Langmuir–Hinshelwood and Eley–Rideal Mechanisms

Introduction

A catalyst brings reactants together at a surface, but the reactants need not arrive in the same way. In a Langmuir–Hinshelwood path both reactants are adsorbed before reacting. In an Eley–Rideal path a gas-phase or solution-phase reactant reacts directly with one already adsorbed. The distinction changes rate equations and can create surprising effects such as inhibition by adding more of one reactant. These names describe elementary mechanistic patterns, not a guarantee that a fitted rate law uniquely proves one pattern.

Core explanation

Consider molecular A and B competing for identical sites, each adsorbing reversibly. Under rapid adsorption equilibrium, θ A = K AP A/D and θ B = K BP B/D, with D = 1 + K AP A + K BP B. If the rate-limiting surface step requires nearby A and B , a simplest mean-field Langmuir–Hinshelwood rate is r = kθ Aθ B = kK AP AK BP B/D² . The constant k includes a site-density convention and the intrinsic surface reaction factor. The product of mean coverages assumes neighbouring pairs are statistically available; real lateral correlations can change it.

At low pressures, D ≈ 1, so r is approximately proportional to P AP B. At very high P A with P B fixed, A covers most sites and θ B falls. The rate can therefore decrease as P A rises: A blocks B even though A itself is a reactant. This is a mechanistically sensible negative apparent order, not necessarily an experimental error. Product adsorption could add another term to D and also inhibit reaction.

For a simple Eley–Rideal scheme, B is adsorbed and A reacts with B directly from gas without occupying a stable site. If B alone follows θ B = K BP B/(1+K BP B), then r = kP Aθ B = kP AK BP B/(1+K BP B) under a proportional incident A flux and rate-limiting reaction assumption. This expression stays first order in P A in its ideal form. If A also adsorbs competitively or mass transport becomes limiting, the observed dependence can differ. Moreover, a similar empirical pressure dependence can arise from more than one microscopic model.

Step-by-step reasoning

Write the elementary steps and identify whether A and B are adsorbed before the slow reaction. Define all site fractions and write their balance. Use justified equilibrium relations for fast adsorption steps, then write the slow-step rate in terms of the relevant coverages or gas pressure. Substitute and simplify only after site balance is solved. Check low- and high-pressure limits, and compare predicted reaction orders with observations while remembering that kinetic fit alone is not a complete structural proof.

Visual explanation

Draw two adjacent occupied sites A and B meeting to produce P for Langmuir–Hinshelwood. Draw a second panel with B on the surface and gas A striking it directly for Eley–Rideal. Beneath, plot an illustrative LH rate against P A at fixed P B: it rises at first and can fall when A crowds B off the surface. Label the two regions as reactant supply and site blocking.

Real-world analogy

In one workshop, both materials must be placed on two adjacent benches before assembly. If every bench is filled with material A, no space remains for B. In another workshop, B waits on a bench while A is delivered directly into the worker's hands. The analogy shows why extra A can inhibit the first scheme but not necessarily the ideal second one.

Real-world example

Surface oxidation of CO on metal catalysts is often analysed by considering CO and oxygen-derived adsorbates. CO can bind strongly enough to block sites needed for oxygen activation, causing inhibition at some conditions. An actual catalyst may involve dissociative O₂ adsorption, multiple site types or an oxide surface, so a textbook two-molecular-adsorbate equation is a simplified teaching model rather than a universal CO-oxidation law.

Why?

Why does a squared denominator appear in the simple LH expression? Both θ A and θ B contain the same vacancy-based denominator D because both compete for the same site pool. Multiplying their coverages gives D². In a simple Eley–Rideal expression, only B needs a vacancy to adsorb, so the ideal denominator appears once.

Common misconception

"A negative order in A means A is not a reactant" is false in surface chemistry. Strong A adsorption can exclude B and reduce the reaction rate. Another mistake is to declare a microscopic mechanism solely because one rate equation fits data: competing models, transport limits and surface heterogeneity must be considered.

Worked example

Question: In the simple LH model K AP A = 2 and K BP B = 1. Find θ A, θ B and r/k.

Reasoning: D = 1+2+1 = 4. Thus θ A = 2/4 = 0.50 and θ B = 1/4 = 0.25; vacant fraction is 1/4. With r = kθ Aθ B, r/k = (0.50)(0.25) = 0.125. Direct substitution into the full expression gives 2×1/4² = 0.125, which checks the result.

Answer: θ A = 0.50, θ B = 0.25 and r/k = 0.125 under the mean-field assumptions.

Quick check

1. Which mechanism has both reactants adsorbed before their surface reaction? Answer: The Langmuir–Hinshelwood mechanism.

Exam focus

Draw and label the two mechanisms before deriving rates. Distinguish partial pressures from coverages and use a complete site balance. Explain why a competitor or strongly adsorbed reactant can inhibit a reaction. State any rapid-equilibrium and rate-limiting-step assumptions; a fitted rate expression without those assumptions has limited mechanistic meaning.

Advanced insight

Mean-field kinetics replace actual neighbour-pair probabilities with products of average coverages. On surfaces with clustering, repulsion or multiple site types this can fail, even when average θ values are measured correctly. Transient isotope labelling, surface spectroscopy and first-principles-informed microkinetic models can provide additional evidence for which species meet in the rate-controlling step.

Summary

Langmuir–Hinshelwood reactions involve two surface-bound reactants; Eley–Rideal reactions involve one bound reactant and one incoming reactant. Combining ideal adsorption coverages with a slow surface step yields different denominators and pressure dependencies. Competitive occupation can create inhibition at high reactant pressure. Mechanistic assignments require more than a good rate-law fit.

Practice questions

1. What does a simple LH rate become at very low P A and P B? Answer: Approximately kK AP AK BP B, first order in each pressure under the stated model. 2. Why can raising P A lower an LH reaction rate at fixed P B? Answer: A can occupy most sites, reducing θ B and the number of A –B reactive pairs. 3. What species are present in an ideal Eley–Rideal collision? Answer: One reactant is incoming from gas or solution and the other is already adsorbed. 4. Does fitting r ∝ P Aθ B prove an Eley–Rideal pathway? Answer: No. Other rate-limiting steps or coverage effects can yield similar dependence; independent mechanistic evidence is needed.

Primary terminology and research: IUPAC Langmuir–Hinshelwood, IUPAC Langmuir–Rideal and pair-site kinetic analysis.