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Combustion Chemistry Hai Wang Stanford University 2015 Princeton-CEFRC Summer School On Combustion Course Length: 3 hrs June 22 – 26, 2015 Copyright ©2015 by Hai Wang This material is not to be sold, reproduced or distributed without prior written permission of the owner, Hai Wang. Lecture 1 1. THERMOCHEMISTRY Thermodynamics is the foundation of a large range of physical science problems. It provides us with a basic understanding about the driving force of a physical process and the limits of such a process. The development of thermodynamic theory was intimately related to combustion. In particular, the second law of thermodynamics was conceived largely to prove that it is impossible to operate a perpetual motion machine. 1.1 Thermodynamic First Law The first law of thermodynamics states that the energy is conserved when this energy is transformed from one form to another. In the context of combustion analysis, we state that for a control mass (or the working fluid), Q −W = ΔU (1.1) where Q (kJ or kcal) is the heat transferred from the working fluid to the surrounding, W (kJ or kcal) is the work done by the working fluid to the surrounding, and U (kJ or kcal) is the internal energy of the working fluid. It is important to mention that energy transformation always involves a process that has a initial state (1) and a final state (2). ΔU = U2 – U1 is therefore the change of internal energy of the working fluid from the initial state to the final state. Thermodynamic analysis follow the convention that if the working fluid gives off heat to the surrounding, Q < 0 , and if the working fluid receives heat from the surrounding, Q is positive. For example, in a simple cooling process, a fluid loses its internal energy to the surrounding (i.e., lowering its temperature) and thus ΔU < 0. Assuming that no work is done (W = 0), then Q < 0. Likewise, if the working fluid does net work to the surrounding, W is positive, and if the surrounding does net work to the working fluid, W < 0 (e.g., for adiabatic compression (Q = 0), the work done by the surrounding to the working fluid serves to raise the internal energy of the working fluid, since –W = ΔU > 0. The symbol U designates the internal energy of a given mass of a substance. Internal energy is a measure of the total energy of the control mass. For example, excluding nuclear energy the internal energy of air is a sum of the kinetic energy of each atom. Therefore, the internal energy can be made a material property if it is defined as the internal energy per mass, an intensive property, denoted here as u (kJ/kg) or u (kJ/kmol). In this course, we shall follow the notation that intensive properties are expressed in lower cases. Assuming that we are running a thermodynamic process that the only work done during the process is that associated with boundary work under a constant pressure P1 = P2 = P (e.g., a piston work), the work done may be calculated from 1-1 Stanford University Version 1.2 ©Hai Wang 2 W = ∫ PdV = P (V2 − V1 ) . 1 (1.2) Putting Eq. (1.2) into Eq. (1.1), we have Q = (U 2 + P2V2 ) − (U 1 + P1V1 ) , (1.3) where V is the volume of the working fluid. In this case, the heat transferred during the process corresponds to a net change of the controlled mass in the quantity U + PV between the initial and final states. We find it convenient to define a new thermodynamic property, the enthalpy H = U + PV , h = u + Pv , h = u + Pv . (1.4a) (1.4b) (1.4c) Here v and v are the specific volumes, having the units (m3/kg) and (m3/kmol) respectively. Clearly, these specific volumes are related to the mass density ρ and molar density (or concentration) c, respectively, i.e., v = 1/ρ and v = 1/c. In general, the internal energy u and enthalpy h depend on only two independent properties that specifying the thermodynamic state, e.g., (T, P), (T, v), or (P, v). For a low-density gas like air or combustion gases, T, P, and v are related by the ideal gas law or the equation of state, Pv = R 'T , Pv = Ru T , (1.5a) (1.5b) where Ru is the universal gas constant (8.314 kJ/kmol-K), R’ is the specific gas constant and equal to Ru/MW, and MW is the molecular weight of the substance. For a low-density gas, the internal energy is primarily a function of T, i.e., u ≅ u (T ) . This relationship may be expressed by defining a constant-volume specific heat cv (kJ/kmol-K) ⎛ ∂u ⎞ . cv = ⎜ ⎟ ⎝ ∂T ⎠v (1.6) For an ideal gas we have du = cv dT . Likewise, the relationship between enthalpy and temperature may be established by defining a constant-pressure specific heat c p (kJ/kmol-K) " ∂h % cp = $ ' , # ∂T & p 1-2 (1.7) Stanford University Version 1.2 ©Hai Wang and dh = c p dT . In other words, the two specific heats defined above characterize the heat required to raise the temperature of a substance by 1 K. Since for an ideal gas dh = du + d ( pv ) = du + Ru dT and u ≅ u (T ) , we see that h and c p are also function of temperature only. The relation between dh and du also yields c p = cv + Ru . Here it is important to note that the enthalpy discussed thus far involves only the heating or cooling a substance. This type of enthalpy is known as the sensible enthalpy or sensible heat. Later, we will introduce two other types of enthalpy, one of which is critical to combustion problems. The first law of thermodynamics is quite insufficient to describe energy conversion. Equation (1.1) states that it is possible to cool a substance of a given mass spontaneously (i.e., lowering its internal energy U), and transfer this energy to the surrounding. In other words, within the first law of thermodynamics, it is possible to transform heat from a lowtemperature body to a high temperature body. We know that this cannot be true. The second law of thermodynamics, to be discussed below, will address this problem. 1.2 Thermodynamic Second Law and Entropy In contrast to the first law of thermodynamics, the second law is more difficult to understand. The Kelvin-Planck statement of this law is It is impossible to construct a device that will operate in a cycle and produce no effect other than the raising of a weight and the exchange of heat with a single reservoir. In other words, it is impossible to construct a heat engine that (a) receives heat continuously from a heat reservoir, (b) turns the heat transferred entirely to work, (c) without having to leave any marks on the surrounding. Without diverging into a lengthy discussion of the second law of thermodynamics, let us define entropy S (kJ/K) as ⎛δ Q ⎞ . S =⎜ ⎟ ⎝ T ⎠ int rev (1.8) where (δQ)int rev is the heat a control mass received during an infinitesimal, internally reversible process. Based on an analysis of thermodynamic cycles, it may be shown that for a spontaneous process to occur, the entropy of the control mass must be equal to or greater than zero, ΔS = S2 – S1 ≥ 0. (1.9) Neither the Kelvin-Planck statement nor Eq. (1.8) really tells us what entropy is. An understanding of entropy will have to come sometime later when we introduce statistical thermodynamics. Here let us place some discussion about entropy in a non-rigorous fashion. Entropy is a measure of molecular randomness. This randomness may be measured by the predictability of the positions of atoms in a substance. A crystal material would have a small entropy because atoms are more or less “locked” into the crystal lattice. In fact, the third law of thermodynamics states that the entropy of a pure crystalline substance at absolute 1-3 Stanford University Version 1.2 ©Hai Wang zero temperature is zero. In other words, the atoms in a pure crystal are “frozen” (no oscillation) at 0 K. Therefore, their spatial position is completely predictive. In contrast, a gas would have a large entropy because molecules that make up the gas constantly move about in the