Rate Laws and Stoichiometry - ivut.iut.ac.ir · An irreversible reaction is one that proceeds i ......

Rate Laws and Stoichiometry

Success is measured not so much by the position one has reached En life. as by the obstacles one has overcome while q i n g to succeed.

Booker T. Washington

Oveniew. In Chapter 2, we showed that i f we had the rate of reaction as a function of conversion, - r ~ = fo, we could calculate reactor volumes necessary to achieve a specified conversion for flow systems and the time to achieve a given conversion in a batch system. Unfortunately, one is seldom, if ever, given - r ~ = XX) directly fram raw data. Not to fear, in this chapter we will show how to obtain the rate of reaction as a function of conversion. This relationship between reaction rate and conversion will be obtained in two steps. In Step 1, Part 1 of this chapter, we define the rate law, which relates the rate of reaction to the concentrations of the reacting species and to temperature. In Step 2, Par? 2 of this chapter, we define concentrations for flow and batch systems and develop a stoichiometric table so that one can write concentrations as a function of conversion. Combining Steps 1 and 2, we see that one can then write the rate as a function conversion and use the techniques in Chapter 2 to design r e a c h systems.

After completing this chapter. you will be abIe to write the rate of reaction as a function of conversion for both liquid-phase and gas-phase reacting systems,

Rate Laws and Storchiometry Chap.

PART 1 RATE LAWS

3.1 Basic Definitions

A homogeneous reacriotl is one that involves only one phase. A heterogeneou reaction involves more than one phase, and the reaction usually occurs at zh interface between the phases. An irreversible reaction is one that proceeds i only one direction and continues in that direction until the reactants ar

3 p e s exhausted. A ~ver.s ihle seacrion, on the other hand, can proceed in eithe direction, depending on the concentrations of reactants and products relative ts the corresponding equilibrium concentrations. An irreversible reaction behave as if no equilibrium condition exists. Strictly speaking, no chemical reaction i cornpleteEy irreversible. However, for many reactions, the equilibrium poin lies so far to the product side that these reactions are treated as i~evers ib l~ reactions.

The molecularity of a reaction is the number of atoms, ions. or molecule involved (colliding) in a reaction step. The terms unirnolecular, bimoteculat and termoleculnr refer to reactions involving, respectively. one. two, or thre~ atoms (or molecules) interacting or colliding in any one reaction step. Thl most common example of a unimoleculnr reaction is radioactive decay, such a, the spontaneous emission of an alpha parlicIe from uranium-238 to give tho rium and helium:

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

The true bimoleculnr reactions that exist are reactions involving free radical! such as

Br + C,H,-+ HBr +C,H,

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

The probability of a remolecular reaction occurring is almost nonexistent, and in most instances the reaction pathway foIlows a series of bimolecular reactions as in the case of the reaction

The reaction pathway for this "Hall of Fame" reaction is quite interesting and i s discussed in Chapter 7 along with similar reactions that form active intermediate complexes in their reaction pathways.

Sec. 3.1 Basic Definitions

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 stoichiometric coefficients. For Reaction (2-2),

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

Rate of formation of C = C (Rate of disappearance of A) CI

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

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

for which

Reaction or stoicbiometry

For example, in the reaction

we have

If NO? is being formed at a rate of 4 mol/m3/s. i.e.,

rNo2 = 4 mol/m3is

Rate Laws and Stoichiornetry Chap. 3

then the rate of formation of NO is

-2 ZNO + O2 + 2N02 rNo = - rNO2 = -4 rno~rn~is

r N ~ = 4 moVm3/s 2 -r,, = 4 moVm31s -ro, = 2 rnol1~15 the rate of disappearance of NO is

and the rate of disappearance of oxygen, Q2, is

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, -r,, depends on temperature and composition. For many reactions. it can be written as the product of a reaction rcrrP co~rstant k, and a function of the concentrations (activities) of the various species involved in the reaction:

The rate law The algebraic equation that relates -r, to the species concentrations is called gives the r e i a t ~ ~ n -

chip the kinetic expression or rate law. The specific rate of reaction (also called the tlon rate and sale constant). kA, like the reaction rate -r,, always refers to a particular spe-

concentratio*. cies in the reaction and normally should be subscripted with respect to that species. However. for reactions in which the stoichiornetric coefficient is I for all species involved in the reaction. for example.

INaOH + IHCl +J INaCI + 1 N,O

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

3.2.1 Power t aw Models And Elementary Rate Laws

The dependence of €he reaction rate. -r,. on the concentrations of the species present. fn(C,), is almost without exception determined by experimental observation. Although the functional dependence on concentration may be postu- Iated from ~heory, experiments are necerrary to confirm the proposed fornl. One of the most common general forms of this dependence is the power law model. Here the rate law 1s the product of concerlrrations of the individual reacting species. each of which is raised to a power. for example.

Sec. 3.2 The Reaciion Order and the Rate t aw 83

0-31

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.' In Equation (3-31, the reaction is a order with respect to reactant A. and order with respect to relzcranr B. The overall order of the reaction, n, is

Overall reaction n=a+/3

nrder The units af -r, tire always in terms of concentration per uni t time while

the units of the specific reaction rate, k,, will vary with the order of the reaction. Consider a reaction involving only one reactant, such as

A + Products

with a reaction order n. The units of the specific reaction rate constant are

k = on cent ration)^ - " Time

Consequently, the rate laws corresponding to a zero-, first-, second-, and third-order reaction, together with typical units for the corresponding rate constants, are:

( k ) = mol/dm3. s

First-order (n = 1 ): -rA = kACA:

( k ) = s - I (3-5)

( k ) = dm?mol s (3 -6)

I Strictly speaking, the reaction rates should be written in terms of the activities, u,, (a, = y,C,. where y, 1s the activity cwffic~ent). Kline and FogIer, ICIS, 82. 93 (19811: ihid., p. 103: and Ind. Eug. CChern hndompnrols 20, 1 55 ( 198 1 ).

r a p - r A = kAaAaR

However, for many rcacting sysrerns, the actlvlry cocficients, y,. do not change appreciably during Ihe course nf the reaclion, and they are adsorbed into the specific reaction rate:

84 Rate Laws and Stoichiometry Chao

{k) = (drn?lmol)'.s-' (3-7

An elernaltny rtaotioa is one that evolves a single step such as t h ~ bimoIecuIar reaction between oxygen and methanol

Om+ CH,OH+CH,O + OH*

The stoichiometric coefficients in this reaction are identic.nl to the powers ir the rate law. Consequently, the rate law for the disappearance of rnolecula oxygen is

The reaction is first order in molecular oxygen and first order in methanol therefore. we say both the reaction and the rate law are elementary. This fom of the rate law can be derived from Colli.~iiot~ T h e o ~ as shown in the Profes. sion Reference Shelf 3A on the CD-ROM. There are many reactions where the

Reference Shelf stoichiornetric coefficients in the reaction are identical to the reaction orders but the reactions are not elementary owing to such things as parhways involv. ing active intermediates and series reactions. For these reactions that are no1

Collision theory elementary but whose stoichiometric coefficients are identical to the reaction orders in the rate law, we say the reactionfollows nn elernenran. rote Eaw. For example, the oxidation reaction of nitric oxide discussed earlier.

is not elementary but follows the elementary rate taw

Note: the rate constant, k, is defined

with respecr to NO. Another nonelementary reaction that follows an elementary rate law is the gas-phase reaction between hydrogen and iodine

with

In summary, far many reactions involving multiple steps and pathways. the powers in the rate laws surprisingly agree with the stoichiometric coefficients. Conse- quently, to facilitate describing this class of reactions, we say a reaction f01lm.s an elementary rate law when the reaction orders are identical with the stoichiometric coeficients of the reacting species for the reaction ns written. It is i m p o m t to remember that the tate laws are determined by experimenhf observation! 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.

SRC 3 2 The Reactloo Order and !he Rate Law 85

The values of specific reaction rates For these and a number of other reactions can be found in the Dcltc~ Bn.w found on the CD-ROM and on the web.

{Vhere do you tind The activatron energy, frequency factor, and reaction orders for a large rate lawr ' number of gas- and liquid-phase reactions can be: found in the National Bureau

of Standards' circulars and supplements.? Also consult the journals listed at rhe end of Chapter I .

A. First-Order Rate Laws

B. Second-Order Rate Laws

' See Problem P3-13, and Section 9.2.

important ref- Kinetic data for larger number of reactions can be obtained on floppy disks and you CD-ROMs provided by Nntionnl lnstitlrre of Standanis and Technology (NIST). Stan-

should also look rn the other literature dard Reference Data 22t/A320 Gaithersburg, MD 20899; phone: (301) 975-2208.

hcfom going to Additional sources are Tables of Chemical Kinetics: I?'omogeneous Reacrions, the lab National Bureau of Standards CircuIar 510 (Sept. 28, 1951); SuppI. 1 (Nov. 14.

1956); Suppl. 2 (Aug. 5 , I960): Suppi. 3 ISept. 15, 1961) (Washington, D.C.: U.S. Government Printing Office). Cl~emical Kinetics and Photochemicnl Dnta for Use m Stratospheric Modeling, Evaluate No. 10. JPL Pubtication 92-20 (Pasadena, Calif.: Jet Propulsion Laboratories, Aug. €5. 1992).

86 Rate laws and Stoichiometry Chap. 3

C. Nonefernenbry Rate Laws

Cumene (C) j Benzene (8) + Pmpylene (P)

13. Enzymatic Reactions (Urea (U) + Urease (E))

+H7O NH,CONH2 + Urease 4 2NHJ + C02 + Urease

E. Biomass Reactions

Substrare (S) + CelIs 1C) + More Cells + Product

NO!?: The rate constant, k. and activation energies for a number of the reactions in these exam- pIes are given in the Dara Base on the CD-ROM and Summary Notes.

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 Iaws 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,

in which the kinetic rare Inw is

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

Sec. 3.2 The Reachon Order and the Rate Law 87

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 rare law is

Both kNlo and k' are strongly temperature-dependent. When a rare expression such as ihe one given above occurs, we cannot state an overaIl reaction order. Here we can only speak of reaction orders under cestain limiting conditions. For example, at very low concentrations of oxygen, the second term in the denominator would be negligible WIT 1 (1 >> k'Co, ), and the reaction would

Apparent reaction be "apparent" first order with respect to nitrous oxide and first order overall. orden 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. k'Co (ktCO7 >> I), the apparent reaction order would be -1 with respect to oxvE& and first order with respect to nitrous oxide an o\,eralI apparenr zero

lmponant resources order. Rate expressions of this type are very common for liquid and gaseous for m e laws reactions promoted by solid catalysts (see Chapter 10). They also wcor in

homogeneous reaction systems with reactive intermediates (see Chapter 7). 11 is interesting to note that although the reaction orders often correspond

to the stoichiometric coefficients as evidenced for the reaction between hydrogen and iodine, the rate expression for the reaction between hydrogen and another halogen, bromine. is quite complex. This nonefernentary reaction

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

In Chapter 7. we will discuss reaction mechanisms and pathways that lead to nonelementary rate laws such as rate of formation of HBr shown in Equation (3-8).

