Previous: Chapter 4. Clausius Inequality as Equilibrium Criterion
Classical thermodynamics does not include time explicitly, and so there are proposals to rename it as ‘thermostatics’. However, this name does not include the ideal Carnot heat engine, in which the conversion of heat into work was investigated and the maximum efficiency was found. Similarly, thermodynamics allows us to consider an ideal combustion process without heat loss and thereby determine the maximum possible flame temperature. The name ‘thermostatics’ again does not include such possibility.
This chapter considers a simple example: computing adiabatic flame temperature during coal combustion in air. This choice makes it possible to keep equations and calculations as simple as possible, although even in this case further simplifications are required. In any case, this example demonstrates that the notorious drawback of thermodynamics — the lack of explicit time — is advantageous by solution of many practical problems.
Thermodynamics offers a unique opportunity to quickly solve a problem by finding the maximum efficiency. The compilation of thermodynamic properties of substances was mainly driven by the problem of estimating rocket thrust; estimation of adiabatic flame temperature was one of the intermediate tasks along this way.
We start with a general formulation of the problem to compute the adiabatic flame temperature. Then the simplest calculation for a complete reaction is considered. After that a more complex approach is discussed involving computing the equilibrium composition at a given temperature and pressure. Finally, I return to Einstein’s words about phenomenological theories and discuss in this respect the Gibbs formalism that underlies modern chemical thermodynamics.
- General statement of the problem
- Computing flame temperature for the complete reaction
- Computing equilibrium composition at given temperature and pressure
- From Carnot cycle to chemical thermodynamics.
General statement of the problem
Let us write down the combustion reaction of coal in air:
C(coal) + O2 = CO2
Air also contains nitrogen (N2), which doesn’t take part in the reaction, but its presence affects the final temperature. First, let us consider the conceptual model for this chapter; Carnot’s drawing is again helpful to visualize the problem:
The system under study — coal and air — is located in the cylinder beneath the piston. A chemical reaction occurs in the cylinder, and thus, in this chapter, we move beyond individual substances. The external pressure above the piston remains constant — let us take one atmosphere, which corresponds to the standard pressure in the reference book ‘Thermodynamic Properties of Individual Substances‘. At the same time, the cylinder remains unconnected to any heat source — expansion work is possible, but heat exchange is excluded. This process is called adiabatic, and in our case, it corresponds to combustion without heat loss.
Let us compare this with candle combustion. The initial state is the solid candle and air at room temperature (in this chapter, a temperature of 25°C = 298.15 K will be used). It is now assumed that heat losses during combustion are ignored and we consider the final state after combustion is completed. The actual flame temperature is lower than the adiabatic temperature; the difference is due to a part of energy lost during heat exchange with the external environment. Another assumption involves either complete reaction or complete equilibrium. These choice leads to an additional difference between the adiabatic flame temperature and the actual temperature.
The Clausius inequality remains as the criterion for the equilibrium state in an adiabatic process. Spontaneous combustion is accompanied by an increase in entropy at constant enthalpy and pressure:
(dS)H,p > 0
When equilibrium is reached, the entropy of the entire system comes to the maximum value. The difference from an isolated system is in other external quantities — enthalpy and pressure— being constant; this is related to expansion work during the adiabatic process.
However, this criterion is not directly employed below. In the case of the complete reaction, the condition of constant enthalpy is enough, since the condition for the complete reaction defines the final state. In the next step, the effect of carbon dioxide dissociation is considered, but unfortunately the equilibrium criterion above would lead to too complex calculations; for simplicity, I stay at the level of computing equilibrium composition at constant pressure and temperature.
Calculation of flame temperature when the reaction is complete
Let us start with the assumption that the coal combustion reaction is complete; we also choose an initial state such as that the amount of air is stoichiometric to the amount of carbon. The choice of this initial state also affects the final results; estimating the temperature of a real flame requires knowledge of the initial composition. Under assumptions above, the reaction below defines the initial and final states; the number of moles of nitrogen corresponds to the nitrogen-to-oxygen ratio in the air:
C(coal) + O2 + 3.76N2 = CO2 + 3.76N2
The initial temperature of substances is assumed to be 25°C = 298.15 K, and the final temperature is to be found. The total pressure remains equal to 1 atm, but the partial pressures of the gases differ from one atmosphere. All gases are assumed to be ideal and in this case the gas enthalpy is independent of pressure. Therefore, standard enthalpies at 1 atm are used for calculations. The calculation is performed for one mole of coal, but this choice does not affect the final temperature.