space, resulting in small predictability regarding their positions. Moreover, an increase in temperature of the gas leads an increase in the speed of molecular motion and smaller predictability of the molecular positions. In other words, entropy increases with an increase in temperature. In contrast, an increase in pressure leads to closer spacing among molecules. As a result, the molecules become more confined spatially and the entropy is smaller at higher pressures. The dissociation of a chemical substance into gaseous fragments always leads to an increase in entropy since it is harder to predict the spatial positions of the fragments than the molecules of their parent substance. The inequality expressed by Eq. (1.9) basically says that for a spontaneous process to occur, the entropy of the control volume must increase, i.e., natural processes favor more randomness than orderness. Conceptually this makes sense since our experience tells us that a building can spontaneously collapse into a pile of rubble, but a pile of rubble would not spontaneously transform into a building (not without our intervention). Two different gases, say, N2 and O2, would always mix and they never spontaneously separate spatially, leading to better predictability of their positions. The concept of entropy is also deeply rooted in our life. Take the life of a workaholic as an example, the first law of thermodynamics states that it is possible for him/her to receive heat Q in the form of food and hopefully without gaining weight (ΔU = 0), to transform this heat entirely to work W. The second law says that he/she really cannot do this. That is, my office always gets messier over time and I will need to clean it (i.e., not all the heat goes to useful work) as time goes by. Because entropy is a measure of randomness, which in turn, is determined by T and P, it is also a material property. It follows that we can define and denote the entropy of a substance by s (kJ/kg-K) or s (kJ/kmol-K). Although we do not know for the time being how to directly measure entropy, we may develop some relationships that can help us to determine the entropy value. Here we apply the first law to a constant T and P, internally reversible process (e.g., compress a volume immersed in a temperature bath by a piston very slowly), δ Qint rev − δ Wint rev = dU , (1.10) but since δ Qint rev = TdS and δ Wint rev = PdV , we have TdS = dU + PdV (1.11) or ds = du Pdv dT Pdv . + = cv + T T T T 1-4 (1.12) Stanford University Version 1.2 ©Hai Wang Replacing u by h − Ts and rearranging, we obtain ds = dh vdP dT vdP . − = cp − T T T T (1.13) Applying the ideal gas law, we may rewrite equations (12) and (13) as dT dv , + Ru T v dT dP . ds = c p − Ru T P ds = cv (1.14) (1.15) One may integrate the above equations to show that v dT + Ru ln 2 , T v1 2 P dT Δ s = s 2 − s1 = ∫ c p − Ru ln 2 . 1 T P1 2 Δ s = s2 − s1 = ∫ cv 1 (1.16) (1.17) Equation (1.17) states that if c p is a constant, an increase of temperature by ΔT from T causes the entropy to increase by c p ln (1 + ΔT T ) and an increase of pressure by ΔP from P leads to the entropy to decrease by Ru ln (1 + ΔP P ) . Given the third law of thermodynamics, which establish the absolute zero for entropy, the entropy of an ideal gas at a given thermodynamic state (i.e., known T and P) can be easily determined if c p is known. Equation (1.17) also states that unlike enthalpy and internal energy, the entropy of an ideal gas is a function of both temperature and pressure. In application, we define the standard entropy s o as 2 Δ s o = s2 o − s1o = ∫ c p 1 dT ⎡ T2 dT ⎤ ⎡ T1 dT ⎤ = ⎢∫ c p − Ru ln ( P o )⎥ − ⎢ ∫ c p − Ru ln ( P o )⎥ , (1.18) 0 0 T ⎣ T T ⎦ ⎣ ⎦ or T s o = ∫ cp 0 dT − R u ln ( P o ) , T (1.19a) o where P is the standard pressure of 1 atm. Hence, T Δs o = ∫ c p 0 1-5 dT , T (1.19b) Stanford University Version 1.2 ©Hai Wang By tabulating this standard entropy, we may easily determine the entropy change of an ideal gas under an arbitrary condition by s (T , P ) = s o (T ) − Ru ln P . Po (1.20) 1.3 Chemical Reactions Before we apply the above thermodynamic principles to combustion analysis, we need to take a moment to review a few aspects of chemical reactions. From a process point of view, a chemical reaction may be viewed as the conversion of reactants that enter into a reactor (the initial state) to products that leaves the reactor (the final state). For example, methane (CH4) flows into a reactor with air (21%O2 and 79% N2). Suppose the molar ratio of oxygen to methane is 2-to-1. We may write that to burn 1 mole of methane, CH4 + 2 O2 + (2×79/21) N2 → Reactor → CO2 + 2H2O + (2×79/21) N2 . Here the products include 1 mole CO2, 2 moles of H2O and (2×79/21) moles of N2. Of course, in writing the above process reaction, we may neglect the box and simply write CH4 + 2 O2 + (2×79/21) N2 → CO2 + 2H2O + (2×79/21) N2 . (1.21) The above reaction is known as the complete combustion reaction as all the carbon in the fuel is oxidized to CO2 and all the hydrogen is converted to H2O. These compounds are called the complete combustion products. If there is no excess oxygen (i.e., all oxygen is consumed in the oxidation process), the characteristic fuel-to-oxygen molar ratio is known as the stoichiometric ratio (equal to ½ for methane). The stoichiometric ratio for an arbitrary fuel CmHn may be readily determined by writing out the complete, stoichiometric reaction, CmHn + (m+ n 4 ) O2 + (m+ n 4 )×(79/21) N2 → m CO2 + n 2 H2O + (m+ n 4 )×(79/21) N2 . (1.22) which gives the stoichiometric ratio equal to 1/(m+ n 4 ) . In a practical combustion process, however, the fuel-to-oxygen molar ratio needs not to be the stoichiometric ratio. For example, a gasoline engine often runs slightly above the stoichiometric ratio at the cold start, for reasons to be discussed later. To characterize fuelto-oxygen ratio in a practical combustion process, we introduce the equivalence ratio, defined as the molar ratio of fuel-to-oxygen for an actual combustion process by that of stoichiometric combustion: φ= ( moles of fuel moles of oxygen )act. ( moles of fuel moles of oxygen )stoi. 1-6 . (1.23) Stanford University Version 1.2 ©Hai Wang Of course, it may be shown that the equivalence ratio may be calculated using the molar ratio of fuel-to-air or the mass ratio of fuel-to-oxygen or fuel-to-air. By examining the equivalence ratio, we can quickly tell the nature of the fuel/air mixture. That is, if φ = 1, we have stoichiometric reaction; if φ < 1 we have excess oxygen that is not completely used in a reaction process and the combustion is called fuel-lean combustion; if φ > 1 we have excess fuel and the combustion is called fuel-rich combustion. ⎧< 1 fuel lean ⎪ φ = ⎨1 stoichiometric ⎪> 1 fuel rich ⎩ Under the fuel rich combustion, the combustion reaction inherently yields incomplete combustion products, like CO, H2 etc. 1.4 Enthalpy of Formation, Enthalpy of Combustion As we discussed in section 1.1, there are 3 types of enthalpy. The first type is associated with heating or cooling of a substance. The second type is latent enthalpy (or heat). This is the enthalpy associated with the phase change of a substance. For example, the latent heat of evaporation of H2O, hlg , is hlg = h g − hl , (1.24) where h g and hl are the enthalpy of water in its vapor and liquid states, respectively. What is perhaps more important to combustion analysis is the reaction enthalpy. For example, reaction (21) releases an amount of heat due to chemical bond rearrangements. Combining Eqs (1.3) and (1.4a), we have Q = H 2 − H1 . Since state 1 corresponds to the reactants, and state 2 corresponds the products, the above equation states that (a) in a non-adiabatic reactor, the heat released from the reactor is equal to the total enthalpy of the combustion products subtracted by the total enthalpy of the reactant, and (b) since for a combustion process Q < 0, H2 < H1, i.e., the total enthalpy of the products is lower than that of the reactants. The nature of reaction enthalpy is very different from the sensible enthalpy, as the former is due to re-arranging chemical bonds and the latter is simply due to heat and cooling without changing the chemical nature of the substance. To calculate the exact amount of reaction enthalpy and therefore the amount of heat release, we need to first understand the concept and application of enthalpy of formation. 