Heterogeneous Reactions In many pas-solid catalyzed reactions. it histori- cally has been the practice to write the rate law in terms of partial pressures rather than concentrations. An example of a heterogeneous reaction and corresponding rate law is the hydrodcmethylation of toluene (T) to form benzene ( B j and methane (MI carried out over a solid catalyst.

88 Rate Laws and Stoichiometry Chap.

Themcdyriarnic Equilibrium

Relationship

The rate of disappearance of toluene per mass of catalyst, - r ' , , follou Langmuir-Hinshelwood kinetics (Chapter 10). and the rate law was foun experimentally to be

where Kg and KT are the adsorption constants with units of kPa-I (or atm-I and the specific reaction rate has units of

[kl = mol toluene

kg cat - s . k ~ a '

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

The rate of reaction per unit weight catalyst, -rA, (e,g., -r;), and [hi rate of reaction per unit volume, - s, , are related through the bulk density p, (mass of solidlvolume) of the cafalyst particles in the fluid media:

moles = ( mass ) ( moles ) - time. volume volume trme . mass

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

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

3.2.3 Reversible Reactions

A11 rate laws for reversible reactions must reduce to the thermodynamic refa- tionship relating the reacting species concentrations at equilibrium. At equilibrium, the rate of reaction is identically zero for all species (i-e., - r , = O ). That is, for the general reaction

the concentrations ax equilibrium are related by the thermodynamic relationship for the equilibrium constant Kc (see Appendix CI.

Sec. 3.2 The Reaction Order and the Ra:e Law 89

The units of the thermcdynamic equilibrium constant. Kc. are Kc, are ( rn~l /drn?)~ ' ( - h - " ,

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 efementary and reversible:

or, symbolically,

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

defined wn a particurar species. Benzene (B) is being depleted by the forward reaction

2C6H6 ---% CIIHIo+H,

in which the rate of disappearance of benzene is

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

r~,forward = -'BcR (3-1 1)

For the reverse reaction between diphenyl (D) and hydrogen (Hz ).

the rate of formation of benzene is given as

Again, both the rale constants k, and k-, are defined with respect to bencene!!! The net rate of formation of benzene is the sum of the rates of formation

from the forward reactiorl [i.e.. Equation (3- 1 I)] and the reverse reaction [inen, Equation [3-12)]:

" TB, ncr = 'B. fo-d + 'B, reverse

90 Rate Laws and Stoichiometry Chap. 3

EIementary reversi hle

A * B

Multiplying both sides of Equation (3-13) by -1. we obtain the rate law for the rate of disappearance of benzene, -r , :

Replacing the ratio of the reverse to forward rare law constants by the equilibrium constant, we obtain

where

k~ = Kc = Concentration equilibrium constant - k-B

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

This is just stoichiornetn:

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

we can obtain the relationship between the various specific reaction rates. k,, k , :

Comparing Equations (3- 15 ) and (3-1 6). we see the relationship between the specrfic reaction sate with respect to diphenyl and the specific reaction rate with respect to benzene is

Sec. 3.3 The Reaction Rate Constant 91

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

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

At equilibrium- the Now let's look at the rate law. At equilibrium, - r ~ r 0, and the rate law given rate law must reduce

to an equation by Equation (3-1 4) becomes consistent wth

thermodynamic equilibrium.

Rearranging, we obtain, as expected, the equilibrium expression

which is identical to Equation 13-17) 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 beat capacity term, ACp = 0 the Cndolhrnk

I temperature dependence of the concentration equilibrium constant is

Kc(T) = K,(T,) exp - - - - [A?(;, :)I Therefore, if we know the equilibrium constant at one temperature, T, [i.e., Kc (T,)], and the heat of reaction, AHRn, we: can calculate the equilibrium constant at any other temperature T For endothermic reactions, the equilibrium constant, Kc, increases with increasing temperacure: for exothermic reactions, Kc

T decreases with increasing temperature. A further discussion of the equilibrium constant and its themlodynamic relationship is given i n Appendix C.

3.3 The Reaction Rate Constant

The reaction rate constant k is not truly a constant: i t is merely independent of the concentrations of the species involved En 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 depends on wherher or not a catalyst is present, and in gas-phase reactions, it may be a function of total pressure. ln liquid systems i t can also be a function of other parameters, . filch as ionic strengrh and choice of solvent. These other variables normally exhibit much Iess effect on the specific reaction rate than temperature does with the exception of supercritical solvents, such as super critical water.


Arrhenius equation

Consequently. for the pilrposes of the material presented here. it will L. assumed that A-, depends only on temperature. This assumption is valid in rno laboratory and industrial reactions and seems to work quite well.

Ir was the great Swedish chemist Arrhenius who first suggested that tl. temperature dependence of the specific reaction rate, kA, could be correlated b an equation of the type

where A = preexponential factor or frequency factor E = activation energy. J/mol or callmol R = gas constant = 8.3 14 Jlmol - K = 1.987 cailmol - K T = absolute temperature, K

Equation (3-1 8), known as the Arrhetlius eylrution, has been verified empir caily to give the rernperature behavior of most reaction rate constants withi

T(K) experimental accuracy over fairly large temperature ranges. The Arrheniu equation is derived in the Professional Reference Shelf 3.A: Colli.rion Theor on the CD-ROM.

Why is there nn activation energy? If the reactants are free radicals th essentially react imtnediately on collision. there usually isn't an activatio energy. However. for rnort atoms and moleculer undergoing reaction, there i 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 the break them and thus form new bonds.

2. The steric and electron repulsion forces must be overcome as th reacting motecuEes come close together.

The activation energy can be thought of as a barrier to energy transfe (from the kinetic energy to the potential energy) between reacting molecule that must be overcome. One way to view the barrier to a reaction is throug the use of the relaction coordinates. These coordinates denote the potenti: energy of the system as a function of the progress along the reaction path a we go from reactants to an intermediate to products. For the reaction

A + E C A - B - C + A B + C

the reaction coordinate is shown in Figure 3-1. Figure 3- I(a) shows the potential energy of the three atom (or molecule

system. A, 8. and C, as well as the reaction progress as we go from reactar specie5 A and BC to products AB and C. Initially A and BC are far apart an1 the system energy is just the bond energy BC. At the end of the reaction, th products AB and C are far apart, and the system energy is the bond energy AE As we move along the reaction coordinate (x-axis) to the right in Figure 3-l(a' the reactants A and BC approach each other, the BC bond begins to break. an1 the energy of the reaction pair increases vntil the top of the barrier is reachec At the top, the transition srure is reached where the intermolecular distance between AB and between BC are equaI (i.e., A-B-C). As a result. the potenria energy of the initial three atoms (molecules) is high. As the reaction proceed

Sec. 3.3 The Reaction Rate Constant 93

I reaaants pradtists

Reaction wordinate 1.3 2.1 23 2.5 2 7

CH3 -I Bond Distance in Angstroms (b)

Figure 3-1 Progress along reaction path, la) Syrribolic reaction: Ib) Calculnted from computationaI ~oftwnre on the CD-RDSI Chapter 3 14'eb Modutc

further. the distance between A and B decreases, and the AB bond begins to form. As we proceed further, the distance between AB and C increases and the energy of the reacting pair decreases to that of the AB bond energy. The calculations to arrive at Figure 3-l(b) are discussed in the CD-ROM web module, and transition state theory is discussed i n the CD-ROM ProfessionaI Reference Shelf R3.2 Transition State Theory.

We see that for the reaction to occur, the reactants must overcame an energy barrier, Es, shown in Figure 3-1. The energy barrier, EB, is related to the activation energy, E. The energy barrier height, EB, can be catcutated from differences in the energies of formation of the transition state molecule and the energy of formation oT the reactants. that is, -.

(3- I 9)

The energy of formation of the reactants can be found in the literarure while the energy of formation of transition state can be calculated using a number of software programs such as CACHE, Spartan. or Cerius2. The activation energy. EA, is often approximated by the barrier hei%ht, which is a good approximation in the absence as quantwm mechanical tunneling.

Kow that we have the general idea for a reaction coordinate ler's consider another real reaction system:

H + C2Hb HI + C2Hs The energy-reaction coordinate diagram for the reaction between a hydrogen atom and an ethane molecule is shown in Figure 3.2 where the bond distor- -. -

Rzfeference Chef lions, breaking, and forming are identified. One can also view the activation energy in terms of collision theory (Pro-

fessional Reference Shelf R3.1). By increasing the temperature, we increase the kinetic energy of the reactant molecules. This kinetic energy can in turn be transfemd through molecular collisions to internal energy to increase the stretching and bending of rhe bonds. causing them ro reach an activated state, vulnerable to bond breaking and reaction (cf. Figures 3-1 and 3-2).

Rate Laws and Stoichiometrl, Chap. 3

Figure 3-2 A diagram of the orbital dtstortions during the reaction H + CH3CH3 + Hz + CHICH3

The d i a p m shows only the interaction w ~ t h the energy state of ethane (the C-H bond). Other molecular orbitals of the ethane also dirton. [Courtesy of R. Masel. Clzemical Kinetics (McGraw Hill, 2002). p. 594.1

The energy of the individual molecules falls within a distribu~ion of energies where some molecules have mare energy than others. One such distribution is shown in Figure 3-3 whereJE,T) is the energy distribution function for the kinetic energies of the reacting molecules. It is interpreted most easily by recognizing ($. d Q as the fraction of molecules that have an energy 'between E and (E + dm. The activation energy has been equated with a minimum

Re,erence chc!f energy that must k possessed by reacting molecules before the reaction will occur. The fraction of the reacting molecules that have an energy EA or greater is shown by the shaded areas in Figure 3-3. The molecules in the shaded area have sufficient kinetic energy to cause the bond to break and reaction to occur. One observes that, as the temperature increases, more molecules have sufficient

f(E,T) Fracl~on of colltstons

at T2 that have energy

l EA \ Fraction of col~is~ons at T, that

have energy EA or greater

Figure 3-3 Enerr? dirtnbuf~on of rcnctlnp ~noleculec

Sec. 3.3 The Aeaction Rate Constant 95

energy to react as noted by an increase in the shaded area, and the rate of reac- tian, -r,, increases.

Postulation of the Arrhenius equation, Equation (3-Is}, remains the greatest single step in chemical kinetics, and retains its usefulness today, nearly a century later. The activation energy, E, is determined experimentally

Calculation of the by carrying out the reaction at several different temperatures. After taking the natural logarithm of Equation (3-18) we obtain

Ssrndw PIM

(3-20)

CI

0 01 slow=-$ and see that the activation energy can be found from a plot of In k , as a func-

OW250033 tionof (ltn.

