Reaction Kinetics & Mechanisms
CollegeRate laws, order, and the effect of temperature
Chemical kinetics quantifies rate mathematically. For a reaction aA + bB → products, the rate law — determined experimentally, never simply read off the balanced equation — takes the form:
The overall order is m + n. Integrated rate laws let concentration be predicted over time; for a first-order reaction, [A] = [A]₀e−kt, giving a constant half-life t½ = ln2/k independent of concentration — the signature of first-order kinetics (e.g. radioactive decay).
Temperature dependence of k is captured by the Arrhenius equation:
Taking logarithms gives a linear form, ln k = −(Ea/R)(1/T) + ln A, so plotting ln k against 1/T yields a straight line of slope −Ea/R — the standard method for measuring activation energy experimentally.
Mechanisms: the molecular story behind a rate law
A balanced overall equation hides the true molecular sequence of steps — the mechanism — by which reactants actually become products. Most reactions proceed through a series of elementary steps, each a single molecular event whose rate law can be written directly from its stoichiometry (unlike the overall reaction). The slowest elementary step is the rate-determining step, and it alone controls the observed overall rate law — exactly like a multi-lane highway narrowing to one lane: the whole flow rate is set by the bottleneck, not the wider stretches before or after it.
This is why measured rate laws often don't match the overall balanced equation's coefficients: if the rate-determining step involves only one molecule of a reactant that appears with a coefficient of 2 in the overall equation, the rate law will still show only first-order dependence on that reactant. Proposed mechanisms must satisfy two tests: their elementary steps must sum to the overall balanced equation, and the rate law predicted by the rate-determining step must match the experimentally measured rate law.
You can never derive a rate law from a balanced equation alone. Reaction order must be determined by experiment (e.g. the method of initial rates), because it reflects the mechanism, not the stoichiometry.
Worked example — determining rate law by the method of initial rates
For A + B → C, three experiments give: (1) [A]=0.10, [B]=0.10, rate=2.0×10⁻³; (2) [A]=0.20, [B]=0.10, rate=8.0×10⁻³; (3) [A]=0.20, [B]=0.20, rate=1.6×10⁻².
- Compare experiments 1 and 2 (B constant, A doubles): rate goes from 2.0×10⁻³ to 8.0×10⁻³, a 4-fold increase. Since 2m = 4, m = 2 (second order in A).
- Compare experiments 2 and 3 (A constant, B doubles): rate goes from 8.0×10⁻³ to 1.6×10⁻², a 2-fold increase. Since 2n = 2, n = 1 (first order in B).
- Write the rate law: rate = k[A]²[B], overall order = 2 + 1 = 3 (third order).
- Solve for k using experiment 1: 2.0×10⁻³ = k(0.10)²(0.10) = k(1.0×10⁻³), so k = 2.0 M⁻² s⁻¹.
- A plausible mechanism must have a rate-determining step consistent with this — e.g. a slow step involving two molecules of A and one of B, or two fast pre-equilibrium steps feeding a slow termolecular-equivalent step.
A first-order reaction has k = 0.0231 s⁻¹. Find its half-life.
t½ = ln2 / k = 0.693 / 0.0231 = 30.0 s. This half-life is the same regardless of the starting concentration — a defining feature of first-order kinetics.
Why does raising temperature by only 10 °C often roughly double a reaction's rate?
The Arrhenius equation shows k depends exponentially on −Ea/RT. A modest increase in T produces a large increase in the fraction of molecules with kinetic energy exceeding Ea (from the Boltzmann distribution's exponential tail), so even a small temperature rise can substantially — often roughly double — the rate for typical activation energies (~50 kJ mol⁻¹).
If a proposed two-step mechanism's slow step is A + A → C, what rate law does it predict, and is that consistent with an overall reaction 2A → C observed to be first order in A?
An elementary step's rate law follows directly from its stoichiometry: A + A → C predicts rate = k[A]², i.e. second order in A. This is not consistent with an experimentally observed first-order dependence, so this mechanism must be rejected — the rate-determining step must instead involve only one A (possibly after a fast pre-equilibrium).
Why is a catalyst's effect on rate explained kinetically rather than thermodynamically?
A catalyst provides an alternative mechanism with a lower activation energy Ea, increasing k (via the Arrhenius equation) and hence the rate — a purely kinetic effect. It does not alter ΔG, ΔH, or the equilibrium constant K, all thermodynamic quantities depending only on initial and final states, which the catalyst does not change.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1Doubling [A] doubles the rate; doubling [B] leaves it unchanged. Find the rate law.
- Rate proportional to [A]^1 since doubling A doubles rate (order 1 in A).
- Rate independent of [B], so order 0 in B.
- Combine the orders into the rate law.
Example 2For a first-order reaction, what happens to rate if [A] is tripled?
- For first order, rate = k [A].
- Tripling [A] multiplies the rate by 3^1.
- So the rate triples.
Now you try
Work each one out first, then tap to reveal the worked answer.