1-7 Stanford University Version 1.2 ©Hai Wang The enthalpy of formation h f at a given temperature is defined as the heat released from producing 1 mole of a substance from its elements at that temperature. By this definition, the enthalpy of formation is zero for the reference elements. These elements are, for example, graphite [denoted by C(S) hereafter), molecular hydrogen H2, molecular oxygen (O2), molecular nitrogen (N2), and molecular chlorine (Cl2). The enthalpy of formation of CO2, say at 298 K, may be conceptually measured by reacting 1 mole of graphite and 1 mole of O2 at 298 K, producing 1 mole of CO2 at the same temperature: C(S) + O2 → CO2 + Q, where Q is the heat released from the above process (Q = –393.522 kJ). Using Eq. (1.24), we have Q = -393.522 (kJ)=H 2 − H1 = 1× h f ,298 K ( CO2 ) − 1× ⎡⎣h f ,298 K ( C(S)) + h f ,298 K (O2 )⎤⎦ = h f ,298 K (CO2 ) . The enthalpy of formation of CO2 is therefore h fo =–393.522 kJ/mol at 298 K. Likewise the enthalpy of formation of CO is determined by measuring the heat release from C(S) + ½ O2 → CO + Q (–110.53 kJ at 298 K), (1.25) and h fo (CO) = –110.53 kJ/mol at 298 K. The conceptual definition uses the same temperature for the reactor, reactants, and products, and this condition is known as the standard condition. For this reason, we use a superscript “o”, i.e., h fo , to designate this standard condition and the corresponding enthalpy of formation is termed as the standard enthalpy of formation. Obviously if reaction (25) is carried out at a different temperature under the standard condition, we do not necessarily get the same heat release. In other words, the enthalpy of formation is dependent on temperature, yet this temperature dependence is related to sensible heat. To illustrate the relationship of enthalpy of formation of a substance at two different temperatures, we sketch the following diagram: 1-8 Stanford University Version 1.2 ©Hai Wang 1 ∑ elements ΔH = ∑ elements ’ h fo,T ’ 2 ⎡⎣ H (T ) − H ( 298 )⎤⎦ = ⎡⎣h (T ) − h ( 298 )⎤⎦ C(S) + ⎡⎣h (T ) − h ( 298 )⎤⎦ O T ΔH CO2 = ⎣⎡h (T ) − h ( 298 ) ⎦⎤ CO 2 2 1 298 K 2 h fo,298 K Elements (1 mole C(s) + 1 mole O2) Substance (1 mole CO2) Recognizing that enthalpy is a state function, i.e., for an ideal gas the enthalpy of a substance is fully defined if the temperature is known, and the change of enthalpy for a process is independent of the path, we may write H 2' − H 1 = H 2' − H 2 + ( H 2 − H 1 ) = "#h (T ) − h (298)$% CO2 + h f ,298 (CO2 ) = H 2' − H 1' + ( H 1' − H 1 ) = h f ,T (CO2 ) + "#h (T ) − h (298)$% C(S) + "#h (T ) − h (298)$% . O2 It follows that h f ,T (CO2 ) = h f ,298 (CO2 ) + "#h (T ) − h (298)$% CO2 { − "#h (T ) − h (298)$% + "#h (T ) − h (298)$% C(S) O2 } , which may be generalized to h fo,T = h fo,298 + ⎡⎣h (T ) − h(298)⎦⎤ substance − ∑ν elements i ⎡⎣h (T ) − h (298)⎤⎦ i , (1.25a) where ν i is the stoichiometric coefficients of the ith elements in the reaction that forms the substance. Therefore if the specific heat or sensible enthalpy of a substance is known, we only need to know the value of enthalpy of formation at a particular temperature, and in general this temperature is equal to 298 K. In combustion analysis, we often group the first and second term of Eq. (1.25a) by defining a total enthalpy as hT ≡ h f ,298 + #$h (T ) − h(298)%& , substance (1.25b) and h f ,T = hT − ∑ elements ν i "#h (T ) − h(298)$% . i (1.25c) Table 1.2 lists the enthalpy of formation of some important species for combustion analysis. 1-9 Stanford University Version 1.2 ©Hai Wang Table 1.2 Standard enthalpy of formation of key combustion species in the vapor state h fo,298 (kJ/mol) Species Name H 2O CO CO2 CH4 C 2H 6 C 3H 8 C4H10 C8H18 C 2H 4 C 2H 2 CH3OH C 6H 6 H• O• OH• Water Carbon monoxide Carbon dioxide Methane Ethane Propane Butane Octane Ethylene Acetylene Methanol Benzene Hydrogen atom Oxygen atom Hydroxyl radical -241.8 -110.5 -393.5 -74.9 -84.8 -104.7 -125.6 -208.4 52.5 226.7 -201.0 82.9 218.0 248.2 39.0 The standard enthalpy of combustion hc (kJ/mol-fuel) is defined as the heat released from the complete combustion of 1 mole of fuel at 298 K. Consider the complete combustion of methane (Eq. 1.21). We will again utilize the path independent property to illustrate that hc may be determined by the sum of enthalpy of formation of the products (multiplied by the molar ratio of the product-to-fuel) subtracted by the sum of enthalpy of formation of the reactants: CH4 + 2O2 + 2×79/21 N2 hc ΔH 1 CO2 + 2H2O + 2×79/21 N2 ΔH 2 C(S) + 2H2 + 2O2 + 2×79/21 N2 hc (kJ/kmol-fuel) = ΔH1 + ΔH 2 = −1 × h fo,298 ( CH 4 ) + 1 × h fo,298 ( CO2 ) + 2 × h fo,298 ( H 2 O ) . For an arbitrary fuel (CmHn) undergoing combustion (1.22), we determine its enthalpy of combustion as 1-10 Stanford University Version 1.2 ©Hai Wang n hc (kJ/kmol-fuel) = m × h fo,298 ( CO2 ) + × h fo,298 ( H 2 O ) − h fo,298 ( Cm Hn ) . 2 In addition, for an arbitrary reaction given by ∑ν A ⎯⎯→ ∑ν i i react. prod. i' Ai'' , (1.26) where Ai and Ai' are the ith reactants and products, respectively, and ν i are termed as the stoichiometric coefficients, we determine the enthalpy of reaction at an arbitrary temperature T by ΔH ro,T = ∑ ν i ' hT ( Ai ' ) − ∑ ν i hT ( Ai ) prod. react. ⎪⎧ ⎪⎫ = ∑ ν i ' h fo,298 ( Ai ' ) − ∑ ν i h fo,298 ( Ai ) + ⎨ ∑ ν i ' ⎡⎣h (T ) − h (298)⎤⎦ i ' − ∑ ν i ⎡⎣h (T ) − h (298)⎤⎦ i ⎬ prod. react. react. ⎩⎪ prod. ⎭⎪ ⎧⎪ ⎫⎪ = ΔH ro,298 + ⎨ ∑ ν i ' ⎡⎣h (T ) − h (298)⎤⎦ i ' − ∑ ν i ⎡⎣h (T ) − h (298)⎤⎦ i ⎬ ⎪⎩ prod. react. ⎭⎪ Since the total numbers of the elements in the reactants and products are identical, the sensible enthalpy terms for the elements in Eq. (1.25c) are canceled out. If ΔHro,T is positive, the reaction absorbs heat. This type of reactions is known to be endorthermic. If ΔHro,T < 0, the reaction releases heat as it proceeds to completion. This type of reactions is known to be exothermic. Conversely, If ΔHro,T > 0, the reaction requires heat to achieve completion. This type of reactions is known to be endothermic. 1.5 Chemical Equilibrium The complete combustion reactions given by Eqs. (1.21) and (1.22) essentially correspond to maximum heat release. That is, if products other than CO2 and H2O are formed, the enthalpy of reaction will be decidedly lower. In practical combustion processes, a combustion reaction can never reach completion. Rather the products of combustion will acquire the state of chemical equilibrium. Although often than not the products will be dominated by the complete combustion products, incomplete combustion products (CO, H2, soot, NO etc) are inherent to a combustion process. Our experience tells us that a process or reaction would be spontaneous if it releases heat. For example, the combustion of methane spontaneously produces CO2 and H2O ( ΔHro,298 <0), but a mixture of CO2 and H2O would not spontaneously react and produce methane and O2. On the other hand, the entropy of 1 mole of CO2 is decidedly smaller than the entropy for a mixture made of 1 mole of CO and 0.5 mole of O2. Likewise the entropy of 1 1-11 Stanford University Version 1.2 ©Hai Wang mole of H2O is smaller than the entropy for a mixture made of 1 mole of H2 and 0.5 mole of O2. Therefore it may be said that reaction (1.21) CH4 + 2 O2 + (2×79/21) N2 → CO2 + 2H2O + (2×79/21) N2 . (1.21) produces