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Chapter 18: Problem 27

The average bond enthalpies of the \(\mathrm{C}-\mathrm{F}\) and \(\mathrm{C}-\) Cl bonds are 485 \(\mathrm{kJ} / \mathrm{mol}\) and 328 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) What is the maximum wavelength that a photon can possess and still have sufficient energy to break the \(\mathrm{C}-\mathrm{F}\) and \(\mathrm{C}-\mathrm{Cl}\) bonds, respectively? (b) Given the fact that \(\mathrm{O}_{2}, \mathrm{N}_{2},\) and \(\mathrm{O}\) in the upper atmosphere absorb most of the light with wavelengths shorter than \(240 \mathrm{nm},\) would you expect the photodissociation of \(\mathrm{C}-\mathrm{F}\) bonds to be significant in the lower atmosphere?

Short Answer

Expert verified
The maximum wavelengths for breaking the C-F and C-Cl bonds are calculated using Planck's equation, \(E = \dfrac{hc}{\lambda}\), with the given bond enthalpies of 485 kJ/mol and 328 kJ/mol, respectively. After calculations, we compare the maximum wavelength of the C-F bond with 240 nm to determine if the photodissociation of C-F bonds is significant in the lower atmosphere. If the maximum wavelength of the C-F bond is larger than 240 nm, the photodissociation would not be significant in the lower atmosphere.

Step by step solution

01

Calculate maximum wavelength for C-F bond

First, we will calculate the maximum wavelength for the C-F bond. We will use the Planck's equation and set the energy equal to the average bond enthalpy: \(E_{CF} = \dfrac{hc}{\lambda_{CF}}\) Given: \(E_{CF} = 485 kJ/mol\) Convert the energy to joules: \(E_{CF} = 485 \times 10^{3} J/mol\) Now, solve for the maximum wavelength (\(\lambda_{CF}\)): \(\lambda_{CF} = \dfrac{hc}{E_{CF}}\)
02

Calculate maximum wavelength for C-Cl bond

Next, we will calculate the maximum wavelength for the C-Cl bond. We will use the same approach, setting the energy equal to the average bond enthalpy: \(E_{C-Cl} = \dfrac{hc}{\lambda_{C-Cl}}\) Given: \(E_{C-Cl} = 328 kJ/mol\) Convert the energy to joules: \(E_{C-Cl} = 328 \times 10^{3} J/mol\) Now, solve for the maximum wavelength (\(\lambda_{C-Cl}\)): \(\lambda_{C-Cl} = \dfrac{hc}{E_{C-Cl}}\)
03

Determine if photodissociation is significant in the lower atmosphere

We'll compare the maximum wavelength of the C-F bond with 240 nm to determine if the photodissociation of C-F bonds is significant in the lower atmosphere. If the maximum wavelength of the C-F bond is larger than 240 nm, then C-F bonds would not absorb enough energy from light to break, and the photodissociation would not be significant in the lower atmosphere.
04

Putting it all together

We have learned that: 1. To find the maximum wavelength that can break a bond, we use the Planck's equation \(E = \dfrac{hc}{\lambda}\), where E represents the bond energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon. 2. We calculated the maximum wavelengths for both the C-F and C-Cl bonds using the provided bond enthalpies. 3. We compared the maximum wavelength for the C-F bond with the cutoff value of 240 nm to determine if the photodissociation would be significant in the lower atmosphere.

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Most popular questions from this chapter

The atmosphere of Mars is \(96 \% \mathrm{CO}_{2},\) with a pressure of approximately \(6 \times 10^{-3}\) atm at the surface. Based on measurements taken over a period of several years by the Rover Environmental Monitoring Station (REMS), the average daytime temperature at the REMS location on Mars is \(-5.7^{\circ} \mathrm{C}\left(22^{\circ} \mathrm{F}\right),\) while the average nighttime temperature is \(-79^{\circ} \mathrm{C}\left(-109^{\circ} \mathrm{F}\right) .\) This daily variation in temperature is much larger than what we experience on Earth. What factor plays the largest role in this wide temperature variation, the composition or the density of the atmosphere?

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