Langmuir-Hinshelwood and Eley-Rideal mechanisms

Why it matters

Power-law rate expressions often fail for catalytic reactions: the order changes with pressure, and adding more of one reactant can even slow the reaction. Langmuir–Hinshelwood (LH) and Eley–Rideal (ER) mechanisms explain this behaviour from surface coverage and give rate laws that can be fitted to data, extrapolated with confidence and used to size catalytic reactors.

Key ideas

Building a rate law from a mechanism (LHHW approach). Write the elementary steps — adsorption, surface reaction, desorption. Assume one step is rate-controlling and that all the others are at equilibrium. Express surface coverages by Langmuir isotherms, use a site balance (vacant + occupied = total), and substitute into the rate of the controlling step. The resulting forms are called Langmuir–Hinshelwood–Hougen–Watson (LHHW) rate laws.

Assumptions. Uniform surface (all sites equivalent), monolayer adsorption, no interaction between adsorbed species, adsorption equilibria described by Langmuir isotherms, and a single rate-controlling step.

Langmuir–Hinshelwood mechanism. Both reactants adsorb and react as adsorbed neighbours: A + S ⇌ A·S, B + S ⇌ B·S, A·S + B·S → C + 2S. With the surface reaction controlling, the rate is proportional to θA·θB, giving a squared denominator. Key behaviour: at fixed pB the rate rises with pA, passes through a maximum when A starts crowding B off the surface, then falls. The maximum occurs where KA·pA = 1 + KB·pB (+ product terms).

Eley–Rideal mechanism. Only A adsorbs; B reacts directly from the gas phase with adsorbed A: A·S + B(g) → C + S. The rate is proportional to θA·pB, so the denominator is not squared and the rate increases monotonically with pA, levelling off at saturation. A maximum in rate vs pA therefore points to LH rather than ER.

Single reactant on one site (A ⇌ A·S → R·S ⇌ R + S, surface reaction controlling): first order at low pA, zero order at high pA, inhibited by adsorbed product.

Dissociative adsorption (e.g. H₂ → 2H·S) gives √(KA·pA) terms.

Adsorption-controlled case. If adsorption of A is slowest, the rate depends on pA and the "equilibrium" pressure of A in balance with the surface; for A → R with A adsorption controlling, the denominator has no KA·pA term.

Choosing a model. Linearise each candidate and test against data — for example, single-site unimolecular: pA/r = 1/(k·KA) + pA/k. Use initial-rate data at different total pressures: the dependence of initial rate on total pressure distinguishes adsorption, surface-reaction and desorption control. Constants must be physically sensible (positive K values, adsorption constants falling with temperature since adsorption is exothermic).

Apparent activation energy. For the single-site case at low coverage, r ≈ k·KA·pA, so E_app = E_true + ΔH_ads; since adsorption is exothermic, E_app is lower than E_true.

Formulas

r = k·KA·pA/(1 + KA·pA + KR·pR) (A → R, single site, surface reaction controlling)

r = k·KA·KB·pA·pB/(1 + KA·pA + KB·pB + KC·pC)² (LH: A + B → C, both adsorbed)

KA·pA,max = 1 + KB·pB (pA giving maximum LH rate, no product adsorbed)

r = k·KA·pA·pB/(1 + KA·pA) (ER: adsorbed A + gas-phase B)

θA = √(KA·pA)/(1 + √(KA·pA)) (dissociative adsorption)

pA/r = 1/(k·KA) + pA/k (linearised single-site form)

• r: rate per mass or area of catalyst (mol/kg·s or mol/m²·s); k: surface rate constant (same units as r); Ki: adsorption equilibrium constants (kPa⁻¹); pi: partial pressures (kPa).

E_app = E + ΔH_ads,A (low-coverage single-site case; ΔH_ads < 0)

Worked examples

Example 1 (standard, LH rate and maximum). For A + B → C with LH kinetics: k = 0.1 mol/kg·s, KA = 0.4 kPa⁻¹, KB = 0.2 kPa⁻¹, product not adsorbed. Find the rate at pA = 5 kPa, pB = 4 kPa, and the pA that maximises the rate at this pB.

1. KA·pA = 2.0; KB·pB = 0.8; denominator = (1 + 2.0 + 0.8)² = 3.8² = 14.44.
2. Numerator = 0.1 × 2.0 × 0.8 = 0.16.
3. r = 0.16/14.44 = 0.0111 mol/kg·s.
4. Maximum: KA·pA = 1 + KB·pB = 1.8 → pA = 4.5 kPa. Then r = 0.1 × 1.8 × 0.8/(3.6)² = 0.144/12.96 = 0.01111 mol/kg·s.

r = 0.0111 mol/kg·s; maximum at pA = 4.5 kPa (so the operating point is almost at the maximum). At pA = 20 kPa the rate falls to 0.0067 mol/kg·s — excess A blocks the sites B needs.

Example 2 (GATE level, fitting a single-site model). For A → R on a catalyst (product weakly adsorbed), initial rates are r = 0.5 mol/kg·s at pA = 10 kPa and r = 1.0 mol/kg·s at pA = 40 kPa. Fit r = k·KA·pA/(1 + KA·pA).

1. Linear form: pA/r = 1/(k·KA) + pA/k.
2. Data: pA/r = 10/0.5 = 20 and 40/1.0 = 40 kPa·kg·s/mol.
3. Slope 1/k = (40 − 20)/(40 − 10) = 0.6667 → k = 1.5 mol/kg·s.
4. Intercept 1/(k·KA) = 20 − 0.6667 × 10 = 13.33 → KA = 1/(1.5 × 13.33) = 0.05 kPa⁻¹.
5. Check: r(10) = 1.5 × 0.5/1.5 = 0.5 ✓; r(40) = 1.5 × 2/3 = 1.0 ✓.

k = 1.5 mol/kg·s; KA = 0.05 kPa⁻¹.

Common mistakes

• Forgetting to square the denominator for a bimolecular surface reaction on two sites (LH), or squaring it for ER.
• Leaving out product adsorption terms when the product is strongly adsorbed.
• Assuming the rate always rises with reactant pressure — not for LH kinetics past the maximum.
• Inconsistent units of K and p (kPa⁻¹ with Pa, or concentration-based K with pressures).
• Accepting fitted constants that are negative or increase with temperature for adsorption.

For GATE CH

Expect questions that ask you to identify the mechanism from a rate expression, compute the rate or coverage from given constants, find the reactant pressure that maximises an LH rate, deduce limiting orders at high and low pressure, and discriminate models from linear plots. Practise deriving the single-site and dual-site forms from a site balance.

Quick check

1. In which mechanism does the rate show a maximum with increasing pA?
2. What is the limiting order in A of r = k·KA·pA/(1 + KA·pA) at high pA?
3. What term appears for dissociative adsorption of H₂?
4. Why is the apparent activation energy at low coverage smaller than the true one?

Answers: 1. Langmuir–Hinshelwood; 2. zero; 3. √(KH₂·pH₂); 4. because the exothermic adsorption heat is added to it (E_app = E + ΔH_ads with ΔH_ads < 0).