Chapter 17. Reaction Equilibria

We must first speak a little concerning contact or mutual touching, action, passion and reaction.

Daniel Sennert (1660)

Another important aspect of the thermodynamics of multicomponent systems is the rearrangement of atoms within and between molecules, known as chemical reaction. Equilibrium thermodynamic considerations tell us the direction and extent to which a reaction will go. As with phase equilibria, the constraint of minimum Gibbs energy dictates the equilibrium results at a fixed T and P.

We begin this chapter by noting that the material from Section 3.6 is important for this chapter and you may wish to read that section again. There are several steps to understand before the equilibrium conversion is calculated, and some steps may seem very theoretical. We begin in Section 17.1 by relating the reaction coordinate to the minimum in Gibbs energy at equilibrium. Then in Section 17.3 we introduce the standard state Gibbs energy of reaction using the Gibbs energies of formation. Next, we relate the Gibbs energies of the components in the reacting system to the chemical potentials and finally develop the equilibrium constant in terms of the ideal gas law and begin to calculate equilibrium conversions. However, the standard state Gibbs energy used to calculate the equilibrium constant depends on temperature, and thus the equilibrium “constant” also changes with temperature which is discussed in Section 17.7. We then proceed with more advanced topics such as energy balances, use of the Gibbs minimization method, and multiple phases in reaction equilibrium.

Chapter Objectives: You Should Be Able to...

1. Solve for the equilibrium reaction coordinate values and the equilibrium mole fractions for a given K_aT and P for single and multiple reactions.

2. Understand the influences of pressure, nonstoichiometric feed, and inerts on reaction equilbrium.

3. Calculate ΔG^o_298.15 and ΔH^o_298.15 for a given reaction.

4. Calculate ΔG^o_T and K_aT using the van’t Hoff equation.

5. Set up the energy balance for a given feed and equilibrium conversion, testing for closure or solving for heat transfer.

6. Incorporate solid species and liquid components into equilibrium calculations.

7. Understand the Gibbs minimization method for calculating reaction equilibrium.

17.1. Introduction

You have probably performed some reaction equilibrium computations before, usually in high school or freshman chemistry. This chapter shows how the “activities” (partial pressures for ideal gases) of products divided by reactants can be related to a quantity, K_a, that does not depend on pressure or composition, and despite its dependence on temperature, it is called the equilibrium constant. Developing the relationship between activities utilizes the concept of minimizing Gibbs energy and rearranging the basic relation. By study of the derivation we learn how to generalize reaction equilibrium analysis to multiple reactions and simultaneous reaction and phase equilibria.

We begin the chapter with an example to provide an overview of some of the methods developed in the chapter. We have selected an introductory reaction where all species are approximated as ideal gases. For ideal gases, we show in upcoming sections that the relation between equilibrium constraint and partial pressure is written

where the symbol Π designates a product (analogous to the symbol ∑ representing the summation sign), y_iP is the partial pressure (always expressed in bar) of the i^th component, and v_i is the stoichiometric coefficient discussed in Section 3.6. Since stoichiometric coefficients are negative for reactants, the product symbol results in a ratio of products over reactants. The solution primarily requires a mass balance relating the partial pressure to the reaction coordinate (also discussed in Section 3.6). The major steps to solving an equilibrium problem are as follows.

1. Ascertain how many phases are present and the method to be used for the equilibrium calculations. Our initial examples will use only a gas phase and determine equilibrium compositions using an equilibrium constant method. Later we will show how to use liquid and solid phases and how to use the Gibbs energy directly.

2. Use standard state properties to obtain the value of the equilibrium constant at the reaction temperature, or for the Gibbs minimization method find the Gibbs energies of the species. Usually this consists of two substeps:

a. Perform a calculation using the standard state Gibbs energies at a reference temperature and pressure.

b. Correct the temperature (and pressure for Gibbs method) to the reaction conditions.

3. Perform a material balance on the reactant and product species and relate the composition to the equilibrium constant or standard state properties from steps (1) and (2).

4. Solve for the equilibrium compositions.

The steps are made clearer by a series of examples. Steps (1) and (2) are lengthy, and the applications are easier to see by first studying steps (3) and (4) as we show in the next example. This example will help to provide motivation for understanding how to use the standard state Gibbs energies in steps (1) and (2).

This example demonstrates the method to use K_a to calculate the reaction coordinate. Readers should note that the value of K_a is fixed at a given temperature, but the equilibrium value of ζ may vary for different feed conditions and often pressure for gas phase reactions as we will show in other examples. To relate the equilibrium conditions to reaction engineering textbooks, we note that most reaction engineering textbooks use conversion rather than reaction coordinate to track reaction progress. By convention, conversion is tracked for the limiting species (the species used up first at the value of ζ closest to zero in the direction of the reaction). A relation is shown in the footnote of Section 3.6.

Several other concepts are important for a general understanding of calculating reaction equilibria. First, we must understand: fundamental relations between the Gibbs energies, activities, and the equilibrium constant (Sections 17.2–17.4); simplifications that are applied for ideal gases and the effect of pressure and inerts (Section 17.5); and calculations of the temperature dependence of the equilibrium constants (Sections 17.7–17.9). Later sections illustrate the adaptation of the fundamental equations to broader applications like multiple reactions with simultaneous phase equilibria.

17.2. Reaction Equilibrium Constraint

Several sub-steps are involved in the procedure outlined in Section 17.1 steps (1) and (2) to find the equilibrium constant. In this section, we derive the equilibrium constraint, and then show how the thermodynamic properties are used to simplify to Eqn. 17.1. At reaction equilibria, the total Gibbs energy is minimized. If the composition of a system is changing, the change in the Gibbs energy is given by:

The fact that species are being created or consumed by a reaction does not alter this equation. At constant temperature and pressure, the first two terms on the right-hand side drop out:¹

Substituting the definition of reaction coordinate from Eqn. 3.39,

Because G is minimized at equilibrium at fixed T and P, the derivative with respect to reaction coordinate is zero:

Now there is one unknown, ζ, in terms of which we can determine the changes in moles for all of the components. We make a further manipulation before we apply the equilibrium constraint. In phase equilibria, we found fugacity to be a convenient property to use because it simplified to the partial pressure for a component in an ideal gas mixture. We can rewrite Eqn. 17.9 in terms of fugacities. We recall our definition of fugacity dG = RT dln f. Integrating from the standard state to the mixture state of interest (cf. generalizing Eqn. 10.48),

where G_i^o is the standard state Gibbs energy of species i and f_i^o is the standard state fugacity. A standard state is introduced for liquids in Section 11.3, and now we generalize the approach. The standard state is at the reaction temperature, but a specified composition (often pure) and pressure P°. Substitution of Eqn. 17.10 into Eqn. 17.9,

or

We will need to calculate both summations appearing in Eqn. 17.12, and then combine the results. Qualitatively, this equation indicates how atoms should be arranged within molecules to minimize Gibbs energy. The connection between Eqn. 17.12 and Eqn. 17.1 will become obvious as we move through the next sections. Let us work on the two summations separately. We will show that the second summation is related to the product of partial pressures for gas phase species which we define as the equilibrium constant. The first summation will relate to the negative numerical value of the equilibrium constant because the two terms of Eqn. 17.12 are equal and opposite.

17.3. The Equilibrium Constant

We now focus on the second summation of Eqn. 17.12. The ratio appearing in the logarithm is known as the activity, (cf. Eqns. 11.23 for a liquid, but now in a general sense):

The numerator represents a mixture property that changes with composition. We have developed methods to calculate in Eqns. 10.61 (ideal gases), 10.68 (ideal solutions), 11.14 (real solution using γ_i), Eqn. 15.13 (real gases using ). The denominator represents the component at a specific standard state, which includes specification of a fixed composition (which can be pure or a mixture state).