Summary Notes

Tutorials

Example 3-1 Determination of the Activalion Energy

Calculate the activation energy for the decomposition of benzene diazoninm chlo- ride to give chlorobenzene and nitrogen:

using the information in Tahle E?- I . l for this first-order reaction.

We start by recalling Equa~ion (3-20)

We can use the data in TabIe E3-I . I to determine the activa~ion energy, E. and fie- quency factor, A . in two d~fferent ways. One way is to make a semilog plot of k vs. (llr) and determ~ne E from the slope. Another way is to use Excel or Polymath to regress the data. The data In Table E3-1 . I was entered in Excel and i s shown in Fig- ure E3-1.1 which was then used to obtain Figure E3-1.2. G step-by-step tutorial to construct both an Excel and a Polymath spread sheet is given in the Chapter 3 Summary Nates on the CD-R0.V.


0.00355 -5.64 0.00305 0.00717 -4.94 0.00300 v I l l

I Figum E3-1.1 Excel spreadsheet.

(a) Graphical Sol~ition

df -

- k .#A - a

74 -

I* -

Figure E3- 1.2(a) shows the semilog plots from which we can calculate the nctivntio energy. From CD-ROM Appendix D, we show how to rearrange Equation (3-20) i the form

$t-?!!L*nlt 1

--.

k , - - E 1 - 1 log - - - - k , 1.3R (T2 TI)

3 m ~ ~ o o o 3 m a m ~ o r o r m 1 ~ 0 r m ~ s 0 0 a 3 ; 0 0 0 0 1 1 1 WZ '*"' '*11

1~6') 1 T 1 ~ i ' ~ 1

r a) (b)

Figure E3-1.2 ( a ) Excel semilog plot: (b) ExceI normal plot.

Rearranging

To use the decade method, choose l l T , and 11T2 so that k , = O.lk, . Ther log(k,/k,) = 1.

When k, = 0.005: 1 = 0.003025 and when kz = 0.0005: 1 = 0.00319 TI T,

Sec. 3.3 The Reaction Rate Constant

Therefore, E = 2.303R - (2.303) (8.31 4 Nmol . K ) - I T - I T (O.OU319-0.0N3025)JK

= I I6 kJ w 28.7 kcal/rnol mol

I The equation for the best-fit of the data

is also shown in Figure E3-1.2(b). From the slope of the line given in Figure 3- I .2(b)

1 From Figure El-1.2(b) and Equation (U-1.3). r e see

1 taking the antilog

The rate dws not always double For s

temperature There is a rule of thumb that states that the rate of reaction doubles for increase of 10"c. eveq IO°C increase in temperature. However, this is tme only for a specific

combination of activation energy and temperature. For example, if the activn- tion energy is 53.6 kJlmol. 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 Prob- lem P3-7 for the derivation of this relationship.)

The larger the activation energy, the more temperature-sensitive is the rate of reaction. M i l e there are no typical values of the frequency factor and activation energy for a first-order gas-phase reaction, if one were forced ro make a guess, values of A and E might be IOl3 s-I and 200 kJlmoP. However, for fami- lies of reactions (e.g., halogenation), a number of corre1ation.s can be used to

Reference Shelf estimate the activation energy. One such corntation is the Polanyi-Semenov equation, which relates activation energy to the heat of reaction (see Profes- sional Reference Shelf 3.1). Another correIarion relates activation energy to

98 Rate Laws and Stolchiomet~ Chap. 3

differences in bond strengths between products and reactantsn3 While activation energy cannot be currently predicted a priori. significant research efforts are under way to calculate activation energies from first principle^.^ (Also see Appendix J.)

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 To, that is,

and at a ternperaturc T

A most useful form and taking the ratio to obtain of k l T )

This equation says that if we know the specific reaction rate ko(To) at a temperature, "r,,, and we know the activation energy, E. we can find the specific reac~ion 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 showed how it was possible to size CSTRs. PFRs, and PBRs using the design equations in Table 3-2 (page 99) if the rate of disappearance of A is known as a function of conversion. X:

Where are we?

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

- r ~ = [k,(T)l[fnIC,.C,. . . . I (3-2) - r & = fIC,I

+ is called a rurp low. In Part 2, Sections 3.5 and 3.6. we show how the concen-

c, = l & , { X ) [ration of the reacting species may he written in terms of the conversion X.

1 C, = h, (X) (3-22) -'?, = p 1x1

and then we can M. Boudart, Kinrfics of Cl~en~irnl Ptnrrssrx (Upper Saddle River. N.J.: Pi-cntice Hall. design ~rorhermal

rerrctors 1968). p. 168. J. N'. Moore and R. G. Pearlon, Xirretic.7 and Mr,cl~nlrisnts, 3rd ed. (New York: Wiley. 1981 ), p. 199. S. W . Renwn. TherntorEt~micnl Ki t~~rrrs . 2nd ed. (New York: Wiley. 1976).

S. M. Senkan, Defnilrd Cl~rr?~ir.r~l Kirlerir. Murkclirrg: Clicinirnl Hrnrfi(11~ Enginepring of the Fufrrre. Ad\'ances in Chemical Engineering. Vol. 18 (San Diego: Academic Precq. 19921, pp. 95-96.

Sec. 3.4 Present Slatus of Our Approach to Reactor Siz~ng and Design 99

TABLE 3-2. DESIGN EQUATIONS

Diflerential Algebmic lntegml Form Fonn Form

Batch N 4 0 F r - r A V (2-6)

The design Backmix equations v=- tCSTR)

F*" ((2.1 3) - FA

Tubular (2-15) dX ( PFR ) V=F,,( - (?-!6j

. n - r ,

Packed bed (2-17) dX

WBR) W = Fhu (2-18)

A

With these additional relationships, one observes that if the rate law is given and the concentrations can be expressed as a function of conversion. r k ~ n it7

fact we have - r A as a ft~ncrion of X and rhis i s nll ,ha1 is needed io eraluore ?he design .equarions. One can use either the numericai techniques described in Chapter 2, or. as we shall see in Chapter 4, a table of integrals, andlor software programs le.g.. Polymath).

Now that we have shown how the rate law can be expressed as a function of concentrations. we need only express concentration as a function of conversion in order to carry out calculations similar to those presented in Chapter 2 to size reactors. If the rate law depends on more than one species. we must relate the concentrations of the different species to each other. This relationship is most easily established with the aid of a stoichiometric table. This table presents the stoichiometric relationships between reacting molecules for a single reaction. That is. it tells us how many rnolecuIes of one species will be formed during a chemical reaction when a given number of molecules of snorher species disap- pears. These ~Iationships will be developed for the general reac~ion

Recall that we have already used stoichiametr-jl to relate the reIatrve rates of reaction for Equation (2-1 ):

T h ~ c s~oich~ometnc ~lal lonship relating

reaction rates W I I I he used in Pan 1 of

Chapter 4.

100 Rate Laws and Stoichiometrj Chap. :

Components of the stoichiornettic table

In formulating our stoichiornetsic table, we shall take species A as o u ~ basis of calculation (i.e.. limiting reactant) and then divide through by the stoichiometric coefficient of A.

in order to put everything on a basis of "pet mole of A." Next, we develop the stoichiometric relationships for reacting species thal

give the change in the number of moles of each species li.e.. A. B, C. and D).

3.5 Batch Systems

Batch reactors are primarily used for the praduction of specialty chemicals and to obtain reaction rate data in order to determine reaction mte laws and rate law parameters such as k, the specific reaction rate.

Figure 3-4 shows an artist's rendition of a batch system in which we will carry ot~t the reaction given by Equation (2-2). At time t = O, we will open the reactor and place a number of moles of species A, B. C, D, and I (NAO, Ng0, N,,, N,, and N,, respectively) into the reactor.

Species A is our basis of calculation, and NAo is the number of moles of A initially present in the reactor. Of these. NA& moles of A are consumed in the system as a result of the chemical reaction, leaving (NAo - NA& moles of A in the system. That is, the number of moles of A remaining in the reactor after a conversion X has been achieved is

We now will use conversion in this fashion to expresc the number of moles of B, C, and D in terms of conversion.

To determine the number of moles of each species remaining after N,,X moles of A have reacted, we form the stoichiometric table (Table 3-3). This stoichiornetric table presents the foIlowing information:

Column I: the particular species Column 2: the number of moles of each species initially present Column 3: the change in the number of mojes brought about by reaction CoIumn 4: the number of moles remaining in the system at time t

To calculate the number of moles of species B remaining at time t , we recall that at time t the number of moles of A that have reacted is N A o X . For every mole of A that reacts, bla moles of B must react; therefore, the total: number of moles of B that have reacted is

moles B reacted = reacted - rnoIes A reacted moles A reacted

Sec. 3.5 Batch Systems

Figure 3.4 Batch reactor. (Schematic with permission by Renwahr.1

TABLE 3-3. STOICH~OMETRIC TMLE FOR A BATCH SYFTEM

It~irtuil~ CItcrrrge Rennining Specirs (mol) (moll (moll

A N%o - X W, = NAG - ,V+,,,XI

Because B is disappearing from the system, the sign of the "change" is negative. NBO is the number of moles initially in the system. Therefore, the number of moles of B remaining in the system. N, , at a time f, is given in the Iast column of Table 3-3 as

102 Rate Laws and Stoichiomefry Chap. 3

The complete stoichiometric tabIe delineated in Table 3-3 is for a11 species in the general reaction

Let's take a took at the totals in the last column of Table 3-3. The stoichiometric coefficients in parentheses (dla -k c/a - bla - 1) represent the increase in the total number of moles per mole of A reacted. Because this term occurs so often in our calculations, it is given the symbol 8:

The parameter 6 tells us the change in the total number of moles per mole of A reacted. The total number of moles can now be calculated from the equation

We recall from Chapter 1 and Part 1 of this chapter that the Enetic rate law (e.g., - r , = kc:) is a function solely of the intensive properties of the

we reacting system (e-g.. temperature, pressure, concentration, and catalysts, if C, = " , [ X I any). The reaction rate, - r , . usually depends on the concenrration of the reacting species raised to some power. Consequently, to determine the reaction rate as a function of conversion X, we need to know the concentrations of the reacting species as a function of conversion.

3.5.1 Equations for Batch Concentrations

The concentration of A is the number of moles of A per unit volume: Batch

concentration N C , = 1 V

After writing similar equations for B. C, and D. we use ;the stoichiometric table to express the concentration of each component in terns of the conversion X:

Sec. 3.5 Batch Systems 103

We further simplify these equations by defining the parameter O, , which allows us to factor N,, in each of the expressions for concentration:

C, = [OD + ( d / a ) X ] NDO

v , with OD = - NAQ

we to 1% now need only to find volume as a function of conversion to obtain the ohlain C, = I I , ( X ) .

species concentration as a function of conversion.

