Chemical Thermodynamics

Chemical Thermodynamics
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Chemical Thermodynamics 15.1 EINFÜHRUNG (INTRODUCTION) This chapter deals with an application of the laws of thermodynamics on which entire textbooks have been written. Chemists call this topic physical chemistry, and it forms the basis of much of applied chemistry. It is important to engineers, because it provides a fundamental understanding of the combustion process in engines, power plants, fuel cells, and other chemically based energy conversion processes. This chapter has three main goals. Our first goal is to be able to calculate the amount of heat produced in the combustion of an organic fuel. The second is to understand the basic elements of chemical equilibrium and dissociation, and the third is to look at the emerging field of fuel cell technology. To be able to discuss these subjects adequately, we need to define what we mean by the term fuel,weneedtodecidehowthethermo- dynamic properties of the products and the reactants are related through a standard reference state,andweneed to lay the foundations for discussing chemical reaction energy conversion efficiency. Modern Engineering Thermodynamics. DOI: 10.1016/B978-0-12-374996-3.00015-4 © 2011 Elsevier Inc. All rights reserved. 591592 CHAPTER 15: Chemical Thermodynamics IS IT CHEMISTRY OR ALCHEMY? A practical chemical technology can be traced to prehistoric times. Primitive metallurgy, medicine, and food preparation are typical examples. They were purely empirical “recipe” driven processes with no form of chemical theory to explain their results. Before the sixth century BC, it was generally believed that all things were composed of a single primitive element. The Greek philosophers Heraclitus (540–480 BC) and Empedocles (490–430 BC) began a new era when they proposed that, instead of a single element, all matter was made up of four elements—air, earth, fire, and water—and that the continual mixing of these elements formed all the objects of the real world. The Greek philosopher Pythagoras (580–500 BC) is generally credited with recognizing the functional significance of num- bers in quantifying the processes of the real world. Pythagoras established an academy of learning in Crotona, Italy, in about 532 BC. The academy prospered long after his death until its destruction in about 390 BC. It is believed that his disci- ples (known as Pythagoreans) working at the academy during this time developed many of the mathematical discoveries now attributed to him (e.g., the Pythagorean theorem for right triangles). Empedocles adopted Pythagoras’ numerology technique in an attempt to quantify the chemistry of his four elements. For example, according to Empedocles, animal bone consisted of two parts water, two parts earth, and four parts fire. Because he believed that all of his four elements were most thoroughly mixed in blood, he concluded that people think mainly with their blood. From about the second century AD until nearly the 19th century, the world embraced what is considered today to be a scientific and chemical curiosity, alchemy. Alchemy was a combination of the occult, astrology, and primitive chemistry. Even its name, derived from Arabic and introduced in the 12th century, is obscure because the root chem seems to have no relevant etymological meaning. The basic function of alchemy was transmutation, which was concerned with transmuting age to youth, sickness to health, death to immortality. More notorious was its preoccupation with the physical transmutation of base metals into gold (i.e., transmuting poverty to wealth). Its central elements were mercury (quicksilver, the liquid metal), sulfur (the stone that burns), and ammonium chloride (sal ammoniac, a source of hydrochloric acid). Successful alchemists tended to be charla- tans whose work was shrouded in mystery. From the Medieval period forward, the central focus of alchemy was the making of gold (religion had successfully taken over the immortality issue), and many prominent scientists, including Isaac Newton (1643–1726), experimented with it seriously. (Newton is thought to have contracted mercury poisoning in about 1690 as a result of his alchemy experiments.) The false science of alchemy, which appealed primarily to the human weakness of greed, went without serious intellectual challenge for nearly 2000 years. In the 17th century, Johann Jochim Becher (1635–1682), an established alchemist at one time engaged in attempting to transmute Danube River sand into gold, proposed that all substances were made up of the classical alchemical elements of mercury, sulfur, and corrosive salts, plus a new fourth weightless element that was produced by combustion. In 1697, the German physician Georg Ernst Stahl (1660–1734) named this supposed fourth element phlogiston and used it to develop a coherent theory of combustion, respiration, and corrosion. His phlogiston theory quickly won universal scientific approval and was the only scientifically accepted theory of matter for nearly 100 years afterward. In 1774, the English clergyman and scientist Joseph Priestley (1733–1804) described some of his experimental results in removing phlogiston from air (the “dephlogistication” of air) to the French chemist Antione Laurent Lavoisier (1743– 1794), who immediately recognized their importance and subsequently carried out similar experiments himself. Lavoisier soon realized that the dephlogiston that Priestley thought he had been working with was actually a unique chemical and, in 1777, he named it oxygen (from the Greek for acid forming). By the early 19th century Lavoisier’s oxidation theory com- pletely replaced Stahl’s phlogiston theory and the era of modem chemistry had begun. Lavoisier had done for chemistry what a century earlier Newton had done for mechanics; he put the subject on a firm analytical foundation. The unfortunate Joseph Priestley was subsequently driven from his home in England because of his public support of the French Revolu- tion, and he settled in America. WHO KILLED THE GREAT FRENCH CHEMIST LAVOISIER? Because Lavoisier maintained a career in the French government as well as science (he was on the original 1790 weights and measures committee that led to the development of the metric system we now use), he was caught up in the French Revolution and was accused of political crimes (such as stopping the circulation of air in Paris by a city wall erected at his suggestion in 1787). He was convicted and guillotined on the same day in 1794.15.2 Stoichiometric Equations 593 15.2 STOICHIOMETRIC EQUATIONS In the early 19th century, the English chemist John Dalton (1766–1844) devised a system of chemical symbols and determined the relative masses of some elemental atoms. He also formulated a theory that combinations of different chemical elements occur in simple mass ratios, which led him to the development of a way of writing a chemical formula that mathematically represented chemical reactions. For example, if elements A and B com- bine in a two to one ratio by mass to form chemical C, Dalton wrote this as 2atomsof A+1atomof B = 1atomof C (15.1) In modern notation, this would simply be 2A+B C Reactants Product where the equality has been replaced by an arrow that indicates the direction of the reaction. The items on the left side of this equation are called the reactants, and those on the right side are called the products of the reaction. A description of the net combining properties of atoms and compounds that occur in a chemical reaction is known today