The Physics of Energy, page 39
which agrees with the data quoted in Table 9.6. This gives an explicit example of Hess’s law (9.11). The same reasoning applies to the calculation of free energies of reaction.
Hess’s Law
The reaction enthalpy for any reaction is equal to the sum of the enthalpies of formation of the products minus the enthalpy of formation of the reactants,
9.4.3 Higher and Lower Heats of Combustion
Combustion reactions must frequently be considered in comparisons of fossil fuels with one another and with renewable energy sources. There is a small, but sometimes confusing, complication regarding the enthalpy of reaction in combustion reactions: when water is produced in a combustion process, the enthalpy of combustion depends upon whether the water ends up as a vapor or a liquid. The difference may be significant because the latent heat of condensation of water is large and because less pdV work is done when water is produced as a liquid than as a vapor at fixed pressure. Both of these effects work in the same direction: the enthalpy of reaction (combustion) is larger – more heat is given off – when the produced water is a liquid rather than as a gas.
The higher heating value (HHV) or gross calorific value of a fuel is defined by the US Department of Energy [53] as “the amount of heat released by a specified quantity (initially at 25℃) once it is combusted and the products have returned to a temperature of 25℃.” Thus defined, the HHV is the negative of the reaction enthalpy at 25℃ and 1 atm, and therefore includes the latent heat of condensation of the water, which is a liquid under those conditions.
The lower heating value (LHV) or net calorific value of a fuel is a measure of the heat released by combustion when the water vapor is not condensed. Unfortunately, there is more than one definition of LHV in common use. The one used by most sources, which we adopt as well, is the same as HHV except that the water is taken to be a vapor at 25℃. The difference between LHV and HHV is, then, exactly the enthalpy of vaporization of the produced water at 25℃. An explicit calculation of HHV and LHV is worked out in Example 9.2.14 Which to use, HHV or LHV, in a given situation depends on the circumstances and on convention. In an internal combustion engine, for example, condensation of water generally takes place outside of the engine and cannot contribute to useful work, thus it seems logical to use the LHV; in principle, however, the additional enthalpy of condensation of water could be captured by a more sophisticated system. HHV is often used in any case for motor vehicle fuels, and we follow this convention here since this is really the theoretical maximum of energy that could be extracted from the combustion process. The HHV may well be appropriate for an external combustion engine.
Example 9.2 HHV and LHV of Methane
Compute the HHV and LHV for combustion of methane (CH), the dominant component of natural gas, and show how the two heating values are related.
HHV: The chemical reaction is CH 2O CO(g) 2HO(l), where we have denoted that the water is liquid. The standard enthalpies of formation of CH, CO, and HO(l) (at 1 atm and 25℃) are kJ/mol, kJ/mol, and kJ/mol, respectively. The HHV is obtained from Hess’s Law: kJ/mol. So HHV 890 kJ/mol. Note that enthalpies of combustion are generally negative, as energy is released.
LHV: Now assume that the water is created as a vapor. The enthalpy of formation of water vapor (at 25℃) is kJ/mol, so the enthalpy of reaction (still at 25℃ and 1 atm) is kJ/mol.
The difference between the HHV and HLV should be the latent heat of condensation of two moles of water at 25C, which is measured to be . This accounts exactly for the 88.0 kJ/mol difference between the HHV and LHV of methane, as it must.
Note that the difference between the LHV and HHV is a 10% effect, which is not insignificant.
9.5Chemical Thermodynamics: Examples
The fundamentals of chemical thermodynamics introduced in this chapter shed light on many issues that arise when we consider sources of energy and energy systems in subsequent chapters. Here we illustrate the power of these tools with three practical examples, addressing the following questions: first, how does the energy content of ethanol compare to that of gasoline? Second, how does the commonly used energy unit a tonne of TNT get its value of GJ? And third, how are metals refined from oxide ores? Other applications can be found among the problems at the end of the chapter.
9.5.1 Ethanol Versus Gasoline
Ethanol (CHOH), or grain alcohol, is an alternative motor fuel that can be produced from biological crops. In §26 we analyze some of the general aspects of biofuels and examine different methods for producing ethanol and its role as a replacement for fossil fuels. Here, we focus on a specific question that can be addressed using the methods of this chapter: how does the energy density of ethanol compare to that of gasoline? This analysis also leads to a simple comparison of the carbon output from these two fuel sources.
