*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.

**1.** Solve for the equilibrium reaction coordinate values and the equilibrium mole fractions for a given *K** _{aT}* and

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

**3.** Calculate Δ*G*^{o}_{298.15} and Δ*H*^{o}_{298.15} for a given reaction.

**4.** Calculate Δ*G*^{o}* _{T}* and

**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.

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

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*_{i}*P* is the partial pressure (always expressed in bar) of the *i** ^{th}* component, and

**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).

Example 17.1. Computing the reaction coordinate

CO and H_{2} are fed to a reactor in a ratio of 2:1 at 500 K and 20 bar, where the equilibrium constant is *K _{a}* = 0.00581. (We will illustrate how to calculate

Compute the equilibrium conversion of CO.

Solution

In the expression for *K _{a}* we insert each

To relate the composition to the mass balance, we select a basis and use the reaction coordinate. **Basis:** 2 mole CO fed. Note the excess CO at the feed conditions. The **reaction coordinate** and method of selecting a basis have already been introduced in Section 3.6. The stoichiometry table becomes

Note that all *n* values must stay positive, constraining the range for a physically acceptable solution to be 0 ≤ ζ ≤ 0.5. The mole fractions can be written in terms of ζ using the stoichiometry table.

Substituting the mole fractions into the equilibrium constant expression,

A trial-and-error solution is much more robust by using the difference of Eqn. 17.5 rather than the ratio of Eqn. 17.4. We solve by trial and error and substitute to get ζ recalling 0 ≤ ζ ≤ 0.5. A summary of guesses:

At reaction equilibrium for the given feed conditions, equilibrium is represented by ζ = 0.21. Now Eqn. 17.3 may be used to find the *y*’s. The conversion of CO is 0.21/2·100% = 10.5%; conversion of H_{2} is 2(0.21)/1·100% = 42%. Note the conversion is species-dependent with nonstoichiometric feed. Conversion can be increased further by increasing the pressure further, or by changing *T* where *K _{a}* is larger, provided a catalyst is available and kinetics are adequate at that

This example demonstrates the method to use *K _{a}* to calculate the reaction coordinate. Readers should note that the value of

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.

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 d*ln *f.* Integrating from the standard state to the mixture state of interest (*cf*. generalizing Eqn. 10.48),

where *G _{i}*

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.

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

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 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}* =

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}*.

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)} + H_{2}O_{(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.

Example 17.2. Calculation of standard state Gibbs energy of reaction

Butadiene is prepared by the gas phase catalytic dehydrogenation of 1-butene:

Calculate the standard state Gibbs energy of reaction and the equilibrium constant at 298.15 K.

Solution

We find values tabulated for the standard state enthalpies of formation and standard state Gibbs energy of formation at 298.15 K.

The equilibrium constant is determined from Eqn. 17.16;

This reaction is not favorable at room temperature because the equilibrium constant is small.

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,

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.

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.

Henry Louis Le Châtelier (1850–1936) was a French chemist. He was elected to the French Académie des Sciences and the Royal Swedish Academy of Sciences in 1907.

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, Σ*v _{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,. Σ

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 Σ*v _{i}* > 0, adding an inert will increase conversion at a fixed total pressure. However, when Σ

Conversions of specific reactants are influenced using nonstoichiometric feed. In Example 17.1, excess CO was fed to the reactor; conversions were 42% (H_{2}), 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

Example 17.3. Butadiene production in the presence of inerts

Consider again the butadiene reaction of Example 17.2 on page 648. Butadiene is prepared by the gas phase catalytic dehydrogenation of 1-butene, at 900 K and 1 bar.

**a.** In order to suppress side reactions, the butene is diluted with steam before it passes into the reactor. Estimate the conversion of 1-butene for a feed consisting of 10 moles of steam per mole of 1-butene.