the largest heat but with a smaller entropy change, whereas reaction (1.27) produces less heat but with a larger change of entropy upon reaction: CH4 + 2 O2 + (2×79/21) N2 → CO + 2H2 + O2 + (2×79/21) N2 . (1.27) We learned from the second law of thermodynamics that without heat release or absorption, a spontaneous process will proceed in the direction to increase entropy. Therefore a compromise between enthalpy and entropy releases mustofbe Figure 1.1. Variation themade. enthalpy, entropy, and Gibbs an exothermic This compromise is responsible for function chemicalforequilibrium. It reaction. may be quantified by the Gibbs G(T,P) = H(T) - TS(T,P) function or Gibbs free energy, G(T,P) -TS(T,P) H(T) 0 100% Reactants Equilibrium Reaction Progress, ε 1 100% Products G ≡ H − TS , (1.28a) g ≡ h − Ts , (1.28b) g ≡ h − Ts , (1.28c) Figure 1.1 shows a possible scenario for variation of the enthalpy, entropy times temperature, Gibbs function for an exothermic reaction as it progresses to completion at a given T and P. We define here a reaction progress variable ε, such that ε = 0 for pure reactants and ε = 1 for pure products. If the reaction is exothermic, the total enthalpy of the reacting gases (reactants and products) decreases as ε increases. Here the spontaneous release of chemical energy is driving the reaction towards completion. If the reaction is accompanied by a decrease in the entropy (e.g., H2 + 1/2O2 → H2O), the – TS(T,P) function would monotonically increase as reaction progresses. This entropy reduction gives resistance towards the completion of reaction. Overall the Gibbs function must decrease initially, but because of the rise of the –TS term, it eventually will have to go 1-12 Stanford University Version 1.2 ©Hai Wang up as ε increases. In other words, the Gibbs function must reach a minimum at some point. The definition of chemical equilibrium is therefore dG (T , P ) =0, dε (1.29a) dG (T , P ) = 0 . (1.29b) or simply Again, the above equilibrium criterion represents a compromise of H and –ST, since both of them prefer to minimize themselves. Therefore, the driving force of chemical reaction lies in the minimization of the Gibbs function. Now let us consider an arbitrary reaction given by Eq. (1.26). The Gibbs function of the reacting gas may be written as G (T , P ) = ∑ ni g i (T , P ) + ∑ ni ' g i ' (T , P ) , react. (1.30) prod. where ni is the molar number of the ith species. Putting Eq. (1.30) into (1.29a), we obtain, for constant T and P, dG (T , P ) dn dn = ∑ g i ( T , P ) i + ∑ g i ' (T , P ) i ' = 0 , dε d ε prod. dε react. (1.31) Conservation of mass requires that − 1 dn1 1 dn 2 1 dn N 1 dn1' 1 dn 2 ' 1 dn N ' =− ... = − = = = = γ (ε ) , ν1 dε ν 2 dε ν n d ε ν 1' d ε ν 2 ' d ε ν n ' d ε (1.32) where N and N’ are the total numbers of reactants and products, respectively, and γ (ε ) is a function that depends only on ε. Combining equations (1.31) and (1.32), we obtain ⎡ ⎤ γ ( ε ) ⎢ − ∑ ν i g i ( T , P ) + ∑ ν i ' g i ' (T , P ) ⎥ = 0 . prod. ⎣ react. ⎦ (1.33) Since γ (ε ) ≠ 0 , we see that equilibrium state is given by − ∑ν i g i (T , P ) + ∑ν i ' g i ' (T , P ) = 0 . react. prod. The function g i is the Gibbs function of species i, which may be expressed by 1-13 (1.34) Stanford University Version 1.2 ©Hai Wang ⎡ ⎛ P ⎞⎤ g i (T , P ) = h fo (T ) − Tsi (T ) = h fo (T ) − T ⎢ si o (T ) − Ru ln ⎜ o ⎟ ⎥ . ⎝ P ⎠⎦ ⎣ ( (1.35) ) We now define a standard Gibbs function g o T , P o = 1 atm as g o (T ) = h fo (T ) − Tsi o (T ) , (1.36) ⎛P ⎞ g i (T , P ) = g o (T ) + Ru T ln ⎜ o ⎟ . ⎝P ⎠ (1.37) and re-write Eq. (1.35) as Putting Eq. (1.37) into (1.34) and rearranging, we have ∑ν prod. ⎡ ⎛P ⎞ ⎛ P ⎞⎤ g (T ) − ∑ν i g io (T ) = −Ru T ⎢ ∑ν i ' ln ⎜ io' ⎟ − ∑ν i ln ⎜ io ⎟ ⎥ , ⎝ P ⎠ react. ⎝ P ⎠⎦ react. ⎣ prod. o i' i' (1.38) o where Pi is the partial pressure of species i, and of course, P = 1 atm. The left-hand side of the above equation may be defined as the standard Gibbs function change of reaction, ΔGro (T ) ≡ ∑ν i ' g io' (T ) − ∑ν i g io (T ) prod. (1.39) react. The right-hand side of Eq. (1.38) may be re-arranged to yield ⎛ ∏ Piν' i ' ⎜ prod. ΔGro (T ) = −Ru T ln ⎜ Piν i ⎜∏ ⎝ react. ⎞ ⎟ ⎟. ⎟ ⎠ (1.40) or K p (T ) ≡ ∏ Pν i' i' prod. ∏ Pν i react. i ⎡ ΔGro (T ) ⎤ = exp ⎢ − ⎥. Ru T ⎦ ⎣ (1.41) o where Kp(T) is the equilibrium constant of the reaction. Note that by neglecting P in Eqs. (1.40) and (1.41), we have forced Pi to take the unit of atm. The equilibrium constant may also be defined by the concentrations of the reactants and products, 1-14 Stanford University Version 1.2 K c (T ) ≡ ©Hai Wang ∏ cν i' i' prod. ∏c react. νi i ∏ Pν i' = i' prod. ∏P νi ( Ru T ) −Δν = K p (T ) ( R u T ) −Δν , (1.42) i react. where Δν = ∑ v i ' − ∑ v i . prod. react. There are several important facts about the equilibrium constant. (a) While Kp is defined as the pressure ratio of the products and reactants (Eq. 1.41), this equilibrium constant is a function of temperature only. (b) Consider the reaction H 2O = H 2 + ½ O 2. (1.43f) The equilibrium constant for the forward direction of the reaction is PH2 PO122 K p , f (T ) = PH2O . We may also write the reaction in the back direction, H 2 + ½ O 2 = H 2O , (1.43b) and its equilibrium constant K p ,b (T ) = PH2O PH2 PO122 . Obviously, K p , f (T ) = 1 K p ,b (T ) . (c) Reaction (43f) may be written alternatively as 2H2O = 2H2 + O2, with its equilibrium constant K ' p, f (T ) = PH22 PO2 1-15 PH22O . (1.43f’) Stanford University Version 1.2 ©Hai Wang Comparing the equilibrium constants for the two forward reactions, we see that 2 K 'p , f (T ) = ⎡⎣K p , f (T )⎤⎦ . (d) Consider the following two reactions H2 = 2 H•. H2O = H• + OH• (1.44f) (1.45f) (where the • denotes that the species is a free radical). We have K p ,44 f (T ) = P P PH2• and K p ,45 f (T ) = H• OH• . PH2 PH2O A linear combination of reactions (43f-45f) yield H2O = ½ H2 + OH• . (1.46f) A little algebra tells us that (e) While Kp is not a function of pressure, Kc generally is dependent on pressure so long as Δν ≠ 0 . On the other hand, if Δν = 0 , Kc(T) = Kp(T). (f) The equilibrium constant of a given reaction may be determined if the enthalpy of formation and the entropy of reactants and products are known through Eqs. (1.36), (1.39) and (1.41). (g) The definition of Kp tells us that the reaction would be more complete if Kp is larger. A larger Kp may be accomplished with a larger, negative ΔGro (T ) . Combining Eqs. (1.36) and (39), we see that ⎡ ⎤ ⎡ ⎤ ΔGro (T ) ≡ ⎢ ∑ ν i ' h fo, i ' (T ) − ∑ ν i h fo,i (T )⎥ − T ⎢ ∑ ν i ' si o' (T ) − ∑ ν i si o (T )⎥ react. react. ⎣ prod. ⎦ ⎣ prod. ⎦ , (1.47) o o = ΔH r (T ) − T ΔSr (T ) where ΔSro (T ) is termed as the entropy of reaction. Therefore a large, negative ΔHro (T ) (reaction being highly exothermic) favors a large Kp , whereas a large, positive ΔSro (T ) (reaction creating a large amount of entropy) also favors a large Kp or promotes the completion of the reaction. 1-16 Stanford University Version 1.2 ©Hai Wang 1.6 Adiabatic Flame Temperature With the concepts of chemical equilibrium understood, we may now try to calculate the equilibrium composition of a combustion reaction. In doing so, we wish to define the adiabatic flame temperature. Consider an adiabatic combustion process whereby the reactants enters into a combustor at temperature T0, and products exit the combustor at the adiabatic flame temperature Tad. Since the process is adiabatic (Q = 0 ), we have H prod. (Tad ) − H react. (T0 ) = 0 . (1.48) We now expand Eq. (1.48) using the total enthalpy equation for each species (Eq. 1.25b), ∑ν prod. h i ' Tad , i ' ⎡ ⎤ − ∑ν i hT0 , i = ⎢ ∑ν i ' h fo,298, i ' − ∑ν i h fo,298, i ⎥ + ∑ν i ' ⎡⎣h (Tad ) − h ( 298 )⎤⎦ i ' react. react. ⎣ prod. ⎦ prod. .