In Part 1 of this series we described nanoparticles, what they are, how they are made and how they compare to other forms of metal. We also described how metal surfaces serve as catalysts facilitating reactions such as the hydrogenation of an alkene. In this part we will explore how nanoparticles excel due to their large surface area.
As we learned in Part 1, metal-catalyzed reactions take place on the surface of the metal. In the case of palladium, only hydrogen and deuterium can migrate to holes below the surface. For all other substances and almost all other metals, the interactions between metal and reactant occurs only at the surface. Even helium cannot migrate below the surface. For example, in the hydrogenation of adsorbed ethene discussed in Part 1, the reaction takes place at the surface of the palladium even though the hydrogen atoms are not limited to the surface. For this reason, a strategic goal in designing catalysts is to maximize the surface area. No formulation does this better than nanoparticles. In this part we explore the importance of surface area through calculations students can understand and perform themselves.
Smaller is better
In the case of spheroidal nanoparticles, we make our estimates using formulas for the sphere. The volume, and therefore number of palladium atoms, increases as the cube of the radius (Vsphere = 4/3p r3), while the surface area, and therefore number of surface atoms, increases as the square of the radius (Asphere = 4 p r2). The ratio of Area to Volume equals 3/rsphere. As the radius of the sphere increases, the ratio Area / Volume decreases.
Smaller spheres have a larger percent of surface atoms.
Estimating the number of atoms in a sphere
One simple way to estimate the number of atoms within a sphere is to calculate the volume of the sphere from the radius using the volume formula, then multiplying the volume (cm3) by the density of the metal (g/cm3) to give the mass of the nanoparticle in grams. The mass is converted to moles and finally to the number of atoms using Avogadro’s number. For a spheroidal nanoparticle of palladium with radius of 1.0 nm (1.0 x 10-7 cm), we can estimate the moles of palladium and the number of atoms, given the density is 12.023 g/cm3:
Vsphere = 4/3 p r3 = 4/3p (1.0 x 10-7 cm)3 = 4.2 x 10-21 cm3
mPd = volume x density = 4.2 x 10-21 cm3 x 12.023 g cm-3 = 5.0 x 10-20 g
nPd = mass / atomic mass = 5.0 x 10-20 g / 106.42 g mol-1 = 4.7 x 10-22 mol
nPd atoms = 4.7 x 10-22 mol x 6.02 x 1023 atoms mol-1 = 285 atoms
To estimate the number of atoms below the surface layer of the sphere, we start by defining the subsurface radius equal to the nanoparticle radius minus one diameter of the element in question. In the figure, this would be the radius of the circle, shown in red, minus one atomic diameter. Palladium has a covalent radius of 0.138 nm, so a nanoparticle with radius of 1.0 nm would have subsurface radius, rsubsurface = 1.0 nm – 2 x 0.138 nm = 0.724 nm. Repeating the calculation above, but using the subsurface radius, one calculates 108 subsurface atoms. The number of surface atoms is the difference, 285 – 108 = 177 surface atoms.
The ratio of surface atoms to total atoms is called the dispersion or the degree of dispersion.1 The percent of surface atoms equals the dispersion x 100%. The nanoparticle described above, with a radius of 1.0 nm, has a dispersion of 0.62, or 62% of the atoms are surface atoms.
Using an Excel spreadsheet we calculated the percent surface atoms as a function of particle radius as shown in Fig. 1. If carried out to the upper limit of a nanoparticle radius, 50 nm, the percent of surface atoms drops to ~2%. Fig. 1 dramatically shows the leveling off of the fraction of surface atoms as the radius increases.
Fig. 1. Percent of surface Pd atoms decreases with an increase in the radius of the nanoparticle.
In Fig. 2 we plot surface area in m2/g of palladium as a function of nanoparticle radius. Area is calculated using the formula given earlier (Asphere = 4 p r2) and the mass is determined from particle volume and the density (mass = volume x density). Consider again the spheroidal nanoparticle of palladium with a radius of 1 nm, for which we calculated a mass of 5.0 x 10-20 g. The area of this nanoparticle is 1.26 x 10-17 m2. To compare nanoparticles with various radii, we calculate area per gram. For 1.0 nm radius nanoparticle of palladium, this value is 250 m2/g Pd. Most remarkably, the units are square meters per gram. Thus, a 1.0 g sample of this size nanoparticle would have a surface area similar to the floor area of a fairly large house.
Fig. 2. Surface area in m2/g decreases as the radius of the Pd nanoparticles increases.
