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Ch.18 - Chemistry of the Environment
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Brown 14th EditionCh.18 - Chemistry of the Environment문제 19b
18장, 문제 19b

(b) Use the energy requirements of these two pro- cesses to explain why photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about 90 km.

검증된 단계별 안내
1
Understand the two processes: Photodissociation involves breaking a molecule into two or more parts using light energy, while photoionization involves removing an electron from a molecule or atom using light energy.
Consider the energy requirements: Photodissociation generally requires less energy compared to photoionization because breaking a bond typically requires less energy than removing an electron.
Relate energy requirements to altitude: At lower altitudes, the energy available from sunlight is less intense compared to higher altitudes. Therefore, processes that require less energy, like photodissociation, are more likely to occur.
Explain the significance of altitude: Below 90 km, the intensity of ultraviolet light is sufficient to cause photodissociation but not enough to cause significant photoionization due to the higher energy requirement of the latter.
Conclude why photodissociation is more important: Given the lower energy requirement and the available energy at altitudes below 90 km, photodissociation of oxygen is more prevalent than photoionization.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Photodissociation

Photodissociation is the process by which a molecule absorbs light energy and breaks apart into its constituent atoms or smaller molecules. In the case of oxygen, this process occurs when ultraviolet (UV) light from the sun provides enough energy to overcome the bond energy of O2, leading to the formation of individual oxygen atoms. This reaction is crucial in the upper atmosphere, where UV radiation is abundant.

Photoionization

Photoionization refers to the process where an atom or molecule absorbs enough energy from light to remove an electron, resulting in the formation of a positively charged ion. For oxygen, this process requires significantly more energy than photodissociation. As a result, at lower altitudes where UV radiation is less intense, photoionization becomes less favorable compared to photodissociation.

Altitude and Atmospheric Chemistry

The altitude of the atmosphere affects the intensity and type of solar radiation that reaches different layers. Below 90 km, the atmosphere is denser, and the amount of UV radiation decreases, making photodissociation more prevalent than photoionization. Understanding how altitude influences these processes is essential for explaining the relative importance of photodissociation in the lower atmosphere.
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Chemistry Gas Laws
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교과서 질문

The ultraviolet spectrum can be divided into three regions based on wavelength: UV-A (315–400 nm), UV-B (280–315 nm), and UV-C (100–280 nm). (a) Photons from which region have the highest energy and therefore are the most harmful to living tissue? (315–400 nm), UV-B (280–315 nm), and UV-C (100–280 nm).

교과서 질문

(a) Distinguish between photodissociation and photoionization.

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교과서 질문

The wavelength at which the O2 molecule most strongly absorbs light is approximately 145 nm. (b) Would a photon whose wavelength is 145 nm have enough energy to photodissociate O2 whose bond energy is 495 kJ/mol? Would it have enough energy to photoionize O2?

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교과서 질문

The dissociation energy of a carbon-bromine bond is typically about 276 kJ/mol. (b) Which kind of electromagnetic radiation—ultraviolet, visible, or infrared—does the wavelength you calculated in part (a) correspond to?

교과서 질문

The wavelength at which the O2 molecule most strongly absorbs light is approximately 145 nm. (a) In which region of the electromagnetic spectrum does this light fall?

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