3. Rate Laws

Success is measured not so much by the position one has reached in life, as by the obstacles one has overcome while trying to succeed.

—Booker T. Washington

3.1 Basic Definitions

A homogeneous reaction is one that involves only one phase. A heterogeneous reaction involves more than one phase, and the reaction usually occurs at the interface between the phases. An irreversible reaction is one that proceeds in only one direction and continues in that direction until one of the reactants is exhausted. A reversible reaction, on the other hand, can proceed in either direction, depending on the concentrations of reactants and products relative to the corresponding equilibrium concentrations. An irreversible reaction behaves as if no equilibrium condition exists. Strictly speaking, no chemical reaction is completely irreversible. However, for many reactions, the equilibrium point lies so far to the product side that these reactions are treated as irreversible reactions.

Types of reactions

The molecularity of a reaction is the number of atoms, ions, or molecules involved (colliding) in a reaction step. The terms unimolecular, bimolecular, and termolecular refer to reactions involving, respectively, one, two, and three atoms (or molecules) interacting or colliding in any one reaction step. The most common example of a unimolecular reaction is radioactive decay, such as the spontaneous emission of an alpha particle from uranium-238 to give thorium and helium

92U23890Th234 +2He4

The rate of disappearance of uranium (U) is given by the rate law

rU = kCU

The only true bimolecular reactions are those that involve the collision with free radicals (i.e., unpaired electrons, e.g., Br•), such as

Br• + C2H6 → HBr + C2H5

with the rate of disappearance of bromine given by the rate law

rBr• = kCBr•CC2H6

The probability of a termolecular reaction, where three molecules collide all at once, is almost nonexistent, and in most instances the reaction pathway follows a series of bimolecular reactions, as in the case of the reaction

2NO + O2 → 2NO2

The reaction pathway for this “Hall of Fame” reaction is quite interesting and is discussed in Chapter 9, along with similar reactions that form active intermediate complexes in their reaction pathways.

In discussing Rates of Chemical Reaction, we have three rates to consider

• Relative Rates

• Rate Laws

• Net Rates

Relative Rates tell us how fast one species is disappearing or appearing relative to the other species in the given reaction.

Rate Laws are the algebraic equations that apply to a given reaction.

Net Rates of formation of a given species (e.g., A) is the sum of the rate of reactions of A all the reactions in which A is either a reactant or product in the system.

A thorough discussion of net rates is given in Chapter 8 where we discuss multiple reactions.

3.1.1 Relative Rates of Reaction

The relative rates of reaction of the various species involved in a reaction can be obtained from the ratio of the stoichiometric coefficients. For Reaction (2-2),

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we see that for every mole of A that is consumed, c/a moles of C appear. In other words,

Rate of formation of Image (Rate of disappearance of A)

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Similarly, the relationship between the rates of formation of C and D is

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The relationship can be expressed directly from the stoichiometry of the reaction

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for which

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Reaction stoichiometry

or

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For example, in the reaction

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we have

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If NO2 is being formed at a rate of 4 mol/m3/s, i.e.,

rNO2 = 4 mol/m3/s

then the rate of formation of NO is

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the rate of disappearance of NO is

rNO = 4 mol/m3/s

and the rate of disappearance of oxygen, O2, is

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3.2 The Reaction Order and the Rate Law

In the chemical reactions considered in the following paragraphs, we take as the basis of calculation a species A, which is one of the reactants that is disappearing as a result of the reaction. The limiting reactant is usually chosen as our basis for calculation. The rate of disappearance of A, –rA, depends on temperature and concentration. For many irreversible reactions, it can be written as the product of a reaction rate constant, kA, and a function of the concentrations (activities) of the various species involved in the reaction:

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Concept 1. Law of Mass Action. The rate of reaction increases with increasing concentration of reactants owing the corresponding increase in the number of molecular collisions. The rate law for bimolecular collisions is derived in the collision theory section of the Professional Reference Shelf R3.1. A schematic of the reaction of A and B molecules colliding and reacting is shown in Figure R3.1 on page 97.

The rate law gives the relationship between reaction rate and concentration.

The algebraic equation that relates –rA to the species concentrations is called the kinetic expression or rate law. The specific rate of reaction (also called the rate constant), kA, like the reaction rate, –rA, always refers to a particular species in the reaction and normally should be subscripted with respect to that species. However, for reactions in which the stoichiometric coefficient is 1 for all species involved in the reaction, for example,

1 NaOH + 1 HC1 → 1 NaC1 + 1 H2O

we shall delete the subscript on the specific reaction rate, (e.g., A in kA), to let

k = kNaOH = kHCl = kNaCl = kH2O

3.2.1 Power Law Models and Elementary Rate Laws

The dependence of the reaction rate, –rA, on the concentrations of the species present, fn(Cj), is almost without exception determined by experimental observation. Although the functional dependence on concentration may be postulated from theory, experiments are necessary to confirm the proposed form. One of the most common general forms of this dependence is the power law model. Here the rate law is the product of concentrations of the individual reacting species, each of which is raised to a power, for example,

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The exponents of the concentrations in Equation (3-3) lead to the concept of reaction order. The order of a reaction refers to the powers to which the concentrations are raised in the kinetic rate law.1 In Equation (3-3), the reaction is α order with respect to reactant A, and β order with respect to reactant B. The overall order of the reaction, n, is

n = α + β

1 Strictly speaking, the reaction rates should be written in terms of the activities, ai, (ai = γiCi, where γi is the activity coefficient). Kline and Fogler, JCIS, 82, 93 (1981); ibid., p. 103; and Ind. Eng. Chem Fundamentals 20, 155 (1981).

