Kinetics of Surface-Catalysed Reactions

Introduction

A solid catalyst changes reaction pathways by providing sites where molecules can bind, rearrange and leave as products. The measured rate depends not just on gas concentrations but also on how many sites are occupied by each reactant and whether the surface reaction or adsorption is slow. Two common limiting pictures are Langmuir-Hinshelwood, where both partners are adsorbed before reaction, and Eley-Rideal, where a gas-phase partner reacts directly with one adsorbed partner. Their differing rate forms help organise experiments, but a fitted curve alone rarely proves one microscopic mechanism uniquely.

Core explanation

Write for an empty catalytic site, A for adsorbed A and B for adsorbed B. In a simple Langmuir-Hinshelwood (LH) scheme, A(g)+ ⇌ A , B(g)+ ⇌ B , then A +B → product plus regenerated sites. The defining idea is that the reacting pair is adsorbed; the IUPAC LH definition states this directly. In an Eley-Rideal (ER), also called Langmuir-Rideal, scheme, A(g)+ ⇌ A , then B(g)+A → product+ . The IUPAC Langmuir-Rideal definition specifies one gas-phase and one adsorbed reactant. These names classify the location of partners at the product-forming encounter, not the identity of a particular industrial catalyst.

For a simple competitive adsorption model, let θ be the fraction of empty sites and θ A, θ B the fractions occupied by A and B. If A and B occupy the same type of single site and their adsorption is near equilibrium, θ A = K Ap Aθ and θ B = K Bp Bθ , where p i is partial pressure and K i has reciprocal-pressure units. Site conservation gives θ +θ A+θ B = 1, so θ = 1/(1+K Ap A+K Bp B). Thus θ A = K Ap A/(1+K Ap A+K Bp B), and likewise for B.

If neighboring A and B react in a rate-limiting step and pair availability is approximated by θ Aθ B, a schematic LH rate per suitable site ensemble is r = k sθ Aθ B = k sK AK Bp Ap B/(1+K Ap A+K Bp B)². The squared denominator comes from two coverage factors. At low pressures, both coverages are small and r is approximately proportional to p Ap B. At high p A with fixed p B, A can occupy nearly all sites while B coverage falls, so the rate can stop rising and even decrease. That apparent inhibition arises from competition for sites, not necessarily a poisonous chemical reaction.

For a simple ER case where B does not appreciably adsorb and A adsorption follows θ A = K Ap A/(1+K Ap A), write r = k ERp Bθ A = k ERK Ap Ap B/(1+K Ap A). This form is linear in p B under the assumed gas-phase encounter law and saturates in p A as A coverage approaches one. Real ER reactions can have additional collision-energy, orientation or product-inhibition factors, and a gas-phase species may weakly adsorb even if the dominant route is ER.

These formulas require assumptions. A real surface can have different site types, neighboring sites may not be statistically independent, lateral interactions can change adsorption energies with coverage, and adsorbed species may diffuse. Products may bind strongly and block sites. A slow adsorption or desorption step changes the rate law. The ACS study of pair-site assumptions in LH kinetics examines how the common coverage-product form can hide geometric requirements for adjacent sites. A primary kinetic analysis of an ammonia-synthesis catalyst illustrates why the rate-controlling surface step can change with catalyst design.

Surface rates are often reported per catalyst mass, area or number of active sites, not simply per solution volume. State the normalisation before comparing k values. A high measured order in gas A at one pressure does not imply an elementary gas-phase collision with that molecularity; coverage functions can produce changing apparent orders. Likewise, a fitted denominator does not prove every adsorbed intermediate in a proposed mechanism. Spectroscopy, isotope labeling, pressure dependence and site-density measurements can constrain the mechanism more strongly.

The catalyst is regenerated after a full turnover, but individual sites can spend most of their time occupied. Strong adsorption is not always beneficial: if a product or one reactant cannot desorb or leaves too few neighboring sites for the other reactant, turnover slows. This connects kinetics to the common catalyst-design tension between binding reactants sufficiently and releasing products readily. The useful question is which elementary step and coverage pattern controls rate under the measured conditions.

Step-by-step reasoning

1. Identify what occupies each site and write a site-balance equation. 2. Decide which adsorption steps may be treated as quasi-equilibrated and define K i with pressure or activity units. 3. Solve for θ and the required reactant coverages. 4. Write the proposed product-forming surface step and substitute the coverages. 5. Examine low- and high-pressure limits and possible site blocking. 6. Compare the predicted orders with data and seek independent evidence for adsorbed intermediates and active-site identity.

Visual explanation

Draw a row of identical surface squares. In the LH panel, show A and B each binding to a square, moving or occupying neighboring positions, reacting and releasing product while sites reopen. In the ER panel, show A on a square and B arriving from the gas directly at that occupied square. Beside the panels draw θ A and θ B versus increasing p A at fixed p B: θ A rises, θ B falls in the competitive model. The paired curves explain why an LH rate can show a maximum rather than increase forever.

