Biomass Reactions

APPENDIX I HOW TO USE THE CRE WEB RESOURCES 931

E. Biomass Reactions

Substrate (S) + Cells (C) → More Cells + Product

Note: The rate constants, k, and activation energies for a number of the reactions in these exam- ples are given in the Databases in Appendix E.

TABLE 3-1 EXAMPLESOF REACTION RATE LAWS (CONTINUED)

NO₂

Cl+ 2NH₃

NO₂ NH₂

+ NH₄Cl r_ONCB k_ONCBC_ONCBC_NH 3

CNBr CH NH+ _{3 2}⎯ →⎯ CH Br NCNH₃ + ₂ r_CNBr kC_CNBrC_CH

3NH₂

CH3COOC2H5C4H9OH ⎯→←⎯ CH3COOC4H9C2H5OH ⎯→←⎯

A + BB C + D r_Ak C[ _AC_BC_CC_D/K_C]

CH CHO₃ ⎯ →cat⎯ CH₄+CO r_CH

3CHO

kC^{3 2}CH₃CHO

CH(CH₃)₂

+ C₃H₆ Cumene (C) Benzene (B) + Propylene (P)

cat

rC

k P[ _CP_BP_PK_P] 1K_BP_BK_CP_C ---

H₂O

⎯⎯→ rU kC_U K_MC_U ---

r_S

kC_SC_C K_SC_S ---

CO C1 ₂→COC1₂

in which the kinetic rate law is

This reaction is first order with respect to carbon monoxide, three-halves order with respect to chlorine, and five-halves order overall.

Sometimes reactions have complex rate expressions that cannot be sepa- rated into solely temperature-dependent and concentration-dependent portions.

In the decomposition of nitrous oxide,

the kinetic rate law is

Both and are strongly temperature dependent. When a rate expression such as the one given above occurs, we cannot state an overall reaction order.

Here, we can only speak of reaction orders under certain limiting conditions.

For example, at very low concentrations of oxygen, the second term in the denominator would be negligible with respect to 1 (1 >> ), and the reac- tion would be “apparent” first order with respect to nitrous oxide and first order overall. However, if the concentration of oxygen were large enough so that the number 1 in the denominator were insignificant in comparison with the second term, ( >> 1), the apparent reaction order would be –1 with respect to oxygen and first order with respect to nitrous oxide, giving an overall apparent zero order. Rate expressions of this type are very common for liquid and gaseous reactions promoted by solid catalysts (see Chapter 10).

They also occur in homogeneous reaction systems with reactive intermediates (see Chapter 9).

It is interesting to note that although the reaction orders often correspond to the stoichiometric coefficients, as evidenced for the reaction between hydro- gen and iodine, just discussed to form HI, the rate expression for the reaction between hydrogen and another halogen, bromine, is quite complex. This non- elementary reaction

proceeds by a free-radical mechanism, and its reaction rate law is

(3-8)

Apparent First-Order Reactions. Because the law of mass action in collision theory shows that two molecules must collide giving a second-order dependence on the rate, you are probably wondering how rate laws such as Equation (3-8) as well as the rate law for first-order reactions come about. An example of first-order reaction not involving radioactive decay is the decomposition of eth- anol to form ethylene and hydrogen.

r_CO

kC_COC_C1

2 3 2

2N₂O→2N₂O₂

r_N

2O

k_N

2OC_N

2O

1 kC_O 2

---

k_N

2O k

kC_O

2

kC_O

2 kC_O

2

H₂Br₂→2HBr

rBr₂

k_Br

2C_H

2C_Br

2 1 2

k' C_HBr C_Br 2

---

In terms of symbols

Rate laws of this form usually involve a number of elementary reactions and at least one active intermediate. An active intermediate is a high-energy molecule that reacts virtually as fast as it is formed. As a result, it is present in very small concentrations. Active intermediates (e.g., A*) can be formed by colli- sion or interaction with other molecules (M) such as inerts or reactants

