Langmuir Adsorption Isotherm

Picture hydrogen gas striking a platinum foil. Some molecules bounce off. Others stick, dissociate, and sit in fixed pockets on the metal surface. As pressure rises, more sites fill. Eventually the surface saturates, and no amount of extra gas increases the coverage further. This saturating behavior is the experimental signature that the Langmuir isotherm was built to explain. The model matters because it is the simplest quantitative bridge between gas pressure and surface coverage. It underlies how chemists interpret catalytic rate laws, gas-storage capacities, and surface-area measurements. Every more sophisticated isotherm — Freundlich, Temkin, BET — is a correction to the assumptions Langmuir made first. This article derives the isotherm from a simple equilibrium between gas-phase molecules and empty surface sites. You will see the assumptions stated explicitly. Then you will see the algebra worked line by line. Two numerical examples follow, along with common pitfalls and a look at Irving Langmuir himself.

Key takeaways

The Langmuir isotherm models adsorption as a dynamic equilibrium between gas molecules and empty surface sites.
It assumes identical, independent sites, one molecule per site, and no interaction between adsorbed molecules.
Fractional coverage follows $\theta = Kp/(1+Kp)$ , rising linearly at low pressure and saturating at high pressure.
A linearized plot of $p/\theta$ against $p$ gives both the saturation capacity and the equilibrium constant $K$ .
Real surfaces often deviate because sites are not equivalent and adsorbates do interact — motivating Freundlich, Temkin, and BET isotherms.
Key idea: Coverage is governed by a single equilibrium constant balancing the rate of adsorption against the rate of desorption from identical, non-interacting sites.

The core insight

The Langmuir isotherm treats adsorption as a chemical equilibrium, gas molecule plus empty site in equilibrium with occupied site, so the fractional coverage $\theta$ takes the same mathematical form as any single-site binding equilibrium, saturating smoothly toward one monolayer as pressure increases.

The physical picture

Imagine a solid surface divided into a fixed number of adsorption sites, all chemically identical. Gas molecules strike the surface at a rate proportional to pressure. A colliding molecule either bounces away or sticks to an empty site. Once stuck, a molecule can also leave the surface at some fixed rate, returning to the gas phase. At equilibrium, the rate of molecules landing on empty sites exactly balances the rate of molecules leaving occupied sites. This is a two-state statistical picture: each site is either empty or filled by exactly one molecule. No molecule can stack on top of another, and no site's occupancy affects its neighbor. Physisorption (weak, van der Waals-type binding) and chemisorption (strong, bond-forming binding) both fit this framework in principle, though chemisorption more often behaves as an ideal monolayer since it forms real chemical bonds that saturate at one molecule per site. The key experimental variable is fractional coverage, $\theta$ , defined as the fraction of sites occupied, running from 0 (bare surface) to 1 (complete monolayer). As gas pressure $p$ increases, $\theta$ rises from near zero, passes through an intermediate regime, and asymptotically approaches unity. The isotherm is a plot of $\theta$ against $p$ at constant temperature, and its characteristic saturating shape is the experimental fingerprint of monolayer, single-site adsorption.

Figure

Langmuir isotherm shape: coverage θ versus pressure p, showing linear low-pressure rise and high-pressure saturation

The second useful picture is mechanistic: a checkerboard surface where sites fill randomly but each site behaves independently, with adsorption and desorption arrows balancing at equilibrium.

Figure

Microscopic equilibrium: adsorption and desorption at a single site, rate constants ka and kd

The core relations

\theta = \frac{Kp}{1+Kp}

where: $\theta$ is fractional coverage, $p$ is equilibrium gas pressure, $K$ is the Langmuir equilibrium constant (units of inverse pressure).

K = \frac{k_a}{k_d}

where: $k_a$ is the adsorption rate constant and $k_d$ is the desorption rate constant.

\frac{p}{\theta} = \frac{1}{K}+p

where: a plot of $p/\theta$ against $p$ is linear, giving intercept $1/K$ and unit slope — the standard linearization used to extract $K$ from data expressed per unit coverage.

