Problem 18 - Chemistry The Central Science In Si Units

Question

In \(\mathrm{CH}_{3} \mathrm{I}\) the \(\mathrm{C}-\mathrm{I}\) bond-dissociation energy is \(241 \mathrm{~kJ} / \mathrm{mol}\). In \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}\) the \(\mathrm{C}-\mathrm{I}\) bond-dissociation energy is \(280 \mathrm{~kJ} / \mathrm{mol}\). What is the range of wavelengths of photons that can cause \(\mathrm{C}-\mathrm{I}\) bond rupture in one molecule but not in the other?

Step-by-Step Solution

Answer

The wavelength range is 427 nm to 498 nm.

Step 1 - Recall Bond Dissociation Energy Concept

Bond dissociation energy is the energy required to break a particular type of bond in one mole of gaseous molecules. This energy can be related to the photon energy that can break the bond using the formula: \( E = hu = \frac{hc}{\lambda} \), where \( E \) is energy, \( h \) is Planck's constant (\(6.626 \times 10^{-34} \) J s), \( u \) is frequency, \( \lambda \) is wavelength, and \( c \) is the speed of light (\(3.00 \times 10^8\) m/s).

Step 2 - Convert Energy to Joules

The given bond-dissociation energies are in kJ/mol. Convert them to Joules per bond using the conversion: \( 1 \text{ mol} = 6.022 \times 10^{23} \text{ molecules} \) (Avogadro's number). For \( \mathrm{CH}_3\mathrm{I} \), \( 241 \text{ kJ/mol} \times \frac{10^3 \text{ J}}{1 \text{ kJ}} \div 6.022 \times 10^{23} \approx 4.002 \times 10^{-19} \text{ J} \). For \( \mathrm{C}_6\mathrm{H}_5\mathrm{I} \), \( 280 \text{ kJ/mol} \times \frac{10^3 \text{ J}}{1 \text{ kJ}} \div 6.022 \times 10^{23} \approx 4.65 \times 10^{-19} \text{ J} \).

Step 3 - Calculate Wavelength Using Energy

Use the equation \( E = \frac{hc}{\lambda} \) to solve for \( \lambda \), where \( E \) is the energy per bond calculated previously. For \( \mathrm{CH}_3\mathrm{I} \): \( \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^{8} \text{ m/s})}{4.002 \times 10^{-19} \text{ J}} \approx 498 \text{ nm} \). For \( \mathrm{C}_6\mathrm{H}_5\mathrm{I} \): \( \lambda = \frac{(6.626 \times 10^{-34} \text{ J s})(3.00 \times 10^{8} \text{ m/s})}{4.65 \times 10^{-19} \text{ J}} \approx 427 \text{ nm} \).

Step 4 - Determine Wavelength Range

To break the \( \mathrm{C-I} \) bond in \( \mathrm{CH}_3\mathrm{I} \), a wavelength of about 498 nm or less is required. However, to break the bond in \( \mathrm{C}_6\mathrm{H}_5\mathrm{I} \), the wavelength must be 427 nm or less. Therefore, the range of wavelengths that can break the bond in \( \mathrm{CH}_3\mathrm{I} \) but not \( \mathrm{C}_6\mathrm{H}_5\mathrm{I} \) is from 427 nm to 498 nm.

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon, and it is a crucial concept when studying bond dissociation. A photon is a particle of light that can transfer energy to molecules, breaking chemical bonds if the energy is sufficient. The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. This relationship is encapsulated in the formula: \( E = h u \), where \( E \) is the photon energy, \( h \) is Planck's constant, and \( u \) is the frequency of the photon.
One fascinating aspect of photon energy is that it emphasizes the particle-like properties of light despite light also exhibiting wave-like behavior. This dual nature is fundamental to understanding how photons interact with matter. In chemical reactions, particularly those involving bond dissociation, the right energy photon can excite molecules to transition states, breaking bonds and facilitating new configurations.
When dealing with photon energy in a practical situation such as breaking a specific bond, the right amount of energy must match or exceed the bond dissociation energy. For instance, in the exercise provided, photons need a particular energy level to break the \( \text{C-I} \) bond in both \( \text{CH}_3\text{I} \) and \( \text{C}_6\text{H}_5\text{I} \) molecules. This ensures that the bonds receive enough energetic boost from the photons to dissociate.

Planck's Constant

Planck's Constant is a fundamental physical constant that is key to quantum mechanics and its explanation of energy packets or quanta. Represented by the symbol \( h \), its numerical value is approximately \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \). Planck's constant makes equivalently small connections between energy and frequency possible, forming the bedrock for understanding processes at atomic and molecular scales.
In the equation for photon energy, \( E = h u \), Planck's constant essentially scales the frequency of light into the corresponding energy. The constant is pivotal because it assures that energy interactions, particularly at a quantum level, occur in discrete amounts, shaping our broader comprehension of physical systems.

• It has enabled the development of technologies such as lasers, through its role in the quantization of light energy.
• It plays a role in defining the photon energy necessary to cause reactions like bond dissociation.

When you calculate photon energy using \( E = \frac{hc}{\lambda} \), Planck's constant (\( h \)), combined with the speed of light (\( c \)), helps convert mathematical terms into observable and measurable real-world phenomena.

Wavelength Calculation

Calculating the wavelength of a photon involves the rearrangement of its energy formula, allowing for determination of light characteristics necessary to influence molecules. This calculation links the concepts of energy, frequency, and speed of light through the expression \( \lambda = \frac{hc}{E} \). Here, \( \lambda \) stands for wavelength, \( h \) for Planck's constant, \( c \) the speed of light, and \( E \) photon energy.
This relationship helps us understand what kind of light, in terms of its wavelength, is required to affect molecules in significant ways, such as breaking bonds. The calculation process includes:

• Converting chemical bond energies from \( \text{kJ/mol} \) to \( \text{J/bond} \), which gives us \( E \) in usable form.
• Using the energy form to substitute into the wavelength formula, achieving the desired range of wavelengths.

For instance, in the provided exercise, determining wavelengths between which a photon can break \( \text{CH}_3\text{I} \) bonds but not \( \text{C}_6\text{H}_5\text{I} \) bonds involves calculating the limits of usable wavelengths:
\( \lambda \) for \( \text{CH}_3\text{I} \) is approximately 498 nm and for \( \text{C}_6\text{H}_5\text{I} \) is 427 nm. Thus, the energy must impart within this range to impact the molecules, highlighting the precision of such quantum-scale interactions.