Question

\(\mathrm{O}_{2}\) undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, \(1.2 \mathrm{eV}\) more energetic than normal. The dissociation of \(\mathrm{O}_{2}\) into two normal atoms of oxygen requires \(482.5 \mathrm{~kJ} / \mathrm{mol} .\) The maximum wavelength effective for photochemical dissociation of \(\mathrm{O}_{2}\) is \((1 \mathrm{eV}=96.5 \mathrm{~kJ} / \mathrm{mol})\) (a) \(248 \mathrm{~nm}\) (b) \(1033.3 \mathrm{~nm}\) (c) \(1236.2 \mathrm{~nm}\) (d) \(200 \mathrm{~nm}\)

Short answer

The maximum wavelength effective for photochemical dissociation of \(\mathrm{O}_{2}\) is \(248 \mathrm{~nm}\).

Step-by-step solution

Calculate the Total Energy Required for Dissociation

First, determine the total amount of energy required for the dissociation of one mole of \(\mathrm{O}_2\) into one normal oxygen atom and another oxygen atom that is 1.2 eV more energetic. Given that \(1 \mathrm{eV} = 96.5 \mathrm{kJ/mol}\), the extra energy required can be calculated as: \(1.2 \mathrm{eV} \times 96.5 \mathrm{~kJ/mol}\).

Express the Total Energy in Joules per Mole

The total energy required for the dissociation is the sum of the energy to dissociate into two normal oxygen atoms and the extra energy for the one atom. \(482.5 \mathrm{ kJ/mol} + 1.2 \times 96.5 \mathrm{ kJ/mol}\).

Convert the Energy Required to Joules

Since \(1 \mathrm{kJ} = 1000 \mathrm{J}\), convert the total energy found in Step 2 to joules.

Determine the Energy of a Photon using Planck's Equation

Using Planck's equation, \(E = h u\), where \(h\) is the Planck constant \(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\), and \(u\) is the frequency in s^{-1}. The frequency can also be expressed in terms of the wavelength as \(u = \frac{c}{\lambda}\), where \(c\) is the speed of light \(3 \times 10^8 \mathrm{m/s}\).

Calculate the Maximum Wavelength

Substitute the energy \(E\) obtained in joules from Step 3 into the equation \(E = \frac{hc}{\lambda}\) and solve for the maximum wavelength \(\lambda\).

Convert the Wavelength into Nanometers

Finally, convert the wavelength from meters to nanometers (nm) by multiplying by \(10^9\), recognizing that \(1 \mathrm{m} = 10^9 \mathrm{nm}\).

Key concepts

Planck's Equation

Understanding Planck's equation is crucial when dealing with photochemical reactions. This fundamental piece of quantum mechanics relates the energy of a photon to its frequency. Mathematically, the equation is expressed as \[\begin{equation}E = h u, \end{equation}\]where

• \[\begin{equation}E\end{equation}\] is the energy of a photon (in joules),
• \[\begin{equation}h\end{equation}\] is Planck's constant (approximately \[\begin{equation}6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}),\end{equation}\]
• \[\begin{equation}u\end{equation}\] is the frequency of the photon (in s^{-1}).

Frequency is inversely proportional to the wavelength, meaning that photons with shorter wavelengths have higher energies. In terms of wavelength (\[\begin{equation}\lambda\end{equation}\]), Planck's equation can be written as \[\begin{equation}E = \frac{hc}{\lambda}.\end{equation}\]
This is pivotal for calculating the energy of a photon needed to initiate photochemical dissociation and is used to find the maximum wavelength effective for such a process.

Wavelength of a Photon

The wavelength of a photon is a measure of the distance over which the photon's wave-like properties repeat. It's essential in determining the type of electromagnetic radiation we are observing, from radio waves to gamma rays. In photochemistry, the wavelength is particularly important because it dictates the energy supplied to molecules during the absorption of light.The relationship between the wavelength \[\begin{equation}(\lambda)\end{equation}\] and frequency \[\begin{equation}(u)\end{equation}\] is given by \[\begin{equation}\lambda = \frac{c}{u},\end{equation}\]where \[\begin{equation}c\end{equation}\] is the speed of light (approximately \[\begin{equation}3 \times 10^8 \mathrm{m/s}).\end{equation}\]
When we solve photochemistry problems, we use the wavelength to understand which lights can cause a particular photochemical reaction. Shorter wavelengths mean more energy per photon, hence their ability to break molecular bonds, which is essential when considering photochemical dissociation.

Energy Required for Molecular Dissociation

In photochemical dissociation, a molecule absorbs light and subsequently breaks apart into two or more fragments. The minimum energy required to achieve this process is defined by the bond enthalpy of the molecule.

• First, one must know the energy needed to break one mole of the bonds in question (usually given in \[\begin{equation}\mathrm{kJ/mol}\end{equation}\] or \[\begin{equation}\mathrm{eV}).\end{equation}\]
• Second, if additional energetic atoms are specified, their energy contribution is added to the total energy needed for dissociation.

To calculate this energy, we often convert units and use the equation of reaction enthalpy, adding any extra energy specifying variations in the products (such as the case of one oxygen atom being more energetic than normal). Ultimately, this total energy informs us how much light energy is necessary to initiate the reaction, which, using Planck's equation, translates into a specific wavelength of light.

Photochemistry Problems

Solving photochemistry problems involves a good understanding of the interplay between light and matter. These problems typically focus on reactions initiated or affected by light, such as the photochemical dissociation of \[\begin{equation}\mathrm{O}_2\end{equation}\] being discussed in our example. The main challenges in these problems include \begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}\begin{enumerate}

calculating the energies involved using Planck's equation and other related formulas,

understanding the molecular properties such as bond enthalpies,

and applying this knowledge to determine the effect of particular wavelengths of light on a molecule.

These problems often involve converting units, understanding reaction mechanisms, and knowing how to use constants like Planck's constant and the speed of light correctly.

With practice, students can learn to navigate these challenges effectively, making photochemistry a rich field for studying the interaction between chemistry and physics. The aim is to predict and explain how light leads to molecular changes, which has practical implications in a wide range of scientific fields, including solar energy, environmental chemistry, and the development of photodynamic therapies.

At the heart of solving photochemistry problems is the relationship between light's wavelength, its associated energy, and how this energy impacts molecules, resulting in chemical changes. Through exercises such as calculating the maximum wavelength for photochemical dissociation, students gain hands-on experience in applying physical chemistry concepts to real-world scenarios.