The total enthalpy change is zero as the total enthalpy remains constant. This condition can be used to determine the unknown temperature in the final state (T2). Since the enthalpy change is independent of the transition path, we choose the most convenient approach to use thermodynamic tables. The transition path is divided in two steps: 1) the chemical reaction at T 1 = 298.15 K, 2) heating the reaction products from 298.15 K to T2. The sum of the enthalpy change in both steps is zero:
∆rH°(T1) + H°(T2, products) − H°(T1, products) = 0
In this equation, the enthalpy of reaction at T1 = 298.15 K is calculated using the enthalpies of formation of the reaction participants. Nitrogen is not required in this case, since it remains in the same state:
∆rH°(T1) = ∆fH°m(CO2, T1) − ∆fH°m(C, coal, T1) − ∆fH°m(O2, T1)
The enthalpy change during heating is written down in terms of the enthalpy change of each substance:
∆H°(products) = [H°m(CO2, T2) − H°m(CO2, T1)] + 3.76[H°m(N2, T2) − H°m(N2, T1)]
To calculate the enthalpy change of a substance by heating, the integral over the heat capacity is required. For simplicity, the integral is replaced by the product of the average heat capacity and the temperature difference:
∆H°(products) = Cp°m(CO2)(T2 − T1) + 3.76Cp°m(N2)(T2 − T1)
All together gives the final equation for the adiabatic temperature; the enthalpy of reaction at T1 is spent on heating the reaction products from T1 to T2 :
T2 = −∆rH°(T1)/[Cp°m(CO2) + 3.76Cp°m(N2)] + T1
The table below displays the thermodynamic data needed for the calculation: the enthalpy of formation in kJ/mol, entropy, and heat capacity in J/(mol K). In thermodynamic tables there was information for graphite only, so the calculation is performed for it; the difference from coal is small.
| С(gr) | O2 | CO2 | N2 | |
| ∆fH°(298.15 K) | 0 | 0 | -393.5 | 0 |
| Cp°(298.15 K) | 8.5 | 29.4 | 37.1 | 29.1 |
| Cp°(1000 K) | 21.7 | 34.9 | 54.3 | 32.7 |
| Cp°(2000 K) | 24.9 | 37.8 | 60.4 | 36.0 |
| S°(298.15 K) | 5.7 | 205.0 | 213.7 | 191.5 |
| S°(2518 K) | 46.6 | 277.5 | 323.2 | 260.3 |
From the information in the table, an estimate of the adiabatic flame temperature of graphite combustion in a stoichiometric amount of air is obtained using heat capacities at 1000 K as average values:
T 2 = 393500/(54.3 + 3.76*32.7) + 298.15 = 2518 K
Entropy was not required for the calculation, but it can be verified that the entropy of the final state is greater than the entropy of the initial state. Graphite is a solid phase under a total pressure of one atmosphere, and therefore standard entropy can be used for it. Unlike enthalpy, the entropy of an ideal gas depends on pressure:
S(T, p) = S°(T) − R ln (p/p°)
In our case, this affects the entropy of the gases, since the gases together are under pressure of one atmosphere, but in the equation for entropy, the partial pressure of the gas should be used. This can be calculated through the mole fraction – I give the equation for nitrogen:
p(N2) = x(N2)p°
Fortunately, in our problem, the mole fraction of nitrogen does not change, and the mole fraction of CO2 in the final state is equal to the mole fraction of oxygen in the initial state. This means that the necessary corrections for the partial pressure in the initial and final states are the same, and therefore they cancel each other out. Therefore, we can use standard gas entropies for the calculation:
S 2 − S 1 = 323.2+3.76*260.3−(5.7+205+3.76*191.5) = 371.2
This confirms the expected result: the combustion of graphite in air in a closed adiabatic system at constant pressure is a spontaneous process.
