## Photochemistry

Further examples of endothermic reactions include the photochemical reactions. In this case the additional source of energy necessary for the endothermic reaction to proceed is solar radiation which can break the chemical bonds of atmospheric species. In this book we will consider only one photochemical process: photodissociation.1

Physics refresher Solar radiation consists of electromagnetic waves. Electromagnetic radiation has a dual wave-particle nature. This means that electromagnetic radiation exhibits both wave-like and particle-like properties. In its wave form electromagnetic radiation can be thought of as a group of superimposed waves sometimes referred to as an ensemble propagating in vacuum with the speed of light c = 2.998 x 108 m s-1 independent of wavelength. Each wave in this ensemble can be treated as a simple sinusoidal function (see Figure 8.1) with a certain wavelength, frequency, and amplitude. The wavelength, X, is the distance between two successive peaks of the wave. The units of X are meters. The frequency of a wave, f, is the number of cycles that pass an observer in a second. The unit of frequency is the hertz (1 Hz is one oscillation per second). The product of wavelength and frequency for an individual wave is equal to the speed of light (speed is distance divided by time): c = X x f. From this equation one can see that waves with higher frequencies have shorter wavelengths, and waves with lower frequencies have longer wavelengths.

When radiation interacts with atoms or molecules, it can be absorbed or emitted only by certain discrete amounts of energy. In other words, electromagnetic radiation is quantized. The waves may be thought of as a beam of particles called photons carrying discrete amounts or packages of energy. The energy of a photon of

1 Interested readers are referred to Basic Physical Chemistry for the Atmospheric Sciences by Peter V. Hobbs (2000) for more information on photochemical reactions.

Visible (400-700 nm)

1010

f 108

Gamma rays

X-rays

Ultraviolet

Infrared

Microwaves

### Radio waves

10 14 = 10-5nm 10- 13 = 10^nm 10 12 = 10-3nm 10- 11 = 10-2nm 10- 10 = 10- 1nm 10-9= 1 nm 10-8= 10nm 10-7= 100 nm 10-6= 1 |im 10-5= 10 |im 10-4= 100 |im „ 10-3=1mm JÏ

105=100km 106 = 1000 km

Figure 8.2 Electromagnetic spectrum.

frequency f is

where h is Planck's constant, h = 6.62 x 10 -34 J s. If we express f in terms of X we obtain hc

Hence, the shorter wavelength (higher frequency) photons are more energetic. The electromagnetic spectrum from radio waves to gamma rays is illustrated in Figure 8.2. The most energetic photons are gamma rays. As one moves vertically down the spectrum from gamma rays towards radio waves, the energy decreases and so does the frequency, while the wavelength increases. A narrow band of the spectrum corresponds to the visible light. Photons in the visible range (approximately 400-700 nm) can be detected by a human eye. Ultraviolet radiation has shorter wavelength (higher frequency) than the visible part of the spectrum.

To describe the radiation penetrating the atmosphere it is useful to introduce the idea of an energy flux. The energy flux (energy passing per unit area perpendicular to the beam, per unit time) is given by

where n0 is the number of photons per unit volume (number density as in a gas). The energy flux of solar photons at the top of the atmosphere is 1370 W m—2. This parameter is called the solar constant. Solar photons propagating through the atmosphere can be absorbed and/or scattered by atmospheric constituents. Consider the attenuation of a photon flux at wavelength X due to photon absorption assuming normal incidence for simplicity (the sun is at zenith, directly overhead (Figure 8.3)). We denote a photon energy flux at wavelength X as FX; its dimension is energy per unit area, per unit time, and per unit wavelength. If at the top of the atmosphere the flux per unit wavelength is FX(top), the flux at height z, FX(z), is described by

Fx(z) = Fx (top) exp(-r(z)) [attenuation of a vertical solar beam]. (8.12)

fsi(top)

Figure 8.3 Schematic diagram of a solar beam coming from directly overhead with attenuation of the beam's intensity indicated by shading.

