Rate-independent limits on a sulphating, self-renewing oxide catalyst: exact master-equation results with lattice kinetic Monte Carlo checks
- 1Department of Physics
- 2National Institute of Technology Agartala
Abstract
Copper oxide takes up sulphur dioxide as a sorbent because sulphation converts it to inactive copper sulphate. Whether a copper oxide surface that sulphates in service and renews itself by exposing fresh copper can instead act as a catalyst for SO 2 oxidation is a question about site statistics as much as chemistry. We study it with a lattice model whose elementary rates are unknown, and we therefore restrict the main claims to statements that hold for every strictly positive assignment of rate constants. Three such results are established. First, under a dual-site mechanism in which O 2 adsorbs dissociatively onto a vacant nearest-neighbour pair and SO 2 onto a single site, an isolated patch of one or two active sites has a stationary product rate of exactly zero, and every connected patch of three or more sites has a strictly positive rate. This is proved and checked exhaustively on all 996 connected graphs of up to seven sites. Second, on a randomly deactivated lattice the share of surviving sites that can never turn over follows the closed form π· (π) = (1 β π) π§ + π§π(1 β π) π‘2 , which is 86.9% on a square lattice at 90% deactivation. Combined with exact stationary rates of every component shape, the classical cluster expansion gives the rate of a randomly deactivated surface without sampling. It shows that an earlier kinetic Monte Carlo run reporting a dead surface was a fluctuation in too short a window, on a surface that was never dead. Third, when sulphated sites are recovered only by discarding the oxide layer, the product obtained per copper atom consumed is exactly (π π /π π ) π sulphate , bounded by the branching ratio between product release and sulphation. This quantity is the inverse cumulative site-loss selectivity of Foley, Johnson and Bhan, and a chemical regeneration channel is the only route past the bound. All stationary solutions use the Grassmann, Taksar and Heyman reduction, which we show is required. A least-squares solve used earlier was wrong by up to twelve orders of magnitude on small rates.
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BibTeX
@misc{gaur2026rateindependent,
title = {{Rate-independent limits on a sulphating, self-renewing oxide catalyst: exact master-equation results with lattice kinetic Monte Carlo checks}},
author = {Pragyaan Gaur},
howpublished = {ChemRxiv},
year = {2026},
month = oct,
doi = {10.26434/chemrxiv.15009719/v1},
url = {https://chemrxiv.org/doi/full/10.26434/chemrxiv.15009719/v1},
note = {CC BY},
}
RIS
TY - UNPB
TI - Rate-independent limits on a sulphating, self-renewing oxide catalyst: exact master-equation results with lattice kinetic Monte Carlo checks
AU - Pragyaan Gaur
PY - 2026
DA - 2026/10/01
PB - ChemRxiv
DO - 10.26434/chemrxiv.15009719/v1
UR - https://chemrxiv.org/doi/full/10.26434/chemrxiv.15009719/v1
AB - Copper oxide takes up sulphur dioxide as a sorbent because sulphation converts it to inactive copper sulphate. Whether a copper oxide surface that sulphates in service and renews itself by exposing fresh copper can instead act as a catalyst for SO 2 oxidation is a question about site statistics as much as chemistry. We study it with a lattice model whose elementary rates are unknown, and we therefore restrict the main claims to statements that hold for every strictly positive assignment of rate constants. Three such results are established. First, under a dual-site mechanism in which O 2 adsorbs dissociatively onto a vacant nearest-neighbour pair and SO 2 onto a single site, an isolated patch of one or two active sites has a stationary product rate of exactly zero, and every connected patch of three or more sites has a strictly positive rate. This is proved and checked exhaustively on all 996 connected graphs of up to seven sites. Second, on a randomly deactivated lattice the share of surviving sites that can never turn over follows the closed form π· (π) = (1 β π) π§ + π§π(1 β π) π‘2 , which is 86.9% on a square lattice at 90% deactivation. Combined with exact stationary rates of every component shape, the classical cluster expansion gives the rate of a randomly deactivated surface without sampling. It shows that an earlier kinetic Monte Carlo run reporting a dead surface was a fluctuation in too short a window, on a surface that was never dead. Third, when sulphated sites are recovered only by discarding the oxide layer, the product obtained per copper atom consumed is exactly (π π /π π ) π sulphate , bounded by the branching ratio between product release and sulphation. This quantity is the inverse cumulative site-loss selectivity of Foley, Johnson and Bhan, and a chemical regeneration channel is the only route past the bound. All stationary solutions use the Grassmann, Taksar and Heyman reduction, which we show is required. A least-squares solve used earlier was wrong by up to twelve orders of magnitude on small rates.
ER -