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However, for many reacting systems, the activity coefficients, γi, do not change appreciably during the course of the reaction, and they are absorbed into the specific reaction rate constant, kA

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Overall reaction order

The units of –rA are always in terms of concentration per unit time, while the units of the specific reaction rate, kA, will vary with the order of the reaction. Consider a reaction involving only one reactant, such as

A → Products

with an overall reaction order n. The units of rate, –rA, and the specific reaction rate, k are

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Generally, the overall reaction order can be deduced from the units of the specific reaction rate constant k. For example, the rate laws corresponding to a zero-, first-, second-, and third-order reaction, together with typical units for the corresponding rate constants, are

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An elementary reaction is one that involves a single reaction step, such as the bimolecular reaction between an oxygen free radical and methanol molecule

O• + CH3OH → CH3O• + OH•.

The stoichiometric coefficients in this reaction are identical to the powers in the rate law. Consequently, the rate law for the disappearance of molecular oxygen is

rO. =kCO. CCH3OH

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Collision theory

The reaction is first order in an oxygen free radical and first order in methanol; therefore, we say that both the reaction and the rate law are elementary. This form of the rate law can be derived from Collision Theory, as shown in the Professional Reference Shelf R3.1 on the CRE Web site (www.umich.edu/~elements/5e/index.html). There are many reactions where the stoichiometric coefficients in the reaction are identical to the reaction orders, but the reactions are not elementary, owing to such things as pathways involving active intermediates and series reactions. For these reactions that are not elementary but whose stoichiometric coefficients are identical to the reaction orders in the rate law, we say the reaction follows an elementary rate law. For example, the oxidation reaction of nitric oxide discussed earlier

2NO + O2 → 2NO2

is not really an elementary reaction, but follows an elementary rate law; therefore,

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Note: The rate constant, k, is defined with respect to NO.

Another nonelementary reaction that follows an elementary rate law is the gas-phase reaction between hydrogen and iodine

H2 + I2 → 2HI

with

rH2 = kH2 CH2 CI2

In summary, for many reactions involving multiple steps and pathways, the powers in the rate laws surprisingly agree with the stoichiometric coefficients. Consequently, to facilitate describing this class of reactions, we say that a reaction follows an elementary rate law when the reaction orders are identical to the stoichiometric coefficients of the reacting species for the reaction as written. It is important to remember that the rate laws are determined by experimental observation! Chapter 7 describes how these and other rate laws can be developed from experimental data. They are a function of the reaction chemistry and not the type of reactor in which the reactions occur. Table 3-1 gives examples of rate laws for a number of reactions. By saying a reaction follows an elementary rate law as written gives us a quick way to look at the reaction stoichiometry and then write the mathematical form of the rate law. The values of specific reaction rates for these and a number of other reactions can be found in the Database found on the CRE Web site.

Where do you find rate laws?

The rate constants and the reaction orders for a large number of gas- and liquid-phase reactions can be found in the National Bureau of Standards’ circulars and supplements.2 Also consult the journals listed at the end of Chapter 1.

2 Kinetic data for a larger number of reactions can be obtained on CD-ROMs provided by National Institute of Standards and Technology (NIST). Standard Reference Data 221/A320 Gaithersburg, MD 20899; phone: (301) 975-2208. Additional sources are Tables of Chemical Kinetics: Homogeneous Reactions, National Bureau of Standards Circular 510 (Sept. 28, 1951); Suppl. 1 (Nov. 14, 1956); Suppl. 2 (Aug. 5, 1960); Suppl. 3 (Sept. 15, 1961) (Washington, DC: U.S. Government Printing Office). Chemical Kinetics and Photochemical Data for Use in Stratospheric Modeling, Evaluate No. 10, JPL Publication 92-20 (Pasadena, CA: Jet Propulsion Laboratories, Aug. 15, 1992).

We note in Table 3-1 that Reaction Number (3) in the First-Order Rate Laws and Reaction Number (1) in the Second-Order Rate Laws do not follow elementary reaction rate laws. We know this because the reaction orders are not the same as the stoichiometric coefficients for the reactions as they are written.

Just as an aside, Prof. Dr. Sven Köttlov, a prominent resident of Riça, Jofostan, was once jailed and questioned for 12 hours for breaking a reaction rate law. He stated it was second order when experiments clearly showed it was first order.

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TABLE 3-1 EXAMPLES OF REACTION RATE LAWS

Very important references. You should also look in the other literature before going to the lab.

3.2.2 Nonelementary Rate Laws

A large number of both homogeneous and heterogeneous reactions do not follow simple rate laws. Examples of reactions that don’t follow simple elementary rate laws are discussed below.