In these calculations, we assumed a spherical shape for the nanoparticles. As a geometric object the sphere has the smallest of all surface areas. Real nanoparticles may be semi-spheroidal, but they also have every imaginable shape from potato-shaped to pancake-shaped. In Part 1 we noted that nanomaterials may take the form of tubes, wires and sheets. All these shapes have surface areas that are even larger per particle than the ones we calculated for spheres. That is to say a nanotube consisting of 1000 atoms has a larger surface area than a sphere of 1000 atoms.
Comparing surface area of nanoparticles with bulk metal samples
Bulk palladium is sold as wire, sheets, bars and pellets. All of these forms have a negligible percent surface area compared to nanoparticles. Consider a single 1.0 g spherical pellet of palladium. Dividing the 1.0 g mass by palladium’s density (12.023 g/cm3) gives a volume of 0.0832 cm3. Assuming that the pellet is shaped enough like a sphere in order to use the sphere formula, one can solve for radius, rsphere = 0.271 cm, and then area,
Asphere = 0.921 cm2. Comparing this area with the area of 1.0 g of 1.0 nm diameter nanoparticles (250 m2), one calculates that the nanoparticles have a surface area that is 2.7 million times larger than that of a single sphere of the same mass.
Most of the chemical supply companies sell palladium in a foil format. The thinnest foil offered is a mere
0.0025 cm thick — less than 100,000 atoms thick based on a diameter for palladium of 0.276 nm. A 1.0 g sample of this Pd foil has a volume of 0.0832 cm3 — the same as any shape. The formula for the volume of a box is
Vbox = Arectangle x h. From this we get Arectangle = 33 cm2.
However the thin foil has two large sides of surface atoms for a total of ~67 cm2. We can ignore the areas of the four edges.
Arectangle = Vbox / h = 0.0832 cm3 / 0.0025 cm = 33 cm2
= 3.3 x 10-3 m2
Atop + bottom = 2 x 3.3 x 10-3 m2 = 6.7 x 10-3 m2
The area of the foil is ~72 times greater than the area of the 1.0 g sphere. However, when compared to the 250 m2 total area we calculated for the 1.0 g of spherical Pd nanoparticles (radius of 1 nm), the latter is over 37,000 times greater than that of the foil. In terms of surface area where chemical reactions are catalyzed by metals, nanoparticles excel.
In Part 3, we will look more closely at the structure of close-packed metals and describe how a simple model of a nanoparticle can be constructed.
Questions
- We describe how a nanoparticle with radius of 1.0 nm would contain ~285 atoms. Repeat this calculation to see if you get the same answer. Use units as you work to assure you are not making mistakes.
- Continuing on with the same 1.0 nm radius nanoparticle of palladium, repeat the calculations to estimate the number of subsurface atoms and surface atoms. Calculate the degree of dispersion and the percent surface atoms.
- Repeat the calculation for this same nanoparticle size to determine the area of an individual nanoparticle (in nm2) and the area per gram Pd in units of m2/g Pd.
- Repeat calculations described in Questions 1 – 3 for a spheroidal nanoparticle (a) with a radius of 0.5 nm, and (b) with a radius of 2.0 nm.
[Answers: (a) For 0.5 nm
percent surface atoms = 91%
Area per gram = 500 m2/g
(b) For 2.0 nm
percent surface atoms = 36%
Area per gram = 125 m2/g
You can have student check their answers on the Figures given on the previous page.]
- Palladium is also sold as wire with a diameter of 0.5 mm (0.25 mm radius). A wire is essentially a cylinder and the formula for area is
Acylinder = 2 x Acircle + Acylinder surface = 2 p r2 + 2 p r h,
where h is the length of the wire. How long in meters would this wire need to be to equal the area of 1 g of 1 nm radius nanoparticles (250 m2)? Express this length in m and km. Which term is negligible for a long wire: Acircle or Acylinder surface? [Answer: 1.6 x 105 m]
- We assumed the nanoparticles were spherical in order to estimate their surface area. Most have irregular shapes instead. Would an irregularly shaped nanoparticle have a larger or smaller number of surface atoms compared to a spherically shaped nanoparticle of the same mass?
- Create a spreadsheet in Excel to make your own version of the graphs in Fig. 2. Column headings include:
- nanoparticle radius (nm),
- Nanoparticle subsurface radius (calculated by subtracting 2 X covalent radius of metal),
- Nanoparticle volume in nm3,
- Nanoparticle subsurface volume in nm3,
- Number of atoms in nanoparticle,
- Number of surface atoms in nanoparticle,
- Dispersion
- Percent surface atoms,
- Area of each nanoparticle (nm2),
- Mass of each nanoparticle, and
- Area per gram in units of m2/g.
Reference
G. Bond, Metal Catalysed Reactions of Hydrocarbons, Springer, 2005, page 36. This ratio is also called the dispersity by European scientists