**b.** Find the conversion if the inerts were absent and side reactions are ignored.

**c.** Find the total pressure that would be required to obtain the same conversion as in (a) if no inerts were present.

In the earlier example, we determined the value at 298.15 K for . Now we need a value at 900 K. The next section explains how the value at 900 K may be obtained. For now, use the following data for at 900 K and 1 bar:

Solution

**a.** Basis of 1 mole 1-butene feed. Set up reaction coordinate, using *I* to indicate inerts,.

The physical range of the solution is 0 ≤ ξ ≤ 1. *P* = 1 bar ⇒ 1.242 ξ^{2} + 2.42 ξ – 2.662 = 0 ⇒ ξ = 0.784. For the basis of 1 mol 1-butene feed, the conversion is 78.4%.

**b.** *n _{I}* = 0 and the basis of feed is the same and 0 ≤ ξ ≤ 1. The total number of moles is

**c.** Rearranging the equilibrium expression for pressure, *P*^{–1} = ξ^{2} / [0.242 · (1 – ξ) · (1 + ξ)], 0 ≤ ξ ≤ 1.

Inserting a reaction coordinate of ξ = 0.784 gives *P* = 0.152 bar. So the reaction would need to run at a much lower pressure without the inerts to achieve the same conversion. In other words, inerts serve to dilute the fugacities of the products and suppress the reverse reaction since there are more moles of product than reactant.

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 Δ*G _{T}*. Note that this is a different term than the

A reaction with is called **exergonic** and results in *K _{a}* > 1, and a reaction with is called

The propensity for a reaction to go forward or backward under actual conditions is determined by Δ*G _{T}*, not .

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 (

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,

The constant *I* may be evaluated from a knowledge of Δ*G ^{o}*

Example 17.4. Equilibrium constant as a function of temperature

The heat capacities of ethanol, ethylene, and water can be expressed as *C _{P}* =

Solution

Taking 298.15 K as the reference temperature,

The variable *J* may be found with Eqn. 3.46 on page 113 at 298.15 K.

Δ*H*_{298.15}^{o} = –45,625 = *J* + Δ*aT* + (Δ*b*/2)·*T*^{2} + (Δ*c*/3)·*T*^{3} + (Δ*d*/4)*T*^{4} = *J* + (9.014 – 3.806 – 32.24) *T* + [(0.2141 – 0.1566 – 0.0019)/2] *T*^{2} + [(–8.39 + 8.348 – 1.055)(1E-5)/3] *T*^{3} + [(1.373 –17.55 + 3.596)(1E-9)/4] *T*^{4}

= *J* – 27.032 *T* + 0.02779 *T*^{2} – (3.657E-6)*T*^{3} – (3.145E-9)*T*^{4}

Plugging in *T* = 298.15 K, and solving for *J, J* = –39.914 kJ/mole. Using this result in Eqn. 17.28 at 298.15 K will yield the variable *I*.

Δ*G _{T}*

Plugging in Δ*G _{R}*

The resultant formula to calculate at any temperature is

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 Δ*H _{T}^{o}* is independent of temperature. That is, suppose Δ

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 (Δ*H _{T}* < 0) lead to

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 Δ*G*_{Tnear}^{o} and Δ*H*_{Tnear}^{o}. Then the values at *T*_{near} can be used as the reference values in Eqn. 17.29.

Example 17.5. Application of the shortcut van’t Hoff equation

Apply the shortcut approximation to the vapor phase hydration of ethylene. This reaction has been studied in the previous example, and the Gibbs energy of reaction and heat of reaction can be obtained from that example.

Solution

The results are very similar to the answer obtained by the general van’t Hoff equation in Example 17.4.

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.

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 CO_{2} and provide O_{2}. 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 CO_{2} to fuel but requires H_{2}. 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. H_{2} 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.

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.

Example 17.6. Simultaneous reactions that can be solved by hand

We can occasionally come across multiple reactions which can be solved without a computer. These are generally limited to textbook problems, but provide a starting point and test case for applying the general approach. Consider the two series/parallel gas phase reactions:

The reactions are considered series reactions because *C* is a product of the first reaction, but a reactant in the second. They are parallel because *A* is a reactant in both reactions. The pressure in the reactor is 10 bar, and the feed consists of 2 moles of *A* and 1 mole of *B*. Calculate the composition of the reaction mixture if equilibrium is reached with respect to both reactions.

