5.2.9Additional lessons
Reaction Orders (HL) — IB Chemistry HL Revision Notes & Topical Questions
5 min read 5 parts Lesson 104 of 172
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Orders and Rate–Concentration Graphs
The shape of a graph reveals the order of the reaction with respect to one reactant.
Zero order
- Zero order The rate does not depend on the concentration, so a graph of rate against concentration is a horizontal line.
First order
- First order The rate is directly proportional to the concentration, so rate against concentration is a straight line through the origin.
Second order
- Second order The rate is proportional to the square of the concentration, so rate against concentration is a curve that rises steeply.
Gradient
- Gradient For a first-order reaction the gradient of the rate–concentration line equals the rate constant k.
Section 2 of 5Concentration–Time Graphs
Concentration–Time Graphs
Zero order
- Zero order The concentration falls in a straight line, because the rate (the gradient) is constant.
First order
- First order The concentration decays exponentially, and the gradient becomes smaller as the reactant is used up.
Second order
- Second order The concentration falls faster at first and levels off more slowly than a first-order curve.
Half-life
- Half-life For a first-order reaction the half-life is constant, whereas for other orders it changes as the reaction proceeds.
| Order | Rate vs [A] | [A] vs time |
|---|---|---|
| 0 | Horizontal line | Straight line, falling |
| 1 | Straight line from origin | Exponential decay, constant half-life |
| 2 | Upward curve | Steeper decay, half-life not constant |
Section 3 of 5Orders and Mechanisms
1
Many reactions happen in several steps, and the slowest step is called the rate-determining step.
2
Species that take part in or before the rate-determining step appear in the rate equation.
3
Species that react only after the slowest step do not appear, and they have order zero.
4
The rate equation can therefore support or reject a proposed mechanism but cannot prove it.
5
For a first-order reaction the half-life and the rate constant are related by t½ = ln 2 / k.
Section 4 of 5Worked Example
Question
- A first-order reaction has k = 0.0231 s⁻¹.
- Calculate the half-life and the concentration left after 60 s if [A]₀ = 0.80 mol dm⁻³.
- t½ = ln 2 / k = 0.693 / 0.0231 = 30.0 s.
- 60 s is exactly two half-lives, so the concentration halves twice.
- [A] = 0.80 × ½ × ½ = 0.20 mol dm⁻³ (check: 0.80 × e−0.0231×60) = 0.200).
Answer t½ = 30 s; [A] = 0.20 mol dm⁻³
Section 5 of 5Exam Checklist & Tips
5.2.9 – Key Points for the Exam
- Zero order The rate is independent of the concentration of that reactant.
- Half-life The time taken for the concentration of a reactant to fall to half its value.
- Rate-determining step The slowest step in a mechanism, which controls the overall rate.
- Molecularity The number of particles colliding in one elementary step.
Common Mistakes
- Saying that the order is shown by the coefficient in the balanced equation.
- Reading the order from the gradient of a concentration–time graph without checking its shape.
- Believing that half-life is constant for every reaction order.
- Including intermediates in the final rate equation when they must be expressed using the original reactants.
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