\theta = \frac{V}{V_{\text{mon}}} = \frac{Kp}{1+Kp}

where: $V$ is the volume of gas adsorbed at pressure $p$ and $V_{\text{mon}}$ is the volume corresponding to a complete monolayer.

\frac{p}{V} = \frac{1}{K V_{\text{mon}}} + \frac{p}{V_{\text{mon}}}

where: plotting $p/V$ against $p$ gives slope $1/V_{\text{mon}}$ and intercept $1/(KV_{\text{mon}})$ , the practical experimental route to surface area.

Derivation AFrom site equilibrium to the Langmuir isotherm

Goal. Derive $\theta(p)$ for gas molecules adsorbing on a fixed number of identical, independent, single-occupancy sites.

Step 1. State the microscopic model. Consider $N$ total surface sites. Let $\theta$ be the fraction occupied, so $\theta N$ sites hold one adsorbed molecule each, and $(1-\theta)N$ sites are empty. Assume every site is energetically identical and that occupying one site does not change the energy of adsorption at a neighboring site — the "no lateral interaction" assumption.

Step 2. Write the rate of adsorption. A gas molecule strikes the surface at a rate proportional to pressure $p$ , since kinetic theory gives a collision flux proportional to $p$ at fixed temperature. Only collisions with empty sites lead to adsorption, so the rate of adsorption per unit area is

\text{rate}_{\text{ads}} = k_a\, p\,(1-\theta)

Step 3. Write the rate of desorption. Molecules leave occupied sites at a rate proportional to how many are there, with no dependence on neighboring occupancy (independent-site assumption):

\text{rate}_{\text{des}} = k_d\,\theta

Step 4. Impose dynamic equilibrium. At equilibrium the two rates are equal, since $\theta$ is no longer changing with time:

k_a\, p\,(1-\theta) = k_d\,\theta

Step 5. Solve for $\theta$ . Expand and collect terms:

k_a p - k_a p\,\theta = k_d\,\theta \quad\Longrightarrow\quad k_a p = \theta\,(k_d + k_a p)

\theta = \frac{k_a p}{k_d + k_a p}

Step 6. Introduce the equilibrium constant. Define $K = k_a/k_d$ , the ratio of adsorption to desorption rate constants, equivalent to the equilibrium constant for the reaction gas + empty site $\rightleftharpoons$ occupied site. Divide numerator and denominator by $k_d$ :

\boxed{\theta = \frac{Kp}{1+Kp}}

This single expression captures both limits correctly: at low pressure $Kp \ll 1$ so $\theta \approx Kp$ (linear, Henry's-law-like regime), and at high pressure $Kp \gg 1$ so $\theta \to 1$ (complete monolayer, saturation). The same derivation follows equivalently from equilibrium thermodynamics, writing $K = [\text{occupied}]/(p\,[\text{empty}])$ , since dynamic kinetic balance and chemical equilibrium give identical algebra for this simple mechanism.

Worked examples

Example 1 — finding K and monolayer coverage from two data points. Suppose nitrogen gas adsorbs on a catalyst sample at 77 K, and the amount adsorbed is expressed as fractional coverage. Two measurements are available: at $p_1 = 10.0\ \text{kPa}$ , $\theta_1 = 0.300$ ; at $p_2 = 40.0\ \text{kPa}$ , $\theta_2 = 0.600$ . Step 1 — set up two equations. From $\theta = Kp/(1+Kp)$ , rearrange to isolate $K$ :

K = \frac{\theta}{p(1-\theta)}

For point 1:

K_1 = \frac{0.300}{(10.0\ \text{kPa})(1-0.300)} = \frac{0.300}{10.0 \times 0.700\ \text{kPa}} = \frac{0.300}{7.00\ \text{kPa}} = 0.0429\ \text{kPa}^{-1}

For point 2:

K_2 = \frac{0.600}{(40.0\ \text{kPa})(1-0.600)} = \frac{0.600}{40.0\times0.400\ \text{kPa}} = \frac{0.600}{16.0\ \text{kPa}} = 0.0375\ \text{kPa}^{-1}

Step 2 — check consistency. The two values, 0.0429 and 0.0375 kPa⁻¹, agree to within about 13%, consistent with real experimental scatter around a genuine Langmuir constant near $K \approx 0.040\ \text{kPa}^{-1}$ . (If the values had disagreed by a factor of two or more, that would signal the Langmuir model does not fit this system.) Step 3 — predict coverage at a new pressure. Using the averaged $K = 0.0402\ \text{kPa}^{-1}$ , predict $\theta$ at $p = 100\ \text{kPa}$ :

\theta = \frac{Kp}{1+Kp} = \frac{0.0402\times100}{1+0.0402\times100} = \frac{4.02}{5.02} = 0.801

Sanity check. Coverage rose from 0.300 to 0.600 to 0.801 as pressure increased tenfold, decelerating toward saturation — exactly the expected shape, since $\theta$ must approach 1 but never exceed it.

Example 2 — extracting surface area from gas-adsorption volumes. A sample of porous carbon adsorbs nitrogen gas at 77 K. The volume adsorbed (corrected to STP) is measured at several pressures:

$p$ / kPa	$V$ / cm³ g⁻¹
5.00	12.0
10.0	18.0
20.0	24.0
40.0	28.8

Step 1 — linearize. Using

\frac{p}{V} = \frac{1}{KV_{\text{mon}}} + \frac{p}{V_{\text{mon}}}

compute $p/V$ for each row:

$p=5.00$ : $p/V = 5.00/12.0 = 0.4167\ \text{kPa g cm}^{-3}$
$p=10.0$ : $p/V = 10.0/18.0 = 0.5556$
$p=20.0$ : $p/V = 20.0/24.0 = 0.8333$
$p=40.0$ : $p/V = 40.0/28.8 = 1.3889$

Step 2 — fit a line. Using the first and last points to estimate slope (a full least-squares fit would use all four; here we illustrate the method):

\text{slope} = \frac{1.3889-0.4167}{40.0-5.00} = \frac{0.9722}{35.0} = 0.02778\ \text{g cm}^{-3}

This slope equals $1/V_{\text{mon}}$ , so

V_{\text{mon}} = \frac{1}{0.02778} = 36.0\ \text{cm}^3\,\text{g}^{-1}

Step 3 — find the intercept and $K$ . Using point $(5.00, 0.4167)$ :

\frac{1}{KV_{\text{mon}}} = 0.4167 - (0.02778)(5.00) = 0.4167 - 0.1389 = 0.2778\ \text{kPa g cm}^{-3}

K = \frac{1}{(0.2778)(36.0)} = \frac{1}{10.00} = 0.100\ \text{kPa}^{-1}

Step 4 — convert to surface area. At STP, one mole of gas occupies $22{,}414\ \text{cm}^3$ . Moles of N₂ in a full monolayer per gram:

n_{\text{mon}} = \frac{36.0\ \text{cm}^3\text{g}^{-1}}{22{,}414\ \text{cm}^3\text{mol}^{-1}} = 1.606\times10^{-3}\ \text{mol g}^{-1}

Number of molecules per gram:

N_{\text{mon}} = (1.606\times10^{-3}\ \text{mol g}^{-1})(6.022\times10^{23}\ \text{mol}^{-1}) = 9.674\times10^{20}\ \text{g}^{-1}

Taking the cross-sectional area of one N₂ molecule as $0.162\ \text{nm}^2$ (a standard reference value):

A = (9.674\times10^{20}\ \text{g}^{-1})(0.162\times10^{-18}\ \text{m}^2) = 156.7\ \text{m}^2\text{g}^{-1}

Sanity check. A specific surface area near 150–160 m² g⁻¹ is a typical, physically reasonable value for a mesoporous activated carbon — this is the essence of the BET-style surface-area measurement, of which the simple Langmuir plot is the single-layer ancestor.