Solution

The material balance gives:

Note that for a physical solution, 0 ≤ ξ_{1} ≤ 1, 0 ≤ ξ_{2} ≤ ξ_{1} to ensure that all mole numbers are positive. This reaction network is independent of *P* because Σ*v _{i}* = 0. The equilibrium constants are

Solving the first equation for ξ_{1} using the quadratic equation,

Similarly, for the second reaction,

We may now solve by trial and error. The procedure is: 1) guess ξ_{1}; 2) solve Eqn. 17.40 for ξ_{2}; 3) solve Eqn. 17.39 for ; 4) if , go to step 1. The iterations are summarized below.

Further iteration results in no further significant change.

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.

Example 17.7. Solving multireaction equilibria with Excel

Methanol has a lower vapor pressure than gasoline. That can make it difficult to start a car fueled by pure methanol. One potential solution is to convert some of the methanol to methyl ether *in situ* during the start-up phase of the process (i.e., automobile). At a given temperature, 1 mole of MeOH is fed to a reactor at atmospheric pressure. It is assumed that only the two reactions given below take place. Compute the extents of the two simultaneous reactions over a range of temperatures from 200°C to 300°C. Also include the equilibrium mole fractions of the various species.

Solution

A worksheet used for this solution is available in the workbook Rxns.xlsx.

Data for reaction (1) have been tabulated by Reactions Ltd.^{a}—at 473.15 K, Δ*H _{T}* = 96,865 J/mol and ln

Data for reaction (2) can be obtained from Appendix E for MeOH and water. For DME, the values are from Reid et al. (1987).^{b}

The shortcut van’t Hoff equation for this reaction gives:

Writing equations for reaction coordinates for reaction 1:

and for reaction 2:

These two equations are solved simultaneously for ξ_{1} and ξ_{2}. We have rearranged the objective functions to eliminate the ratios of ξ functions and use differences instead because the Excel Solver is much more robust with this mathematical form. The solution is implemented in the worksheet DUALRXN in Rxns.xlsx or Matlab Ex17_07.m. In the example here (see Fig. 17.3), the Δ*C _{P}* for both reactions is neglected. The equations derived above are entered directly into the cells, and the Solver tool is called.

**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.

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 CO

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 CO

Example 17.8. Chemical coupling to induce conversion

Example 17.3(a) considered use of steam as a diluent where the conversion was found to be 78% using 10 moles of steam as diluent and only 44% without the diluent. Consider the conversion by inducing higher conversion by replacing the 10 mole steam with 10 mole CO_{2} which adds the water-gas shift reaction. For the water-gas shift written as , *K _{a}*

Solution

The butadiene reaction has been written in Example 17.3(a) and ξ_{1} will be used for that 1-butene reaction and ξ_{2} will be used for the water-gas shift reaction. The stoichiometry table is,

Physical limits for the reaction coordinates are 0 ≤ ξ_{1} ≤ 1 and 0 ≤ ξ_{2} ≤ ξ_{1}. Solving Eqns. 17.41 and 17.42 simultaneously, we find ξ_{1} = 0.949 and ξ_{2} = 0.792. Reviewing previous examples, the conversion at 1 bar was only 44% without an inert, increased to 78% with an inert, and increased to 95% using CO_{2} to induce conversion by reaction coupling. Note that even though the water-gas shift equilibrium constant is not very large, it makes a significant difference in the conversion of 1-butene. Whether this is implemented depends on the feasibility of economically separating the products.

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 _{a}*s):

The large equilibrium constant of 17.44 forms CH_{3}Cl_{(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.

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.

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 *T ^{in}*.

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

**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.

Example 17.9. Adiabatic reaction in an ammonia reactor

Estimate the outlet temperature and equilibrium mole fraction of ammonia synthesized from a stoichiometric ratio of N_{2} and H_{2} fed at 400 K and reacted at 100 bar. How would these change if the pressure was 200 bar?