Homogeneous Reactions. The overall order of a reaction does not have to be an integer, nor does the order have to be an integer with respect to any individual component. As an example, consider the gas-phase synthesis of phosgene,

CO + C12 → COC12

in which the kinetic rate law is

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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 separated into solely temperature-dependent and concentration-dependent portions. In the decomposition of nitrous oxide,

2N2O → 2N2 + O2

the kinetic rate law is

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Both kN2O and k′ 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>> kCO2), and the reaction 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, kCO2(kCO2>> 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 hydrogen and iodine, just discussed to form HI, the rate expression for the reaction between hydrogen and another halogen, bromine, is quite complex. This nonelementary reaction

H2 + Br2 → 2HBr

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

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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 (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 ethanol to form ethylene and hydrogen.

C2H6→C2H4 + H2

rC2H6 = kCC2H6

In terms of symbols

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rA = kCA

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 collision or interaction with other molecules (M) such as inerts or reactants

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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 Professional Reference Shelf on the CRE Web site for Chapter 3.

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

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 deactivated by collision with another molecule or the activated molecule can decompose to B and C.

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

rA = kACA

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).

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 catalyst that is important, rather than the reactor volume. Consequently, we use Image 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 corresponding rate law is the hydrodemethylation of toluene (T) to form benzene (B) and methane (M) carried out over a solid catalyst

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

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where the prime in Image notes typical units are in per kilogram of catalyst (mol/kg-cat/s), PT, PH2, and PB are partial pressures of toluene, hydrogen, and benzene in (kPa or atm), and KB and KT are the adsorption constants for benzene and toluene respectively, with units of kPa–1 (or atm–1). The specific reaction rate k has units of

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

To express the rate of reaction in terms of concentration rather than partial pressure, we simply substitute for Pi using the ideal gas law

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

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

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 Pi and Image we can write the rate law for the hydromethylation of toluene in terms of concentration and in (mole/dm3) and the rate, –rT in terms of reactor volume, i.e.,

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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 relationship relating the reacting species concentrations at equilibrium. At equilibrium, the rate of reaction is identically zero for all species (i.e., –rA ≡ 0). That is, for the general reaction

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the concentrations at equilibrium are related by the thermodynamic relationship for the equilibrium constant KC (see Appendix C).

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Thermodynamic equilibrium relationship

The units of the thermodynamic equilibrium constant, KC, are (mol/dm3)d + cba.

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 reaction to be elementary and reversible

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or, symbolically,

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

Benzene (B) is being depleted by the forward reaction

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in which the rate of disappearance of benzene is

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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

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For the reverse reaction between diphenyl (D) and hydrogen (H2),

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the rate of formation of benzene is given as

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


Again, both the rate constants kB 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)]

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Net rate

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

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Replacing the ratio of the reverse to forward rate law constants by the reciprocal of the concentration equilibrium constant, KC, we obtain

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where

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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, rD, in terms of the concentrations of hydrogen, H2, diphenyl, D, and benzene, B. The rate of formation of diphenyl, rD, must have the same functional dependence on the reacting species concentrations as does the rate of disappearance of benzene, –rB. The rate of formation of diphenyl is

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Using the relationship given by Equation (3-1) for the general reaction

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Relative rates

we can obtain the relationship between the various specific reaction rates, kB, kD

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

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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

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

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

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Rearranging, we obtain, as expected, the equilibrium expression

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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, ΔCP = 0, the temperature dependence of the concentration equilibrium constant is

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Therefore, if we know the equilibrium constant at one temperature, T1 [i.e., KC(T1)], and the heat of reaction, Image, we can calculate the equilibrium constant at any other temperature T. For endothermic reactions, the equilibrium constant, KC, increases with increasing temperature; for exothermic reactions, KC 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, KC, the reaction behaves as if it were irreversible.

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 molecular 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 kA 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 specific reaction rate, kA, could be correlated by an equation of the type

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Arrhenius equation

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

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.

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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 repulsive 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 discuss 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

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the potential energy surface and the reaction coordinate are shown in Figures 3-1 and 3-2. Here EA, EC, EAB, and EBC are the potential energy surface energies of the reactants A, BC and product molecules (AB and C), and EABC 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.

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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).

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Figure 3-2 Progress along reaction path. (a) Symbolic reaction; (b) Calculated from computational software on the CRE Web site, Chapter 3 Web Module.

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 interacting; RBC 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 potential energy increases up one side of the valley hill (RBC 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., RBC 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(a) 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 reaction 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

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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 minimum energy barrier, EB, shown in Figure 3-2(a). The energy barrier, EB, is related to the activation energy, E. The energy barrier height, EB, can be calculated from differences in the energies of formation of the transition-state molecule and the energy of formation of the reactants; that is,

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The energy of formation of the reactants can be found in the literature or calculated from quantum mechanics, while the energy of formation of the transition 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 approximated by the barrier height, EB, 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 consider 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 distortions, breaking, and forming are identified. This figure shows schematically how the hydrogen molecule, H, moves into CH3–CH3 molecule, distorting the C-H bond and pushing the C-H bond off the methyl hydrogen to arrive at the transition state. As one continues along the reaction coordinate, the methyl hydrogen moves out of the C-H bond into the H-H bond to form the products CH3CH2 and H2.