The second sum of Eqn. 17.12 can be manipulated after inserting the activity notation,

In a reacting mixture and/or a_i will change as the reacting composition moves toward equilibrium. However, at equilibrium, the product term of activities is extremely important. We define the product term at equilibrium as the equilibrium constant K_a with the a subscript to denote that activity is used:

Combining the definition of the equilibrium constant with Eqn. 17.12, the first summation can be used to find the value of the constant:

Note that use of the term constant can be misleading because it depends on temperature. It is constant with respect to feed composition and changing mole numbers of reacting species as we will show below. We use a subscript a on the equilibrium constant to stress that it depends on activities. As we show later, there are other approximations for the equilibrium constant, and subscripts are used to differentiate between different conventions.

The Equilibrium Constant for Ideal Gases

The activity is a general property defined by Eqn. 17.13. We have seen it applied to liquids in Section 11.5. For ideal gases, the numerator of the activity is . We complete the formula for activity by selecting the standard state. For gaseous reacting species, the convention is to use a standard state of the pure gas at P° = 1 bar. For an ideal gas, f_i^o = P^o (Eqn. 9.29). Thus, f_i^o = 1 bar. The fugacity ratio (activity) is dimensionless provided that we always express the partial pressure in bar. The second sum of Eqn. 17.12 for ideal gases simplifies to

where the first two equalities are general, but the last is restricted to ideal gases. We will later reevaluate the fugacity ratio for nonideal gases, liquids, and solids. Now let us examine the first sum of Eqn. 17.12 which will give us the value of K_a.

17.4. The Standard State Gibbs Energy of Reaction

The first term on the right side of Eqns. 17.12 and 17.16, , is called the standard state Gibbs energy of reaction at the temperature of the reaction, which we will denote . The standard state Gibbs energy of reaction is analogous to the standard state heat of reaction introduced in Section 3.6. The standard state Gibbs energy for reaction can be calculated using Gibbs energies of formation.

As an example, for CH_4(g) + H₂O_(g) → CO_(g) + 3H_2(g)

It may be helpful to think of the sum as representing a path via Hess’s law where the reactants are “unformed” to the elements and then “formed” into the products. The signs of the formation Gibbs energies of the products are positive and the signs for the reactants are negative. Thus,

The Gibbs energies of formation are typically tabulated at 298.15 K and 1 bar, and special calculations must be performed to calculate at other temperatures—the calculations will be covered in Section 17.7. Like the enthalpy of formation, the Gibbs energy of formation is taken as zero for elements that naturally exist as molecules at 298.15 K and 1 bar, and the same cautions about the state of aggregation apply. Gibbs energies of formation are tabulated for many compounds in Appendix E at 298.15 K and 1 bar. Note that for water, the difference between and is the Gibbs energy of vaporization at 298 K. The difference is nonzero because liquid is more stable. (Which phase will have a lower Gibbs energy of formation at 298.15 K and 1 bar?)

The standard state Gibbs energy of reaction is related to the equilibrium constant through Eqn. 17.16,

Once the value of the equilibrium constant is known, equilibrium compositions can be determined, as shown in Example 17.1. The next example illustrates calculation of the Gibbs energy of reaction and the equilibrium constant.

Composition and Pressure Independence of K_a

The use of standard states for calculating has important implications on the composition and pressure independence of K_a. The standard state is at a fixed pressure, P°. Thus, is independent of pressure. The standard states are also at fixed composition (often pure), and thus is independent of equilibrium composition. Looking at Eqn. 17.20, we conclude that because is independent of equilibrium composition and pressure, K_a is independent of equilibrium composition and pressure. One important point is that the state of aggregation in the standard state is important and the values of and K_a do depend on the state of aggregation in the standard state; this point will be clarified in later sections.

We have now demonstrated steps 2(a), 3, and 4 for the procedure given in Section 17.1. The concepts have been demonstrated, but we must correct the temperature before doing calculations at temperatures other than 298.15 K. The butadiene reaction of Example 17.2 becomes more favorable with a larger K_a at higher temperatures. We will discuss some important aspects of the effects of pressure and inerts and also discuss reaction spontaneity before showing the calculation of temperature corrections.

17.5. Effects of Pressure, Inerts, and Feed Ratios

At a given temperature, equilibrium values of the reaction coordinate are affected by pressure, inerts, and feed ratios. The principle that changing the quantities affects equilibrium conversions is known as Le Châtelier’s principle in honor of Henry Louis Le Châtelier who first characterized the phenomenon. An understanding of Le Châtelier’s principle is important for operating industrial reactions. Two important modifications led to significant hydrogen conversions in Example 17.1 even though the equilibrium constant was small—use of pressure and nonstoichiometric feed.

Pressure Effects

Pressure has little effect on the activities of condensed species (e.g., the Poynting correction is typically small) and thus it has a primary significance only for reactions with gas phase components. Pressure has important effects when both 1) gas species are involved in reactions and 2) the stoichiometric numbers of gas species are different for reactants and products. When the stoichiometric moles of gas species are the same for reactants and products, P has no effect by the ideal gas approximation, and for nonideal gases only indirect effects due to fugacity coefficients.

The equilibrium constants for ideal gases can be written

This form makes the pressure effect more obvious. As mentioned above, when the stoichiometric number of gas moles is the same for products and reactants, Σv_i = 0 and the pressure effect drops out. When the stoichiometric numbers of vapor reactant moles is greater than the stoichiometric numbers of product vapor moles, an increase in pressure will drive the reaction to higher conversions,. Σv_i < 0. When the stoichiometric gas mole ratios are reversed, a decrease in pressure will help drive the reaction to higher conversions, Σv_i > 0. In Example 17.1 the pressure of 20 bar was important to yield significant conversions. It can be helpful to consider that qualitatively the pressure “squeezes” the reaction towards the side with fewer gas moles. As an exercise, determine the reaction coordinate for the same feed when the pressure is 1 bar.

Inerts

A component that does not participate in a reaction is called inert. Inert gas components often have an indirect, but important effect on the equilibrium reaction coordinate when gas phase species are present. Inerts change the overall mole fractions and thus mitigate the pressure effects. When Σv_i > 0, adding an inert will increase conversion at a fixed total pressure. However, when Σv_i < 0, the mitigation of the pressure effect is undesirable and inerts should be avoided. Qualitatively, the presence of an inert decreases the “squeezing” effect mentioned above.

Nonstoichiometric Feed

Conversions of specific reactants are influenced using nonstoichiometric feed. In Example 17.1, excess CO was fed to the reactor; conversions were 42% (H₂), and 10.5% (CO). Generally, an excess of one reactant will tend to increase conversion of the other reactant. The effect can be seen qualitatively using Eqn. 17.2. For a given K_a, at a certain value of y_CH3OH, a higher value of y_CO results in a lower value of y_H2. When using stoichiometric feed (CO:H2 = 1:2) in Example 17.1, the equilibrium conversions of H₂ and CO are equal (40.6%). Example 17.1 includes both excess CO and a pressure effect. The excess CO in Example 17.1 is high enough to mitigate the beneficial pressure effect in a manner similar to an inert gas. The feed ratio giving highest H₂ conversion for the specified conditions uses less excess CO, (CO:H₂ = 1:1), which results in conversions of 45.2% (H₂) and 22.6% (CO). Use of nonstoichiometric feed is common in industrial reactions because in some cases it helps avoid side reactions in addition to effects on equilibrium.

17.6. Determining the Spontaneity of Reactions

In our preliminary examples, we have assumed rather idealized cases where none of the products are present in the inlet. However, in some cases, products may be present and then the reaction direction may not be as we anticipate. We can look at the reaction thermodynamics in a slightly different way to determine the direction of the reaction under given compositions, T and P. Starting from Eqn. 17.10, we may add by weighting with the stoichiometric numbers, resulting in

The term on the left side is called the Gibbs energy of reaction and is given the symbol ΔG_T. Note that this is a different term than the standard state Gibbs energy of reaction (the second term) that uses the superscript °. Thus, we can write,

A reaction with is called exergonic and results in K_a > 1, and a reaction with is called endergonic, resulting in K_a < 1. This provides an indication of whether the equilibrium favors products or reactants, but does not mean that reactions with small values of K_a cannot be conducted industrially. For example, Example 17.1 involved a small K_a (thus endergonic with ), yet the conversion of H₂ was 42%. The propensity for the reaction to go forward or backward depends instead on the Gibbs energy of reaction ΔG_T at the concentrations represented by the fugacity ratios. If the conditions provide ΔG_T < 0, then the Gibbs energy is lowered when the reaction proceeds in the forward direction. If we evaluate conditions and ΔG_T > 0, then the reaction goes in the reverse direction than what we have written. In either case, the concentrations adjust until the system reaches the equilibrium condition, ΔG_T = 0, and then Eqn. 17.12 applies. In summary, the direction a reaction proceeds is determined by ΔG_T, not by . Note when evaluating the fugacity term for determining spontaneity that the actual conditions are used, not the equilibrium conditions. At the feed conditions of Example 17.1, y_CH3OH = 0, which ensures ΔG_T < 0 at the feed conditions even though .

