Surface Catalysis and Adsorption Kinetics

Ammonia synthesis, catalytic converters, and hydrogenation of vegetable oils all share one hidden bottleneck: gas molecules must first stick to a metal surface before anything chemical can happen. That sticking step obeys its own kinetics, distinct from anything in gas-phase or solution chemistry. Coverage, not concentration, becomes the currency of rate. This matters because surface reaction rates often plateau, or even fall, as pressure rises — behavior that flatly contradicts naive rate laws. Understanding why requires a proper adsorption isotherm and a mechanism for what happens once molecules are bound. This article builds the Langmuir isotherm from first principles. It then develops the Langmuir–Hinshelwood and Eley–Rideal mechanisms for bimolecular surface reactions. You will see how fractional coverage controls the observed order in each regime. Two full numerical examples show how real rate data are fitted and interpreted.

Key takeaways

The Langmuir isotherm gives fractional surface coverage $\theta$ as a function of gas pressure, assuming identical, independent sites and single-layer adsorption.
At low pressure $\theta \propto P$ (first order); at high pressure $\theta \to 1$ and the rate saturates (zero order).
Langmuir–Hinshelwood kinetics: both reactants adsorb, then react on the surface — rate depends on the product of two coverages.
Eley–Rideal kinetics: one reactant adsorbs, the other reacts directly from the gas phase — rate is first order in gas-phase partial pressure.
Competitive adsorption of a product or inert species can poison a catalyst by occupying sites, lowering the observed rate.
Key idea: heterogeneous catalytic rates are controlled by surface coverage, not bulk concentration, and coverage saturates because the number of active sites is finite.

The core insight

Because a catalytic surface offers a fixed number of adsorption sites, the fraction of sites occupied, $\theta$ , rises with pressure but saturates at unity; every heterogeneous rate law is built by substituting this bounded, pressure-dependent $\theta$ (from the Langmuir isotherm) into an elementary surface-reaction step, which is why apparent reaction orders shift from first to zero as pressure increases.

The physical picture

Picture a metal surface as a two-dimensional lattice of $N_s$ identical adsorption sites, each capable of holding at most one adsorbate molecule. Gas-phase molecules strike the surface, and a fraction of collisions with empty sites result in a sticking (adsorption) event. Simultaneously, molecules already adsorbed can leave (desorb) with some probability per unit time. At equilibrium these two fluxes balance, and a definite fraction of sites, $\theta$ , stays occupied for a given gas pressure $P$ and temperature $T$ . This is fundamentally different from solution kinetics. In solution, a reagent's concentration can rise without bound as you add more of it. On a surface, the number of sites is fixed. Once every site is occupied, adding more gas cannot increase the amount reacting per unit time — the rate saturates. This saturation is the qualitative fingerprint of heterogeneous catalysis, and it is precisely what the Langmuir isotherm quantifies. Once coverage is known, the catalytic step itself can proceed by one of two limiting mechanisms. In the Langmuir–Hinshelwood (LH) mechanism, both reactants first adsorb onto neighboring sites, diffuse across the surface, and react when they meet. In the Eley–Rideal (ER) mechanism, only one reactant adsorbs; the second reacts directly from the gas phase by colliding with the adsorbed species, with no need to itself find a vacant site. Most industrial catalytic reactions (CO oxidation on Pt, N₂ + H₂ on Fe) are LH-type; ER-type collisions are less common but well documented, for example in some H atom recombination reactions on surfaces.

Figure

Langmuir isotherm: fractional coverage θ versus pressure, for two values of the adsorption equilibrium constant K

Figure

Two limiting mechanisms on a catalytic surface: Langmuir–Hinshelwood versus Eley–Rideal

The core relations

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

where: $\theta$ is fractional coverage, $P$ is gas pressure, $K = k_{ads}/k_{des}$ is the adsorption equilibrium constant.

\theta_A = \frac{K_A P_A}{1 + K_A P_A + K_B P_B}

where: competitive Langmuir isotherm for two adsorbates A and B sharing the same sites.

