Reaction–Diffusion Coupling

Introduction

A reaction can only happen when reactants meet. If the chemical step is slow, reactants meet many times before they react, and the measured rate reflects the chemistry. If the chemical step is very fast, almost every encounter leads to reaction and the rate is set by how quickly molecules can diffuse together or reach a surface. Many measured rates lie between these extremes. Recognising when transport is disguising the true kinetics is essential in solution chemistry, electrochemistry and heterogeneous catalysis.

Core explanation

Two steps in series. For a bimolecular reaction in solution, A and B first diffuse together to form an encounter pair, then either react or separate:

A + B ⇌ {AB} → products

with diffusion rate constant k d, separation constant k₋d and reaction constant k r. The steady-state approximation for {AB} gives

k obs = k d k r ÷ (k₋d + k r)

If k r is much larger than k₋d, k obs ≈ k d: the reaction is diffusion-controlled . If k r is much smaller than k₋d, k obs ≈ (k d/k₋d)k r: the reaction is activation-controlled .

The diffusion limit. The Smoluchowski treatment gives k d = 4πR DN A, where R is the reaction distance and D the sum of diffusion coefficients. A useful estimate is k d ≈ 8RT/(3η), where η is the solvent viscosity. For water at 298 K (η ≈ 0.89 mPa s) this gives about 7 × 10⁹ dm³ mol⁻¹ s⁻¹. Measured rate constants near 10⁹–10¹⁰ dm³ mol⁻¹ s⁻¹ indicate diffusion control, and such rates fall when viscosity is increased.

Reactions at surfaces. At an electrode or catalyst particle, reactant must cross a diffusion layer of thickness δ. The transport flux is k m(c bulk − c surface), where the mass-transfer coefficient k m = D/δ. The surface reaction consumes reactant at k s c surface. Equating the two fluxes at steady state gives

1/k obs = 1/k m + 1/k s

These behave like resistances in series: the slower process dominates. In electrochemistry this is exactly why current approaches the limiting current at large overpotential, where k s becomes very large and transport alone controls the rate.

Porous catalysts. Inside a porous pellet, reactant is consumed as it diffuses inward. If reaction is fast, only the outer shell works, and the effectiveness factor (observed rate divided by rate with no transport limitation) falls below 1. Transport limitation also distorts the apparent activation energy, which drops to roughly half the true value because diffusion depends only weakly on temperature.

Diagnostic tests. Transport control is revealed if the rate changes with stirring speed, electrode rotation rate, particle size or solvent viscosity. Chemical control is indicated when these changes have no effect.

Formulae

Encounter model: k obs = k d k r ÷ (k₋d + k r).

Diffusion limit: k d ≈ 8RT/(3η).

Surface reaction: 1/k obs = 1/k m + 1/k s, with k m = D/δ.

Damköhler number: Da = k s/k m; Da much greater than 1 means transport control.

Step-by-step reasoning

To decide what controls an observed rate:

1. Estimate the diffusion-limited rate constant or mass-transfer coefficient. 2. Compare it with the measured rate constant. 3. If they are similar, suspect transport control. 4. Test by changing stirring, rotation rate or viscosity. 5. If needed, correct the data using 1/k s = 1/k obs − 1/k m.

Visual explanation

Draw two resistors in series: one labelled transport (1/k m) and one labelled reaction (1/k s). The total resistance sets the flow of reactant. Then sketch a concentration profile near a surface: flat in the bulk, falling linearly across the diffusion layer, reaching almost zero at the surface when reaction is fast.

Real-world analogy

A fast cashier cannot serve customers any faster than they can reach the till through a crowded shop. If the aisles are clear, the cashier's speed controls throughput; if the aisles are packed, walking time dominates, and hiring faster cashiers makes no difference.

Real-world example

The neutralisation H₃O⁺ + OH⁻ → 2H₂O has a rate constant of about 1.4 × 10¹¹ dm³ mol⁻¹ s⁻¹, one of the fastest known, because the ions attract each other and protons move by rapid hopping through the hydrogen-bond network. Catalytic converters are also designed with thin washcoats to limit internal diffusion resistance.

Why?

Why do the resistances add as reciprocals of rate constants? At steady state the flux through each step must be equal, just as current through resistors in series is the same. Each step needs its own concentration drop to drive that flux, and the drops add up.

Common misconception

"A measured activation energy always reflects the chemical step." Under transport control the measured value mostly reflects the temperature dependence of diffusion, typically 10–20 kJ mol⁻¹ in water, and says little about the chemical barrier.

Worked example

Question: At an electrode, D = 1.0 × 10⁻⁹ m² s⁻¹ and δ = 20 μm. The surface rate constant k s = 2.0 × 10⁻⁴ m s⁻¹. Find k m and k obs.

Reasoning: k m = D/δ = 1.0 × 10⁻⁹ ÷ 2.0 × 10⁻⁵ = 5.0 × 10⁻⁵ m s⁻¹. 1/k obs = 1/(5.0 × 10⁻⁵) + 1/(2.0 × 10⁻⁴) = 20 000 + 5000 = 25 000 s m⁻¹.

Answer: k obs = 4.0 × 10⁻⁵ m s⁻¹; the process is mainly transport-limited, since Da = 4.

Quick check

1. Why does increasing solvent viscosity slow a diffusion-controlled reaction but hardly affect an activation-controlled one? Answer: Viscosity lowers diffusion coefficients, which set the rate only when encounters, not the chemical step, are limiting.

Exam focus

Derive the encounter-pair expression and state its two limits. Use 1/k obs = 1/k m + 1/k s for surface processes. Quote diagnostic tests for transport control and explain why apparent activation energies fall under diffusion limitation.

Advanced insight

For ions, electrostatic attraction or repulsion changes the encounter rate; the Debye correction multiplies k d by a factor that can exceed 5 for oppositely charged ions in water. At very short times after reactants are created, the rate constant also depends on time, because nearby pairs react before a steady diffusion profile forms.

Summary

Observed rates combine transport and chemistry. The encounter model gives k obs = k dk r/(k₋d + k r), with diffusion control at about 10⁹–10¹⁰ dm³ mol⁻¹ s⁻¹ in water. At surfaces, transport and reaction resistances add in series. Stirring, rotation, particle size and viscosity tests reveal which process limits the rate.

Practice questions

1. State the approximate diffusion-limited rate constant for neutral molecules in water at 298 K. Answer: About 7 × 10⁹ dm³ mol⁻¹ s⁻¹, from k d ≈ 8RT/(3η). 2. If k m = 1.0 × 10⁻⁵ m s⁻¹ and k s = 1.0 × 10⁻³ m s⁻¹, which process controls the rate and what is k obs? Answer: Transport controls; 1/k obs = 100 000 + 1000 = 101 000 s m⁻¹, so k obs ≈ 9.9 × 10⁻⁶ m s⁻¹. 3. An observed rate rises when a rotating disc electrode is spun faster. What does this show? Answer: Mass transport contributes to rate control, because faster rotation thins the diffusion layer and raises k m. 4. Why does internal diffusion in a porous catalyst lower its effectiveness factor? Answer: Reactant is consumed near the outer surface before it can diffuse to inner sites, so interior sites contribute little.