transformation of the reactants through highly reactive intermediates (active centers) to final products.

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1 LETUE. ETIN MEHNISMS ND EVLUTIN F TE FMS We have already seen that single reactions, in general, do not occur in one step; ut that it is a particular sequence o elementary reactions (reaction mechanism) that leads to the overall reaction. The nowledge o the reaction mechanism is extremely desirale since then the overall rate can e determined rom irst principles o the law o mass action which applies to every elementary step in the sequence. I such an attempt results in too complex algeraic expressions, at least experiments can e designed to test whether certain steps o the mechanism are rate limiting, which helps in inding a simpliied rate orm. nowledge o the expected rate orm helps in designing the right experiments or rate determination. eaction Mechanism - this sequence o elementary events represents the detailed pathway o transormation o the reactants through highly reactive intermediates (active centers) to inal products. ased on the structure o this sequence we can draw a more appropriate distinction etween catalytic and noncatalytic processes. n open sequence is one in which an active center is not reproduced in other steps o the sequence. The reaction is noncatalytic. closed sequence is one in which an active center is reproduced so that a cyclic pattern repeats itsel, and a large numer o molecules o products can e made rom only one active center. eaction is catalytic. Let us consider some examples: Mechanism : + l + l active center + + F F verall stoichiometry: l + F F + l noncatalytic + ; open sequence;

2 Mechanism : E ES E + P catalytic + S ES active center ES; closed sequence; verall stoichiometry: S P E This second mechanism aove is the Michaelis-Menten mechanism or enzyme catalyzed reactions such as isomerization o glucose, etc. We should e particularly careul to spot a closed sequence loop within a larger mechanistic path. For example, a typical chain propagation reaction ollows Mechanism elow: M + M + M M M + M M Steps and o the aove mechanism orm a closed (catalytic) sequence in which the active # centers (intermediates, radicals) and are repeatedly regenerated. asic ule: Whenever a mechanism is hypothesized or a single reaction consisting o N elementary steps the weighted sum o the N steps (i.e. the sum o all steps each o them multiplied y an integer,, etc.) must result in the overall stoichiometry or the reaction under consideration. I it does not, the mechanism is not consistent with stoichiometry and should e discarded. The aove asic ule has to e applied udiciously. It is certainly true or an open sequence mechanism. s a matter o act, it can e used then to discard the steps that are incompatile. However, in mechanisms containing a closed sequence it is the steps o the closed sequence that must lead to the overall stoichiometry. In our mechanism aove, the stoichiometry is M M + M. The act that termination o active centers may lead to M 4 or M 5 is immaterial since those would e present in an ininitesimal amount and would not aect the n n n mass alance o the single reaction under consideration i.e.,.

3 However, the "impurities" (e.g., M 4 and M 5 ) resulting rom such a mechanism may e important rom an environmental standpoint and we should strive to understand what they are and in what amounts they could orm. Finding the multipliers (stoichiometric numers), y which certain steps o the mechanism have to e multiplied in order to lead to the overall stoichiometry upon summation, is a trivial matter or relatively simple mechanisms consisting o two to three steps. For example, i a single reaction stoichiometry is given y and the mechanism is Mechanistic Step Stoichiometry o the Step Stoichiometric Numer verall Stoichiometry: it is clear that the irst step should e repeated twice (stoichiometric numer v ) in order to get the overall stoichiometry. For lengthy complex mechanisms simple inspection may not e the est way to proceed. I the reaction stoichiometry is given in such a way that all stoichiometric coeicients are integers (which can always e done y multiplying through with an appropriate numer) then the reaction stoichiometry may e written as: S v 0 () I the mechanism consists o N steps, and the stoichiometric coeicient o species in step i is v i, then the steps o the mechanism can e presented y a set o linear equations: S vi 0 or i,,... N () In order or the mechanism to e consistent with stoichiometry we must e ale to ind a set o stoichiometric numers (multipliers) v i i,,... N not all zero rom the ollowing set o linear equations:

4 N i v i v i v or,,... S () where S is the total numer o species including active intermediates, i.e., S S + I where I is the total numer o active intermediates. Naturally the overall stoichiometric coeicients v or active intermediates S +, S +, S +..., S, S must all e zero. Example: Mechanism: verall Stoichiometry: : v + 0 v : 0 + v v I: v v 0 Stoichiometry o each step: Thus we have a set o S + I + i N. The nonzero solution or v i exists i the matrix o v i has ran N. I that is the case select N equations and solve or S equations ( S ) with N unnowns ( ) i v. v ( ) In the aove simple example we have one intermediate species ( ) so that I and we have two stale species ( and ) so that S and hence S S + I. The mechanism consists o two steps so that N and we need to determine the two stoichiometric numers v and v y which step and step, respectively, should e multiplied to arrive at the overall stoichiometry. I we numer, and as species,, and, in that order, equation () can e written in the ollowing matrix orm: 0 ν (a) 0 ( ν ) The matrix o stoichiometric coeicient ( ν i ) is: 0 0 4

5 and the nonzero solution or the ( ν ) ν vector is guaranteed i that matrix has ran which it clearly does. The matrix multiplication in eq (a) results in rows that are presented aove which generate the solution or ν and ν. See any text on linear algera or matrices or details i you encounter a prolem o this nature. good reerence or chemical engineers is mundson N.., "Mathematical Methods in hemical Engineering-Matrices and Their pplications", Prentice Hall, 966. When a consistent mechanism is proposed, the overall rate o reaction can e derived ased on one o the ollowing two assumptions:. Pseudo-steady state assumption (PSS), also called Quasi-steady state assumption (QSS). ate limiting step assumption (LS) 5

6 6.. PSEUD-STEDY STTE SSUMPTIN (PSS) Since the intermediates (active centers) which may e o various nature, (e.g., ree radicals, ions, unstale molecules etc.) appear only in the mechanism ut not in the overall stoichiometry or the reaction it can e saely assumed that their concentration is at all times small. I this was not so, then they would e detectale and would appear in the overall stoichiometry since a sustantial portion o the reactants would e in that orm at some point in time. Now, reasoning leads us to conclude that then the net rate o ormation o these active intermediates must also always e small. I this was not true, and i at some point in time the net rate o ormation was high, then the quantity o this intermediate would have to rise and we have already concluded that this cannot happen. Thus, the net rate o ormation o all the active intermediates involved in the mechanism, ut not appearing in the overall stoichiometry or the reaction, must e very small. This is the asic hypothesis o the PSS: The net rates o ormation o all active intermediates are negligily small and hence are approximately zero. This requirement or a mechanism o N steps including S stale species that appear in overall reaction stoichiometry and I intermediates S S + I can e ormally written as: N v i i r i 0 or S +,... S (4) From these I linear expressions we can evaluate the I concentrations o the intermediates in terms o the rate constants and the S concentrations o stale species. Sustituting these intermediate concentrations into the expression or the rate o a desired component N vi ri (5) i,,... S we otain the required rate orm or that component. 6

7 Thus, the procedure or applying the PSS to a single reaction can e outlined as ollows:. Write down the hypothesized mechanism and mae sure that it is consistent with stoichiometry.. ount all the active intermediates.. Set up the net rate o ormation or every active intermediate. ememer, since the steps in the mechanism are elementary, law o mass action applies in setting up rates o each step. Mae sure to include the contriution o every step in which a particular intermediate appears to its net rate o ormation. 4. Set all the net rates o active intermediates to e zero and evaluate rom the resulting set o equations the concentrations o all active intermediates. 5. Set up the expression or the rate o reaction using a component which appears in the ewest steps o the mechanism (in order to cut down on the amount o algera). Eliminate all the concentrations o active intermediates that appear in this rate orm using the expressions evaluated in step 4. Simpliy the resulting expression as much as possile. 6. I you need the rate o reaction or another component simply use the rate orm evaluated in step 5 and the relationship etween the rates o various components and their stoichiometric coeicients. 7. I there is solid theoretical or experimental inormation which indicates that certain terms in the rate orm otained in step 5 or 6 may e small in comparison to some other ones simpliy the rate orm accordingly. 8. hec the otained rate orm against experimental data or the rate. 9. ememer that even i the derived rate orm agrees with experimental data that still does not prove the correctness o the proposed mechanism. However, your rate orm may e useul especially i the reaction will e conducted under conditions similar to those under which experimental data conirming the rate orm were otained. Let us apply this to an example. 7