We begin by computing the HHV for each fuel. We assume complete combustion of ethanol and gasoline (we take gasoline here for simplicity to be 100% octane, CH; a more detailed discussion of the composition and properties of actual gasoline mixtures is given in §11 and §33), so we have the chemical reactions
(9.12)
Hess’s law gives
(9.13)
(9.14)
Thus, octane produces far more energy per mole than ethanol. Octane, however, also has a greater molar mass: 114 g/mol versus 46 g/mol for ethanol. If we divide the enthalpy of combustion by the molar mass, we get the specific energy density for the fuels (which is positive by convention). The results are
So ethanol stores less energy than octane per kilogram. Burning ethanol produces only % as much energy per unit mass as burning octane.
Now let us consider the CO production from each of these fuels. Burning one mole of ethanol produces two moles of CO, whereas one mole of octane produces eight moles of CO. From the point of view of minimizing CO output for a given amount of energy use, the relevant quantity to compare is the CO intensity, defined as the amount of CO per unit of energy. In this respect, ethanol and octane are almost identical: 1.46 mol(CO)/MJ for ethanol and 1.48 mol(CO)/MJ for octane. In fact, as we discuss further in §33, the common fuel with the lowest CO intensity is methane (CH), which produces 1.12 mol(CO)/MJ when burned.
We see from their almost identical CO intensities that the relative environmental impact of gasoline versus ethanol depends entirely on how these two fuels are produced. Gasoline, refined from petroleum, is a fossil fuel. Burning it introduces net CO into the atmosphere. Bioethanol, as it grows, uses photosynthesis to capture CO from the atmosphere. Therefore burning it would appear to be neutral with respect to the CO content of the atmosphere. The complete story is more complex, however, since the energy inputs and consequent carbon footprint associated with ethanol production vary widely depending upon how the ethanol is produced. We explore these questions in more detail in §26.
9.5.2 A Tonne of TNT?
TNT, or trinitrotoluene, is a commonly used explosive with the chemical formula CH(NO)CH. TNT has given its name to one of the standard units of energy, the tonne of TNT, 1 tTNT 4.184 GJ, often used to characterize the destructive power of weapons of war.
When TNT explodes, it does not burn in the ambient atmosphere. Such a process would occur too slowly to generate an explosion. Instead, the pressure wave created by a detonator causes the TNT to decompose, liberating gases that expand violently. The “tonne of TNT” refers to the enthalpy liberated when TNT disintegrates, not to TNT’s enthalpy of combustion.
Combustion of TNT is described by the reaction
(9.15)
Using kJ/mol [22] and other enthalpies of formation from Table 9.8, Hess’s law gives
Table 9.8 Standard enthalpy and free energy of formation for some useful chemical compounds, all at 25℃ and 1 atm(from [50]).
Compound Chemical
formula (kJ/mol) (kJ/mol)
Diatomic gases H, O, Cl,… 0* 0*
Methane CH(g) –75 –51
Water vapor HO(g) –242 –229
Water liquid HO(l) –286 –237
Octane (l) C(l) –250 +6
Ethanol (g) COH(g) –235 –168
Ethanol (l) COH(l) –278 –175
Carbon dioxide (g) CO(g) –394 –394
Carbon monoxide (g) CO –111 –137
Calcium oxide CaO –635 –604
Iron ore (hematite) Fe –824 –742
Calcium carbonate CaCO –1207 –1129
Glucose C –1268 –910
Ammonia NH –46 –16.4
* by definition
The specific energy density of TNT (molecular mass 227 gm/mol) is then
(9.16)
Clearly, a “tonne of TNT” is much less than the enthalpy of combustion of one tonne of TNT.