Solution

For a rough estimate we will use the shortcut approximation of temperature effects. Furthermore, we will assume *K*_{ϕ} ≈ 1. (Is this a good approximation or not?^{a}) Therefore we obtain,

**Basis:** Stoichiometric ratio in feed.

For the purposes of the example, the shortcut van’t Hoff equation will be used to iterate on the adiabatic reactor temperature. However, the full van’t Hoff method will be used to obtain and at an estimated nearby temperature *T*_{near} = 600K as suggested in Section 17.8. Then the shortcut van’t Hoff equation will be used over a limited temperature range for less error. The energy balance will also use ; we will create an energy balance path through *T*_{near} = 600K rather than 298.15K. We will compare the approximate answer with the full van’t Hoff method at the end of the example.

For ammonia, , . Since the reactants are in the pure state, the respective reactant formation values are zero, and therefore the formation values for ammonia represent the standard state values for the reaction. Inserting the formation values along with the heat capacities into the detailed van’t Hoff equation—one of the *K _{a}* calculators highlighted in the margin note to Example 17.4 on page 653 is used—at an assumed temperature of 600 K, the values obtained are and

From an assumed value of *T,* this equation will provide the equilibrium constant. Some manipulation is necessary to obtain the material balance from *K _{a,T}*. Plugging the mole fraction expressions into Eqn. 17.17, and collecting the fractions 1/2 and 3/2,

defining

Applying the quadratic formula,

The strategy will be to guess *T*, and calculate *K _{a,T}*,

Heat capacity integrals and the energy balance have been entered in the workbook Rxns.xlsx. At the initial guess of 600 K, the *F*(*T*) of Eqn. 17.49 is 19.4 kJ. A converged result is found at 699 K shown in Fig. 17.4 and the = 0.33, conversion of feed is 33%. At 200 bar, the answer is 739 K, and conversion is 38%.

The detailed van’t Hoff is available in the same workbook and results in 698 K and 33% conversion at 100 bar, and 737 K and 37% conversion at 200 bar.

**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.

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 *T ^{in}*.

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

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.

Example 17.10. Oligomerization of lactic acid

Lactic acid is a bio-derived chemical intermediate produced in dilute solution by fermentation. Lactic acid is an α-hydroxy carboxylic acid. As an aqueous solution of lactic acid is concentrated by boiling off water, the carboxylic acid on one molecule reacts with a hydroxyl on another forming a dimer and releasing water. Denoting the “monomer” as L_{1} and a dimer as L_{2},

The dimer has a hydroxyl and carboxylic acid that can react further to form trimer L_{3},

As more water is removed, the chain length grows, forming oligomers. The oligomerization can be represented by a recurring reaction for chain formation. Each liquid phase reaction that adds a lactic acid molecule can be modeled with a universal temperature-independent value of *K _{a}* = 0.2023 and the solutions may be considered ideal.

**a.** Determine the mole fractions and wt% of species in a 50 wt% lactic acid solution in water where the distribution is approximated by only reaction 17.51.

**b.** Repeat the calculations for an 80 wt% lactic acid solution in water where both reactions are necessary to approximate the distribution.

Solution

**a.** Basis: 100 g total, 50 g of L_{1} = (50 g)/(90.08 g/mol) = 0.555 mol initially, 50 g = (50 g)/(18.02 g/mol) = 2.775 mol water initially, and 3.330 mol total. The equilibrium relation is *K _{a}* = 0.2023 =

Solving, we find, ξ = 0.0193, *x*_{L1} = (0.555 – 2(0.0193))/3.33 = 0.155, *x*_{L2} = 0.0193/3.33 = 0.006, *x*_{H2O} = (2.775 + 0.0193)/3.33 = 0.839. Note that although the mole fraction of L_{2} seems small, converting to wt%, the water content is (2.775 + 0.0193)(18.02 g/mol)/(100 g)·100% = 50.4 wt%, L_{1} is (0.555 – 2(0.0193))(90.08 g/mol)/(100 g)·100% = 46.5 wt%, and L_{2} is 0.0193(162.14 g/mol)/(100 g) · 100% = 3.1 wt%.