3.5.2 Constant-Volume Batch Reaction Systems

Some significant simplifications in the reactor design equations are possible when the reacting syrtem undergoes no change in volume as the reaction progresses. These systems are called constant-volume. or constant-density, because of the invariance of either voIume or density during rhe reaction process. This situation may arise from several causes. In gas-phase batch systems, the reactor is usually a sealed constant-volume vessel with appropriate instsu- ments to measure pressure and temperature within the reactor. The volunle within this vessel is fixed and will not change. and is therefore a constant-\.olume system (V = V,,). The laboratory bomb calorimeter reactor i s a typical example of this type of reactor.

Another example of a constant-volume gas-phase isothermal reaction occurs when the number of moles of products equals the number of moles of reactants. The water-gas shtft reaction. important in coal gasification and many other processes, is one of these:

In this reaction, 2 mol of reactant forms 2 mol of product. %'hen the number of reactant molecules forms an equal number of product molecules at the sm~ie temperature and pressure, the volume of the reacting mixture will not change if the conditions are such that the ideal gas law is applicable. Qr if the com- pressihiliry factors of the products and reactants are appmximateIy equal.

For liquid-phase reactions taking place in solution. the solvent usually dominates the situarron. As a rewlt. changes in the denrity of the rolute do not

104 Rate Laws and Stolchiometry Chaw

affect the overall density of the solution significantly and therefore i t is essen tially a constant-volume reaction process. Most liquid-phase organic reaction do not change density during the reaction and represent still another case t which the constant-volume simplifications apply. An important exception t this general rule exists for polymerization processes.

For the constant-volume systems described earlier, Equation (3-25) ca be simplified to give the following expressions relating concentration and con version:

Concentration as a function of CB = NAo f ( N ~ ~ / N ~ ~ ) - (b/a)Xl- - 40 [@,-(blalXl = & ( O B - : X ) (

conversion when v~ vn no volume chanee

occurs with reactron CC = NAO

[ ( N , , / ~ , , 3 + (c/a)Xl yo

To summarize for liquid-phase reactions (or as we will soon see for isotherma and isobaric gas-phase reactions with no change in the total number of moles) we can use a rate law for reaction (2-2) such as -r, = kACACB to obtair

l-----l - r , =AX)% that is,

- r , = kCACB = kc:*( 1 - X)

Substituting for the given parameters k. CAO, and OB, we can now use the tech x niques in Chapter 2 to size the CSTRs and PFRs for liquid-phase reactions.

Example 3-2 Expressing = hj(X) for a Liquid-Phase Reaction

Soap consists of the sodium and potassium salts of various fatty acids such as oleic stearic, patmitic, lauric, and my~isttc acids. The saponification for the formation o soap from aqueous caustic soda and glyceryl stearate is

Letting X represent the conversion of sodium hydroxide (the moles of sodiurr hydroxide reacted per mole of sodium hydroxide initially present), set up a stoichio metric table expressing the concentration of each species in terms of its initial con centration and the conversion X.

Sec 3.5 Batch Systems 1 05

Choosing a bacis of calculation

Stoichiomerric table (batch)

Because we are taking sodlurn hydroxide as our basis, we divide through by the stoichiometric coefficient of sodium hydmx~de to put the reaction expression in the form

We may then perform the calculations shown in Table E3-2.1. Because this ir a liquid-phase reaction, the density p is considered to be constant; therefore, V = I.', .

TABLE E3-2.1. S~Y)ICHIOMETRIC TABLE FOR LIQUID-PHASE SOAP REALTION

Specie1 Symbol Initially C h a n ~ e Remaining Concentration

NaOH A NAO -maax N ~ o ( l - x ) C ~ o ( 1 - x )

Water (inen) I % - - Nto &lo Af T,, 0 NT = NTO

/ Example 3-3 What i s Be Limiting Reactant?

Waving set up the stoichiometric table i n Example 3-2, one can now readily use i t to calculate the concentrations at a given conversion. If the initial mixrute consists solely of sodium hydroxide at a concentmtion of 10 rnol/dmJ (i.e., 10 rnol/L or 10 kmollrn3 5 and of plyceryl stearate at a concentration of 2 molldm3, what is the concentration of glycerine when the co'nversion of sodium hydroxide i s (a) 20% and (b) 90%?

[ Solution

Only the reactants NaOH and (Cl,H35COO)3C3FE5 are initidly present: therefore. 0, = 0, = 0.

Rate Laws and Stoichiometry Chap. 3

1 (a) For 20% conversion of NaOH:

(h) For 909 conversion of NaOH:

] Let us find C,:

3,6 Flow Systems

The bass or calculal'on chould

he Ihe limirinf reaclant.

The form of the stoichiometric table for a continuous-flaw system (see Figure 3-5) is vir~ually identical to that for a batch system (Table 3-3) except (hat we replace N!o by q;:, and N, by F, (Table 3-4). Taking A as the basis. divide Equation i2-1) through by the stoichiornetric coefficient of A to obtain

Oops!! Negative concentration-impossible! What went wrong? Ninety percent conversion of NaOH is not possible. because glyceryl stearate

is the limiting reactant. Consequently, all the glyceryl stearate i s used up before 90% of the NaOH could be reacted. It i s irnprtant to choose the Ilrnlting reactant as the basis of calcuIation.

Entering

F~~ 1

Figure 3-5 Flou- reactor.

Sec. 3.6 flow Systems 107

F ~ e d Rate lo Change wirhin Reacfox Reactor Eflueni Rote fmm Reocro r

Species (molltime) (molltirm) (mu1 {time)

A F A 0 - FAOX FA = FA0 ( I - X)

B F~~ = @ B ~ A O -- b FAOX a

Sro~chiometric table (Row) C Fco " @cFm : FAOX

where

and Bc, OD. and 8, are defined similarIy.

3.6.1 Equations for Concentrations in Flow Systems

For a flow system, the concentration C, at a given point can be determined from the molar flow rate F A and the volumetric flow rate v at that point:

Definition of concentration for a

flow system

Units of u are typically given in terms of liters per second, cubic deci- meters per second, or cubic feet per minute. We now can write the concenrra- tions of A, B, C, and D for t h e general reaction given by Equation (2-21 in terms of their respective entering molar flow rates (F,,, F R O . F,,. F,,), the conversion X. and the volumetric flow rate, v .

108 Rate Laws and Stoichiometry Chap. :

3.6.2 Liquid-Phase Concentrations

For liquids, volume change with reaction is negligible when no phase changer are taking place. Consequently, we can take

For IfquidS Then Cq =C*o(I - X I

Therefore, for a given rate law we have

- r A = ( X ) C,=C,, @,--X etc. ( :I

Consequentiy, btsing arty one oj' the rare lows in Port I of this chapr~l; cve con now Jnd -r, = AX) for liquid-phase reactions, However, for pas-phase reactions the volumetric flow rate most often changes during the course of the reaction because of a change in the total number of moles or in temperature or pressure. Hence, one cannot always use Equation (3-29) to express concentration as a function of conversion for gas-phase reactions.

3.6.3 Change in the Total Number of Moles with Reaction in the Gas Phase

In our previous discussions, we considered primarily systems in which the reaction volume or volumetric flow rate did not vary as the reaction pro- gressed. Most batch and liquid-phase and some gas-phase systems fall into this category. There are other systems, though, in which either V or u do vary. and these will now be considered.

A situation where one encounters a varying flow rate occurs quite frequently in gas-phase reactions that do not have an equal number of product and reactant moles. For example, in the synthesis of ammonia,

4 rnol of reactants gives 2 mol of product. In ffow systems where this type of reaction occurs, the molar flow rate will be changing ar the reaction progresses. Because equal numbers of moles occupy equaI volumes in the gas phase at the same temperature and pressure. the volumetric flow rate wilI also change.

Another variable-volume situation, which occurs much less frequently, is in batch reactors where volume changes with time. Everyday examples of this situation are the combustion chamber of the internal-combustion engine and the expanding gases within the breech and barrel of a fiream as it is fired.

In the stoichiometric tables presented on the preceding pages, it was not necessary to make assumptions concerning a volume change in the first four coIumns of the table (i.e.. the species, initial number of moles or molar feed

See. 3.6 Flow Systems 109

rate, change within the reactor, and the remaining number of moles or the molar effluent rate). All of these columns of the stoichiometric table are independent of the volume or density. and they are irfenticnl for constant-volume (constant-density) and varying-volume (varying-density) situations. Only when concentration is expressed as a function of conversion doer variable dens~ty enter the picture.

Batch Reactors with Variable Volume Although variable volume batch reactors are seldom encountered because they are usually solid steel containers. we wiIl develop the concentrations as a function of conversion because (1) they have been used to collect reaction data for gas-phase reactions, and (2) the development of the equations that express volume as a function of conversion w~i l facilitate analyzing flow systems with variable volumetric flow rates.

Individual concentrations can be determined by expressing the volume V for a batch system, or volumetric flow rate v for a flow system, as a function of conversion using the following equation of state:

Equation of state PV = ZN,RT (3-30)

i n which V = volume and N , = total number of moles as before and

T = temperature. K P = total pressure, atm &Pa; t atm = 101.3 kPa) Z = compressibility factor

R = gas constant = 0.08206 dm" aatmtmol - K Thi s equation is valid at any point i n the system at any time t . At time

r = 0 (i.e., when the reaction is initiated). Equation 13-30) becomes

Dividing Equation (3-30) by Equation (3-3 1 ) and rearranging yields

We now want to express the volume V as a function of the conversion X. Recalling the equation for the total number of moles In Table 3-3,

where

S = Change in total number of moles Mole of A reacted


We divide Equation (3-33) through by N,:

Then

- 'T = I + E X (3-34) NTQ

Rc[a l i~nsh i~ between where yAo is the mole fraction of A initially present, and where 8 and e

E = ~ + n - ~ - ~ ) ~ = J b A o ~ N, (3-351

Equation (3-35) holds for both batch and flow systems. To interpret E, let's rearrange Equation (3-34)

Interwiati*n of at complete conversion, (i .e.. X = I and N , = NTf)

- - Change in total number of moles for complete conversion Total moles fed

If all species in the generalized equation are in the gas phase. we can substitute Equation (3-34) with Equation (3-32) to arrive at

In the gas-phase systems that we shall be studying, the temperatures and pressures are such that the compressibility factor will not change significantly during the course of the reaction: hence Z,=Z. For a hatch system, the volume of gas at any time I is

Volume of gas for a variable volume

batch reaction

Equation (3-38) applies only to a 1~arinble-1~r)fu?nr hatch reactor, where one can now substitute Equation (3-38) into Equation (3-25) to express r, =PX). HOW- ever, if the reactor is a rigid steel container of constant volume, then of course

Sac. 3.6 FIOW Systems Ill

V = V,. For a constant-volume container, V = I],, and Equation 13-38] can be used to calculate the gas pressure inside the reactor as a function of temperature and conversion.

