Problem 16 - Chemistry: The Central Science

Question

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?

Step-by-Step Solution

Answer

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 1: 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

Step 2: 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

Step 3: 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

Step 4: 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 \(55 \mathrm{nm}\).

Key Concepts

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.