Question

Naturally occurring uranium ore is \(0.7 \%\) by mass fissionable \({ }^{235} \mathrm{U}\) and \(99.3 \%\) by mass nonfissionable \({ }^{238} \mathrm{U}\). For its use as nuclear reactor fuel, the amount of \({ }^{235} \mathrm{U}\) must be increased relative to the amount of \({ }^{238} \mathrm{U}\). Uranium ore is treated with fluorine to yield a gaseous mixture of \({ }^{235} \mathrm{UF}_{6}\) and \({ }^{238} \mathrm{UF}_{6}\) that is pumped through a series of chambers separated by porous barriers; the lighter \({ }^{235} \mathrm{UF}_{6}\) molecules \((\mathscr{A}=349.03 \mathrm{~g} / \mathrm{mol}\) ) effuse through each barrier faster than molecules of \({ }^{238} \mathrm{UF}_{6}(\mathscr{A}=352.04 \mathrm{~g} / \mathrm{mol}),\) until the final mixture obtained is \(3-5 \%\) by mass \({ }^{235} \mathrm{UF}_{6} .\) This process generated \(33 \%\) of the world's enriched uranium in 2008 but has now been replaced with a less expensive centrifuge process. Calculate the ratio of the effusion rates of \({ }^{235} \mathrm{UF}_{6}\) to \({ }^{238} \mathrm{UF}_{6}\).

Short answer

The ratio of the effusion rates of \({ }^{235} \mathrm{UF}_{6} \) to \({ }^{238} \mathrm{UF}_{6} \) is approximately 1.0043.

Step-by-step solution

Identify the formula for effusion rates

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: \[ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \] where \(M_1\) and \(M_2\) are the molar masses of the gases.

Note the molar masses

The molar masses given are: - \( M_1 = 349.03 \) g/mol for \({ }^{235} \mathrm{UF}_{6} \) - \( M_2 = 352.04 \) g/mol for \({ }^{238} \mathrm{UF}_{6} \)

Plug the molar masses into the formula

\[ \frac{\text{Rate}_{ \({ }^{235} \mathrm{UF}_{6} \)}}{\text{Rate}_{ \({ }^{238} \mathrm{UF}_{6} \)}} = \sqrt{\frac{352.04}{349.03}} \]

Calculate the ratio

Calculate the square root: \[ \sqrt{\frac{352.04}{349.03}} \approx \sqrt{1.00861} \approx 1.0043 \] Therefore, the ratio of effusion rates is approximately 1.0043.

Key concepts

Effusion Rates

Effusion is the process by which gas molecules escape through a tiny hole into a vacuum. Graham's law of effusion helps us compare how quickly different gases effuse. It states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means lighter gases effuse faster than heavier ones. The formula can be expressed as \( \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \). The exercise tells us to calculate the effusion rates of two uranium hexafluoride isotopes (\

Molar Masses

The molar mass of a substance is the mass of one mole of that substance. It’s expressed in grams per mole (g/mol). In the given problem, we have two isotopes of uranium hexafluoride: \( { }^{235} \mathrm{UF}_6 \) and \( { }^{238} \mathrm{ UF}_6 \). Their respective molar masses are: 349.03 g/mol for \( { }^{235} \mathrm{UF}_6 \) and 352.04 g/mol for \( { }^{238} \mathrm{UF}_6 \).
Molar masses are crucial for calculating effusion rates because of their direct relationship in Graham's law. By comparing the square roots of the molar masses, we can determine how much faster one gas effuses compared to another. Calculate it as follows: \[ \frac{\text{Rate}_{ { }^{235} \mathrm{UF}_6 }}{\text{Rate}_{ { }^{238} \mathrm{UF}_6 }} = \sqrt{\frac{352.04}{349.03}} \]

Uranium Enrichment

Uranium enrichment is the process of increasing the concentration of \( { }^{235} \mathrm{U} \) isotopes in uranium. Naturally occurring uranium contains only about 0.7% \( { }^{235} \mathrm{U} \) and 99.3% \( { }^{238} \mathrm{U} \). However, for nuclear reactors, the concentration of \( { }^{235} \mathrm{U} \) needs to be increased to 3-5%.
The process described in the problem involves converting uranium ore to a gaseous state by reacting it with fluorine to form uranium hexafluoride \( (\mathrm{UF}_6 ) \). Then, using effusion, the lighter \( { }^{235} \mathrm{UF}_6 \) molecules pass through porous barriers faster than the heavier \({ }^{238} \mathrm{UF}_6\) molecules, gradually enriching the mixture.
Though this gaseous diffusion method was prominent in the past, it has largely been replaced by centrifuge processes, which are more cost-effective and efficient.

Isotopes

Isotopes are atoms of the same element that have different numbers of neutrons, and thus different atomic masses. Both \( { }^{235} \mathrm{U} \) and \( { }^{238} \mathrm{U} \) are isotopes of uranium, meaning they have the same number of protons but different numbers of neutrons.
In nuclear chemistry, different isotopes have different nuclear properties, which makes some suitable for fuel in nuclear reactors, like \( { }^{235} \mathrm{U} \), and others less so, like \( { }^{238} \mathrm{U} \).
The atomic mass of each isotope affects their behavior in processes like effusion, as demonstrated in the exercise. The slight difference in atomic mass (and consequently molar mass) between \( { }^{235} \mathrm{UF}_6 \) and \( { }^{238} \mathrm{UF}_6 \) is significant enough to be exploited for uranium enrichment.

Nuclear Fuel

Nuclear fuel refers to materials that can be used to generate nuclear energy. In most nuclear reactors, uranium or plutonium isotopes are used. \( { }^{235} \mathrm{U} \) is a fissionable isotope of uranium, meaning it can sustain a nuclear chain reaction, releasing a large amount of energy. This energy is harnessed for electricity generation in nuclear power plants.
The uranium must be enriched to increase the proportion of \( { }^{235} \mathrm{U} \) for it to be effective as nuclear fuel. Without enrichment, the natural concentration is too low for most reactor designs.
Enriched uranium is then fabricated into fuel rods or pellets that are loaded into the reactor. As the reactor operates, these fuel rods undergo fission, releasing energy that heats water to produce steam, which drives turbines to generate electricity.