Question

In \(\mathrm{CH}_{3} \mathrm{I}\) the \(\mathrm{C}\) - I bond-dissociation energy is \(241 \mathrm{~kJ} / \mathrm{mol}\). In \(\mathrm{C}_{6} \mathrm{H}_{5}\) I the \(\mathrm{C}-\) 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?

Short answer

The range of wavelengths of photons that can cause C-I bond rupture in one molecule but not in the other is given by: \( \lambda_{range} = (\lambda_{CH_{3}I}, \lambda_{C_{6}H_{5}I}) \) Where: \( \lambda_{CH_{3}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{241 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \) \( \lambda_{C_{6}H_{5}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{280 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \)

Step-by-step solution

Convert bond dissociation energies to J/molecule

First, we need to convert the given bond dissociation energies from kJ/mol to J/molecule. We will use Avogadro's number (6.022 x 10^23) for this purpose. For CH3I: \( 241 \ \frac{kJ}{mol} * \frac{10^3 J}{1 kJ} * \frac{1 mol}{6.022\times10^{23} \ molecules} \) For C6H5I: \( 280 \ \frac{kJ}{mol} * \frac{10^3 J}{1 kJ} * \frac{1 mol}{6.022\times10^{23} \ molecules} \)

Calculate energy of the photons for both molecules

Now that we have the energy of the bonds, we can find the photon energy responsible for breaking the bond. This can be done using Planck's equation: \( E = h\nu \) Where h is Planck's constant, \(6.626\times10^{-34} \ Js\), and ν is the frequency of the photon. Since the frequency (ν) and the wavelength (λ) are related by the speed of light (c) as follows: \( \nu = \frac{c}{\lambda} \) We can rewrite the Planck's equation as: \( E = hc \times \frac{1}{\lambda} \) Now, we can find the wavelength (λ) for each molecule by rearranging the equation: \( \lambda = \frac{hc}{E} \) Substitute the energy and Planck's constant values for both molecules: For CH3I: \( \lambda_{CH_{3}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{241 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \) For C6H5I: \( \lambda_{C_{6}H_{5}I} = \frac{6.626 \times 10^{-34} Js \cdot 3\times10^8 \frac{m}{s}}{280 \times 10^3 \frac{J}{mole} \times \frac{1 mole}{6.022\times10^{23} \ molecules}} \)

Determine the range of wavelengths

Now that we have found the wavelengths that cause the bonds to rupture, we can find the range of wavelengths that can cause bond rupture in one molecule but not in the other. The range of wavelengths is between the wavelengths calculated for CH3I and C6H5I: \( \lambda_{range} = (\lambda_{CH_{3}I}, \lambda_{C_{6}H_{5}I}) \) This range represents the photons that can cause bond rupture in one molecule but not in the other. Make sure to compare the values obtained for each molecule and order them accordingly to find the correct wavelength range.

Key concepts

Photon Energy

Photon energy is a crucial concept in understanding how light interacts with matter, like molecules. When we talk about photon energy, we are referring to the energy carried by a single photon, which is a tiny packet of light. This concept is vital in chemical reactions such as bond dissociation, where photons provide the energy needed to break a chemical bond.

Imagine light as a stream of these photons. Each photon carries a specific amount of energy, which is determined by its frequency. Higher frequency photons have higher energy, while lower frequency photons have lower energy. This differentiation is significant in determining which photons can break certain chemical bonds.

The energy of a photon is expressed in units like joules (J), and it is this energy that we calculate when trying to understand how molecules like \( ext{CH}_3 ext{I}\) and \( ext{C}_6 ext{H}_5 ext{I}\) behave under light exposure. Knowing the energy is the first step toward calculating the range of wavelengths that can affect the molecules.

Planck's Equation

Planck's Equation connects the energy of a photon with its frequency. This equation is fundamental in the field of quantum mechanics and is expressed as:

\( E = h u \)

Here, \(E\) represents the energy of the photon, \(h\) is Planck's constant \(6.626 \times 10^{-34} \ ext{Js}\), and \(u\) is the frequency of the photon. This relationship helps us understand how changing the frequency affects photon energy.

Planck's constant is a fundamental value that remains unchanged and acts as the proportionality constant between energy and frequency. This equation provides a bridge from the abstract quantum world into practical applications, such as determining if a photon's energy is sufficient to break a specific chemical bond.

Understanding Planck's Equation allows chemists and physicists to predict how light will interact with materials. By rearranging this equation using the relationship between frequency and wavelength, we can calculate the wavelengths needed for certain chemical processes.

Wavelength Calculation

Wavelength calculation is essential in establishing which particular photons will interact with a given molecule. After determining the photon's energy using Planck's Equation, we can find the wavelength \(\lambda\) using the following relation derived from the speed of light equation:

\[ \lambda = \frac{hc}{E} \]

Here, \(c\) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\). This equation highlights the inverse relationship between wavelength and energy: as energy increases, wavelength decreases, and vice versa.

Using the formula, after substituting known values such as Planck's constant and the calculated energy for the photons that affect the molecules \(\text{CH}_3\text{I}\) and \(\text{C}_6\text{H}_5\text{I}\), we determine their corresponding wavelengths.

The calculated wavelengths indicate the range of light that can be used to break the chemical bond. For an exercise focusing on bond dissociation energy, finding this wavelength helps identify the appropriate light sources or lasers to use in experiments or industrial processes. Thus, understanding wavelength calculation bridges the gap from theoretical equations to practical implementations.