Real-world analogy

Imagine a workbench with a limited number of stations. If two parts must both be placed at adjacent stations, filling every station with part A leaves no room for part B even though A is abundant. That resembles LH site blocking. If part B can arrive from outside and react with A already on a station, the ER picture is more apt. The analogy only illustrates site occupancy; molecular adsorption and surface reaction are not human assembly operations.

Real-world example

An investigator varies gas partial pressures over a solid catalyst while measuring initial product formation and surface spectra. At low p A and p B, rate rises with both. At high p A, the rate declines while spectroscopy shows A-rich coverage and little adsorbed B. A competitive LH model could explain that pattern. A plausible ER model might predict a different dependence on p B. The team tests both against coverage data and product inhibition before claiming which elementary route dominates.

Why?

Why can adding more of a reactant slow a catalytic surface reaction? In the competitive LH model, extra A increases θ A but decreases the fraction of free sites available for B adsorption. If product formation requires neighboring A and B , the product θ Aθ B can fall once A occupies most sites. The decrease is a coverage effect; the catalyst has not necessarily been permanently deactivated.

Common misconception

“A catalyst surface has unlimited sites, so higher reactant pressure always raises rate.” Site conservation imposes finite capacity and can produce saturation or inhibition. Another mistake is equating a fitted LH-shaped law with unique proof that both reactants adsorb. Different mechanisms can yield similar apparent orders over a narrow range, especially if adsorption and reaction steps have comparable rates. Normalise surface rates consistently and seek direct adsorption evidence.

Worked example

At one condition, suppose K Ap A = 1.0 and K Bp B = 1.0 under the competitive equal-site model. The denominator is 1+1+1 = 3, so θ = θ A = θ B = 1/3. If the ideal LH surface-step coefficient is k s = 9.0 turnovers per suitable site ensemble per second, the simple rate factor is r = k sθ Aθ B = 9.0(1/3)(1/3) = 1.0 in the same normalised turnover-rate units. If K Ap A rises to 9.0 while K Bp B stays 1.0, θ A = 9/11 and θ B = 1/11; r = 9.0 × 9/121 ≈ 0.67. More A has lowered the modeled rate by crowding B off the surface.

Quick check

1. Which mechanism has both reacting partners adsorbed at the product-forming step? Answer: Langmuir-Hinshelwood, in which both adsorbed species react on neighbouring surface sites.

Exam focus

Define LH and ER by adsorption status of both reactants. Write a site balance before deriving any denominator. For competitive ideal adsorption, use θ i = K ip i/(1+ΣK jp j). Explain why the LH rate can fall at high pressure of one reactant and why the ER toy form has a different denominator. State the assumptions of identical sites, near-equilibrium adsorption and a selected rate-controlling surface step.

Advanced insight

Microkinetic models solve coupled balances for many adsorbed intermediates and can identify a degree of rate control for each elementary step instead of assuming one fixed rate-determining step. Spatial correlations may matter because two adsorbates must occupy neighboring sites; a mean-field product θ Aθ B neglects those correlations. Surface restructuring or coverage-dependent adsorption energies can change apparent orders as conditions vary. In such cases, kinetic Monte Carlo or operando spectroscopy can add information that a single algebraic rate law cannot supply.

Summary

Surface-catalyzed rate laws depend on adsorption and finite site coverage. Langmuir-Hinshelwood reactions involve two adsorbed reactants; Eley-Rideal reactions involve one adsorbed reactant and one arriving from the gas. Simple competitive Langmuir coverages generate denominators that explain saturation and site-blocking inhibition. These are mechanistic models to test with kinetics and surface evidence, not automatic labels for any fitted curve.

Practice questions

1. For competitive adsorption with K Ap A = 3 and K Bp B = 1, find θ A and θ B. Answer: Denominator is 5, so θ A = 3/5 and θ B = 1/5.

2. Why may an LH rate decrease when p A becomes very large at fixed p B? Answer: A occupies most sites and suppresses B coverage, reducing the number of reactive A –B pairs.

3. What gas-pressure dependence does the simple ER model predict for nonadsorbing B at fixed θ A? Answer: Rate is proportional to p B under the stated direct gas–surface encounter assumption.

4. Give one observation beyond rate-versus-pressure fitting that can support an adsorbed-intermediate mechanism. Answer: Operando spectroscopy showing A or B coverage, isotopic labeling, or measured site-density dependence can provide additional evidence.

5. In the simple competitive model, what is θ when K Ap A = 2 and K Bp B = 1? Answer: θ = 1/(1+2+1) = 1/4.