Calculation of equilibrium composition at given temperature and pressure
However, the entropy change estimated above is not at the maximum value. At high temperatures, gas dissociation becomes noticeable and it means that the assumption of the complete reaction is not correct. In our case, it is sufficient to consider only the dissociation of carbon dioxide into carbon monoxide and oxygen, since the concentrations of the other components are negligible:
| CO2 | = | СO | + | 0.5O2 |
| 1 − ξ | ξ | 0.5ξ |
Let us take one mole of CO2 as the initial state; the chemical variable ξ indicates the number of moles of dissociated CO2 and, simultaneously, the number of moles of dissociation products that have been formed. To compute the equilibrium composition is to find the value of the chemical variable that corresponds to the state of equilibrium. Nitrogen is not included in the chemical reaction, but it must be considered, as its presence shifts the equilibrium position.
Unfortunately, using the maximum entropy criterion leads to complex calculations — it requires finding two unknowns (temperature and a chemical variable), as well as to take into account the dependence of entropy on temperature. Therefore, I limit myself to explain the calculation of the equilibrium composition at constant temperature and pressure. The final state of the reaction is taken, one mole of carbon dioxide plus 3.76 moles of nitrogen at 2518 K and 1 atm (ξ = 0), and the goal is to compute the value of the chemical variable at equilibrium.
I start with the fundamental inequality from Chapter 4, ‘Clausius Inequality as Equilibrium Criterion‘. It defines the direction of a spontaneous process in a system at constant external pressure and temperature:
dU − TexdS + pexdV < 0
Thermal and mechanical equilibrium of the system with its environment is now assumed (T = Tex and p = pex). Then the left-hand side can be replaced by the change in the Gibbs energy of the system:
d(U − TS + pV) < 0 => (dG)T,p < 0
Thus, the Gibbs energy of a system is a criterion for the direction of the spontaneous processes at constant temperature and pressure and, as a result, a criterion for the equilibrium state – the minimum value of the Gibbs energy of the system under given conditions.
The system under consideration consists of a gaseous solution (nitrogen, carbon dioxide, oxygen, and carbon monoxide). The ideal gas solution is a good approximation and in this case, the total Gibbs energy is the sum over the components and it contains one unknown — the chemical variable:
G(ξ) = 3.76 Gm(N2) + (1 − ξ)Gm(CO2) + ξGm(CO) + 0.5ξGm(O2)
However, the Gibbs energy of an individual component depends on the partial pressure of the component, which is the product of the mole fraction and the total pressure:
Gm,i = G°m,i + RT ln[xi(ξ) p°]
The mole fraction, like the number of moles, is a function of the chemical variable. Substituting into the original expression and discarding terms that are independent of the chemical variable yields the final expression:
G(ξ) = ξΔrG° + RT[(1 − ξ) ln p(CO2) + ξ ln p(CO) + 0.5ξ ln p(O2) + 3.76 ln p(N2)]
Here the partial pressures are functions of the chemical variable, and ΔrG° is the standard Gibbs energy of the carbon dioxide dissociation reaction, which can be calculated from thermodynamic tables for a given temperature: ΔrG°(2518 К) = 67.4 kJ/mol. A plot of the function, made in Gnuplot (see the script in Appendix), is shown below. It allows us to estimate the value of the chemical variable for which the minimum value of the Gibbs energy is achieved (the exact value is ξ = 0.213).
The graph also shows the difference between local and global equilibrium. Any value of the chemical variable characterizes a gas solution in local thermal and mechanical equilibrium — all substances are at a given temperature, and the sum of the partial pressures is equal to the external pressure. At the same time, chemical equilibrium is achieved only at the minimum values of the Gibbs energy.
The standard enthalpy of the carbon dioxide dissociation reaction at this temperature is ΔrH°(2518 К) = 275.1 kJ/mol. The equilibrium value of the chemical variable shows that a part of the enthalpy released during graphite combustion (393.5 kJ) equal to 275.1*0.213 = 58.6 kJ is spent on the dissociation of CO2. This leaves less energy available for heating the components, which in turn leads to a smaller adiabatic flame temperature.