17.7. Temperature Dependence of K_a

Always remember that depends on the standard state, which changes with temperature. Comparing Examples 17.2 and 17.3, at 298 K (K_a = 1E-14), but decreases to at 900 K (K_a = 0.242). In order to calculate , it may seem that we need to know ΔG_f^o for each compound at all temperatures. Fortunately this is not necessary because the can be determined from the Gibbs energy for the reaction at a certain reference temperature (usually 298.15 K) together with the enthalpy for the reaction and the heat capacities of the species.

Suppose we have a table of standard energies of formation at 298.15 K but we would like the value for at some other temperature. We can account for temperature effects by applying classical thermodynamics to the change in Gibbs energy with respect to temperature using the Gibbs-Helmholtz relation,

which results in the van’t Hoff equation:

For accurate calculations, we must recognize that the heat of reaction depends on temperature. We have developed the standard heat of reaction in Section 3.6 and discussed the temperature dependence there. We show later that an assumption of a temperature-independent heat of reaction results in a short-cut approximation that is often close to the full calculation. We first show the full calculation. Substituting into the van’t Hoff equation (Eqn. 17.26) and integrating again,

where we previously described finding J in Eqn. 3.46 on page 113. If desired, all values at T_R can be lumped together in a constant, I.

The constant I may be evaluated from a knowledge of ΔG^o₂₉₈ by plugging in T = 298.15 on the right-hand side as illustrated below.

17.8. Shortcut Estimation of Temperature Effects

Recall Eqn. 17.25, which we refer to as the general van’t Hoff equation:

We can make rapid estimates of the equilibrium constant when we make the approximation that ΔH_T^o is independent of temperature. That is, suppose ΔC_P = Δa = Δb = Δc = Δd = 0, which means the sensible heat effects for the reactants and products are the same. This is most closely approximated when all species are about the same molecular size and the same state of aggregation. With this approximation, in Eqn. 17.27, or we can integrate Eqn. 17.25 directly to obtain what we refer to as the shortcut van’t Hoff equation:

This equation enables rapid screening for the effects of temperature and the detailed van’t Hoff can be used as a follow-up calculation. As a particular observation, we take special note from the above equation that exothermic reactions (ΔH_T < 0) lead to K_a decreasing as temperature increases, and endothermic reactions (ΔH_T > 0) lead to K_a increasing as temperature increases. This means that equilibrium conversion (for a specified feed) decreases with increasing temperature for exothermic reactions, and increases for endothermic reactions. This effect is illustrated in Fig. 17.1. The emphasis is placed on equilibrium because reaction rates increase with temperature. Industrial application of exothermic reactions are almost always run at elevated temperatures even though the equilibrium constant decreases at high temperature. For all reactions, the reaction rate will approach zero as equilibrium is approached. The benefit of faster kinetics typically outweighs the smaller equilibrium constant for exothermic reactions when economics are considered. There connection between equilibrium constants and kinetic rates that approach zero is explained in Section 17.15.

Figure 17.1. Qualitative behavior of equilibrium conversion for exothermic and endothermic reactions.

The approximate results of the shortcut van’t Hoff equation should be followed with the detailed van’t Hoff for critical applications. To improve shortcut estimates, the detailed van’t Hoff can be used at T_near within 100 K of the temperatures of interest to calculate ΔG_Tnear^o and ΔH_Tnear^o. Then the values at T_near can be used as the reference values in Eqn. 17.29.

17.9. Visualizing Multiple Equilibrium Constants

Plots of equilibrium constants provide a rapid method to visualize the gross trends and orders of magnitude. Fig. 17.2 illustrates how several reactions can be illustrated in a single graph. The equilibrium constants are calculated with the full temperature dependence. Note that the plots are nearly linear as would be approximated by the short-cut van’t Hoff. Exothermic reactions have a negative slope and endothermic reactions have a positive slope. When dealing frequently with a set of reactions, such a graph can serve as a “road map” where the optimal temperature window of operation maximizes desired products while minimizing by-products and potential coupling of reactions.

Figure 17.2. Graphical analysis of competing reactions.

The reactions of Fig. 17.2 are typically involved in many high-profile applications including: combustion, chemical-vapor infiltration, reforming, coking during reforming, space station gas management, electrolysis, and the hydrogen economy. Several of these reactions have common names which are listed below.

To illustrate interpretation of the graph, consider an application of the above reactions to material management in space station gas management, where an objective is to remove CO₂ and provide O₂. There are many ways that the reactions could be combined. On a space station sunlight is relatively abundant. Therefore, high temperatures and solar cells are available, but food must be imported. Note that the Sabatier reaction would convert waste CO₂ to fuel but requires H₂. Fig. 17.2 shows that the equilibrium constant is favorable below 900 K. The Bosch reaction also favors products at temperatures below 900 K. The Bosch reaction produces graphitic carbon, which can be collected in dense form and conveniently disposed. The hydrogen required for the Bosch reaction could be generated by water decomposition, which could be achieved with electrolysis or pyrolysis, with the benefit of co-producing oxygen for respiration. A small extrapolation of Fig. 17.2 shows that water pyrolysis is favorable above 2300 K.² Coupling the Sabatier reaction with methane pyrolysis has been suggested. Methane pyrolysis is favorable above 700 K. This would produce hydrogen for other use. Hydrogen production could also be achieved by the syngas reaction, if graphitic carbon was available. H₂ could be enhanced and CO removed by the water-gas shift. Catalysts can selectively alter the kinetics to minimize undesired products, although they cannot alter the equilibrium constraints. Nevertheless, all combinations are constrained by material balances, which dictate the overall reactions.

This kind of reaction network analysis is typical of many applications. For example, some simple economic considerations show why producing hydrogen by steam reforming of methane (natural gas) is the preferred method compared to electrolysis. The energetic cost of water electrolysis raises serious doubts about electrolysis feasibility on Earth. With abundant electrical energy, it might be more appropriate to operate electric vehicles. It is not practical to articulate all the ways that this kind of network analysis can be applied to modern problems, but these illustrations should suggest the manner of proceeding for many such analyses. Noting that energies of reaction are an implicit part of the analysis, a tremendous wealth of information is implied by a single graph like Fig. 17.2. Later, in Section 17.11, we demonstrate how combining an unfavorable reaction with a strongly favorable reaction can help to drive the unfavorable reaction.

17.10. Solving Equilibria for Multiple Reactions

When the equilibrium state in a reacting system depends on two or more simultaneous chemical reactions, the equilibrium composition can be found by a direct extension of the methods developed for single reactions. Each reaction will have its own reaction coordinate in which the compositions can be expressed. Some of the products of one reaction may act as reactants in another reaction, but the amount of that substance can still be written in terms of the extents of the reactions. Eventually, the material balances lead us to a system of N nonlinear equations in terms of N unknowns. We illustrate a solution by hand and then demonstrate how numerical solvers can be used.