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Figure 3-3 A diagram of the orbital distortions during the reaction
H• + CH3CH3 → H2 + CH2CH3
The diagram shows only the interaction with the energy state of ethane (the C-H bond). Other molecular orbitals of the ethane also distort. Courtesy of Richard I. Masel, Chemical Kinetics and Catalysis, p. 594 (Wiley, 2001).

Concept 3. Fraction of Molecular Collisions That Have Sufficient Energy to React. Now that we have established a barrier height, we need to know what fraction of molecular collisions have sufficient energy to cross over the barrier and react. To discuss this issue, we consider reactions in the gas phase where the reacting molecules will not have only one velocity, U, but a distribution of velocities, f(U, T). Some will have high velocities and some will have low velocities as they move around and collide. These velocities are not defined with respect to a fixed coordinate system; these velocities are defined with respect to the other reactant molecules. The Maxwell-Boltzmann distribution of relative velocities is given by the probability function, f(U, T)

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We usually interpret f(U, T) with dU, i.e., f(U, T)dU = fraction of reactant molecules with velocities between U and (U + dU).

Rather than using velocities to discuss the fraction of molecules with sufficient energy to cross the barrier, we convert these velocities to energies using the equation for kinetic energy in making this conversion

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Using this substitute, the Maxwell Boltzmann probability distribution of collision with energy e (cal/molecule) at temperature T is

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In terms of energy per mole, E, instead of energy per molecule, e, we have

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where E is in (cal/mol), R is in (cal/mol/K), and f(E, T) is in mol/cal.

This function is plotted for two different temperatures in Figure 3-4(a). The distribution function f(E, T) is most easily interpreted by recognizing that [f(E, T)dE] is the fraction of collision with energies between E and E + dE.

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For example, the fraction of collisions with energies between 0.25 kcal/mol and 0.35 kcal/mol would be

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This fraction is shown by the shaded area in Figure 3-4(a) and is approximated by the average value of f(E,T) at E = 0.3 kcal/mole is 0.81 mol/kcal.

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Figure 3-4(a) Energy distribution of reacting molecules.

Thus 8.1% of the molecular collisions have energies between 0.25 and 0.35 kcal/mol.

We can also determine the fraction of collision that have energies greater than a certain value, EA

Image

This fraction is shown by the shaded area for EA = 2.5 kcal/mole for T = 300 K (heavier shade) and for T = 500K (lighter shade). One can easily see that for T = 500K a greater fraction of the collisions that cross the barrier with EA = 2.5 kcal/mol and react, which is consistent with our observation that the rates of reaction increase with increasing temperature.

For EA > 3RT, we can obtain an analytical approximation for the fraction of molecules of collision with energies greater than EA by combining Equations (3-21) and (3-22) and integrating to get

Image

Figure 3.4(b) shows the fraction of collisions that have energies greater than EA as as function of the choose EA at two different temperatures. One observes for an activation energy EA of 20 kcal/mol and a temperature of 300 K the fraction of collisions with energies greater than 20 kcal/mol is 1.76×10–14 while at 600 K, the fraction increases to 2.39×10–7, which is a 7 orders of magnitude difference.

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Figure 3-4(b) Fraction of collision with energies greater than EA.

To summarize this discussion on the three concepts

Concept 1 The rate increases with increasing reactant concentration,

Concept 2 The rate is related to the potential barrier height and to the conversion of translational energy into potential energy, and

Concept 3 The rate increases with the increasing fraction of collisions that have sufficient energy to cross over the barrier and form products.

To carry this discussion to the next level is beyond the scope of this text as it involves averaging over a collection of pairs of molecules to find an average rate with which they pass over the transition state to become products.

Recapping the last section, the energy of the individual molecules falls within a distribution of energies some molecules have more energy than others. One such distribution is the Boltzmann distribution, shown in Figure 3-4, where f(E, T) is the energy distribution function for the kinetic energies of the reacting molecules. It is interpreted most easily by recognizing the product [f(E, T)dE] as the fraction of molecular collisions that have an energy between E and (E + dE). For example, in Figure 3-4(a), the fraction of collisions that have energies between 0.25 kcal/mol and 0.35 kcal/mol is 0.081, as shown by the shaded area on the left. The activation energy has been equated with a minimum energy that must be possessed by reacting molecules before the reaction will occur. The fraction of the molecular collisions that have an energy EA or greater is shown by the shaded areas at the right in Figure 3-4(a). Figure 3-4(b) shows the fraction of collisions with energies greater than as a function of EA at two different temperatures.

3.3.2 The Arrhenius Plot

Postulation of the Arrhenius equation, Equation (3-18), remains the greatest single advancement in chemical kinetics, and retains its usefulness today, more than a century later. The activation energy, E, is determined experimentally by measuring the reaction rate at several different temperatures. After taking the natural logarithm of Equation (3-18), we obtain

Image
Image

We see that the activation energy can be found from a plot of ln kA as a function of (1/T), which is called an Arrhenius plot. The larger the activation energy, the more temperature-sensitive the reaction. That is, for large E, an increase in just a few degrees in temperature can greatly increase k and thus increase the rate of reaction.

