Chemistry Labs

Physical chemistry

Complex reaction mechanisms, catalysis

Building rate laws from elementary steps, steady-state and rate-determining-step approximations, chain reactions, and how catalysts open alternative pathways.

1 Lower secondary2 Upper secondary3 Undergraduate4 Advanced5 Research

IntuitionIntuition: many steps hidden under one equation

A balanced equation is a receipt, not a recipe. Most reactions run as sequences of elementary encounters — some fast, one slow enough to throttle everything — and catalysts work by supplying a different, cheaper route through the same net transformation. Watching the intermediates come and go, and how the rate depends on each species, is how the hidden mechanism is reconstructed.

Switch between a catalytic cycle, the Lindemann–Hinshelwood scheme and the H₂ + Br₂ radical chain; the labelled arrows are elementary steps.

SchoolSchool: elementary steps and the rate-determining step

Definition: Elementary step and mechanism

An elementary step is a single microscopic event — one collision, one bond rearrangement — whose rate law follows directly from its molecularity. A mechanism is a set of elementary steps whose sum reproduces the overall equation. The slowest necessary step (rate-determining) sets the overall rate and imposes the observed orders.

Three standard tools turn a mechanism into a prediction. The rate-determining-step picture assumes all earlier steps are fast equilibria. The steady-state approximation treats a low-concentration intermediate as roughly constant: its formation and consumption balance. Chain mechanisms add carriers that regenerate themselves, so tiny amounts of initiator sustain many cycles; termination steps eventually destroy the carriers.

A→MkX1,  kX−1AX∗→kX2Pv=k1k2[A]k−1[M]+k2(Lindemann)\ce{A ->[k_1,\ k_{-1}][M] A^{\ast} ->[k_2] P}\qquad v=\frac{k_1 k_2[\ce{A}]}{k_{-1}[\ce{M}]+k_2}\quad(\text{Lindemann})

In the Lindemann–Hinshelwood picture of gas-phase unimolecular reactions, a molecule is first energised by collision (A + M ⇌ A + M), then the excited A either reacts (k₂) or loses its energy (k₋₁M). The steady-state rate law above explains the experimental puzzle: at high pressure the reaction is first order (every excited molecule survives to react), while at low pressure it becomes second order (activating collisions are the bottleneck).

Example: Rate-determining step arithmetic

The mechanism for 2NO₂Cl → 2NO₂ + Cl₂ is proposed as (1) NO₂Cl ⇌ NO₂ + Cl (fast equilibrium, K₁) and (2) Cl + NO₂Cl → NO₂ + Cl₂ (slow, k₂). Predict the rate law.

Solution

The slow step gives v = k₂[Cl][NO₂Cl]. From the fast equilibrium [Cl] = K₁[NO₂Cl]/[NO₂], so v = k₂K₁[NO₂Cl]²/[NO₂]: second order in substrate, order −1 in the product NO₂ — an unusual rate law that only a multistep mechanism can produce, and indeed the one observed.

UndergraduateUniversity: catalysis lowers the effective barrier

A catalyst accelerates forward and reverse reactions equally by opening a path of lower activation free energy; it is regenerated each cycle, so it does not appear in the stoichiometry and cannot change K. Homogeneous catalysts live in the same phase (enzymes, organometallic complexes): the catalytic cycle shown in the simulation alternates substrate binding, transformation and product release. Heterogeneous catalysts act on a surface; autocatalysis makes the product itself a catalyst, producing sigmoid conversion curves.

v=kcat[E][S]KM+[S](Michaelis–Menten)v=\frac{k_{\mathrm{cat}}[\ce{E}][\ce{S}]}{K_M+[\ce{S}]}\qquad(\text{Michaelis--Menten})

Enzyme kinetics is the most familiar steady-state mechanism: E + S ⇌ ES → E + P. At low [S] the rate is proportional to substrate and enzyme; at saturation every enzyme is occupied and the rate plateaus at v_max = k_cat[E] — a zeroth-order regime in substrate. The turnover number k_cat can reach 10⁵–10⁷ s⁻¹ for carbonic anhydrase or catalase.

Example: Effective barrier from rate ratios

A catalyst speeds a reaction by a factor 10⁶ at 300 K, with equal pre-exponential factors. By how much does it lower the activation free energy?

Solution

k_cat/k = exp(ΔΔG‡/RT) = 10⁶, so ΔΔG‡ = RT ln 10⁶ = 8.314 × 300 × 13.8 ≈ 34.5 kJ mol⁻¹. A “mere” 35 kJ mol⁻¹ of barrier lowering buys a million-fold rate — the arithmetic behind all catalysis.

AdvancedAdvanced: selectivity, oscillations and network thinking

Catalysis is about selectivity as much as speed: the best catalyst lowers the barrier to the desired product more than to side products, which is why Sabatier’s principle (intermediates bound neither too weakly nor too strongly) organises catalyst design. Kinetic networks can also misbehave beautifully: with autocatalysis and feedback, reactions like Belousov–Zhabotinsky oscillate or form propagating fronts; near a steady state, small parameter changes can flip the dominant pathway entirely.

ResearchResearch frontier

References

  • Chemical Kinetics · K. J. Laidler, 1987
  • Discussion on “The Radiation Theory of Chemical Action” · F. A. Lindemann et al., 1922
  • Fundamental Concepts in Heterogeneous Catalysis · J. K. Nørskov, F. Studt, F. Abild-Pedersen, T. Bligaard, 2014