These equations were amenable to the quadratic formula, but in general equilibrium criteria can be more complicated. Fortunately, standard programs available that are formulated to solve numerically multiple nonlinear systems of equations, so we can concentrate on applying the program to thermodynamics instead of developing the numerical analysis. Many software packages like Mathematica, Mathcad, MATLAB, and even Excel offer the capability to solve nonlinear systems of equations. Excel provides an especially convenient basis for illustrating the methods presented here.

a. These data are slightly different from values calculated using tabulated properties from Appendix E, but such variations are common in thermochemical data. The equilibrium compositions are about the same if the example is reworked using data from Appendix E.

b. Reid, R., Prausnitz, J.M., Poling, B. 1987. The Properties of Gases and Liquids, 4th ed. New York: McGraw-Hill.

c. See the online supplement for an introduction to Solver.

Figure 17.3. Worksheet DUALRXN from workbook Rxns.xlsx for Example 17.7 showing converged answers at several temperatures.

17.11. Driving Reactions by Chemical Coupling

Frequently, one may encounter a reaction that is not favored by K_a, and manipulation of temperature or pressure or feed composition provides only limited benefit for the desired conversion. In these cases, it may be possible to couple the reaction to another, more favorable, reaction to drive the overall production forward. Biological systems use coupling extensively. The building of sugars and biological tissue from CO₂ and water is thermodynamically unfavorable. Carbon is fully oxidized and it must be reduced to create carbohydrates, and the reactions are endergonic at room temperature. These reactions are achieved by coupling an unfavorable carbon reduction with a strongly exergonic reaction.

To illustrate the principles of chemical coupling with a simple set of reactions, let us consider the production of butadiene from butene dehydrogenation at 900 K. We have investigated this reaction in Example 17.3 where we showed that the reaction is endergonic: K_a = 0.242 is small. The example showed that conversion is improved by diluting with steam. Consider instead if CO₂ is fed to the reactor and a catalyst is provided for the water-gas shift reaction (c.f. Eqn. 17.32). The CO₂ could then react with H₂ product of the dehydrogenation, inducing higher conversion for the dehydrogenation. The hydrogen product is removed by Le Châtelier’s principle and the dehydrogenation reaction is pulled forward.

Chemical coupling can be classified in three ways: (1) induction, where a second reaction “pulls” a desired reaction by removing a product as in Example 17.8; (2) pumping, where the second reaction creates additional reactant for the desired reaction to “pump”; or (3) complex, where both induction and pumping are operative.³ An example of chemical pumping starts with the reaction of methyl chloride and water to form methanol and hydrochloric acid.

By adding the methyl chloride synthesis reaction,

This overall reaction becomes (adding the reactions, and take the product of the K_as):

The large equilibrium constant of 17.44 forms CH₃Cl_(g) readily, to pump reaction 17.43 via Le Châtelier’s principle. Through chemical coupling, the prospects of developing a feasible reaction network are virtually endless.

17.12. Energy Balances for Reactions

We have previously introduced the energy balance in Section 3.6 and also discussed adiabatic reactors. In this section we consider that there may be a there is a maximum possible value of ξ (outlet conversion) due to chemical equilibrium. Equilibrium may affect both adiabatic and nonadiabatic reactors, but we cover adiabatic reactors, and the extension to nonadiabatic should be obvious with the inclusion of the heat term.

Adiabatic Reactors

The energy balance for a steady-state adiabatic flow reactor is given in Eqn. 3.53 on page 118. The variables T^out and from the energy balance also appear in the equilibrium constraint that will govern maximum conversion. Earlier, in Chapter 3, we considered the reaction coordinate to be specified. However, in a reaction-limited adiabatic reactor, we must solve the energy balance together with the equilibrium constraint to simultaneously determine the maximum conversion and adiabatic outlet temperature. Using the energy balance from Eqn. 3.53, do the following.

1. Write the energy balance, Eqn 3.53. Calculate the enthalpy of the inlet components at Tⁱⁿ.

2. Guess the outlet temperature, T^out. Calculate the enthalpy of the outlet components at T^out.

3. Determine at T^out using the chemical equilibrium constant constraint.

4. Calculate for this conversion.

5. Check the energy balance for closure.

6. If the energy balance does not close, go to step 2.

As you might expect, this type of calculation lends itself to numerical solution, such as the Solver in Excel.

a. We can evaluate this assumption by calculating the reduced temperatures at the end of our calculation and estimating the virial coefficients, then fugacity coefficients.

Graphical Visualization of the Energy Balance

The energy balance is presented in Fig. 3.6 on page 119. The difference here is that the appropriate curve from Fig. 3.6 is superimposed on the plot and the outlet conversion and outlet temperature are determined by the intersection of the energy balance line and the equilibrium line. Fig. 17.5 illustrates an exothermic reaction. In the event that the reaction does not reach equilibrium because of kinetic limitations, the reaction coordinate must be located along the energy balance line below the equilibrium value. For the case of the ammonia reaction, the equilibrium constraint curve could be generated by inserting various temperatures in Eqn. 17.47 and then determining the reaction coordinate from Eqn. 17.48. The energy balance is plotted using Eqn. 3.55. The dot in the figure represents the point where the energy balance and equilibrium constraint are both satisfied. Note that an endothermic reaction will have an energy balance with a negative slope, and the equilibrium line will change shape as shown in Fig. 3.6, making the plot for an endothermic reaction a mirror image of Fig. 17.5 reflected across a vertical line at Tⁱⁿ.

Figure 17.5. Approximate energy balance for an exothermic reaction. The dot simultaneously represents the equilibrium outlet conversion and reaction coordinate value at the adiabatic outlet temperature. The plot for an endothermic reaction will be a mirror image of this figure as explained in the text.

17.13. Liquid Components in Reactions

When a liquid component is involved in a reaction, the fugacity ratio for activity in Eqn. 17.15 is typically expressed using activity coefficients. Thus,

where P is expressed in bar, and the Poynting correction is often negligible, as shown. Another important change in working with liquid components is that in determining K_a liquid phase values are used for , not the ideal gas values. Frequently these values are not available in the literature, so it is common to express equilibrium in terms of temperature-dependent correlations for K_a as described in Section 17.18.

Equilibrium constants calculated using liquid phase species are different from the equilibrium constants for the same reactants and products in the gas phase. Consider that for the vapor phase. If a vapor phase reaction is simultaneously in phase equilibrium with a liquid phase with the same components, by modified Raoult’s law we could replace . However, for the liquid equilibrium constant we should use only the activity (Eqn. 17.50) without the vapor pressure. Thus, we conclude that the equilibrium constants for the liquid phase reaction must be different from the equilibrium constant for the same vapor phase reaction, and also that the standard state Gibbs energy change for the same reaction must be different.

a. Vu, D. T., Kolah, A.K., Asthana, N.S., Peereboom, L., Lira, C.T., Miller, D.J. 2005. “Oligomer distribution in concentrated lactic acid solutions.” Fluid Phase Equil. 236:125–135.

If a vapor state coexists with a liquid phase during a reaction, the phase equilibria and reaction equilibria are coupled. Reactions need to be considered in only one of the two phases, and the equilibrium compositions will be consistent with compositions that would have been determined by the same reaction equilibria in the other phase. Similarly, some reaction equilibria constants may be known for only one or the other of the phases, and the equilibria can be solved by using reaction equilibria for whichever phase is most convenient. Simultaneous reaction and phase equilibria can be extremely useful for driving reactions in preferred directions as we illustrate in Example 17.15.

17.14. Solid Components in Reactions

When a solid component is involved in a reaction, the fugacity ratio for activity in Eqn. 17.15 is typically expressed using activity coefficients. For a solid solution,

where P is expressed in bar, P^sat represents the solid sublimation pressure, and the Poynting correction is often negligible. Commonly, multiple solids exist as physical mixtures of pure crystals as discussed in Section 14.10 on page 556. When the solids are immiscible, Eqn. 17.56 simplifies to

Similar to working with liquids, solid phase data are used for , not the ideal gas values. When these values are not available in the literature, it is common to express equilibrium in terms of temperature-dependent correlations for K_a as described in Section 17.18.

Consider the reaction:

The carbon formed in this reaction comes out as coke, a solid which is virtually pure carbon and separate from the gas phase. What is the activity of this carbon? Since it is pure, a_C = 1. Would its presence in excess ever tend to push the reaction in the reverse direction? Since the activity of solid carbon is always 1 it cannot influence the extent of this reaction. How can we express these observations quantitatively? Eqn. 17.15 becomes

To compute ΔG^o_T as a function of temperature, we apply the usual van’t Hoff procedure. This means that C_P,c can be treated just like C_P of the gaseous species.