as the stoichiometry of the reaction. Dalton’s chemical equation notation provides a shorthand mathe- matical version of such a description, and the numerical values that precede the chemical symbols in these equa- tions are called the stoichiometric coefficients of the reaction. These coefficients represent the number of atoms or molecules involved in the reaction, and since mass is conserved in ordinary chemical reactions, the number of atoms of each chemical element must be the same in both the reactants and the products. Therefore, a chemical equation can be balanced (i.e., mass balanced) by requiring stoichiometric coefficients that produce the same num- ber of atoms of each chemical species on both sides of the reaction equation. For example, the reaction that occurs in burning hydrogen to completion in a pure oxygen atmosphere can be written in modern notation as aðH Þ+bðO Þ cðH OÞ 2 2 2 where a, b,and c are the stoichiometric coefficientsfor the reaction. Anindividualatomic species balance now gives AtomichydrogenðÞ H balance:2a = 2c AtomicoxygenðÞ O balance:2b = c thus producing two equations in the three unknowns, a, b,and c. Since such reactions are usually of interest per unit mass of fuel supplied, we can arbitrarily set a=1: then, the atomic hydrogen and oxygen balances gives c=1 1 and b = =0.5.Therefore,our finalbalanced equation would read 2 H +0:5ðO Þ H O (15.2) 2 2 2 After an extensive period of experimentation, the Italian chemist Count Amado Avogadro (1776–1856) proposed in 1811 that equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules. WHAT DOES THE WORD STOICHIOMETRY MEAN? The term stoichiometry comes from the Greek words stoicheion (component) and metron (measure). It was introduced in 1792 by the German chemist Jeremias Benjamin Richter, when he suggested that substances react chemically according to relations that resemble mathematical formulae. WHAT IS AVOGADRO’SLAW? Avogadro’s law states that equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules. The Avogadro constant, N (originally called Avogadro’snumber), is the number of atoms in exactly 12 kg of A carbon-12. The 2006 value is 26 26 N = 6:0221×10 atoms/kgmole = 2:7316×10 atoms/lbmole A Although Avogadro introduced this law in 1811, it was not generally accepted by the scientific community until after 1858. Incidentally, André Marie Ampère (1775–1836) popularized the term molecule for an assembly of atoms in about 1814.594 CHAPTER 15: Chemical Thermodynamics With this proposition, he was able to show that hydrogen, oxygen, nitrogen, and the like exist as diatomic molecules in nature. Modern experiments have determined that the actual number of molecules in any sub- 26 stance whose mass in kilograms is equal to its (relative) molecular mass M is 6.022 × 10 . Therefore, the mass of one molecule of this substance is M kilograms: 26 6:022×10 The acceptance of Avogadro’s law soon led to the development of the mole (sometimes abbreviated mol,both being a contraction of the German word molekul) as a convenient chemical mass unit. Originally a “mole” was defined as a mass in grams that was equal to the molecular mass M of a substance (i.e., originally a “mole” of carbon-12 contained 12 grams of carbon-12). Today, we need to recognize that this “mole” is really a gram-mole, or gmole,becausethe “mole” unit is used in so many different units systems today. Now, it must carry a prefix showing the units system being used, such as gmole, kgmole,or lbmole. But remember that these “moles” are not equal to each other, since 1 kgmole=1000 gmole=2.2046 lbmole. Since equal “moles” of different substances contain the same number of molecules, the stoichiometric coefficients, which initially represented only individual molecules, can also be used to represent the number of moles of each element present. Therefore, Eq. (15.1) can be written in the equivalent form 2gmoleof A+1gmoleof B = 1gmoleof C or 2kgmoleof A+1kgmoleof B = 1kgmoleof C or 2lbmoleof A+1lbmoleof B = 1lbmoleof C and Eq. (15.2) can be interpreted as a reaction between 1 kgmole of H and 0.5 kgmole of O producing 2 2 1 kgmole of H O or 1 lbmole of H and 0.5 lbmole of O producing 1 lbmole of H O and so forth. 2 2 2 2 Most combustion processes occur in air, not pure oxygen. The composition of air used to determine its thermo- dynamic properties on a molar or volume basis is 78.09% nitrogen, 20.95% oxygen, 0.93% argon, and 0.03% carbon dioxide and trace elements. For convenience, we round this off to 79.0% N and 21.0% O.Indoing 2 2 this, we are essentially dividing air into two components: pure oxygen and a mixture of noncombustibles (N , Ar, CO ). The noncombustible group, called atmospheric nitrogen, has a mole fraction composition of 2 2 n N 78:09 78:09 2 χ = = = = 0:9879 = 98:79% N 2 n +n +n 78:09+0:93+0:03 79:05 N Ar CO 2 2 n 0:93 Ar χ = 0:0118 = 1:18% = = Ar n +n +n 79:05 N Ar CO 2 2 and n CO 0:03 2 χ = = = 0:00038 = 0:038% CO 2 n +n +n 79:05 N Ar CO 2 2 where n , n , and n are the number of moles of nitrogen, argon, and carbon dioxide present in the mixture. N2 Ar CO2 Then, from Eq. (12.11), the equivalent molecular mass of this mixture is ðMÞ =∑χ M =ð0:9879Þð28:016Þ+ð0:0118Þð39:944Þ+ð0:00038Þð44:01Þ i i atmospheric nitrogen = 28:16kg/kgmole = 28:16lbm/lbmole From this point on, we refer to the atmospheric nitrogen mixture as simply nitrogen, and we assume that air has a molar composition of 21.0% oxygen and 79.0% nitrogen, where the molecular mass of oxygen is still 32.00, but the molecular mass of nitrogen is now 28.16 instead of 28.016. The equivalent molecular mass of air is still 28.97, since the argon and carbon dioxide are now merely grouped with the nitrogen. For this air composition, each mole of oxygen is accompanied by 79:0/21:0 = 3:76 moles of nitrogen. Thus, if the hydrogen combustion reaction described by Eq. (15.2) were carried out in air instead of pure oxygen, it would be written as H +0:5½O +3:76ðN Þ H O+1:88ðN Þ 2 2 2 2 2 In this equation, the nitrogen is assumed to be inert and therefore passes through the reaction unchanged.15.2 Stoichiometric Equations 595 CHEMICAL REACTION EQUATIONS AND SIGNIFICANT FIGURES We have a problem using significant figures in the equations for chemical reactions. For example, in the reaction of hydro- gen and oxygen to form water, H +0.5O → H O, we do not want to have to write it as 1.00 H + 0.500 O → 1.00 H O. 2 2 2 2 2 2 So, in this chapter, we assume that the coefficients in an equation for a chemical reaction have at least three significant figures without actually writing them as such in the reaction equations. The amount of air or oxygen used in a combustion process can be described as 1. The percent of theoretical air or oxygen required to carry out the reaction. 2. The percent of excess or deficit air or oxygen actually used in the reaction. 3. The air/fuel (A/F) or fuel/air (F/A) ratio used in the reaction measured on either a mass or a mole basis. One hundred percent theoretical air is the minimum amount of air that supplies enough oxygen to carry out complete combustion. The percentage of excess air is simply the percentage of theoretical air supplied minus 100, and the percentage of deficit air is 100 minus the percent of theoretical air supplied. The air/fuel ratio (A/F) is the amount of air used per unit of fuel consumed, and it can be expressed either in mass or mole units. The relation between the mass and molar air/fuel ratios is given by  Molecularmassof air,M air ðA/FÞ =½ðA/FÞ inmolesof air/moleof fuel× mass molar Molecularmassof fuel,M fuel The fuel/air ratio (F/A) is simply the inverse of the air/fuel ratio. 