Dissociative and Competitive Adsorption

Introduction

The basic Langmuir model treats one gas molecule as occupying one site. Many catalytic surfaces are more complicated: H₂ may split into two adsorbed H atoms, while CO and O₂ can compete for sites in a reacting mixture. Modifying the site balance changes the pressure dependence in predictable ways. Deriving these forms is more reliable than memorising a long list of isotherms, and it makes clear which new chemical assumption each formula contains.

Core explanation

For ideal dissociative adsorption, write A₂(g) + 2 ⇌ 2A . Suppose each A atom occupies one identical site and neighbouring vacant sites are available independently. Let θ be the fraction covered by A atoms. An equilibrium relation has K = θ²/[P A₂(1−θ)²], so θ/(1−θ) = √(KP A₂) and θ = √(KP A₂)/[1 + √(KP A₂)] . At low pressure, coverage grows approximately with the square root of pressure, not linearly. This square-root result depends on the two-site stoichiometry and independent-site approximation; real dissociation can require adjacent vacancies whose probability is not simply (1−θ)².

For two molecular gases A and B each using one shared site, let θ A, θ B and θ be their fractions. The site balance is θ + θ A + θ B = 1. Ideal equilibria give θ A = K A P A θ and θ B = K B P B θ . Substitution yields θ = 1/(1 + K A P A + K B P B) , θ A = K A P A/(1 + K A P A + K B P B) and the analogous expression for B. Raising P B lowers θ A at fixed P A because both species compete for the same vacancies. A high affinity can partly offset a lower partial pressure.

The formulas assume common sites, no lateral interactions, equilibrium, and compatible one-site stoichiometry for the mixture. A real porous sorbent may have different sites for different gases, size exclusion, adsorbate interactions, pore filling or chemically reactive surfaces. Pure-component fitted K values may therefore predict mixture composition poorly. Direct mixture-adsorption measurements are valuable, especially for gas separation design.

Step-by-step reasoning

Write the balanced adsorption step before selecting a formula. Count sites consumed per incoming molecule. For dissociation, express the equilibrium quotient and take a square root only after checking both product and vacancy powers. For mixtures, define one vacancy fraction shared by all competitors, write θ i = K iP iθ for each one-site component, then solve the site balance. Verify that all coverages are between zero and one and sum to one.

Visual explanation

Draw a pair of neighbouring empty boxes and an A₂ molecule above them. An arrow leads to one A atom in each box. Below, plot θ against P and compare a square-root rise with the ordinary molecular Langmuir curve. In a second sketch, color a row of sites as vacant, A-covered or B-covered; increasing B partial pressure replaces some A and some vacancies rather than creating new sites.

Real-world analogy

A standard parking space holds one car, while a long vehicle might need two adjacent spaces. Its chance of finding space then depends on pairs of vacancies, not just the total number of empty spaces. Two kinds of cars also compete for the same lot. This analogy captures site counting, although molecules can move and interact in ways a parking model omits.

Real-world example

H₂ dissociation on a metal catalyst produces surface H atoms used in hydrogenation. The ideal square-root isotherm is a useful starting point for predicting how surface H population changes with H₂ pressure. In gas-separation materials, CO₂ and CH₄ can compete for adsorption sites. Measured binary mixtures on activated carbon show competitive uptake, but realistic models may need pore and interaction effects beyond the simplest shared-site expression.

Why?

Why does dissociative adsorption have a square root? One gas molecule produces two adsorbed fragments and consumes two sites. The equilibrium quotient therefore contains squared occupied and vacant fractions. Taking the square root to solve for θ produces √P. Why does a competitor appear in the denominator? Its occupancy reduces the common vacancy fraction that all species require.

Common misconception

"More total pressure always increases coverage of every component" is false in a mixture. If total pressure rises because only a competitor's partial pressure rises, a target component can lose sites. Another mistake is to use the molecular Langmuir form for a molecule that dissociates into two adsorbed fragments without re-deriving the site balance.

Worked example

Question: On a shared-site surface, K A = 2.0 bar⁻¹, P A = 0.50 bar, K B = 4.0 bar⁻¹ and P B = 0.25 bar. Calculate θ A, θ B and θ .

Reasoning: K AP A = 1.0 and K BP B = 1.0. The denominator is 1 + 1 + 1 = 3. Therefore θ A = 1/3, θ B = 1/3 and θ = 1/3. Notice that B has twice A's K but half its pressure, so the two products and coverages match. The fractions sum to one, checking the calculation.

Answer: Each component occupies one-third of sites and one-third remain vacant in this ideal model.

Quick check

1. At low H₂ pressure, what pressure dependence does the ideal two-site dissociative adsorption model predict for H-atom coverage? Answer: Coverage is approximately proportional to the square root of H₂ pressure.

Exam focus

Start from the adsorption stoichiometry and site balance. Keep partial pressures, not total pressure, in the competitive equations. Use K iP i as dimensionless terms and verify that coverages sum to one. State assumptions whenever applying pure-gas parameters to a mixture.

Advanced insight

The square-root model can fail when adsorption requires a specific adjacent pair, atoms interact laterally, or the surface reconstructs. Competitive adsorption can be transiently different from its equilibrium composition because species have different arrival and departure rates. A fast weak binder may occupy sites early and then be displaced by a slower strong binder; equilibrium equations alone cannot describe that time path.

Summary

Dissociation changes the site stoichiometry and gives a square-root pressure dependence in the simplest two-site model. Two one-site gases competing on the same ideal surface share a denominator of 1 + ΣK iP i. Both models expose how vacancies control uptake, while real surfaces may require mixture measurements and treatment of interactions or site heterogeneity.

Practice questions

1. How many sites does one A₂ molecule consume after dissociating into two singly bound A atoms? Answer: Two sites. 2. If K BP B rises while K AP A stays fixed, what happens to θ A in the ideal shared-site model? Answer: It decreases because the common denominator grows. 3. What is θ A if K AP A = 2 and K BP B = 1? Answer: θ A = 2/(1+2+1) = 0.50. 4. Why might pure-gas Langmuir fits fail to predict a real mixture? Answer: Gases can experience different pores, nonideal interactions, size exclusion or site heterogeneity not represented by shared identical sites.

Primary studies: binary-gas adsorption on activated carbon and kinetic competition model.