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This lesson covers collision theory, the Maxwell-Boltzmann distribution, rate equations, orders of reaction, the rate constant, rate-determining steps, half-life, and the Arrhenius equation. Kinetics is the study of the speed of chemical reactions and the factors that affect it, and it is examined extensively at A-Level across all specifications.
Key Definition: The rate of reaction is the change in concentration of a reactant or product per unit time, usually measured in mol dm⁻³ s⁻¹.
At GCSE, you learnt that rate is affected by temperature, concentration, surface area, and catalysts. At A-Level, we quantify these effects mathematically using rate equations and the Arrhenius equation.
For a reaction to occur, particles must collide with sufficient energy (equal to or greater than the activation energy, Ea) and with the correct orientation. Not all collisions lead to reaction — only those that meet both criteria are called successful collisions.
Key Definition: The activation energy (Ea) is the minimum energy that colliding particles must possess for a reaction to occur.
Increasing temperature, concentration, or surface area increases the frequency of collisions and/or the proportion of collisions with sufficient energy, thereby increasing the rate.
The Maxwell-Boltzmann distribution shows the distribution of molecular kinetic energies in a gas at a given temperature.
The graph has molecular kinetic energy on the x-axis and the number (or fraction) of molecules with that energy on the y-axis. The curve starts at the origin (no molecules have zero energy), rises steeply to a peak (the most probable energy), then tails off asymptotically to the right — the curve never touches the x-axis because there is no maximum energy. The area under the curve represents the total number of molecules and remains constant.
A vertical line is drawn at the activation energy (Ea). The area under the curve to the right of this line represents the number of molecules with energy ≥ Ea — these are the molecules capable of reacting.
At a higher temperature, the distribution curve shifts to the right and flattens — the peak is lower and shifted to a higher energy. The total area under the curve stays the same (same number of molecules). Crucially, the area to the right of Ea is significantly larger, meaning a much greater proportion of molecules have energy ≥ Ea. This is why even a small temperature increase can cause a large increase in rate.
A catalyst provides an alternative reaction pathway with a lower activation energy (Ea'). The distribution curve itself does not change — the same molecules are present at the same temperature. However, the Ea line shifts to the left (to Ea'), so a larger proportion of molecules have energy ≥ Ea'. More molecules can react per unit time, so the rate increases.
Exam Tip: When drawing Maxwell-Boltzmann distributions, ensure the curve starts at the origin, the peak of the higher-temperature curve is lower and further right, and the total area under both curves is the same. Never let the curve touch the x-axis on the right-hand side. Common mark scheme points: the curve at higher T is flatter, broader, and shifted right; the area to the right of Ea is significantly greater.
For a reaction A + B → products, the rate equation takes the form:
rate = k[A]ᵐ[B]ⁿ
where:
Key Definition: The order of reaction with respect to a reactant is the power to which the concentration of that reactant is raised in the rate equation. It can only be determined experimentally.
The rate equation can only be determined experimentally — it cannot be deduced from the stoichiometric equation.
The order with respect to a reactant describes how the rate changes when the concentration of that reactant changes:
Orders can be determined from initial rates experiments:
flowchart TD
A["Compare two experiments<br/>where only ONE reactant<br/>concentration changes"] --> B["How does the rate change<br/>when concentration doubles?"]
B -->|"Rate unchanged<br/>(×1)"| C["Zero Order<br/>(m = 0)"]
B -->|"Rate doubles<br/>(×2)"| D["First Order<br/>(m = 1)"]
B -->|"Rate quadruples<br/>(×4)"| E["Second Order<br/>(m = 2)"]
The reaction A + B → C was studied. The following data were obtained:
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.20 | 1.6 × 10⁻³ |
Determine the rate equation and calculate k.
Solution:
Step 1: Find the order with respect to A.
Compare experiments 1 and 2 (only [A] changes):
[A] doubles (0.10 → 0.20), rate quadruples (2.0 × 10⁻⁴ → 8.0 × 10⁻⁴).
Rate multiplier = 8.0 × 10⁻⁴ / 2.0 × 10⁻⁴ = 4 = 2²
Therefore, order with respect to A = 2 (second order).
Step 2: Find the order with respect to B.
Compare experiments 2 and 3 (only [B] changes):
[B] doubles (0.10 → 0.20), rate doubles (8.0 × 10⁻⁴ → 1.6 × 10⁻³).
Rate multiplier = 1.6 × 10⁻³ / 8.0 × 10⁻⁴ = 2 = 2¹
Therefore, order with respect to B = 1 (first order).
Step 3: Write the rate equation.
rate = k[A]²[B]
Overall order = 2 + 1 = 3 (third order).
Step 4: Calculate k using data from any experiment (using experiment 1).