1 ðF/AÞ = molar ðA/FÞ molar and 1 ðF/AÞ = mass ðA/FÞ mass With these definitions, it is easy to see that the hydrogen combustion described in the reaction H + 2 0:5½O +3:76ðN Þ H O+1:88ðN Þ uses 100% theoretical air and no excess or deficit air is involved. The 2 2 2 2 molar air/fuel ratio for this reaction is 0:5×ð1+3:76Þmolesof air n air ðA/FÞ = = = 2:38molesof air/moleH 2 molar n 1moleof fuel fuel 1 and the mass air/fuel ratio is  Molecularmassof air,M air ðA/FÞ =½ðA/FÞ molesof air/moleof fuel × mass molar Molecularmassof fuel,M fuel 28:97Massof air/moleof air = 2:38molesof air/moleof fuelðH Þ × 2 2:016Massof fuel/moleof fuel = 34:2gramsof air/gramof H = 34:2kgof air/kgof H = 34:2lbmof air/lbmof H 2 2 2 Then, the molar and mass fuel/air ratios are simply −1 ðF/AÞ =ðA/FÞ = 0:42molesH /moleair 2 molar molar and −1 ðF/AÞ =ðA/FÞ = 0:029gramH /gramair 2 mass mass = 0:029kgH /kgair 2 = 0:029lbmH /lbmair 2 1 While the mole is in fact a unit of mass, in this section we use the term mass to indicate nonmolar units.596 CHAPTER 15: Chemical Thermodynamics 2 If this reaction is carried out with 150% theoretical air (i.e., 50% excess air), it has the form H +1:5 × ð0:5Þ½O +3:76ðN Þ H O+0:25ðO Þ+2:82ðN Þ 2 2 2 2 2 2 with a molar air/fuel ratio of 1:5×ð0:5Þð1+3:76Þ n air ðA/FÞ = = = 3:57molesair/moleH 2 molar n 1 fuel Similarly, if it is carried out at 75% theoretical air (i.e., 25% deficit air), it is H +0:75 × ð0:5Þ½O +3:76ðN Þ 0:75ðH OÞ+0:25ðH Þ+1:41ðN Þ 2 2 2 2 2 2 with a molar air/fuel ratio of 0:75×ð0:5Þð1+3:76Þ n air ðA/FÞ = = = 1:785molesair/moleH 2 molar n 1 fuel and so forth. 15.3 ORGANIC FUELS The term organic has been used in chemistry since the late 18th century and originally referred only to materials occurring in or derived from living organisms.Today,thistermisusedtorepresent all compounds of carbon, whether derived from living organisms or not. Other elements frequently found in organic compounds are hydrogen, oxygen, nitrogen, sulfur, and phosphorus. Since the number of organic compounds is very large, they are subdivided into groups having similar properties. Hydrocarbons, alcohols, carbohydrates, proteins, and fats are typical organic compound subdivisions. Many common organic fuels are made up of only carbon and hydrogen atoms and are consequently called hydrocarbons. The hydrocarbon class of organic molecules can be further subdivided into the groups shown in Figure 15.1. Using the atomic mass balance technique discussed at the beginning of this chapter, it is easily shown that the stoichiometric reaction equation for the combustion of a typical hydrocarbon of the form C H using 100.% n m theoretical air is C H +ðÞ n+m/4½O +3:76ðN Þ nðCO Þ+ðÞ m/2ðH OÞ+3:76ðÞ n+m/4ðN Þ (15.3a) n m 2 2 2 2 2 Hydrocarbons Aliphatic Aromatic Alkanes Alkenes Alkynes Benzoid Nonbenzoid (C H ) (C H ) (C H ) (C H ) n 2n+2 n 2n n 2n+2 2 n 2n+2 6 Ethylene Acetylene Benzene Naphthalene Methane (C H ) (C H ) (CH ) (C H ) (C H ) 10 8 4 2 4 2 2 6 6 Propylene Propyne Anthracene Ethane Toluene (C H ) (C H ) (C H ) (C H ) 3 6 (C H ) 2 6 3 4 14 10 7 8 etc. etc. Propane etc. Xylene (C H ) (C H ) 3 8 8 10 Butane etc. (C H ) 4 10 etc. FIGURE 15.1 Classification of hydrocarbons. 2 Remember, in this chapter we assume each coefficient in an equation for a chemical reaction has at least three significant figures without actually writing them as such in the reaction equation.15.3 Organic Fuels 597 ANSWERS SOMETIMES COME IN DREAMS In the early part of the 19th century, chemists were puzzled by the fact that it was possible to construct two seemingly different compounds that had dissimilar physical properties yet identical chemical formulae. For example, ethanol and dimethyl ether both have the same chemical formula, C H O, yet ethanol boils at 79°C while dimethyl ether boils at 2 6 −24°C. Materials that have the same chemical formula but dissimilar physical properties are called isomers. The isomer puzzle was solved by the German chemist Friedrich August Kekulé (1829–1896) in a dream while dozing on the top deck of a horse-drawn bus in London. In his dream, he realized that the atoms of a molecule could be arranged in different geo- metric structures; and in 1858, he introduced the schematic notation, still used, in which bonds between atoms are repre- sented by lines drawn between their corresponding chemical symbols (e.g., H as H—H). He was then able to show that 2 isomers were simply the result of different bonding patterns. For example, butane (C H ) has the isomers n-butane, which 4 10 boils at−0.5°C, and isobutane, which boils at−12°C. (The n prefix is always used to denote the normal, or chainlike, struc- ture, whereas the iso prefix is used to denote the branched structure.) These two isomer bonding patterns are shown in Figures 15.2 and 15.3. H HH H H HH H C CC C H H C CC H H H HH H HH H H H HH C C C C HC C H H CH H H HH H Ethane (C H ) Ethylene (C H ) Acetylene (C H ) n-Butane Isobutane 2 6 2 4 2 2 (normal butane) (C H ) (alkane, saturated) (alkene, unsaturated) (alkyne, unsaturated) 4 10 (C H ) 4 10 FIGURE 15.3 FIGURE 15.2 Ethane (alkane, saturated), ethylene (alkene, unsaturated), and n-Butane and isobutane. acetylene (alkyne, unsaturated). Another 19th century hydrocarbon curiosity was the existence of the two classes, aliphatic (fatty) and aromatic (fragrant) compounds. Aromatic hydrocarbons always had at least six carbon atoms and a smaller proportion of hydrogen atoms than the aliphatic hydrocarbons. In 1865, Kekulé again found the solution in a dream. He envisioned a six-carbon chain closing on itself to form a ring, like a snake biting its own tail, and he concluded that the aromatic compounds contain such rings whereas the aliphatic compounds contain only straight chains. Within the aliphatic group, the alkanes are characterized by having carbon atoms with single bonds between them, while the alkenes have carbon atoms with double bonds between them, and alkynes have carbon atoms with triple bonds between them. If all the bonds within an organic compound are single, then the compound is said to be saturated; but if multiple bonds exist between any two carbon atoms in the compound, it is said to be unsaturated. Thus, the alkanes are all saturated hydrocarbons, while all the remaining hydrocarbons are unsaturated. Also, its combustion with excess air using 100.(x) percent theoretical air (i.e., 100.(x−1) percent excess air), where x≥ 1.0, is C H +xnðÞ +m/4½O +3:76ðN Þ nðCO Þ+ðÞ m/2ðH OÞ n m 2 2 2 2 +ðx−1Þðn+m/4ÞðO Þ (15.3b) 2 +xð3:76Þðn+m/4ÞðN Þ 2 And its combustion in deficit air using 100.(y) percent theoretical air (i.e., 100.(1−y) percent deficit air), where ð2n+mÞ÷ð4n+mÞ≤y≤1:0, is C H +ynðÞ +m/4½O +3:76ðN Þ nð2y−1Þ−mð1−yÞ/2 ðCO Þ n m 2 2 2 (15.3c) +ðÞ 2n+m/2ð1−yÞðCOÞ+ðÞ m/2ðH OÞ 2 +yð3:76ÞðÞ n+m/4ðN Þ 2 In Eq. (15.3c), it has been assumed that the hydrogen is much more reactive than the carbon, so it will take up all the oxygen it needs to be converted into water. This leaves only the carbon subject to incomplete combustion.598 CHAPTER 15: Chemical Thermodynamics EXAMPLE 15.1 Determine the stoichiometric reaction equation for methane (CH ) burned in 4 a. 100.% theoretical air. b. 150.% excess air. c. 20.0% deficit air. Solution We could solve this problem for methane, CH =C H by setting n=1and m=4inEqs.