Question

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbondioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 A. (a) Estimate how many platinum atoms would fit into a 2.0 -nm sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \hat{\mathrm{A}}=1 \times 10^{-10} \mathrm{m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of 2.8 A. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0 -nm nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

Short answer

For a 2.0 nm platinum nanoparticle, approximately 278 Pt atoms are present with around 205 atoms on the surface. This results in approximately 73.74% of Pt atoms being on the surface. For a 5.0 nm platinum nanoparticle, you will need to repeat the calculations with the appropriate radius. The nanoparticle with a higher percentage of surface atoms will be more catalytically active, as surface atoms are responsible for catalytic reactions due to their higher reactivity.

Step-by-step solution

Calculate the volume of a platinum atom

To find the volume of one platinum atom, we first convert the edge length of its face-centered cubic arrangement from angstroms to meters and then calculate its volume. Edge length: \(3.924 \hat{A} = 3.924 \times 10^{-10} \mathrm{m}\) The volume of the cubic unit cell is edge length cubed. Volume of cubic unit cell: \(V_c = (3.924\times 10^{-10})^3 = 6.037\times 10^{-29} \mathrm{m}^3\) Since there are 4 platinum atoms in a face-centered cubic arrangement, Volume of one platinum atom: \(V_{Pt} = \frac{6.037 \times 10^{-29}}{4} = 1.509\times 10^{-29} \mathrm{m}^3\)

Calculate the volume of the nanoparticle

Find the volume of the 2.0 nm nanoparticle using the formula for the volume of a sphere: \(V_{nano} = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (1\times10^{-9})^{3} = 4.19\times 10^{-27} \mathrm{m}^3\)

Estimate the number of platinum atoms in the nanoparticle

Divide the volume of the nanoparticle by the volume of one platinum atom: Number of atoms: \(\frac{V_{nano}}{V_{Pt}} = \frac{4.19\times 10^{-27}}{1.509\times 10^{-29}} \approx 278\) Part (b)

Calculate surface area of the nanoparticle

Using the formula for surface area of a sphere: \(A_{nano} = 4 \pi r^2 = 4 \pi (1\times 10^{-9})^2 = 1.26\times 10^{-17} \mathrm{m}^2\)

Find the area covered by one platinum atom on the surface

Using the atomic diameter of platinum: Diameter of one atom: \(2.8 \hat{A} = 2.8 \times 10^{-10} \mathrm{m}\) The "footprint" area of one atom: \(A_{Pt} = \pi (1.4\times 10^{-10})^{2} = 6.154\times 10^{-20} \mathrm{m}^2\)

Estimate the number of surface platinum atoms

Divide the surface area of the nanoparticle by the area of one platinum atom: Number of surface atoms: \(\frac{A_{nano}}{A_{Pt}} = \frac{1.26\times 10^{-17}}{6.154\times 10^{-20}} \approx 205\) Part (c)

Calculate the percentage of surface platinum atoms

Divide the number of surface atoms by the total number of atoms and multiply by 100: Percentage: \(\frac{205}{278} \times 100 \approx 73.74\%\) Part (d) For this part, you need to repeat the same steps using 5.0 nm as the diameter of the nanoparticle. So the radius will be \(2.5\times 10^{-9}\) now, and you will follow the same steps done before to get estimations for a 5.0 nm nanoparticle. Part (e) For the comparison, you will have to compare the percentages of surface atoms obtained for the 2.0 nm and the 5.0 nm nanoparticles. The higher percentage of surface atoms would indicate more catalytic activity, as surface atoms are responsible for catalytic reactions due to their higher reactivity.

Key concepts

Cubic Unit Cell Volume

The cubic unit cell volume is a fundamental aspect of crystallography and materials science that describes how entities—typically atoms, molecules, or ions—are packed in a crystal structure. In face-centered cubic (fcc) arrangements, such as the one in which platinum naturally crystallizes, the unit cell is shaped like a cube, with atoms at all eight corners and one at the center of each face.

Understanding the volume of the cubic unit cell (\(V_c\)) is crucial because it allows us to calculate how many atoms are contained within that space. Since the unit cell of platinum has an edge length of 3.924 angstroms (\(3.924 \times 10^{-10}\text{m}\)), we can find its volume by cubing this value: \(V_c = (3.924 \times 10^{-10})^3\text{m}^3\). In an fcc unit cell, we have four whole platinum atoms as each corner atom counts as \frac{1}{8}\ of an atom (due to sharing with adjacent cells) and each face atom counts as \frac{1}{2}\ (due to sharing with one neighboring cell). Dividing the unit cell volume by four gives us the volume occupied by a single platinum atom.Having this knowledge facilitates a deeper comprehension of the material's density and the packing efficiency within its crystalline structure. These insights are particularly important when considering applications of materials on the nanoscale, such as in the creation of platinum nanoparticles for catalysis.

Nanoparticle Surface Atoms

Nanoparticle surface atoms play a pivotal role in defining the properties of a nanoparticle, especially in catalysis. Their significance arises from the fact that, unlike atoms inside the bulk material, surface atoms have unsatisfied bonds which make them more reactive. This characteristic is crucial in catalytic processes where the reactivity of atoms is exploited to facilitate chemical reactions.

For nanoparticles such as those made of platinum, which has a face-centered cubic arrangement, the number of atoms residing on the surface can be estimated by dividing the total surface area by the 'footprint' of a single atom. The surface-area-to-volume ratio is higher for smaller particles, which means that a larger proportion of atoms are exposed on the surface in comparison to those that are inside the particle.In the specific case of a platinum nanoparticle with a diameter of 2.0 nm, by calculating its surface area and assuming a single platinum atom 'footprint' based on its atomic diameter, one can estimate how many atoms are residing on the surface. This provides insights into the activated sites available for chemical reactions, making surface atoms a vital factor in evaluating the potential catalytic efficacy of the nanoparticle.

Catalytic Activity of Nanoparticles

The catalytic activity of nanoparticles is fundamentally influenced by their physical dimensions and structural characteristics. Nanocatalysts, like platinum nanoparticles, exhibit unique and often enhanced properties compared to their bulk counterparts. A key factor driving this enhanced catalytic activity is the high number of surface atoms relative to the total number of atoms.

Catalysis occurs at the surface where reactant molecules interact with the catalyst's atoms. Hence, nanoparticles with a high surface-area-to-volume ratio provide more active sites for reactions. The percentage of surface atoms can be directly correlated with the catalytic activity potential: the greater the proportion of surface atoms, the more active the nanoparticle can be.Considering two platinum nanoparticles of different sizes, the smaller nanoparticle with a higher percentage of surface atoms will typically be more catalytically active than the larger one. This surface-dominated behavior is a critical concept when optimizing catalysts for reactions such as the oxidation of carbon monoxide to carbon dioxide. For educational and practical applications, grasping the relationship between nanoparticle size, surface atoms, and catalytic activity enables chemists and materials scientists to effectively design and employ catalysts in various chemical processes.