**b.** Basis: 100 g total, 80 g of L_{1} = (80 g)/(90.08 g/mol) = 0.888 mol initially, 20 g = (20 g)/(18.02 g/mol) = 1.110 mol water initially, and 1.998 mol total. Moles are conserved in both reactions. The equilibrium relations are *K _{a}*

Solving simultaneously, ξ_{1} = 0.0907, ξ_{2} = 0.009, *x*_{L1} = (0.888 – 2(0.0907) – 0.009)/1.998 = 0.349, *x*_{L2} = (0.0907 – 0.009)/1.998 = 0.041, *x*_{L3} = 0.009/1.998 = 0.0045, *x*_{H2O} = (1.110 + 0.0907 + 0.009)/1.998 = 0.6055. The weight fractions are: (1.110 + 0.0907 + 0.009)(18.02 g/mol)/(100 g)·100% = 21.8 wt% water, (0.888 – 2(0.0907) – 0.009)(90.08 g/mol)/(100 g)·100% = 62.8 wt% L_{1}, and (0.0907 – 0.009)(162.14 g/mol)/(100 g)·100% = 13.2 wt% L_{2}, 0.009(234.2 g/mol)/(100 g)·100% = 2.1 wt% L_{3}.

**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.

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 Δ*G ^{o}_{T}* as a function of temperature, we apply the usual van’t Hoff procedure. This means that

Example 17.11. Thermal decomposition of methane

A 2-liter constant-volume pressure vessel is evacuated and then filled with 0.10 moles of methane, after which the temperature of the vessel and its contents is raised to 1273 K. At this temperature the equilibrium pressure is measured to be 7.02 bar. Assuming that methane dissociates according to the reaction , compute *K _{a}* for this reaction at 1273 K from the experimental data.

Solution

We can calculate the mole fractions of H_{2} and CH_{4} as follows. Since the temperature is high, the total number of moles finally in the vessel can be determined from the ideal gas law (assuming that the solid carbon has negligible volume): *n* = *PV*/*RT* = 0.702·2000/(8.314)(1273) = 0.1327. Now assume that ξ moles of CH_{4} reacted. Then we have the following total mass balance: *n _{T}* = 0.10 + ξ. Therefore, ξ = 0.0327 and

Note that the equilibrium constant indicates that significant decomposition will occur (the reaction is exergonic, *K _{a}* > 1) and that graphite forms. Such behavior is known as “coking” and is common during industrial catalysis. Industrial application of catalysis often includes consideration of “regeneration’” of the catalyst by burning off the coke and using the heat of combustion elsewhere in the chemical plant.

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

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 _{A}P*/

where in this case, Σ*v _{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.

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}* =

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.

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.

Example 17.12. Butadiene by Gibbs minimization

Review Example 17.3(a) where steam is used to enhance conversion for 1-butene dehydrogenation. Gibbs energies of formation at 900 K for the hydrocarbons are summarized in that example.

The Gibbs energy of formation for water at 900 K is –198.204 kJ/mol. Vary conversion by selecting values of the reaction coordinate, calculating the Gibbs energy by Eqn. 17.69, and plotting the total Gibbs energy as a function of reaction coordinate. Demonstrate that Gibbs energy is minimized. Compare the equilibrium composition with that found in Example 17.3(a).

Solution

The initial moles of feed are 1 mol of 1-butene and 10 moles of steam. As an example calculation, select ξ = 0.1. Then the material balance provides, *n*_{C4H8} = 0.9, *n*_{C4H6} = *n*_{H2} = 0.1, *n*_{H2O} = 10. The mole fractions are *y*_{C4H8} = 0.9/(0.9 + 2(0.1) + 10) = 0.08108, *y*_{C4H6} = *y*_{H2} = 0.0090, y_{H2O} = 0.9009, and inserting the quantities into Eqn. 17.69, gives (inserting components in the order given above)

Repeating the calculation at various extents of reaction results in the following plot:

Careful analysis would show that the minimum is at ξ = 0.784 as in the earlier example. Note the changes in values are a small percentage of the absolute values of the total Gibbs energy, a numerical observation that is important in setting up convergence in Excel.

