Question

As discussed in the "Chemistry Put to Work" box in Section \(10.8\), enriched uranium can be produced by effusion of gaseous \(\mathrm{UF}_{6}\) across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, \(\mathrm{U}(\mathrm{g})\). Calculate the ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{223} \mathrm{U}\), and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

Short answer

The effusion rate ratio for \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.038, while the ratio for \(\mathrm{UF}_{6}\) is 1.0043. In comparison, the rate of effusion for the uranium isotopes is larger than that of \(\mathrm{UF}_{6}\). This means that pure uranium isotopes would separate more efficiently than the associated \(\mathrm{UF}_{6}\) isotopes due to their higher effusion rate ratio.

Step-by-step solution

1. Identify the molar masses of \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\)

The gaseous uranium isotopes have the following molar masses: - Molar mass of \({ }^{235}\mathrm{U}\): 235 g/mol - Molar mass of \({ }^{223}\mathrm{U}\): 223 g/mol

2. Apply Graham's Law of Effusion

Using Graham's Law of Effusion, we can find the ratio of effusion rates: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{Molar\:mass_{\:^{223}\mathrm{U}}}}{\sqrt{Molar\:mass_{\:^{235}\mathrm{U}}}} \]

3. Calculate the effusion rate ratio

Next, we substitute the given molar masses into the formula: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{223\:g/mol}}{\sqrt{235\:g/mol}} \] Calculate the square root of the molar masses: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{223}}{\sqrt{235}} \] Finally, divide the square root of molar masses: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = 1.038 \] The effusion rate ratio for \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.038. Now compare this ratio to the ratio for \(\mathrm{UF}_{6}\) given in the essay. According to the essay, the effusion rate ratio for \(\mathrm{UF}_{6}\) containing \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.0043.

4. Comparison

As per the calculation, the effusion rate ratio for the uranium isotopes is 1.038, while the ratio for \(\mathrm{UF}_{6}\) is 1.0043. In comparison, the rate of effusion for the uranium isotopes is larger than that of \(\mathrm{UF}_{6}\). This means that pure uranium isotopes would separate more efficiently than the associated \(\mathrm{UF}_{6}\) isotopes due to their higher effusion rate ratio.

Key concepts

Effusion Rate

Effusion rate is a measure of how quickly gas molecules escape through a small hole into a vacuum. In the context of uranium isotope separation, understanding effusion is crucial. According to Graham's Law of Effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier ones.
Graham's Law is expressed mathematically as:

• \( \frac{Rate_1}{Rate_2} = \frac{\sqrt{Molar\:mass_2}}{\sqrt{Molar\:mass_1}} \)

The law was applied to calculate the effusion rate ratio for uranium isotopes \( ^{235}U \) and \( ^{223}U \), resulting in a ratio of 1.038. This tells us that \( ^{235}U \) effuses slightly faster than \( ^{223}U \). Understanding these rates helps in processes like uranium enrichment, where isotopes need to be separated efficiently. By knowing the effusion rates, scientists can develop better methods for isotope separation.

Uranium Isotopes

Uranium isotopes are variations of the uranium element with different numbers of neutrons. The two isotopes discussed here, \(^{235}U\) and \(^{223}U\), are both forms of gaseous uranium used in various scientific and industrial applications. Here's a quick look at each:

• \( ^{235}U \) is widely used in nuclear reactors and weapons due to its fissile properties.
• \( ^{223}U \) is less common and less stable.

Uranium isotopes are vital in nuclear chemistry and play a significant role in energy production. Their separation via effusion takes advantage of their slight differences in molar mass, thereby helping in creating concentrated forms like enriched uranium. Understanding these isotopes allows scientists to innovate and improve techniques such as gaseous diffusion for isotope separation.

Molar Mass

Molar mass is a critical concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's essential for calculating how substances interact in reactions and is particularly useful in separating isotopes like uranium.For uranium isotopes, the molar masses are:

• \( ^{235}U \): 235 g/mol
• \( ^{223}U \): 223 g/mol

These differences in molar mass are crucial when using Graham's Law of Effusion to separate isotopes. The square root of the molar mass is used to calculate effusion rates, with lighter isotopes effusing more rapidly.
Understanding molar mass allows chemists to predict how substances will behave under different conditions and to tailor processes such as isotope separation for optimal efficiency. This knowledge is fundamental in applications across nuclear energy and other technological fields that utilize isotopes.