This equation follows from the radiative transfer theory. 2 The coefficient t in the radia the exponent is called the optical depth:

The optical depth t is proportional to the vertically integrated column density f™ N (z)dz where N is a concentration of atmospheric species absorbing at X. The integration from z to to reflects the path the photons travel from the top of the atmosphere to height z. The coefficient of proportionality a is called the absorption cross-section. This parameter describes the ability of a particular gaseous species to absorb photons; it is measured in m2 (often in the literature as cm2). Absorption cross-sections can be measured in the laboratory. When the optical depth gets close to unity, the flux is attenuated by a factor of roughly three (e ~ 2.7). For example, for X between 240 and 300 nm (ultraviolet range) t reaches unity due to the absorption by ozone approximately at heights of 30-38 km. This means that the solar photons in this range are absorbed by ozone in the stratosphere and do not reach the troposphere. At shorter wavelengths, between 175 and 200 nm, radiation is absorbed by oxygen at heights of 40-80 km. At wavelengths greater than 310 nm, most photons penetrate into the troposphere and reach the surface. If the sun has zenith angle 0 = 0, then cos 0 has to be added in the formula for the flux attenuation (Figure 8.4):

Fx (z) = Fx (top) exp(-t(z)/cos 0) [attenuation at zenith angle 0]. (8.14)

The larger the zenith angle, the stronger the attenuation at a given height z.

The photons in the ultraviolet and visible ranges are energetic enough to break molecules apart. This process is called photodissociation. Photodissociation plays a very important role in the troposphere and the stratosphere. For example, a key reaction in the troposphere is the photodissociation of ozone by ultraviolet radiation:

where the notation hf denotes a photon with frequency f. This notation emphasizes that the energy carried by the photon is the frequency times Planck's constant. The formation of tropospheric ozone is due to photodissociation of NO2:

2 A beam is attenuated in a distance interval dz by an amount proportional to the incoming beam's flux and to the amount of attenuating material in the interval. The result is dFx = -AFx dz where A is proportional to the amount of attenuating material per unit volume. Integration of this equation leads to exponential decay along the path, known as Beer's Law.

The atomic oxygen then leads to the formation of ozone by recombination with O2:

The formation of the ozone layer is also caused by photodissociation. In the stratosphere, ultraviolet radiation with X < 240 nm photodissociates molecular oxygen O2 creating atomic oxygen:

Ozone is then formed by recombination of atomic and molecular oxygen (reaction (8.17)).

Example 8.3 Consider the photodissociation of an oxygen molecule that creates two ground state oxygen atoms:

What photon energy is required for this reaction to proceed? What part of the electromagnetic spectrum corresponds to this energy?

Answer: First let us examine the standard enthalpy of this reaction:

AH° = 2 x 249.17 - Na hf = 498.34 kJ mol-1 - Na hf (8.20)

where NA is Avogadro's number. To find the minimum energy of a photon required to break one O2 molecule we have to equate the standard enthalpy of this reaction to zero. Only photons with energy hf greater than this minimum energy are able to break an O2 molecule apart: hf > 498.34/NA kJ. This inequality gives the value of the smallest frequency required: f > 498.34 x 103/(6.022 x 1023 x 6.62 x 10-34) = 12.49x 1014 Hz, or the largest wavelength. X < 0.24x10-6m = 240 nm. Therefore, radiation with X < 240 nm, which corresponds to the ultraviolet part of the spectrum, is needed for the reaction (8.19) to proceed. □

We can find the energy of a photon necessary for a certain reaction to proceed without examination of the enthalpy, if we know the energy of dissociation of a chemical bond.

Example 8.4 During the daytime an important source of NO in the stratosphere is the dissociation of NO2 molecules:

Find the maximum wavelength of electromagnetic radiation required for this reaction, if the energy of dissociation of an NO2 molecule is 5.05 x 10-19J. Energy of dissociation is often given in electronvolts:3

(this reaction is also important in polluted urban air, since it is a source of tropospheric ozone).

Answer: Photons with energy hf > 5.05 x 10-19 J are needed to dissociate an NO2 molecule. Then, f > 5.05 x 10-19/(6.62 x 10-34) = 7.6 x 1014 Hz. Finally, X < 2.998 x 108/7.62 x 1014 = 0.39 x 10-6m = 390 nm, which is at the boundary between the visible and ultraviolet parts of the spectrum. □

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