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.

Example 17.13. Direct minimization of the Gibbs energy with Excel

Apply the Gibbs minimization method to the problem of steam cracking of ethane at 1000 K and 1 bar where the ratio of steam to ethane in the feed is 4:1. Determine the distribution of C_{1} and C_{2} products, neglecting the possible formation of aldehydes, carboxylic acids, and higher hydrocarbons.

Solution

The solution is obtained using the worksheet GIBBSMIN contained in the workbook Rxns.xlsx (see Fig. 17.6).

One fundamental problem and one practical problem remain to be faced; there are several constraints that must be respected during the minimization process. These are the atom balances. We must keep in mind that we are not destroying matter, only rearranging it. So the number of carbon atoms, say, must be the same at the beginning and end of the process. Atom balance constraints must be written for every atom present. The atom balances are given straightforwardly by the stoichiometry of the species. For this example the balances are as follows.

O-balance: *n _{O}^{feed}* – 2

H-balance: *n _{H}^{feed}* –

C-balance: *n _{C}^{feed}* –

The practical problem that remains is that the numerical solver often attempts to substitute negative values for the prospective species. This problem is easily treated by solving for the log(*n _{i}*) during the iterations and determining the values of

In order to apply Gibbs minimization, the Gibbs energy of formation is required for each component at the reaction temperature. This preliminary calculation is the same type of calculation as performed in Example 17.4 on page 653, but is not shown here. For example, the Gibbs energy of methane is simply the Gibbs energy of the formation reaction C_{(s)} + at 1000 K. The values are embedded in the calculation of *G _{i}* shown in the worksheet in Fig. 17.6.

The primary product of this particular process is hydrogen. Fracturing hydrocarbons is a common problem in the petrochemical industry. This kind of process provides the raw materials for many downstream processes. The extension of this method to other reactions is straightforward. Some examples of interest would include several systems with environmental applications: carbon monoxide and NO* _{x}* from a catalytic converter, or by-products from catalytic destruction of chlorinated hydrocarbons. Evaluating the equilibrium possibilities by this method is so easy that it should be a required preliminary calculation for any gas phase process reaction study.

Example 17.14. Pressure effects for Gibbs energy minimization

Apply the Gibbs energy minimization method to the methanol synthesis reaction using stoichio-metric feed at 50 bar and 500 K. The reaction has been discussed in Example 17.1 on page 643, and in Section 17.5 on page 649.

Solution

It is convenient to first find Δ*G ^{o}_{f,}*

For a basis of 1 mole CO: *n _{C,feed}* = 1 =

Note: The objective function changes weakly with mole numbers near the minimum, so tighten the convergence criteria or re-run the Solver after the first convergence. Convergence is sensitive to the initial guess. An initial guess which works is log(n* _{i}*) = –0.1 for all

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

It is not difficult to imagine situations in which reactions take place in the presence of multiple phases. Absorption of CO_{2} into NaOH solution involves a reaction of the CO_{2} 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 H_{2}. 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 H_{2}, 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.

Example 17.15. The solvent methanol process

In a process being considered for methanol synthesis, a heavy liquid phase is added directly to the reactor to absorb the heat of reaction. The liquid is then circulated through an external heat exchanger. Usually, the catalyst is slurried in the liquid phase. An alternative to be considered is putting the catalyst in a fixed bed and adding just enough liquid so that a fairly small amount of vapor is left at the end of the reaction. Supposing naphthalene was used as the heavy liquid phase, use the Peng-Robinson equation to obtain approximate vapor-liquid *K*-value expressions of the form

for each component at a temperature of 200–250°C and pressures from 50–100 bar.^{a} In the worksheet computations, you may assume the *K*-value of naphthalene to be negligible.

Solution

A worksheet used for this solution is available in the workbook Rxns.xlsx.