8 onsider the decomposition o ozone. We want to now its rate. The overall stoichiometry is (6). The proposed mechanism consists o steps: + (7a) + + (7) The mechanism is oviously consistent with the overall stoichiometry, the stoichiometric numer o oth steps eing one. We have already assumed that the last step is irreversile since it is highly unliely that two oxygen molecules would spontaneously "collide" to produce an active oxygen atom and ozone.. ounting the active intermediates we ind one, namely,.. The net rate o ormation o is: 0 (8) 4. ccording to PSS the aove rate must e zero. From the aove equation we get + (9) 5. Since oth and appear in oth steps o the mechanism we can set up the rate o disappearance o ozone directly + (0) Sustituting the expression or, and inding the common denominator, leads to: 8

9 () This is a complete rate expression ased on the hypothesized mechanism. 7. oncentration o ozone is orders o magnitude smaller than the concentration o oxygen. It also appears that the second step in the mechanism is slow. Thus, when >> the rate simpliies to: () 8. The experimentally ound rate under these conditions o low is. greement seems to exist etween the proposed mechanism and data. onsider another example. The overall reaction stoichiometry is given as: + () and the experimentally determined rate is (4) It is desired to determine whether the mechanism outlined elow is consistent with the oserved rate.. + I (5a) I + + (5). The mechanism is oviously consistent with the overall stoichiometry since a straight orward addition o steps leads to it. The numer o intermediates is one, I. 9

10 . The net rate o ormation o the intermediate is: 0 (6) I I 4. nd according to PSS is equal to zero. The concentration o I is then I (7) 5. Since appears in oth steps o the mechanism and only in one, set up the rate o ormation o or convenience (8) I Eliminating I we get (9) 6. From stoichiometry (0) 7. The rate agrees with the experimentally determined one. (ompare eq. (0) and (4)). The mechanism may e right. NTE: In writing the net rate o ormation o I, I, in step we used the equivalent rate and thus ased the rate constant on I. I we were to write the rate o disappearance o, () I notice that the irst term has to e multiplied y the stoichiometric coeicient o in the st step, which is two, since the rate o disappearance o in the st step is twice the rate o appearance o I in that step. I we ollowed a dierent convention we would have ased on the let hand side rom where the arrow originates i.e. we would have ased it on a component. In that case a actor/ would have appeared with in step and there would e no ut in ront o in the expression or. This seemingly trivial point oten causes a lot o errors 0

11 and grie. Just notice that i we orgot the proper stoichiometric relationship, and the expression or in eq. () did not have a ut, while I was as given, they would e identically equal to each other and thus asserting 0 would e asserting that 0, which is asurd since I which is a respectale expression. This demonstrates the importance o not orgetting the stoichiometric coeicients in setting up rates. t the end it should e mentioned that the PSS ortunately does not only rest on the veral arguments presented at the eginning o this section (can you ind any ault with this?) ut has solid mathematical oundations originating in the theory o singular perturations. This will e illustrated in an ppendix since the same theory is applicale to may other situations. The application o PSS to multiple reactions is a straightorward extension o the aove procedure.

12 6. TE LIMITING STEP SSUMPTIN (LS) This approach is less general than the previous one, in the sense that we must now more aout the mechanism than ust its orm in order to apply it. The rate expression otained rom LS represents thus a limiting case o the one that could e otained using PSS. The asic hypothesis o the LS is as ollows:. ne step in the mechanism is much slower than the others and thus that rate limiting step determines the overall rate.. With respect to the rate limiting step all other steps may e presumed in equilirium. This is est explained ased on an example. onsider a reaction +. Suppose that we want to ind its rate orm ased on the mechanism shown elow. In addition, it is nown that the st step is the slowest. I we then depict the magnitude o the orward and reverse rates or each step with, and the magnitude o the net orward rate y a solid arrow, and i we eep in mind that the net rate must e o the same magnitude in all the steps (which is required y PSS) ecause otherwise the active intermediates would accumulate someplace, we get the picture presented to the right o the hypothesized mechanism elow learly the magnitude (length)o the arrows indicating orward and reverse rates isthe smallest in step, this step is the slowest and limits the rate. t the same time the magnitude o the net rate orward is almost / o the total rate orward in step and that step clearly is not in equilirium while the magnitude o the net rate orward ( ) in comparison to the total rate orward ( ) and total reverse rate ( ) in steps and is negligile. Thus, in these two steps rates orward and acward are approximately equal, and in comparison to step these two steps have achieved equilirium. Thus, the procedure or applying the LS to a single reaction can e outlined as ollows:

