Catalyst particle size and accessible area ($S=3V_s/r$)

Consider an original geometric model of $N$ separate nonporous solid spheres, each with radius $r$. Hold their total volume $V_s$ fixed. The sphere formulas giveAt fixed mass density, fixed $V_s$ also means fixed total mass. Halving $r$ requires eight times as many particles and doubles their total external surface area $S$; this follows by eliminating $N$, without any assumption about chemical kinetics.

For example, take one sphere of radius $1.00\mathrm{mm}$, so $V_s=4\pi /3\mathrm{m}{\mathrm{m}}^3$. Eight spheres of radius $0.50\mathrm{mm}$ have the same $V_s$. Their total surface areas are respectively $4\pi \mathrm{m}{\mathrm{m}}^2$ and $8\pi \mathrm{m}{\mathrm{m}}^2$, approximately $12.57$ and $25.13\mathrm{m}{\mathrm{m}}^2$. Sixty-four particles of radius $0.25\mathrm{mm}$ would instead give $16\pi \mathrm{m}{\mathrm{m}}^2$, four times the original area.

To obtain a kinetic prediction, add assumptions explicitly. Let $f$ be the accessible fraction of external surface area, $\sigma$ the amount of substance of catalytic surface sites per unit accessible area, and $\nu$ the mean number of completed conversions per site per unit time under the chosen conditions. Then the model's reaction product formation rate isHere $\nu$ has units ${\mathrm{s}}^{-1}$ and $\sigma$ has units $\mathrm{mol}{\mathrm{m}}^{-2}$, giving $\mathrm{mol}{\mathrm{s}}^{-1}$ overall. Halving $r$ doubles this predicted rate only if $f$, $\sigma$ and $\nu$ remain unchanged. Particle contact can hide area; pores can add area; diffusion, altered surface structure and catalyst deactivation can change the other factors. Consequently, twice the calculated external area is not a universal promise of twice the measured reaction rate. The geometric argument changes contact, not the net Gibbs free energy or the equilibrium constant of the same overall chemical reaction.