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Chapter 14: Problem 113

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?

Short Answer

Expert verified
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 by step solution

01

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, \(392.4 \mathrm{pm} = 392.4 \times 10^{-12} \mathrm{~m}\)
02

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: \(N = \frac{V_{sphere}}{V_{atom}}\)
03

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: \(N_{surface} = \frac{S}{A_{Pt}}\)
04

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: \(Percentage\,of\,surface\,atoms = \frac{N_{surface}}{N} \times 100\%\)
05

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.
06

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.

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