Computing the *K*-value would normally require calling the Peng-Robinson equation during every flash and reaction iteration. This approximate correlation enables you to use Excel to perform the calculations since it is independent of any external programming requirements. The correlation should be suitably accurate if you “guess” compositions for developing the correlation that are reasonably close to the compositions at the outlet of the reactor. We suggest a guess for feed composition of {0.02, 0.10, 0.02, 0.035, 0.005, 0.82} for {CO, H_{2}, CO_{2}, methanol, water, and naphthalene}.

As an example of a way to develop a synthetic data base, perform flash calculations at 75 bars and temperatures of {200, 210, 220, 230, 240, 250} and the suggested feed composition. Tabulate the *K*-values for each component and plot them logarithmically with reciprocal temperature on the abscissa. Select a set of points, then select “add trendline” from the Chart menu. Select the options for a logarithmic fit, and displaying the equation on the chart. The coefficients of the equation give the *a* and *b* for the local “shortcut” correlation. (This step simplifies the implementation in Excel, but would be unnecessary if you were writing a dedicated program with access to a Peng-Robinson subroutine.)

Designating the solvent as component 6, the vapor-liquid *K*-values can be estimated as follows.

Solve for the simultaneous reaction and phase equilibria at 240°C and *P* = 100 bars considering the following two reactions:

Add moles of the heavy liquid until 9 moles of liquid is obtained for every mole of vapor *output*. The gases are fed in proportions 2:7:1 CO:H_{2}:CO_{2}.

Applying the shortcut van’t Hoff equation (calculated at 503 K using ln*K _{a}* and Δ

**Stoichiometry**

Imagine performing a flash at each new extent of conversion:

Writing objective functions:

This worksheet is called SMPRXN. An example of the output from a feed of 2,7,1,0,0 mole each of CO, H_{2}, CO_{2}, CH_{3}OH, H_{2}O is shown in Fig. 17.7.

The method of solving this problem is extremely similar to the DUALRXN problem. The only significant addition is an extra constraint equation which specifies that the vapor mole fractions must sum to unity. Note that ξ_{1} is greater than unity. This is because we have 2 moles of CO in the feed, so 1.3 moles converted is about 65%.

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

Example 17.16. NO_{2} absorption^{a}

The strength of concentrated acid which can be produced is limited by the back pressure of NO_{2} over the acid leaving the absorbers. The overall reaction, obtained by adding reactions (*a*) and (*b*), is shown as (*c*). Here we assume that N_{2}O_{4} is equivalent to 2NO_{2}.

The gas entering the bottom plate of a nitric acid absorber contains 0.1 mole of NO per mole of mixture and 0.25 mole of NO_{2} per mole mixture. The entering gas also contains 0.3 bar partial pressure of oxygen, in addition to inert gas. The total pressure is 1 bar. The acid made by the absorption operation contains 50% by weight of HNO_{3}, and the operation is isothermal at 86°F. Estimate the composition of the gas entering the second plate and the strength of the gas leaving the second plate.

Solution

(Basis: 1 mole gaseous feed)

Assume *y _{w}* =

For liquid:

For vapor:

We can determine the mole fractions from the weight fractions:

Noting from the *CRC Handbook*^{b} the vapor pressure of HNO_{3} is 64.6 mmHg, we can estimate the activity coefficients of HNO_{3} and water from the *x-y* data in *The Chemical Engineers’ Handbook*.^{c}

Gibbs energies of formation are available in Appendix E for all but nitrogen dioxide, and Reid et al.^{d} give the standard Gibbs energy of formation as 52 kJ/mol and the standard heat of formation as 33.87 kJ/mol. Performing a shortcut calculation using Kcalc.xlsx, the equilibrium constant at 303.15 K is *K _{a}* = 0.0054.

At *P* = 1 bar:

Substituting for the reaction coordinate:

Solving the cubic equation, ξ = 0.0431.

