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One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advantages and disad- vantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

Short Answer

Expert verified
The relative rates of effusion for \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\) are 1, \(\frac{\sqrt{28}}{\sqrt{29}}\), and \(\frac{\sqrt{28}}{\sqrt{30}}\), respectively. Using carbon dioxide for oxygen isotope separation can offer advantages like increased safety and lower cost, but it might also result in slower effusion and lower separation efficiency due to its higher molecular mass and formation of various chemical species.

Step by step solution

01

Obtain Molecular Masses

First, we need to determine the molecular masses of the three carbon monoxide isotopes: \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). \({ }^{12} \mathrm{C}^{16} \mathrm{O}\): Molecular mass = 12 (for carbon) + 16 (for oxygen) = 28 amu. \({ }^{12} \mathrm{C}^{17} \mathrm{O}\): Molecular mass = 12 (for carbon) + 17 (for oxygen) = 29 amu. \({ }^{12} \mathrm{C}^{18} \mathrm{O}\): Molecular mass = 12 (for carbon) + 18 (for oxygen) = 30 amu.
02

Apply Graham's Law of Effusion

We can now use Graham's law of effusion formula to calculate the relative rates of effusion for the isotopes. 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: Rate of effusion = \(k \times \frac{1}{\sqrt{molar\ mass}}\) Here, k is a proportionality constant, which will be the same for all three isotopes in consideration.
03

Calculate Relative Rates of Effusion

We will calculate the relative rates of effusion for each isotope by choosing \({ }^{12} \mathrm{C}^{16} \mathrm{O}\) as a reference. Relative rate of effusion for \({ }^{12} \mathrm{C}^{16} \mathrm{O}\): \(\frac{Rate_{^{12}C^{16}O}}{Rate_{^{12}C^{16}O}} = \frac{\frac{k}{\sqrt{28}}}{\frac{k}{\sqrt{28}}}\)= 1 Relative rate of effusion for \({ }^{12} \mathrm{C}^{17} \mathrm{O}\): \(\frac{Rate_{^{12}C^{17}O}}{Rate_{^{12}C^{16}O}} = \frac{\frac{k}{\sqrt{29}}}{\frac{k}{\sqrt{28}}}\)= \(\frac{\sqrt{28}}{\sqrt{29}}\) Relative rate of effusion for \({ }^{12} \mathrm{C}^{18} \mathrm{O}\): \(\frac{Rate_{^{12}C^{18}O}}{Rate_{^{12}C^{16}O}} = \frac{\frac{k}{\sqrt{30}}}{\frac{k}{\sqrt{28}}}\)= \(\frac{\sqrt{28}}{\sqrt{30}}\)
04

Discuss Advantages and Disadvantages of Carbon Dioxide for Isotope Separation

Using carbon dioxide instead of carbon monoxide for oxygen isotope separation can have advantages and disadvantages. Advantages: 1. Carbon dioxide is a more stable and less toxic molecule compared to carbon monoxide, which makes the process safer. 2. Carbon dioxide is more abundant in nature and easier to obtain compared to carbon monoxide, which could result in a more cost-effective process. Disadvantages: 1. Carbon dioxide has a higher molecular mass than carbon monoxide (44 amu for COâ‚‚ vs 28-30 amu for CO), which could lead to slower effusion and lower separation efficiency when it comes to separating isotopes. 2. Carbon dioxide can also form a variety of chemical species with different masses, which could complicate the separation process. In summary, the relative rates of effusion for \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\) are 1, \(\frac{\sqrt{28}}{\sqrt{29}}\), and \(\frac{\sqrt{28}}{\sqrt{30}}\), respectively. There are advantages and disadvantages to using carbon dioxide for oxygen isotope separation instead of carbon monoxide, which should be considered when determining the best method for the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isotope Separation
Isotope separation is a critical process used in various scientific and industrial applications to distinguish isotopes of an element. Isotopes are atoms of the same element, containing the same number of protons but differing in the number of neutrons they possess.
An example of isotope separation is separating oxygen isotopes using gaseous diffusion or effusion methods. Through these methods, atoms of different masses (due to the differing neutron count) are separated based on their rates of effusion.
The separation is necessary in fields like nuclear power generation and medical imaging, where specific isotopes are required for efficacy and safety. The separation process itself relies on physical properties of the isotopes rather than chemical interactions, making it a unique and precise technique.
Molecular Mass
Molecular mass, or molecular weight, is the sum of the atomic masses of all atoms in a molecule. It is usually expressed in atomic mass units (amu).
For example, in the case of carbon monoxide (CO) isotopes:
  • extbf{\(^{12}C^{16}O\)} has a molecular mass of 28 amu (12 for carbon + 16 for oxygen).
  • extbf{\(^{12}C^{17}O\)} has a molecular mass of 29 amu (12 for carbon + 17 for oxygen).
  • extbf{\(^{12}C^{18}O\)} has a molecular mass of 30 amu (12 for carbon + 18 for oxygen).
Knowing the molecular mass is crucial for predicting and comparing the behavior of these molecules in physical processes, such as effusion. The subtle differences in isotopic masses guide the separation technique's efficiency and mechanism.
Effusion Process
The effusion process describes the movement of gas molecules through a small opening from a container without any external force except their own kinetic energy. It is a random yet measurable event, closely described by 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:\[Rate \ of \ effusion = k \times \frac{1}{\sqrt{molar \ mass}}\]Here, \(k\) is a constant and does not change across different isotopes involved in the comparison.
This principle is applied in separating isotopes of a gas, such as carbon monoxide. Isotopes with lower mass effuse more quickly than those with higher mass, enabling separation. By calculating and comparing the rates, scientists can effectively isolate lighter isotopes from heavier ones.
Carbon Monoxide
Carbon monoxide (CO) is a colorless and odorless gas commonly used in laboratory and industrial settings. It comprises one carbon atom and one oxygen atom.- Carbon monoxide is often utilized in processes such as isotope separation due to its simple composition and reactive properties. Its relatively low molecular mass (28 amu for \(^{12}C^{16}O\) base form) aids in efficient effusion and effective separation of isotopes.
However, CO is a highly toxic gas. It requires careful handling and specific safety measures when used in applications such as isotope separation processes. The choice to use it over other molecules, like carbon dioxide, often depends on its efficiency and the comparative simplicity of its molecular structure.
Carbon Dioxide
Carbon dioxide (COâ‚‚) is a naturally occurring gas composed of one carbon and two oxygen atoms. It has a molecular mass of about 44 amu, making it heavier than carbon monoxide. - In the context of isotope separation, COâ‚‚ presents certain advantages and disadvantages compared to CO:
  • It is less toxic and more stable, providing a safer environment for operations.
  • It is more readily available and less costly, allowing for potentially more accessible industrial-scale separation processes.
  • Despite these benefits, COâ‚‚ has higher molecular mass which might slow down the effusion process, reducing the efficiency in separating isotopes.
  • Additionally, the possibility of forming numerous chemical species could complicate separation efforts, making CO a more straightforward choice for certain processes due to its simpler form.
The decision to use CO or CO₂ often hinges on balancing these factors—particularly safety, cost, and process efficiency.

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Most popular questions from this chapter

Calculate the pressure exerted by \(0.5000 \mathrm{~mol} \mathrm{~N}_{2}\) in a \(10.000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 107 .

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