Limiting cases & common traps

Regime or trap	What happens
Low pressure, $Kp \ll 1$	$\theta \approx Kp$ , a linear "Henry's law" regime for adsorption
High pressure, $Kp \gg 1$	$\theta \to 1$ , coverage saturates regardless of further pressure increase
Dissociative adsorption (e.g. H₂ → 2H)	Requires two adjacent empty sites; correct form is $\theta = \sqrt{Kp}/(1+\sqrt{Kp})$
Multilayer physisorption at high $p/p^{\circ}$	Langmuir underestimates uptake; BET isotherm is needed
Strongly heterogeneous surface sites	Real $K$ varies site to site; Freundlich or Temkin isotherms fit better
Lateral interactions (attractive or repulsive)	Violates the independent-site assumption; causes S-shaped or shifted isotherms

The chemist behind it

Irving Langmuir developed this isotherm in 1916–1918 while investigating gas adsorption on tungsten filaments for General Electric's incandescent lamp research. He recognized that treating the surface as a fixed lattice of equivalent, independent binding sites — rather than a continuum — gave a mechanistic derivation matching experimental saturation curves far better than earlier empirical power laws. His broader work on surface chemistry, including monolayers on liquids and the concept of adsorbed film structure, earned him the 1932 Nobel Prize in Chemistry, making him the first industrial chemist so honored.

Origins · 1 minIrving Langmuir and how this was discovered

Irving Langmuir

1881–1957 · American chemist and physicist at General Electric who founded modern surface chemistry

Langmuir's isotherm reframed adsorption as site-limited chemical equilibrium, a conceptual leap that still underlies catalysis, gas sensors, and surface-area measurement today.

Ask the tutor: How does the Langmuir isotherm change when two gases compete for the same surface sites?

Why does the Langmuir isotherm assume exactly one molecule per site

This reflects the chemisorption picture the model was built on: a gas molecule forms a localized bond to a specific surface atom or site, and once that bond forms, the site cannot accept a second molecule. It is a direct extension of simple binding equilibria, like a ligand binding one protein site. Physisorption sometimes stacks additional layers on top, which is why multilayer adsorption needs the BET isotherm instead, a natural generalization covered separately.

How is the Langmuir isotherm related to enzyme kinetics equations

The mathematical form is identical to the Michaelis–Menten equation and to simple one-site ligand-binding equilibria. All three describe a species binding reversibly to a fixed population of independent, saturable sites. The variables differ (pressure and coverage versus substrate concentration and reaction rate), but the algebra of deriving a rectangular-hyperbola relationship from equilibrium is exactly the same, which is why the isotherm is often taught alongside enzyme saturation kinetics.

Can the Langmuir isotherm be derived from statistical thermodynamics instead of kinetics

Yes. Treating each site as an independent two-state system (empty or occupied) and applying the grand canonical ensemble gives the same result, with $K$ expressed in terms of the adsorption energy and partition functions of the gas and the adsorbed state. The kinetic derivation shown here is more intuitive for a first encounter, but the thermodynamic route confirms the same equation without assuming a specific rate mechanism.

Why does real experimental data often deviate from the ideal Langmuir prediction

Real surfaces frequently have energetically inequivalent sites, since edges, defects, and terraces bind molecules with different strengths. Adsorbed molecules can also interact laterally, either repelling each other (making further adsorption harder) or attracting each other (promoting island formation). These effects violate the core assumptions and are the reason the Freundlich isotherm, the Temkin isotherm, and the BET multilayer model were later developed as more general or more specific corrections.

What is a practical way to test whether a system truly obeys the Langmuir isotherm

Measure coverage or adsorbed volume across a wide pressure range, from well below to well above the expected saturation region. Plot $p/\theta$ or $p/V$ against $p$ : genuine Langmuir behavior gives a straight line across the entire range with a single consistent slope and intercept, not just a local patch. Consistency of the extracted $K$ value across multiple pressure pairs, as in the worked example, is a quick diagnostic before trusting the model.