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In \(\mathrm{CF}_{3} \mathrm{Cl}\) the \(\mathrm{C}-\mathrm{Cl}\) bond- dissociation energy is \(339 \mathrm{~kJ} / \mathrm{mol} .\) In \(\mathrm{CCl}_{4}\) the \(\mathrm{C}-\mathrm{Cl}\) bond-dissociation energy is \(293 \mathrm{~kJ} / \mathrm{mol}\). What is the range of wavelengths of photons that can cause \(\mathrm{C}-\mathrm{Cl}\) bond rupture in one molecule but not in the other?

Short Answer

Expert verified
The range of wavelengths of photons that can cause C-Cl bond rupture in one molecule but not in the other is from \(353 \mathrm{nm}\) to \(408 \mathrm{nm}\) with a range of \(55 \mathrm{nm}\).

Step by step solution

01

Relate bond dissociation energy and photon energy

We can relate the energy of a photon to its wavelength with the equation \(E = hf\) where \(E\) is the photon's energy, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), \(f\) is the frequency of light, and \(c\) is the speed of light (\(3.0 \times 10^{8} \mathrm{m/s}\)). Furthermore, we know that \(c = \lambda f\), where \(\lambda\) is the wavelength. Combining both relations, we have \(E = \dfrac{hc}{\lambda}\). Step 2: Convert bond dissociation energy to photon energy
02

Convert bond dissociation energy to photon energy

First, we need to convert the bond dissociation energies given in kJ/mol to J/photon. We will use the Avogadro's constant (\(N_{A} = 6.022 \times 10^{23} \mathrm{mol^{-1}}\)) to do that. For \(\mathrm{CF}_{3} \mathrm{Cl}\): \(E_{1} = \dfrac{339 \times 10^{3} \mathrm{J/mol}}{6.022 \times 10^{23} \mathrm{mol^{-1}}} = 5.63 \times 10^{-19} \mathrm{J}\) For \(\mathrm{CCl}_{4}\): \(E_{2} = \dfrac{293 \times 10^{3} \mathrm{J/mol}}{6.022 \times 10^{23} \mathrm{mol^{-1}}} = 4.87 \times 10^{-19} \mathrm{J}\) Step 3: Find the corresponding wavelengths
03

Find the corresponding wavelengths

Now, we will use the formula \(E = \dfrac{hc}{\lambda}\) to find the wavelengths corresponding to the photon energies we just calculated. For \(\mathrm{CF}_{3} \mathrm{Cl}\): \(\lambda_{1} = \dfrac{hc}{E_{1}} = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(3.0 \times 10^{8} \mathrm{m/s})}{5.63 \times 10^{-19} \mathrm{J}} = 3.53 \times 10^{-7} \mathrm{m} = 353 \mathrm{ nm}\) For \(\mathrm{CCl}_{4}\): \(\lambda_{2} = \dfrac{hc}{E_{2}} = \dfrac{(6.626 \times 10^{-34} \mathrm{Js})(3.0 \times 10^{8} \mathrm{m/s})}{4.87 \times 10^{-19} \mathrm{J}} = 4.08 \times 10^{-7} \mathrm{m} = 408 \mathrm{ nm}\) Step 4: Find the range of wavelengths
04

Find the range of wavelengths

Finally, we can find the range of wavelengths of photons that can cause C-Cl bond rupture in one molecule but not in the other by subtracting the smaller wavelength from the larger one. Range of wavelengths: \(408 \mathrm{nm} - 353 \mathrm{nm} = 55 \mathrm{nm}\) Thus, the range of wavelengths of photons that can cause C-Cl bond rupture in one molecule but not in the other is from \(353 \mathrm{nm}\) to \(408 \mathrm{nm}\) with a range of \(55 \mathrm{nm}\).

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Key Concepts

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

Wavelength of Photons
The wavelength of a photon is a measure of the distance between consecutive peaks of the electromagnetic wave associated with the photon. It is inversely proportional to the photon's energy, which is why photons with higher energy, such as gamma rays and X-rays, have shorter wavelengths than those with lower energy, like radio waves.

To understand the relationship between energy and wavelength, we can recall the formula,
\[E = \frac{hc}{\lambda}\]
where \(E\) is the photon energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. In the context of bond dissociation, the wavelength of photons necessary to break a chemical bond should be in the range that provides enough energy to overcome the bond dissociation energy. Knowing the specific wavelengths that can cause bond rupture allows chemists to target and break specific bonds in complex molecules using controlled light sources, such as lasers.
Photon Energy
Photon energy is the energy carried by a single photon and can be calculated using its wavelength or frequency. The energy of a photon is fundamental in understanding phenomena such as the photoelectric effect, fluorescence, and even the breaking of chemical bonds.

The energy of a photon can be expressed through the formula,
\[E = hf = \frac{hc}{\lambda}\]
with \(E\) representing photon energy, \(h\) signifying Planck's constant, \(f\) as the frequency, and \(\lambda\) as the wavelength. Photon energy is particularly useful when considering how light can initiate chemical reactions, as specific bond dissociation energies require photons of corresponding energies to be broken.
Planck's Constant
Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. Designated as \(h\), it has a value of approximately \(6.626 \times 10^{-34} \mathrm{Js}\).

Planck's constant appears in the formula for photon energy,
\[E = hf = \frac{hc}{\lambda}\]
This constant is a cornerstone of quantum theory and is crucial for understanding the quantized nature of energy exchange in the atomic and subatomic realms. It is also a part of the calculations for determining the energy required to break chemical bonds, as seen in the exercise on bond dissociation energies.
Avogadro's Constant
Avogadro's constant, denoted by \(N_{A}\), is the number of constituent particles (usually atoms or molecules) in one mole of a substance. Its approximate value is \(6.022 \times 10^{23} \mathrm{mol^{-1}}\).

Avogadro's constant is employed to bridge the gap between the atomic scale and the macroscopic scale. In the context of our problem, it is used to convert bond dissociation energy, which is typically given per mole, to the energy per individual photon. This conversion is crucial to ascertain the energy that individual photons need to have to break a single molecular bond. This constant underscores the immense number of atoms or molecules in just one mole of substance, linking the microscopic world with quantities we can work with in the laboratory.

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Most popular questions from this chapter

A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone: $$\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g)$$ At \(298 \mathrm{~K}\) the rate constant for this reaction is \(4.8 \times 10^{5} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Based on the units of the rate constant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not. (c) From the magnitude of the rate constant, would you expect the activation energy of this reaction to be large or small? Explain. (d) Use \(\Delta H_{f}^{\circ}\) values from Appendix \(C\) to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

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