= Catalyst particle size and accessible area
{title2=$S=3V_s/r$}
Consider an original geometric model of $N$ separate nonporous <solid> <spheres>, each with <radius> $r$. Hold their total <volume (thermodynamics)> $V_s$ fixed. The <sphere> formulas give
$$
V_s=N\frac{4\pi r^3}{3},\qquad S=N4\pi r^2=\frac{3V_s}{r}.
$$
At 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 $V_s=4\pi/3\,\mathrm{mm^3}$. Eight <spheres> of <radius> $0.50\,\mathrm{mm}$ have the same $V_s$. Their total <surface areas> are respectively $4\pi\,\mathrm{mm^2}$ and $8\pi\,\mathrm{mm^2}$, approximately $12.57$ and $25.13\,\mathrm{mm^2}$. Sixty-four particles of <radius> $0.25\,\mathrm{mm}$ would instead give $16\pi\,\mathrm{mm^2}$, four times the original area.
\Image[/catalyst-particle-size-and-area.png]
{title=Smaller separate <catalyst> particles expose more <surface area> at fixed total <solid> <volume (thermodynamics)>}
{id=image-catalyst-particle-size-and-area}
{description=Original exact <sphere> model. The circles are particle silhouettes: one of <radius> $r_0$ and eight of <radius> $r_0/2$ represent equal three-dimensional <solid> <volumes>, with twice the external <surface area> in the second case. The right-hand curve is $S/S_0=r_0/r$, with the quarter-radius example marked too. It is a geometric prediction, not a measured <reaction rate>, a peroxide experiment, or a time series.}
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 is
$$
\frac{dn_{\mathrm B}}{dt}=\nu\sigma fS=\frac{3\nu\sigma fV_s}{r}.
$$
Here $\nu$ has units $\mathrm{s^{-1}}$ and $\sigma$ has units $\mathrm{mol\,m^{-2}}$, giving $\mathrm{mol\,s^{-1}}$ 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>.
Back to article page