Instead, when TNT explodes, it disintegrates into water, carbon monoxide, nitrogen and left over carbon,
(9.17)
(Other pathways can occur, including 2(TNT) 5 H + 12 CO + 2 C+3 N2, but for an estimate, we consider only the first reaction.) A calculation similar to the first gives kJ/mol and a specific energy density of = 4.1 GJ/t (see Problem 9.15), which is within 2% of the defined value. Note that the tonne of TNT is a small fraction of the specific energy of oil (41.9 GJ/t) or coal (29.3 GJ/t). Unlike TNT, however, the decomposition of familiar fuels such as methane, ethanol, or gasoline is endothermic. They require oxygen to liberate energy (by combustion) and do not explode unless their vapors are mixed with air.
9.5.3 Energy and Metals
Mining and refining of metals represents a small but significant contribution to world primary energy consumption. In the US, for example, mining and refining of metals accounted for 582 PJ or about 2.6% of US industrial primary energy consumption in 2007 [12, 54]. The discoveries of metals label major divisions of human history such as the Bronze Age and the Iron Age, and metals continue to play an essential role in modern societies.
With this in mind, our last example of chemical thermodynamics comes from the field of metallurgy, the chemistry and material science of metals. Metals are often obtained from ore deposits, where they are found in relatively simple mineral compounds, mixed with commercially worthless rock referred to as gangue. Sulfide minerals, such as iron pyrite (FeS), chalcopyrite (FeCuS), and galena (PbS), are valuable sources of iron, copper, and lead. Oxides such as cassiterite (SnO), hematite (FeO), and alumina (AlO, produced, e.g., from bauxite ore) are primary sources of tin, iron, and aluminum. The first step in refining a sulfide ore is generally to convert it to an oxide. This can be done by roasting – heating in an oxidizing environment so that oxygen replaces the sulfur – a reaction with negative reaction free energy (Problem 9.21).
In order to recover the pure metal from an oxide, it is necessary to drive off the oxygen. At first sight, the situation seems analogous to the calcination reaction described above, in which heat is used to drive CO out of limestone. This turns out to be impossible for most oxides, however, because oxygen binds too strongly to metals. Consider, for example, cassiterite. Tin was one of the first metals refined by man – (adding tin to copper initiated the Bronze Age in the fourth millennium BC), which suggests that it is relatively easy to refine. The free energy of formation of SnO is negative at 25℃ (see Table 9.9), so the opposite reaction, SnO Sn + O, will not occur spontaneously at that temperature. Recall that calcination is performed by raising the temperature high enough to change the sign of the free energy of reaction, so that the reaction proceeds spontaneously. It is natural then to ask, at what temperature does SnO decompose spontaneously? As shown in Example 9.3, due primarily to the large enthalpy of formation of SnO, cassiterite does not spontaneously decompose until temperatures of order 2800 K are realized. This is high – even for modern technology.
Table 9.9 Free energies and enthalpies of formation (kJ/mol) and molar entropies (J/mol K) for Sn, Fe, and Al and their oxides at 25℃ (data from [50]).
Example 9.3 Refining Tin From Cassiterite
Estimate the temperature at which oxygen can be spontaneously driven out of SnO, (a) SnO Sn O, and compare this with the temperature at which the reaction (b) SnO 2 CO Sn 2CO can spontaneously occur.
The basic idea is similar to the calcination reaction described in §9.4.1: although the reactions do not proceed spontaneously at room temperature, the reaction entropy is positive, so the reaction free energy decreases with increasing temperature and eventually goes negative, allowing the reaction to proceed spontaneously at an elevated temperature. We estimate this temperature by assuming that the reaction free energy and reaction entropy are independent of temperature. The first step is to compute the entropy of formation of SnO using , which yields J/mol K. Then, combining the data in Tables 9.8 and 9.9, wecompute the reaction free energy and reaction entropy and construct the free energy as a function of T,
From these we estimate that reaction (a) can proceed spontaneously above K, while reaction (b) can proceed spontaneously above K. Thus, it is not practical to obtain tin from cassiterite simply by driving off the oxygen. In contrast, since tin melts at 505 K, a stream of carbon monoxide, such as might be available from a smoldering campfire, would easily reduce cassiterite in surrounding rocks to a puddle of molten tin!