(15.3a–c).However,since 4 n m methane is a simple compound, it is more enlightening to carry out the individual atomic balances to obtain the correct reaction equations. a. The general combustion equation for 1 mole of methane in 100% theoretical air is CH +a½O +3:76ðN Þ bðCO Þ+cðH OÞ+dðN Þ 4 2 2 2 2 2 3 The element balances are Carbon (C) balance: 1=b Hydrogen (H ) balance: 2=c 2 Oxygen (O ) balance: a = b+c/2 = 1+2/2 = 2 2 Nitrogen (N ) balance: að3:76Þ = d = 2ð3:76Þ = 7:52 2 The resulting stoichiometric equation for 100.% theoretical air is CH +2½O +3:76ðN Þ CO +2ðH OÞ+7:52ðN Þ 4 2 2 2 2 2 b. The 150.% excess air corresponds to 250.% theoretical air. The reaction equation now has O in the products and 2 consequently has the form CH +2:5ð2Þ½O +3:76ðN Þ aðCO Þ+bðH OÞ+cðO Þ+dðN Þ 4 2 2 2 2 2 2 and again element balances can be used to find that a=1, b=2, c=3, and d=2.5(2)(3.76)=18.8; so that, for 150.% excess air, we have CH +5½O +3:76ðN Þ CO +2ðH OÞ+3ðO Þ+18:8ðN Þ 4 2 2 2 2 2 2 c. The 20.0% deficit air corresponds to 80.0% theoretical air. Again, assuming all the hydrogen reacts to water, the reaction now has CO in the products and has the form CH +0:8ð2Þ½O +3:76ðN Þ aðCO Þ+bðCOÞ+cðH OÞ+dðN Þ 4 2 2 2 2 2 and again the element balances can be used to yield the coefficients a = 0:2,b = 0:8,c = 2:0, and d = 6:016: Note that these results correspond to the same coefficients one would obtain using Eq. (15.3c). The final reaction equation for 20.0% deficit air is CH +1:6½O +3:76ðN Þ 0:2ðCO Þ+0:8ðCOÞ+2ðH OÞ+6:016ðN Þ 4 2 2 2 2 2 Exercises 1. Determine the number of kgmoles of water produced in the reaction of Example 15.1 per kgmole of methane burned in 200.% excess air. Answer: 2 kgmoles per kgmole of CH . 4 2. Determine the molar and mass A/F ratios for parts a, b, and c in Example 15.1. Answer: a. (A/F) =9.52 lbmole air/lbmole CH =9.52 kgmole air/kgmole CH molar 4 4 (A/F) =17.24 lbm air/lbm CH =17.24 kg air/kg CH mass 4 4 b. (A/F) =23.8 lbmole air/lbmole CH =23.8 kg air/kg CH molar 4 4 (A/F) =43.1 lbm air/lbm CH =43.1 lbm air/lbm CH mass 4 4 c. (A/F) =7.616 lbmole air/lbmole CH =7.616 kg air/kg CH molar 4 4 (A/F) =13.79 lbm air/lbm CH =13.79 lbm air/lbm CH mass 4 4 3. Rework Example 15.1 for combustion in 200% theoretical air. Answer:CH + 4O + 3.76(N )→ CO + 2(H O) + 4 2 2 2 2 2(O ) + 15.04(N ). 2 2 3 In modern chemistry, the term element refers to the stable form of a substance composed of only one kind of atom. The chemically stable forms of carbon, hydrogen, oxygen, and nitrogen in these reactions are C, H,O , and N , rather than C, H, O, and N. So, even though H,O , and N are 2 2 2 2 2 2 really diatomic molecules, they are considered to be the proper forms for these elements in common chemical reactions.15.4 Fuel Modeling 599 Hydrocarbon fuels refined from petroleum normally contain a mixture of many organic components. Gasoline, for example, is a mixture of over 30 compounds. It is, however, convenient to model these fuels as a single average hydrocarbon compound of the form C H , as discussed in the following section. n m 15.4 FUEL MODELING It is fairly easy to obtain accurate composition analysis of combustion products with modern gas chromato- graphy or mass spectroscopy techniques. With an accurate combustion analysis of a fuel that is in reality a com- plex mixture of hydrocarbons, an equivalent or average hydrocarbon model of the form C H can be determined n m from a chemical element balance. For example, if the combustion products contain only CO,CO,O,H O, 2 2 2 and N , then Eqs. (15.3a–c) could be used to determine the composition parameters n and m when the stoichio- 2 metric coefficients of the products are measured. Since the fuel model formula C H represents an average of all n m the different hydrocarbon compounds present in the fuel mixture, n and m usually do not turn out to be inte- gers and the resulting model does not represent any real hydrocarbon (except possibly when n and m are rounded to integers). EXAMPLE 15.2 A new hydrocarbon fuel is being developed that consists of 1.00 kgmole of methane (CH)mixedwith3.00kgmolesof 4 propane (C H ). Determine the hydrocarbon fuel model for this mixture. 3 8 Solution This mixture is assumed to produce 1.00 kgmole of the fuel model C H ,so n m 1CH +1C H =1C H 4 3 8 n m An element balance for this equation gives Carbonbalance:1+3ð3Þ = 10 = n Hydrogenbalance:4+3ð8Þ = 28 = m Consequently, the hydrocarbon fuel model is C H . 10 28 Exercises 4. Determine the hydrocarbon fuel model in Example 15.2 if only 2.00 kgmoles of propane are used in the mixture. Answer: Hydrocarbon fuel model=C H . 7 20 5. Determine the hydrocarbon fuel model in Example 15.2 if ethane (C H ) is used in place of the methane. Answer: 2 6 Hydrocarbon fuel model=C H . 11 30 6. 1.00 lbmole each of methane (CH ), propane (C H ), and butane (C H ) are mixed to form 1.00 lbmole 4 3 8 4 10 of a new super fuel. Determine the hydrocarbon fuel model for this mixture. Answer: Hydrocarbon fuel model=C H . 8 22 Often, you do not know the exact hydrocarbon composition of a fuel. However, you can analyze the products of the fuel’s combustion and deduce the fuel model. Several modern instruments produce an accurate exhaust gas analysis. For example, gas chromatography and mass spectrometry are commonly used in exhaust gas analysis today. But perhaps the quickest, simplest, and most inexpensive method of obtaining an approximate combustion analysis is with an Orsat analyzer. This gas analyzer uses a chemical absorption technique to determine the volume fractions (which are equivalent to the mole fractions) of CO,CO,andO in the exhaust gas (see Figure 15.4). 2 2 Since it cannot measure the H O content, the exhaust gas sample is always cooled to room temperature, or below 2 the dew point of any water vapor present, so that most of the water in the combustion products condenses out. Therefore, the Orsat technique is said to produce a dry products analysis. Also, the Orsat technique cannot detect unburned hydrocarbons (typically CH ) and free hydrogen (H ) in the exhaust gas. These are usually small and 4 2 can normally be neglected. However, studies have shown that the mole fractions of methane and hydrogen in the combustion products of a hydrocarbon can be approximated as χ ≈0:0022 andχ ≈0:5ðx Þ, and these rela- CO CH H 4 2 tions can be used with an Orsat analysis if necessary.600 CHAPTER 15: Chemical Thermodynamics Sample Valve E Measuring C B A D chamber CuCl C H O KOH 6 6 3 Leveling bottle CO O CO 2 2 Absorbed Absorbed Absorbed (a) (b) FIGURE 15.4 A typical Orsat analyzer. (a) Schematic: Vessel A contains a potassium hydroxide (KOH) solution that absorbs CO . Vessel B contains 2 a pyrogallic acid (1,2,3-trihydroxybenzene, C H O ) solution that absorbs O . Vessel C contains a cuprous chloride (CuCl) solution 6 6 3 2 that absorbs CO. The remaining gas is assumed to be N . Vessel D is the measuring chamber, and vessel E is the leveling bottle. 