Here, the activation occurs when translational kinetic energy is transferred into energy stored in internal degrees of freedom, particularly vibrational degrees of freedom.³ An unstable molecule (i.e., active intermediate) is not formed solely as a consequence of the molecule moving at a high velocity (high-translational kinetic energy). The translational kinetic energy must be absorbed into the chemical bonds where high-amplitude oscillations will lead to bond ruptures, molecular rearrangement, and decomposition. In the absence of photochemical effects or similar phenomena, the transfer of translational energy to vibrational energy to produce an active intermediate can occur only as a consequence of molecular collision or interaction. Collision theory is discussed in the Profes- sional Reference Shelf on the CRE Web site for Chapter 3.

As will be shown in Chapter 9, the mole A becomes activated to A* by collision with another molecule M. The activated molecule can become deacti- vated by collision with another molecule or the activated molecule can decom- pose to B and C.

Using this mechanism, we show in Section 9.1.1 that at high concentrations of M the rate law for this mechanism becomes

In Chapter 9, we will discuss reaction mechanisms and pathways that lead to nonelementary rate laws, such as the rate of formation of HBr shown in Equation (3-8).

3W. J. Moore, Physical Chemistry (Reading, MA: Longman Publishing Group, 1998).

⎯ →⎯ + C H_{2 6} C H_{2 4} H₂

r_C

2H₆

kC_C

2H₆

A ⎯⎯→ B + C r_A

kC_A

+ ⎯ →⎯ + A M ^k1 A^* M

+ ← ⎯⎯ →⎯⎯ +

↓ +

k

A M A M

B C

k k

*

3

1 2

r_A

k_AC_A

Heterogeneous Reactions. Historically, it has been the practice in many gas-solid catalyzed reactions to write the rate law in terms of partial pressures rather than concentrations. In heterogeneous catalysis it is the weight of cata- lyst that is important, rather than the reactor volume. Consequently, we use in order to write the rate law in terms of mol per kg of catalyst per time in order to design PBRs. An example of a heterogeneous reaction and corre- sponding rate law is the hydrodemethylation of toluene (T) to form benzene (B) and methane (M) carried out over a solid catalyst

The rate of disappearance of toluene per mass of catalyst, , i.e., (mol/mass/time) follows Langmuir-Hinshelwood kinetics (discussed in Chapter 10), and the rate law was found experimentally to be

where the prime in notes typical units are in per kilogram of catalyst (mol/kg-cat/s), P_T, , and P_B are partial pressures of toluene, hydrogen, and benzene in (kPa or atm), and K_B and K_T are the adsorption constants for ben- zene and toluene respectively, with units of kPa^–1 (or atm^–1). The specific reac- tion rate k has units of

You will find that almost all heterogeneous catalytic reactions will have a term such as (1 + K_AP_A + … ) or (1 + K_AP_A + … )² in the denominator of the rate law (cf. Chapter 10).

To express the rate of reaction in terms of concentration rather than par- tial pressure, we simply substitute for P_i using the ideal gas law

(3-9) The rate of reaction per unit weight (i.e., mass) catalyst, (e.g., ), and the rate of reaction per unit volume, , are related through the bulk density ρb (mass of solid/volume) of the catalyst particles in the fluid media:

In fluidized catalytic beds, the bulk density, ρb, is normally a function of the volumetric flow rate through the bed.

Consequently, using the above equations for P_i and we can write the rate law for the hydromethylation of toluene in terms of concentration and in (mole/dm³) and the rate, in terms of reactor volume, i.e.,

rA

C₆H₅CH₃H₂ cat C→ ⁶H₆CH₄

rT

rT

kP_H

2P_T 1K_BP_BK_TP_T ---

rT

P_H

2

k

{ } mol toluene kg-cat s kPa² ---

P_iC_iRT

rA

rT

r_A

r_A

( )b (rA)

Relating rate per unit volume and rate by per unit mass of catalyst

moles time volume

--- mass volume ---

⎝ ⎠

⎛ ⎞ moles

time mass ---

⎝ ⎠

⎛ ⎞

r_T r_T

or as we will see in Chapter 4 leave it in terms of partial pressures.