\text{rate}_{LH} = k\,\theta_A\theta_B = \frac{k K_A K_B P_A P_B}{\left(1 + K_A P_A + K_B P_B\right)^2}

where: $k$ is the surface rate constant for the bimolecular surface reaction step.

\text{rate}_{ER} = k\,\theta_A P_B = \frac{k K_A P_A}{1 + K_A P_A}\,P_B

where: B reacts directly from the gas phase with adsorbed A; only A obeys a Langmuir isotherm.

Derivation ADeriving the Langmuir isotherm from a dynamic-equilibrium site balance

Goal: obtain $\theta(P)$ for non-dissociative, single-layer adsorption on identical, independent sites.

Step 1. Define the two competing fluxes. Adsorption requires a gas molecule to strike an empty site. The rate of adsorption per unit area is proportional to the pressure (collision frequency with the surface) and to the fraction of sites currently empty, $(1-\theta)$ :

\text{rate}_{ads} = k_{ads} P (1-\theta)

Desorption requires no collision; it depends only on how many sites are occupied:

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

Step 2. Impose dynamic equilibrium. At constant $P$ and $T$ , $\theta$ becomes constant when the two fluxes balance:

k_{ads} P (1-\theta) = k_{des}\,\theta

Step 3. Solve for $\theta$ . Expand and collect terms in $\theta$ :

k_{ads}P - k_{ads}P\theta = k_{des}\theta \quad\Longrightarrow\quad k_{ads}P = \theta\left(k_{des} + k_{ads}P\right)

Divide both sides by $k_{des}$ and define $K \equiv k_{ads}/k_{des}$ :

\theta = \frac{k_{ads}P}{k_{des} + k_{ads}P} = \frac{KP}{1+KP}

Step 4. Check the two limits. As $P \to 0$ , the denominator $\to 1$ , so $\theta \to KP$ : coverage rises linearly, and any surface rate law becomes first order in $P$ . As $P \to \infty$ , $KP \gg 1$ , so $\theta \to 1$ : every site is occupied, and any surface rate law becomes independent of $P$ (zero order in the gas-phase reactant).

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

Worked examples

Example 1 — Extracting $K$ and the monolayer capacity from adsorption data (foundational). A gas adsorbs non-dissociatively on a catalyst sample. The volume of gas adsorbed (at STP, per gram of catalyst) is measured at 298 K:

P (kPa)	V (cm³ g⁻¹)
10	3.5
30	6.3
60	8.0
100	9.2

Assume Langmuir behavior, $V = V_m KP/(1+KP)$ , where $V_m$ is the monolayer capacity. Step 1 — linearize. Invert the isotherm:

\frac{P}{V} = \frac{1}{V_m K} + \frac{P}{V_m}

A plot of $P/V$ against $P$ should be a straight line with slope $1/V_m$ and intercept $1/(V_m K)$ . Step 2 — compute $P/V$ for each point.

$P=10$ : $P/V = 10/3.5 = 2.857\ \text{kPa g cm}^{-3}$
$P=30$ : $P/V = 30/6.3 = 4.762\ \text{kPa g cm}^{-3}$
$P=60$ : $P/V = 60/8.0 = 7.500\ \text{kPa g cm}^{-3}$
$P=100$ : $P/V = 100/9.2 = 10.870\ \text{kPa g cm}^{-3}$

Step 3 — fit a line. Using the first and last points to estimate slope (a full least-squares fit gives essentially the same numbers):

\text{slope} = \frac{10.870 - 2.857}{100 - 10} = \frac{8.013}{90} = 0.0890\ \text{g cm}^{-3}

So $V_m = 1/0.0890 = 11.24\ \text{cm}^3\,\text{g}^{-1}$ . Intercept: using the point at $P=10$ ,

\text{intercept} = 2.857 - (0.0890)(10) = 2.857 - 0.890 = 1.967\ \text{kPa g cm}^{-3}

Then

K = \frac{1}{V_m \times \text{intercept}} = \frac{1}{(11.24)(1.967)} = \frac{1}{22.11} = 0.0452\ \text{kPa}^{-1}

Step 4 — sanity check. At $P=100$ kPa, predicted coverage fraction:

\theta = \frac{KP}{1+KP} = \frac{(0.0452)(100)}{1+(0.0452)(100)} = \frac{4.52}{5.52} = 0.819

Predicted $V = \theta V_m = (0.819)(11.24) = 9.20\ \text{cm}^3\text{g}^{-1}$ , matching the measured 9.2 cm³ g⁻¹ almost exactly — the fit is self-consistent.

