Question

The concentration of \(\mathrm{H}_{2} \mathrm{O}\) in the stratosphere is about \(5 \mathrm{ppm}\). It undergoes photodissociation according to: $$ \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}(g)+\mathrm{OH}(g) $$ (a) Write out the Lewis-dot structures for both products and reactant. (b) Using Table \(8.4,\) calculate the wavelength required to cause this dissociation. (c) The hydroxyl radicals, OH, can react with ozone, giving the following reactions: $$ \begin{array}{l} \mathrm{OH}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{HO}_{2}(g)+\mathrm{O}_{2}(g) \\ \mathrm{HO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{OH}(g)+\mathrm{O}_{2}(g) \end{array} $$ What overall reaction results from these two elementary reactions? What is the catalyst in the overall reaction? Explain.

Short answer

a) The Lewis-dot structures: - H₂O: Oxygen atom forms two single bonds with two hydrogen atoms and has one lone pair. - H: Hydrogen atom has one valence electron and no additional electrons. - OH: Oxygen atom forms a single bond with the hydrogen atom and has two lone pairs. b) The wavelength required for photodissociation can be calculated using the dissociation enthalpy provided in Table 8.4 and the formula: \(\lambda = \frac{h \times c}{E}\), where h and c are Planck's constant and the speed of light, respectively. c) The overall reaction is: OH(g) + O₃(g) + O(g) → OH(g) + 2O₂(g). The catalyst in the overall reaction is OH(g), as it participates in the reaction but is not consumed or changed.

Step-by-step solution

a) Lewis-dot structures

The Lewis-dot structures show the arrangement of valence electrons in molecules. We have to draw them for H2O, H, and OH. 1) Water molecule (H₂O) has two hydrogen atoms and one oxygen atom. - The oxygen atom has six valence electrons and each hydrogen atom has one valence electron. - The oxygen atom forms two single bonds with the two hydrogen atoms, utilizing four electrons. - The remaining two electrons on oxygen form a lone pair. 2) Hydrogen atom (H) has only one valence electron and no additional electrons. 3) Hydroxyl radical (OH) has one hydrogen atom and one oxygen atom. - The oxygen atom has six valence electrons and the hydrogen atom has one valence electron. - The oxygen atom forms a single bond with the hydrogen atom, utilizing two electrons. - The remaining four electrons on oxygen form two lone pairs.

b) Calculate the wavelength

To calculate the wavelength required to cause the photodissociation of H₂O, we should refer to Table 8.4, which must provide the necessary dissociation enthalpy. The dissociation enthalpy can be calculated using the Planck's equation: \[E = h \times \nu\] Where E is the dissociation energy, h is the Planck's constant (6.626 × 10⁻³⁴ J·s), and ν is the frequency of light. The relationship between the wavelength (λ) and frequency (ν) of light is given by: \[\nu = \frac{c}{\lambda}\] Where c is the speed of light (3.0 × 10⁸ m/s). By combining these two equations, we can find the wavelength: \[E = h \times \frac{c}{\lambda}\] Rearrange to solve for λ: \[\lambda = \frac{h \times c}{E}\] Now use the dissociation enthalpy provided in Table 8.4 and plug in the values to calculate the wavelength.

c) Identify the overall reaction and catalyst

We have two elementary reactions: Reaction 1: OH(g) + O₃(g) → HO₂(g) + O₂(g) Reaction 2: HO₂(g) + O(g) → OH(g) + O₂(g) To find the overall reaction, we need to add these elementary reactions and cancel out any intermediate species that appear on both the reactants and products sides. Here, HO₂(g) is the intermediate species. Add Reaction 1 and Reaction 2: OH(g) + O₃(g) + HO₂(g) + O(g) → HO₂(g) + O₂(g) + OH(g) + O₂(g) Cancel out any intermediate species (HO₂): OH(g) + O₃(g) + O(g) → OH(g) + 2O₂(g) Observe that OH(g) appears on both sides of the reaction as a reactant and product. This indicates that OH(g) is a catalyst since it participates in the reaction but is not consumed or changed in the overall reaction.

Key concepts

Lewis-dot structures

Understanding Lewis-dot structures is fundamental for visualizing how atoms share or transfer electrons in a molecule. As the exercise demonstrates, drawing these structures for molecules and radicals such as water (H₂O), hydrogen atom (H), and hydroxyl radical (OH), can illuminate the connective patterns of bonding.

In simple terms, a Lewis-dot structure consists of the element's symbol surrounded by dots, which represent its valence electrons. Elements form bonds to achieve a stable electron configuration, often resembling that of the nearest noble gas. In the water molecule, each hydrogen atom shares its electron with oxygen, forming single covalent bonds. With two lone pairs, oxygen achieves an octet, reflecting the desired electron stability.

For radicals like OH, which play a critical role in reactions such as photodissociation, their Lewis-dot structures show unpaired electrons, indicating their high reactivity. This visualization makes it easier to predict how such reactive species might interact in subsequent chemical reactions.

Wavelength calculation in photodissociation

The wavelength of light necessary to induce photodissociation is a key factor in understanding the process. The exercise points to Table 8.4, which would provide the dissociation enthalpy needed to break the bonds in H₂O.

To calculate this wavelength, we use the Planck's equation in conjunction with the speed of light. It’s a way of linking the energy required to break a molecular bond with the property of light that carries that energy. In doing so, we can see that the wavelength is inversely proportional to the dissociation enthalpy. Shorter wavelengths have higher energy photons capable of initiating photodissociation.

These calculations are not just academic exercises; they're essential for understanding how solar radiation can lead to the formation of reactive species in the Earth's atmosphere, such as the hydroxyl radical (OH), and their implications for atmospheric chemistry and environmental science.

Stratospheric ozone chemistry

The role of hydroxyl radicals in stratospheric ozone chemistry is a complex but crucial topic in atmospheric science. As outlined in the exercise, OH radicals can react with ozone (O₃) to form oxygen (O₂) and other intermediates.

Through a series of reactions, these intermediates help regulate the concentration of ozone, which is vital for protecting life on Earth from harmful ultraviolet radiation. The step-by-step reactions show the interplay of various species in the stratosphere and highlight the delicate balance of chemical reactions occurring high above the Earth’s surface.

The balance between ozone production and destruction is critical, and understanding the chemistry involved is important for assessing the impact of human activities, such as the emission of chlorofluorocarbons, on the ozone layer.

Chemical catalysts

Chemical catalysts play an indispensable part in many reactions, as seen in the example involving hydroxyl radicals and stratospheric ozone. A catalyst is a substance that accelerates a chemical reaction without being consumed in the process.

In the exercise, it becomes clear that the hydroxyl radical (OH) serves as a catalyst in the overall reaction. It shows up at the start and is regenerated at the end, ready to engage in another reaction cycle. Catalysts like OH can dramatically increase the rate of a reaction, leading to significant changes in chemical dynamics, even with very low concentrations.

This concept forms the backbone of many industries and environmental processes, making the understanding of catalysts vital not just for chemistry students, but for anyone interested in the broader implications of chemical reactions in our world.