Chemical, Process & Catalysis Engineering
Langmuir-Hinshelwood Bimolecular Surface Kinetics Formula & Self-Inhibition Peak
Calculate competitive bimolecular Langmuir-Hinshelwood catalytic reaction rate r = k_s * K_A * P_A * K_B * P_B / (1 + K_A * P_A + K_B * P_B)^2 and optimal reactant partial pressure P_A,opt = (1 + K_B * P_B) / K_A.
Why feeding MORE CO to a Pt catalyst can crash oxidation rate — Langmuir-Hinshelwood self-inhibition!
In Irving Langmuir (1921) and Cyril Hinshelwood’s (1926) competitive bimolecular surface mechanism (such as CO + O* -> CO2 on noble metals or alkene hydrogenation), both reactants A and B compete for the same active site ensemble (*). When surface reaction A* + B* -> Products is rate-determining and adsorption is quasi-equilibrated, fractional coverages theta_A = K_A * P_A / D and theta_B = K_B * P_B / D (where D = 1 + K_A * P_A + K_B * P_B) multiply to give r_LH = k_s * K_A * P_A * K_B * P_B / (1 + K_A * P_A + K_B * P_B)^2!
Because the denominator is squared, increasing P_A at fixed P_B initially raises r_LH (+1st order at low coverage), hits a sharp maximum at exact site parity K_A * P_A,opt = 1 + K_B * P_B (so theta_A = 0.50!), and then decreases as ~1 / P_A (-1st apparent order!) as strongly bound A crowds B off the catalyst surface!
| P_A Regime (k_s = 2.50, K_A = 4.0 bar^-1, K_B = 1.5 bar^-1, P_B = 0.80 bar) | Coverage theta_A | Coverage theta_B | Catalytic Rate r_LH |
|---|---|---|---|
| Lean P_A = 0.10 bar | 0.1538 | 0.4615 | 0.1775 mol/(g·s) |
| Near-Optimal P_A = 0.50 bar | 0.4762 | 0.2857 | 0.3401 mol/(g·s) |
| Exact Volcano Peak P_A,opt = 0.55 bar | 0.5000 | 0.2727 | 0.3409 mol/(g·s) |
| Self-Poisoned P_A = 3.00 bar (-1st Order Tail) | 0.8451 | 0.0845 | 0.1785 mol/(g·s) |
Competitive Bimolecular Langmuir-Hinshelwood Rate r_LH
Evaluates k_s * K_A * P_A * K_B * P_B / (1 + K_A * P_A + K_B * P_B)^2 for surface rate constant B2, adsorption constants B3 & B5, and partial pressures B4 & B6.
=IF(OR(B2 <= 0, B3 <= 0, B4 < 0, B5 <= 0, B6 < 0), "Requires Positive Rate/Adsorption Constants & P >= 0", ROUND((B2 * B3 * B4 * B5 * B6) / ((1 + B3 * B4 + B5 * B6) ^ 2), 4))Optimal Reactant Partial Pressure P_A,opt = (1 + K_B * P_B) / K_A Maximizing Rate
Solves dr_LH / dP_A = 0 at fixed P_B, yielding the exact partial pressure where theta_A = 0.50.
=IF(OR(B3 <= 0, B5 <= 0, B6 < 0), "Requires Positive Adsorption Constants & P_B >= 0", ROUND((1 + B5 * B6) / B3, 4))Try the example
Use the original values or change them to test your understanding.