In summary on reaction orders, they cannot be deduced from reaction stoichiometry. Even though a number of reactions follow elementary rate laws, at least as many reactions do not. One must determine the reaction order from the literature or from experiments.

3.2.3 Reversible Reactions

All rate laws for reversible reactions must reduce to the thermodynamic rela- tionship relating the reacting species concentrations at equilibrium. At equilib- rium, the rate of reaction is identically zero for all species (i.e., ).

That is, for the general reaction

(2-1) the concentrations at equilibrium are related by the thermodynamic relation- ship for the equilibrium constant K_C (see Appendix C).

(3-10) The units of the thermodynamic equilibrium constant, K_C, are . To illustrate how to write rate laws for reversible reactions, we will use the combination of two benzene molecules to form one molecule of hydrogen and one of diphenyl. In this discussion, we shall consider this gas-phase reac- tion to be elementary and reversible

or, symbolically,

The forward and reverse specific reaction rate constants, and , respectively, will be defined with respect to benzene.

Benzene (B) is being depleted by the forward reaction

in which the rate of disappearance of benzene is

ρ

{ }

( )

− = ⎡⎣ ′ ⎤⎦

+ + ⎛ •

⎝⎜ ⎞

⎠⎟

= •

⎛

⎝⎜

⎞

⎠⎟

r k RT C C

K RTC K RTC s

k s

1

mol dm dm

mol

b k

T

2

H T

B B T T

3

3

2

r_A0 aAbB ⎯⎯→←⎯⎯ cCdD

Thermodynamic equilibrium

relationship K_C C_Ce^c C_De^d

C_Ae^a C_Be^b ---

mol/dm³

( )^dcba

2C₆H₆ ⎯⎯→←⎯⎯ ^kB C₁₂H₁₀H₂

kB

2B ⎯⎯→←⎯⎯ ^kB D H ₂

kB

k_B k_B

2C₆H₆ ⎯⎯→ ^kB C₁₂H₁₀H₂

r_{B forward,}

k_BC_B²

If we multiply both sides of this equation by 1, we obtain the expression for the rate of formation of benzene for the forward reaction

(3-11) For the reverse reaction between diphenyl (D) and hydrogen ( ),

the rate of formation of benzene is given as

(3-12) Again, both the rate constants k_B and k_–B are defined with respect to benzene!!!

The net rate of formation of benzene is the sum of the rates of formation from the forward reaction [i.e., Equation (3-11)] and the reverse reaction [i.e., Equation (3-12)]

(3-13) Multiplying both sides of Equation (3-13) by 1, and then factoring out k_B, we obtain the rate law for the rate of disappearance of benzene,

Replacing the ratio of the reverse to forward rate law constants by the recipro- cal of the concentration equilibrium constant, K_C, we obtain

(3-14) where

The equilibrium constant decreases with increasing temperature for exothermic reactions and increases with increasing temperature for endothermic reactions.

Let’s write the rate of formation of diphenyl, r_D, in terms of the concen- trations of hydrogen, H₂, diphenyl, D, and benzene, B. The rate of formation of diphenyl, r_D, must have the same functional dependence on the reacting species concentrations as does the rate of disappearance of benzene, –r_B. The rate of formation of diphenyl is

(3-15) r_{B forward,} kBC_B²

H₂ C₁₂H₁₀H₂ ⎯⎯→ ^kB 2C₆H₆

The specific reaction rate constant, k_i, must be defined with respect to a particular species.

