Question

Why is the photodissociation of \(\mathrm{N}_{2}\) in the atmosphere relatively unimportant compared with the photodissociation of \(\mathrm{O}_{2} ?\)

Short answer

The photodissociation of N2 in the atmosphere is relatively unimportant compared to the photodissociation of O2. This is because the energy required to dissociate N2 is higher (\(9.8 \, \mathrm{eV}\)) than that needed for O2 (\(5.2 \, \mathrm{eV}\)). The solar radiation at Earth's atmosphere contains a larger fraction of photons with energy equal to or higher than 5.2 eV, making the photodissociation of O2 more likely to occur than N2, which requires a photon energy of 9.8 eV or higher.

Step-by-step solution

Understand Photodissociation

Photodissociation is a process where a molecule absorbs a photon (a particle of light) and subsequently breaks apart into smaller fragments. In the context of this exercise, we are comparing photodissociation of two molecules, nitrogen (N2) and oxygen (O2).

Energy Required for Photodissociation of N2 and O2

Let's look at the energies required for the photodissociation of N2 and O2 molecules. For nitrogen (N2): The bond dissociation energy of an N-N triple bond in a nitrogen molecule (N2) is about \(942 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}\). For this reaction to occur: \[ \mathrm{N}_{2} + h\nu \longrightarrow 2\mathrm{N} \] the energy required per photon can be estimated by converting this value to electron-volt (eV) using the relation \(1 \, \mathrm{eV} \approx 96.5 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}\): \[ \frac{942 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}}{96.5 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1} \cdot \mathrm{eV}^{-1}} \approx 9.8 \, \mathrm{eV} \] For oxygen (O2): The bond dissociation energy of an O=O double bond in an oxygen molecule (O2) is about \(498 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}\). For the reaction: \[ \mathrm{O}_{2} + h\nu \longrightarrow 2\mathrm{O} \] the energy required per photon can also be estimated by converting this value to electron-volt (eV), which is: \[ \frac{498 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1}}{96.5 \, \mathrm{kJ} \cdot \mathrm{mol}^{-1} \cdot \mathrm{eV}^{-1}} \approx 5.2 \, \mathrm{eV} \]

Comparing Solar Radiation Absorption

The sun emits radiation with a wide range of energies. However, not all of these energies can reach the Earth's atmosphere due to various factors, including absorption and scattering by other atmospheric components. When comparing the energy required for the photodissociation of N2 and O2, N2 requires a more energetic photon (around 9.8 eV) compared to O2 (around 5.2 eV). The solar radiation at Earth's atmosphere consists of a comparatively smaller fraction of photons with energy equal to or higher than 9.8 eV as required for the photodissociation of N2. On the other hand, there is a larger fraction of photons carrying energy equal to or higher than 5.2 eV, sufficient for the photodissociation of O2.

Relative Importance of Photodissociation of N2 and O2

The relative importance of photodissociation of N2 and O2 in the atmosphere can now be explained by considering the energies required for their dissociation and the available solar radiation. Since a much larger fraction of solar radiation consists of photons with energy equal to or higher than the energy required to dissociate O2 (5.2 eV), the photodissociation of O2 is more likely to occur in the Earth's atmosphere compared to N2 (which requires 9.8 eV). This makes the photodissociation of O2 more important in the Earth's atmosphere than the photodissociation of N2.