How, then, did the ancients obtain tin from cassiterite? The trick was to use carbon, or more likely carbon monoxide (CO), to steal away the oxygen from the ore. As shown in Example 9.3, the reaction can proceed spontaneously at . So tin can be obtained from by reacting it with carbon monoxide at temperatures that are easily achievable with “campfire technology.” The use of carbon monoxide as a reducing agent to remove oxygen from metal-oxide ores remains widespread today. It is, for example, the basic chemical process used to refine iron within a blast furnace. The method, however, does not work for aluminum (Problem 9.22), which is why aluminum was not isolated and used until modern times.
Discussion/Investigation Questions
9.1 The heat capacity per molecule of carbon monoxide at constant pressure is shown in Figure 9.8 from 175 K to over 5000 K. Explain why for small T and explain its subsequent increase with T.
Figure 9.8 The heat capacity per molecule of CO at constant pressure as a function of temperature [24].
9.2 HO has a triple point near 0℃, where all three phases, solid (ice), liquid (water), and gas (water vapor), can coexist. So there are three possible phase transitions, solid liquid (melting), water gas (vaporization), and solid vapor (sublimation). Each has an associated enthalpy of transition. Can you suggest a relation among , , and ? Check your conjecture by looking up data. Be sure to get data at the same temperature.
9.3 The Iron Age began around 1200 BCE, more than 2000 years after the first appearance of bronze in human history. It may come as a surprise, therefore, that the reaction FeO+3CO 2Fe +3CO is exothermic ( kJ/mol) and has negative reaction free energy ( kJ/mol) at NTP. So hematite can be quite easily reduced to iron by the CO produced in a hot, oxygen starved fire. Why, then, was iron not discovered earlier?
Problems
9.1 As described in §21, the universe is filled with electromagnetic radiation left over from the Big Bang. This radiation has a characteristic temperature of K. At what wavelength does the power spectrum of this radiation peak? What is the energy of a photon with this wavelength?
9.2 A blue supergiant star may have a surface temperature as high as 50 000 K. Estimate the wavelength at which it emits maximum power. Why does the star appear blue?
9.3 When the spin of the electron in a hydrogen atom flips from parallel to antiparallel to the direction of the proton’s spin, a photon of energy J is emitted. What is the photon’s energy in electron volts? What is its frequency? Its wavelength? Above roughly what temperature would you expect this transition to be thermally excited? This emission line of neutral hydrogen plays a central role in modern astronomy.
9.4 [T] The cubic growth of the heat capacity of a solid as a function of temperature continues only up to roughly one tenth of the Debye temperature associated with the shortest wavelength excitations in the solid. This may seem surprising, since this implies that at this point the heat capacity becomes sensitive to the absence of modes whose probability of excitation is on the order of . A derivation of the Debye model parallel to that of the photon gas in §22.3 shows that the heat capacity in the Debye model takes the form
where x is a rescaled energy parameter. At large x, gives the Boltzmann suppression of states with increasing energy, while measures the increase in the number of states available with increasing energy. Compute (approximately) the minimum temperature at which the factor exceeds one for some x in the range of integration.
9.5 The power output of air conditioners is measured in “tons,” an ancient nomenclature dating back to the days when air was cooled by blowing it over blocks of ice. A ton of air conditioning is defined to be the energy required to melt one ton (2000 pounds) of ice at 32℃ distributed over a period of one day. What is the power equivalent (in kilowatts) of one ton of air conditioning?
9.6 Find the frequency of the lowest vibrational mode of a diatomic molecule in terms of the parameters of the Morse potential, eq. (9.5).
9.7 The Morse potential parameters for oxygen are given by eV, Å, and Å [55]. Using the result from the previous problem, estimate the energy necessary to excite the oxygen molecule out of its ground state. Compare with your answer to Problem 7.10.
9.8 [T] Another commonly used model for the potential energy of a diatomic molecule is the Lennard–Jones (LJ) potential . Using the results of Problem 9.6, find the choice of the parameters and ϵ that gives the same equilibrium separation between atoms and the same harmonic oscillator frequency as the Morse potential. Unlike the Morse potential, the depth of the LJ potential is not an independent parameter. What is the relation between the depth , equilibrium separation , and harmonic oscillator frequency ω of the LJ potential?