2 (b) Photograph of a typical Orsat analyzer. EXAMPLE 15.3 The exhaust gas of a gasoline-fueled automobile engine is cooled to 20.0°C and subjected to an Orsat analysis. The results (on a volume or mole basis) are CO = 7:10% 2 CO = 0:800% O = 9:90% 2 N = 82:2% 2 Total = 100:% Determine a. The hydrocarbon model (C H ) of the fuel. n m b. The composition of the fuel on a molar and a mass basis. c. The air-fuel ratio on a molar and a mass basis. d. The % of theoretical air used in the combustion process. Solution a. Since the Orsat analysis is carried out at 20.0°C, we assume that virtually all the water of combustion has condensed out and therefore the composition given is on a dry basis. However, the water term must be left in the chemical reaction15.4 Fuel Modeling 601 equation, since it results from the oxidation of the hydrogen in the fuel. For convenience, we write the combustion reaction for 100. moles of dry product formed by burning 1.00 mole of the fuel model C H . Using the given n m combustion analysis, we have C H +a½O +3:76ðN Þ 7:10ðCO Þ+0:800ðCOÞ+9:90ðO Þ+bðH OÞ+82:2ðN Þ n m 2 2 2 2 2 2 The element balances are Carbon (C) balance: n=7.10 + 0.800=7.90 Hydrogen (H) balance: m = 2b Nitrogen (N ) balance: 3:76a = 82:2, or a = 82:2/3:76 = 21:9 2 Oxygen (O ) balance: a = 21:9 = 7:10+0:800/2+9:90+b/2, or b = 9:00: 2 Then, from the preceding hydrogen balance, m = 2b = 18:0: Consequently, the fuel model is C H , which is 7.90 18.0 approximately octane, C H . The final reaction equation is 8 18 C H +21:9½O +3:76ðN Þ 7:10ðCO Þ+0:800ðCOÞ+9:90ðO Þ+9:00ðH OÞ+82:2ðN Þ 7:90 18:0 2 2 2 2 2 2 b. On a molar basis, 1.00 mole of fuel contains 7.90 moles of C and 18.0 moles of H, and on a molar percentage basis, this becomes 7.90/(7.90 + 18.0)(100.)=31.0% C and 18.0/(7.90 + 18.0)(100)=69.0% H. The molecular mass of the fuel in this model is M = 7:90ð12Þ+18:0ð1Þ = 113kg/kgmole = 113lbm/lbmole fuel and so the fuel’s composition on a mass basis is ðÞ 7:90kgmoleC/kgmolefuelðÞ 12:0kgC/kgmoleC /ðÞ 113kgfuel/kgmolefuel = 0:840kgC/kgfuel = 0:840lbmC/lbmfuel and ðÞ 9:00ðÞ 2:016 /113 = 0:161kgH/kgfuel = 0:161lbmH/lbmfuel Therefore, the fuel can be said to consist of 31% carbon and 69% hydrogen on a molar basis or 84% carbon and 16% hydrogen on a mass basis. c. Referring to the final combustion equation determined in part a, the air/fuel ratio on a molar basis is 21:9×ð1+3:76Þmolesof air n air ðA/FÞ = = = 104molesair/molefuel molar n 1moleof fuel fuel and on a mass basis it is  28:97kgair/kgmoleair ðA/FÞ =ðÞ 104kgmoleair/kgmolefuel × mass 113kgfuel/kgmolefuel = 26:7kgair/kgfuel = 26:7lbmair/lbmfuel d. To determine the percent of theoretical air used, we must first determine the minimum air required for complete combustion. The reaction for 100.% theoretical air has the form C H +a½O +3:76ðN Þ bðCO Þ+cðH OÞ+dðN Þ 7:90 18:0 2 2 2 2 2 The element balances are Carbon (C) balance: 7.90=b Hydrogen (H) balance: 18.0=2c,or c=11.0 Oxygen (O ) balance: a = b+c/2 = 7:90+9:00/2 = 12:4 2 Nitrogen (N)balance:3:76a = d = 3:76ð12:4Þ = 46:6 2 Then the theoretical molar air/fuel ratio (for 100% theoretical air) is 12:4ð1+3:76Þ = = 59:0moleair/molefuel ðÞ A/F molar 1 theoretical finally, the percent of theoretical air used in the actual combustion process is " , %of theoreticalair =ðÞ A/F ðÞ A/F × 100 molar molar actual theoretical = 104/59:0 ð100Þ = 177% or 77.0% excess air. (Continued)602 CHAPTER 15: Chemical Thermodynamics EXAMPLE 15.3 (Continued) Exercises 7. Determine the hydrocarbon fuel model in Example 15.3 if the CO and O concentrations are both 0.00%, and the CO 2 2 and N concentrations are 17.0% and 83.0%, respectively, in the Orsat analysis. Answer: Hydrocarbon fuel model= 2 C H . 17 20.3 8. Determine the molar and mass air/fuel ratios in Example 15.3 if the CO and O concentrations are both 0.00%, and the 2 CO and N concentrations are 17.0% and 83.0%, respectively, in the Orsat analysis. Answer: A/F=10.15 kgmole air/ 2 2 kgmole fuel=1.31 kg air/kg fuel. 9. If the CO concentration in Example 15.3 is 0.00% and the N concentration is 83.0%, with all the other concentrations 2 unchanged, determine the hydrocarbon fuel model and % of excess air used in the combustion process. Answer: Hydrocarbon fuel model=C H and % of excess air=74.2%. 7.1 22.28 In the previous example, we assume that nearly all the water produced by the combustion process condensed out by the time the combustion products cooled to 20.0°C. For this to be a valid assumption, the dew point of the combustion products must be at 20.0°C or higher. The determination of the dew point temperature for this reaction is illustrated in the next example. EXAMPLE 15.4 Determine the dew point temperature of the combustion products given in Example 15.3 if the total pressure of the mixture is 14.7 psia. Solution From Eq. (12.23) of Chapter 12, the volume fractions, mole fractions, and partial pressure ratios are all equal for a mixture of ideal gases. Exhaust products at atmospheric pressure are sufficiently ideal to allow us to determine the water vapor par- tial pressure at its condensation temperature (i.e., dew point) from this relation. The total number of moles of product, from part a of Example 15.3, is 109 moles. Then, using Eq. (12.23) wherein p is the total pressure of the mixture gives m 9:00 π = p /p =ψ =χ = = 0:0826 H O H O m 2 2 H O H O 2 2 109 whereπ is the partial pressure ratio,ψ is the volume fraction, andχ is the mole fraction. So, p = 0:0826ðÞ 14:7 = 1:21psia H O 2 The saturation temperature of water vapor at this pressure is defined to be the dew point temperature. By interpolation in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that TðÞ 1:21psia = T = 108°F = 42:3°C sat DP Thus, the exhaust products must be cooled to 108°Fð42:3°CÞ or below to condense the water of combustion and have an essentially dry exhaust gas. Exercises 10. Determine the partial pressure of the water vapor in Example 15.4 if the mixture total pressure is 0.150 MPa. Answer: p = 12.4 kPa. H O 2 =178 psia. 11. If the dew point temperature in Example 15.4 is 212°F, what is the mixture total pressure? Answer: p m 12. Determine the partial pressure and dew point temperature of the water vapor present in the 100.% theoretical air combustion process given in part d of Example 15.3. Assume the mixture total pressure is 14.7 psia. Answer: p = H O 2 2.08 psia, T =128°F. DP WHY DO AUTOMOBILE EXHAUST SYSTEMS RUST? Water condenses in an automobile’s exhaust system and drips out the tailpipe until the entire exhaust system has been heated above the dew point temperature by the exhaust gases. This water promotes corrosion and causes the exhaust sys- tem to rust out sooner if the vehicle is used for short trips than trips long enough (a half hour or more) to dry out the exhaust system.15.5 Standard Reference State 603 If moisture enters the combustion process as humidity in the inlet air, this moisture is carried through the reaction as an inert element and adds to the combustion water in the products. This has the net effect of raising the dew point temperature. This is illustrated in the next example. EXAMPLE 15.5 During the automobile engine fuel combustion test discussed in Example 15.3, the dry bulb and wet bulb temperatures of the inlet air were measured to be 90.0°F and 75.0°F, respectively. Determine (a) the amount of water carried into the engine in the form of inlet humidity and (b) the new dew point temperature of the exhaust products. Assume the exhaust is at a total pressure of 14.7 psia. Solution From the psychrometric chart, Figure D.6a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, for T =90°Fand T =76°F, the relative humidityϕ=50% and the humidity ratio,ω=(105 grains of H O per lbm of DB WB 2 dry air) × (1 lbm/7000. grains)=0.0150 lbm H O/lbm dry air. On a molar basis, the humidity ratio is 2  28:97lbmdryair/lbmoledryair ω = 0:0150lbmH ðÞ O/lbmdryair 2 18:016lbmH O/lbmoleH O 2 2 = 0:0241lbmoleH O/lbmoledryair 2 From the balanced reaction equation of part a of Example 15.3, we find that the amount of dry air used per mole of fuel is 21.9(1 + 3.76)=104 moles, and this now carries with it 0.0241(104)=2.51 moles of water. Assuming this water passes through the reaction unchanged, the total amount of water now in the exhaust is 9:00+2:51 = 11:5 moles per mole of fuel. Consequently, the total moles of product are 111.5, the mole fraction of water vapor in the exhaust is now n H O 2 χ = = 11:5/111:5 = 0:103 H O 2 n total and Eq. (12.23) gives the partial pressure of the water vapor in the exhaust as p = 0:103ð14:7Þ = 1:52psia H O 2 Again, interpolating in Table C.1a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find TðÞ 1:52psia = T = 116°F = 46:5°C sat DP Exercises 13. What happens to the water vapor in the engine’s exhaust in Example 15.5 if the surrounding air temperature is 20.0°C? Answer: It condenses into liquid water, since the surrounding air temperature is less than the dew point temperature of the water vapor. 