r_{B, reverse} k_BC_DC_H 2

Net rate r_Br_{B, net}r_{B, forward}r_{B, reverse} r_B k_BC_B² k_BC_DC_H

2

r_B

Elementary reversible A ⎯→←⎯ B rA

k CA C_B K_C ---

⎝ ⎠

⎛ ⎞

r_B

k_BC_B² k_BC_DC_H

2 k_B C_B² k_B k_B --- CDC_H 2

⎝ ⎠

⎜ ⎟

⎛ ⎞

r_B

k_B C_B² C_DC_H

2

K_C ---

⎝ ⎠

⎛ ⎞

k_B k_B

---K_CConcentration equilibrium constant

r_D k_D C_B² C_DC_H

2

K_C ---

⎝ ⎠

⎛ ⎞

Using the relationship given by Equation (3-1) for the general reaction (3-1) we can obtain the relationship between the various specific reaction rates, ,

(3-16) Comparing Equations (3-15) and (3-16), we see the relationship between the specific reaction rate with respect to diphenyl, k_D, and the specific reaction rate with respect to benzene, k_B, is

Consequently, we see the need to define the rate constant, k, with respect to a particular species.

Finally, we need to check to see if the rate law given by Equation (3-14) is thermodynamically consistent at equilibrium. Applying Equation (3-10) (and Appendix C) to the diphenyl reaction and substituting the appropriate species concentration and exponents, thermodynamics tells us that

(3-17) Now let’s look at the rate law. At equilibrium, –r_B ≡ 0, and the rate law given by Equation (3-14) becomes

Rearranging, we obtain, as expected, the equilibrium expression

that is identical to Equation (3-17) obtained from thermodynamics.

From Appendix C, Equation (C-9), we know that when there is no change in the total number of moles and the heat capacity term, ΔC_P = 0, the temperature dependence of the concentration equilibrium constant is

(C-9)

Relative rates r_A

a --- r_B

b --- r_C

---c r_D ---d

k_B k_D r_D

---1 r_B ²

--- kB C_B² C_DC_H 2K_C

[ ]

2

--- k_B

--- C2 _B² C_DC_H

2

K_C ---

k_D k_B ---2

K_C C_DeC_H

2e

C²Be

---

At equilibrium, the rate law must reduce to an equation consistent wth thermodynamic

equilibrium. r_B0 k_B C²Be

C_DeC_H

2e

K_C ---

K_C C_DeC_H

2e

C²Be

---

K_C( )T K_C( )T₁ exp HRx

---R 1 T₁ --- 1

T---

⎝ ⎠

⎛ ⎞

Therefore, if we know the equilibrium constant at one temperature, T₁ [i.e., K_C (T₁)], and the heat of reaction, , we can calculate the equilibrium constant at any other temperature T. For endothermic reactions, the equilibrium constant, K_C, increases with increasing temperature; for exothermic reactions, K_C decreases with increasing temperature. A further discussion of the equilibrium constant and its thermodynamic relationship is given in Appendix C. For large values of the equilibrium constant, K_C, the reaction behaves as if it were irre- versible.

3.3 Rates and the Reaction Rate Constant

There are three molecular concepts relating to the rate of reaction we will discuss:

Concept 1. Law of Mass Action. The rate of reaction increases with increasing concentration of reactants owing to the increased number of molec- ular collisions at the higher reactant concentrations. In Section 3.2, we have just discussed Concept 1: rate laws, and the dependence of the rate reactant concentration. We now discuss Concept 2, potential energy surfaces and energy barriers, and Concept 3, the energy needed for crossing the barriers.

3.3.1 The Rate Constant k

The reaction rate constant k is not truly a constant; it is merely independent of the concentrations of the species involved in the reaction. The quantity k is referred to as either the specific reaction rate or the rate constant. It is almost always strongly dependent on temperature. It also depends on whether or not a catalyst is present, and in gas-phase reactions, it may be a function of total pressure. In liquid systems it can also be a function of other parameters, such as ionic strength and choice of solvent. These other variables normally exhibit much less effect on the specific reaction rate than does temperature, with the exception of supercritical solvents, such as supercritical water.

Consequently, for the purposes of the material presented here, it will be assumed that k_A depends only on temperature. This assumption is valid in most laboratory and industrial reactions, and seems to work quite well.