Using the maximum entropy criterion allows us to simultaneously determine the equilibrium composition and temperature, but, as already mentioned, the calculations are quite complex. In Additional Information there is a link to a paper to compute the freezing of supercooled water, in which this criterion was used for a simpler problem. Alternatively, an iterative solution can be proposed. A correction is made for the value of the enthalpy due to dissociation. This gives a new estimate of the adiabatic flame temperature. The equilibrium composition can now be computed at this temperature, the new correction for the enthalpy due to dissociation can be applied, and so on until convergence is achieved.
From Carnot cycle to chemical thermodynamics
Currently, chemical thermodynamics includes a detailed consideration of solutions, as well as chemical and phase transformations — all this are beyond the scope of this book. The formalism of modern chemical thermodynamics was developed by physicist Josiah Willard Gibbs in his classic work ‘On the Equilibrium of Heterogeneous Substances‘ in 1876-1878. It took a considerable time for chemists to recognize the importance of Gibbs’s work, but it is this work that forms the foundation of modern chemical thermodynamics. Of course, we should not forget the experimental research necessary to create a database of thermodynamic properties — without this, it would be impossible to use of the formalism to solve practical problems. This part of research is correctly called experimental chemical thermodynamics and is the result of the collaborative work of physicists and chemists.
Let us recall Einstein’s characterization of phenomenological physics as a science of ‘concepts which are close to experience’. Classical thermodynamics is now called phenomenological exactly in this sense. This attitude is well expressed in Werner Heisenberg’s 1966 article ‘The Role of Phenomenological Theories in the System of Theoretical Physics‘ – note the comparison of phenomenological thermodynamics with Ptolemaic theory:
‘Phenomenological theories understandably always develop where the observed phenomena cannot yet be attributed to general laws of nature. The reason for this impossibility may lie either in the high degree of complication of the phenomena in question, which does not yet allow such a connection because of the mathematical difficulties, or in the ignorance of the relevant natural laws themselves. … Examples of the second case: … in the first half of the 19th century Faraday’s reflections on electricity and phenomenological thermodynamics; in ancient astronomy — Ptolemaic theory of cycles and epicycles in planetary motion.’
However, the introduction of entropy during the research on Carnot’s ideal heat engine and the Gibbs formalism to compute chemical equilibria do not fit into the framework of ‘concepts which are close to experience’. I would say that the Gibbs formalism is rather unexpected consequences of the laws of classical thermodynamics. Similarly, the general laws of nature discussed by Heisenberg and that are examined in the next part of the book allow us to estimate the thermodynamic properties of substances from molecular constants. Yet, they do not allow us to derive an equilibrium criterion as they do not have the arrow of time.
I start with a brief historical digression that will allow me better to discuss this issue. The course of chemical reactions was explained by chemists by means of chemical affinity. Kinetic theory in the 19th century did not have concepts to explain chemical reactions; it could not explain combustion at all. Calorimetry of chemical reactions has provided the first quantitative indicator to assess chemical affinity – the reaction heat. In the second half of the 19th century, chemists began to position reaction heat as a criterion for the spontaneous chemical process (the Berthelot-Thomsen principle).
Gibbs laid the foundations of modern chemical thermodynamics, but he was not interested to apply his formalism to solve chemical problems. Pierre Duhem correctly recognized the power of Gibbs’s method and, in his 1884 dissertation, demonstrated the fallacy of Berthelot’s principle. Duhem also proposed the use of the thermodynamic potential in other areas of physical chemistry. However, Duhem’s work irritated Berthelot and his supporters, leading to difficulties in Duhem’s career as a physicist. At the same time, chemists did not pay attention to Dugem’s work – for them there was too much incomprehensible physics and mathematics.
Elements of thermodynamics were introduced into chemistry in the late 19th and early 20th centuries by Jacob van’t Hoff, Svante Arrhenius, and Wilhelm Ostwald. Their simplified approach was more appealing to chemists, and step by step, thermodynamics permeated chemistry. Further development brought back the use of the original rigorous Gibbs formalism, but this process took considerable time. Currently, the advantages of the Gibbs method are most evident in the case of phase diagrams (see CALPHAD, CALculation of PHAse Diagrams), where thermodynamics allows us to employ joint processing of diverse experiments to determine the most accurate Gibbs energies of phases and after that to use them in calculating phase diagrams.