14. If the inlet air in Example 15.5 contains 3.214 moles of water vapor per mole of fuel burned (instead of 2.50 moles of H O per mole of fuel), determine the new dew point temperature. Answer: T =118°F. 2 DP 15. If the inlet air in Example 15.5 has a relative humidity of 100.% and a dry bulb temperature of 90.0°F, what is the new exhaust dew point temperature? Answer: T =123°F. DP By comparing the results of Examples 15.4 and 15.5, we see that combustion air with 50.0% relative humidity has a dew point temperature 7.6°F (4.2°C) higher than that of dry combustion air. 15.5 STANDARD REFERENCE STATE Because we deal with a variety of elements and compounds in combustion reactions, it is necessary to define a common thermodynamic reference state for all these substances. Recall that, in developing the steam tables, we chose the triple point of water as the reference state and arbitrarily set the specific internal energy of liquid water equal to zero at that point. Therefore, the values of u and h in the steam tables are not the actual specific internal energies and enthalpies of steam, they are only relative values. This is sufficient, since most of our formulae use u − u or h − h for changes occurring within a system and the effect of the reference 2 1 2 1 state cancels out in the subtraction process. In the case of the gas tables, we take 0 K and 1 atm as the thermodynamic reference state and arbitrarily set the specific internal energy equal to zero at this state. How- ever, in the case of combustion processes, a more pragmatic thermodynamic reference state of 25.0°Cand 0.100 MPa (approximately 1 atm) is chosen. This is called the standard reference state (SRS) for combustion reactions. However, since most of the calorimeters used to study combustion processes are steady state, steady flow, open systems, it is more convenient to set the specific enthalpy rather than the specific internal energy of the elements equal to zero at this state.604 CHAPTER 15: Chemical Thermodynamics WHAT IS THE STANDARD REFERENCE STATE? The standard reference state (SRS) is defined by the following temperature and pressure: SRStemperature = T°= 25:0°C = 298K = 77:0°F = 537R SRSpressure = p°= 0:100MPa = 14:5psia≈1atm Consequently, the specific internal energies of the elements at the SRS are always negative and computed from u°= −p°v°,where p°= 0:100 MPa and v° is the corresponding specific volume of the element in question. Thermo- dynamic properties atthe standardreferencestate are alwaysdenoted bya superscript °. 15.6 HEAT OF FORMATION When a reaction gives off or liberates heat, the reaction is said to be exothermic, and when it absorbs heat, it is said to be endothermic. Our sign convention for heat transport of energy requires that Q 0whereas exothermic Q 0. The heat of formation of a compound is the heat liberated or absorbed in the reaction when the endothermic compound is formed from the stable form of its elements at the standard reference state. For example, if the elements and the resulting compound are both at the standard reference state, then we can write  ° ElementsðÞ attheSRS CompoundðÞ attheSRS + q f compound  ° where q is the molar heat of formation of the compound at the standard reference state. f compound In 1840, the Swiss chemist Germain Henri Hess (1802–1850) discovered that the total amount of heat liberated or absorbed during a chemical reaction is independent of the thermodynamic path followed by the reaction. This is known as Hess’s law or the law of constant heat sums. It allows us to determine heats of formation for com- pounds that cannot be synthesized directly from their elements. For example, the complete combustion of a hydrocarbon compound of the form C H in pure oxygen, wherein n m the reactants and the products are both maintained at the standard reference state, can be written as C H +aðO Þ nðCO Þ+ðÞ m/2ðH OÞ+HHV n m 2 2 2 C H n m where HHV is the higher heating value of the hydrocarbon (defined later, see Tables 15.2 and 15.3). We also have the following carbon dioxide and water formation reactions: C+O CO −393:5MJ/kgmoleCO 2 2 2 and H +ð1/2ÞðO Þ H O−285:8MJ/kgmoleH O 2 2 2 2 Now Hess’s law states that the heats liberated or absorbed in these reactions are independent of the reaction path, so we can rearrange them as CO C+O +393:5MJ/kgmoleCO 2 2 2 WHAT IS HESS’S LAW? Using caloric theory, Henri Hess tried to extend Dalton’s interpretation of chemical reactions by attempting to find exam- ples of the combination of caloric with chemical elements in simple mass ratios. He discovered that, for a given reaction, the total amount of caloric (heat) involved was always the same, independent of the number of intermediate steps con- tained within the reaction. Today, we know that this is really true only for aergonic, steady state, steady flow, open systems and for isobaric, closed systems where the heat of reaction equals the change in total enthalpy (because enthalpy is a point function and therefore independent of the actual chemical path taken by the reaction).15.6 Heat of Formation 605 and H O H +ð1/2ÞðO Þ+285:8MJ/kgmoleH O 2 2 2 2 Then, the combustion equation for the compound C H can be written as n m C H +aðO Þ nðC+O +393:5Þ+ðÞ m/2ðÞ H +ðÞ 1/2 O +285:8 +HHV n m 2 2 2 2 C H n m nðCÞ+ðÞ m/2ðÞ H +ðÞ n+m/4ðÞ O 2 2 + nðÞ 393:5 +ðÞ m/2ðÞ 285:8 +HHV C H n m Now, an oxygen balance on the original compound combustion equation gives a=n + m/4, so the O terms in 2 the previous equation cancel, and again using Hess’s law to rearrange this equation, we get nðÞ C +ðÞ m/2ðÞ H C H − nðÞ 393:5 +ðÞ m/2ðÞ 285:8 +HHV 2 n m C H n m  (15.4) ° C H + q n m f C H n m where the HHV is in MJ/kgmole of compound. Consequently, the heat of formation in MJ/kgmole of the hydro- carbon fuel C H at the standard reference state is n m  ° q = −½nð393:5Þ+ðm/2Þð285:5Þ+HHV inMJ/kgmole (15.5) C H f n m C H n m The use of this relation is illustrated in the following example. EXAMPLE 15.6 To prevent the Universe from collapsing in a deadly hypergeometric spiral, you must quickly determine the heat of forma- tion of methane gas CH (g) at the standard reference state. Normally, you would react carbon and hydrogen gas in your 4 laboratory to form methane and measure its heat of formation during the reaction. Unfortunately, there is no known reac- tion by which you can form methane by reacting solid carbon with hydrogen gas. How will you save the Universe? Solution Even though we do not know how to form CH (g) from a direct reaction of solid carbon and hydrogen gas, we can use 4 Eqs. (15.4) and (15.5) with Tables 15.2 and 15.3 to calculate the heat of formation of CH (g) at the standard reference 4 state. Tables 15.2 and 15.3 give the higher heating value (HHV) of CH as−890.4 MJ/kgmole. Then Eq. (15.5) gives 4 ° ðq Þ = −½nð393:5Þ+ðm/2Þð285:5Þ+HHV  C H f CH n m 4 = −½1ð393:5Þ+ð4/2Þð285:5Þ+HHV  CH 4 = −½393:5+2ð285:8Þ+ð−890:4Þ = −74:7MJ/kgmoleof CH 4 and then Eq. (15.4) becomes CðsÞ+2½H ðgÞ CH ðgÞ−74:7MJ/kgmoleof CH 2 4 4 In this example, we denote the physical state of the substances in parentheses as solid (s), liquid (ℓ), or gas (g). Note that the negative sign on the HHV of methane indicates that the combustion of methane is an exothermic (heat-producing) reaction. Exercises 16. The methane in Example 15.6 is replaced by acetylene gas. You must now determine the heat of formation of acetylene ° gas, C H (g), at the standard reference state to save the Universe. Answer: ðq Þ =+227MJ=kgmole. 