As Albert Einstein realized, light can be considered not only a wave phe- nomenon but also to have particle-like properties in that it is absorbed (or emitted) by matter only in finite packets, now called photons. The quan- tity of energy, E, associated with each photon is related to the frequency, , and the wavelength, , of the light by the formulas
E h or E hc/ since c
Here, h is Planck’s constant (6.626218 1034 J s) and c is the speed of light (2.997925 108 m s1). From the equation, it follows that the shorter the wavelength of the light, the greater the energy it transfers to matter when
Review Questions 1–4 at the end of this chapter refer to the material covered above.
absorbed. Ultraviolet light is high in energy content, visible light is of inter- mediate energy, and infrared light is low in energy. Furthermore, UV-C is higher in energy than UV-B, which in turn is more energetic than is UV-A.
For convenience, the product hc in the equation above can be evaluated on a molar basis to yield a simple formula relating the energy absorbed by 1 mole of matter when each of its molecules absorbs one photon of a particu- lar wavelength of light. If the wavelength is expressed in nanometers, the value of hc is 119,627 kJ mol1 nm, so the equation becomes
E 119,627/ where E is in kJ mol1 if is expressed in nm.
The photon energies for light in the UV and visible regions are of the same order of magnitude as the enthalpy (heat) changes, H°, of chemical reactions, including those in which atoms dissociate from molecules. For example, it is known that the dissociation of molecular oxygen into its mona- tomic form requires an enthalpy change of 498.4 kJ mol1:
O29: 2 O H° 498.4 kJ mol1
In general, we can calculate enthalpy changes for any reaction by recalling from introductory chemistry that for any reaction, H° equals the sum of the enthalpies of formation, Hf°, of the products minus those of the reactants:
H° ∑Hf° (products) ∑Hf° (reactants) In the case of the reaction above,
H° 2 Hf° (O, g) Hf° (O2, g)
From data tables, we find that Hf° (O, g) 249.2 kJ mol1, and we know that Hf° (O2, g) 0 since O2 gas is the stablest form of the element.
By substitution,
H° 2 249.2 0 498.4
To a good approximation, for a dissociation reaction, H° is equal to the energy required to drive the reaction. Since all the energy has to be supplied by one photon per molecule (see below), the corresponding wavelength for the light is
119,627 kJ mol1 nm/498.4 kJ mol1 240 nm
Thus any O2 molecule that absorbs a photon from light of wavelength 240 nm or shorter has sufficient excess energy to dissociate.
O2 UV photon ( 240 nm) 9: 2 O
Reactions that are initiated by energy in the form of light are called photochemical reactions. The oxygen molecule in the above reaction is variously said to be photochemically dissociated or photochemically decomposed or to have undergone photolysis.
In terms of photon energy, UV-C UV-B UV-A visible infrared.
Stratospheric Chemistry: The Ozone Layer 15
Atoms and molecules that absorb light (in the ultraviolet or visible region) immediately undergo a change in the organization of their electrons.
They are said to exist temporarily in an electronically excited state, and to denote this, their formulas are followed by a superscript asterisk (*). How- ever, atoms and molecules generally do not remain in the excited state, and therefore do not retain the excess energy provided by the photon, for very long. Within a tiny fraction of a second, they must either use the energy to react photochemically or return to their ground state—the lowest energy (most stable) arrangement of the electrons. They quickly return to the ground state either by themselves emitting a photon or by converting the excess energy into heat that becomes shared among several neighboring free atoms or molecules as a result of collisions (i.e., molecules must “use it or lose it”).
M photon M photon
reaction M heat M*
Consequently, molecules normally cannot accumulate energy from several photons until they receive sufficient energy to react; all the excess energy required to drive a reaction usually must come from a single pho- ton. Therefore light of 240 nm or less in wavelength can result in the disso- ciation of O2 molecules, but light of longer wavelength does not contain enough energy to promote the reaction at all, even though certain wave- lengths of such light can be absorbed by the molecule (see Figure 1-3). In the case of an O2 molecule, the energy from a photon of wavelength greater than 240 nm can, if absorbed, temporarily raise the molecules to an excited state, but the energy is rapidly converted to an increase in the energy of motion of it and of the molecules that surround it.
O2 photon ( 240 nm) 9: O2* 9: O2 heat
O2 photon ( 240 nm) 9: O2* 9: 2 O or O2 heat
PROBLEM 1-1
What is the energy, in kilojoules per mole, associated with photons having the following wavelengths? What is the significance of each of these wave- lengths? [Hint: See Figure 1-2.]
(a) 280 nm (b) 400 nm (c) 750 nm (d) 4000 nm ●
PROBLEM 1-2
The H° for the decomposition of ozone into O2 and atomic oxygen is 105 kJ mol1:
O39: O2 O PROBLEM1-1
PROBLEM1-2
What is the longest wavelength of light that could dissociate ozone in this manner? By reference to Figure 1-2, decide the region of sunlight (UV, visible, or infrared) in which this wavelength falls. ●
PROBLEM 1-3
Using the enthalpy of formation information given below, calculate the maximum wavelength that can dissociate NO2 to NO and atomic oxygen.
Recalculate the wavelength if the reaction is to result in the complete dis- sociation into free atoms (i.e., N 2 O). Is light of these wavelengths avail- able in sunlight?
Hf° values (kJ mol1): NO2: 33.2; NO: 90.2; N: 472.7; O: 249.2 ● Of course, in order for a sufficiently energetic photon to supply the energy to drive a reaction, it first must be absorbed by the molecule. As you can infer from the examples of the absorption spectra of O2 and O3 (Figures 1-3 and 1-4), there are many wavelength regions in which mole- cules simply do not absorb significant amounts of light. Thus, for example, because ozone molecules do not absorb visible light near 400 nm, shining light of this wavelength on them does not cause them to decompose, even though 400-nm photons carry sufficient energy to dissociate them to atomic and molecular oxygen (see Problem 1-2). Furthermore, as discussed above, just because molecules of a substance absorb photons of a certain wavelength and such photons are sufficiently energetic to drive a reaction does not mean that the reaction necessarily will occur; the photon energy can be diverted by a molecule into other processes undergone by the excited state. Thus the availability of light with sufficient photon energy is a necessary but not a sufficient condition for reaction to occur with any given molecule.