Calculation of the activation energy

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Figure 3-5 Calculation of the activation energy from an Arrhenius plot.


Example 3–1 Determination of the Activation Energy

Calculate the activation energy for the decomposition of benzene diazonium chloride to give chlorobenzene and nitrogen

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using the information in Table E3-1.1 for this first-order reaction.

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TABLE E3-1.1 DATA

Solution

We start by recalling Equation (3-20)

Image

We can use the data in Table E3-1.1 to determine the activation energy, E, and frequency factor, A, in two different ways. One way is to make a semilog plot of k vs. (1/T) and determine E from the slope (–E/R) of an Arrhenius plot. Another way is to use Excel or Polymath to regress the data. The data in Table E3-1.1 was entered in Excel and is shown in Figure E3-1.1, which was then used to obtain Figure E3-1.2.

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Figure E3-1.1 Excel spreadsheet.

Tutorials

A step-by-step tutorial to construct both an Excel and a Polymath spreadsheet is given in the Chapter 3 Summary Notes on the CRE Web site.

The equation for the best fit of the data

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Image

Figure E3-1.2 (a) Excel semilog plot; (b) Excel normal plot after taking ln(k).

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is also shown in Figure E3-1.2(b). From the slope of the line given in Figure 3-1.2(b) and Equation (3-20), we obtain

Image

From Figure E3-1.2(b) and Equation (E3-1.1), we see

ln A = 37.12

Taking the antilog, we find the frequency factor to be

A = 1.32 × 1016 s–1

Image

Analysis: The activation energy, E, and frequency factor, A, can be calculated if we know the specific reaction rate, k, at two temperatures, T1 and T2. We can either use the Arrhenius equation (3-18) twice, once at T1 and once at T2, to solve two equations for the two unknowns, A and E, or we can take the slope of a plot of (ln k) as a function of (1/T); the slope will be equal to (–E/R).


The rate does not always double for a temperature increase of 10°C.

Image

There is a rule of thumb that states that the rate of reaction doubles for every 10°C increase in temperature. However, this rule is true only for specific combinations of activation energies and temperatures. For example, if the activation energy is 53.6 kJ/mol, the rate will double only if the temperature is raised from 300 K to 310 K. If the activation energy is 147 kJ/mol, the rule will be valid only if the temperature is raised from 500 K to 510 K (see Problem P3-7B for the derivation of this relationship).

Image

The larger the activation energy, the more temperature-sensitive is the rate of reaction. While there are no typical values of the frequency factor and activation energy for a first-order gas-phase reaction, if one were forced to make a guess, values of A and E might be 1013 s–1 and 100 kJ/mol. However, for families of reactions (e.g., halogenation), a number of correlations can be used to estimate the activation energy. One such correlation is the Polanyi-Semenov equation, which relates activation energy to the heat of reaction (see Professional Reference Shelf R3.1). Another correlation relates the activation energy to differences in bond strengths between products and reactants.4 While the activation energy cannot be currently predicted a priori, significant research efforts are under way to calculate activation energies from first principles.5

4 M. Boudart, Kinetics of Chemical Processes (Upper Saddle River, NJ: Prentice Hall, 1968), p. 168. J. W. Moore and R. G. Pearson, Kinetics and Mechanisms, 3rd ed. (New York: Wiley, 1981), p. 199. S. W. Benson, Thermochemical Kinetics, 2nd ed. (New York: Wiley, 1976).

5 R. Masel, Chemical Kinetics and Catalysis, New York: Wiley, 2001, p. 594.

One final comment on the Arrhenius equation, Equation (3-18). It can be put in a most useful form by finding the specific reaction rate at a temperature T0; that is,

k(T0) = AeE/RT0

and at a temperature T

k(T) = AeE/RT

and taking the ratio to obtain

Image

A most useful form of k(T)

This equation says that if we know the specific reaction rate k(T0) at a temperature, T0, and we know the activation energy, E, we can find the specific reaction rate k(T) at any other temperature, T, for that reaction.

3.4 Present Status of Our Approach to Reactor Sizing and Design

In Chapter 2, we combined the different reactor mole balances with the definition of conversion to arrive at the design equation for each of four types of reactors, as shown in Table 3-2. Next we showed that if the rate of disappearance is known as a function of the conversion X

rA = g(X)

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Where are we?

then it is possible to size CSTRs, PFRs, and PBRs operated at the same conditions under which –rA = g(X) was obtained.

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TABLE 3-2 DESIGN EQUATIONS

The design equations

In general, information in the form –rA = g(X) is not available. However, we have seen in Section 3.2 that the rate of disappearance of A, –rA, is normally expressed in terms of the concentration of the reacting species. This functionality

Image

is called a rate law. In Chapter 4, we show how the concentration of the reacting species may be written in terms of the conversion X

Image

Image
and then we can design isothermal reactors


With these additional relationships, one observes that if the rate law is given and the concentrations can be expressed as a function of conversion, then in fact we haverA as a function of X and this is all that is needed to evaluate the isothermal design equations. One can use either the numerical techniques described in Chapter 2 or, as we shall see in Chapter 5, a table of integrals, and/or software programs (e.g., Polymath).