13 . Write down the hypothesized mechanism and mae sure that it is consistent with stoichiometry.. Determine which is the rate determining step. This should e done ased on experimental inormation. ten various steps are tried as rate limiting due to the lac o inormation in order to see whether the mechanism may yield at all a rate orm compatile with the one ound experimentally.. Set up the net rates o all the steps, other than the rate limiting one, to e zero, i.e. set up equilirium expressions or all other steps. 4. Set up the rate orm ased on the law o mass action or the rate determining step, and eliminate all concentrations o intermediates using the expressions evaluated in step. 5. Using the stoichiometric numer o the rate determining step relate its rate to the desired rate o a particular component. 6. hec the otained rate orm against the experimentally determined rate. 7. ememer that the agreement or disagreement etween the derived and experimental rate orm does not prove or disprove, respectively, the validity o the hypothesized mechanism and o the postulated rate limiting step. I the two rate orms disagree try another rate limiting step and go to step. I the two rate orms agree use the rate orm with caution. The greatest limitation o the rate orms ased on LS is that they are much less general than those developed rom PSS and they do not test the mechanism in its entirety. The rate orms ased on LS may e valid only in narrow regions o system variales (e.g., T, concentrations) since with the change in variales (i.e. concentrations, T) the rate limiting step may switch rom one step in the mechanism to another. learly, the magnitude o our arrows representing the rates depends on concentration levels, temperature, pressure etc. and may change rapidly as conditions change. n example in the shit o the rate limiting step is the previously covered Lindemann's-hristensen mechanism or unimolecular reactions.

14 Suppose that we want to evaluate the rate o ormation o or the reaction given aove and under the assumptions made.. We irst test the compatiility o the proposed mechanism with stoichiometry. y inspection we ind: v (a) v () v (c). Step is rate limiting (given). Set up equilirium expressions or step and (a) () 4. Set up the rate or the rate limiting step r l (4) Using eq. () and eliminating we get rl (4a) 4

15 5 5. The stoichiometric numer o the rate limiting step is s v l as per (eq. a). Thus, since l l r v v r l (5) 6. Inormation not availale. Now suppose that in the same mechanism step was rate limiting (i.e. arrows or the rate o step now eing much shorter than those in step and ). We can quicly write the equilirium relationships or step and : (6a) (6) and rom these otain the concentrations o the intermediates (7a) (7) The rate or the rate limiting step (step ) is: r l (8) e r (8a) r e since l v (9)

16 learly this is an entirely dierent rate orm than otained previously ased on step eing rate limiting. NT ENE : It would e quite tedious to ind the rate orm or the aove mechanism ased on PSS since when the net rates o ormation o intermediates are set to zero we get nonlinear equations due to the product resulting rom the rate orward in step. Try it anyway or an exercise. NT ENE : I the irst step ( v ) was rate limiting our rate orm is given y (5) t equilirium 0 so that (0) ecall that (a) and that or the reaction + the equilirium (concentration units) constant y is given () Sustituting eqs. (a) and () into eq. (0) we get () 6

17 7 ecall that s P where s is the stoichiometric numer o the rate determining step. Here v s. Hence, () I the nd step ( ) v is rate limiting the rate orm is given y eq. (8a) (8a) so that at equilirium 0 ' () and () Let us go ac to the reaction o decomposition o ozone 0 0. The mechanism was given eore and let us assume that the nd step is rate limiting. Then (4a) (4) r l (5)

18 Now v l ut v Thus, (6) Notice that this is the rate orm otained when using PSS ater certain additional assumptions were made. The rate orm generated y the use o PSS without additional assumptions is much more general. The aove expression is its limiting case. For multiple reactions, rate orms can e otained y LS y the straightorward extension o the aove rules. 8

19 6. HLF LIFE ND HTEISTI ETIN TIME ased on either PSS or LS we are usually ale to derive a rate orm or a particular reaction. ten n-th order rate orm is otained, say or reactant : n ( ) Then the characteristic reaction time is deined as τ and it is the time that it n o would tae in a close system (atch) or the concentration to decay to e o its original value. o d The alance or reactant in a closed system is. So when the reaction is not n-th dt o order the characteristic reaction time can e deined as τ where the rate o is evaluated o at initial conditions. Hal-lie is the time needed or the reactant (species) concentration to e reduced to hal o its original value in a closed system. Integration o the species alance yields t t o dt o o o d o d τ d where c o and o. The integration is perormed y sustituting into the aove expression all concentrations in terms o using the stoichiometric relations. For an n-th order reaction we get t n n n ( n ) ( n ) o τ For a irst order process t ln τ ln 9

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