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 advan- tages and disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

Short Answer

Expert verified
The relative rates of effusion of \({ }^{12} \mathrm{C}^{16}\mathrm{O}\), \({ }^{12} \mathrm{C}^{17}\mathrm{O}\), and \({ }^{12} \mathrm{C}^{18}\mathrm{O}\) are approximately 1:0.965:0.931. Using carbon dioxide instead of carbon monoxide for the separation process has some advantages, such as being less toxic and more stable, but might also lead to a less efficient separation due to smaller differences in relative effusion rates.

Step by step solution

01

Find the molar masses of the isotopes

Calculate the molar masses of the three isotopes of carbon monoxide by summing the atomic masses of the carbon and oxygen atoms. We will use the atomic mass unit (u) for the calculation: - \({ }^{12} \mathrm{C}^{16}\mathrm{O}\): \(12 u + 16 u = 28 u\) - \({ }^{12} \mathrm{C}^{17}\mathrm{O}\): \(12 u + 17 u = 29 u\) - \({ }^{12} \mathrm{C}^{18}\mathrm{O}\): \(12 u + 18 u = 30 u\)
02

Apply Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of two gases is inversely proportional to the square root of their molar masses: \(\frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}}\). We can use this formula to compare the rates of effusion of the three isotopes: Relative rate of effusion of \({ }^{12} \mathrm{C}^{16}\mathrm{O}\) to \({ }^{12} \mathrm{C}^{17}\mathrm{O}\): \(\frac{Rate_{16}}{Rate_{17}} = \sqrt{\frac{29u}{28u}} \approx 1.035\) Relative rate of effusion of \({ }^{12} \mathrm{C}^{16}\mathrm{O}\) to \({ }^{12} \mathrm{C}^{18}\mathrm{O}\): \(\frac{Rate_{16}}{Rate_{18}} = \sqrt{\frac{30u}{28u}} \approx 1.069\)
03

Advantages and disadvantages of using carbon dioxide

Separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide could have the following advantages and disadvantages: Advantages: 1. Carbon dioxide is less toxic, making the process safer for workers. 2. Carbon dioxide is a more stable molecule, which could lead to fewer complications or side reactions during the process. Disadvantages: 1. The molar masses of the isotopes would be higher in the case of carbon dioxide, which could lead to smaller differences in relative effusion rates, making the separation process more challenging or less efficient. 2. Carbon dioxide might be less reactive with certain materials, making it potentially more difficult to separate the isotopes after the diffusion process. In conclusion, the relative rates of effusion of \({ }^{12} \mathrm{C}^{16}\mathrm{O}\), \({ }^{12} \mathrm{C}^{17}\mathrm{O}\), and \({ }^{12} \mathrm{C}^{18}\mathrm{O}\) are approximately 1:0.965:0.931. While using carbon dioxide instead of carbon monoxide for the separation process has some advantages, it might also lead to a less efficient separation due to smaller differences in relative effusion rates.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -mL nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another \(100.0\) -mL nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

Consider separate \(1.0\) -L gaseous samples of \(\mathrm{H}_{2}, \mathrm{Xe}, \mathrm{Cl}_{2}\), and \(\mathrm{O}_{2}\) all at STP. a. Rank the gases in order of increasing average kinetic energy. b. Rank the gases in order of increasing average velocity. c. How can separate \(1.0\) -L samples of \(\mathrm{O}_{2}\) and \(\mathrm{H}_{2}\) each have the same average velocity?

Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at \(273 \mathrm{~K} ?\) Do all molecules in a 1 -mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at \(546 \mathrm{~K} ?\) Explain.

Which of the following statements is(are) true? For the false statements, correct them. a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules. b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules. c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly. d. As temperature increases, the effect of interparticle interactions on gas behavior is increased. e. At constant \(V\) and \(T\), as gas molecules are added into a container, the number of collisions per unit area increases resulting in a higher pressure. f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and moles of gas.

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(\mathrm{a}\) pressure of \(2.00\) atm. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free