Flow Reactors with Variable Volumetric Flow Rate. An expression sirnifar to Equation (3-38) for a variable-volume batch reactor exists for a variable-volume fiow system. To derive the concentrations of each species in terns of conversion for a variable-volume flew system, we shall use the relationships for the total concentration. The total concentration, CT. at any point in the reactor is the total molar flow rate, 6, divided by volumetric flow rate v [cf. Equation (3-27)J. In the gas phase, the total concentration is also found from the gas law. Cr = PER1 Equating these two ~Iationships gives

P c , = F ' = - v ZRT

At the entrance to the reactor,

Taking the ratio of Equation (3-40) to Equation (3-39) and assuming negligible changes in the compressibility factor, we have upon rearrangement

We can now express the concentration of species j for a flow system in terms of its flow rate, 5, the temperature, T, and total pressure. P.

Use th~s form for membrane reactop

(Chapter 4) and for rnuEiipfe

reaction.: (Chapter 61

The total moiar flow rate is just the sum of the molar flaw rates of each of the species in the system and is

11 2 Rate Laws and Stoichiometry Chap

One of the major objectives of this chapter is to learn how to express any giv~ rate law - r , as a function of conversron. The schematic diagram in Figu 3-6 helps to summarize our discussion on this point. The concentration of t l key reactant. A (the basis of our calculations), is expressed as a function conversion in both flow and batch systems, for various conditions of temper ture. pressure, and voIume.

Flow

v NO Phase Change

J NO Phase Change

OR NO Sern~penneable Membranes

4 Isothermal

Ifr Neglect Pressure Drop

c, = c*,(e, - $ x )

1 +EX

Flgure 3-6 Expressing concentration as a function of conversion.

Gns-phase volumetric flow

rate

Sec. 3.6 Flow Svsterns 113

We see that conversion is not used in this sum. The molar flow rates, F,, are found by solving the mole balance equations. Equation 11-42) wiIl be used for measures ohher than conversion when we discuss membrane reactors (Chapter 4 Pan 2) and multiple reactions (Chapter 6). We wiil use this form of the concenfration equation for multiple gas-phase reactions and for membrane reactors.

Now let's express the concentration in terms of conversion for gas flow systems. From Table 3 4 the total niolx Row rate can be written in terms of conversion and is

FT = F f f l + F,40 8 X 13-43}

Substituting for F , in Equation (3-41) gives

U = U o FTO + f,, 5X P, T Fm (F) E

The concentration of species j is

The molar flow rate of species j is

where v, is the stoichiometric coefficient, which is negative for reactants and positive for products. For example, for the reaction

v, = -1, v, = -bla, v, = c / a . v D = d / a , and O j = FplFAw

Substituting for v using Equation (3-42) and for F,, we have

Gasphase concmtration as a

function of conversion

114 Rate Laws and Stoichiometry C h a ~ . 3

Rearranging

Recall that yAo = F,,/F,, , C,, = y,,Cm, and E from Equation (3-35) (i.e., E = ?'A06).

The stoichiornettic table for the gas-phase reacrion (2-2) is given jn Table 3-5.

Wc now have v

C, = h , ( X , and - FC - F,, I (-), + ( r ln)Xl - - FA(, + ( I . / N J X I TO P @ , + ( r / a ) X T, p

c - y - - r , = R ( X ) u V ~ ( I + ~ X ) ( ) I + E X )F[g)

for variable-volume gas-phase reaction< -FD - F 4 ~ [ 8 ~ C ( d ! u ) X l = Q, f ( d l a ) X 7, p D - - -

L'

Exumple 3 4 Maniprlatiotr of the Equation for C, = hj (XI

1 Show under H hat condirions and manipulation the expression for CB for a gas Row system reduces to that given in Tahle 3-5.

Soll~tinrr

For a flow system the concentration is dcj i~~cd as

From Tahle 3-3, the molar Row raIe and conversion are related by

Cornhininy Eqi~ations (E.1-4.1) and (E3-4.2) yields

Sec. 5.6 Flow Systems

This equalion for u Using Equation (3-45) gives us is only for a gas- I

phase reaction I

to substitute for the volumetric flow rate gives

( which is identical to the concentration expression for a variable-volume batch reactor.

I Example 3-5 Determining Cj = hi (XI for a Gas-Phase Reaction

A mixture of 28% SO, and 72% air is charged to a flow reactor in which SO, is oxidized.

2so2 + 0, ----4 2S0,

First. set up a stoichiometric table using only the symbols (i.e.. O , , F , ) and then pzpare a second stoichiometric table evaluating numericalry as many symbols as possible for the case when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227'C.

Taking SO: as the basis of calculation. we divide the reaction through by the stoichiometric cmfficient of Our chosen basis of calculation:

I SOz + f02 - SO?

The initial stoichiometric table is given as Table E3-5.1. Initially, 72% of the total number of moles i s air containing (21% O2 and 79% N 2 ) along with 2 8 8 SO?.

From the definition of conversion, we ~ubstitute not only for the molar flow rate of SO, (A ) in tenns of conver~ion hut a150 for the volumetric flow rate as a function of conversion.

116 Rate Laws and Stoichiometry Chap. ;

Species S m h f Initially Chmge Remnining

so2 A FA, - F A U X F A F , = F , , , ( l - X )

NegIecting pressure

drop. P = Po

isotherma[ operation, T = To

SO, C 0 +FAfiX Fc = FAOX

Recalling Equation (3-451, we have

Neglecting pressure drop in the reaction. P = P,, yields

If the reaction is also carried out isothermally. T = To . we obtain

The concentration of A initially is equal ta the mole fraction of A initially multiplied by the total concentration. The total concentration can be calculated from an eaua- tion of state such as the ideal gas taw. Recall that y~~ = 0.28, To = 500 K. and Po= 1485 Wa.

Sec, 3.6 Flow Systems

I The total concentration i s

The concentration of the inert is nor

constant!

I We now ewlunte e . .

The concentntions of different species at various conversions are calculated in Table E3-5.2 and plotted in Figure E3-5.1. Note that the concentntion of N2 is changing even though i t is an inert species in this reaction!!

TABLE E3-5.1. COXCESTRATIOY 45 A F U Y ~ I O N OF CONVERSIOS

C, (molldm')

Species X=O.O X = 0 . 2 5 X = 0 . 5 X=8.75 X = 1.0

SO, C, = 0.100 0078 0.054 0.028 0.000

a? C, = 0.054 0.043 0.031 0.018 0.005 SO, C, = 0.000 0.026 0.054 0.084 0.1 16

We are now in a position to express - r , as a function of X. For example, ifthe nte law for this reaction were first order in SO, (i.e., A) and in 0, (i+e., S), with k = 280 dm3/mol - s , then the rate law becomes


Nore: Because the voiurnetic Row rate varies with conversion, the

concentration of inert5 (N2) i s not

constant.

Now use techniques presented

i n Chapter 2 to size reaclors.

Figure E3-5.1 Concentration as a function of conversion.

- 1 Taking the reciprocal of - r , yields

We see that we could ~ite a variety of combinations of i s o r h e m i reactors using the

1 techniques discussed in Chapter 2.

Thus far in this chapter, we have focused mostly on irreversible reactions. The procedure one uses for the isothermal reactor design of reversible reactions is virtually the same as that for irreversible reactions. with one nota-

Need to first cllrculate xr ble exception. First calculate the maximum conversion that can be achieved at

the isothermal reaction rernperature. This value is the equilibrium conversion. In the following example it will be shown how our algorithm for reactor design is easily extended to reversible reactions.

I Example 3-6 Calcuhting the Equilibrium Conversion

The reversible gas-phase decomposition of nitrogen tetroxide, N,O,. to nitrogen dioxide, NO2,

Sec. 3.6 flm Systems f19

is to be carried out at constant temperature. The feed consists of pure NzO, at 340 K and 202.6 kFa (2 am). The concentration equilibrium constant. Kc. at 340 K is 0.1 molldm". (a) Calculate the equilibrium conversion of N,O, in a constant-volume batch

reactor. (b) CalcuIate the equilibrium conversion of N20, in a flow reactor. (c) Assuming the reaction is elementary, express the rate of reaction soleIy as a

function of conversion for a Row system and for a batch system. Id) Determine the CSTR volume necessary to achieve 80% of the equilibrium

conversion.

At equilibrium the concentrations of the reacting species are relaled by the relationship dictated by thermodynamics [see Equation (3-10) and Appendix C]

( (a, Batch system-constant volume, V = Y o . See %Me E3-6.1.

Living Example Problea

For batch systems C, = N, / V ,

C - !'~oPo , ( I ) ( ? atm) *' RT,, (0.082 atm.dm3/rnol +K)(340 K)

At equilibrium. X= J,,. and we substitute Equations (E3-6.2) and (E3-6.31 illto Equation (E3-6. I ) .


There is a PoIymath tutorial in the

summary Notes of Chapter 1

We will use Polymath to solve for the equilibnum conversion and let xeb repfese the equilibrium conversion in a constant-volume batch reactor. Equation (E3-6.m written in Polymath format becomes

f (xeb) = xeb - [kc*(l - xeb)/(?*cao) J "0.5

The Polymath program and solution are given in Table E3-6.1.

When looking at Equation (E3-6.4). you probably asked yourself. "Why not use tl quadratic formula to solve for the equilibnum convesston in both batch and flo syrterns?' That is,

I Batch: X, = -[(-I + JF + l6CAOIKc)/(CA,J K c ) ] 8

[(E- I ) + J(G- I ) ~ * ~ I F + ~ C ~ ~ / K ~ ) I Flow: X, =

2 ( ~ + 4 C A o l K c )

The answer is that future problems will be nonlinear and require Polymath solution: and I wanted to increase the reader's ease in using Polymath.

TABLE E3-6.2. POLYMATH PROGRAM AND SOLLTEOY FOR BOTH BATCH 4x0 FLOW SYSTEMS

m i a b l e f (XI Ini G u w Xeb 4.078E-08 0 . 5 Xe f 2.622E-10 0 . 5 Kc 0.1

NLES Report Wenewt)

Nonlinear equations . I: f(Xeb) = Xeb-(Kcm(l -Xeb)l(4"Cao))"O.S = 0 i i : f(Xef) = Xef-(Kc'(1-Xef)'(t+eps*Xef)I(CCao)~.S = O

Explicit equations i - ] Kc=O.l if.; Cao=0.07174 :3; eps= 1

The equilibrium conversion in a constant-volume batch reactor is

Sec. 3.6 Flow Systems 121

At equilibrium, X = X,. and we can substitute Equations (E3-6.5) and (E3-6.6) into Equation (E3-6.1) to obtain the expression

R,lymath ~ ~ ~ ~ ~ i ~ l Chapter I

I Simplifying giver

Nore: A tutorial of Polymath can be found in thc summary notes uf Chapter I .

