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 \({ }^{238} \mathrm{U}\), and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

Short answer

The ratio of effusion rates for ${ }^{235} \mathrm{U}$ and ${ }^{238} \mathrm{U}$ is approximately \(1.0064\). To compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay, we can calculate the percentage difference or simply compare the numerical values.

Step-by-step solution

Recall Graham's law of effusion

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, the relationship can be written as: \[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \] where Rate₁ and Rate₂ are the effusion rates of two gases, and M₁ and M₂ are their respective molar masses.

Calculate the ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\)

Using Graham's law of effusion, we can calculate the ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\). We know the molar masses of these isotopes: M₁ = 235 u, M₂ = 238 u (where u is the atomic mass unit). Plug the values into the equation: \[ \frac{Rate_{235}}{Rate_{238}} = \sqrt{\frac{238}{235}} \] Now, calculate the ratio: \[ \frac{Rate_{235}}{Rate_{238}} = \sqrt{\frac{238}{235}} \approx 1.0064 \]

Compare the ratio to the ratio for \(\mathrm{UF}_{6}\)

The question states that we need to compare the calculated ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{238} \mathrm{U}\) to the ratio for \(\mathrm{UF}_{6}\) given in the essay. We have calculated the ratio to be approximately 1.0064 for the uranium isotopes. Now, let's assume that the ratio for \(\mathrm{UF}_{6}\) given in the essay is R. To compare, we can either find the percentage difference between the two ratios or simply compare the numerical values. Depending on the context and the value of R, we will be able to draw conclusions on how the ratio for uranium isotopes differs from the ratio for \(\mathrm{UF}_{6}\).

Key concepts

Effusion Rates

Effusion rates describe how quickly gas particles pass through a small opening into a vacuum. This concept is crucial in both chemistry and physics because it helps us understand how different gases behave under specific conditions. • **Effusion Process**: In effusion, gas particles gradually escape through tiny pores of a membrane. This process is often used industrially to separate isotopes of gases.
• **Graham's Law of Effusion**: According to Graham's law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means that lighter gases move faster than heavier ones due to their lower molar mass. The mathematical expression for Graham's law is: \[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} \] where \(Rate_1\) and \(Rate_2\) are the effusion rates of two gases, \(M_1\) and \(M_2\) are their respective molar masses.

For instance, to calculate the ratio of effusion rates for isotopes, we apply this formula to their molar masses. In the context provided, the calculation involves uranium isotopes, highlighting the practical application of Graham's law in separating isotopes.

Molar Mass

The molar mass of a substance is defined as the mass of one mole of its particles measured in grams per mole (g/mol). It's essentially the "weight" of a molecule or an atom that determines how it interacts in various chemical processes. • **Basic Calculation of Molar Mass**: For any substance, molar mass can be calculated by summing the masses of all the atoms in its chemical formula. For example, you would add atomic masses of each atom in a compound to get its molar mass.
• **Significance in Effusion**: The molar mass plays a significant role in the effusion process. Because lighter molecules (lower molar mass) effuse more quickly than heavier ones, knowing the molar mass of substances is vital for predicting and comparing their behaviors.

For uranium isotopes such as \( ^{235}U \) and \( ^{238}U \), the molar masses are approximately 235 u and 238 u, respectively. These slight differences affect their effusion rates as predicted by Graham's law. This subtle difference is crucial for processes like uranium enrichment, where separating these isotopes efficiently is essential.

Isotopes

Isotopes are variants of the same chemical element that differ in their number of neutrons, giving them different masses. While isotopes have almost identical chemical behaviors, their physical properties, such as mass, can differ significantly. • **Understanding Isotopes**: All atoms of a given element have the same number of protons, but isotopes can have different numbers of neutrons. Consequently, they have different mass numbers. For example, \( ^{235}U \) and \( ^{238}U \) are both isotopes of uranium, differing only in the number of neutrons.
• **Role in Nuclear Chemistry**: In nuclear chemistry, understanding isotopes is crucial. Isotopes like \( ^{235}U \) are fissile, meaning they can sustain a nuclear reaction, which has applications in nuclear reactors and atomic bombs.

Their different masses make isotopes behave differently during physical processes like effusion and diffusion. This difference is significant in isotope separation techniques—processes vital in various industrial and scientific sectors.