It was the great Nobel Prize–winning Swedish chemist Svante Arrhenius (1859–1927) who first suggested that the temperature dependence of the spe- cific reaction rate, k_A, could be correlated by an equation of the type

(3-18) where A = pre-exponential factor or frequency factor

E = activation energy, J/mol or cal/mol

R = gas constant = 8.314 J/mol • K = 1.987 cal/mol • K T = absolute temperature, K

Equation (3-18), known as the Arrhenius equation, has been verified empiri- cally to give the correct temperature behavior of most reaction rate constants within experimental accuracy over fairly large temperature ranges. The Arrhe- nius equation is derived in the Professional Reference Shelf R3.A: Collision Theory on the CRE Web site.

HRx

Arrhenius equation k_A( )T Ae^E/RT

T(K) k

Additionally, one can view the activation energy in terms of collision theory (Professional Reference Shelf R3.1). By increasing the temperature, we increase the kinetic energy of the reactant molecules. This kinetic energy can in turn be transferred through molecular collisions to internal energy to increase the stretching and bending of the bonds, causing them to reach an activated state, vulnerable to bond breaking and reaction.

Why is there an activation energy? If the reactants are free radicals that essentially react immediately on collision, there usually isn’t an activation energy. However, for most atoms and molecules undergoing reaction, there is an activation energy. A couple of the reasons are that in order to react

1. The molecules need energy to distort or stretch their bonds so that they break and now can form new bonds.

2. The molecules need energy to overcome the steric and electron repul- sive forces as they come close together.

The activation energy can be thought of as a barrier to energy transfer (from kinetic energy to potential energy) between reacting molecules that must be overcome. The activation energy is the minimum increase in potential energy of the reactants that must be provided to transform the reactants into products. This increase can be provided by the kinetic energy of the colliding molecules.

In addition to the concentrations of the reacting species, there are two other factors that affect the rate of reaction,

• the height of the barrier, i.e., activation energy, and

• the fraction of molecular collisions that have sufficient energy to cross over the barrier (i.e., react when the molecules collide).

If we have a small barrier height, the molecules colliding will need only low kinetic energies to cross over the barrier. For reactions of molecules with small barrier heights occurring at room temperatures, a greater fraction of molecules will have this energy at low temperatures. However, for larger barrier heights, we require higher temperatures where a higher fraction of colliding molecules will have the necessary energy to cross over the barrier and react. We will dis- cuss each of these concepts separately.

Concept 2. Potential Energy Surfaces and Energy Barriers. One way to view the barrier to a reaction is through the use of potential energy surfaces and the reaction coordinates. These coordinates denote the minimum potential energy of the system as a function of the progress along the reaction path as we go from reactants to an intermediate to products. For the exothermic reaction

A + BC A – B – C ⎯→ AB + C

the potential energy surface and the reaction coordinate are shown in Figures 3-1 and 3-2. Here E_A, E_C, E_AB, and E_BC are the potential energy surface energies of the reactants A, BC and product molecules (AB and C), and E_ABC is the energy of the complex A–B–C at the top of the barrier.

Figure 3-1(a) shows the 3–D plot of the potential energy surface, which is analogous to a mountain pass where we start out in a valley and then climb up

⎯→←⎯

to pass over the top of the pass, i.e., the col or saddle point, and proceed down into the next valley. Figure 3-1(b) shows a contour plot of the pass and valleys and the reaction coordinate as we pass over the col from valley to valley.

Energy changes as we move within the potential energy surfaces.

At point X in Figure 3-1(b), species A and BC are far apart and are not inter- acting; R_BC is just the equilibrium bond length. Consequently, the potential energy is just the BC bond energy. Here, A and BC are in their minimum potential energy in the valley and the steep rise up from the valley to the col from X would correspond to increases in the potential energy as A comes close to BC. If the BC bond is stretched from its equilibrium position at X, the

Energy

Transition state

3 1

RBC, B–C Distance (angstroms) 1.5

2 2.5

1 1.5 2 2.5 3

R_AB, A–B Distance (angstroms) (b)

(a)

X YR_AB

, A–B Distance R^BC

, B–C Di stance

X

Y

Figure 3-1 A potential energy surface for the H + CH3OH ⎯→H2 + CH2OH from the calculations of Blowers and Masel. The lines in the figure are contours of constant energy.