(b) Flvw system. The stoich~ometric table is the same as that for a batch s y ~ t e m except that the number of moles of each species, .V, , is replaced by the tnofar flow rate of that species. F,. For constant temperature and pressure, the volumetric f f o ~ rate is u = vo / l + e x ) , and the resuIting concentrations of species A and B are

1 Rearnnging to use Polymath yields

For a flow system with pure N,O, feed, e = y , ~ , 6 = 1 (2 - I ) = I . We shall let Xef represent the egullibriurn conversion in a flow system. Equa-

tion (E3-6.8) written In the Polymath format becomes

I f(Xef) = Xef - [kc*(l - Xef)*[ l + eps*Xe~/4/cao]"0.5

This solution is alro shown in Table E3-6.2 (X,, = 0.51). Note that the equilibrium conversion in a flow reactor tine.. X, = 0.5 1 ). with

negligible pressure drop, is greater than the equilibrium convenion in a constant-volume batch reactor (X, = 0.44 ). Recalling Le Chitelier's principle, can you suggest nn explanation for this difference in X, ?

(c) Rate laws. Assuming that the reaction fallows an eIementar)r rate law, then

] 1. For a constant volume ( V = Vo) batch system.

122 Rate Laws and Stokhiomatry Chap. 3

- r , = J-(X) for a flow reactor

Here C, = N A I V, and C, = & 1 Vo. Substituting Equations (E3-6.2) and (E3-6.3) into the rate law, we obtain the rate of disappearance of A as a function of conversion:

2. For a flow system. Here C, = FAlv and Ca = Fs/u with v = vU (I t. m. Consequently. we can substitute Equations (E3-6.5) and (E3-6.6) ~nto Equation (B-6.9) 10 oblain

As expected, the dependence of reaction rate on conversion for a constant- volume batch system [i.e., Equation (E3-6. lo)] is different than that fur a Row system [Equation (E3-6.11)] for gas-phase reactions.

I f we substitute the values for CAO, KC, E, and k = 0.5 min-' in Equation (E3-6.11). we obtain -rA solely as a function of X for the flow system.

We can now form our Levenspiel plot. W e see (I/-r,) goes to infinity as X approaches X,.

I Figure E3-6.1 Levencpiel plot Tor a flow %!\tern.

5ec. 3.6 Flow Systems 123

The CRE Algorithm mole Balance. Ch 1 .Rate Law. Ch 3 mSto~chiorneiry. Ch 3 *Combine. Ch 4 'Evaluate, Ch 4 ' E n e ~ y Balance. Ch 8

(d) CSTR volume. Just for fun let's calcuIate the CSTR reactor voIume necessary to achieve 80% of the equilibrium conversion of 50% {i.e., X = 0.8X,) X = 0.4 for a feed rate of 3 rnollmin.

1 The CSTR volume necessary to achieve 40% conversion is 1.7 1 m3.

Closure. Having completed this chapter you should lx able to write the rate law in terns of concentration and the Arrhenius temperature dependence. The next step is to use the stoichiometric table to write the concentrations in terms of conversion to finally amve at a relationship between the rate of reaction and conversion. We have now completed the fiat three basic building blocks in our algorithm to study isothermal chemical reactions and reactors.

S toichiornetry

Rate Law

I Mole Balance 1)

In Chapter 4. we will focus on the curnbine and evaluation building blocks which will then complete our algorithm for isothermal chemical reactor design.

Rate Laws and Stoichiometry Chao.

S U M M A R Y

P A R T 1

1. Relative rates of reaction for the generic reaction:

The relative rates of reaction can be written either as

1. Re(rc~crion o d r is determined from experimental observation:

The reaction in Equation (S3-3) is a order with respect to species A and order with respect to species B , whereas the overall order, n. is a + p. Rea tion order is determined From experimental observation. XF a = I and P = we would say that the reaction is first order with respect to A, second ord with respect to B, and overall third order. We say a reaction follows an el m e n t q sate law if the reaction orders a g m with the stoichiornetric coeA cients for the reaction as written.

3. The temperature dependence of a specific reaction rate is given by the Arrh nius equntion,

where A is the frequency factor and E the activation energy. If we know the specific reaction rate, k, at a temperature, To, and the acl

vation energy, we can find k at any temperature. T,

Similarly from Appendix C, Equation (C-9), if we know the equilibrium co~ stant at a temperature, TI. and the heat of reaction, WF can find the equilil riurn constant at any other temperature

Chap. 3 Summary

P A R T 2

4. The sroichiomerric table for the reaction given by Equation 6 3 - 1 ) being carried out in a flow system is

Species Entering Chunge h a v i n g

A F~ - F~&f FAo( 1 - A')

d e b where 6 =

5. In the case of ideal gases, Equations (S3-6) and (S3-7) relate volume and voI- umetric flow rate to conversion.

Batch constant volume: V = vo (S3-6,

Flow systems: Gas: T

Liquid: v = u,

For the general reaction given by (S3-11, we have

6 = Change in total number of moles Mole of A reacted

Definitions of 6 and and E

e = Change in totaI number of moles for complete conversion Total number of moles fed to the reactor

t26 Rate Laws and Stoichiometry Chap. 3

6. Far pas-phase reactions, we use the definition of concentration (C, = FJu) along with the stoichiometric table and Equation {S3-7) to write the concentration of A and C in terms of conversion.

Q,irh @, = & - 'c" - 3 FA, CAO ! . ~ n

7. For incompressible liquids. the concentrattons of species A and C in the reaction gtven by Equatlon rS3-1) can be writ~en as

c, = C,, Oc + - X i ..I FZuations (S?- 17 ) and (S3-13) also hold for gas-phase reactions carried out at constant volume in batch systems

8. I n terms ul' pas-phace molar Row rates, the concentration of species r is

Fi F To Ci = C," - - - FT" Po T

C D - R O M M A T E R I A L

Learning Resuurce 1 . Sun119ran' N o ~ s fc~r Cliupicr 3 2 We11 Mvdtt /e .~

summa^:, Nmes A. Cooking a P o ~ u The chemical reaclion eng~ncrring is 3ppIred Fa cookin: a polato

k Starch (c~stal l ine)-+ Starch amorphous

- Solved 7rob!ems

Chap. 3 CD-ROM Material 127

3. MolecuIar Reaction Engineering Molecular simulators (Spiman. Ceriuq21 are used to make predictions of the activation energy. The fundamentals of density fi~nctional are presented along with specific examples.

3. Inreracfi1,e Crnrprrrrr Modrrler A. QUIZ Show I1

Snh~rd Pmhler~ir A. CDP.1-A, Actirnt~nn Energ! h r a Beetle Pw<hing a Ball of dun^ B CDP1-R, Micrnelccrronic\ Induqrr! and the Sln~chlorne~ric Tible

'rrtluently Asked Qucstion~J-In Upd;i~eslFAit' I C O ~ \eciion

Rate Laws and Stoichiometry Char

Professional Reference ShcIf R3.1. Coiii~ioti Thmy

In h i s section, the fundamentals of collision theory

Schematic of collision cross wclion

are applied to the reaction

A + B + C + D

to arrive at the following rate law

The activation energy, E,, can be estimated from the Pofanyi equation

R3.2. Transition Srclre Theoy In this section, the nte law and rate Iaw pmmeters are derived for the reacti

using transition state theory. The Figure P3B- 1 shows the energy of the ma ecules along the reaction cmrdinate which measures the progress of the reactic

Figure P3B-I Reaction coordinate for (a) SH2reaction, and (b) generalized reaction. (c) 3-1 energy surhce for generalized reaction.

We will now use statistical and quantum mechanics to evaluate KAC to arri at the equation

m

Chap. 3 CD-ROM Material 129

where q' r:, overnIl the partilion function per unit volume and i< the product or tranklational, vibrat~on. rotational. and electnc partition functions: that is.

4' = q',q'vq',q',

The indivrdual panitron Functions to bc evaluated areS

Translatiou Reference Shelf

Rotation

The Evring Fquatioq

R3.3. MokcuIur Dynamics 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."

Nonreactive Traiectom

R. Masel. Chemical Kinetics: (New York McGmw Hill, 2002), p. 594. M. KarpIus. R.N. Porter. and R.D. S h m a , 1 Chem. Phys., 43 (9). 3259 (3965),


.-.-. . -' Reference Shelf

'1 t ~ m e

From these trajectories. one can cnlcufate the following reaction crosq section. ST. shown for the case where both the tibrational and rotarronal quantum nuinhers are zero:

The specific reaction m e can then he calculated fmm first principle for simple mnlecules.

R3.4. ,I.IEn~icres Orher TI~otl Cnrn.eniorr Gut Plrrr~p

OVorr: T h i s toptc will be covered in Chapter 4 but for r h o ~ e who want to use 11 now. look un the CD-ROM.) For membrane reactors and gas-phase rntllti- ple reactions. it is much more cc~nvenient to work In term5 of the number of molec bV4. hrT,) or molar Row rates (FA. F,, etc.) rather than con\er\inn.

R3.5. Rcurtiotrs ~ 7 1 1 1 Co~t(irt~ \(rtio~t We now concider a pas-pira~e reaction in which uundensntron occurs. An example of thic clav of reaclions is

Here we uill deielnp our \toichiometric table for reaction5 nith phase chanre When one nf (he product5 conrlene:. durrnp the c o i t r ~ of a reaction. ciilculaticln of the change In inlumc n t ~olumctric AON raw nutht be undcr- tnhen in a \ l~gh~ ly r l ~ t l ' c r c ~ ~ ~ manner. Plois of the lnular Ron mlc< of cnndcn- \.!te D I h t tol;il, togelher nlih tht' recrprt~al rate. are huwn hcre a\ :I

fu~lutinn of cnnir.r\ioi~

Chap. 3 Ouestions and Problems

Q U E S T I O N S A N D P R O B L E M S :---,/: ,' @ The subscript to each of the problem numbers indicates the level of difficulty: A, least

difficult; D. most difficult. Ciomcwork 7-oblems A = . B=. C = + D=++

Creatlvs Thinking (0

Is) P3-2, (a)

List the impmiant concepts that you learned from this chapter. What concepts are you not clear about? Explain the strategy to evaluate reactor design equations and how this chapter expands on Chapter 2. Choose a FAQ from Chapers 1 through 3 and say why it was the most helpful. Listen to the audios a on the (23. Sclect a topic and explain it.