The CRE Algorithm

• Mole Balance, Ch 1

• Rate Law, Ch 3

• Stoichiometry, Ch 4

• Combine, Ch 5

• Evaluate, Ch 5

• Energy Balance, Ch 11


Summary

1. Relative rates of reaction for the generic reaction:

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The relative rates of reaction can be written either as

Image

2. Reaction order is determined from experimental observation:

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The reaction in Equation (S3-3) is α order with respect to species A and β order with respect to species B, whereas the overall order, n, is (α + β). If α = 1 and β = 2, we would say that the reaction is first order with respect to A, second order with respect to B, and overall third order. We say a reaction follows an elementary rate law if the reaction orders agree with the stoichiometric coefficients for the reaction as written.

Examples of reactions that follow an elementary rate law:

Irreversible reactions

Image

Examples of reactions that follow nonelementary rate laws:

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3. The temperature dependence of a specific reaction rate is given by the Arrhenius equation

Image

where A is the frequency factor and E the activation energy.

If we know the specific reaction rate, k, at a temperature, T0, and the activation energy, we can find k at any temperature, T

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Concept 1 The rate increases with increasing reactant concentrations,

Concept 2 The rate is related to the potential barrier height and to the conversion of translational energy into potential energy, and

Concept 3 The rate increases as the fraction of collisions that have sufficient energy to cross over the barrier and form products increases.

Similarly from Appendix C, Equation (C-9), if we know the partial pressure equilibrium constant KP at a temperature, T1, and the heat of reaction, we can find the equilibrium constant at any other temperature

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CRE Web Site Materials

Expanded Material

1. Collision Theory

2. Transition State

3. Molecular Simulation

4. How to Estimate the Activation Energy from the Polanyi Equation

5. Web P3-1A Puzzle ProblemWhat’s Wrong with this Solution?”

6. Additional Homework Problems

Learning Resources

1. Summary Notes for Chapter 3

2. Web Modules

A. Cooking a Potato

Chemical reaction engineering is applied to cooking a potato

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with

k = AeE/RT

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B. Molecular Reaction Engineering

Molecular simulators (e.g., Gaussian) are used to make predictions of the activation energy.

FAQs (Frequently Asked Questions)

Professional Reference Shelf

R3.1. Collision Theory

In this section, the fundamentals of collision theory

Image

Figure R3.1 Schematic of collision cross section.

are applied to the reaction

A + B → C + D

to arrive at the following rate law

Image

The activation energy, EA, can be estimated from the Polanyi equation

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R3.2. Transition-State Theory

In this section, the rate law and rate-law parameters are derived for the reaction

A + BC Image ABC# → AB + C

using transition-state theory. Figure PW3B-1 shows the energy of the molecules along the reaction coordinate, which measures the progress of the reaction.

Image

Figure PW3B-1 Reaction coordinate for (a) SN2 reaction, and (b) generalized reaction. (c) 3-D energy surface for generalized reaction.

R3.3. Molecular Dynamics Simulations

The reaction trajectories are calculated to determine the reaction cross section of the reacting molecules. The reaction probability is found by counting up the number of reactive trajectories after Karplus.6

6 M. Karplus, R.N. Porter, and R.D. Sharma, J. Chem. Phys., 43 (9), 3259 (1965).

Nonreactive Trajectory

Image

Reactive Trajectory

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Questions and Problems

The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.

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Questions

Q3-1c

(a) List the important concepts that you learned from this chapter. What concepts are you not clear about?

(b) Explain the strategy to evaluate reactor design equations and how this chapter expands on Chapter 2.

(c) Choose a FAQ from Chapters 1 through 3 and say why it was the most helpful.

(d) Listen to the audios Image on the CRE Web site. Select a topic and explain it.

(e) Read through the Self-Tests and Self-Assessments in Chapters 1 through 3 in the Summary Notes on the CRE Web site. Select one and critique it.

(f) Which example on the CRE Web site’s Summary Notes for Chapters 1 through 3 was most helpful?

Problems

P3-1A

(a) Example 3-1. Make a plot of k versus T and ln k versus (1/T) for E = 240 kJ/mol and for E = 60 kJ/mol. (1) Write a couple of sentences describing what you find. (2) Next, write a paragraph describing the activation, how it affects chemical reaction rates, and what its origins are.

(b) Collision Theory—Professional Reference Shelf. Make an outline of the steps that were used to derive

rA = AeE/RTCACB

(c) The rate law for the reaction (2A + B → C) is Image with kA = 25(dm3/mol)2/s. What are kB and kC?

P3-2B Molecular collision energies—refer to Figure 3-3.

(a) What fraction of molecular collisions have energies less than or equal to 2.5 kcal at 300 K? At 500 K?