17.15. Rate Perspectives in Reaction Equilibria

We have avoided discussing rate effects until now with the rationale that most coverage for reaction kinetics will occur in a course focused on reactor design. Nevertheless, there is overlap between the topics of reaction equilibria and reaction rates that can serve as a bridge between the two subjects. In all equilibrium phenomena, it is important to recognize that the balance achieved is dynamic, not static. For example, the molecules at the interface between a vapor and liquid are not stationary; they are perpetually exchanging between the vapor and liquid. Application of thermodynamics helps us understand the conditions where the balance occurs. Similarly, under conditions of chemical reaction equilibrium, the species are continuously interconverting with equal rates for the forward and reverse reactions.

From a thermodynamic perspective, the true driving force for chemical reaction is the activity. When the activities are balanced as given by Eqn. 17.15 the reaction reaches equilibrium and the forward are reverse rates are equal. The activities are directly proportional to the concentrations for liquids and vapors. Thus, it is common to use concentrations instead of activities for simple kinetic models. Consider the vapor-phase reaction,

For example, if two components, A and B, react to form C and D, then the rate of accomplishing the reaction must depend on the probability of the two components colliding with each other. This probability decreases as the concentration of one of the components diminishes. By convention the rates are typically written for the stoichiometrically limiting component. Also, they are typically written for the rate of formation per volume of reacting mixture.⁴ If A is the limiting reactant, the rate of the formation of A due to reaction of A with B would then be,

where the minus sign acknowledges that A is disappearing rather than forming, and the subscript f indicates reaction in the forward direction and k_f is known as the forward rate constant. When the exponents on the rate equation match the stoichiometric coefficients, the reaction is called an elementary reaction. When a reaction is equilibrium-limited, it is considered kinetically reversible. Recognizing that A is formed by reaction of C and D, the reverse reaction rate for formation of A is

The net rate of formation of A must be zero at reaction equilibrium,

Recognizing that concentration of a gas phase component is related to partial pressure, [A] = y_AP/RT, and similarly for other components. Rearranging Eqn. 17.63 and inserting the partial pressure results in

where in this case, Σv_i = –1, but the general expression is written to help readers remember the general relation for gas phase reactions. Note the manner in which the forward and reverse reaction rate constants are related to the equilibrium constant. This means that if the forward rate constant is measured in an experiment when the product concentrations are low, then the reverse rate constant can be determined from the equilibrium constant. Note that similar relations can be written for liquid-phase elementary reactions.

Certainly many reactions have rate expression more complicated than the elementary reactions discussed here. For example, enzyme catalyzed reactions often involve a binding step that is not represented by the simple statistical concept of the elementary reaction. Many reactions involve intermediate species that must be included in the mechanism and kinetic rate law. Understanding more complex rate laws is an important skill covered in reaction engineering courses. Our intention here was to show the relation for elementary reactions and to communicate the concept of forward and reverse rates approaching each other at reaction equilibrium. Note that this means that if the reaction approaches equilibrium in the forward direction, the overall rate of disappearance of A will slow and become zero. At slow rates, the reactor volume must be large to achieve meaningful change in reaction coordinate, so it is rarely economical to run commercial reactors all the way to equilibrium. However, the calculation of the equilibrium condition is important for any reactor design in order to know the limiting conversion, and usually avoid the conditions! Often, the equilibrium constant is used to calculate the reverse rate constant from the forward rate constant as discussed above.

17.16. Entropy Generation via Reactions

When introducing entropy and reversibility in Section 4.11 on page 175, we made a general statement that spontaneous reactions generate entropy. Then, in Section 4.12 on page 177 we derived relations between availability and entropy generation. In that section, we treated a single nonreactive stream. For a reaction in a steady-state open system, Eqn. 4.54 becomes

B^out = H^out – T_oS^out and Bⁱⁿ = Hⁱⁿ – T_oSⁱⁿ involve H and S evaluated at the respective T^out and Tⁱⁿ. Enthalpies of mixed streams were first introduced in Chapter 3 and entropies for mixed streams in Chapter 4.⁵ Models for departures functions and excess properties in Chapters 11–15 can be added to improve the mixture property values. The concepts for proper choice of a reference state for properties is important as discussed in Chapter 3 in the section Energy Balances for Reactions on page 113. The flow terms in Eqn. 17.65 are not the same as the Gibbs energy, but the availability will decrease with a spontaneous reaction. Therefore, both sides of the equation will be negative, and shaft work will not be obtained (a normal situation in an industrial reactor), and then entropy is generated. Note that entropy generation can be decreased for a spontaneous reaction only if work is produced. For an electrochemical redox reaction (Chapter 18), the process can produce some electrical work, which is one reason that fuel cells are of much current interest. For a closed-system process, the analysis is similar to the open-system process. The relation is seen most readily if T_o = T (and additional work can be obtained using a heat engine between T and T_o), and at constant pressure, the closed system balance (Eqn. 4.57) becomes

We conclude that production of nonexpansion/contraction work equal to changes in the Gibbs energy is necessary to eliminate entropy generation. Note that it is possible to relate the total work from a reaction to Helmholtz energy using Eqn. 4.58.

17.17. Gibbs Minimization

A remarkably simple technique can be applied to solve for the equilibrium compositions of species. It is most effective when only a gas phase is present. This technique recognizes the simplicity of the fundamental problem of minimizing the Gibbs energy at equilibrium. By expressing the total Gibbs energy of the mixture in terms of its ideal solution components, we can simply request that the value of the Gibbs energy be minimized. The Gibbs energy of the mixture is calculated by Eqn. 10.42 and the needed chemical potential (partial molar Gibbs energy) is given by Eqn. 10.59:

where the last equality assumes all components are ideal gases. If we take the reference state as the elements in their natural form at the standard state, then, at the standard state pressure, . However, frequently the reactions are not at standard state pressure. The pressure effect on Gibbs energy is given by Eqn. 9.17. When the pressure effect is added,

Combining Eqns. 17.67 and 17.68 results in

To find equilibrium compositions, we just need to minimize Eqn. 17.69 by varying the mole numbers n_i of each component while simultaneously satisfying the atom balance. Note that the mole fractions in the equation will also change as the mole numbers are varied. We do not need to explicitly write out the reactions. This method assumes that equilibrium is reached by whatever system of reactions is necessary. Most process simulators provide Gibbs minimization as a process unit.

The previous example is somewhat contrived because the reaction was specified and the reaction coordinate was varied. The method is applicable without specifying reactions as long as the atom material balances are satisfied when selecting mole numbers.

17.18. Reaction Modeling with Limited Data

Frequently chemical engineers must model chemical reactions without information on the Gibbs energy of formation. Although it can be estimated, often a better method is to conduct experiments to find equilibrium conditions and calculate K_a directly from compositions. Experiments can be performed over a temperature range, resulting in an empirical representation of Eqn. 17.28. In fact, methods like this are used to experimentally determine Gibbs energies of formation. The data can be fitted to determine constants I and J. If the data are scattered, it is better to use the simplified van’t Hoff Eqn. 17.29 in the form lnK_a = a + b/T, where a and b are fitted to the experimental data, and b is an indirect measurement of –ΔH_R^o/R. Thus, equilibrium data for new compounds or for liquid or solid reactions (discussed next) are often presented in the literature in the form of an equation for K_a instead of providing Gibbs energies of formation.

17.19. Simultaneous Reaction and VLE

It is not difficult to imagine situations in which reactions take place in the presence of multiple phases. Absorption of CO₂ into NaOH solution involves a reaction of the CO₂ as it dissolves to form sodium bicarbonate and sodium carbonate. Hydrogen bonding in “pure” fluids implies reaction and phase equilibrium at saturation conditions. The production of methyl t-butyl ether (MTBE) as an oxygenated fuel additive is an interesting process in which catalyst is placed on the trays of a distillation column. As catalysts are developed which are active at progressively lower temperatures, multiphase reactions should become even more common. In low-temperature methanol synthesis (~240°C), it can be advantageous to add a liquid phase to absorb the heat of reaction, as described in the example below. Biological pathways are also in development for many products, which frequently involve multiple phases.