The lines are spaced 5 kcal/mol. Richard I. Masel, Chemical Kinetics and Catalysis, p. 370, Fig.7.6 (Wiley, 2001).

Transition state Products

Reactants

CH₃ -I Bond distance in angstroms

Ground state energy

CH₃I + Cl CH₃CI + I

1.9 2.1 2.3 2.5 2.7 23

25 27 29 31 33 36 39 41

Potential energy

Reaction coordinate Products Reactants

X Y

Energy barrier A-B-C

A+BC

AB+C E_B

H_RX

(E_AB + E_C) (E_A + E_BC) E_A-B-C

(a) (b)

Energy (kcal/mol)

RBC

(

A^○

)

Equilibrium BC position

(c)

Figure 3-2 Progress along reaction path. (a) Symbolic reaction; (b) Calculated from computational software on the CRE Web site, Chapter 3 Web Module. (c) Side view at point X in Figure 3-1(b) showing the valley.

(c)

potential energy increases up one side of the valley hill (R_BC increases). The potential energy is now greater because of the attractive forces trying to bring the B–C distance back to its equilibrium position. If the BC bond is compressed at X from its equilibrium position, the repulsive forces cause the potential energy to increase on the other side of the valley hill, i.e., R_BC decreases at X.

At the end of the reaction, point Y, the products AB and C are far apart on the valley floor and the potential energy is just the equilibrium AB bond energy.

The sides of the valley at Y represent the cases where AB is either compressed or stretched causing the corresponding increases in potential energy at point Y and can be described in an analogous manner to BC at point X.

We want the minimum energy path across the barrier for converting the kinetic energy of the molecules into potential energy. This path is shown by the curve X→Y in Figure 3-1 and also by Figure 3-2(a). As we move along the A–B distance axis in Figure 3-2(a), A comes closer to BC and B begins to bond with A and to push BC apart such that the potential energy of the reac- tion pair continues to increase until we arrive at the top of the energy barrier, which is the transition state. In the transition state the molecular distances between A and B and between B and C are close. As a result, the potential energy of the three atoms (molecules) is high. As we proceed further along the arc length of the reaction coordinate depicted in Figure 3-1(a), the AB bond strengthens and the BC bond weakens and C moves away from AB and the energy of the reacting pair decreases until we arrive at the valley floor where AB is far apart from C. The reaction coordinate quantifies how far the reaction has progressed. The commercial software available to carry out calculations for the transition state for the real reaction

CH₃I Cl → CH₃Cl I

shown in Figure 3-2(b) is discussed in the Web Module Molecular Modeling in Chemical Reaction Engineering on the CRE Web site.

We next discuss the pathway over the barrier shown along the line Y–X.

We see that for the reaction to occur, the reactants must overcome the mini- mum energy barrier, E_B, shown in Figure 3-2(a). The energy barrier, E_B, is related to the activation energy, E. The energy barrier height, E_B, can be calcu- lated from differences in the energies of formation of the transition-state mol- ecule and the energy of formation of the reactants; that is,

(3-19) The energy of formation of the reactants can be found in the literature or cal- culated from quantum mechanics, while the energy of formation of the transi- tion state can also be calculated from quantum mechanics using a number of software packages, such as Gaussian (http://www.gaussian.com/) and Dacapo (https://wiki.fysik.dtu.dk/dacapo). The activation energy, E, is often approxi- mated by the barrier height, E_B, which is a good approximation in the absence of quantum mechanical tunneling.

Now that we have the general idea for a reaction coordinate, let’s con- sider another real reaction system

H• + C₂H₆→ H₂ + C₂H₅•

The energy-reaction coordinate diagram for the reaction between a hydrogen atom and an ethane molecule is shown in Figure 3-3 where the bond distor-

E_BEfAB C (EfAEfB C )