium Approximation

    When a fast reversible step comes before a slow step, the intermediate's concentration can be written in terms of reactants using the equilibrium expression of the fast step.

  10. 5.10

    Multistep Reaction Energy Profile

    A multistep energy profile has one peak per elementary step, with intermediates in the valleys; the step with the highest barrier is rate-determining.

  11. 5.11

    Catalysis

    A catalyst provides a different mechanism with a lower activation energy; it is consumed and then regenerated and does not change ΔH or the equilibrium constant.

Must know

  • Rate law: rate = k[A]ᵐ[B]ⁿ; overall order = m + n
  • Zero order: [A] = [A]0 − kt (linear [A] vs t)
  • First order: ln[A] = ln[A]0 − kt (linear ln[A] vs t)
  • Second order: 1/[A] = 1/[A]0 + kt (linear 1/[A] vs t)
  • First-order half-life: t½ = 0.693/k (independent of concentration)
  • Units of k: zero order M·s−1, first order s−1, second order M−1·s−1
  • Rate-determining step = slowest step; intermediates must not appear in the final rate law
  • Catalyst: lowers Eₐ, is regenerated, and does not change ΔH or K

Common mix-ups

  • The exponents in the rate law equal the coefficients in the balanced overall equation.

    Orders come from experiment (or from the rate-determining step of a mechanism); coefficients only match for elementary steps.

  • Raising the temperature speeds up a reaction mainly because there are more collisions.

    The main reason is that a much larger fraction of collisions have energy ≥ Eₐ; the increase in collision frequency is small by comparison.

  • A catalyst is not involved in the reaction.

    A catalyst takes part in the mechanism (it is consumed in one step and regenerated in a later one) but it does not appear in the overall equation.

Practice questions

AP-style multiple choice. Every answer comes with the reasoning, including why the tempting wrong answer is wrong.

1 / 6

For A + B → C: Trial 1, [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10−3 M/s. Trial 2, [A] = 0.20 M, [B] = 0.10 M, rate = 8.0 × 10−3 M/s. Trial 3, [A] = 0.10 M, [B] = 0.20 M, rate = 2.0 × 10−3 M/s. What is the rate law?

Free-response practice

The decomposition 2N2O5(g) → 4NO2(g) + O2(g) is studied at a constant temperature. A plot of ln[N2O5] versus time is a straight line with a slope of −6.2 × 10−4 s−1. (a) Write the rate law and justify it. (b) Calculate the half-life. (c) If [N2O5]0 = 0.100 M, calculate [N2O5] after 2000. s.

Flashcards

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See it in real life

Everyday reactions that use Unit 5 ideas, animated atom by atom.