2 2 acetylene f 17. Oops, it is not methane or acetylene gas in Example 15.6, it is ethylene gas. So now, to save the Universe, you must determine the heat of formation of ethylene gas, C H (g), at the standard reference state. Answer: 2 4 ° ðq Þ = −52:4MJ=kgmole. ethylene f 18. Well, I bet you are surprised to find out that, at the last minute, it was ethane gas that was actually used to generate the deadly hypergeometric collapse of the Universe. To save all life in the Universe, you must now determine the heat of ° formation of ethane gas, C H (g), at the standard reference state. Answer: ðq Þ = −84:5MJ=kgmole: 2 6 f ethane Notice that, in this example, we do not take into account the heat of formation of H from atomic hydrogen H 2 or O from atomic oxygen O. This is because the elements used in the formation of a compound must be in 2 their stable molecular forms at the standard reference state. In the case of methane, the elements are solid carbon (graphite), C, and diatomic hydrogen gas, H . 2606 CHAPTER 15: Chemical Thermodynamics Consider a chemical reaction occurring in the steady state, steady flow, System boundary aergonic, open system shown in Figure 15.5. The energy rate balance (ERB) applied to this system yields Reaction Reactants Products vessel _ _ _ _ _ _ _ Q =∑ðmhÞ−∑ðmhÞ =∑ðnhÞ−∑ðnhÞ = H −H P R r P R P R _ where Q is the exothermic or endothermic heat transfer rate of the reac- r Q r _ _ tion, and H and H are the total enthalpy rates of the products and reac- P R tants, respectively. FIGURE 15.5 A steady state, steady flow, aergonic reaction vessel. If this reaction is to be used to determine the standard reference state heat of formation of a compound, then the temperature and pressure of the reactants and the products must be maintained by sufficient heat transfer at 25.0°C and 0.100 MPa. In this case, the specific enthalpies of all the reactant elements are zero (by _ _ _ _ definition), so H ° ° = H = H = H = 0, and the previous equation reduces to R R elements elements _ _ _° _ ° °= H = H Q°= Q P compound r f _ ° where Q is the standard reference state heat rate of formation of the compound. In this case, the heat rate of for- f mation, the heat transfer rate of the reaction, and the total enthalpy rate of the products are all equal. We now define the molar specific enthalpy of formation, h°, of a compound at the standard reference state as f    ° _ _ °/ ðh°Þ = H°/n_ = Q n_ = q (15.6) f f compound compound compound f compound ° where n_ and q are the molar flow rate and the molar heat of formation of the compound, respectively. f Because of our sign convention that heat energy entering the system is positive while that leaving the system is negative, the heats and enthalpies of formation of exothermic reactions are always negative,whilethoseof endothermic reactions are always positive. Table 15.1 gives the specific molar enthalpies (heats) of formation for some common compounds. Heats of formation can also be estimated from the atomic bond energies of the compound. Table 15.1 Molar Specific Enthalpy of Formation at 25.0°C (77.0°F) and 0.100 MPa h° f Substance M kg/kgmole or lbm/lbmole MJ/kgmole Btu/lbmole Carbon monoxide, CO(g) 28.011 −110.529 −47,522 Carbon dioxide, CO (g) 44.011 −393.522 −169,195 2 Sulfur dioxide, SO (g) 64.07 −296.83 −127,622 2 Water, H O(g) 18.016 −241.827 −103,973 2 Water, H O(ℓ) 18.016 −285.838 −122,896 2 Methane, CH (g) 16.043 −74.873 −32,192 4 Acetylene, C H (g) 26.038 +226.731 +97,483 2 2 Ethylene, C H (g) 28.054 +52.283 +22,479 2 4 Ethane, C H (g) 30.070 −84.667 −36,403 2 6 Propane, C H (g) 44.097 −103.847 −44,649 3 8 Butane, C H (g) 58.124 −126.148 −54,237 4 10 Benzene, C H (g) 78.114 +82.930 +35,656 6 6 Octane, C H (g) 114.23 −208.447 −89,622 8 18 Octane, C H (ℓ) 114.23 −2411.952 −107,467 8 18 Carbon, C(s) 12.011 0 0 Oxygen, O (g) 32.00 0 0 2 Hydrogen, H (g) 2.016 0 0 2 Nitrogen, N (g) 28.013 0 0 2 Note: Here, (g) indicates gas or vapor state and (ℓ) indicates liquid state. Source: Van Wylen, G. J., Sonntag, R. E., 1976. Fundamentals of Classical Thermodynamics, SI Version, second ed. Wiley, New York, p. 496 (Table 12.3). Copyright © 1976 John Wiley & Sons. Reprinted by permission of John Wiley & Sons.15.7 Heat of Reaction 607 EXAMPLE 15.7 The formation of 1 mole of water by the combustion of its elements can be written as H +0:5×O H O+q 2 2 2 r where q is the heat transfer that occurs per mole of water formed. Suppose we wanted to make exactly 0.160 kg of liquid r water by this combustion reaction. Determine the heat transfer required to keep both the reactants (H and O ) and the pro- 2 2 ducts (H O) at the standard reference state (25.0°C and 0.100 MPa) while this reaction takes place. 2 Solution For the isothermal (25.0°C) and isobaric (0.100 MPa) combustion of 1.00 kgmole of hydrogen gas with 0.500 kgmole of oxygen gas, Eq. (15.6) gives ° ° ðq Þ = q = h f r SRS f and since the reaction is at 25.0°C, the water formed is in the liquid state. Then, from Table 15.1, we find that ðh°Þ = f H2OðlÞ ° 285.838 MJ/kgmole. So, ðq Þ =285.838 MJ/kgmole, and the heat transfer required for this reaction on a mass rather f H2OðlÞSRS than a molar basis is q r q = r M where M is the molecular mass. Finally, the total heat transfer required is Q =mq,or r r  q −285:838MJ/kgmole r Q = mq = m =ð0:160kgÞ = −2:54MJ r r M 18:016kg/kgmole Exercises 19. Rework Example 15.7 and determine the heat transfer required for the formation of 0.160 kg of water vapor H O(g) at 2 the SRS rather than liquid water. Answer: ðQ Þ = −2:15MJ: r H OðgÞ 2 20. Repeat Example 15.7 and determine the heat transfer required for the formation of 1.00 kg of methane gas CH (g) when 4 both the reactants and the products are at the standard reference state. Answer: ðQ Þ = −4:67MJ: r CH ðgÞ 4 21. Use the technique of Example 15.7 to determine the heat transfer required for the formation of 1.00 gallon (6.25 lbm) of liquid octane C H . Answer: ðQ Þ = −5880Btu: 8 18(ℓ) r C H ðℓÞ 8 18 15.7 HEAT OF REACTION In the previous section, we saw that the heat of formation of a compound was the same as the heat of reaction when that compound was formed from its elements at the standard reference state. An oxidation reaction of a fuel is normally called combustion; therefore, the heat of reaction of the oxidation of a fuel in air or pure oxygen is also known as the heat of combustion or the heating value of the fuel. A bomb calorimeter is a closed, rigid (constant volume) vessel that can be used to determine the heat of reaction of a liquid or solid fuel sample (see Figure 15.6). When the final temperature in the bomb has been reduced to its initial standard state temperature of 25.0°C by the water bath, the resulting energy balance on the bomb is Q = mðu −u Þ = nðu −u Þ = U −U (15.7) r P R P R P R Then, from the definition of enthalpy, we can write H −H = U −U +ðpVÞ −ðpVÞ P R P R P R and, since the reactants are likely to be solids or liquids with a small volume and a low pressure, we can ignore them and set ðpVÞ ≈0, then R H −H = U −U +ðpVÞ = Q +ðnℜTÞ P R P R r P P which provides a convenient relation between the constant volume heat of reaction measured by the bomb