Read through the ~ e l f ~ c s t s and Self Assessments for LRctures 1 through 4 on the CD-ROM. Select one and critique it. (See Preface.) Which example on the CD-ROM Lecture nnres for Chapters I throuyh 3 was most helpful? Which of the JCMs for the first three chapters was the most fun? Example 3-1. Make a plot of k versus T. On this plot also sketch k versus f l l 7 l for E = 240 kllmol, for E = 60 kllmol. Write a couple of sentences describing what you find. Next write a paragraph describtng the actlva- tion, how it affects chemical reaction rates. and what its origins are? Example 3-2. Would the example be correct if water was considered an ~nert? Example 3-3. How would the answer change if the initial concentration or glyceryl sterate were 3 molldm3? Example 3-4. What is the srnallesr value of QB = (MBl,dNA,) for which the concentration of B will no, become the Iimiting reactant? Example 3-5. Under what conditions will the concentration of the inen nitrogen be constant'? Plor Equation (E3-5.2) of (I/-r,) as a function of X up to value of X = 0.94. What did you find? Example 3-6. Why is the equilibrium conversion lower for the batch sys- rem than the flow system? Wlll this always be the case for constant volume batch systems? For the case in which the total concentrarion Cm is to remain constant as the inerts are varied, plot the equilibrium conversion as a function nf mole fraction of inerts for both a PFR and a constant-volume batch reactor. The pressure and temperature are constant at 2 atm and 340 K. Only N:O, and inen 1 are to he fed. Collision Theory-Profeqsional Reference Shelf. Make an outline of the steps !hat were u ~ e d to derive

(h) The rote law for the reaction (7.4 + B -+ C) i s - r ~ = k , ~ : C, with k, = 1S(drnYmol)~ls. What arc k , and A,?

(i) At low temperatures the rate law for rhc reartian (;A + ;B + C ) ir

- r , = kACACR. If the reaction is reversible a t htgh temperatures, what is Klncur r L'hallrnpc I I the rate law?

P3-3, h a d the interactive Computer Mudthe (ICMI Kinetrc Challenge from the ?#<I XRI ?IXI CD-ROM. R u n the moduIc. and then record yot~r performance rlumber for

\he lnodule uhich indrcate\ your mastering of the mater~nl. Your professor has

f 32 Rate Laws and Stoichlometry Chap.

the key to decode your performance number. ICM Kinetin Chnlleng Performance

P3-4, The frequency of flashing of fireflies and the frequency of chirping of cricket

Ear crickets:

as a funchon of temperature follow [J. Clrc~l~. ELILI~ .. 5. 333 ( 1972) Reprinte by permission.].

The running speed of ants and the flight speed of honeybees aa a function c temperature are given below [Source: B. Heinrich, Thr Hot-Bloorlrd Insecl (Cambridge, Mass.: Harvasd University Press. 1993)j.

For fireflies:

For ants:

T ("C) 21.0 2 5 . 0 300

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

(b) What i c the velocity of the honeybee at 40°C? At -YC? (c) Do the bees, ants, crickets. and fireflies have anything in common? If st

what is it? You may also do n pairwise comparison. (d) Would more data help clarify the relationships among frequency, speec

and temperature? If so. in what temperature rhould the data be obtained Pick an insect, and expIain how you would carry out the experiment t obtain more data. [For an alternative to thi? problem. see CDP3-A,.]

P3-5, Troubleshooting. Corrosion of high-nickel stamless steel plates was found t occur in 3 distillation column used at DuPont to separate HCN and water. Sur furic acid i s always added at the top of the column to prevent polyrnerizatio of HCN. Water collects at the bottom of the coiumn and HCN at the top. Th amount of corrosion on each tray is shown in Figure P3-5 as a function r plate location in the column.

The bottom-most temperature of the column is approximately 125"l and the topmast is IWC. Tne cormsion rate is n function of temperature an the concenrration of an HCN-H2S04 complex. Suggest an explanation for th observed corrosion plate profile in the column. What effect wouid the colurn operating conditions have on the corrosion profile?

For honeybees: T ("C) 2.5 30 35 30

V ( c d s ) 0.7 1.8 3 1

Chap. 3 Questions and Problems

Too I

mills

HCN b$o,

Figure P3-5

P3-6, Inspector Sgt A m k c m b y of htland 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 unware she had coated her glass ~ i t h an antidote before she fiIled both grasses wrth the poisoned brandy. Bonnie married her second husband, Clyde, and some years later when she hod tired of him. she called him one day to tell him of her new promoziofi 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 sug. gested that instead of brandy. they celebrate with ice cold Russian vodka and they down i t Cossack style, in one gulp. She agreed and decided to follow her prev~ously successful plan and to put the poison in the vodka and the antidote in her glass. The next d ~ y , both were found dead. Sgr. Amhercromby arrives. What were the first three questions he asks? What are two possible explana- tions? Bnsed on what you learned from this chapter. what do you feel Sgt, Arnbercmmby suggested as the most logical explanation?

[Professor Flavio Marin FIores, ITESM. Monterrey, Mexico] P3-7, (a) The rule of thumb that the rate of reactran 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 ~ h i c h the rule of thumb holds. Nglecr any variation of concentration with tempenlure.

(b) Determine the activation e n e w and frequency factor from the followtng data:

(c) Write a paragraph explaining activation energy, E. and how it affects the chemical reaction rate. Refer to Section 3.3 and especially the Profes- sional Reference Shelf sections R3.1, R3.2. and R3.3 if n e c e s s q

P3-8, Air bags contain a mixture of NaN,. KNO,, and SiO?. When ignited. the fol- Towing reactions take place:

(1) Z N a , 4 2Na + 3?1? (2) lONa + 2 m O 3 + K20 -+ 5Na20 + N, ( 3 ) K,O + Na,O + SiO, + ahl ine silicate glass

134 Rate Laws and Stolchiometry Chap. 3

Reactions (2) and (3) are necessary to handle the toxic scdjum reaction product from detonation. Set up a stoachiome?ric table solely in terms of NaN, (A), KNOj (B), etc., and the number of moles initially. IF 150 g of sodium azide are present in each air bag, how many grams of KN03 and SiOz must be added to make the reaction products safe In the form of alkaline silicate glass after the bag has inflated. The sodium azide is in itself toxic. How would you propose to handle all the undetonatd air bags in cars piling up in the nation's junkyards.

P3-9, Hot Potato. Review the "Cooking a Potato" web module on the CD-ROM or on the web. (a) It took the potato described on the web 1 hour to cmk at 350°F. Builder

Bob suggests that the potato can be cooked in half that time if the oven temperature is raised to 60Q°F. What do you think?

(b) Buzz Lightyear says, "No Bob," and suggests that i t would he quicker to boil the potato in water at 100°C because the heat transfer coefficient is 20 times greater. What are the rradeoffs of oven versus boiling?

(cj Ore Ida Tater Tots is a favorite of one of the ProctorslGraden in the class, Adam Cole. Tater Tots are 1112 the size of SI whole potato but approximately the same shape. Estimate how long it would take to cook a Tater Tot at 400°F? At what time would it be cooked half way through? (rlR = 037

P3-10, (a) Write the rate law for the following reactions assuming each reaction follows an elementary rate law.

(b). Write the rate law for the reaction

if the reaction [ I ) is second order in B and overall third order. 12) is zero order i n A and first order in R, ( 3 ) is zero order in both A and B. and (4) is first order in A and overall zero order.

(c) Find and uprite the rate laws for the followrng reactions ( 1 ) Hz + BT? -4 ?HBr (2) H, + I2 + ?HI

P3-11, Sel up a stuichiometric table for each of the following reactions and express the concenlratton of each species in the reacrion as a function of conversion

Chap. 3 auestions and Probrems 135

evaluating all constants {e.g., E. Q). Then, assume the reaction follows an elementary rate law. and write the reaction rate solely as a function of conversion. i.e., -rA = fl). (a) For the liquid-phase reaction

the initial concentra~ions of ethylene oxide and water are I Ib-rnol/fk3 and 3.47 Ib-mollf~l 162.4I Iblft? + 18), res~xctive3y. If k = 0 1 drn'lmol . s a1 300 K with E = 12,500 cal/mol. calculate the space-lime volume for 90G conversion at 3OQ K and at 3.50 K.

Ib) For the isothermal, isobaric gas-phase pyrolysis

pure ethane enters the flow reactor at 6 at111 and 1 I(H) K, How nould your equation for the concentration and reaction rale change if the reaction were to be carried out in a constant-volume batch reactor?

(c) For the isothermal, isobaric, catalyttc gas-phase oxidation

the feed enters a PER at h atm and I?fiO°C and is a ~toichiometric nithture of only oxygen and ethylene.

(d) For the isothermal, isobaric. catalytic €a<-phase reaction is carried out In

a PBR

the feed enters a PBR a1 6 nrtn and 170°C and i s n s~o~cliinmetl.ic ~ n i r - ture. Whdl catalyst weight is required to reach 80% conrervon In a Ru- idized CSTR at 1 70rC and 27floC? The rate conctarlt I \ detined wrt benzene and vn = 50 dm'hnin

k B = 53 mol at 200 K u i th E = 80 !d/niol

kgcat . min atm

P3-12, There were 5430 million pounds of ethylene oxrde produced in the United Stares in 1995. The flowcheet for the camniercial production of ethylene oxide (EO) hy oxidation of ethylene is shown In Eigilre Pl-I?. We note that the process essentrally conqists of two systems. a reaction system and a sepn- rillion kyqtem. Discuss the flowcheet and how your anhwer' to P3-E 1 (c) would change i f alr IS wed In a \to~cliiometric reed. Th15 reactlon I \ studled Further in Chapter 4.

Rate Laws and Stoichiometry Chap.

EO EO €0 Light-ends €0 reactor absorber stripper rejection refiner

Sta~t

EOtwaler fssdwabr

gfycol plant

Orbased EO reaction EO fernvery and refining

Figure P3-12 ED plant Rowsheel. IAdapred from R. A. Meyers, ell.. Hor~dbool, r$ C/~ernicul Ptrnii~crir~n PPrnte.f~cr. C/~eler~ricrrl Pmc.e\s Trrhnolrrx! H<rrrrlb~~ok Series. N e a York: McGraw-Hill. 1983, p. 1.5-5. iSBN 0-67-04 1-765-1.1

P3-13, The formation of nitroanalyine (an important intermediate in dyes+alle fast orange) is formed From the reaction of nrthon~trochlorobenzene (ONCE and aqueous ammonia. (See Table 3-1 and Example 9-2.)

The liquid-phase reaction is first order in both ONCB and ammonia wit! k = 0.00 I7 d k m o l . min at 188'C with E = 1 1.173 caltinol. The iniria enterlng concentrations of ONCB and ammonia are 1.8 krnollrnJ and 6.1 kmollrn'. respectively (more on this reaction in Chapter 9).