(1.49) − ∑ν i ⎡⎣h (T0 ) − h ( 298 )⎤⎦ i = 0 react. Obviously the first term on the right-hand side of Eq. (1.49) is the standard enthalpy of reaction ΔHro,298 . The second term determines the sensible heat needed to heat the products from 298 K to the adiabatic flame temperature Tad. To simplify our analysis, we shall assume that the reactants enter into the reactor at T0 = 298 K so the third term becomes 0. Rearranging Eq. (1.49), we see that −ΔHro,298 = ∑ν i ' ⎡⎣h (Tad ) − h ( 298 )⎤⎦ i ' . (1.50) prod. In other words, the adiabatic flame temperature is obtained when all the heat released from a combustion reaction is used to raise the product temperature from 298 to Tad. The existence of chemical equilibrium makes the calculation of this adiabatic flame temperature a bit more involved. Specifically, while the values of ν i are always well defined, ν i' is not since it is dependent on the equilibrium composition of the products. Consider the combustion of 1 mole of carbon (graphite) in 1 mole of oxygen at a pressure of 1 atm. 1 mole C(S) + 1 mole O2 → x CO2 + yCO + zO2. (1.51) The reactant temperature is 298 K. The principle of chemical equilibrium states that the products cannot be entirely CO2. Rather, a small amount of CO (y moles) must be produced along with z moles of O2 unused. These products are in equilibrium at the adiabatic flame temperature among themselves through 1-17 Stanford University Version 1.2 ©Hai Wang CO2 = CO + ½ O2 , (1.52) with its equilibrium constant given by K p (Tad ) = PCOPO122 PCO2 ⎞ yz 1 2 ⎛ P = ⎜ x ⎝ x + y + z ⎟⎠ 12 ( ) = exp −ΔGr0 R u Tad . (1.53) (We need to recognize that the products of a combustion process cannot be in equilibrium with the reactants of the process. Rather it is the products that are in equilibrium among themselves.) Since there are four unknowns in Eq. (1.53) (i.e., x, y, z and Tad), we need to provide three more equations to solve this problem. Two of these equations come from mass conservation: Carbon: x + y = 1 mol , Oxygen: 2x + y + 2z = 2 mol . (1.54) (1.55) The last equation is given by Eq. (1.50), which may be expanded to give − ⎡⎣ xh fo,298 (CO2 ) + yh fo,298 (CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO 2 + y ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO + z ⎡⎣h (Tad ) − h ( 298 )⎤⎦ O .(1.56) 2 Clearly the coupled Eqs (1.53-1.56) cannot be solved analytically. We shall defer to section 1.8 and use Excel to solve the equation. Any more realistic combustion equilibrium problems will have to be solved numerically by a computer—a topic we will discuss also in section 1.8. Figure 1.2 shows the variation of the adiabatic flame temperature as a function of equivalence ratio for the combustion of methane, propane, ethylene and benzene in air at 1 atm pressure. As expected, the flame temperature peaks at an equivalence ratio around unity, slightly on the fuel rich side. The decrease of Tad towards smaller φ is caused by dilution of unused oxygen as well as the greater amount of nitrogen brought into the combustor with oxygen. The decrease of the flame temperature towards larger φ is because of oxygen deficiency, which does not allow the fuel to be fully oxidized. Figure 1.3 shows the effect of pressure on the adiabatic flame temperature. It is seen that as pressure increases, (a) the adiabatic flame temperature becomes higher and (b) the peak shifts towards φ = 1. Here it may be noted that the pressure serves to reduce the extent of dissociation of CO2 and H2O, and in doing so, force the reaction towards better completion. To explain the variation of Tad as a function of pressure, we plot in Figure 1.4 the equilibrium composition at φ = 1. It is seen that as pressure decreases, the extent of CO2 and H2O dissociation into H2, O2, CO, and even highly unstable free radical species, including H•, O•, and OH• becomes more and more significant. 1-18 Stanford University Version 1.2 ©Hai Wang Adiabatic Flame Temperature, Tad (K) 2500 Figure 1.3. Adiabatic flame temperature ofof Figure 1.4. 1.2.Mole fractions Figure Adiabatic propane combustion inflame air equilibrium productsofcomputed for temperature methane, at several pressures. propane propane, combustion ethyleneinand air (benzene φ=1). at 1 atm pressure. 2000 ethylene 1500 benzene propane methane 1000 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Adiabatic Flame Temperature, Tad (K) Equivalence Ratio, φ 100 atm 10 atm 2300 1 atm 2200 0.1 atm 2100 0.85 0.90 0.95 1.00 1.05 1.10 1.15 Equivalence Ratio, φ H2O Mole Fraction 10-1 CO2 CO 10-2 O2 OH H2 10-3 H O 10-4 10-5 0.1 1 10 100 Pressure, P (atm) 1-19 1.20 Stanford University Version 1.2 ©Hai Wang 1.7 Tabulation and Mathematical Parameterization of Thermochemical Properties Key thermodynamic or thermochemical properties discussed so far include the constantpressure specific heat, sensible enthalpy, enthalpy of formation, and entropy under the standard condition. It is important to recognize that under the ideal gas condition, c p is not a function of pressure, but it is a function of temperature. A common method is to tabulate, among others, c p (T ) , s o (T ) , h(T)–h(298) and h fo (T ) as a function of temperature. Such table is known as the JANAF table.1 Appendix 1.A gives a truncated version of these tables for species listed in Table 1.2. In research publications, these tables are usually condensed to something that looks like Table 1.3. Table 1.3. Thermochemical properties of selected species in the vapor state. Species h fo ( 298K ) s o ( 298K ) (kJ/mol) (J/mol-K) c p (T ) (J/mol-K) 298 500 1000 1500 2000 2500 C(S) 0 5.730 8.523 14.596 21.624 23.857 25.167 25.976 H2 0 130.663 28.834 29.297 30.163 32.358 34.194 35.737 O2 0 205.127 29.377 31.082 34.881 36.505 37.855 38.999 H 2O -241.8 188.810 33.587 35.214 41.294 47.333 51.678 54.731 CO -110.5 197.640 29.140 29.811 33.163 35.132 36.288 36.917 CO2 -393.5 213.766 37.128 44.628 54.322 58.222 60.462 61.640 CH4 -74.6 186.351 35.685 46.494 73.616 90.413 100.435 106.864 C 2H 6 -83.9 229.051 52.376 77.837 122.540 144.761 158.280 165.774 C 3H 8 -103.8 270.141 73.530 112.409 174.614 204.334 222.359 232.305 C4H10 -125.8 309.686 98.571 148.552 227.379 264.424 286.823 299.106 C8H18 -208.7 466.772 187.486 286.282 431.399 494.910 534.404 557.447 52.3 219.156 42.783 62.321 93.860 109.190 118.563 123.799 C 2H 4 C 2H 2 227.4 200.892 43.989 54.715 67.908 75.906 81.045 84.262 -200.9 239.785 44.030 59.526 89.656 105.425 113.891 118.560 82.8 269.020 82.077 138.240 210.948 240.242 257.667 267.023 H• 218.0 114.706 20.786 20.786 20.786 20.786 20.786 20.786 O• 249.2 161.047 21.912 21.247 20.924 20.846 20.826 20.853 39.3 183.722 29.887 29.483 30.694 32.948 34.755 36.077 CH3OH C 6H 6 OH• The thermochemical data of a large number of species has been compiled by the National Institute of Standards and Technology. These data may be found at http://webbook.nist.gov/chemistry/. Another web-based source of data is the Active Tables (ATcT): https://cmcs.ca.sandia.gov/cmcs/portal/user/anon/js_pane. 1 Chase, M. W., Jr., J. Phys. Chem. Ref. Data, 4th Edition, Mono. 9, Suppl. 1 (1998a). 1-20 Stanford University Version 1.2 ©Hai Wang For combustion calculations, a very good source of thermochemical data is: Alexander Burcat and Branko Ruscic “Third Millennium Thermodynamic Database for Combustion and Air-Pollution Use,” 2005 (http://www.technion.ac.il/~aer0201/ or http://garfield.chem.elte.hu/Burcat/burcat.html). The database is the result of extensive work by Professor Alexander Burcat of Technion University, Israel over the last 20 years. In this database, the thermochemical data are expressed in the form of polynomial function and are thus more compact than the JANAF table. A typical record of thermochemical data may look like the following: Species name TLow, Thigh, Tmid Composition Reference source Phase CO2 L 7/88C 1O 2 0 0G 200.000 6000.000 1000. 