(b) What fraction of molecular collisions have energies between 3 and 4 kcal/mol at T = 300 K? At T = 500 K?

(c) What fraction of molecular collisions have energies greater than the activation energy EA = 25 kcal at T = 300 K? At T = 500K?

P3-3B

(a) Use Figure 3-1(b) to sketch the trajectory over the saddle point when the BC and AB molecules vibrate with the minimum separation distance being 0.20 Angstroms and the maximum separation being 0.4 Angstroms.

(b) At point Y, RAB = 2.8 Angstroms, sketch the potential energy as a function of the distance RBC noting the minimum at valley floor.

(c) At Point X, RBC = 2.8 Angstroms, sketch the potential energy as a function of RAB noting the minimum on the valley floor.

P3-4B Use Equation (3-20) to make a plot of f(E,T) as a function of E for T = 300, 500, 800, and 1200 K.

(a) What is the fraction of molecules that have sufficient energy to pass over a energy barrier of 25 kcal at 300, 500, 800, and 1200 K?

(b) For a temperature of 300 K, what is the ratio of the faction of energies between 15 and 25 kcal to the same energy range (15-25 kcal) at 1200 K?

(c) Make a plot of f(E > EA, T) as a function of (EA/RT) for (EA/RT) > 3. What is the fraction of collisions that have energies greater than 25 kcal/mole at 700 K?

(d) What fraction of molecules have energies greater than 15 kcal/mol at 500 K?

(e) Construct a plot of f(E > EA, T) versus T for EA = 3, 10, 25, and 40 kcal/mole. Describe what you find.

P3-5A The following figures show the energy distribution function at 300 K for the reaction A + B → C

Image

Figure P3-5(a)

Image

Figure P3-5(b)

For each figure, determine the following:

(a) What fraction of the collisions have energies between 3 and 5 kcal/mol?

(b) What fraction of collisions have energies greater than 5 kcal/mol? (Ans.: Figure P3-5 (b) fraction = 0.28)

(c) What is the fraction with energies greater than 0 kcal/mol?

(d) What is the fraction with energies 8 kcal/mol or greater?

(e) If the activation energy for Figure P3-5(b) is 8 kcal/mol, what fraction of molecules have an energy greater than EA?

(f) Guess what (sketch) the shape of the curve f(E, T) versus E shown in Figure P3-5(b) would look like if the temperature were increased to 400 K. (Remember: Image)

P3-6A The frequency of fireflies flashing and the frequency of crickets chirping as a function of temperature follow. Source: Keith J. Laidler, “Unconventional applications of the Arrhenius law.” J. Chem. Educ., 5, 343 (1972). Copyright (c) 1972, American Chemical Society. Reprinted by permission.

Image
Image

The running speed of ants and the flight speed of honeybees as a function of temperature are given below. Source: B. Heinrich, The Hot-Blooded Insects (Cambridge, MA: Harvard University Press, 1993).

Image

(a) What do the firefly and cricket have in common? What are their differences?

(b) What is the velocity of the honeybee at 40°C? At–5°C?

(c) Nicolas wants to know if the bees, ants, crickets, and fireflies have anything in common. If so, what is it? You may also do a pair-wise comparison.

(d) Would more data help clarify the relationships among frequency, speed, and temperature? If so, in what temperature should the data be obtained? Pick an insect, and explain how you would carry out the experiment to obtain more data. For an alternative to this problem, see CDP3-AB.

(e) Data on the tenebrionid beetle whose body mass is 3.3 g show that it can push a 35-g ball of dung at 6.5 cm/s at 27° C, 13 cm/s at 37° C, and 18 cm/s at 40° C.

(1) How fast can it push dung at 41.5° C? Source: B. Heinrich. The Hot-Blooded Insects (Cambridge, MA: Harvard University Press, 1993). (Ans.: 19.2 cm/s at 41.5° C)

Image

(2) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.

P3-7B Troubleshooting. Corrosion of high-nickel stainless steel plates was found to occur in a distillation column used at DuPont to separate HCN and water. Sulfuric acid is always added at the top of the column to prevent polymerization of HCN. Water collects at the bottom of the column and HCN collects at the top. The amount of corrosion on each tray is shown in Figure P3-5B as a function of plate location in the column.

Image

Figure P3-5B Corrosion in a distillation column.

The bottom-most temperature of the column is approximately 125°C and the topmost is 100°C. The corrosion rate is a function of temperature and the concentration of an HCN–H2SO4 complex. Suggest an explanation for the observed corrosion-plate profile in the column. What effect would the column operating conditions have on the corrosion-plate profile?

P3-8B Inspector Sgt.Ambercromby of Scotland Yard. It is believed, although never proven, that Bonnie murdered her first husband, Lefty, by poisoning the tepid brandy they drank together on their first anniversary. Lefty was unaware she had coated her glass with an antidote before she filled both glasses with the poisoned brandy. Bonnie married her second husband, Clyde, and some years later when she had tired of him, she called him one day to tell him of her new promotion at work and to suggest that they celebrate with a glass of brandy that evening. She had the fatal end in mind for Clyde. However, Clyde suggested that instead of brandy, they celebrate with ice-cold Russian vodka and they down it Cossack style, in one gulp. She agreed and decided to follow her previously successful plan and to put the poison in the vodka and the antidote in her glass. The next day, both were found dead. Sgt. Ambercromby arrives. What are the first three questions he asks? What are two possible explanations? Based on what you learned from this chapter, what do you feel Sgt. Ambercromby suggested as the most logical explanation?