Example 2 — Predicting the rate law for CO oxidation on Pt (Langmuir–Hinshelwood, applied). The reaction $2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2$ on platinum follows an LH mechanism: both CO and O₂ adsorb competitively, then react on adjacent sites. Given $K_{CO} = 2.0\ \text{kPa}^{-1}$ , $K_{O_2} = 0.05\ \text{kPa}^{-1}$ , $k = 4.0\times10^{-3}\ \text{mol m}^{-2}\text{s}^{-1}$ , at 500 K, find the rate at (a) $P_{CO}=5$ kPa, $P_{O_2}=20$ kPa, and (b) $P_{CO}=50$ kPa, $P_{O_2}=20$ kPa. Step 1 — write the rate law.

\text{rate} = \frac{k\,K_{CO}K_{O_2}\,P_{CO}P_{O_2}}{\left(1+K_{CO}P_{CO}+K_{O_2}P_{O_2}\right)^2}

Step 2 — case (a), low CO pressure. Denominator:

1 + (2.0)(5) + (0.05)(20) = 1 + 10 + 1.0 = 12.0

Numerator:

(4.0\times10^{-3})(2.0)(0.05)(5)(20) = (4.0\times10^{-3})(2.0)(0.05)(100)

Compute stepwise: $2.0 \times 0.05 = 0.10$ ; then $0.10 \times 100 = 10.0$ ; then $4.0\times10^{-3}\times 10.0 = 4.0\times10^{-2}$ .

\text{rate}_a = \frac{4.0\times10^{-2}}{(12.0)^2} = \frac{4.0\times10^{-2}}{144} = 2.78\times10^{-4}\ \text{mol m}^{-2}\text{s}^{-1}

Step 3 — case (b), high CO pressure (CO now poisons the surface). Denominator:

1 + (2.0)(50) + (0.05)(20) = 1 + 100 + 1.0 = 102.0

Numerator:

(4.0\times10^{-3})(2.0)(0.05)(50)(20)

Compute: $2.0\times0.05 = 0.10$ ; $50\times20 = 1000$ ; $0.10 \times 1000 = 100$ ; $4.0\times10^{-3}\times100 = 0.40$ .

\text{rate}_b = \frac{0.40}{(102.0)^2} = \frac{0.40}{10404} = 3.85\times10^{-5}\ \text{mol m}^{-2}\text{s}^{-1}

Step 4 — interpret. Increasing $P_{CO}$ tenfold decreased the rate by roughly a factor of 7. This is textbook CO self-poisoning: CO binds so strongly ( $K_{CO} = 2.0\ \text{kPa}^{-1}$ ) that it floods the surface, leaving too few free sites for O₂ to adsorb, and $\theta_{O_2}$ collapses even though $P_{O_2}$ is unchanged. This negative order in one reactant is impossible in ordinary gas-phase kinetics but is routine in LH surface catalysis.

Limiting cases & common traps

Regime	Behavior	Common trap
$KP \ll 1$ (low pressure / weak adsorption)	$\theta \approx KP$ ; rate first order in $P$	Assuming first order holds at all pressures
$KP \gg 1$ (high pressure / strong adsorption)	$\theta \approx 1$ ; rate zero order in $P$	Concluding the surface reaction step itself is "slow" when the surface is simply full
LH mechanism	rate $\propto \theta_A\theta_B$ , can show negative order	Forgetting that a strongly adsorbing species can suppress another's coverage even as its own pressure rises
ER mechanism	rate $\propto \theta_A P_B$ , always increases with $P_B$	Confusing ER with LH because both reactants "appear" in the rate law
Dissociative adsorption (e.g. H₂ → 2H(ads))	$\theta \propto \sqrt{KP}$ , not $KP$	Applying the non-dissociative isotherm formula unmodified