Let us be back at the end of 19th century and see what Ludwig Boltzmann meant by ‘phenomenological theory’. He expressed his views in the lecture ‘On the Development of Methods of Theoretical Physics in Modern Times‘ in 1899. Boltzmann begins to discuss phenomenological theories with Maxwell’s electromagnetic theory:
‘Maxwell wished his theory to be regarded as a mere picture of nature, a mechanical analogy as he puts it, which at the present moment allows one to give the most uniform and comprehensive account of the totality of phenomena. We shall see how influential this Maxwellian position was on the further development of his theory. Through his practical successes he quickly helped these theoretical ideas to victory.’
Moreover, Boltzmann connects the success of Hertz’s experiments, confirming Maxwell’s theory, with the subsequent success of the development of phenomenological theories:
‘indeed just as a pendulum goes beyond its rest position to the opposite side, extremists went so far as to brand all conceptions of classical physical theory as misguided.’
By ‘classical physical theory’ Boltzmann means physics of the first half of 19-th century with the ideal of to find ‘true nature of things’. Thus, he characterizes phenomenological theories as follows:
‘So that the sole task of physics consisted in using trial and error to find the simplest equations that satisfied certain required formal conditions of isotropy and so on, and then to compare them with experience. This is the most extreme form of phenomenology, which I should like to call mathematical, whereas general phenomenology seeks to describe every group of facts by enumeration and by an account of the natural history of all phenomena that belong to that area, without restriction as to means employed except that it renounces any uniform conception of nature, any mechanical explanation or other rational foundation.’
Hence in this lecture Boltzmann considers phenomenological thermodynamics to be comparable with Maxwell’s electromagnetic theory. Moreover, according to the Boltzmann’s lecture, the success of phenomenological thermodynamics at the end of the 19th century was related to the success of Maxwell’s theory. I would agree with such meaning of ‘phenomenological theory’ in respect to thermodynamics, but nowadays the meaning of the term ‘phenomenological theory’ is quite different. No one anymore compares the development of classical thermodynamics to that of Maxwell’s electromagnetic theory.
Next: Chapter 6. Entropy of Non-equilibrium States
References
Thermodynamic Properties of Individual Substances (In Russian). In four volumes, third edition, 1978–1982.
Werner Ηeisenbeгg, Die Rolle der phänomenologischen Theorien im System der theoretischen Physik, in Preludes in Theoretical Physics : in Honor of V. F. Weisskopf, 1966, p. 166-169.
Photis Dais, The penetration of thermodynamics into chemistry: The birth of chemical thermodynamics in Europe and America, PhD Thesis, 2021.
Ludwig Boltzmann, Theoretical Physics And Philosophical Problems Selected Writings, 1974, On the Development of the Methods of Theoretical Physics in Recent Times (1899), p. 77-100.
Additional information
Freezing of Supercooled Water: The thermodynamics of freezing supercooled water in an isolated system is examined. A spontaneous transition occurs to a heterogeneous equilibrium of ice and water at the equilibrium temperature of the phase transition.
Appendix
Gnuplot code for plotting:
set terminal png enhanced size 500,400
set output 'C_02_C02.png'
R = 8.31441
ptot = 1
T = 2518
DelG = 67400
pN2(ksi) = 3.76/(3.76+1+0.5*ksi)*ptot
pCO2(ksi) = (1-ksi)/(3.76+1+0.5*ksi)*ptot
pCO(ksi) = ksi/(3.76+1+0.5*ksi)*ptot
pO2(ksi) = 0.5*ksi/(3.76+1+0.5*ksi)*ptot
G(ksi) = ksi*DelG+R*T*(3.76*log(pN2(ksi))+(1-ksi)*log(pCO2(ksi))+ksi*log(pCO(ksi))+0.5*ksi*log(pO2(ksi)))
set xlabel 'ksi'
set ylabel 'G'
plot [x=0.18:0.25] G(x) title 'G'Discussion