*x*_{HNO3} = (2ξ/(*W* + ξ) = 2·0.0431/(*W* + 0.0431) = 0.222 tells us that

*W* = –0.0431 + 2·0.0431/0.222 = 0.345 moles

So the composition of gas entering the second stage is *y*_{NO} = 0.157; *y*_{NO2} = 0.132; *y*_{O2} = 0.328; *y*_{I} = 0.383. Computations for further stages would be similar.

**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.

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.

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}* =

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.

**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 ln*K* 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.

**P17.1.** An equimolar mixture of H_{2} and CO is obtained by the reaction of steam with coal. The product mixture is known as “water-gas.” To enhance the H_{2} content, steam is mixed with water-gas and passed over a catalyst at 550°C and 1 bar so as to convert CO to CO_{2} by the reaction:

Any unreacted H_{2}O is subsequently condensed and the CO_{2} is subsequently absorbed to give a final product that is mostly H_{2}. This operation is called the water-gas shift reaction. Compute the equilibrium compositions at 550°C based on an equimolar feed of H_{2}, CO, and H_{2}O.

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: N_{2} + C_{2}H_{2} 2HCN. The feed to a reactor in which the above reaction takes place contains gaseous N_{2} and C_{2}H_{2} 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: C_{4}H_{8} C_{4}H_{6} + H_{2}. 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 Δ*G*^{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, Δ*H ^{f}*

**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 H_{2}O 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} + CO_{2}? (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.

C_{2}H_{4}O* _{(g)}* + H

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 2*A*_{1} *A*_{2}. 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

**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:

N_{2} + 3H_{2} 2NH_{3}

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, N_{2} + 3H_{2} 2NH_{3} equilibrated with excess N_{2} such that N_{2}′s equilibrium mole fraction is 0.55. Does adding more N_{2} to the equilibrated mixture result in more NH_{3}? 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% N_{2}.

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

C_{6}H_{5}CH = CH_{2}* _{(g)}* +H

The system initially contains 3 mol H_{2} 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,

C_{3}H_{8(g)} C_{2}H_{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: C_{2}H_{4} + H_{2}O C_{2}H_{5}OH. 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% N_{2}, 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)} CH_{3}OH_{(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% SO_{2}, 20 mol% O_{2}, and 65 mol% N_{2} enters a catalytic reactor operating and forms SO_{3} at 480°C and 2 bar.

**a.** Determine the equilibrium conversion of SO_{2}.

**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% H_{2}, 12 mol% CO, 8 mol% CO_{2}, and 5 mol% N_{2}. The system comes to equilibrium at 550 K and 100 bar with respect to the following reactions:

2H_{2(g)} + CO_{(g)} CH_{3}OH_{g)}

H_{2(g)} + Co_{2(g)} CO_{(g)} + H_{2}O_{(g)}

Assuming ideal gases, derive the equations that would be solved simultaneously for ξ_{1}, ξ_{2} where 1 refers to the first reaction listed. Provide numerical values for the equilibrium constants. Determine ξ_{1} and ξ_{2} 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 (CH_{2}=CHCH_{2}CH_{3}). 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.

**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 + H_{2} C_{(s)} + H_{2}O. 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:

C_{3}H_{8} + 3H_{2}O 3CO + 7H_{2}

CO + H_{2}O CO_{2} + H_{2}

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 NO_{2}, N_{2}O, N_{2}O_{4}, and NH_{3}. Estimate the compositions of all species at each temperature {400°C, 500°C, 600°C} and plot the ratio of NH_{3}/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^{o}C. 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 (CaCO_{3}) 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 cm

**17.21.** One suggestion for sequestering CO_{2} 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 CO_{2} 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

**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 CO_{2} 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 Δ*C _{P}* =

**a.** The heat of reaction is represented by Δ*H* = Δ*H*_{R} + Δ*C _{P}*(

**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.

**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, Δ*H* = 540 kJ/mol and Δ*S* = 1.54 kJ/mol.

**c.** Evaluate the contributions to Δ*G* 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 Δ

**f.** At pH = 1, *T _{m}* = 43°C. Provide a plot of plot [

**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 Δ*V.* 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, Δ*C _{P}*

**b.** Calculate Δ*S*_{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

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