Deriving the Langmuir Isotherm

Introduction

Gas uptake on a solid often rises sharply at low pressure and then approaches a plateau. The Langmuir isotherm gives a compact model for this shape. Its value lies as much in its transparent assumptions as in its equation: every adsorption site is treated as equivalent, one molecule occupies one site, and adsorbed molecules do not change their neighbours' binding energies. This page derives the relation from both a rate balance and a site equilibrium, then shows what a fitted plateau does and does not prove.

Core explanation

Let θ be the fraction of sites occupied by gas molecules A, so the vacant fraction is 1 − θ. In the simplest kinetic model, adsorption rate per site is k a P(1 − θ), and desorption rate per site is k d θ. At equilibrium they are equal: k a P(1 − θ) = k d θ. Rearranging gives θ = KP/(1 + KP) with K = k a/k d. If P is expressed in bar, this version of K has units bar⁻¹; the product KP must be dimensionless. The loading is q = q max θ, hence q = q max KP/(1 + KP) .

The same result can be expressed as the site reaction A(g) + ⇌ A , where is vacant and A occupied. Under an ideal independent-site model, the ratio θ/(1 − θ) is proportional to gas activity or pressure. Solving for θ recovers the same functional form. A thermodynamic equilibrium constant should itself be made dimensionless by using activity such as P/P°, even though the practical fitted K in a pressure form commonly carries inverse-pressure units. Keeping those conventions distinct avoids impossible units inside a logarithm.

At low pressure, KP ≪ 1, so θ ≈ KP and q ≈ q max KP: uptake is nearly linear. At high pressure, KP ≫ 1, so θ approaches one and q approaches q max. At P = 1/K, θ = 0.5, providing a convenient interpretation of the fitted affinity. A plateau may indicate site saturation within this model, but real materials can also show pore filling, multilayer adsorption or a limited measured pressure range. Fitting a Langmuir curve does not prove that the microscopic surface is perfectly uniform.

The model assumes a single adsorbate on identical sites, no adsorbate–adsorbate interactions, one molecule per site and reversible equilibrium. It may work as a useful approximation despite imperfect assumptions. Chemisorption with dissociation, competitive mixtures, and heterogeneous surfaces require modified expressions. Slow uptake also means a measured short-time curve may not be an equilibrium isotherm at all.

Step-by-step reasoning

Define θ and the vacant fraction. Write the adsorption term proportional to gas pressure and vacancy, then the desorption term proportional to occupancy. Set them equal only at equilibrium and solve algebraically. Replace θ by q/q max if the instrument reports amount adsorbed. Check limits P → 0 and P → ∞ and locate half coverage at P = 1/K. Finally compare the assumptions with what is known about the actual adsorbent.

Visual explanation

Draw a row of identical empty boxes for surface sites. At low pressure only a few contain A; at high pressure nearly all are filled. Plot θ against P as a rising curve that flattens below one, and mark the point P = 1/K, θ = 0.5. Beside the plot, draw opposing arrows A(g)+ ⇌ A and label them k aP(1−θ) and k dθ.

Real-world analogy

Imagine identical numbered seats in a small theatre. People arrive more often when the lobby is crowded, but only empty seats can be occupied. Seated people leave at a rate proportional to how many are seated. At a steady crowd, arrivals to empty seats balance departures. The analogy clarifies the algebra while hiding the molecular details of adsorption energy and transport.

Real-world example

An activated surface exposed to a gas may show uptake that approaches a finite loading. Researchers can fit q max and K to compare materials or estimate a useful pressure range. If the same sample has multiple pore environments, an excellent fit over one pressure window may still yield an effective q max rather than a literal count of identical atomic sites. Measurements at several temperatures and additional structural evidence can test the interpretation.

Why?

Why does the rate of adsorption include 1 − θ? Occupied sites cannot accept another molecule in the one-layer model. As coverage rises, fewer vacancies remain, naturally slowing further adsorption. Why does desorption include θ? Only occupied sites have a molecule to leave. The plateau follows from site exclusion, not from a universal maximum amount that every surface must have.

Common misconception

"K is always dimensionless" is unsafe when the equation is written with pressure in units. A fitted pressure-form K has inverse-pressure units; a dimensionless thermodynamic constant uses a standard-state activity. Another error is to claim the plateau proves Langmuir assumptions. A fit describes data over its measured domain but cannot uniquely identify microscopic structure.

Worked example

Question: For a one-site gas model, K = 2.0 bar⁻¹ and q max = 0.80 mmol g⁻¹. Calculate θ and q at P = 0.50 bar.

Reasoning: KP = (2.0 bar⁻¹)(0.50 bar) = 1.0, so θ = 1/(1 + 1) = 0.50. Then q = 0.80 × 0.50 = 0.40 mmol g⁻¹. Since P = 1/K, half coverage is expected. At much higher pressure the model would approach 0.80 mmol g⁻¹ without exceeding it.

Answer: θ = 0.50 and q = 0.40 mmol g⁻¹.

Quick check

1. Why does the simplest Langmuir adsorption rate decrease as coverage approaches one? Answer: The fraction of vacant sites, 1 − θ, decreases, so fewer sites can accept new molecules.

Exam focus

Derive the equation rather than recalling it without assumptions. Show the rate balance, define K and check its units, then write q = q maxθ. Use the low- and high-pressure limits to catch algebra errors. Mention that a real adsorption plateau can have other causes and that fitted parameters depend on the selected model and pressure range.

Advanced insight

The ideal site model resembles a two-state statistical system: each independent site is empty or occupied, and the relative statistical weight is KP. Normalising those weights gives 1/(1+KP) for vacancy and KP/(1+KP) for occupancy. This viewpoint clarifies why interactions between neighbours break the simple denominator and why a distribution of site energies produces a broader, more complicated isotherm.

Summary

Balancing adsorption to vacant sites against desorption from occupied sites gives θ = KP/(1+KP). Multiplying by monolayer capacity gives q. Half coverage occurs at P = 1/K, and saturation is approached at high pressure. The equation assumes equivalent independent one-molecule sites at equilibrium; a good numerical fit does not by itself establish those microscopic assumptions.

Practice questions

1. What is θ when KP = 3? Answer: θ = 3/(1+3) = 0.75. 2. What is the low-pressure approximation for loading? Answer: q ≈ q max KP when KP is much smaller than one. 3. What pressure gives half coverage if K = 0.40 kPa⁻¹? Answer: P = 1/K = 2.5 kPa. 4. Why might a measured curve deviate from the Langmuir form? Answer: Sites may differ in energy, adsorbates may interact, adsorption may be multilayered, or equilibrium may not have been reached.

Primary definitions and measured-isotherm context: IUPAC adsorption terminology and gas-adsorption calorimetry and isotherms.