0.46365111E+01 0.27414569E-02-0.99589759E-06 0.16038666E-09-0.91619857E-14 -0.49024904E+05-0.19348955E+01 0.23568130E+01 0.89841299E-02-0.71220632E-05 0.24573008E-08-0.14288548E-12-0.48371971E+05 0.99009035E+01-0.47328105E+05 ai (i = 1,7) for Tmid < T < Thigh 1 2 3 4 ai (i = 1,7) for Tlow < T < Tmid The polynomial fits are made for two separate temperature ranges (Tlow ≤ T ≤ Tmid and Tmid ≤ T ≤ Thigh). There are 7 polynomial coefficients, ai (i,=1,..7), for each temperature range. The thermochemical data are calculated from these fits as c p Ru = a1 + a 2T + a 3T 2 + a 4 T 3 + a 5T 4 , (1.57) s Ru = a1 ln T + a2T + a3 T 2 + a4 T 3 + a5 T 4 + a7 . hT Ru = a1T + a2 T 2 2 + a3 T 3 3 + a4 T 4 4 + a5 T 5 5 + a6 (1.58) o 2 3 4 (1.59) where hT is the total enthalpy (Eq. 1.25b), s o is the standard entropy (1 atm). Note that these equations are in strict agreement with known relationships among the thermochemical properties, i.e., hT (T ) = ∫ T c p dT + h fo,298.15 , (1.60) c p d ln T + s o ( 298.15 ) . (1.61) 298.15 s o (T ) = ∫ T 298.15 To calculate and tabulate the thermochemical data properties in the JANAF type form, c p and s o are directly calculated with Eqs. (1.57) and (1.58). The sensible enthalpy is determined as h (T ) − h ( 298.15 ) = hT (T ) − hT ( 298.15 ) , (1.62) where hT is calculated using Eq. (1.59). For enthalpy of formation, we use h fo (T ) = hT (T ) − ∑ ν h (T ) i T ,i elements 1-21 (1.63) Stanford University Version 1.2 ©Hai Wang where ν i represents the molecular composition of the substance. For example, a CxHyOz species has ν C = x , ν H2 = y 2 and ν O2 = z 2 . An EXCEL spreadsheet has been prepared for the JANAF like tabulation. The file may be downloaded from the course web site http://melchior.usc.edu/public/AME579/Week_2/NASA_poly to JANAF.xls. Burcat’s database may also be downloaded in text form http://melchior.usc.edu/public/AME579/Week_2/Burcat Thermo.txt, and in Excel form http://melchior.usc.edu/public/AME579/Week_2/Burcat Thermo.xls. 1.8 Solution of an Equilibrium and Adiabatic Flame Temperature Problem We shall now return to the problem of carbon (graphite) oxidation in section 1.6. We wish to calculate the adiabatic flame temperature for combustion of 1 mole of carbon (graphite) in 1 mole of oxygen at a pressure of 1 atm. The initial temperature is 298 K. The four equations are ⎧x + y = 1 ⎪2 x + y + 2z = 2 ⎪ 12 ⎪ yz 1 2 ⎛ ⎞ P ⎪ = exp ( −ΔGr0 Ru Td ) . ⎨ x ⎜⎝ x + y + z ⎟⎠ ⎪ ⎪− ⎡⎣ xh fo,298 ( CO2 ) + yh fo,298 ( CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO2 ⎪ ⎪+ y ⎡h (Tad ) − h ( 298 )⎤ + z ⎡h (Tad ) − h ( 298 )⎤ ⎦ CO ⎣ ⎦ O2 ⎩ ⎣ Solution of the above problem is provided in an Excel sheet downloadable from http://melchior.usc.edu/public/AME579/Week_2/carbon oxidation.xls. Note that to run the Excel solver requires the user to download the thermochemical property tables from http://melchior.usc.edu/public/AME579/Week_2/NASA_poly to JANAF.xls. This file should be placed in the same directory as the carbon oxidation.xls file. The solution of this set of nonlinear algebraic equations gives ⎧x = 0.233 mol ⎪ y = 0.767 mol ⎪ ⎨ ⎪z = 0.384 mol ⎪ ⎩Tad = 3537 K 1-22 Stanford University Version 1.2 ©Hai Wang Now suppose that the carbon is burned in air, instead of pure oxygen, the set of nonlinear algebraic equations may be revised by including the molar number of N2 (=1 mole O2×79/21) and the sensible enthalpy required to heat up the nitrogen, ⎧x + y = 1 ⎪2 x + y + 2z = 2 ⎪ 12 ⎪ yz 1 2 ⎛ ⎞ P ⎪ = exp ( −ΔGr0 Ru Td ) . ⎨ x ⎜⎝ x + y + z + 79 21 ⎟⎠ ⎪ ⎪− ⎡⎣ xh fo,298 ( CO2 ) + yh fo,298 ( CO )⎤⎦ = x ⎡⎣h (Tad ) − h ( 298 )⎤⎦ CO2 ⎪ ⎪+ y ⎡h (Tad ) − h ( 298 )⎤ + z ⎡h (Tad ) − h ( 298 )⎤ + 79 21 ⎡h (Tad ) − h ( 298 )⎤ ⎦ CO ⎣ ⎦ O2 ⎣ ⎦ N2 ⎩ ⎣ The solution is ⎧x = 0.893 mol ⎪ y = 0.107 mol ⎪ ⎨ ⎪z = 0.053 mol ⎪ ⎩Tad = 2312 K Comparing the two sets of solution, we find that (a) the adiabatic flame temperature is notably lower when air is used, and (b) the reaction is less complete when pure oxygen is used because the higher adiabatic flame temperature forces a greater extent of CO2 dissociation into CO and O2. Two commonly used equilibrium solvers are Stanjan (or the equilibrium solver – EQUIL of the ChemKin suite of package) and the NASA Equilibrium code (cec86). We will use the equilibrium solver of the ChemKin suite of package for the current class. Instructions about the computer code can be found on p. 30. 1-23 Stanford University Version 1.2 ©Hai Wang Appendix A1. Truncated JANAF tables Graphite (C(S)) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 8.523 8.592 11.832 14.596 16.836 18.559 19.835 20.794 21.624 22.169 22.660 23.102 23.499 23.857 24.177 24.465 24.724 24.957 25.167 25.358 25.531 25.691 25.838 25.976 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) (kJ/mol) 0.000 0.017 1.042 2.368 3.944 5.717 7.640 9.674 11.795 13.985 16.227 18.515 20.846 23.214 25.616 28.048 30.508 32.992 35.498 38.025 40.569 43.131 45.707 48.298 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 5.730 5.787 8.713 11.659 14.526 17.257 19.823 22.217 24.451 26.538 28.489 30.320 32.047 33.681 35.231 36.705 38.111 39.454 40.740 41.972 43.156 44.295 45.391 46.449 1-24 h fo (T ) g of (T ) (kJ/mol) 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 Stanford University Version 1.2 ©Hai Wang Hydrogen (H2) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 28.834 28.850 29.277 29.297 29.254 29.344 29.615 29.970 30.163 30.634 31.089 31.527 31.950 32.358 32.752 33.132 33.499 33.853 34.194 34.525 34.843 35.151 35.449 35.737 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) (kJ/mol) 130.663 130.856 139.229 145.768 151.105 155.619 159.553 163.062 166.232 169.130 171.815 174.320 176.672 178.891 180.992 182.989 184.893 186.714 188.459 190.135 191.749 193.305 194.807 196.260 0.000 0.058 2.969 5.900 8.827 11.755 14.702 17.681 20.690 23.730 26.817 29.948 33.122 36.337 39.593 42.887 46.219 49.586 52.989 56.425 59.893 63.393 66.923 70.483 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1-25 h fo (T ) g of (T ) (kJ/mol) 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 Stanford University Version 1.2 ©Hai Wang Oxygen (O2) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 29.377 29.387 30.120 31.082 32.080 32.990 33.747 34.355 34.881 35.232 35.569 35.893 36.205 36.505 36.795 37.074 37.343 37.604 37.855 38.098 38.334 38.562 38.784 38.999 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) (kJ/mol) 205.127 205.323 213.870 220.692 226.448 231.463 235.919 239.930 243.578 246.919 249.999 252.859 255.530 258.038 260.404 262.643 264.769 266.795 268.731 270.584 272.361 274.070 275.716 277.304 0.000 0.059 3.031 6.090 9.249 12.503 15.842 19.248 22.710 26.216 29.756 33.329 36.934 40.569 44.234 47.928 51.649 55.396 59.169 62.967 66.789 70.634 74.501 78.390 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 1-26 h fo (T ) g of (T ) (kJ/mol) 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000 Stanford University Version 1.2 ©Hai Wang Water vapor (H2O(v)) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 33.587 33.596 34.268 35.214 36.320 37.508 38.733 39.986 41.294 42.659 43.941 45.145 46.275 47.333 48.324 49.251 50.117 50.925 51.678 52.380 53.034 53.641 54.206 54.731 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 188.810 189.034 198.783 206.527 213.044 218.731 223.819 228.453 232.733 236.734 240.501 244.066 247.454 250.683 253.770 256.727 259.567 262.299 264.930 267.469 269.921 272.292 274.587 276.811 0.000 0.067 3.458 6.930 10.506 14.197 18.008 21.944 26.007 30.206 34.536 38.991 43.563 48.244 53.027 57.907 62.876 67.928 73.059 78.262 83.533 88.867 94.260 99.707 1-27 h fo (T ) (kJ/mol) -241.821 -241.841 -242.849 -243.836 -244.767 -245.632 -246.435 -247.182 -247.859 -248.454 -248.979 -249.442 -249.847 -250.199 -250.504 -250.766 -250.989 -251.178 -251.336 -251.468 -251.576 -251.664 -251.735 -251.792 g of (T ) (kJ/mol) -228.585 -228.496 -223.896 -219.043 -213.996 -208.798 -203.481 -198.066 -192.571 -187.013 -181.404 -175.753 -170.070 -164.359 -158.626 -152.875 -147.110 -141.334 -135.548 -129.756 -123.957 -118.154 -112.348 -106.539 Stanford University Version 1.2 ©Hai Wang Carbon monoxide (CO) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 29.140 29.143 29.375 29.811 30.415 31.133 31.894 32.606 33.163 33.637 34.069 34.460 34.813 35.132 35.418 35.674 35.903 36.107 36.288 36.447 36.589 36.713 36.822 36.917 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 197.640 197.835 206.246 212.844 218.330 223.071 227.278 231.077 234.543 237.726 240.672 243.414 245.981 248.394 250.671 252.826 254.872 256.818 258.675 260.449 262.148 263.778 265.342 266.848 0.000 0.058 2.982 5.940 8.950 12.027 15.178 18.404 21.694 25.035 28.420 31.847 35.311 38.808 42.336 45.891 49.470 53.071 56.691 60.327 63.979 67.645 71.321 75.009 1-28 h fo (T ) (kJ/mol) -110.529 -110.518 -110.105 -110.002 -110.147 -110.472 -110.912 -111.423 -111.985 -112.588 -113.214 -113.862 -114.531 -115.220 -115.926 -116.651 -117.392 -118.149 -118.922 -119.710 -120.514 -121.332 -122.166 -123.014 g of (T ) (kJ/mol) -137.155 -137.334 -146.344 -155.421 -164.495 -173.529 -182.509 -191.428 -200.288 -209.089 -217.835 -226.527 -235.168 -243.761 -252.308 -260.810 -269.268 -277.685 -286.062 -294.400 -302.699 -310.962 -319.189 -327.381 Stanford University Version 1.2 ©Hai Wang Carbon dioxide (CO2) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 37.128 37.218 41.286 44.628 47.359 49.589 51.425 52.970 54.322 55.268 56.124 56.899 57.596 58.222 58.782 59.281 59.724 60.116 60.462 60.765 61.031 61.263 61.464 61.640 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 213.766 214.015 225.297 234.881 243.268 250.741 257.487 263.635 269.288 274.510 279.356 283.880 288.122 292.118 295.894 299.473 302.874 306.114 309.206 312.164 314.997 317.715 320.326 322.839 0.000 0.074 4.006 8.307 12.911 17.762 22.816 28.038 33.404 38.884 44.454 50.106 55.831 61.623 67.473 73.377 79.328 85.320 91.349 97.411 103.501 109.616 115.753 121.908 1-29 h fo (T ) (kJ/mol) -393.505 -393.506 -393.572 -393.655 -393.786 -393.963 -394.171 -394.389 -394.606 -394.821 -395.033 -395.243 -395.453 -395.665 -395.882 -396.104 -396.334 -396.573 -396.823 -397.086 -397.362 -397.653 -397.960 -398.285 g of (T ) (kJ/mol) -394.372 -394.377 -394.658 -394.920 -395.162 -395.378 -395.566 -395.727 -395.865 -395.980 -396.076 -396.154 -396.217 -396.264 -396.297 -396.316 -396.322 -396.314 -396.294 -396.262 -396.216 -396.157 -396.086 -396.001 Stanford University Version 1.2 ©Hai Wang Methane (CH4) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 35.685 35.760 40.530 46.494 52.730 58.650 63.998 68.850 73.616 77.713 81.404 84.728 87.720 90.413 92.839 95.028 97.007 98.802 100.435 101.929 103.302 104.573 105.756 106.864 186.351 186.590 197.488 207.161 216.190 224.769 232.957 240.778 248.277 255.489 262.412 269.061 275.452 281.597 287.511 293.206 298.695 303.989 309.099 314.036 318.809 323.430 327.906 332.246 c p (T ) 0.000 0.071 3.871 8.217 13.179 18.752 24.889 31.535 38.656 46.226 54.185 62.495 71.120 80.029 89.194 98.589 108.192 117.984 127.947 138.066 148.329 158.724 169.241 179.872 1-30 h fo (T ) (kJ/mol) -74.594 -74.655 -77.704 -80.544 -83.013 -85.070 -86.749 -88.096 -89.114 -89.814 -90.269 -90.510 -90.564 -90.454 -90.202 -89.828 -89.348 -88.775 -88.124 -87.403 -86.622 -85.788 -84.908 -83.986 g of (T ) (kJ/mol) -50.544 -50.383 -41.830 -32.527 -22.685 -12.462 -1.970 8.711 19.525 30.425 41.378 52.360 63.353 74.344 85.323 96.282 107.217 118.122 128.994 139.833 150.635 161.401 172.130 182.821 Stanford University Version 1.2 ©Hai Wang Ethane (C2H6) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 52.376 52.642 65.627 77.837 89.040 99.101 107.983 115.743 122.540 127.823 132.662 137.081 141.106 144.761 148.069 151.053 153.735 156.138 158.280 160.183 161.866 163.347 164.644 165.774 229.051 229.402 246.340 262.313 277.511 292.007 305.833 319.010 331.564 343.495 354.827 365.623 375.932 385.794 395.244 404.311 413.023 421.400 429.465 437.234 444.725 451.954 458.934 465.679 c p (T ) 0.000 0.105 6.024 13.205 21.558 30.974 41.338 52.533 64.455 76.977 90.005 103.495 117.408 131.704 146.348 161.307 176.549 192.045 207.768 223.693 239.797 256.059 272.460 288.982 1-31 h fo (T ) (kJ/mol) -83.854 -83.956 -88.822 -93.084 -96.664 -99.580 -101.902 -103.711 -105.061 -106.039 -106.753 -107.232 -107.503 -107.589 -107.516 -107.305 -106.978 -106.553 -106.050 -105.486 -104.877 -104.236 -103.579 -102.916 g of (T ) (kJ/mol) -31.884 -31.534 -13.313 6.072 26.250 46.975 68.077 89.438 110.975 132.629 154.359 176.139 197.949 219.771 241.593 263.407 285.204 306.981 328.733 350.459 372.156 393.825 415.466 437.079 Stanford University Version 1.2 ©Hai Wang Propane (C3H8) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 73.530 73.949 94.097 112.409 128.644 142.749 154.863 165.312 174.614 181.688 188.162 194.072 199.452 204.334 208.751 212.732 216.308 219.508 222.359 224.889 227.124 229.090 230.809 232.305 270.141 270.634 294.707 317.709 339.671 360.587 380.460 399.318 417.224 434.203 450.294 465.592 480.174 494.104 507.435 520.212 532.475 544.257 555.590 566.502 577.016 587.157 596.944 606.397 c p (T ) 0.000 0.147 8.564 18.906 30.976 44.563 59.459 75.480 92.483 110.303 128.801 147.917 167.598 187.791 208.449 229.527 250.982 272.776 294.872 317.237 339.840 362.653 385.649 408.807 1-32 h fo (T ) (kJ/mol) -103.842 -103.977 -110.281 -115.638 -120.004 -123.452 -126.111 -128.108 -129.506 -130.416 -130.989 -131.262 -131.268 -131.042 -130.612 -130.009 -129.259 -128.389 -127.422 -126.380 -125.285 -124.155 -123.008 -121.861 g of (T ) (kJ/mol) -23.472 -22.932 5.058 34.533 64.993 96.111 127.667 159.515 191.554 223.707 255.927 288.184 320.450 352.709 384.947 417.151 449.316 481.436 513.507 545.529 577.499 609.418 641.288 673.110 Stanford University Version 1.2 ©Hai Wang Normal butane (C4H10) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 98.571 99.112 125.064 148.552 169.281 187.205 202.523 215.684 227.379 236.202 244.275 251.642 258.344 264.424 269.920 274.871 279.314 283.287 286.823 289.958 292.722 295.148 297.267 299.106 309.686 310.347 342.467 372.947 401.906 429.383 455.407 480.039 503.378 525.470 546.373 566.221 585.119 603.154 620.398 636.913 652.752 667.962 682.585 696.656 710.210 723.277 735.884 748.058 c p (T ) 0.000 0.198 11.425 25.128 41.043 58.890 78.397 99.323 121.485 144.670 168.700 193.502 219.006 245.150 271.872 299.116 326.829 354.963 383.472 412.314 441.451 470.847 500.470 530.291 1-33 h fo (T ) (kJ/mol) -125.767 -125.926 -133.355 -139.607 -144.631 -148.521 -151.440 -153.544 -154.914 -155.689 -156.057 -156.064 -155.751 -155.158 -154.322 -153.280 -152.063 -150.705 -149.234 -147.677 -146.061 -144.409 -142.742 -141.081 g of (T ) (kJ/mol) -16.535 -15.802 22.056 61.658 102.403 143.898 185.882 228.181 270.675 313.275 355.927 398.595 441.252 483.876 526.452 568.969 611.420 653.800 696.104 738.333 780.486 822.565 864.572 906.509 Stanford University Version 1.2 ©Hai Wang Normal octane (C8H18) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 187.486 188.566 240.137 286.282 326.349 360.219 388.302 411.539 431.399 446.430 460.225 472.856 484.394 494.910 504.469 513.136 520.974 528.044 534.404 540.110 545.217 549.775 553.836 557.447 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 466.772 468.029 529.457 588.108 643.939 696.868 746.862 793.980 838.389 880.221 919.665 957.009 992.480 1026.265 1058.515 1089.363 1118.919 1147.279 1174.529 1200.743 1225.989 1250.328 1273.813 1296.497 0.000 0.376 21.852 48.222 78.905 113.285 150.756 190.783 232.950 276.852 322.195 368.858 416.730 465.703 515.680 566.567 618.279 670.736 723.865 777.596 831.867 886.621 941.805 997.373 1-34 h fo (T ) (kJ/mol) -208.731 -209.011 -221.937 -232.547 -240.815 -246.981 -251.414 -254.468 -256.354 -257.332 -257.701 -257.523 -256.861 -255.773 -254.313 -252.534 -250.484 -248.210 -245.755 -243.159 -240.461 -237.695 -234.894 -232.087 g of (T ) (kJ/mol) 16.269 17.780 95.386 175.994 258.516 342.252 426.749 511.715 596.959 682.343 767.789 853.243 938.664 1024.023 1109.297 1194.470 1279.530 1364.470 1449.285 1533.973 1618.536 1702.974 1787.291 1871.491 Stanford University Version 1.2 ©Hai Wang Ethylene (C2H4) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 42.783 42.992 53.098 62.321 70.529 77.685 83.843 89.153 93.860 97.500 100.836 103.885 106.664 109.190 111.479 113.546 115.406 117.074 118.563 119.889 121.062 122.097 123.005 123.799 219.156 219.443 233.213 246.070 258.174 269.597 280.383 290.572 300.213 309.332 317.961 326.154 333.956 341.403 348.524 355.346 361.889 368.174 374.218 380.036 385.640 391.045 396.261 401.299 c p (T ) 0.000 0.086 4.897 10.676 17.327 24.746 32.831 41.487 50.641 60.212 70.131 80.369 90.899 101.694 112.729 123.982 135.431 147.057 158.840 170.764 182.813 194.972 207.228 219.569 1-35 h fo (T ) (kJ/mol) 52.298 52.234 49.173 46.440 44.085 42.100 40.445 39.075 37.968 37.079 36.342 35.742 35.262 34.890 34.610 34.410 34.276 34.198 34.164 34.163 34.185 34.222 34.265 34.306 g of (T ) (kJ/mol) 68.280 68.387 74.241 80.832 87.938 95.408 103.141 111.063 119.123 127.283 135.517 143.807 152.138 160.500 168.884 177.282 185.691 194.105 202.522 210.940 219.358 227.774 236.188 244.601 Stanford University Version 1.2 ©Hai Wang Acetylene (C2H2) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 43.989 44.134 50.268 54.715 58.110 60.916 63.429 65.775 67.908 69.787 71.517 73.108 74.568 75.906 77.130 78.248 79.269 80.199 81.045 81.816 82.516 83.153 83.733 84.262 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 200.892 201.187 214.772 226.492 236.780 245.954 254.254 261.862 268.905 275.467 281.614 287.402 292.874 298.065 303.004 307.714 312.216 316.527 320.662 324.636 328.458 332.140 335.692 339.121 0.000 0.088 4.825 10.086 15.734 21.688 27.907 34.369 41.055 47.941 55.008 62.240 69.625 77.150 84.802 92.572 100.449 108.423 116.486 124.629 132.847 141.131 149.475 157.875 1-36 h fo (T ) (kJ/mol) 227.397 227.393 227.170 226.848 226.417 225.896 225.322 224.737 224.172 223.638 223.135 222.659 222.209 221.782 221.375 220.986 220.612 220.249 219.897 219.552 219.211 218.873 218.535 218.194 g of (T ) (kJ/mol) 209.884 209.766 203.922 198.146 192.444 186.822 181.279 175.808 170.402 165.051 159.748 154.485 149.258 144.062 138.894 133.751 128.631 123.530 118.449 113.385 108.338 103.306 98.288 93.285 Stanford University Version 1.2 ©Hai Wang Methanol vapor (CH3OH(v)) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 44.030 44.155 51.400 59.526 67.334 74.146 79.807 84.681 89.656 93.611 97.125 100.239 102.994 105.425 107.570 109.459 111.124 112.593 113.891 115.042 116.067 116.984 117.810 118.560 239.785 240.080 253.735 266.074 277.628 288.534 298.817 308.502 317.675 326.410 334.709 342.608 350.140 357.330 364.204 370.783 377.088 383.136 388.945 394.530 399.906 405.086 410.083 414.908 c p (T ) 0.000 0.088 4.852 10.397 16.746 23.830 31.536 39.764 48.475 57.642 67.183 77.054 87.219 97.642 108.294 119.148 130.178 141.366 152.691 164.139 175.696 187.349 199.089 210.908 1-37 h fo (T ) (kJ/mol) -200.932 -201.005 -204.576 -207.747 -210.407 -212.581 -214.360 -215.827 -216.987 -217.843 -218.487 -218.953 -219.269 -219.463 -219.556 -219.571 -219.523 -219.429 -219.301 -219.151 -218.987 -218.817 -218.647 -218.482 g of (T ) (kJ/mol) -162.241 -161.981 -148.427 -134.013 -119.008 -103.597 -87.902 -72.003 -55.958 -39.811 -23.597 -7.336 8.955 25.264 41.583 57.905 74.226 90.543 106.854 123.158 139.455 155.744 172.025 188.300 Stanford University Version 1.2 ©Hai Wang Benzene (C6H6) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 s o (T ) h(T)-h(298) (J/mol-K) (J/mol-K) (kJ/mol) 82.077 82.718 112.650 138.240 159.404 176.423 189.948 200.995 210.948 217.961 224.361 230.183 235.466 240.242 244.545 248.408 251.863 254.939 257.667 260.074 262.187 264.033 265.637 267.023 269.020 269.572 297.567 325.536 352.677 378.577 403.052 426.081 447.775 468.215 487.458 505.651 522.905 539.317 554.961 569.905 584.203 597.905 611.052 623.683 635.832 647.528 658.800 669.673 c p (T ) 0.000 0.165 9.968 22.550 37.468 54.292 72.636 92.198 112.798 134.249 156.370 179.101 202.388 226.178 250.421 275.072 300.089 325.432 351.065 376.955 403.070 429.383 455.869 482.503 1-38 h fo (T ) (kJ/mol) 82.818 82.707 77.627 73.463 70.145 67.540 65.507 63.930 62.775 61.966 61.377 60.985 60.767 60.701 60.766 60.940 61.203 61.538 61.925 62.349 62.791 63.237 63.673 64.085 g of (T ) (kJ/mol) 129.707 130.023 146.585 164.325 182.823 201.817 221.144 240.698 260.405 280.209 300.077 319.986 339.918 359.860 379.803 399.738 419.660 439.566 459.452 479.318 499.163 518.988 538.794 558.582 Stanford University Version 1.2 ©Hai Wang Hydrogen Atom (H•) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 20.786 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 114.706 114.845 120.825 125.463 129.253 132.457 135.232 137.681 139.871 141.852 143.660 145.324 146.865 148.299 149.640 150.900 152.088 153.212 154.278 155.293 156.260 157.184 158.068 158.917 0.000 0.042 2.120 4.199 6.277 8.356 10.435 12.513 14.592 16.670 18.749 20.828 22.906 24.985 27.063 29.142 31.221 33.299 35.378 37.456 39.535 41.614 43.692 45.771 1-39 h fo (T ) (kJ/mol) 217.995 218.007 218.630 219.243 219.859 220.473 221.078 221.667 222.241 222.801 223.340 223.858 224.355 224.830 225.284 225.718 226.131 226.525 226.900 227.257 227.596 227.920 228.227 228.520 g of (T ) (kJ/mol) 203.281 203.183 198.146 192.955 187.639 182.220 176.714 171.133 165.488 159.785 154.032 148.235 142.399 136.529 130.627 124.698 118.743 112.766 106.769 100.754 94.722 88.675 82.614 76.541 Stanford University Version 1.2 ©Hai Wang Oxygen atom (O•) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 21.912 21.900 21.462 21.247 21.138 21.062 20.987 20.923 20.924 20.903 20.885 20.870 20.857 20.846 20.838 20.832 20.828 20.826 20.826 20.827 20.831 20.837 20.844 20.853 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 161.047 161.194 167.427 172.190 176.053 179.306 182.113 184.581 186.785 188.778 190.596 192.267 193.813 195.252 196.597 197.860 199.051 200.177 201.245 202.261 203.230 204.156 205.043 205.894 0.000 0.044 2.209 4.344 6.462 8.572 10.675 12.770 14.861 16.953 19.042 21.130 23.216 25.301 27.385 29.469 31.552 33.635 35.717 37.800 39.883 41.966 44.050 46.135 1-40 h fo (T ) (kJ/mol) 249.171 249.185 249.864 250.469 251.008 251.491 251.924 252.316 252.677 253.014 253.329 253.626 253.905 254.169 254.418 254.653 254.875 255.085 255.283 255.471 255.647 255.813 255.968 256.113 g of (T ) (kJ/mol) 231.743 231.626 225.669 219.548 213.312 206.991 200.604 194.165 187.684 181.168 174.623 168.052 161.458 154.846 148.216 141.571 134.913 128.242 121.561 114.871 108.171 101.464 94.750 88.029 Stanford University Version 1.2 ©Hai Wang Hydroxyl radical (OH•) T (K) 298 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 c p (T ) (J/mol-K) 29.887 29.878 29.573 29.483 29.534 29.684 29.916 30.240 30.694 31.174 31.641 32.093 32.529 32.948 33.349 33.730 34.091 34.433 34.755 35.056 35.339 35.602 35.848 36.077 s o (T ) h(T)-h(298) (J/mol-K) (kJ/mol) 183.722 183.922 192.469 199.056 204.434 208.997 212.975 216.516 219.724 222.672 225.405 227.955 230.350 232.608 234.748 236.781 238.719 240.572 242.346 244.049 245.686 247.263 248.784 250.252 0.000 0.060 3.030 5.982 8.931 11.892 14.871 17.878 20.923 24.017 27.157 30.344 33.575 36.850 40.165 43.519 46.910 50.336 53.796 57.287 60.806 64.354 67.926 71.523 1-41 h fo (T ) (kJ/mol) 39.346 39.348 39.376 39.333 39.240 39.108 38.945 38.759 38.569 38.389 38.217 38.052 37.895 37.744 37.599 37.459 37.324 37.193 37.065 36.939 36.813 36.687 36.560 36.432 g of (T ) (kJ/mol) 34.631 34.599 33.010 31.422 29.848 28.292 26.758 25.246 23.754 22.282 20.825 19.382 17.952 16.533 15.124 13.723 12.331 10.946 9.568 8.196 6.830 5.470 4.116 2.767