The Physics of Energy, page 38
While thermal ionization is not relevant for most terrestrial systems, photons – electromagnetic radiation – with energy above 13.6 eV can be found in nature or produced by artificial means. By Planck's relation, , 13.6 eV corresponds to a wavelength of 91.2 nm, which is well into the ultraviolet region. About 5 parts per billion of the energy of sunlight (as it arrives at the top of the atmosphere) is carried by photons with energies above 13.6 eV, so hydrogen and other atoms high in Earth's atmosphere can be ionized by sunlight. Also, medical X-rays and environmental radiation from cosmic rays and from radioactive nuclei often include energetic photons that can ionize atoms. In fact, for the purposes of discussions of radiation safety (§20), electromagnetic radiation is divided according to whether it is energetic enough to ionize atoms. Ionizing radiation, usually defined as radiation whose quanta have energy greater than roughly 20 eV, is known to be an environmental hazard, while lower-energy (longer wavelength) radiation is believed to be less of a concern.6
It takes 13.6 eV to remove hydrogen's one electron, but much more energy is required to completely ionize heavier elements like oxygen. The binding of a single electron to a nucleus with charge Z is , so if seven of the eight electrons in oxygen were removed, the last electron would be bound by eV, well into the domain of X-rays (Figure 9.2). The energies required to ionize each of the eight electrons in oxygen are listed in Table 9.5. The first (outermost) electron is the easiest to remove, with an ionization energy close to that of the single electron in hydrogen. The binding energy grows quickly with successive ionizations. If we add up the energy necessary to ionize the two hydrogen atoms and one oxygen atom in HO, we find7
(9.8)
The electromagnetic radiation associated with ionization ranges from about 4 eV/photon, which can ionize the most weakly bound electron in the cesium atom and corresponds to a wavelength of 300 nm, up to the ionization energy of the innermost electron in a heavy element like uranium, which requires a jolt of over 100 keV, corresponding to a wavelength of 1/100 nm, at the upper limit of X-rays.
Table 9.5 Ionization energies for the eight electrons in the oxygen atom.
Electron Ionization energy (kJ/mol) eV/molecule
1st 13 149 13.62
2nd 3 388 35.12
3rd 5 301 54.93
4th 7 469 77.41
5th 10 990 113.9
6th 13 327 138.1
7th 71 330 739.4
8th 84 090 871.4
Ionization Energies
Removing the electrons from an atom takes energies in the range of 5–100 keV. The outermost electron is most easily removed, and the last electron is the hardest, with a binding energy of eV. At atmospheric pressure a significant fraction of the hydrogen and oxygen atoms from water are ionized at temperatures exceeding 10 000 K.
Nuclear Binding
The energy necessary to remove a proton or neutron from a typical nucleus is roughly 8 MeV. This sets the scale for the energy available per nucleus in nuclear reactions. To thermally separate the protons and neutrons in the oxygen nuclei from water, a temperature on the order of K would be needed.
9.3.4 Nuclear Binding and Beyond
The binding of protons and neutrons into nuclei is a subject for a later chapter (§17), but for completeness we mention it here. Protons and neutrons are bound by about 8 MeV per particle in a typical nucleus with many protons and neutrons.8 The binding energy per particle increases with the number of protons Z in the nucleus up to , which is iron ( – 26 protons and 30 neutrons), and then decreases slowly thereafter. So it is energetically favorable for nuclei with Z significantly below 26 to combine (fuse) together and for nuclei with Z significantly above 26 to break apart (fission). These processes play important roles in nuclear power systems described in later chapters. The total binding energy of oxygen – the total energy given off if 8 protons and 8 neutrons could be brought together to form – is J/mol. Although this is not a practical nuclear reaction, it gives some idea of the immense energies available when nuclear processes are involved. The temperature needed to thermally separate an oxygen nucleus into its component protons and neutrons is around K. Such temperatures occurred in the very early moments of our observable universe, less than one second after the big bang.
At even higher temperatures, the neutrons and protons themselves dissociate into more fundamental objects called quarks. And at even higher temperatures, yet more fundamental constituents may be revealed that are at present unknown. Some of the concepts of subnuclear phenomena are introduced in §14 and §21. But for the purposes of terrestrial energy applications, all relevant aspects of internal energy have been explored by the time the ice has dissociated into individual neutrons, protons, and electrons.
Finally, the ultimate energy content of matter is dictated by Einstein's famous relation, . If we could somehow liberate all its mass as energy, one gram of matter would emit of energy. We cannot liberate these energies, however, in any practical way, because all matter (at least in our part of the universe) is made using particles like neutrons and protons, and there is no observed naturally occurring (anti-)matter composed of antineutrons or antiprotons. Furthermore, the total number of baryons (§14), including neutrons, protons, and some other exotic types of matter, is conserved. So there is no way to produce antimatter without producing corresponding matter at the same time. Thus, as far as we know, protons and neutrons cannot be made to disappear into pure energy without using at least as much energy to produce antiprotons and antineutrons in the first place. In particular, no matter how much we heat the block of ice, at least at temperatures that can be understood using existing physical theories, there is no point at which the matter composing the ice will completely annihilate into pure energy.
