Mohammad Hosseini

Have you measured the SSA by H2 chemisorption? If so, there is a link between SSA and particle size. The Sauter Mean Diameter (D[3, 2]) and SSA are related by:

D[3, 2] = 6/SSA.

If you work with SSA's in m2/g (thus correcting for the density of Ni), then the calculated size is given in micrometers (microns; 10-6 m). I recommend the book J R Anderson Structure of Metallic Catalysts Academic Press (1975) page 365

I would like to point out, that the relation of D = 6/SSA is only valid for monodisperes samples with a narrow sizedistribution.

Philipp Schreier

Yes, in the same way that (pi/6)*d^3 is the volume of a sphere... All single numbers, such as averages, will ignore the usual situation that there’s a distribution. The ’6’ is not universal - in catalysis one may use 5/SSA on the basis that we have a cube where only 5 faces are active and the 6th is tied to the substrate. I recommend J R Anderson’s book as above.

Do you have a better formula? If so, please share it.

Alan F Rawle There is no better formular for the given information. I just wanted to mention the assumptions we use with this formular. It is always good to be aware of the assumptions, that are made within the formulars that one uses.

Philipp Schreier

I can't let this one go. The formula (SSA = 6/D[3,2]) is accurate and theoretically perfect. A simple dimensional analysis shows the link too:

• SSA in m2/cm3 has dimensions of L2/L3 = L-1
• Size is simply L1 (in the D[3, 2], Sauter Mean Diameter, protocol; L3/L2 = L)

Recipricol SSA is size. Thus SSA is in inverse proportion (as expected) to the size. The constant of proportionality is 6 (and this comes from a consideration of 3-fold symmetry; we find '6' in a lot of formulae - the Stokes' equation, for example. In particle size, volume of sphere = [π*d]/6 or ~ 0.523*d; or perhaps you prefer (π/6)*d - it's the same). This shows one advantage of using diameters rather than radii).

In the same way that F = ma, then the SSA = 6/D[3,2] equation is exact by definition. We could argue in the F = ma case that particles of different masses have different accelerations for the same applied force. It brings no further understanding of the system.

What is likely to be of most importance is the assumed stoichiometry of the chemisorption process. Is it 1:1 (Ni-H) or 1:2 (H-Ni-H)? That is, does the H dissociate in the chemisorption process? With CO chemisorption, we have to distinguish between bridged and linear forms of CO.

Alan F Rawle

d[3,2] = 6/SSA is an approximation from the definition of SSA = Sges/Vges = 4 Pi (d/2)² / 4/3 Pi (d/2)³ ~ 6/d[3,2] d[3,2] = 6/SSA but the approximation assumes, that the diameter for the total volume and the total surface is the same otherwise you couldnt cancle out d²/d³ to 1/d, which is only true for a monodisperse sample. If you have anything other, the approximation is not true anymore.

Philipp Schreier

There is obviously some confusion here so we must return to the basics – see attached. It’s N, the total number of particles in the system, that cancel in the numerator and denominator, not d (the diameter of any particle). Otherwise the D[3,2], Sauter mean diameter, the surface area weighted moment mean, would be identical to the D[4,3], De Brouckère, or volume/mass weighted moment mean. Even a simple Wiki search will show you this: https://en.wikipedia.org/wiki/De_Brouckere_mean_diameter

So, in the distribution of 3 particles of sizes 1, 2, and 3 units, the D[3,2] will be (13 + 23 + 33)/(12 + 22 + 32) ~ 2.57 and the D[4,3] will be (14 + 24 +34)/(13 + 23 + 33) ~ 2.72. The D[3,2] is an average (yes, with all its consequences) – not a term in which the diameters cancel out…

An early derivation is shown in J M DallaValle 'Micromeritics The Technology of Fine Particles' 2nd Edition (1948) Pitman pages 44 & 45 and similarly shown by Heywood in 1947 (my bold indications):

“Thus, if N is the number of particles per unit-weight considered as spheres, r the density of the particles, and d their mean diameter:

pi/6d3Nr = 1

And since the specific surface (surface of the particles per unit weight) is N pi d2 = Sw substituting this value of Sw in the above equation we have Sw = (6/r)*(1/d) Eq (3-6)”

Ideally, we should replace weight by mass, but otherwise the derivation is exact for a distribution of particles.

It is useful to work in units of m2/g (or m2/cm3 before real density - in g/cm3 - is taken into account as (for unit density) as then the units of Sauter Mean Diameter size (reciprocal SSA) are directly expressed in micrometers (microns, 10-6 m). Thus:

1 μm sized particles provide a surface area of 6m2/g.

100 nm sized particles provide a surface area of 60m2/g.

10 nm sized particles provide a surface area of 600m2/g.

1 nm sized particles provide a surface area of 6000m2/g

In our example above if the 1, 2, and 3 unit spheres were in microns, the calculated SSA is the reciprocal of the Sauter Mean Diameter (2.57 as we calculated above). Thus, SSA = 6/2.57 ~ 2.33 m2/cm3

Note the averaging taking place to get the SMD.

See: Chapter 7 'Characterization of Nanomaterials' in book: Metrology and Standardization of Nanotechnology DOI: 10.1002/9783527800308.ch7 (2017)

Thank you a lot.

Alan F Rawle