Problem 113 - Chemistry: The Central Science in SI Units, Global Edition

Question

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .(\mathbf{a})\) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .(\mathbf{b})\) Estimate how many platinum atoms are on the surface of a 2.0-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 \(280 \mathrm{pm}\) (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) 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?

Step-by-Step Solution

Answer

For a 2.0-nm platinum nanoparticle, we find the volume of the sphere to be approximately \(4/3\pi \times 10^{-27} \mathrm{m^3}\). Using a face-centered cubic arrangement, we calculate the volume of one platinum atom and estimate the number of atoms in the sphere (N). Next, we find the surface area of the sphere and the area of one platinum atom to estimate the number of atoms on the surface (N_surface). The percentage of Pt atoms on the surface is then calculated as \(\frac{N_{surface}}{N} \times 100\%\). Repeating this process for a 5.0-nm platinum nanoparticle, we can compare the catalytic activity of both sizes. In general, a higher percentage of surface atoms results in higher catalytic activity, as more atoms are exposed and available for catalytic processes.

Step 1: Converting units into meters

First, we need to convert the given lengths into meters. The diameter of the platinum nanoparticles is given to be approximately 2nm. We have: \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m}\) So, \(2 \mathrm{nm} = 2 \times 10^{-9} \mathrm{~m}\) Similarly, the platinum edge length is given to be 392.4 pm. We have: \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) So, \(392.4 \mathrm{pm} = 392.4 \times 10^{-12} \mathrm{~m}\)

Step 2: Calculate volume and number of atoms in the sphere

The volume of a sphere can be found using the formula: \(\frac{4}{3} \pi r^{3}\) where r is the radius of the sphere. Since the diameter is given, we have: \(r = \frac{diameter}{2} = \frac{2 \times 10^{-9}}{2} = 10^{-9} \mathrm{m}\) Now, plug the value of r into the formula for the volume: \(V = \frac{4}{3}\pi(10^{-9})^3 = \frac{4}{3}\pi(10^{-27}) = \frac{4}{3}\pi \times 10^{-27} \mathrm{m^3}\) Now, we need to calculate the volume of one platinum atom. Since platinum crystallizes in a face-centered cubic arrangement, we know that there are 4 atoms per unit cell. The volume of a unit cell can be calculated by: \(V_{unit\,cell} = (edge\,length)^3\) Now, we can calculate the volume of one platinum atom: \(V_{atom} = \frac{V_{unit\,cell}}{4\,atoms}\) Now, we can estimate the number of platinum atoms in the sphere: \(N = \frac{V_{sphere}}{V_{atom}}\)

Step 3: Calculate the surface area and number of atoms on the surface of the sphere

The surface area of a sphere can be found using the formula: \(S = 4\pi r^2\) where r is the radius of the sphere. We have calculated the radius in the previous step. Plug the value of r into the surface area formula: \(S = 4\pi(10^{-9})^2 = 4\pi \times 10^{-18} \mathrm{m^2}\) Next, we need to approximate the area of one platinum atom. The "footprint" of one Pt atom can be estimated by treating it as square of the length equal to its atomic diameter of 280 pm. So, we have the area: \(A_{Pt} = (280 \times 10^{-12})^2 = 78400 \times 10^{-24} \mathrm{~m^2}\) Now, we can estimate the number of platinum atoms on the surface of the 2.0-nm Pt sphere: \(N_{surface} = \frac{S}{A_{Pt}}\)

Step 4: Calculate the percentage of Pt atoms on the surface of the nanoparticle

To calculate the percentage of Pt atoms on the surface of the nanoparticle, we can use the formula: \(Percentage\,of\,surface\,atoms = \frac{N_{surface}}{N} \times 100\%\)

Step 5: Repeat calculations for 5.0-nm platinum nanoparticles

We need to repeat the process to find the number of atoms inside the sphere, the number of atoms on the surface, and the percentage of atoms on the surface for a 5.0-nm platinum nanoparticle.

Step 6: Compare catalytic activity of 2.0-nm and 5.0-nm nanoparticles

Based on the calculated percentage of surface atoms, we can now compare the catalytic activity of 2.0-nm and 5.0-nm nanoparticles. In general, a higher percentage of surface atoms would lead to higher catalytic activity, as more atoms are exposed and available for catalytic processes.

Key Concepts

Catalysis

Catalysis is a fundamental process in chemistry where a substance, known as a catalyst, speeds up a reaction without being consumed by it. It's like having a helper who makes a task easier without getting worn out.
The catalyst offers an alternative pathway with lower activation energy. So, reactions that might take a long time without a catalyst can happen quickly. This is crucial in industrial processes where time is money. For platinum nanoparticles, they serve as catalysts in the reaction that turns carbon monoxide into carbon dioxide.

• This is vital in applications like vehicle emissions control where toxic CO needs to be efficiently removed.
• Platinum, due to its unique properties, provides an excellent surface for this conversion process.

The role of catalysis cannot be overstated. It not only enhances the speed but can also affect the yield and selectivity of reactions. Catalysts are critical in many chemical reactions that are part of renewable energy applications, such as solar fuel generation and fuel cell technologies.

Atomic Structure

The atomic structure of platinum is key to understanding how it behaves as a nanoparticle. Platinum crystallizes in a face-centered cubic (FCC) lattice.
This specific arrangement is not random but provides a highly organized structure of atoms.

• In an FCC lattice, each platinum atom is surrounded by others, making it a dense and stable structure.
• The repeating pattern of atom locations in a unit cell defines the properties and stability of the material.

Platinum's atomic structure determines its physical and chemical properties. This includes its high density, excellent thermal stability, and catalytic efficiency. The regular arrangement in the crystal facilitates the uniform exposure of reactive sites on the nanoparticle's surface, which is critical for its catalytic applications.

Nanotechnology

Nanotechnology deals with materials and devices on the scale of nanometers (one billionth of a meter). At this tiny scale, materials can demonstrate drastically different properties from their bulk counterparts.
Platinum nanoparticles illustrate this well. They are about 2-5 nanometers in diameter and have unique optical, electrical, and catalytic properties due to their small size.

• Nanoparticles have a larger surface area-to-volume ratio, leading to more active sites available for chemical reactions.
• Their small size can lead to quantum effects, impacting how the material behaves.

In the realm of platinum nanoparticles, nanotechnology enables new applications and efficiencies, particularly in catalysis. This technology is advancing fields like medicine and energy, providing solutions that were previously impossible with larger materials.

Catalytic Activity

The catalytic activity of a substance depends primarily on the number of atoms available on its surface to participate in chemical reactions.
In the case of platinum nanoparticles, their small size means that a significant proportion of their atoms are on the surface. This makes them incredibly effective catalysts.

• The more surface atoms, the greater the catalytic activity, because more atoms can directly interact with reactants.
• This is why nanoparticles are often considered when optimizing catalytic processes.

For instance, in a 2-nm platinum nanoparticle, a larger percentage of atoms are exposed compared to in bigger particles like 5-nm ones. Therefore, smaller particles generally show higher catalytic activity, which is advantageous in applications requiring efficient chemical transformations.