9.4Internal Energy, Enthalpy, and Free Energy in Chemical Reactions
It is time to return to earth, so to speak, and explore in further detail the flow of energy and entropy in matter in the course of chemical reactions. Our main aims here are to explain how to compute the energy released or absorbed in a chemical reaction and to establish the criteria that determine whether a chemical reaction can proceed spontaneously.
The concepts of enthalpy (§5.5) and free energy (§8.8) were originally developed to describe the flow of energy in chemical reactions. In most cases of interest to us, chemical reactions occur at fixed pressure, so the reactants can do pdV work during the reaction. This means that some of the heat that flows in or out of the reactants goes into pdV work and the rest goes into the internal energy of the system. Heat equates to enthalpy under these conditions. This explains some common terminology: chemists refer interchangeably to “enthalpy of reaction” and “heat of reaction.” We use the former term throughout. The enthalpy of reaction is thus defined to be the amount of energy that is absorbed (if ) or released (if ) as heat in the course of a reaction that takes place at constant pressure. Since enthalpy is a state function, also refers to the change in value of the enthalpy of the system as it goes through the reaction.
The entropy audit for a reaction that takes place at temperature T must include both the change in entropy of the reactants and the entropy delivered to the surroundings, . The total entropy change of the system and environment combined is related to the change in the Gibbs free energy (eq. (8.68))
(9.9)
The sign of determines whether the reaction can occur spontaneously or not. The Gibbs condition states that if is negative, then the entropy of the reactants plus the environment increases and the reaction may occur spontaneously. If is positive, it cannot.9
9.4.1 Enthalpy and Free Energy of Reaction: An Example
To make the discussion concrete, we first look at a specific example of a simple chemical reaction. In the next section (§9.4.2) we set up a general approach. The reaction we choose to follow is known as calcination.10 In this reaction, limestone (calcium carbonate = CaCO) is heated to drive off carbon dioxide, leaving behind quicklime (calcium oxide = CaO). This reaction is one of the essential steps in making cement, which is in turn used in the production of concrete. The enormous amount of concrete produced worldwide is currently responsible for about 5% of all the CO added to the atmosphere by human activity; roughly 60% of this CO can be attributed to calcination in concrete production.
The first step in analyzing calcination is to write a correctly balanced chemical equation,
(9.10)
Table 9.6 gives all of the relevant thermodynamic information about this reaction. Let us examine the implications of the numbers in this table. First, U, H, G, and S are all state functions. Thus, the table lists the changes in the properties of the system. For example, the enthalpy of the system increases by 178.3 kJ/mol when this reaction takes place, which makes eq. (9.10) an endothermic reaction: heat must be supplied to make it go.
Table 9.6 Thermodynamic information about the reaction CaCO CaO +CO (at T = 25℃ and 1 atm).
Quantity Value
Reaction internal energy +175.8 kJ/mol
Reaction enthalpy +178.3 kJ/mol
Reaction entropy +160.6 J/mol K
Reaction free energy +130.4 kJ/mol
Next, since is positive, the reaction does not occur spontaneously. In fact, the reverse reaction does occur spontaneously: quicklime, left in the presence of CO, will slowly absorb it, forming calcium carbonate.11 Note that is positive. The entropy of the reactants increases in this reaction, so one might have thought it would go spontaneously. However so much heat is taken out of the environment that , given by eq. (9.9), is negative and the reaction does not occur spontaneously at 25℃.
Finally, note that the variables are related (Problem 9.10) by (1) and (2) . The last equality uses the ideal gas law, , to replace by . Thus, when one mole of limestone undergoes calcination, mole of CO is emitted.
How can we use the numbers from Table 9.6? First, they tell us the energy required to drive this reaction. A heat input of 178.3 kJ/mol is needed to drive the CO out of limestone. Conversely, if quicklime reacts with carbon dioxide to make calcium carbonate, 178.3 kJ/mol is given off.12 Calculations such as this enable us to compute the energy consumption of manufacturing processes and compare the energy content of different fuels.
Another extremely useful application of the information in Table 9.6 is to estimate the temperature to which limestone must be raised to allow this reaction to occur spontaneously. Notice that the second term in depends explicitly on the temperature, and is positive. Thus, as the temperature increases, decreases, and at a high enough temperature, is negative and the reaction can occur spontaneously. This is an essential feature of calcination reactions and other reactions of the form solid solid gas. When a gas is produced from a solid, the reaction entropy change is typically positive. This means that, even when the reaction is endothermic, raising the temperature enough may drive the free energy of reaction to be negative, so that the reaction can occur spontaneously.
Reaction Enthalpy and Free Energy
Reaction enthalpy determines how much heat is given off () or absorbed () during a constant pressure chemical reaction.
Reaction (Gibbs) free energy is related to the total entropy change in a reaction (at constant pressure). When , the entropy of the reactants plus environment increases and a reaction can (but may not) proceed spontaneously; if , the reverse reaction may proceed spontaneously.