Source: Professor Flavio Marin Flores, ITESM, Monterrey, Mexico. [Hint: View the YouTube video (www.youtube.com) made by the chemical reaction engineering students at the University of Alabama, entitled The Black Widow. You can access it from the CRE Web site (www.umich.edu/~elements/5e) using the YouTube Videos link under By Concepts on the home page.]

P3-9B Activation Energy

(a) The rule of thumb that the rate of reaction doubles for a 10°C increase in temperature occurs only at a specific temperature for a given activation energy. Develop a relationship between the temperature and activation energy for which the rule of thumb holds. Neglect any variations in concentrations.

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(b) Write a paragraph explaining activation energy, E, and how it affects the chemical reaction rate. Refer to Section 3.3 and especially the Professional Reference Shelf sections R3.1, R3.2, and R3.3 if necessary.

P3-10C The initial reaction rate for the elementary reaction

2A + B → 4C

was measured as a function of temperature when the concentration of A was 2 M and that of B was 1.5 M.

Image

(a) What is the activation energy?

(b) What is the frequency factor?

(c) What is the rate constant as a function of temperature using T = 300K as the base case?

P3-11B Determine the rate law for the reaction described in each of the cases below involving species A, B, and C. The rate laws should be elementary as written for reactions that are either of the form A → B or A + B → C.

(a) The units of the specific reaction rate are Image: Rate Law _______

(b) The units of the specific reaction rate are Image: Rate Law _______

(c) The units of the specific reaction rate are Image: Rate Law ________

(d) The units of a nonelementary reaction rate are Image: Rate Law ________

P3-12A

(a) Write the rate law for the following reactions assuming each reaction follows an elementary rate law. Give the units of kA for each, keeping in mind some are homogeneous and some reactants are heterogeneous.

Image

P3-13A

(a) Write the rate law for the reaction

2A + B → C

if the reaction

(1) is second order in B and overall third order, –rA = ________

(2) is zero order in A and first order in B, –rA = ________

(3) is zero order in both A and B, –rA = ________

(4) is first order in A and overall zero order. –rA = ________

(b) Find and write the rate laws for the following reactions

(1) H2 + Br2 → 2HBr

(2) H2 + I2 → 2HI

Image

P3-14B The rate laws for each of the reactions listed below were obtained at low temperatures. The reactions are highly exothermic and therefore reversible at high temperatures. Suggest a rate law for each of the reactions [(a), (b), and (c)] at high temperatures, which may or may not be elementary.

(a) The reaction

A → B

is irreversible at low temperatures and the rate law is

rA = kCA

(b) The reaction

A + 2B → 2D

is irreversible at low temperatures and the rate law is

rA = kCA1/2CB

(c) The gas-solid catalyzed reaction

Image

is irreversible at low temperatures and the rate law is

Image

In each case, make sure that the rate laws at high temperatures are thermodynamically consistant at equilibrium (cf. Appendix C).

P3-15B Solar Chemical Energy Storage. The major ways to utilize, capture, or store the sun’s energy are solar thermal (see P8-16B), solar voltaic, biomass conversion, solar water splitting (P10-13B), and solar chemical. Solar chemical refers to processes that harness and store solar energy by adsorbing light in a reversible chemical reaction; see http://en.wikipedia.org/wiki/Solar_chemical. For example, the photodimerization of anthracene absorbs and stores solar energy that can be released when the reverse reaction takes place.

Image

Figure P3-15.1 Anthracene dimerization.

Another reaction of interest is the Norbonadiene-Quadricyclane (NQ) couple, where solar energy is adsorbed and stored in one direction and released in the other.

Image

Figure P3-15.2 The Norbonadiene-Quadricyclane (NQ) couple is of potential interest for sunlight energy storage.

(a) Suggest a rate law for the reversible photodimerization of anthracene.

(b) Suggest a rate law for the reversible storage of energy by the NQ couple.

Supplementary Reading

1. Two references relating to the discussion of activation energy have already been cited in this chapter. Activation energy is usually discussed in terms of either collision theory or transition-state theory. A concise and readable account of these two theories may or may not be found in

BURGESS, THORNTON W., The Adventures of Reddy Fox. New York: Dover Publications, Inc., 1913.

LAIDLER, K. J. Chemical Kinetics. New York: Harper & Row, 1987, Chap. 3.

MASEL, R., Chemical Kinetics and Catalysis. New York: Wiley, 2001, p. 594.

2. Prof. Dr. Sven Köttlov of Jofostan University has pointed out that in addition to the books listed above, the rate laws and activation energies can be found in NBS circulars, such as those shown below Table 3-1 on page 75. Also check chemistry literature, e.g., Journal of Physical Chemistry.

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