2.0 × 10⁻⁴ = k × (0.10)² × (0.10)
2.0 × 10⁻⁴ = k × 1.0 × 10⁻³
k = 2.0 × 10⁻⁴ / 1.0 × 10⁻³
k = 0.20 dm⁶ mol⁻² s⁻¹
The shape of a rate–concentration graph depends on the order:
The rate constant k varies with temperature. Its units depend on the overall order of the reaction:
| Overall order | Units of k |
|---|---|
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | dm³ mol⁻¹ s⁻¹ |
| 3 | dm⁶ mol⁻² s⁻¹ |
Exam Tip: To work out the units of k, substitute the units of rate (mol dm⁻³ s⁻¹) and concentration (mol dm⁻³) into the rate equation and cancel. This is a common exam question and easy marks if you show your working clearly.
Key Definition: The half-life (t½) of a reaction is the time taken for the concentration of a reactant to fall to half its initial value.
For a first-order reaction, the half-life is constant and is independent of the initial concentration:
t½ = ln 2 / k = 0.693 / k
This equation can be rearranged to find k from a measured half-life, or to find the half-life from a known k value.
For zero-order reactions, t½ = [A]₀ / 2k (half-life depends on initial concentration — it gets shorter as concentration decreases).
For second-order reactions, t½ = 1 / (k[A]₀) (half-life also depends on initial concentration — it gets longer as concentration decreases).
The rate-determining step (RDS) is the slowest step in a multi-step reaction mechanism. The rate equation gives information about the rate-determining step:
For example, if rate = k[A][B], then one molecule of A and one molecule of B are involved in the rate-determining step.
If rate = k[A]² (and B is a reactant), then the rate-determining step involves two molecules of A and no molecules of B. B must react in a subsequent fast step.
The Arrhenius equation quantifies how the rate constant k varies with temperature:
k = Ae^(−Eₐ/RT)
where:
Taking natural logarithms of both sides gives:
ln k = ln A − Eₐ/(RT)
This is in the form y = mx + c, so a graph of ln k (y-axis) against 1/T (x-axis) gives a straight line with gradient −Eₐ/R and y-intercept ln A.
This allows the activation energy to be determined graphically from experimental data.
The rate constant for a reaction is 1.24 × 10⁻³ s⁻¹ at 300 K and 5.48 × 10⁻² s⁻¹ at 350 K. Calculate the activation energy.
Solution:
Using the two-point form of the Arrhenius equation:
ln(k₂/k₁) = (Eₐ/R) × (1/T₁ − 1/T₂)
ln(5.48 × 10⁻² / 1.24 × 10⁻³) = (Eₐ / 8.314) × (1/300 − 1/350)
ln(44.19) = (Eₐ / 8.314) × (0.003333 − 0.002857)
3.789 = (Eₐ / 8.314) × (4.762 × 10⁻⁴)
Eₐ / 8.314 = 3.789 / 4.762 × 10⁻⁴
Eₐ / 8.314 = 7957
Eₐ = 7957 × 8.314 = 66,100 J mol⁻¹ = 66.1 kJ mol⁻¹
Exam Tip: When using the Arrhenius equation, always check your units. Eₐ must be in J mol⁻¹ (not kJ mol⁻¹) when using R = 8.314 J K⁻¹ mol⁻¹, and T must be in kelvin. A common error is to forget to convert kJ to J. When using the two-temperature form, be careful with the order of subtraction for 1/T₁ − 1/T₂.
A catalyst provides an alternative reaction pathway with a lower activation energy. This increases the rate constant k (since Eₐ appears in the Arrhenius equation as e^(−Eₐ/RT) — a lower Eₐ gives a larger k). More molecules have sufficient energy to react at a given temperature. Crucially, a catalyst does not change the position of equilibrium, the value of Kc/Kp, or the enthalpy change — it only affects the rate, causing equilibrium to be reached more quickly.
Homogeneous catalysts are in the same phase as the reactants (e.g. H⁺ ions catalysing ester hydrolysis). They typically work by forming an intermediate with one reactant, then reacting with the second reactant to regenerate the catalyst.
graph TD
A["Reactant A + Catalyst"] --> B["Intermediate<br/>(Catalyst-A complex)"]
B --> C["Intermediate + Reactant B"]
C --> D["Products + Catalyst<br/>(regenerated)"]
D -->|"Catalyst recycled"| A
Heterogeneous catalysts are in a different phase from the reactants (e.g. iron in the Haber process, vanadium(V) oxide in the Contact process). They work by adsorbing reactant molecules onto their surface, weakening bonds and bringing reactants into close proximity with the correct orientation.
graph LR
A["Reactant molecules<br/>approach surface"] --> B["Adsorption<br/>onto catalyst<br/>surface"]
B --> C["Bonds weakened<br/>Reactants in correct<br/>orientation"]
C --> D["Reaction occurs<br/>on surface"]
D --> E["Products<br/>desorb from<br/>surface"]