The thermodynamic analysis of this seemingly complex kind of process is actually very similar to the analysis of multireaction equilibria. The extent of formation of a second phase is analogous to a reaction coordinate. The easiest way to illustrate the formulation of the problem is to consider an example. Suppose the methanol synthesis reaction was carried out at 75 bars and 240°C such that the gas phase mole fractions were 0.25 CO, 0.25 MeOH, and 0.50 H₂. Based on a stoichiometric feed composition, the conversion would then be 50%. Now suppose this gas phase was placed in contact with a liquid phase with a nonvolatile solvent. The K-ratios at these conditions are about 10 for CO and H₂, and about 1.0 for MeOH. What would be the composition in the liquid phase and the extent of conversion if only liquid was removed? The composition of the liquid would be 0.025 CO and 0.25 MeOH. Therefore, the extent of conversion would be 0.25/0.275 = 91%. Thus, the addition of a liquid phase greatly enhances conversion of this process. The example below elaborates on these findings in a much more formal manner.

a. Note: The symbols a and b are simply regression coefficients, not the equation of state parameters a and b.

a. S. Lee, Personal Communication, 1993.

b. Weast, R.C. 1979. CRC Handbook of Chemistry and Physics, 60th ed, Boca Raton, FL: CRC Press, p. D-224.

c. Perry, R.H., Chilton, C.H. 1986. The Chemical Engineers’ Handbook, 6th ed. New York, NY: McGraw-Hill, p. 3–70.

d. Reid, R.C., Prausnitz, J.M., Poling, B. 1987. The Properties of Gases & Liquids, 4th ed, New York, NY: McGraw-Hill.

17.20. Summary

We have greatly enlarged the scope of our coverage of engineering thermodynamics with very little extension of the conceptual machinery. All we really did conceptually was recall that the Gibbs energy should be minimized. The provision that atoms can be moved from one chemical species to another with commensurate changes in energy and entropy simply means that reference states must be assigned to elemental standard states instead of standard states based on pure components, such that the free energies of all the components can be compared. In this sense we begin to comprehend in a new light the broad range of applications mentioned in Einstein’s quote at the end of Chapter 1. Instead of conceptual challenges, reaction equilibria focus primarily on the computational aspects of setting up and solving the problems. Notably, equation solvers can provide multidimensional capability. These are tools that can be adapted to many problems, even those beyond the scope of thermodynamics. You should familiarize yourself with such tools and build the expertise that will permit you to enhance your productivity.

Important Equations

The shortcut van’t Hoff equation provides a rapid method for screening the effect of temperature on the equilibrium constant. For best results, the temperature range should be limited as suggested in Section 17.8.

When combined with the equilibrium constraint Eqn. 17.15, reaction conversions can be estimated for many common scenarios.

The most common scenario is for ideal gases, because most industrial reactions are conducted at high temperature (to accelerate rates) and low pressure (to minimize cost).

For liquid phase components, we showed that the activity should be written as a_i = x_iγ_i. For solid species, we showed that the activity is unity when the solid is pure. The relevant Gibbs energy of formation must be used when liquids or solids appear in the equilibrium constant expression.

For simultaneous reaction and phase equilibria, we showed that the reaction could be treated in any one of the phases. Since the phase equilibrium constraint asserts equality of fugacity between phases, conceiving of the reaction in any particular phase is inconsequential, providing that the activities and Gibbs energies of formation use the same standard state.

Test Yourself

1. Reproduce the values given for the Gibbs energy of formation at 900 K for the component in Example 17.3. What values are given by the shortcut van’t Hoff equation?

2. Explain why a plot of lnK versus 1/T exhibits slight curvature.

3. When T is raised, the equilibrium constant is found to increase. Is the reaction endothermic or exothermic?

4. Explain why formation of a pure solid product does not inhibit a gas-solid reaction from going forward.

5. A reaction occurs with all components present in liquid and gas phases. Compositions in both phases are measured and the activities are calculated. The equilibrium constant value calculated using vapor phase activities is different from the equilibrium constant calculated using liquid phase activities. Is something wrong?

6. Describe the behavior of the reaction rate as equilibrium is approached. Explain why most industrial reactors are not run near equilibrium. Explain why calculation of an equilibrium constant is a good idea whenever designing a reactor.

17.21. Practice Problems

P17.1. An equimolar mixture of H₂ and CO is obtained by the reaction of steam with coal. The product mixture is known as “water-gas.” To enhance the H₂ content, steam is mixed with water-gas and passed over a catalyst at 550°C and 1 bar so as to convert CO to CO₂ by the reaction:

Any unreacted H₂O is subsequently condensed and the CO₂ is subsequently absorbed to give a final product that is mostly H₂. This operation is called the water-gas shift reaction. Compute the equilibrium compositions at 550°C based on an equimolar feed of H₂, CO, and H₂O.

Data for 550°C:

P17.2. One method for the production of hydrogen cyanide is by the gas-phase nitrogenation of acetylene according to the reaction: N₂ + C₂H₂ 2HCN. The feed to a reactor in which the above reaction takes place contains gaseous N₂ and C₂H₂ in their stoichiometric proportions. The reaction temperature is controlled at 300°C. Estimate the product composition if the reactor pressure is: (a) 1 bar; (b) 200 bar. At 300°C, .

P17.3. Butadiene can be prepared by the gas-phase catalytic dehydrogenation of 1-butene: C₄H₈ C₄H₆ + H₂. In order to suppress side reactions, the butene is diluted with steam before it passes into the reactor.

a. Estimate the temperature at which the reactor must be operated in order to convert 30% of the 1-butene to 1,3-butadiene at a reactor pressure of 2 bar from a feed consisting of 12 mol of steam per mole of 1-butene.

b. If the initial mixture consists of 50 mol% steam and 50 mol% 1-butene, how will the required temperature be affected?

P17.4. Ethylene oxide is an important organic intermediate in the chemical industry. The standard Gibbs energy change at 298 K for the reaction mole. This large negative value of ΔG^o_T indicates that equilibrium is far to the right at 298 K. However, the direct oxidation of ethylene must be promoted by a catalyst selective to this reaction to prevent the complete combustion of ethylene to carbon dioxide and water. Even with such a catalyst, it is thought that the reaction will have to be carried out at a temperature of about 550 K in order to obtain a reasonable reaction rate. Since the reaction is exothermic, an increase in temperature will have an adverse effect on the equilibrium. Is the reaction feasible (from an equilibrium standpoint) at 550 K, assuming that a suitable catalyst selective for this reaction is available? For ethylene oxide, ΔH^f₂₉₈ = -52.63 kJ/mol. Heat capacity equations (in J/mole-K) for the temperature range involved may be approximated by C_P,C2H4O = 6.57 + 0.1389 T(K); C_P,C2H4 = 15.40 + 0.0937 T(K); C_P,O2 = 26.65 + 0.0084 T(K)

P17.5. The water-gas shift reaction is to be carried out at a specified temperature and pressure employing a feed containing only carbon monoxide and steam. Show that the maximum equilibrium mole fraction of hydrogen in the product stream results when the feed contains CO and H₂O in their stoichiometric proportions. Assume ideal gas behavior.

P17.6. Assuming ideal gas behavior, estimate the equilibrium composition at 400 K and 1 bar of a reactive gaseous mixture containing the three isomers of pentane. Standard formation data at 400 K are

P17.7. One method for the manufacture of synthesis gas depends on the vapor-phase catalytic reaction of methane with steam according to the equation . The only other reaction which ordinarily occurs to an appreciable extent is the water-gas shift reaction. Gibbs energies and enthalpies for the problem are tabulated below in kJ/mol.

Compute the equilibrium compositions based on a 1:1 feed ratio at 600 K and 1300 K and 1 bar and 100 bars.