(a) Wrire the rate Taw for the rate of disappearance of ONCB in terms o concentration.

(b) Set up a stoichiometric table for this reaction for a flow system. (c) Explain how part (a) would be different for a batch system. (d) Write -rA solely as a function of conver~ion. -r, = (e) What is the initiaE rate of reaction (X = 0) at 188"C? -r,* =

at 25°C:' -rA = at 288"C? -r, =

If) What is the rate of reaction when X = 0.90 at 18S°C? -7, = at 2S°C? -rA = at 28X°CT -r, =

(g) What would be the corresponding CSTR reactor volume at 2S°C tr achieve 90% conversion at 25°C and at 288°C for a molar feed rate of: moUmin

at25"C7 V = at 28R°C? Y =

Chap. 3 Ouestions and Probtsms 1 37

P3-1dR Atlnpted from M. L. Shuler and E Kargi, Bioyrncr.~~ En~irrecr in~. Prenticr Hall (20021. Cell grouth tithes place in b~ore:~ctors called chemostnt?.

A substrate such as glucose is used to grow cells and produce a product.

Cell% Substrare ---4 More Cells Ibiomass) + Prcduct

A gcneric molecule formuIn for the biomass is C41H73NUM01 Consider the growth of n generic organism on glucose

Experimentally, it was shown that for this organism, the cells convert 213 of carbon st~bstrate to btomass. (a) Calculate the staichiornetric coefficients a, b, c, d, and e (Hint: c a w 0111

atom balanceh [Atlr: c = O 911). (b) Calculate the yjeId coefficients Yo (g cells/g substrate) and Y,,,,

(g cellslg 0:). The grom of cells are dry weight (no water-gdwj IAns: Y, ,,% = 1.77 gdw cellslg 021 (gdw = gnms dry weight).

P3-ISa The gas-phase reaction

!N1+iH, 4 NH,

i s to be carried out isothermally. The molar feed is 50% H, and SO% N?, at a pressure of 16.4 atm and 227". (a) Construct a complete stoichiornetric table. (b) What are C,. 8. and E ? Calculate the concentrations of ammonia and

hydrogen when the conversion of H2 1s 60%. (Ans: C,*= 0.1 rnol/dm3) (c) Suppose by chance the reaction is elementary with k,; = 40 dmVmolls.

Wrtre the rate of reaction F O I P ~ as a function of conveision for ( I ) a flow system and (2) a constant volume batch system.

P3-1fin Calculate the equilibrium codversion and concentrations for each of the following reactions. (a) The liquid-phase reaction

with C,40 = Cgo = 2 rnoIldrnS and Kc = 10 dm3/mol.

(b) The gas-phase reaction

carried out in a flow reactor with no pressurn drop. PureA enters at a temperature of JOa K and 10 atm. At this temperature, Kc = 0.Z5(dm3/mol)~.


(c) The gas-pha~e reaction in part (b) carried out in a constant-volume batch reactor,

(dl The gas-phase reaction in part (b) carried out in a constant-prescure batch reaction.

P3-17, Consider a cylir~drical barch reacror thar has one end fitted with a frictionless piston attached to a spring (Figure P3- 17). The reaction

with the rate expression

is laking place in this type of reactor.

I I Reoclion occurs in hare

(a) Write the rate law solely as a function of convenioli. numerically evaluating all posslbIe symbols. (Atls.: -r, = 5.03 X 10-" I( i - X )3/( 1 + 3X)'"] Ib niollft7.s )

(h) What is the conversion and rate of reaction when I' = 0.2 ft3'? (Arts.: X = 0.259. -rA = 8.63 X lo-"' lb rnollft?-s.)

Addifionnl irtfnr~t~nriiir~: Equal moles of A and B are present at I = 0 Initial \?olume: 0.15 ft3 Value of k,: 1.0 ~ft ' l lb rno1):-s-I The relationship between the volume of the reactor and pressure within the reactor 1s

V = (0.1)(P) (Vin ft3. P i n atml

Temperature of ayrtem (considered constant): 140°F G3\ constant: 0.73 ftl.atm/lh rn~l. 'R

P3-18, Read the section\ o n Colliqion Theory. Transition Stale Theory. and Molecu- lar nynamics in the Professional Reference Shelf on the CD-ROM. la) Use colIlsicln theory to outline the derivation of the Polanyi-Sementlv

equation. ~{h ich can he uxed to estimate activation energles from the heats of reac~ion, AH,, according 10 the equation

thE Why i q thiq 3 reasonable correla~inn? (Hitlt: See Pnlfewinnal Reference Shelf 3R. 1 : Cnlliriort T l t e n ~ )

(c) Concider the ToHon ~ n g fanilly of reactions:

Chap. 3 Questions and Problems

Estimate the activation energy for the reaction

Web Mtnt

CH, # + RBr CH,Br + R #

which has an exothermic heat of reaction of 6 kcallrnoi (i.e., AH, = -6 kcallmol).

(d) What parameters v a y over the widest range in calcularing the reaction rate constant using transition state theory ( i t . , which ones do you need to focus your anention on to get a good approximation of the specific reaction rate)? (See Professional Reference Shelf R3.2.)

(e) List the assumptions made in the molecular dynamics simulation R3.3 used to calculate the activation energy for the hydrogen exchange reaction.

(T) The volume of the box used to calculate the translational partition function for the activated complex was taken as I dmq. True or False?

(g) Suppose the disrance between two atoms of a linear rnolecuIe in the transition state was set at half the true value. WouId the rate constant increase or decrease over that of the true value and by how much (i.e., what factar)?

(h) List the parameters you can obtain from Cerius2 to calculate the molecular partition functions.

P3-19D Use Spartan, CACHE, Cerius2. Gaussian. or some other chemical computa- tional software package to calculate the heats of formation of the reactants, products, and transition stare and the activation barrier, EB. for the foIlowing reactions: ( 8 ) CH30H + 0 4 CH30 + OH (b) CH,Br + OH --+ CH,OH + Bt

P3-20B It is proposed 20 produce ethanol by one of two ~eactions:

C2HsCI + OH- w ClH50H + C1- (1)

Use Spartan (see Append~x J) or some other software package to ansuw ik foIlwhg: (a) What is he ratio of the rates of reaction at 25'C? IOO°C? SOO°C? (b) Wh~ch reactran scheme would you choose to make ethanol? (Hint: Consult

Cltemicol Markfir~g Reporrer or w~.~:clnemn~cek.com for chemical prices). fProkssor R. Baldwin. Colorado School of M~nes l

Additinna! Homework Problems on CD-ROM

Temperature Effects

CDP3-An Estimate how fast a Tenebrionid Beetle can push a ball of dung at 41.5"C. (Solution Included.)

CDM-Bn Use the Polanyi equatron ro calculate activation energies. (3rd Ed. Solved F:oblcms P3-?OR]

CDP3-CR G1ve11 the irreversible rate law at low temperature, write the reverhihle rate law at high temperature. [3rd Ed. P3-]OR]

Rate Laws and Stoichiornetry Chap.

Stoichiornetry

Hall of Fame

CDP3-D, Set up a stoichiometric table for

in terns of molar Row rates. (Solution included.) CDP3-E, Set up n stoichiometric table for the reaction

[Znd Ed. P3-10,) CDP3-Fa The elemenfary reaction A(gj + B(1) P C(g) takes place in

square duct containing liquid B, which evaporates into the gas react ing with A. [2nd Ed. P3-20,]

Reactions with Phase Change

CDP3-GB Silicon is used in the manufacture of micro~lectronics devices. Set ul a stochiometric table for the reaction (Solution included.)

[2nd Ed. P3-16,] CDP3-H, Reactions with condensation between chlorine and methane. [3rd Ec

P3-2 1 ,I. CDP3-I, Reactions with condensation

C,H,(g) -+ 2Br(g) CZH4Br:(1, g) + 2HBtfg)

[3rd Ed. P3-22,] CDP3-J, Chemical vapor deposition

3SiH,(g) + 4NH3(g) --t SiN,(s) + l ZH,lgl

[3d Ed. P3-23B] 0 P 3 - K B Condensation occurs in the gas-phase reaction:

CzM&) + 2CI(g) d CH2C12(g, 0 + 2HCl(g)

[2nd Ed. P3-17,]

New Problems on the Web

CDP3-New From time to time new problems relating Chapter 3 material to weq day interests or emerging technologies will be placed On the wet Solutions to these problems can be obtained by emailing the author.

Chap 3 Suoplementary Reading

S U P P L E M E N T A R Y R E A D I N G

1 . Two references relating to the discuwion of activation energy have already been ctted in this chapter. Acrikation energy i s usually discursed in terms nf either collision theory or trans~tion-state theory. A concise and readable account of there two theorle~ can be found in

Masel, R.. Chemical Kinerics, New York: McGraw W i l l , 2002. p. 594. LAIDLER. K. J. Chrreicnl Kitlctics. New York: Harper & Row. 1985. Chap. 3

An expanded but stjlI elenlentary presentation can be found in

MOORE, J. W., and R. G. PEARSOY, Kinetics nttd Mecl~nrrism, 3rd ed. New York: Wiley. 198 1. Chaps. 4 and 5.

A more advanced treatise of activation energies and collision and transition-state theories is

B E W O K , S . W., Tire Fowldutions of ChemicnI Kinetics. New York: McGraw- Hill. 1960.

STEIXFELD. J. 1.. J. S. FRANCISCO, W. L. HASE, Chemrcal Kinetics and Dynam- ics, 2nd ed. New Jersey: Prentice Hail, 1999.

2. The books listed nbove also give the rate laws and activation energiec for or number of reactions: in addition, as mentioned earher in this chapter, an extensive listing of rate Iaws and activation energies can be found in NBS circulars:

Kinetic data for larger number of reactions can be obtained on Floppy Disks and CD-ROMs provided by National Institute of Standards and Technology (NIST). Standard Reference Data 221fA320 Gaithersburg, MD 20899: ph: (301) 975-2208 Additional sources are Tables of Chemical Kinetics: Homogeneous Reactions. Nntional Bureau of Standards Circular 510 (Sept. 28, 1951): Suppl. 1 (Kov. 14. I956): Suppl. 2 (Aug. 5. 1960): Suppl. 3 (Sept. 15. 1961) (Washington. D.C.: U.S. Government Printing Office). Chem~cal Kinetics and Photochemical Data for Use in Srratosphenc Modeling. Evaluate No, LO, JPL Publication 92-20. Aug. 15. 1992. Jet Propulsion Labontories. Pasadena Calrf.

3. Also consult the current chemistry literature for the appropriate algebraic form of the rate law for a given reaction. For example, check the JournoE of Physical Chemist. in addition to the journals listed in Section 4 of the Supplementary Reading section in Chapter 4.

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