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Chapter 5: Problem 121

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 disadvantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.

Short Answer

Expert verified
The relative rates of effusion for the given isotopes are: \( \frac{Rate_{(^{12}C^{16}O)}}{Rate_{(^{12}C^{17}O)}} = \sqrt{\frac{29}{28}}\) and \( \frac{Rate_{(^{12}C^{16}O)}}{Rate_{(^{12}C^{18}O)}} = \sqrt{\frac{30}{28}}\) Advantages of using Carbon Dioxide (CO2) instead of Carbon Monoxide (CO) include safer handling due to CO2 being non-toxic and non-flammable. Disadvantages include slower diffusion rates due to higher molar mass and the potential for solid hydrate formation at low temperatures, complicating the process.

Step by step solution

01

Understand 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, it can be written as: \[ \frac{Rate_1}{Rate_2} = \sqrt{\frac{Molar \ Mass_2}{Molar \ Mass_1}}\] Where, Rate_1 and Rate_2 are the rates of effusion of the two gases, and Molar Mass_1 and Molar Mass_2 are their respective molar masses.
02

Calculate the molar masses of the given isotopes

Given isotopes of Carbon Monoxide: \(^{12}C^{16}O\), \(^{12}C^{17}O\), and \(^{12}C^{18}O\) Molar mass of \(^{12}C^{16}O\): 12 (Carbon) + 16 (Oxygen) = 28g/mol Molar mass of \(^{12}C^{17}O\): 12 (Carbon) + 17 (Oxygen) = 29g/mol Molar mass of \(^{12}C^{18}O\): 12 (Carbon) + 18 (Oxygen) = 30g/mol
03

Calculate the relative rates of effusion

Using Graham's law, we will find the relative rates of effusion for the given isotopes. Relative rate of \(^{12}C^{16}O\) and \(^{12}C^{17}O\): \[ \frac{Rate_{(^{12}C^{16}O)}}{Rate_{(^{12}C^{17}O)}} = \sqrt{\frac{29}{28}}\] Relative rate of \(^{12}C^{16}O\) and \(^{12}C^{18}O\): \[ \frac{Rate_{(^{12}C^{16}O)}}{Rate_{(^{12}C^{18}O)}} = \sqrt{\frac{30}{28}}\]
04

Advantages and Disadvantages of using Carbon Dioxide

Let's discuss some advantages and disadvantages of using Carbon Dioxide instead of Carbon Monoxide for the gaseous diffusion process. Advantages: 1. Carbon Dioxide (CO2) is a non-toxic and non-flammable gas, unlike Carbon Monoxide (CO) which is highly toxic and flammable. Hence, handling CO2 is safer than handling CO. Disadvantages: 1. The molar mass of CO2 is higher than that of CO, which could lead to slower diffusion rates. 2. Carbon Dioxide can form solid hydrates at low temperatures, which might complicate the gaseous diffusion process.

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