P17.8. Is there any danger that solid carbon will form at 550°C and 1 bar by the reaction 2CO = C_s + CO₂? (ANS. Yes)

P17.9. Calculate the equilibrium percent conversion of ethylene oxide to ethylene glycol at 298 K and 1 bar if the initial molar ratio of ethylene oxide to water is 3.0.

C₂H₄O_(g) + H₂O_(l) = (CH₂OH)₂ (1 M aq solution)

To simplify the calculations, assume that the gas phase is an ideal gas mixture, that γ_w = 1.0, and that the shortcut K value is applicable for ethylene oxide and ethylene glycol.

P17.10. Acetic acid vapor dimerizes according to 2A₁ A₂. Assume that no higher-order associations occur. Supposing that a value for K_a is available, and that the monomers and dimers behave as an ideal gas, derive an expression for y_A1 in terms of P and K_a. Then develop an expression for PV/n_oRT in terms of y_A1, where n_o is the superficial number of moles neglecting dimerization. Hint: Write n_o/n_T in terms of y_A1 where n_T = n₁ + n₂.

17.22. Homework Problems

17.1. For their homework assignment three students, Julie, John, and Jacob, were working on the formation of ammonia. The feed is a stoichiometric ratio of nitrogen and hydrogen at a particular T and P.

Julie, who thought in round numbers of product, wrote:

John, who thought in round numbers of nitrogen, wrote:

N₂ + 3H₂ 2NH₃

Jacob, who thought in round numbers of hydrogen, wrote:

a. How will John’s and Jacob’s standard state Gibbs energy of reactions compare to Julie’s?

b. How will John’s and Jacob’s equilibrium constants be related to Julie’s?

c. How will John’s and Jacob’s equilibrium compositions be related to Julie’s?

d. How will John’s and Jacob’s reaction coordinate values be related to Julie’s?

17.2. The simple statement of the Le Châtelier principle leads one to expect that if the concentration of a reactant were increased, the reaction would proceed so as to consume the added reactant. Nevertheless, consider the gas-phase reaction, N₂ + 3H₂ 2NH₃ equilibrated with excess N₂ such that N₂′s equilibrium mole fraction is 0.55. Does adding more N₂ to the equilibrated mixture result in more NH₃? Why?

17.3. The production of NO by the direct oxidation of nitrogen occurs naturally in internal combustion engines. This reaction is also used to produce nitric oxide commercially in electric arcs in the Berkeland-Eyde process. If air is used as the feed, compute the equilibrium conversion of oxygen at 1 bar total pressure over the temperature range of 1300–1500°C. Air contains 21 mol% oxygen and 79% N₂.

17.4. The following reaction reaches equilibrium at the specified conditions.

C₆H₅CH = CH_2(g) +H_2(g) C₆H₅C₂H_5(g)

The system initially contains 3 mol H₂ for each mole of styrene. Assume ideal gases. For styrene, , .

a. What is K_a at 600°C?

b. What are the equilibrium mole fractions at 600°C and 1 bar?

c. What are the equilibrium mole fractions at 600°C and 2 bar?

17.5. For the cracking reaction,

C₃H_8(g) C₂H_4(g) + CH_4(g)

the equilibrium conversion is negligible at room temperature but becomes appreciable at temperatures above 500 K. For a pressure of 1 bar, neglecting any side reactions, determine:

a. The temperature where the conversion is 75%. [Hint: conversion = amount reacted/amount fed. Relate ξ to the conversion.]

b. The fractional conversion which would be obtained at 600 K if the feed to a reactor is 50 mol% propane and 50 mol% nitrogen (inert). (Consider the reaction to proceed to equilibrium.)

17.6. Ethanol can be manufactured by the vapor phase hydration of ethylene according to the reaction: C₂H₄ + H₂O C₂H₅OH. The feed to a reactor in which the above reaction takes place is a gas mixture containing 25 mol% ethylene and 75 mol% steam.

a. What is the value of the equilibrium constant, K_a, at 125°C and 1 bar?

a. Provide an expression to relate K_a to ξ. Solve for ξ.

17.7. Ethylene is a valuable feedstock for many chemical processes. In future years, when petroleum is not as readily available, ethylene may be produced by dehydration of ethanol. Ethanol may be readily obtained by fermentation of biomass.

a. What percentage of a pure ethanol feed stream will react at 150°C and 1 bar if equilibrium conversion is achieved?

b. If the feed stream is 50 mol% ethanol and 50 mol% N₂, what is the maximum conversion of ethanol at 150°C and 1 bar?

17.8. The catalyzed methanol synthesis reaction, CO_(g) + 2H_2(g) CH₃OH_(g), is to be conducted by introducing equimolar feed at 200°C. What are the mole fractions and the temperature at the outlet if the system is adiabatic at 10 bar and the catalyst provides equilibrium conversion without any competing reactions?

17.9. A gas stream composed of 15 mol% SO₂, 20 mol% O₂, and 65 mol% N₂ enters a catalytic reactor operating and forms SO₃ at 480°C and 2 bar.

a. Determine the equilibrium conversion of SO₂.

b. Determine the heat transfer required per mole of reactor feed entering at 295 K and 2 bar.

17.10. The feed gas to a methanol synthesis reactor is composed of 75 mol% H₂, 12 mol% CO, 8 mol% CO₂, and 5 mol% N₂. The system comes to equilibrium at 550 K and 100 bar with respect to the following reactions:

2H_2(g) + CO_(g) CH₃OH_g)

H_2(g) + Co_2(g) CO_(g) + H₂O_(g)

Assuming ideal gases, derive the equations that would be solved simultaneously for ξ₁, ξ₂ where 1 refers to the first reaction listed. Provide numerical values for the equilibrium constants. Determine ξ₁ and ξ₂ ignoring any other reactions.

17.11. The 10/25/93 issue of Chemical and Engineering News suggests that the thermodynamic equilibrium in the isomerization of n-butene (CH3CH=CHCH3, a mix of cis and trans isomers) is reached at a temperature of 350°C using a zeolite catalyst. The products are isobutene and 1-butene (CH₂=CHCH₂CH₃). The isobutene is the desired product, for further reaction to MTBE. Determine the equilibrium composition of this product stream at 1 bar.

17.12. Acrylic acid is produced from propylene by the following gas phase reaction:

A significant side reaction is the formation of acetic acid:

The reactions are carried out at 310°C and 4 bar pressure using a catalyst and air as an oxidant. Steam is added in the ratio 8:1 steam to propylene to stabilize the heat of reaction. If 50% excess air is used (sufficient air so that 50% more oxygen is present than is needed for all the propylene to react by the first reaction), calculate the equilibrium composition of the reactor effluent.

17.13.

a. As part of a methanol synthesis process similar to problem 17.10, one side reaction that can have an especially unfavorable impact on the catalyst is coke formation. As a first approximation of whether coke (carbon) formation would be significant, estimate the equilibrium extent of coke formation based solely on the reaction: CO + H₂ C_(s) + H₂O. Conditions for the reaction are 600 K and 100 bar.

b. Is coke formation by the reaction from part (a) expected at the conditions cited in problem 17.10?

17.14. Hydrogen gas can be produced by the following reactions between propane and steam in the presence of a nickel catalyst:

C₃H₈ + 3H₂O 3CO + 7H₂

CO + H₂O CO₂ + H₂

Neglecting any other competing reactions:

a. Compute the equilibrium constants at 700 K and 750 K.

b. What is the equilibrium composition of the product gas if the inlet to the catalytic reactor is propane and steam in a 1:5 molar ratio at each of the temperatures and 1 bar?

17.15. Write and balance the chemical reaction of carbon monoxide forming solid carbon and carbon dioxide vapor. Determine the equilibrium constant at 700 K and 750 K. Will solid carbon form at the conditions of problem 17.14?