The crossover temperature at which a reaction can proceed spontaneously occurs when . To estimate for calcination we first assume that and do not change significantly with temperature. Then, setting , we find K ℃. When the T dependence of and is included, the precise value of is closer to 890℃ [52]. Thus, we can expect the calcination reaction to occur when limestone is heated above ℃.
Making quicklime from limestone generates CO in several ways. Since the reaction is endothermic, kJ/mol must be supplied to the limestone, usually by burning a fossil fuel. The reactants must furthermore be heated above 890℃ for the reaction to take place, again requiring heat from fossil fuels (though in principle this energy could be recaptured as the quicklime and CO cool, and could be reused as heat energy as is done in cogeneration (§13.5.3)). Finally, the reaction itself releases one molecule of CO for each molecule of CaO created. Given the huge amounts of concrete used in the modern world, it is no wonder that 5% of anthropogenic CO comes from making it.
9.4.2 Hess's Law and Enthalpy/Free Energy of Formation
The example of making quicklime from limestone illustrates the importance of the enthalpy and free energy of reaction. Chemical reactions are ubiquitous in energy-related applications, such as the refining of ores to obtain metals, the manufacture of fertilizer, and the refining of crude oil into more useful products. Combustion, however, is particularly important, since it is the principal way in which we harvest energy from fossil fuels (and biofuels). In most familiar circumstances, the process of combustion is burning a substance in air. More generally, combustion pertains to a class of exothermic reactions in which a compound reacts completely with an oxidant.13 We deal here with combustion primarily in the context of organic compounds reacting with oxygen. In an idealized complete combustion reaction of this sort, every carbon atom in the original compound ends up in CO, every hydrogen atom ends up in HO, and every nitrogen in N. Note that nitrogen is, in general, not oxidized, and the fate of other, less common elements must be specified if it is important. In practice, combustion is usually not complete. Some carbon and other elements may be left as ash, and some carbon may be partially oxidized to carbon monoxide (CO).
Chemists do not tabulate and for every possible chemical reaction. Instead, for all compounds of interest, they tabulate the enthalpy of formation and the free energy of formation, denoted and . These quantities are defined using the specific reaction that builds up a chemical compound out of its constituents. For any more general reaction, and can be used to compute and , using the fact that H and G are state functions and therefore depend only on the state of a substance and not on how it is made.
The formation reaction for a given compound begins with the constituent elements in the form that they are found at standard values of temperature and pressure, usually 25℃ and 1 atm. Elements that are gases under these conditions are assumed to be provided in the gaseous state. Only bromine and mercury are liquids at 25℃ and 1 atm. Note that gases that are found as diatomic molecules are taken in the molecular state. This applies to O, H, N, Cl, etc. Table 9.7 lists a few formation reactions in standard form.
Table 9.7 Examples of formation reactions. These reactions illustrate the conventions of formation reactions: elements are assumed to be supplied in the phase (solid, liquid, or gas) that is stable under standard conditions and in the most stable molecular form. In particular, those that form diatomic molecules, O, H, N, F, Cl, Br, and I, are assumed to be supplied as such.
Compound Formation reaction
Water H(g) O(g) HO (l)
CO C O(g) CO(g)
Methane C 2H(g) CH(g)
Sodium bromide Na Br(l) NaBr(s)
Glucose 6C + 6H(g) 3O(g) CHO(s)
Notice that the reaction equations in Table 9.7 are normalized to produce exactly one molecule of the desired compound. This may require, for example, reacting “half a molecule” of a diatomic gas. Note also that the state (solid, liquid, gas) of the resulting compound must be indicated. The enthalpy of formation for a compound in a gaseous state is greater (less negative) than a liquid because some extra pdV work must be done when the gas is formed. Compare, for example, the enthalpies of formation of liquid and gaseous ethanol, or of water liquid and vapor, in Table 9.8, where a more extensive list of and for energy-related chemical compounds is given.
Given the enthalpies of formation for various compounds, the strategy for computing the enthalpies of reaction is simple. Because enthalpy is a state function, the enthalpy of reaction for a given reaction does not depend upon how the reaction actually occurs. We can choose to view the reaction as a two-step process in which the reactants are first taken apart into their constituent elements, and then these elements are formed into the products. The reaction enthalpy, therefore, is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants,
(9.11)
This is the essence of Hess’s law. Example 9.1 illustrates Hess’s law for the calcination reaction considered above. The free energy of reaction can be computed in a precisely analogous fashion.
Example 9.1 Hess’s Law Example: Calcination
Consider again the decomposition of limestone into CaO and CO:
We can view this as a sequence of two steps. In the first step, CaCO is “unformed,” with enthalpy of reaction equal to minus its enthalpy of formation (Table 9.8),
In the second step, the products are combined to make CaO and CO2,
Even though these are impractical reactions, we can imagine making CaO and CO from CaCO this way, and that is all that is necessary to compute the enthalpy of reaction. The overall heat of reaction is clearly the sum of the enthalpies of reaction for the three reactions above,