calorimeter and the total enthalpy change of the reaction occurring inside the bomb calorimeter. The heat of reaction of gases, liquids, and some solids is more often measured in a steady state, steady flow, aer- gonic calorimeter, similar to that shown in Figure 15.5. An energy rate balance on this type of calorimeter gives the heat transfer rate of the reaction as _ _ _ Q = H −H P R r608 CHAPTER 15: Chemical Thermodynamics Mixer 110 V motor Thermometer Fuse wire Test sample d Bomb (close rigid vessel) Insulation Water FIGURE 15.6 An adiabatic, constant volume bomb calorimeter. _ and dividing through by the fuel molar flow rate n gives the molar heat of reaction q as fuel r _ _ q = Q /n = h −h = h (15.8) fuel P R RP r r In this equation, we define the quantities _ _ h = H /n_ =∑ðÞ n/n_ h =∑ðÞ n/n h (15.9) R R fuel i fuel i i fuel i R R and _ h = H /n_ =∑ðÞ n_ /n_ h =∑ðÞ n/n h : (15.10) P P fuel i fuel i i fuel i P P where i=1, 2, 3, … , n, and n is the number of reactants or products. In these equations, h and h are the total R P reactant and product enthalpies per unit mole of fuel consumed. Combining Eqs. (15.9) through (15.11) gives the molar heat of reaction as n n i i q =∑ h −∑ h (15.11) r i i n n fuel fuel P R The heating value of a fuel is the heat of reaction produced by the complete combustion of a unit mole (or mass) of the fuel when both the reactants and the products are maintained at the standard reference state (SRS). When the fuel contains hydrogen, the combustion products contain water that can be in either the liquid or vapor phase. The higher heating value (HHV) is produced when the water in the combustion products is condensed into the liquid state, and the lower heating value (LHV) occurs when this water is in the vapor state: HHV = qðÞ attheSRSandliquidH O 2 r LHV = qðÞ attheSRSandH Ovapor 2 r The relation between these two heating values is simply   n n H O H O MJ 2 2 HHV = LHV− ðh°Þ = LHV− 44:00 (15.12) fg H O 2 n n kgmoleH O fuel fuel 2 where both the HHV and the LHV are in MJ/kgmole of fuel. In this equation, n /n is the number of moles H O fuel 2 of water produced per mole of fuel burned, and ðh°Þ = 44:00MJ=kgmole is the phase change molar specific fg H O 215.7 Heat of Reaction 609 CRITICAL THINKING In the calculation of the higher and lower heating values, HHV and LHV, both the reactants and the products are at the SRS. In the HHV calculation, all the water in the combustion products is condensed into a liquid, but in the LHV calcula- tion, all this water is in the vapor phase. Wait a minute, how can you have all the LHV water in the vapor phase at the SRS temperature of 25.0°C, which is clearly below the dew point temperature of water at the SRS pressure of 0.100 MPa (T = DP T (0.100 MPa)=100°C)? The answer is that you cannot, so the LHV state does not really exist. But, if the LHV can never sat be achieved, then what good is it? The answer is that the LHV is merely a benchmark value used for evaluating combustion processes. For example, the HHV and LHV are most commonly used in the calculation of combustion efficiency η , combustion defined as Actual heat produced by a combustion process η = combustion HHVorLHV In this calculation, the HHV is used if the temperature of the combustion products can be reasonably lowered below 100.°C (as in industrial power plant boilers or domestic home furnaces), and the LHV is used when this is not possible (as in inter- nal combustion engines). When the temperatures of the combustion products are necessarily above 100.°C, the LHV makes a more reasonable benchmark for the calculation of a combustion efficiency (note that, since the LHV is less than the HHV, the combustion efficiency is larger when the LHV is used). enthalpy of water at the standard reference state temperature (25.0°C or 77.0°F). Note that both the HHV and the LHV are negative for heat-producing combustion reactions. If the reactants and the products are both at the standard reference state, then Eq. (15.11) gives the standard refer- ence state heat of reaction, q°,as r   q°=∑ðÞ n/n h° −∑ðÞ n/n h° (15.13) i fuel f i fuel f r i i P R This equation can be used to determine the HHV and LHV heats of combustion tabulated in Tables 15.2 and 15.3, as illustrated in the following example. Table 15.2 Molar Based Higher Heating Values (HHV) for Common Fuels Where Both the Reactants and the Products Are at the SRS and the Water in the Combustion Products Is in the Liquid Phase (HHV) molar Fuel MJ/kgmole Btu/lbmole Hydrogen, H (g) −285.84 −122,970 2 Carbon, C(s) −393.52 −169,290 Carbon monoxide, CO(g) −282.99 −121,750 Methane, CH (g) −890.36 −383,040 4 Acetylene, C H (g) −1299.60 −559,120 2 2 Ethylene, C H (g) −1410.97 −607,010 2 4 Ethane, C H (g) −1559.90 −671,080 2 6 Propylene, C H (g) −2058.50 −885,580 3 6 Propane, C H (g) −2220.00 −955,070 3 8 n-Butane, C H (g) −2877.10 −1,237,800 4 10 Benzene, C H (g) −3270.0 −1,406,000 6 6 n-Octane, C H (g) −5454.5 −2,345,000 8 18 n-Decane, C H (g) −6754.7 −2,904,000 10 22 Methyl alcohol, CH OH(g) −764.54 −328,700 3 Ethyl alcohol, C H OH(g) −1409.30 −606,280 2 5 Source: Reprinted by permission of the publisher from Holman, J. P., 1980. Thermodynamics, third ed. McGraw-Hill, New York, p. 466 (Table 11-1).610 CHAPTER 15: Chemical Thermodynamics Table 15.3 Mass Higher Heating Values (HHV) for Common Fuels Where Both the Reactants and the Products Are at the SRS and the Water in the Combustion Products Is in the Liquid Phase (HHV) mass Fuel MJ/kg Btu/lbm Hydrogen, H (g) −142.9 −61,485 2 Carbon, C(s) −32,79 −14,108 Carbon monoxide, CO(g) −10.11 −4,348 Methane, CH (g) −55.65 −23,940 4 Acetylene, C H (g) −49.98 −21,505 2 2 Ethylene, C H (g) −50.39 −21,679 2 4 Ethane, C H (g) −52.00 −22,369 2 6 Propylene, C H (g) −49.01 −21,085 3 6 Propane, C H (g) −50.45 −21,706 3 8 n-Butane, C H (g) −49.61 −21,341 4 10 Benzene, C H (g) −41.92 −18,026 6 6 n-Octane, C H (g) −47.85 −20,570 8 18 n-Decane, C H (g) −47.57 −20,451 10 22 Methyl alcohol, CH OH(g) −23.89 −10,272 3 Ethyl alcohol, C H OH(g) −30.64 −13,180 2 5 Source:Reprinted bypermission ofthe publisher from Holman, J.P., 1980. Thermodynamics, third ed. McGraw-Hill, New York,p.466 (Table 11-1). EXAMPLE 15.8 Determine the higher and lower heating values of methane. Note that, for the determination of the HHV and LHV, the com- bustion reaction must occur with 100.% theoretical air and both the reactants and the products must be at the standard reference state. For the HHV calculation, the water in the combustion products must be in the liquid phase, and for the LHV calculation, it must be in the vapor phase. Solution For 100.% theoretical air, the combustion equation for methane is CH +2:00 O +3:76ðÞ N CO +2:00ðÞ H O +7:52ðÞ N 4 2 2 2 2 2 Since both the reactants and the products are at the standard reference state, we can use Eq. (15.13) to find the heat of combustion, which is either the HHV or the LHV, depending on how the water term is handled. Then,     ° ° ° ° ° h =∑ðÞ n/n h = h +2:00 h +7:52 h R i fuel f f f f i CH4 O2 N2 R and, from Table 15.1, we find that  ° h = −74:873MJ/kgmoleCH f 4 CH 4   ° ° Since O and N are the elements of the compound CH , in their standard states (see Table 15.1), h = h = 0: Then, f f 2 2 4 O N 2 2 h°= −74:873MJ/kgmoleCH : Similarly, R 4     ° ° ° ° h =∑ðÞ n/n h = h +2:00 h +7:52 h° N P i fuel f f f f 2 i CO H OðÞ ℓ 2 2 P  ° From Table 15.1, we find that h = 0, and f N2  ° h = −393:522MJ/kgmoleCO f 2 CO 2  ° h = −241:827MJ/kgmoleH Ovapor f 2 H OgðÞ 2  h° = −285:838MJ/kgmoleH Oliquid f 2 H OðℓÞ 2

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