17.16. Catalytic converters on automobiles are designed to minimize the NO and CO emissions derived from the engine exhaust. They generally operate between 400°C and 600°C at 1 bar of pressure. K.C. Taylor (1993. Cat. Rev. Sci. Eng. 35:457.) gives the following compositions (in ppm, molar basis) for typical exhaust from the engine:

The additional products of the effluent stream include NO₂, N₂O, N₂O₄, and NH₃. Estimate the compositions of all species at each temperature {400°C, 500°C, 600°C} and plot the ratio of NH₃/CO as a function of temperature. (Note: Use the options of the Solver software to set the precision of the results as high as possible.)

17.17. Styrene can be hydrogenated to ethyl benzene at moderate conditions in both the liquid and the gas phases. Calculate the equilibrium compositions in the vapor and liquid phases of hydrogen, styrene, and ethyl benzene at each of the following conditions:

a. 3 bar pressure and 298 K, with a starting mole ratio of hydrogen to styrene of 2:1

b. 3 bar pressure and 423 K, with a starting mole ratio of hydrogen to styrene of 2:1

c. 3 bar pressure and 600 K, with a starting mole ratio of hydrogen to styrene of 2:1

17.18. Habenicht et al. (1995. Ind. Eng. Chem. Res., 34:3784) report on the reaction of t-butyl alcohol (TBA) and ethanol (EtOH) to form ethyltertiary-butyl ether (ETBE). The reaction is conducted at 170^oC. A typical feed stream composition (in mole fraction) is:

Isobutene is the only significant by-product. Assuming that equilibrium is reached in the outlet stream, estimate the minimum pressure at which the reaction must be conducted in order to maintain everything in the liquid phase. Do isobutene or ETBE exceed their liquid solubility limits at the outlet conditions?

17.19. Limestone (CaCO₃) decomposes upon heating to yield quicklime (CaO) and carbon dioxide. At what temperature does limestone exert a decomposition pressure of 1 bar?

17.20. Two-tenths of a gram of CaCO_3(s) is placed in a 100 cm³ pressure vessel. The vessel is evacuated of all vapor at 298 K and sealed. The reaction CaCO_3(s) = CaO_(s) + CO_2(g) occurs as the temperature is raised. At what temperature will the conversion of CaCO₃ be 50%, and what will the pressure be?

17.21. One suggestion for sequestering CO₂ is to synthesize carbonate polymers. Polycarbonate is well known for its strength and transparency. To gauge the feasibility of this approach, consider the synthesis of dimethyl carbonate (DMC) from methanol and CO₂ at 350 K.

a. Write a balanced stoichiometric equation for this reaction. Highlight any by-products.

b. Estimate the K_a value for the reaction. What pressure is required to achieve P·K_a = 0.1?

c. Methanol and DMC are both liquids at this temperature. Explain how to estimate their partial pressures at a given extent of conversion. You may neglect the CO₂ in the liquid for this question.

17.22. Ethyl acetate is to be produced by a liquid phase reaction.

a. Use the shortcut van’t Hoff equation to calculate the expected conversion of HOAc for equimolar feeds of EtOH and HOAc in a batch reactor at 80°C.

b. Repeat part (a) with a 3:1 ratio of EtOH to HOAc at 80°C.

17.23. Hamilton, et al.,⁶ have studied the binding of DNA chromosomes to proteins WT1 and EGR1. WT1 uses a zinc binding site to suppress a certain type of kidney tumor. EGR1 binds to regulate cell proliferation. There may be an important regulatory link between the two proteins.⁷

a. Use the binding data above to determine the standard state Gibbs energies of reaction at each temperature, the heat of reaction over the temperature range, and the entropy of reaction over the temperature range.

b. The binding constants are about the same size. Evaluate the relative magnitudes of the enthalpy and entropy of formation for each binding type. Does the thermodynamic analysis imply that the binding types are similar or different?

17.24. The enthalpy of reaction for many biological reactions and surfactants is strongly temperature-dependent, but instead of using full heat capacities, the temperature dependence can be characterized by differences in heat capacities ΔC_P = C_P,prod – C_P,react, typically assumed to be independent of temperature.⁷

a. The heat of reaction is represented by ΔH = ΔH_R + ΔC_P(T – T_R). Show that

b. Provide an equation consistent with part (a) for the temperature dependence of the entropy of reaction.

17.25. Lysozyme (MW = 14.313 kDa) undergoes a phase transition from a native folded (N) to unfolded state (U) that can be considered a reversible reaction, N_(aq) U_(aq), where the subscript (aq) indicates that the protein is in an aqueous solution. At high temperature the protein is in state U and at low temperature, it is in state N. The melting temperature (T_m) is where the concentrations are equal. The melting temperature can be changed by pH, ionic strength (added salt), denaturant (solvents), or pressure. The results of problem 17.24 apply. The protein unfolding has been studied by Cooper, et al.⁸ The heat to unfold 1 mg/mL of protein in 2 mL of pH = 4 solution at T_m = 78°C is 0.0755 J, ΔC_P = 6.3 kJ/mol-K. The ΔV = –40 mL/mol.⁷

a. What is the Gibbs energy change for the unfolding at the melting temperature? (Hint: The answer requires conceptual thought, not many calculations.)

b. Show that at the melting temperature at pH = 4, ΔH = 540 kJ/mol and ΔS = 1.54 kJ/mol.

c. Evaluate the contributions to ΔG at 23°C and pH = 4. Is the process driven by enthalpy or entropy under these conditions?

d. Show that the relation for [N]/[U] is For a solution of overall concentration 1 mg/mL, plot [U] as a function of temperature for 20 ≤ T ≤ 110 °C, and provide a tabular summary of [U] at 60°C and 90°C.

e. When pH is varied, the net charges on the protein change (as will be explained in Chapter 18), resulting in different T_m and ΔH. In the range 20 ≤ T_m ≤ 78 °C, the relation is ΔH(kJ/mol) = 48.6 + 6.3 T_m(°C). Derive the relation between ΔS and T_m.

f. At pH = 1, T_m = 43°C. Provide a plot of plot [U] as a function of temperature for 20 ≤ T ≤ 110 °C, and provide a tabular summary of [U] at 25°C and 55°C.

g. (advanced) Lysozyme has four disulfide bonds that are important in the folding behavior. The authors created a mutant protein with only three disulfide bonds. The relation in (e) still holds. At pH = 2.5, T_m = 23°C for the mutant. Evaluate the contributions of enthalpy and entropy to the folding.

h. (advanced) Provide an interpretation for the sign of ΔV. At pH = 4, how much pressure is needed to lower the melting temperature 5°C to 73°C?⁹

17.26. Surfactants clump together to form organized structures called micelles that can be spheres, rods, and so forth. The formation of the clump can be modeled as a “chemical” reaction, though there are no chemical bonds formed or broken. When surfactants are in solution below the critical micelle concentration, CMC, the surfactant molecules are almost all “free” in solution. At the CMC, micelles start to form. Above the CMC, the amount of free surfactant is almost constant in solution and as more surfactant is added to the solution, more micelles form. It is conventional to provide the property changes of surfactants per mol of surfactant molecules, not per mole of micelle.

a. Given the data below for nonylglucoside (NG) demicellization in water, calculate the micellization value, ΔC_Pmic, as a function of temperature. Is the heat capacity of a micelle greater than or less than the heat capacity of the molecules that make up the micelle?

b. Calculate ΔS_mic and for the surfactant as a function of temperature. Is the overall solution (including water) more ordered or less ordered after micellization?

NG data:

17.27. Micelle formation in surfactants is described in problem 17.26. Solve the problem using the data for sodium docecyl sulfate, SDS in water.

SDS data:

17.28. For nonylglucoside, NG, thermodynamic data for demicellization in water are presented in problem 17.26. Model the micelle reaction as nS M_n where S is free surfactant and M_n is a micelle. Treat the solution as an ideal solution. Vary the total concentration of NG from 0 up to 20 mmol/L. Water is 55.5 mol/L. Calculate mole fractions and molar concentrations (mmol/L) of the free surfactant (in mM) and the micelles (in μM) of NG at T=285 K, where n = 58. Plot curves for the concentrations. Identify the approximate CMC in mmol/L. Hint to simplify calculations: The mol/L added is very small relative water molarity (55.5 mol/L). The density can be assumed to be constant.