The chemist behind it

Irving Langmuir developed his adsorption isotherm around 1916–1918 while investigating gas adsorption on tungsten filaments for General Electric, work motivated by improving incandescent light bulbs. He recognized that adsorption is fundamentally a chemical bonding phenomenon confined to a single molecular layer, a sharp break from earlier empirical isotherms like Freundlich's, which had no clear physical basis or saturation limit. Langmuir's site-balance argument — adsorption flux versus desorption flux — is the same dynamic-equilibrium logic used throughout kinetics, and it earned him the 1932 Nobel Prize in Chemistry for his broader work in surface chemistry.

Origins · 1 minIrving Langmuir and how this was discovered

Irving Langmuir

1881–1957 · American chemist and physicist who founded quantitative surface science and modeled monolayer gas adsorption from a kinetic site balance

Ask the tutor: How would the Langmuir–Hinshelwood rate law change if the reaction instead required three adjacent empty sites for adsorption, as in dissociative N₂ chemisorption on iron?

Why does the Langmuir isotherm assume identical, independent sites?

The derivation treats every site as chemically equivalent and assumes an adsorbed molecule does not affect its neighbors. This makes the adsorption and desorption rate constants, $k_{ads}$ and $k_{des}$ , single fixed numbers rather than functions of $\theta$ . Real surfaces often violate this — step edges, defects, and lateral interactions between adsorbates mean the site-balance derivation is an idealization. It remains an excellent starting model because it captures the essential saturation physics correctly, and deviations are usually treated as corrections (Freundlich, Temkin isotherms) layered on top.

How can a reaction show negative order in a reactant, as in CO oxidation on Pt?

Negative order arises specifically in Langmuir–Hinshelwood kinetics when one adsorbate binds far more strongly than the other. As $P_{CO}$ increases, $\theta_{CO}$ approaches 1 and crowds out oxygen, so $\theta_{O_2}$ falls faster than the CO term in the rate expression rises. The net rate, proportional to $\theta_{CO}\theta_{O_2}$ , decreases. This is impossible in elementary gas-phase kinetics, where adding more of a reactant cannot lower its own reaction rate, making it a hallmark diagnostic of competitive surface adsorption.

How do you distinguish Langmuir–Hinshelwood from Eley–Rideal experimentally?

The clearest test is the pressure dependence of one reactant while holding the other's surface coverage saturated. In LH kinetics, raising the pressure of a strongly adsorbing species can suppress the rate (negative order) because it blocks sites needed by the co-reactant. In ER kinetics, the gas-phase reactant never needs a site, so the rate keeps rising (or plateaus only through the adsorbed species' own saturation) and never turns over. Isotopic labeling and molecular beam scattering experiments have also directly confirmed ER events for specific systems, such as H-atom recombination.

Does the Langmuir isotherm apply to liquids or only gases?

The same site-balance logic extends to adsorption from solution, though "pressure" is replaced by solute concentration. The isotherm becomes $\theta = Kc/(1+Kc)$ , used widely in analyzing adsorption of solutes onto activated carbon or chromatographic stationary phases. The physical assumptions (identical sites, monolayer, no lateral interaction) carry over unchanged. Deviations at high concentration, where multilayer adsorption or site heterogeneity becomes significant, are treated with BET theory or empirical isotherms rather than the simple Langmuir form.

Why does catalysis matter if it does not shift the equilibrium of the reaction?

A catalyst, homogeneous or heterogeneous, only speeds up the approach to equilibrium; it cannot change the equilibrium constant or the thermodynamics, since it does not appear in the overall reaction. Surface catalysis works by lowering the activation energy pathway, typically by breaking bonds during adsorption so weaker, easier-to-rearrange surface-bound fragments react instead of intact gas-phase molecules. This is why the Langmuir–Hinshelwood and Eley–Rideal frameworks focus entirely on rate laws and coverage, never on shifting the position of equilibrium.